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You can Download Chapter 9 Hydrogen Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 9 Hydrogen

1st PUC Chemistry Hydrogen One Mark Questions and Answers

Question 1.
What is the strength of
(a) 10 volume H₂O₂
(b) 20 volume H₂O₂?
Answer:
(a) 3%
(b) 30%.

Question 2.
Name the most abundant element in the universe.
Answer:
Hydrogen.

Question 3.
Which isotope of hydrogen is radioactive?
Answer:
Tritium (³₁H)  is radioactive.

Question 4.
Name the isotope of hydrogen which used in nuclear reactor.
Answer:
²₁H (deuterium)

Question 5.
What is meant by hard water?
Answer:
Hard water is water containing bicarbonates, sulphates and chlorides of Ca²⁺ and Mg²⁺.

Question 6.
Why does elemental hydrogen react with other substances slowly at room temperature ?
Answer:
It is because it has high bond dissociation energy (436 kJ mol^-1)

Question 7.
What is the importance of heavy water with regard to nuclear power generation ?
Answer:
It is used as coolant and moderator i.e., to slow down the fast moving neutrons.

Question 8.
Why does water has a high boiling point and a high melting point as compared to H₂S?
Answer:
Water molecules are associated with intermolecular H-bonding whereas H₂S is not.

Question 9.
When sodium hybride is electrolysed, hydrogen is liberated at which electrode? Give equation.
Answer:
At anode;

at anode 2H^– – 2e^– → H₂(g)

Question 10.
Which gas is evolved when Mg3N₂ (Magnesium nitride) is treated with H2O? Give chemical reaction.
Answer:
NH₃ gas is evolved. Mg₃N₂ +6H₂O → 3Mg(OH)₂ +2NH₃

Question 11.
How is heavy water produced from ordinary water?
Answer:
It is obtained by repeated electrolysis of ordinary water.

Question 12.
What are the ways in which water molecules are bonded to an anhydrous salt to form a hydrate?
Answer:
Co-ordinate bond and H-bond

Question 13.
Why is H₂ more reactive than D₂?
Answer:
H – H has less bond dissociation energy than D-D

Question 14.
Which of the substances present in water cause permanent hardness of water?
Answer:
CaSO₄, MgSO₄, CaCl₂ and MgCl₂ cause permanent hardness of water.

Question 15.
Permanent hardness cannot be removed by:
(a) Adding washing soda to water,
(b) Boiling water,
(c) Adding sodium polymetaphosphate to water,
(d) Passing water through ion exchange resins.
Answer:
(b) Boiling water

Question 16.
How is pure H₂ obtained ?
Answer:
Pure H₂ and O₂ are obtained by electrolysis of acidified H₂O.

Question 17.
What is source of sun’s energy ?
Answer:
Nuclear fusion reaction.

Question 18.
Which type of bond is presents in water molecule?
Answer:
Covalent bond and inter molecular Hydrogen bond.

Question 19.
Which type of oxide is water?
Answer:
H₂O is an amphoteric oxide.

Question 20.
Which isotope of hydrogen does not have neutron?
Answer:
¹₁H does not have neutron. It is called protium or ordinary hydrogen.

Question 21.
Which out of nascent hydrogen and dihydrogen is more reactive?
Answer:
Nascent hydrogen (newly bom hydrogen) [H] is more reactive than molecular hydrogen.

Question 22.
What is molar mass of heavy water?
Answer:
Molar mass of heavy water is 20 g mol^-1

Question 23.
How does CaC₂ react with heavy water?
Answer:
CaC₂+2D₂O → Ca(OD)₂ calcium deuteroxide + C₂D₂ Deuteroethyne

Question 24.
Name two compounds which retard decomposition of H₂O₂.
Answer:
Glycerol and acetanilide retard decomposition of H₂O₂.

Question 25.
Give two examples of interstitial hydrides.
Answer:
CuH, FeH are examples of interstitial hydrides.

Question 26.
What is the tradename of sodium hexa-metaphosphate?
Answer:
Calgon.

Question 27.
In a reaction of F₂ and H₂O, what is the role of water?
Answer:
2F₂+2H₂O → 4HF + O₂ water is acting as reducing agent.

Question 28.
What is the nature of H₂O₂ ?
Answer:
H₂O₂ is acidic in nature.

Question 29.
Does hydrogen peroxide act as

• Strong acid,
• Bleaching agent ?
• Answer:
• No, Weak acid,
• Yes

Question 30.
Old paintings of lead are generally washed with dilute solution of hydrogen peroxide in order to regain its colour. Why?
Answer:
Black colour in paintings is due to lead sulphide (PbS). This is oxidized to lead sulphate by hydrogen peroxide.

Question 31.
Melting point, enthalpy of vapourisation and viscosity data of H₂O and D₂O is given below:

On the basis of this data explain in which of these liquid intermolecular forces are stronger ?
Answer:
Intermolecular forces of attraction are stronger in D₂O than in water.

Question 32.
Write the Lewis structure of hydrogen peroxide.
Answer:

Question 33.
Why is water molecule polar ?
Answer:
Due to difference in electronegativity between O and H and angular shape

Question 34.
Write redox reaction between fluorine and water.
Answer:
2F₂(g) + 2H₂O(l) → 4H^–(aq) + 4F^–(aq) + O₂

Question 35.
Concentrated sulphuric acid cannot be used drying H₂. Why?
Answer:
Cone. H₂SO₄ on absorbing water from moist H₂ produces so much heat that H₂ catches fire.

Question 36.
Why is dihydrogen gas not preferred in balloons ?
Answer:
H₂ is combustible in nature. Therefore it may react with oxygen violently. Thus it is not used in balloons.

Question 37.
Give one method of preparing deuterium.
Answer:
It is prepared by electrolysis of heavy water (D₂O) 2D₂O → 2D₂ + O₂

Question 38.
Which salts present in water mark it permanent hard?
Answer:
Calcium and Magnesium chlorides and sulphates.

Question 39.
Name a process which can remove both temporary and permanent hardness of water.
Answer:
Permutite.

Question 40.
Complete the reaction : Fe(s) + H₂O(g) →
Answer:
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

Question 41.
What is water gas and how it is obtained?
Answer:
Water gas is mixture of CO and H₂. It is obtained by passing super heated steam over hot coke.

Question 42.
A sample of hard water is allowed to pass through an anion exchanger. Will it produce lather with soap easily?
Answer:
No. Ca²⁺ and Mg²⁺ ions are still present and these will react with soap to form curdy white ppt. Therefore it will not produce lather with soap solution easily.

Question 43.
Anhydrous BaO₂ is not used for preparing H₂O₂. Why ?
Answer:
BaSO₄ formed dining the reaction of BaO₂ with H₂SO₄ forms a protective layer around unreacted BaO₂ and the reaction stops after sometime.

Question 44.
Find the volume strength of 2N H₂O₂ solution.
Answer:
Volume strength = 5.6 × Normality
Volume strength = 5.6 × 2
= 11.2 volumes

Question 45.
Give an example each of an ionic hydride and a covalent hydride.
Answer:
Ionic : NaH or CaH₂
Covalent: H₂O,B₂H₆,CH₄ etc.

Question 46.
Can distilled water be called as deionised water ?
Answer:
Yes, distilled water does not contain any cations and anions and hence can be called as deionized water.

Question 47.
Explain why oxide ion is called a hard ion?
Answer:
Oxide ion is very small in size and thus cannot be easily polarized and hence it is called a hard ion.

Question 48.
Anhydrous BaO₂ is not used for preparing H₂O₂. Why?
Answer:
Anhydrous BaO₂ is not used because the BaSO₄ formed during the reaction forms a protective layer around unreacted BaO₂ and the reaction stops after some time.

Question 49.
How is D₂O₂ prepared ?
Answer:
D₂O₂ is prepared by distillation of potassium per sulphate (K₂S₂O₈) with D₂O

Question 50.
What is hydride gap ?
Answer:
The region of the periodic table from groups 7-9 which do not form hydrides is called the hydride gap.

Question 51.
What is perhydrol ?
Answer:
Perhydrol is the trade name for H₂O₂ which is used as an antiseptic for washing wounds, teeth and ears.

Question 52.
What is syngas ? Why is it called so ?
Answer:
Mixtures of CO and H₂ are called syngas. It is so named because it is used for the synthesis of methanol and a number of hydrocarbons.

Question 53.
What you mean by degree of hardness ?
Answer:
Degree of hardness is defined as the number of parts of calcium carbonate or equivalent to various calcium and magnesium salts present in a million parts of water by mass. It is expressed in ppm.

Question 54.
For which work H.C. Very get Noble prize in 1934.
Answer:
Heavy hydrogen or deuterium was separated from liquid hydrogen by fractional evaporation by H.C. Urey. For this great contribution, he was awarded the Noble Prize in Chemistry in 1934.

1st PUC Chemistry Hydrogen Two Marks Questions and Answers

Question 1.
Write one chemical reactions for the preparation of D₂O₂.
Answer:
D₂O₂ can be prepared by the reaction of D₂SO₄ dissolved in water over BaO₂.
BaO₂ + D₂SO₄ → BaSO₄ + D₂O₂

Question 2.
H₂O₂ is a better oxidizing agent than water. Explain.
Answer:
H₂O₂ is better oxidising agent than water because
(a) It oxidizes an acidified KI solution to I₂ which gives blue colour with starch solution but water does not.
(b) H₂O₂ turns black PbS to white PbSO₄ but water does not.

Question 3.
Dihydrogen reacts with dioxygen (O₂) to form water. Write the name and formula of the product when the isotope of hydrogen which has one proton and one neutron in its nucleus is treated with oxygen? Will the reactivity of both the isotope be the same towards oxygen? Justify your answer.
Answer:
The isotope of hydrogen containing one proton and one neutron is deuterium (D).
2D₂(g) + O₂(g) → 2D₂O Deuterium oxide
Since D-D bond is stronger than H – H bond, therefore, D₂ is less reactive towards oxygen than H₂.

Question 4.
Rohan heard that instructions were given to the laboratory attendant to store a particular chemical i.e., keep it in the dark room, add some urea in it, and keep it away from dust. This chemical acts as an oxidizing as well as a reducig agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents.

• Write the name of this compound.
• Explain why such precautions are taken for storing this chemical?

Answer:

• Hydrogen peroxide, H₂O₂.
• H₂O₂ decomposes in the presence of light and dust particles. Therefore, it is stored in wax-lined glass or plastic vessels in the presence of stabilizers like urea.

Question 5.
Why is the ionization enthalpy of hydrogen higher than that of sodium?
Answer:
The size of hydrogen is smaller than that of sodium and therefore, the ionization enthalpy of H is higher (1312 kJ mol^-1) than that of Na (496 kJ mol^-1).

Question 6.
Basic principle of hydrogen economy is transportation and storage of energy in the form of liquid or gaseous hydrogen. Which property of hydrogen may be useful for this purpose?
Answer:
Basic property of hydrogen economy is that it can be converted into liquid by cooling under high pressure and therefore, can be transported.

Question 7.
Why can dilute solution of hydrogen peroxide not be concentrated by heating?
Answer:
H₂O₂ cannot be concentrated simply by heating because it decomposes much below its boiling point. Therefore, concentration of H₂O₂ is carried out in a number of stages.

Question 8.
Why is hydrogen peroxide stored in wax-lined bottles?
Answer:
Hydrogen peroxide is decomposed by the rough surfaces of glass, alkali oxides present in it and light. Therefore, to prevent its decomposition, H₂O₂ is usually stored in coloured paraffin wax coated plastic or Teflon bottles.

Question 9.
Phosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides. Why ?
Answer:
H₃SO₄ acts as a catalyst for decomposition of H₂O₂. Therefore, some weaker acid such as H₂PO₄ is preferred over H₂SO₄ for preparing H₂O₂ from peroxide.

Question 10.
Atomic hydrogen combines with almost all elements but molecular hydrogen does not. Explain.
Answer:
Atomic hydrogen is highly unstable and hence very reactive. Therefore, it combines with almost all the elements, However, molecular hydrogen has large bond dissociation enthalpy (458.5 kJ mol,m^-1) and therefore, less reactive.

Question 11.
A coulourless liquid ‘A’ contains H and O elements only. It decomposes slowly on exposure to light. It is stabilized by mixing urea to store in the
presence of light,
(i) Suggest possible structure of A.
(ii) Write chemical equations for its decomposition reaction in light.
Answer:
The liquid A is hydrogen pexoxide (H₂O₂)

Question 12.
An ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with Al₂Cl₆.
Answer:
Since the ionic hydride of alkali metal has significant covalent character, therefore, it is LiH. Since LiH is very stable, therefore, it is almost unreactive towards O₂ and Cl₂. It reacts with Al₂Cl₂ form lithium aluminium hydride.
8LiH + Al₂Cl₆ → 2LiAlH₄+6LiCl

Question 13.
Hydrogen is not prepared by the reaction of concentrated sulphuric acid on zinc. Explain.
Answer:
Concentrated sulphuric acid is not used for the preparation of hydrogen because it reacts with zinc to form SO₂ gas instead of hydrogen.
Zn + 2H₂SO₄ → ZnSO₄ + 2H₂O + SO₂↑

Question 14.
Which element can oxidize water to O₂?
Answer:
Fluorine being more electronegative than oxygen can oxidize water to oxygen. In this, water acts a reducing agent and hence itself gets oxidized to O₂.
2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F_–m(aq) + O₂(g)

Question 15.
Dilute solution of hydrogen peroxide cannot be heated strongly for its concentration. Explain.
Answer:
Dilute solution of H₂O₂ cannot be concentrated by heating because it decomposes into H₂ and O₂ on heating.

Question 16.
Can sodium bicarbonate make water hard?
Answer:
Sodium bicarbonate cannot make water hard because soaps themselves are sodium salts of fatty acids which are soluble in water.

Question 17.
Which of the two hydrogen or deuterium undergoes reactions more rapidly and why?
Answer:
Hydrogen undergoes reactions more rapidly than deuterium. This is because of mass differences. For example, reaction between hydrogen and chlorine is about 14 times faster than between deuterium and chlorine.

Question 18.
Can we remove completely temporary hardness due to Mg(HCO₃)₂ by boiling?
Answer:
Temporary hardness of water due to Mg(HCO₃)₂ can be completely removed by boiling because soluble Mg(HCO₃)₂ is converted into insoluble MgCO₃ which can be easily removed by filtration.

Question 19.
H2O2 acts as an oxidizing agent as well as reducing agent. Why ?
Answer:
In H₂O₂ , oxygen has -1 oxidation state which lies between maximum (0 or +2 in OF₂) . and minimum -2. Therefore, oxygen can be oxidized to O₂ (zero oxidation state) acting as reducing agent or can be reduced to H₂O or OH- ( -2 oxidation state) acting as an oxidizing agent.

Question 20.
Why do lakes freeze from top towards bottom ?
Answer:
The upper surface of lake is in contact with the cold environment and starts to freeze. The density of ice is less than that of liquid water. Therefore, ice floats on the surface. Thus, the ice layer at lower temperature floats over the water below it. Due to this the freezing of water into ice occurs continuously from top towards bottom.

Question 21.
A mixture of H₂O₂ and hydrazine with copper (II) is used as a rocket propellant. Why ?
Answer:
The reaction between hydrazine (NH₂NH₂)and H₂O₂ in the presence of Cu(II) is highly exothermic and is accompanied by a large increase in energy as well as in volume of the products. Therefore, this mixture is used as a rocket propellant.

Question 22.
Water cannot be used to extinguish petrol fires. Why?
Answer:
Water is used to extinguish most fires because it lowers the temperature of burning material. However, in case of petrol fires, petrol being lighter than water, floats over water and hence fire spreads instead of being extinguished.

Question 23.
Ferric chloride is reduced when zinc and hydrochloric acid are added to its solution and not by passing H₂ gas through its solution. Explain.
Answer:
Ordinary H₂ is less reactive and therefore, it does not reduce acidified FeCl₃ solution. However, when zinc is added to acidified FeCl₃ solution, nascent hydrogen is produced which has enormous energy. It is more reactive than ordinary H₂ and reduces acidified FeCl₃ solution.
FeCl₃ + H₂ → No reaction; Zn + H₂SO₄ → ZnSO₄ + H₂

Question 24.
Hard water is softened before using in boilers. Explain
Answer:
Hard water on boiling forms precipitates of MgCO₃, CaCO₃ and CaSO₄ which form scales in the boilers. As a result of these scales in the boilers. The boiler gets deteriorated due to over heating. Moreover, these scales are non-conducting and therefore, more fuel is consumed. Therefore, in order to prevent the formation of scales, hard water softened before using in boilers.

Question 25.
What are the uses of Heavy Water?
Answer:
(a) Heavy water is used as a coolant and moderator for neutrons in nuclear reactors.
(b) Heavy water is used in the preparation of deuterium and deuterated compounds.
(c) It is used as a tracer element in metabolic processes.

Question 26.
Presence of water is avoided during the preparation of H₂O₂ from Na₂O₂ Why?
Answer:
Water present during the reaction, reacts with Na₂O₂ form NaOH which tends to decompose H₂O₂.
Na₂O₂ + 2H₂O → H₂O₂ +2NaOH
2H₂O₂ → 2H₂O + O₂

Question 27.
Discuss the position of hydrogen in the periodic table is not justified.
Answer:
Its position in periodic table is not justified because it resembles both alkali metals and halogens.

• Its electronic configuration is similar to alkali metals.
• It is non-metal like halogens.
• Its ionization energy is high like halogens.
• It can lose electrons to form H+ like alkali metals.

Question 28.
Distinguish clearly between salt-like and covalent hydrides.
Answer:
Salt-like hydrides are also called ionic hydrides. They are formed by group 1 and 2 elements with H₂. They are basic in nature. They dissolve in water forming alkali and give H₂ gas, .e.g., NaH, CaH₂, KH etc.
Covalent hydrides are formed between less electropositive or electronegative elements.
Ex : B₂H₆, AlH₃, BeH₂, They are insoluble in water.

Question 29.
Describe the industrial use of hydrogen which depends on
(a) the heat liberated when it burns
(b) its ability to react with vegetable oil in the presence of a catalyst
(c) its ability to unite with N₂.
Answer:
(a) It is used in oxy-hydrogen flame for welding purpose.
(b) It is used to manufacture vanaspati ghee.
(c) It is used for manufacture of NH₃.

Question 30.
(a) How would you prepare dihydrogen from water by using a reducing agent?
(b) How would you prepare dihydrogen from a substance other than water?
(c) How would you prepare very pure H₂ in the laboratory?
(d) How would you prepare heavy hydrogen in the laboratory?
Answer:
(a) 2Na + 2H₂O → 2NaOH + H₂
(b) Zn + 2HCl → ZnCl₂ + H₂
(c) It is obtained by electrolysis of acidified water.
(d) It is obtained by electrolysis of heavy water.

Question 31.
What are the products of the following reactions ? Write balanced equation
(i) FeO(s) + H₂(g) →
(ii) MnO₄ + H₂O₂ →
Answer:

(ii) 6H⁺ + 2MnO₄ + 5H₂O₂ > 2Mn²⁺ + 8H₂O + SO₂

Question 32.
State whether each of the following compounds is a acidic anhydride. Write a balanced equation for the reaction with water in each : Na₂O, P₄O₆ SO₂, Al₂O₃
Answer:
Na₂O is base, Na₂O + H₂O → 2NaOH
P₄O₆ is basic, P₄O₆ + 6H₂O →4H₃PO₃

AlO₃ is amphoteric oxide (acidic as well as basic)

Question 33.
Describe the structure of common form of ice.
Answer:
In the normal hexagonal ice, each oxygen atom is tetrahedrally surrounded by four other oxygen atoms, there being an hydrogen atom in between each pair of oxygen. Each hydrogen is covalent bonded to one oxygen and linked to the other oxygen by a covalent bond. Such an arrangement leads to packing having large open spaces. The density of ice is therefore less than liquid water.

Question 34.
Distinguish clearly between (a) hard and soft water, (b) temporary hardness and permanent hardness.
Answer:

Hard Water	Soft Water
1. water which contains soluble of Ca2+, Mg2+.	1. It does not contain Ca2+ and Mg2+.
2. It does not form lather with soap easily. It forms precipate of insoluble substances with soap	2. It forms lather with soap.

Temporary hardness	Permanent hardness
1. It is caused by bicarbonates of Ca2+ and Mg2+.	1. It is due to chlorides and sulphates of Ca2+ and Mg2+.
2. It is removed by boiling.	2. It is removed by heating with washing soda.

Question 35.
How is H₂O₂ manufactured ?
Answer:
Hydrogen peroxide is manufactured by electrolysis of 50% H₂SO₄.

Question 36.
What is the structure of H₂O₂? Draw a schematic diagram indicating the shape of the molecule clearly.
Answer:
Hydrogen peroxide has open-book structure. It has non-planar structure. In the crystal, the dihedral angle 111.5° reduces to 90.2° on account of hydrogen bonding. There is single bond between two oxygen atom in hydrogen peroxide.

Question 37.
Show how H₂O₂ functions as a as an oxidizing agent with two examples.
Answer:
Oxidising agent:

• PbS + 4H₂O₂ → PbSO₄ + 4H₂O
• 21^– + H₂O₂ + 2H^– → I₂+2H₂O

Question 38.
Show how H₂O₂ functions as oxidising agent with two examples.
Answer:
Oxidising agent: PbO₂ + H₂O₂ → PbO + H₂O + O₂
H₂O₂ + Ag₂O →H₂O + O₂ + 2Ag

Question 39.
H₂O₂ is used to restore the colour of old paintings containing lead sulphide. Write a balanced equation for the reaction that takes place in this process.
Answer:

Question 40.
What is the mass of H₂O₂ present in 1 litre of a 2M solution ? Calculate the volume of oxygen at STP liberated upon the complete decomposition of 100 cm3 of the above solution.
Answer:
1 litre of 1M solution of H₂O₂ contains 34g of H₂O₂ [(2H = 2) + (20 = 32) = 34]
I litre of 2 M solution of H₂O₂ contains 34 × 2 = 68g of H₂O₂

1000 ml of 2M solution contains 68g of H₂O₂
100 ml of 2M solution contains 6.8g of H₂O₂
68g of H₂O₂ will produce 22.4 litre of O₂ at STP
6.8 g of H₂O₂ will produce = 2.24 litre of O₂ to STP.

Question 41.
Ionisation energy of hydrogen is much higher than those of alkali metals. In case of former combines with latter, which type of compound will be formed? Explain when H₂ reacts with C at high temperature, what type of hydride will be formed. Give chemical equation.
Answer:
Ionic hydride will be formed, e.g.,

Methane gas a covalent hydride is formed :

Question 42.
Fill in blanks:

1. Hydrogen shows character as well as …………………. character.
2. H+ ion in water exists ……………………. ion called ……………………….. ion.
3. Hydrogen is liberated at …………………………….. when NaH is electrolysed.
4. Hydrogen shows ………………………. and ………………………… oxidation state in compounds.

Answer:

1. Electropositive, electronegative,
2. H3O⁺, hydronium,
3. anode,
4. +1 and – 1

Question 43.
How does H2O2 react with KMnO4 in acidic medium ? Write four uses of H₂O₂.
Answer:
2KMnO₂ + 3H₂SO₄ + 5H₂O₂ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
Uses :

• It is used as oxidizing agent,
• It is used as reducing agent.
• It is used as bleaching agent,
• It is used as rocket propellant.

Question 44.
Sea water can’t be used in boiler. Explain given chemical equations.
Answer:
Sea water contains soluble salts of calcium and magnesium, which on boiling change to insoluble salts and get deposited as a crust on the inner sides of a boiler. This crust acts as a bad conductor of heat and prevent the flow of heat to water. A large quantity of fuel gets wasted.
Chemical Reaction
\(\begin{array}{l}{\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{CaCO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}} \\ {\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{MgCO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}}\end{array}\)

1st PUC Chemistry Hydrogen Three Marks Questions and Answers

Question 1.
Sample of hydrogen peroxide solution is 1.5 M. If it is to be labelled ‘X’ volume; what is the value of ‘X’
Answer:

Mass of H₂O₂ per liter = 1.5 × 34 = 51 g

68 g of H₂O₂ at STP gives 22.4 L of O₂
51 g of H₂O₂ gives at STP = \(\frac{22.4 \times 51}{68}\)
= 16.8 L of H₂O₂ that has 51g of H₂O₂ will give 16.8 L of O₂
Hence value of X is 16.8 L.
∴ The given sample of H₂O₂ is 16.8 Vol.

Question 2.
Explain with examples (i) Polymeric hybrids (ii) Complex hydrides.
Answer:
Polymeric hydrides: These are formed by elements having electro negativity in the range 1.40 – 2.0. These usually exist in the polymeric form in which the monomer molecules are held together in two or three dimensions by hydrogen bridges. Some common examples are : (BeH₂)_n, (AlH₃)_n, (InH₃)_n , (GaH₃)_n , (SpH₄)_n
These are amorphous solids which decompose above 523 K to evolve H₂ gas.
Structure. Beryllium hydride exits in the polymeric form involving three-center-two electron bonds as shown below :

Complex hydrides: In these hydrides, the hydride ion (H^–) acts as the ligand and is attached to the central metal atom by coordinate bonds. These are formed both by transition elements and non-transition elements. Among the non-transition elements, the most important complex hydrides are formed by elements of group 13. These are sodium borohydride (NaBH₄), lithium borohydride (LiBH₄) and lithium aluminium hydride (LiAlH₄). These are versatile reducing agents and are widely used for reduction of organic compounds.

Question 3.
In what respects does hydrogen resemble alkali metals ? How does it resemble halogen ?
Answer:
(i) Resemble with alkali metals :

• It has one valence electron in s-orbital like alkali metals
• It can lose one electron to form H+ ion like alkali metals.
• It is liberated at cathode during electrolysis of compounds like H₂O, HCl etc.
• It shows +1 oxidation state like alkali metals.
• It is a strong reducing agent like other alkali metals.

(ii) Resemblance with halogen :

• Hydrogen is non-metal like halogens.
• Hydrogen forms diatomic molecule like halogens.
• It has ionization energy high like halogens.
• Its electronegativity is similar to halogens.
• It can gain one electron to form H- ion.
• When NaH is electrolysed, hydrogen is liberated at anode like halogens.

Question 4.
In what respects H₂ differs from alkali metals and halogens ?
Answer:
Difference with alkali metals.

• Hydrogen is a non-metal whereas alkali metals are most electropositive elements.
• Hydrogen form diatomic molecules whereas alkali metals do not.
• Hydrogen mostly forms covalent compounds whereas alkali metals mostly from ionic compounds.
• The ionization energy of hydrogen is much higher than that of alkali metals.
• The size of H+ is smaller than those of alkali metals.

Difference with halogens

• It has less tendency to form hydrides (H^–) as compare to halogens.
• It does not have lone pair of electrons like halogens.
• It is less reactive than halogens due to high bond dissociation energy.

Question 5.
Complete the following reaction
(i) Na + H₂ →
(ii) N₂ + H₂ →
(iii)

(iv) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \stackrel{\text { Sunlight }}{\longrightarrow}\)
(v) H₂+Br_{2 →}
(vi) \(\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2} \stackrel{\text { heat }}{\longrightarrow}\)
Answer:
(i) Na + H₂ → 2NaH
(ii) N₂(g) + 3H₂(g) →2NH₃(g)

1st PUC Chemistry Hydrogen Four Marks Questions and Answers

Question 1.
Make a comparison of atomic hydrogen with nascent hydrogen any two. How the active hydrogen and heavy hydrogen are formed ?
Answer:
(a) Comparison of atomic and nascent hydrogen
Main points of differences are :

1. Nascent hydrogen can be produced even at room temperature but atomic hydrogen is produced only at very high temperature.
2. Nascent hydrogen can never be isolated, but atomic hydrogen can be isolated.
3. Reducing power of atomic hydrogen is much greater than that of nascent hydrogen.
4. Reactivity of the three forms of hydrogen increases in the order.
   Molecular hydrogen (H₂) < Nascent hydrogen < Atomic hydrogen.

(b) 1. Active hydrogen: It is obtained by subjecting stream of molecular hydrogen at ordinary temperature to silent electric discharge at about 30,0000 volts. It is very reactive in nature (half-life = 0.33 second) and combines directly at ordinary temperatures with Pb and S forming their hydrides.

2. Heavy hydrogen: It is manufactured by the electrolysis of heavy water containing a little of H₂SO₄ or NaOH to make the solution conducting.

In the laboratory, it can be prepared by the action of heavy water on sodium metal.
2D₂O(1) + 2Na(s) → 2NaOD(aq) + D₂ (g)

Question 2.
Describe the different method by which concentration of Hydrogen peroxide.
Answer:
Concentration of hydrogen peroxide. Hydrogen peroxide obtained by any method is always in the form of a dilute solution. A great care is to be taken for concentrating its solution because it is unstable and decomposes on heating.
2H₂O₂ → 2H₂O + O₂
The decomposition of H₂O₂ is catalysed by the ions of heavy metals present as impurities. The solution of H₂O₂ is concentrated by following methods.
(1)By careful evaporation on a water bath. A dilute solution of H₂O₂ is taken in evaporation dish and is heated at 313-323 K. Water evaporates slowly and hydrogen peroxide solution of about 45-50% strength is obtained.

(2) By dehydration in a vacuum desiccators. The dilute (50%) solution of H₂O₂ obtained is further concentrated by placing the same in a vacuum desiccators containing concentrated H₂SO₄ as a dehydrating agent. Here water vapours are absorbed by concentrated sulphuric acid. This is shown in the diagram.

(3) By distillation under reduced pressure. The solution of hydrogen peroxide is further concentrated by subjecting it to distillation under reduced pressure. The solution is distilled at 308-313 K under a reduced pressure of 15 mm Hg. Water present in the solution distills over leaving behind about 98-99% concentrated solution of hydrogen peroxide.

(4) By crystallization. The last traces of water present in H₂O₂ are removed by freezing it in a freezing mixture of solid CO₂ and ether. The crystals of hydrogen peroxide separate out. These crystals are removed, dried and then remelted to obtained 100% pure hydrogen peroxide.

(5) Storage of hydrogen peroxide. In order to check the decomposition of hydrogen peroxide, a small amount of acetanilide (i.e., negative catalyst) is added to it before storing the hydrogen peroxide.
Hydrogen peroxide cannot be concentrated by distillation at ordinary pressure because it undergoes decomposition into water and oxygen as it is highly unstable liquid. It decomposes even on long standing or on heating.

Question 3.
What different methods are used for softening the hard water ? Explain the principle used in each method ? (Write any two)
Answer:
Hard water can be softened by the following methods depending upon the nature of hardness.
(a) Temporary hardness :
1. By boiling. It can be removed by merely boiling the water. Boiling decomposes the bicarbonates to give carbon dioxide and insoluble carbonates, which can be removed by filtration.

2. Clark’s process : Temporary hardness can be removed, by the addition of a calculated amount of lime, where magnesium carbonate or calcium carbonate is precipitated.
Ca(HCO₃)₂ + Ca(OH)2 → 2CaCO₃+2H₂O
Mg(HCO₃)₂ +Ca(OH)₂ → CaCO₃+MgCO₃ +2H₂O

(b) Permanent hardness: With sodium carbonate. On treatment with washing soda, Ca²⁺ and Mg²⁺ in hard water are precipitated. The precipitate of the insoluble carbonates thus formed is removed by filtration.

The anion may remain in solution but in this form it is not dangerous.

(ii) Ion-exchange method. The common substances used for this process is zeolite which is hydrated sodium aluminium silicate, NaAl(SiO₂)₂. The exchange occurs when on passing over the zeolite bed, sodium ions from zeolite are replaced by calcium and magnesium ions. Thus .
2NaZe + Ca²⁺ → (Ze)₂ Ca + 2Na+ ; 2NaZe + Mg²⁺ → (Ze)₂ Mg + 2Na⁺
When all the sodium ions of zeolite has been replace, the zeolite is said to be exhausted. It can be regenerated by treatment with strong solution of sodium chloride.
2Na + (Ze)₂ Ca → 2ZeNa + Ca²⁺

Question 4.
How can you show that H₂O₂ works both as oxidizing and reducing agent by giving examples.
Answer:
Oxidising properties. H₂O₂ has a tendency to accept electrons in chemical reactions and thus behaves as an oxidizing agent in both acidic and alkaline medium.
(a) In acidic medium
H₂O₂ → H₂O + O
H₂O₂ + 2H⁺ + 2e^– → 2H₂O
Example:
2Fe²⁺ + 2H⁺ + H₂O₂ → 2Fe³⁺ + 2H₂O
b) In alkaline medium H₂O₂ + OH^– + 2e^–
Example:
\(3 \mathrm{Cr}^{3+}+4 \mathrm{H}_{2} \mathrm{O}_{2}+10 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_{4}^{2-}+8 \mathrm{H}_{2} \mathrm{O}\)

Reducing properties. H₂O₂ can give electrons in a few reactions and thus behaves as a reducing agent.
a) In acidic medium, H₂O₂ → O₂ + 2H⁺ + 2e^–
\(2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{M}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{O}_{2}\)

b) In alkaline medium, H₂O₂ + 2OH^– → 2H₂O + O₂ + 2e^–
2Fe³⁺ + H₂O₂ + 2OH^– → 2Fe²⁺ + O₂ + 2H₂O

Question 5.
A colourless and odourless gas is used for the hydrogenation of vegetable oils, name the gas. Also name the process when the gas reacts with dinitrogen at 673 K, 200 atm and in presence of Fe as catalyst; to produce another gas that gives brown precipitate with Nessler’s reagent.
Answer:
The agent is H₂ (dihydrogen)
It is a Haber’s process. The reaction is

Question 6.
(a) What do you understand by metallic hydrides?
(b) Give four uses of hydrogen peroxide.
Answer:
(a) Metallic hydrides are hydrides formed by transition metals with hydrogen. They are non-stoichiometric compounds, i.e., they do not have fixed composition.

(b)

• It is used as oxidizing agent and reducing agent.
• Its dilute solution is used as mild antiseptic.
• It is used as bleaching agent.
• Cone. H₂O₂ is used as rocket fuel.

Question 7.
Write balanced equations for the following:
(a) Hydrated Barium peroxide reacts with orthphosphoric acid.
(b) H₂O₂ is allowed to react with PbS (lead sulphide)
Answer:
(a) 3BaO₂.8H₂O + 2H₃PO₄ → Ba₃ (PO₄)₂ + 2H₂O₂ + 23H₂O
(b) PbS + 4H₂O₂ →PbSO₄ +4H₂O₂

Question 8.
How is hydrogen peroxide prepared industrially ? Explain why it is stored in coloured wax-lined glass or plastic bottles ?
Answer:
Industrially, H₂O₂ is prepared by the auto oxidation of 2-ethylanthraquinol to 2-ethylanthraquinone. H₂O₂ is stored in wax-lined glass or plastic bottles so that it does not decompose into H₂O and O₂ .
Because it is decomposed to H₂O and O₂ when comes in contact with sunlight.

Question 9.
(a) Distinguish between temporary hardness and permanent hardness.
(b) What happens when lead sulphide is reacted with hydrogen peroxide solution ?
Answer:
(a)

Temporary Hardness	Permanent Hardness
1. It is caused by bicarbonates of Ca2+ and Mg2+	1. It is caused by sulphates and chlorides of Ca2+    and Mg2+
2. It is removed by boiling.	2. It is removed by adding washing soda.

(b) PbS04 is formed which is white PbS + 4H₂O₂ → PbSO₄ + 4H₂O

Question 10.
(a) What is the importance of heavy water with regard to nuclear power generation?
(b) What happens when acidified ferrous sulphate solution is reacted with hydrogen peroxide solution ?
Answer:
(a) It is used as coolant and moderator in nuclear reactor.
(b) 2FeSO₄ +H₂SO₄ +H₂O₂ → Fe₂(SO₄)₃ +2H₂O(I), have hydrogen peroxide acts as oxidizing agent.

Question 11.
Why hard water does not form lather with soap? What advantage has soap over detergents?
Answer:
It is because Ca²⁺ and Mg²⁺ ions present in hard water react with soap to form Ca²⁺ and Mg²⁺ salts of fatty acids which are insoluble in water.
Advantages of soap over detergent; soap are biodegradable where as detergents are non-biodegradable.

Question 12.
Advantage of soap over detergent : Soaps are biodegradable whereas detergents are non-biodegradable.
Difference in chemical behaviour of compound of hydrogen with elements of atomic number 17 and 20.
Answer:
Atomic number 17 = Chlorine, and Altomic number 20 = Calcium.
(b) H₂ + Cl₂ → 2HCl
Cl₂ is oxidizing agent, H₂ is reducing agent.
Ca + H₂ → CaH₂

Question 13.
Reaction of steam on hydrocarbon or coke at high temperature in presence of catalyst yields a mixture of gases.
(a) Name the gases produced.
(b) Write a reaction of the above mentioned process.
(c) What is a specific name given to the mixture of gases?
(d) Write the reaction when the mixture of gases is reacted with steam in presence of iron chromate as catalyst.
Also give the special name of this reaction.
Answer:
(a) Carbon monoxide and dihydrogen

(c) Synthetic gas or syngas or water gas.

Water-gas shift reaction is the special name of this reaction.

Question 14.
The mixture of hydrazine and hydrogen peroxide with copper (II) catalyst is used as a rocket propellant. Why ? Write the reactions involved.
Answer:
When hydrazine and hydrogen peroxide react in presence of Cu²⁺ Catalyst the following reaction is takes place.
\(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \stackrel{\mathrm{Cu}^{2+}}{\longrightarrow} \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)
Since the formation of nitrogen and steam is accompanied with evolution of large amount of heat thus act as a rocket propellant.

Question 15.
A gas, A obtained by heating aluminum with sodium hydroxide was passed over a black solid, B at 773K when a grey metal, C and a colourless liquid, D were obtained. The metal C displaces copper from copper sulphate solution and becomes passive on treatment with cone. IINO3. Identity A, B, C and D and explain all the reactions.
Answer:
(i) 2Al(s) + 2Na0H)(aq) + 2H₂O(l) → 2NaAlO₂(aq) + 3H₂(g) Dihydrogen(A)

(iii) Fe(s) + CuSO₄(aq) → FeSO₄(sq) + Cu(s)

Iron is rendered passive due to the formation of ferrosoferric oxide or magnetic oxide (black solid (B))

Question 16.
Hydrogen peroxide acts both as an oxidizing agent and as a reducing agent in alkaline solution towards certain first row transition metal ions. Illustrate both these properties of H2O2 using chemical equations.
Answer:
Oxidising agent: 2Cr(OH)₃ + 4NaOH + 3H₂O₂ → 2Na₂CrO₄ + 8H₂O
Here, Cr³⁺ gets oxidized to Cr⁶⁺
Reducing agent: 2K3[Fe(CN)₆] + 2KOH + H₂O₂ → 2K₄ [Fe(CN)₄ ] + 2H₂O + O₂
Here, Fe³⁺ gets reduced to Fe²⁺.

Question 17.
A white is either Na₂O or Na₂O₂. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid
1. Identify the substance and explain with balanced equation.
2. Explain what would happen to the red litmus if the white solid were the other compound.
Answer:
1. Na₂O₂ + 2H₂O → 2NaOH + H₂O₂
H₂O₂ thus produced turns red litmus paper white due to its bleaching action.

2. Na₂O + H₂O → 2NaOH
NaOH thus produced will turn red litmus blue.

Question 18.
The process \(\frac { 1 }{ 2 }\) H₂(g) + e^– → H^– (g) is endothermic ((AH =+151kJmor^-1), yet salt like hydrides are known. How do you account for this?
Answer:
It is true that formation of hydride(H^–) ion is an endothermic process, yet alkali and alkaline earth metals from salt like hydrides. This is due to the reason that high lattice energy released (energy released during the formation of solid metal hydrides from their corresponding ions, i.e., M⁺ and H^–) more than compensates the energy needed for the formation of H^– ions from H₂ gas.

1st PUC Chemistry Hydrogen Numerical Solved Examples:

Question 1.
Calculate the volume strength of a 3% solution of H₂O₂ .
Answer:
100 mL of solution contains H₂O₂ = 3g
1000 mL of solution will contain H₂O₂ = \(\frac { 3 }{ 100 }\) x lOOO = 30g
H₂O₂ decomposes:

68g of H₂O₂ give at NTP = 22.4L
30g of H₂O₂ will give O₂ at NTP = \(\frac { 22.4 }{ 68 }\) x 30 = 9.88L = 9880mL
But 30g of H₂O₂ is present in 1000 mL of H₂O₂
1000 mL of H₂O₂ solution gives O₂ at NTP = 9880 mL
lmL of H₂O₂ solution will give O₂ at NTP = \(\frac { 9880 }{ 1000 }\) = 9.88 mL
Hence volume strength of 3% H₂O₂ = 9.88

Question 2.
30 mL of H₂O₂ solution after acidification required 30 mL of N/10 KMnO₄ solution for complete oxidation. Calculate the percentage and volume strength of H₂O₂ solution.
Answer:
To calculate the normality of H₂O₂ solution, apply normality equation

For H₂O₂ , V₁ = 40mL, N₁ =.? ; For KMnO₄, V₂ = 30, N₂ = \(\frac { N }{ 10 }\)
N₁ × 30 = \(\frac { N }{ 10 }\) x 30 N₁ = \(\frac { N }{ 10 }\) x \(\frac { 30 }{ 30 }\) = \(\frac { 1 }{ 10 }\) N = 0.1N
Strength = Equivalent wt x Normality = 17 × .01 = 1.7 g/L
% of strength = \(\frac { 1.7 }{ 1000 }\) × 100 = 0.17%
H₂O₂ decomposes as :

68g of H₂O₂ give O₂ at NTP = 22.4L
1.7g of H₂O₂ will give O₂ at NTP = \(\frac { 22.4 }{ 68 }\) × 1.7 = 0.56L

Question 3.
Calculate the normality of 20 volume solution of H₂O₂.
Answer:
20 volume of H₂O₂ solution means that 1L of this solution will liberate 20L of O₂ at NTP

22.4L of O₂ at NTP produced from H₂O₂ = 68g
20L of O₂ at NTP is produced from H₂O₂ \(\frac { 68 }{ 22.4 }\) × 20 = 60.71g
Strength of H₂O₂ = 60.71 g/L
Gram equivalents of H₂O₂ = \(\frac { 60.71 }{ 17 }\) = 3.57
Normality = \(\frac { 3.57 }{ 1 }\) =3.57 N

Question 4.
Calculate the amount of Hydrogen peroxide presents in 10 ml of 25 volume solution of H₂O₂ .
Answer:

22.4L of O₂ at NTP is produced from H₂O₂ = 68g
25 L of O₂ at NTP is produced from H₂O₂ = \(\frac { 68 }{ 22.4 }\) × 24 = 75.9g
Strength of 25 volume H₂O₂ = 75.9 g/L
Amount of H₂O₂ present in lOmL = \(\frac{75.9}{\mathrm{H}_{2} \mathrm{O}_{2}}\) × 10 = 0.759g

Question 5.
10 ml of given H₂O₂ solution contains 0.91 g of H₂O₂. Express its strength in volumes.
Answer:

Amount of H₂O₂ in 1L \(\frac{0.91 \times 1000}{10}\) = 91g
68g of H₂O₂ give O₂ at NTP = 22.4L
91 g of will give O₂ at NTP = \(\frac{22.4}{68}\) × 91 = 30
Since 91 g of is present in 1L, Volume strength = 30 volume

Question 6.
Calculate the strength of 5 volume H₂O₂ solution.
Answer:
5 volume H₂O₂ solution means that 1L of 5 volume H₂O₂ on decomposition gives 5L of O₂ at NTP.

22.4 of O₂ at NTP is produced from H₂O₂ = 68 g
5 L at O₂ at NTP is produced from H₂O₂ = \(\frac { 68 }{ 22.4 }\) x 5 = 15.18 g
But 5L of O₂ at NTP is produced from 1L of 5 volume H₂O₂
Strength of H₂O₂ in 5 volume H₂O₂ = 15.18 g/L
Percentage strength of H₂O₂ solution = \(\frac { 15.18 }{ 1000 }\) xlOO = 1.518%

Question 7.
What mass of hydrogen peroxide will be present in 2 litres of a 5 molar ‘ solution ? Calculate the mass of oxygen which will be liberated by the decomposition of 200 ml. of this solution.
Answer:
Molar mass of H₂O₂ = 34 g mol^-1
1L of 5M solution of H₂O₂ will contain H₂O₂ = 34 × 5 g
2L of 5M solution of H₂O₂ will contain H₂O₂ =2 × 34 × 5 = 340 g
200 mL of 5M solution will contain H₂O₂ = \(\frac { 340 }{ 2000 }\) × 200 = 34 g
2H₂O₂ → H₂O + O₂
68g of H₂O₂ on decomposition will give O₂ = 32g
34g of H₂O₂ on decomposition will give O₂ = \(\frac { 32 }{ 68 }\) × 34=16g

Question 8.
To a 25 mL H₂O₂ solution, excess of acidified solution of K1 was added. The iodine liberated required 20.0mL of 0.3 N Na₂S₂O₃ solution. Calculate the volume strength of H₂O₂ solution.
Answer:
Applying normality equation,

N₁ × 25 = 0.3 × 20

Strength of H₂O₂ = 0.24 × 17 = 4.08 g/L

68 g of H₂O₂ produce at NTP = 22400mL
4.08 g of H₂O₂ will produce O₂ at NTP = \(\frac { 22400 }{ 68 }\) × 4.08 = 1344 mL
Now 4.08 g of H₂O₂ are present in 1000 mL of solution.
1000 mL of H₂O₂ solution 1344 mL of O₂ at NTP
1 mL of H₂O₂ solution will give = \(\frac { 1344 }{ 1000 }\) = 1.344 mL of O₂ of NTP.

Question 9.
Calculate the volume of 10 volume H₂O₂ required to neutralize 200 mL of 2 N KMn04 in acidic medium.
Answer:

22.4 L of O₂ at NTP is produced from H₂O₂ = 68g
10L of O₂ at NTP is prodced from H₂O₂

Gram equivalents of H₂O₂ = \(\frac { 30.357 }{ 17 }\) =1.79
Normality of H₂O₂ = \(\frac { 1.79 }{ 1 }\) =1.79 N

1. 79 × V₁ = 2 × 200

Question 10.
0.5L each of three samples of H₂O₂ labelled 10 vol, 15 vol and 20 vol are mixed and then diluted with equal volume of water. What is the strength of resultant H₂O₂ solution ?
Answer:
Volume strength of H₂O₂ solution = 5.6 × Normality
Normality = \(\frac{\text { Vol. strength }}{5.6}\)
Normality of 10 vol of H₂O₂ = \(\frac { 10 }{ 5.6 }\) N
Normality of 10 vol of H₂O₂ = \(\frac { 15 }{ 5.6 }\) N
Normality 20 vol of H₂O₂ = \(\frac { 20 }{ 5.6 }\) N
Let 500 mL of each solution is mixed and then total volume of mixture becomes = 1500 mL. Since it is diluted with equal volume of water, total volume becomes 3000 mL.
Applying normality equation N₁V₂ + N₂V₂ + N₃V₃ = N₄V₄
\(\begin{array}{l}{\frac{10 \times 500}{5.6}+\frac{15 \times 500}{5.6}+\frac{20 \times 500}{5.6}=\mathrm{N}_{4} \times 3000} \\ {\mathrm{N}_{4}=\frac{45 \times 500}{5.6 \times 3000}=1.339 \mathrm{N}}\end{array}\)

Question 11.
Volume strength of resulting solution = 1.339 × 5.6 = 7.5 the degree of hardness of a given sample of hard water is 60 ppm. If the entire hardness is due to MgSO₄, how much of MgSO₄ is present per kilogram of hard water ?
Answer: Degree of hardness of water = 60ppm
Since degree of hardness is the number of parts of calcium carbonate or equivalent to calcium and magnesium salts present in a million parts of water by mass.
10⁶ g of water contain 60 g of CaCO₃
Now 1 mol of CaCCO₃ = 1 mol of MgSO₄
100g of CaCOa = 120g of MgSO₄
10⁶ g of water contain MgSO₄ = \(\frac{60 \times 200}{100}=72 \mathrm{g}\) = 72 g
10³ g of water will contain MgSO₄ = \(\frac{72}{10^{6}} \times 10^{3}\) = 0.072 g
1 kg of water contain MgSO₄ = 72 mg

Question 12.
Find the volume strength of 1.6N NH₂O₂ solution.
Answer:
Strength = Normality × Eq. wt. Eq. wt. of H₂O₂ =17
∴ Strength of 1.6 NH₂O₂ solution = 1.6 × l7gL^-1
Now 68g of H₂O₂ gives 22400 mL O₂ atNTP/STP
∴ 1.6 × 17g of H₂O₂ will give = \(\frac { 22400 }{ 68 }\) × l.6 × l7 = 8960 mL of O₂ at STP
But 1.6 × 17g of H₂O₂ are present in 1000 mL of H₂O₂ solution.
Hence 1000 mL of H₂O₂ solution gives 8960 mL of O₂ at STP
1 mL of H₂O₂ will give = 8.96 mL of O₂ at STP
Hence the volume strength of 1.6 N H₂O₂ solution is = 8.96 volume.

Question 13.
Calculate the volume of 10 volume H₂O₂ solution that will react with 200 mL of 2N KMnO₄ in acidic medium.
Answer:
Normality 10 volume H₂O₂ = \(\frac{10 \times 68}{22.4 \times 17}=\frac{10}{5.6} \mathrm{N}\)
Applying normality equation
\(\begin{aligned} \mathrm{N}_{1} \mathrm{V}_{1} &=\mathrm{N}_{2} \mathrm{V}_{2} \\\left(\mathrm{H}_{2} \mathrm{O}_{2}\right) &\left(\mathrm{KMnO}_{4}\right) \\ \frac{10}{5.6} \times \mathrm{V}_{1} &=2 \times 200 \\ \mathrm{V}_{1} &=\frac{2 \times 200 \times 5.6}{10}=224 \mathrm{cm}^{3} \end{aligned}\)

Question 14.
Calculate the percentage strength and strength in grams per litre of 10 volume hydrogen peroxide solution.
Answer:
Hydrogen peroxide decomposes on heating according to the equation :

From the equation,
22.4 litres of O₂ at N.T.P. are obtained from 2 × 34 or 68 g of H₂O₂
∴ 10 ml of O₂ at N.T.P. will be obtained from \(\frac { 68 }{ 22400 }\) × 10g of H₂O₂
But 10 ml of O₂ at N.T.P. are produced from 1 ml of 10 volume H₂O₂ solution
Thus 1 ml of 10 volume H₂O₂ solutioin contains \(\frac { 68 }{ 22400 }\) × 10g of H₂O₂
∴ 100 ml of 10 volume H₂O₂ solution will contain \(\frac { 68 }{ 22400 }\) × \(\frac { 10 }{ 1 }\) x 100 = 3.036g
Thus, a 10 volume H₂O₂ solution is approx. 3% alternatively, 1000 ml of 10 volume
H₂O₂ will contain \(\frac { 68 }{ 22400 }\) × 10 × 1000 = 30.36g
Therefore, strength of H₂O₂ in 10 volume
H₂O₂ = 30.36gL^-1

Question 15.
Calculate the normality of 20 volume hydrogen peroxide solution.
Answer:
To calculate the strength in g/1 of 20 volume H₂O₂ solution
By definition, 1 litre of 20 volume H₂O₂ solution on decomposition gives 20 litres of oxygen at NTP
Consider the chemical equation,

Now 22.4 litres of O₂ at NTP will be obtained from H₂O₂ = 68g

Thus, the strength of 20 volume H₂O₂ solution = 60.7 g/1

Step 2 : To calculate equivalent weight of H₂O₂

From the above equation, 32 parts by wt. of oxygen are obtained from 68 parts by wt. of H₂O₂
8 parts by wt of oxygen will be obtained from \(\frac { 68 }{ 32 }\) x 8 = 17 parts by wt. of H₂O₂
Eq. wt of H₂O₂ = 17

Step 3: To calculate the normality of 20 volume H₂O₂ solution
Now we know that, Normality = \(=\frac{\text { Strength }}{\text { Eq.wt }}=\frac{60.7}{17}=3.57\)
Hence the normality of 20 volume H₂O₂ solution = 3.57N

Question 16.
Find the volume strength of 1.6 N H₂O₂ solution.
Answer:
We know that strength = Normality × Eq. wt; and Eq. wt of H₂O₂ = 17
Strength of 1.6 N H₂O₂ solution = 16 × 17 g/1
Now 68 g of H₂O₂ gives 22400 ml O₂ at NTP
16 × 17 g of H₂O₂ will give \(\frac { 22400 }{ 68}\) × 1.6 × 17 = 8960 ml of O₂ at NTP
But 1.6 × 17 g of H₂O₂ are present in 1000 ml of H₂O₂ solution.
Hence 1000 ml of H₂O₂ solution gives 8960 ml of O₂ at STP
lml of H₂O₂ solution will give = \(\frac { 8960 }{ 1000}\) = 8.96 ml of O₂ at NTP
Hence the volume strength of 1.6N H₂O₂ solution = 8.96 volume

Question 17.
Calculate the volume strength of a 35% solution of H₂O₂.
Answer:
Step 1: To calculate the amount of H₂O₂ present in one litre of 3% solution.
100 ml of H₂O₂ solution contain H₂O₂ = 3g
1000 ml of H₂O₂ solution will contain H₂O₂ = \(\frac { 3 }{ 100}\) x 1000 = 30 g

Step 2 : Calculate the volume strength. Consider the chemical equation.

Now 68 g of H₂O₂ give O₂ at NTP = 22.4 litres
30g of H₂O₂ will give O₂ at NTP = \(\frac { 22.4 }{ 68}\) × 30 = 9.88 litres = 9880 ml
But 30g of H₂O₂ are present in 1000 ml of H₂O₂
Hence 1000 ml of H₂O₂ solution gives O₂ at NTP = 9880 ml
1 ml of H₂O₂ solution will give O₂ at NTP = \(\frac { 9880 }{ 1000}\) = 9.88 ml
Hence the volume strength of 3% H₂O₂ solution = 9.88

Question 18.
What is the mass of hydrogen peroxide present in 1 litre of 2M solution ? Calculate the volume of oxygen at STP liberated upon complete decomposition of 100 cm3 of the above solution.
Answer:
Step 1: To calculate the mass of H₂O₂ in 1 litre of 2M solution.
Molecular mass of H₂O₂ = 2 × 1 + 2 × 16 = 34 amu
By definition, I litre of 1M H₂O₂ contains 34 g of H₂O₂
1 litre of 2M H₂O₂ will contain H₂O₂ = 34 × 2 = 68 g

Step 2 : To calculate the volume of O₂ liberated at STP fromlOO cm³ of 2M solution.
1 litre 2M H₂O₂ solution contains H₂O₂ = 68 g
100 cm³ of 2M H₂O₂ solution will contain \(\frac { 68 }{ 1000}\) × 100 = 6.8g of H₂O₂
The equation representing the decomposition of H₂O₂ is
2H₂2O 2 × 34 = 68g → 2H²O + O₂2 2400cm³ at STP
Now 68 g of H₂O₂ at STP give O₂ = 22400 cm³ k of O₂
6.8 g of H₂O₂ at STP will evolve O₂ = \(\frac { 22400 }{ 68}\) x 6.8 = 2240cm³ = 2.24 mL reg

Question 19.
30 ml of a H₂O₂ solution after acidication required 30 ml of N/10 KMnCh solution for complete oxidation. Calculate the percentage and volume strength of H₂O₂ solution.
Answer:
Step 1: To determine the normality of H₂O₂ solution. From the given solution.
For H₂O₂ V₁ = 30ml, N₁ = ? For KMnCh, V₂ = 30ml, N₂ = N/10
Applying normality equation N₁V₁ = N₂V₂
i.e., 30 × N₁ = 30 × 1/10
Thus, the normality of H₂O₂ solution = 0.1N

Step 2: To determine the percentage strength of H₂O₂ solution.
We know that, H₂O₂ → 2H⁺ + O₂ + 2e-_–
Eq. wt. of H₂O₂ = 30 / 2 = 17
Hence strength of H₂O₂ solution = Normality × Eq. wt = 0.1 × 17 = 1.7 g/litre
Percentage strength of H₂O₂ = \(\frac{1.7 \times 100}{1000}=0.17 \%\)

Step 3 : To determine the volume strength of H₂O₂ solution.
Consider the chemical equation 2H₂O₂68g → 2H₂O + O₂22400ml atNTP
Now 68 g of H₂O₂ give O₂ at NTP = 22400 ml
1.7g of H₂O₂ will give O₂ = \(\frac{22400}{68} \times 1.7=560 \mathrm{ml}\)
But 1.7 g of H₂O₂ are present in 1000 ml of H₂O₂ solution.
Hence 1000 ml of H₂O₂ solution gives 560 ml of O₂ at NTP
1ml of H₂O₂ solution gives = \(\frac { 560 }{ 1000 }\) = 0.56 ml of O₂at NTP
Volume strength of H₂O₂ solution = 0.56

Question 20.
One litre of a sample of hard water contains Img of CaCl₂ and 1 mg of MgCl₂ Find out the total hardness in terms of parts of CaCO₃ per 10⁶ parts of water by mass, (i) Mol. Mass of CaCl₂ = 111
Answer:
Now lllg CaCl₂ = lOOg of CaCO₃
1 mg of CaCl₂ = \(\frac { 100 }{ 111 }\) × 1 mg of CaCO₃= 0.9 mg of CaCO₃

(ii) Mol. Mass of MgCk = 95
Now 95g of MgCl₂ = lOOg of CaCO₃
1 mg of MgCl₂ = \(\frac { 100 }{ 95 }\) × 1 mg of CaCO3 = 1.05 mg of CaCO₃
Thus, 1 litre of hard water contain = 0.90 + 1.05 = 1.95 mg of CaCO₃.

Question 21.
A 5.0 cm³ solution of H₂O₂ liberates 0.50 g of iodine from an acidified K1 solution. Calculate the strength of H₂O₂ solution in terms of volume strength at STP.
Answer:
(a) 2K1 + H₂SO₄ + H₂O → K₂SO₄ + 2H₂O +I₂
From the above equation, H₂O₂ s I₂ (both are required)
34g of H₂O₂ = 254g of I₂
∴ 0.508 g of I₂ will be liberated from H₂O₂ = \(\frac { 34 }{ 254 }\) x 0.508 = 0.068g

(b) The decomposition of H₂0₂ occurs as
2H₂O₂2X34 = 68g → 2H₂O + O₂22400 cm³ at NTP
∴ 0.068 of H₂O upon decomposition will give O₂ = \(\frac { 22400 }{ 68 }\) x 0.068 = 22.4ml

(c) Now 5.0 cm³ of H₂O₂ solution gives O₂ = 22.4cm³ at NTP
∴ 1.0 cm³ of H₂O₂ solution gives O₂ = \(\frac { 22.4 }{ 5 }\) = 4.48 cm³ at NTP
Thus volume strength of give H₂O₂ solution = 4.48

Question 22.
Calculate molarity and normality of 20 volume of H₂O₂. What is its mass percentage?
Answer:
vol. of H₂O₂ means 1 ml of H₂O₂ gives 20ml of O₂ at STP

22400 ml of O₂ of STP is given by 68g of H₂O₂
20 ml of O₂ at STP is given by \(\frac { 68 }{ 22400 }\) x 20g of H₂ = 0.0606 H₂O₂
1000 ml of H₂O₂ contains 0.0606 × 1000 = 60.6 = 60.6
∴ Molarity = \(\frac{60.6}{\text { Mol. wt of } \mathrm{H}_{2} \mathrm{O}_{2}}\)
∴ M = \(\frac { 60.6 }{ 34.0 }\) = 1.78 \(\left[\because \mathrm{Eqwt}=\frac{\mathrm{mol} \mathrm{wt}}{2}\right]\)

Question 23.
20 ml of solution having 0.2 g of impure H₂O₂ reacts with 0.316 of KMnO₄ in acidic medium. Calculate (a) Purity of H₂O₂ (b) Volume of dry oxygen evolved at 0° C and 750 mm pressure.
Answer:
2MnO^–₄(aq) + 6H⁺ (aq) + 5H₂O₂(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 50(g)
158 x 2 g of KMnO₄ reacts with 34 × 5 g of H₂O₂
0.316 g of KMnO₄ will react with \(\frac { 170 }{ 0.2 }\) x 0.316 g of H₂O₂ = 0.169 g of H₂O₂
=> Percentage purity of H₂O₂ = \(\frac { 0.169 }{ 0.2 }\) x 100 = 85%
Also 316 g of KMn04 at STP produces 5 × 22 L of O₂

Conversion to 750 mm pressure 760 x 0.112 = x × 750 X = 0.113 L

You can Download Chapter 3 Motion in a Straight Line Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 3 Motion in a Straight Line

1st PUC Physics Motion in a Straight Line TextBook Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:

1. a railway carriage moving without jerks between two stations.
2. a monkey sitting on top of a man cycling smoothly on a circular track.
3. a spinning cricket ball that turns sharply on hitting the ground.
4. a tumbling beaker that has slipped off the edge of a table.

Answer:

1. This is an example of uniform linear motion. Hence the carriage can be considered as a point object.
2. The monkey also undergoes motion smoothly on the circular track. Hence it can be heated as a point object.
3. The ball is spinning and undergoes changes in the plane of its motion when it hits the ground. Various parts of the ball experience different forces when the ball hits the ground. It is a rigid body and cannot be treated as a point object.
4. Different parts of the beaker experience different magnitude of forces during its motion. Hence it cannot be treated as a point object.

Question 2.
The position-time (x-i) graphs for two children A and B returning from their school! O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;

1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/ different) time.
5. (A/B) overtakes (B/A) on the road (once/twice).

Answer:
1. The distance between P and O (the school) is less then the distance between Q and O. (from the graph)
⇒ A lives closer to the school than B.
2. From the graph, we see that the line for A starts before the line for B.
⇒ A starts from the school earlier than B.
3. Velocity = Slope of the x-t graph. Clearly, the slope for B > slope for A.
⇒ B walks faster than A
4. From the graph, we see that both lines end at the same time.
⇒ A and B reach home at the same time.
5. From the graph, we see that the lines intersect once. Also, B walks faster.
⇒ B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Time for the woman to reach her office = \(\frac{2.5 \mathrm{km}}{5 \mathrm{km} \mathrm{h}^{-1}}\)
= 30 min.
∴ She reaches her office at 9.30 am. Time for the woman to return home = \(\frac{2.5 \mathrm{km}}{25 \mathrm{km} \mathrm{h}^{-1}}\)
= 0.1 h = 6 min.
∴ She reaches her home at 5.06 pm

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:

From the above table, see that every 8 s the drunkard moves 2 m forward.

By looking at the shape of the graph, we can see that the graph for entire motion will be like this:-

From the graph, we see that the drunkard will fall into the pit at 137 s.
Analytically:-
The drunkard covers 2 m every 8 s.
∴ He will cover 8 m in 32 s. When he takes 5 steps forwards, he will fall into the pit.
∴ Total time = 32 + 5 = 37 s.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Speed of the exhaust with respect to an observer on the ground = Speed of exhaust with respect to the plane – Speed of plane with respect to the ground, (minus sign because the plane and exhaust move in opposite directions)
= (1500 – 500) km h^-1
= 1000 km h^-1

Question 6.
A car moving along a straight highway with a speed of 126 km h^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer: v₀ = 126 km h^-1
= 126 × \(\frac{5}{18}\) m s^-1
= 35 m s^-1
Stopping distance = (x – x₀) = 200 m
Since the car is stopped , v = 0.
v₂ = v₀² + 2 a (x – x₀)
0 = 35² + 2a (200)
⇒ a = – 3.06 m s^-2
Retardation = 3.06 m s^-2
v = v₀ + at
⇒ 0 = 35 – 3.06 t
⇒ t = 11.4 s
Time for car to stop = 11.4 s.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B Just brushes past the driver of A, what was the original distance between them?
Answer:
Initially,

Relative Acceleration a = 1 ms^-2
v₀ =72 km h^-1 = 72 × \(\frac{5}{18}\) m s^-1
= 20 ms^-1
Relative displacement = x
Relative velocity initially = V₀ – V₀ = 0
X = 1/2 at²
= 1/2 × 1 × 50² = 1250 m
Original distance between the trains = 1250 m.

Question 8.
On a two-lane road, car A Is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:

V_A= 36 km h^-1 = 10 m s^-1
VB = Vc = 54 km h^-1 = 15 m s^-1
Velocity Of B with respect to A
= V_BA = V_B – V_A
= 15m -10 = 5 m s^-1
Velocity of C with respect to A .
V_CA = V_C – V_A
= 15 – ( -10) = 25 m s^-1
S = V_CA t
1000 m = 25 × t
⇒ t = 40 s
C will take 40 s to cover AC. In that time, B must cover AB to overtake and avoid collisions.
s = V_BA t + 1/2 at²
1000 = 5 × 40 + 1/2 × a × 40²
⇒ a = 1 ms^-2
Acceleration of B = 1 ms^-2

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving In either direction every T minutes. A man cycling with a speed of 20 km h^-1 In the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period Tof the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Let the speed of the bus be v km h^-1
Distance covered by one bus before the next one leaves the same town is vT For buses going in the same direction as the cyclist, frequency of the bus is 18 min i.e.,
\(\frac{v T}{\text { relative velocity of the bus w.r.t. the cyclist }}\) = 18 min
\(\frac{v T}{v-20}\) = 18……………………… (1)
Similarly, for buses going in the opposite direction, frequency of the bus = 6 min
\(\frac{v T}{\text { relative velocity of the bus w.r.t. the cyclist }}\) = 6 min
\(\frac{v T}{v+20}\) = 6 ……………………… (2)
\((1) \div(2)\)
⇒ \(\frac{v+20}{v-20}\)
Substitute for v in (1)
\(\frac{40 \times T}{40-20}\) =18 ⇒ T = 9 min.

Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.

1. What is the direction of acceleration during the upward motion of the ball?
2. What are the velocity and acceleration of the ball at the highest point of its motion?
3. Choose the X = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of X-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
4. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s^-2 and neglect air resistance).

Answer:
1. The direction of acceleration is vertically downwards (towards the Earth)
2. At the highest point, velocity = 0 m s^-1 acceleration = g = 9.8 m s^-2 vertically downwards.
3. During upward motion:

• position – positive
• velocity – negative
• acceleration – positive

During downward motion:

• position – positive
• velocity – positive
• acceleration – positive

4. v = 0 at the highest point
v₀ = 29.4 m s^-1
g = -9.8 m s^-2
v² = v₀² + 2 g h
0 = 29.4² = 2 × (-9.8) × 4
⇒ h = 44.1m
Height of ascension = 44.1 m
v = v₀ + g t
0 = 29.4 – 9.8 t
t = 3 s
Time of ascension = Time of descension
∴ Time for the ball to return = 6 s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

1. with zero speed at an instant may have non-zero acceleration at that instant
2. with zero speed may have non-zero velocity,
3. with constant speed must have zero acceleration,
4. with positive value of acceleration must be speeding up.

Answer:
1. True:
Consider a ball thrown vertically, upward. At the highest point, the speed is zero but the acceleration of the ball is non-zero       ( 9.8ms^-2 vertically downwards). Acceleration does not depend on instantaneous speed.

2. False:
Since the magnitude of velocity is speed, a body with zero speed must have zero velocity.

3. True:
In the case of a body rebounding with the same speed, the acceleration at the time of impact is infinite, which is not practical physically.

4. False:
This depends on the chosen positive direction. The statement is true when the direction of motion and acceleration are along the chosen positive direction.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Velocity of the ball when it hits the ground = v
v₀ = 0, h = 90 m, g = 9.8m s^-2
v² = v₀² + 2 g h
v² = 0² + 2 × 9.8 × 90
⇒ v = 42 ms^-1
Time for descent = t
v = v₀ + g t
42 = 0 + 9.8 t
t = 4.3 s
During the rebound, initial speed of the ball = 9/10 of v
= 9/10 × 42
= 37.8 ms^-1
Total time for ascent and descent = t₁
Net displacement = 0
0 = 37.8 t – 1/2 gt²
37.8 t = 1/2 × 9.8 × t² ⇒ t = 7.7 s
Time at which maximum height is achieved = 4.3 + 1/2 7.7
= 8.15 s

Question 13.
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show In both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Answer:
(a) Consider a person who leaves home for work in the morning and returns home in the evening.
The magnitude of the displacement of the person during the interval is zero but the total length of path covered = 2 × distance between the home and the workplace.
Magnitude of displacement is the magnitude of the shortest length between the two points. The second quantity is clearly always greater than or equal to the first quantity. The equality holds if the body has not moved at all (path length = magnitude of displacement = 0)

(b) Average velocity = \(\frac{\Delta x}{\Delta t}\). If Δ X is zero (like in the example explained in (a)) then average velocity = 0.
Average speed = \(\frac{\text { Total path length }}{\text { Time taken }}\)
2 × \(\frac{\text { Distance between the home and the work place }}{\text { Time taken to traverse this distance }}\)
= Clearly average speed ≥ magnitude of average velocity. Equality exists when body does not undergo any motion.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the

1. magnitude of average velocity, and
2. average speed of the man over the interval of time

(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 tp 40 min?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer:
v₁ = 5 km h^-1, v₂ = 7.5 km h^-1, d = 2.5 km
Time for the man to reach the market
= t₁ = d/v₁= 2.5/5 = 0.5 h = 30 min
Time for the man to return from the market
= t₂= d/v₂ = 2.5/7.5 = 1/3 h= 20 min

1. To find the average velocity,
Average velocity = \(\frac{\text { Total path length }}{\text { Total time taken }}\)

• During the interval between 0 to 30 min,
  Displacement = 2.5 km
  time = 30 minutes
  ∴ Average velocity = \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}\) = 5 km h^-1
• interval between 0 and 50 minutes,
  Displacement = 0
  ∴ Average velocity = 0
• During between 0 to 40 minutes,
  Displacement = 2.5 km – 7.5 kmh^-1 × 10 min = 1.25 km
  time = 40 minutes
  ∴ Average velocity = \(\frac{1.25 \mathrm{km}}{40 \mathrm{min}}\) = 1.875 km h^-1
  

2. To find the average speed:
Average speed = \(\frac{\text { Total path length }}{\text { Total time taken }}\)

• During the interval between 0 to 30 min
  Path length = 2.5 km
  time = 30 minutes
  ∴ Average speed = \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}\) = 5 km h^-1
• During the interval between 0 and 50 minutes,
  Path length = 2.5 km + 2.5 km
  Time taken = 50 minutes
  ∴ Average speed = \(\frac{2.5 \mathrm{km}+2.5 \mathrm{km}}{50 \mathrm{min}}\) = 6 km h^-1
• During 0 to 40 minutes
  Path length = 5 kmh^-1 × 30 min + 7.5 kmh^-1 × 10 min
  2.5 km + 1.25 km = 3.75 km
  time = 40 minutes
  ∴ Average velocity = \(\frac{3.75 \mathrm{km}}{40 \mathrm{min}}\) = 5.625 km h^-1

Question 15.
In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous speed is always equal to the magnitude of the instantaneous velocity because for very small instants of time the length of the path is equal to the magnitude of displacement.

Question 16.
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:
(a) From the graph, we see that for certain instants of time, the particle has 2 positions, which is not possible. Hence this graph cannot possibly represent one-dimensional motion of a particle.

(b) From the graph, we see that for certain instants of time, the particle has 2 velocities, which is not possible. Hence this graph cannot possibly represent one-dimensional motion of a particle.

(c) From the graph, we see that for some instants of time, the particle has negative speed. But speed is always a non-negative quantity. Hence this graph cannot possibly represent one-dimensional motion of a particle.

(d) From the graph, we see that for a time interval the total path length is decreasing. But total path length is always a nondecreasing quantity. Hence this graph cannot possibly represent one-dimensional motion of a particle.

Question 17.
Figure 3.21 shows the X- t plot of onedimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
No, it is wrong to make such statements about the trajectory of the particle because an x-t graph does not show the particle’s trajectory.
Context:
Consider a body dropped from the top of a tower at time t = 0. If the vertically downward direction is chosen as the positive direction, then the body’s x -t graph would resemble the one given in the question.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a thief’s car speeding away In the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m s^-1, with what speed does the bullet hit the thief’s car?
Answer:
Speed of the police van = v_p
= 30 km h^-1 = 25/3 ms^-1.
Speed of the thief’s car is
v_t = 192 km h^-1 = 160/3 ms^-1
Speed of the bullet = v_b = 150 ms^-1
Speed of the bullet with respect to the police car = v_b + v_p = 150 + 25/3
= 475/3 msv^-1
Speed with which the bullet hits the thief’s car = Speed of the bullet with respect to the thief’s car
= 475/3 – v_t = 475/3 – 160/3 =105 ms^-1

Question 19.
Suggest a suitable physical situation for each of the following graphs (Fig 3.22):


Answer:
(a) consider a ball which is pushed at some time t < 0 towards a wall. Upon rebounding from this wall it hits the opposite wall and comes to a stop. If x = 0 for the initial position, then this context may have the given x -t graph.

(b) Consider a ball thrown vertically upwards, with the vertically upward direction chosen as the positive direction. Each time it hits the ground, it loses a fraction of its velocity and finally comes to rest. The ball may be represented the given v -t graph in this context.

(c) Consider a cricket ball moving with a uniform velocity which is hit by the bat and then turns back. In this case, the a – t graph for the ball may be similar to the one given above.

Question 20.
Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t= 0.3 s, 1.2 s, -1.2 s.

Answer:
Let the maximum amplitude of the sine wave be A, where A is positive. We can see that the particle obeys the
equation x = – A sin \(\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right)\)
Where T = 2 s = period of the sine wave
∴ position = x = – A sin (π t)
velocity – v = \(\frac{dx}{dt}\) = – A_π cos (π t)
acceleration = a = \(\frac{dv}{dt}\) = A_π² sin(π t)
At t = 0.3 s
x = – A sin (0.3_π) = negative
v = – A cos (0.3_π) = negative
a = A _π² sin (0.3_π) = positive
Since sin(0.3_π) > 0 and cos(0.3_π) > 0
At t = 1.2 s
x = – A sin (1.2_π) = – A sin (1.2_π) = positive
v = – A _π cos (1.2_π) = – A cos (1.2_π) = positive
a = A_π² sin (1.2_π) = negative
since sin(1.2_π)<0 and cos(1.2_π) < 0
At t = – 1.2 s
x = – A sin (- 1.2_π) = A sin (1.2_π)= negative
v = – A it cos(- 1.2_π) = – A_π cos (1.2_π) = positive.
a = A_π² sin (- 1.2_π) = – A_π 2sin(1.2_π) = positive
Since sin(-θ) = – sinθt cos(-θ) = cosθ sin(1.2_π) <0 and cos(1.2_π) < 0

Question 21.
Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each Interval.

Answer:
The magnitude of the slope of the x – t graph is highest in 3 and least in 2. Hence, the average speed is greatest in 3 and least in 2. Also, the sign of the slope of the x -t graph is positive for 1 and 2 and negative for 3. Therefore sign of the average velocity ‘V’ is:-
v > 0 for intervals 1 and 2
v < 0 for intervals 3.

Question 22.
Figure 3.25 gives a speed-time graph of a particle In motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and an in the three Intervals. What are the accelerations at the points A, B, C and D?

Answer:
The magnitude of the slope of the speed-time graph is greatest for interval 2. Hence average acceleration is greatest in magnitude in interval 2. The value of speeds in interval 3 is the highest. Hence the average speed in interval 3 is highest. Sign of a = sign of the slope of the speed-time graph, a > 0 for 1 and 3 a < 0 for 2. At points, A, B, C and D, the tangent to the curve will be parallel to the time axis. Hence slope at the points is zero,
∴ a = 0 at points A, B, C and D.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1, 2, 3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:
Initial velocity u = 0.
Acceleration a = 1 ms^-2.
Acceleration time = 10 s
Let s_n be the total displacement (or distance in this case) in time ‘n’ seconds.
s_n = ut + 1/2 at²
s_n = 0(n) + 1/2 a(n²) = 1/2 an² for n ≤ 10
Distance covered in ‘n’^th second d_n = s_n – s_n – 1
= 1/2 an² – 1/2 a(n-1)²
= an – 1/2 a (for n ≤ 10)
= (n – 1/2) m
At the end of 10 seconds, velocity acquired v = u + at = 0 + a (10)
10 a = 10 ms^-1
Distance covered in the ‘n’^th second = 10m for n >10
∴ d_n = {(n-1/2)m n ≤ 10,
10m n > 10
The plot is a straight line inclined to the time axis for uniformly accelerated motion. (It is a straight line parallel to the time axis for uniform motion).

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^-1. How much time does the bail take to return to his hands? if the lift starts moving up with a uniform speed of 5 m s^-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer:
Initial velocity n = 49 ms^-1
Acceleration g = -9.8 ms²
Displacement s = 0
s = ut + 1/2 at²
0 = 49 t + 1/2 (-9.8) t²
⇒ t = 0 or t = 10 s
t = 0 represents initial time. The required ‘t’ here is t = 10 s. In the case where the lift also moves, the relative velocity of the ball with respect to the boy remains the same, i.e, 49 ms^-1. The relative displacement is also 0. Hence the time required in this case is also 10 s.

Question 25.
On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the

1. speed of the child running in the direction of motion of the belt?.
2. speed of the child running opposite to the direction of motion of the belt?
3. time taken by the child in (1) and (2)?

Which of the answers alter If motion is viewed by one of the parents?

Answer:
1. Speed of the child running in the direction of motion of the belt = Speed of the child with respect to the belt + Speed of the belt with respect to the ground
= 9 + 4 = 13 km h^-1.
2. Speed of the child running opposite to the direction of motion of the belt = speed of the child with respect to the belt speed of the belt with respect to the ground
= 9 – 4 = 5 km h^-1.
3. Time taken by the child in both cases is the same because the speed of the child with respect to the belt does not change.
Speed v = 9 km h^-1 = 2.5 ms^-1
Distance d = 50 m
Time t = \(\frac{\text { Distance }}{\text { Speed }}\) = \(\frac{50}{2.5}\) = 20 s
The answer to (a) and (b) when the motion is viewed by one of the parents is 9 km h^-1. This is because the parents are also on the belt and are moving with respect to the ground. They only see the motion of the child with respect to the belt. In (c), the answer remains unaltered because the speed of the child with respect to the belt does not depend on the speed of the parents.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with Initial speeds of 15 m s^-1 and 30 m s^-1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take
g = 10 ms^-2. Give the equations for the linear and curved parts of the plot.

Answer:
Initial velocity of stone 1 is u₁ = 15 ms^-1
Initial velocity of stone 2 = u₂ = 30 ms^-1
Acceleration = g = -10ms^-2 (vertically upwards direction is chosen as positive)
x₁ = 200 + u₁t + 1/2 at²
= 200 + 15 t – 5 t²
x₂ = 200 + u₂t + 1/2 at²
= 200 + 30 t – 5 t².
Also x₁ = 0 for t = 8 s and
x₂ = 0 for t = 10s.
∴ For 0 ≤ t ≤ 8 s
x₂ – x₁ = 15 t
At t = 8 s,
x₂ – x₁ = 120 m
For 8 < t; t ≤ 10 s
x₂ – x₁ = 200 + 30 t – 5 t² – 0
= 200 + 30 t – 5 t².
For t > 10 s,
x₂ – x₁ = 0
∴ The given graph is correct and because it matches with the equations obtained.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown In Fig. 3.28. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s.
(b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?.
Answer:
(a) Distance covered = Area under the speed – time graph
= 1/2 × 10 × 12 = 60 m
Average speed = \(\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{60 m}{10 s}\) = 6 ms^-1

(b) For 0 ≤ t ≤ 5 s,
Speed v = \(\frac{12}{5}\) t ms^-1
For 5 < t ≤ 10 s,
Speed v = 12 – \(\frac{12}{5}\) (t – 5) ms^-1
∴ Speed at t = 2 s is 4.8 ms^-1 and speed at t = 6 s is 9.6 ms^-1
Distance covered = Area of trapezium ABEF + Area of trapezium BCDE

= 1/2 (4.8 + 12) × 3 + 1/2 (12 + 9.6) × 1
= 36 m
Average speed = \(\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{36}{6-2}\) = 9 ms^-1

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂

1. x(t₂) = x(t₁ + v (t₁) (t₂ – t₁ + (1/2) a (t₂– t₁)²
2. v(t₂) = v(t₁) + a(t₂ – t₁)
3. v_average = (x(t₂) – x(t₁)/(t₂-t₁)
4. average = (v(t₂ – v(t₁))/(t₂ – t₁)
5. x(t₂) = x(t₁) + v_average (t₂ – t₁) + (1/2) a_average (t₂ – t₁)²
6. x(t₂) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer:

1. Wrong, because it is not known whether the acceleration ‘a’ is constant.
2. Wrong, because it is not known whether the acceleration ‘a’ is constant.
3. Correct. By definition.
4. Correct. By definition
5. Wrong. The formula should contain v₁ (t) instead of v_average.
6. Correct By definition.

1st PUC Physics Motion in a Straight Line One Mark Questions and Answers

Question 1.
When do you say that a body is in motion?
Answer:
A body is said to be in motion when it changes its position with respect to time and surrounding.

Question 2.
What is a particle?
Answer:
A particle is a geometrical mass point.

Question 3.
Define displacement.
Answer:
The change of position of a body in a particular direction is called displacement.

Question 4.
Can the displacement of a moving body be zero?
Answer:
Yes

Question 5.
Define speed.
Answer:
The rate of change of position of a body in any direction is known as speed.

Question 6.
Define velocity.
Answer:
The rate of change of position of a body in a particular direction is called velocity or Rate of displacement of a body is called velocity.

Question 7.
Define uniform velocity.
Answer:
If a body covers equal distances in equal intervals of time in a given direction, however small the intervals maybe then its velocity is said to be uniform.

Question 8.
Define variable velocity.
Answer:
If the velocity of a body either changes in magnitude or in direction or both, then its velocity is said to be variable.

Question 9.
Define average velocity.
Answer:
The average velocity of a body is defined as the ratio of its total displacement to the total time.

Question 10.
Define acceleration.
Answer:
Rate of change of velocity of a body is called acceleration.

Question 11.
Define uniform acceleration.
Answer:
If the velocity of a body changes by an equal amount in equal interval of time, however, small the interval maybe then its acceleration is said to be uniform.

Question 12.
What is a position-time graph?
Answer:
A graph drawn by taking time along the x-axis and displacement along the y-axis is called position-time graph.

Question 13.
What does the slope of the position-time graph indicate?
Answer:
The slope of the position-time graph indicates velocity.

Question 14.
Draw the position-time graph of a particle which is at rest.
Answer:

Question 15.
Draw the position-time graph of an object starting from rest and moving with uniform velocity.
Answer:

Question 16.
What is a velocity-time graph?
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph.

Question 17.
What does the slope of the velocity-time graph indicate?
Answer:
The slope of the velocity-time graph indicates the acceleration.

Question 18.
What does the area under the velocity-time graph indicate?
Answer:
Area under velocity-time graph indicates the displacement.

Question 19.
An object is moving with uniform velocity. What Is its acceleration?
Answer:
Zero

Question 20.
How do you determine the instantaneous velocity of a particle from the position-time graph?
Answer:
Instantaneous velocity at any point is given by the slope of the tangent drawn to the position-time graph at that point.

Question 21.
Draw the velocity-time graph of a particle moving with

1. uniform velocity
2. variable velocity.

Answer:
1)

2)

Question 22.
Draw the velocity-time graph of an object starting from rest and moving with uniform acceleration.
Answer:

Question 23.
Draw the velocity-time graph of an object moving with constant retardation.
Answer:

Question 24.
Draw the position time of particle which is initially at rest and moving with uniform acceleration.
Answer:

Question 25.
What is the acceleration time graph?
Answer:
A graph drawn by taking time along x-axis and acceleration along the y-axis is called velocity-time graph.

Question 26.
Write the expression for the distance travelled by a body during the nth second of its motion.
Answer:
Distance travelled by a body during the n^th second of motion is given by
s_n = u + a(n – 1/2)
where u is the initial velocity and a is the uniform acceleration.

Question 27.
Write the expression for the acceleration in terms of distance travelled in two consecutive intervals of time.
Answer:
Acceleration is given by,
a = \(\frac{s_{2}-s_{1}}{t^{2}}\)
where s₁ and s₂ are the distances travelled in two consecutive intervals of time ‘t’ seconds each.

Question 28.
What is relative velocity?
Answer:
The velocity of a particle in motion relative to another particle is called relative velocity.

Question 29.
When is the relative velocity of two bodies maximum?
Answer:
Relative velocity of two bodies is maximum when they are moving in opposite directions.

Question 30.
What does the area of a-t graph indicate?
Answer:
Area of a-t graph represents the change in velocity of the body in a given time interval.

Question 31.
Draw the a-t graph of a body moving with uniform acceleration.
Answer:

Question 32.
Why is it not necessary for a body following another, to stop, to avoid collision?
Answer:
If the relative velocity is zero, no collision will occur.

Question 33.
If in case df a motion, displacement is directly proportional to the square of the tune elapsed, what do you think about its acceleration, le, constant or variable? Explain why?
Answer:
x ∝ t²
⇒ x = kt² Where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{d}{d t}\) (2 k t)
= 2k = constant
∴ Acceleration is constant.

Question 34.
Why does the earth Impact the same acceleration to all bodies?
Answer:
Since acceleration is the force on unit mass, the acceleration due to gravity is constant.

Question 35.
What will be the nature of the velocity-time graph for a uniform motion?
Answer:
The v -t graph will be a line parallel to the time axis.

Question 36.
The position coordinate of a moving particle is given by x = 6 + 18t + 9t2(x is in metres and t is in seconds). What is its velocity at t s 2s?
Answer:
x = 6 + 18 t + 9t²
v= \(\frac{\mathrm{d} x}{\mathrm{dt}}\) = 18 + 18 t
v at t = 2 s = 18 + 18 (2)
= 54 ms^-1

Question 37.
A ball is thrown straight up. What is its velocity and acceleration at the top?
Answer:
At the top, v = 0, a = g.

1st PUC Physics Motion in a Straight Line Two Mark Questions and Answers

Question 1.
Distinguish between speed and velocity.
Answer:

1. Speed is the rate of change of position of a body in any direction while velocity is the rate of change of position of a body in a particular direction.
2. Speed is a scalar quantity whereas velocity is a vector quantity.

Question 2.
What is a velocity-time graph? What Is Its importance?
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph. The velocity-time graph can be used

1. To find the nature of the motion of a body.
2. To determine the velocity of a body at any instant.
3. To calculate the displacement of a body in a given time.

Question 3.
If V₁ and v₂ are the velocities of two bodies then what is their relative velocity when they move in the

1. same and
2. opposite direction?

Answer:

1. When the bodies move in the same direction, the relative velocity of 1^st body w.r.t 2^nd is v₁₂ = v₁ – v₂
2. When the bodies move in the opposite directions, the relative velocity of 1^st body w.r.t 2^nd is v₁₂ = v₁ + v₂.

Question 4.
A car moving with a uniform velocity 54kmph is brought to rest in travelling a distance of 5m What is the retardation produced by brake?
Answer:
Initial velocity of the car.,
u = 54km/hr
= \(\frac{54 \times 1000}{3600}\) = 15 ms^-1
Final velocity of the car, v =0.
Let a be the retardation of the car. Then using the relation.
v² = u² + 2as
0² = 15² + 2a × 5
a = \( -\frac{15 \times 15}{2 \times 5}\) = \(\frac{225}{10}\) = – 22.5 ms^-2.

Question 5.
An automobile moving with a uniform velocity of 500ms^-1 is brought to rest in travelling a distance of 5m. What is the acceleration produced by the brakes?
Answer:
Initial velocity of the car, u = 500ms^-1
Final velocity of the car v = 0
Let a be the acceleration of the car.
Then using the relation.
v² = u² + 2aS
0² = 500² + 2a × 5
a = \( -\frac{500^{2}}{10}\) = – 25 × 10³mS^-2.

Question 6.
The equation of motion of a body is given by S = 2t + 2t², where S is in metre and time in second. What is the acceleration of the body?
Answer:
S = 2t + 2t². This is of the standard form S = ut + 1/2 at²
By comparison, 1/2 a = 2 or a = 4m/s².

Question 7.
Draw a velocity-time graph for a particle in the following situations:

1. Starts from rest and moving with uniform acceleration
2. Moving with uniform retardation.

Answer:
1. Body moving the uniform acceleration.

2. Body moving with uniform retardation.

Question 8.
Define velocity and acceleration.
Answer:
Velocity is defined as the rate of displacement. Acceleration is known as the rate of change of velocity.

Question 9.
Two parallel nail tracks run North-South. Train A moves due north with a speed of 54 km h^-1 and train B moves due south with a speed of 90 km h^-1. What is the relative speed of B with respect to A in ms^-1?
Answer:
V_A = 54 km h^-1
V_B = – 90 km h^-1
(north direction is chosen as positive)
V_BA = V_B – V_A
= – 90 – 54 = -144 km h^-1
= – 40 ms^-1

Question 10.
Differentiate between average speed and instantaneous speed of an object.
Answer:
The distance covered in unit time is called average speed.
V_avg = \(\frac{x\left(t_{2}\right)-x\left(t_{1}\right)}{t_{2}-t_{1}}\)
The speed at any instant of time is called instantaneous speed.
V_inst= \(\Delta \mathrm{t} \rightarrow 0 \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\)

Question 11.
Velocity time graph of a moving object is shown below. What is the acceleration of the object? Also, draw displacement time graph for the motion of the object.

Answer:
Acceleration = Zero (because velocity is constant) Displacement time graph is shown below

Question 12.
Two straight lines are drawn on the same displacement time graph make angles 30° and 60° with the time axis respectively in the figure. Which line represents greater velocity? What is the ratio of the velocity of line A to that of line B?

Answer:
Velocity = Slope of x -t graph
V_A = tan 30° = \(1_{\sqrt{3}}\)
V_B = tan 60° = Clearly, V_B > V_A.
\(\frac{V_{A}}{V_{B}}\) = \(\frac{1_{\sqrt{3}}}{\sqrt{3}}\) = \(\frac{1}{3}\)

1st PUC Physics Motion in a Straight Line Three Mark Questions and Answers

Question 1.
The distance x travelled by a body In a straight line is directly proportional to t². Decide on the type of motion associated. If x ∝ t³ what change will you observe?
Answer:
x ∝ t² = kt² where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (2 k t )
= 2 k
∴ Acceleration is constant / uniform.
If x ∝ t³ where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (3 k t²)
= 6 k t
∴ Acceleration is non-uniform

Question 2.
Draw the following graph for an object under free-fall:

1. Variation of acceleration with respect to time.
2. Variation of velocity with respect to time
3. Variation of distance with respect to time.

Answer:
1.

2.

Line 2 appears only when the body bounces.
3.

Question 3.
A body starts from rest accelerates uniformly along a straight line at the rate of 10 ms^-2 for 5 seconds. It moves for 2 seconds with a uniform velocity of 50 ms^-1. Then it retards uniformly and comes to rest in 3s. Draw the velocity-time graph of the body and find the total distance travelled by the body.
Answer:

Total distance travelled = Area under the v – t graph
= 1/2 × 50 × (10 + 2)
= 300 m

Question 4.
The displacement (in metre) of a particle moving along x-axis given by x = 18 t + 5t². Calculate:-

1. the instantaneous velocity at t = 2s
2. average velocity between ta 2 s and 3 s
3. instantaneous acceleration

Answer:
x = 18 t + 5 t²
v(t) = \(\frac{d x}{d t}\) = 18 + 10 t,
a(t) = \(\frac{d v}{d t}\) = 10 ms^-2
1. v (2) = 18 + 10 (2) = 38 ms^-1
2. v (2) = 38 ms^-1

v(3) = 18 + 10 (3) = 48 ms^-1
Since acceleration is constant,
V_avg \(\frac{v(3)+v(2)}{2}\) = \(\frac{48+38}{2}\)
= 43 ms^-1.

3. Instantaneous acceleration a = 10 ms^-2.

1st PUC Physics Motion in a Straight Line FourFive Mark Questions and Answers

Question 1.
Derive an expression for velocity of the particle after time ‘t’ or Derive v = u + at using v-t graph.
Answer:
Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its initial velocity and v be its velocity after a time t. The velocity-time graph AB of this particle is shown in the figure.

Acceleration of the particle is given by the slope of the v-t graph.
From the graph,
slope of the line AB = \(\frac{B C}{A C}=\frac{B C}{O D}=\frac{B D-C D}{O D}\)
But, BD = v, the final velocity,
OA = u, the initial velocity,
OD = t, the time.
∴ Slope of the graph AB is,
a = \(\frac{B D-C D}{O D}\)
= \(\frac{v-u}{t}\)
Thus a = \(\frac{v-u}{t}\) or v = u + at.

Question 2.
What is the v-t graph? Derive an expression for distance covered by the particle in time ‘t’ or Derive the equation s = ut + 1/2 at² using v-t graph.
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle in t seconds is given by,
s = area under v-t graph
= area of the triangle ABC + area of the rectangle ABDO
= 1/2 BC × AB + OA × OD From the graph,
BC = CD – BD = v – u,
AB = OD = t and OA = u
∴ s=1/2 ( v-u) t + ut …….. (1)
But, a = Slope of the graph = \(\frac{v-u}{t}\)
∴ v – u = at
On substituting in (1) we have,
s = 1/2 at.t + ut
s = ut + 1/2 at².

Question 3.
Derive an expression for the velocity of the particle after covering distance ‘s’ OR Derive the equation v² = u² + 2as using the velocity-time graph.
Answer:
particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle is given by,
s = area under v-t graph
= area of the trapezium OACD
= 1/2 (OA+CD) × OD
But OA = u, CD = v and OD = t
∴ s = 1/2 (u + v) t or
2s = (u+v) t ………(1)
But, a = slope of the graph = \(\frac{v-u}{t}\)
∴ t = \(\frac{v-u}{a}\)
On substituting in (1),
2s = \((u+v)\left(\frac{v-u}{a}\right)\)
i.e., v² – u² = 2as
or v² = u² + 2as.

Question 4.
Derive an expression for distance travelled during the n^th second of motion.
OR
Derive s_n= u + a(n – 1/2)
Answer:
Consider a particle moving with uniform acceleration ‘a’.Let u be the initial velocity of the particle.
Distance travelled during n^th second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds
s_n = un + 1/2 an² – [ u(n-1) + 1/2 a(n-1)²]
= un + 1/2 an² – [ un-u + 1/2 a(n²-2n +1)]
= un + 1/2 an² – [ un-u + 1/2 an²-an + 1/2 a]
= un + 1/2 an² – un+u – 1/2 an²+an – 1/2 a
= u + an – 1/2 a
s_n =u + a(n – 1/2)

Question 5.
Derive an expression for acceleration in terms of distance travelled in two successive equal interval of time.
OR
Derive the equation a = \(\frac{s_{2}-s_{1}}{t^{2}}\)
Answer:
Consider a particle moving with uniform acceleration a. Let it start from a point with initial velocity ‘u’ and covers the distances s₁ and s₂ in two successive equal interval of time ‘t’ each. Distance covered in the first interval
s₁ = ut + 1/2 at² ………. (1)
Distance covered at the end of time 2t,
s₁ + s₂ = u(2t) + 1/2 a (2t)²
s₁ + s₂ = 2ut + 2at²……………. (2)
(s₁ + s₂ ) – 2s₁ = (2ut + 2at²) – (2ut + at²)
s₂ – s₁ = at²
a = \(\frac{s_{2}-s_{1}}{t^{2}}\)

Question 6.
Define uniform velocity obtain v² – u² = 2as from the v – t graph, where the symbols have the usual meaning.
Answer:
If a body undergoes equal displacements in equal intervals of time, however small interval of time may be then velocity is said to be uniform velocity.
particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t= 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelles by the particle is given by,
s = area under v-t graph
= area of the trapezium OACD
= 1/2 (OA+CD) × OD
But OA = u, CD = v and OD = t
∴ s = 1/2 (u + v) t or 2s = (u+v) t ………(1)
But, a = slope of the graph = \(\frac{v-u}{t}\)
∴ t = \(\frac{v-u}{a}\)
On substituting in (1), 2s = \((u+v)\left(\frac{v-u}{a}\right)\)
i.e., v² – u² = 2as
or v² = u² + 2as.

Question 7.
A particle is moving with uniform acceleration covers a distance of 45 m in 6^th second and 75 m in 12^th second during its motion. Calculate the displacement of the particle after 20s?
Answer:
S₆ = 45m, S₁₂=75m
using S_n = u + a (n – 1/2 )
i.e., S₆  = u + a or 45 = u + 5.5a ……… (1)
S₁₂= u + a(12 – 1/2) or 75 = u + a(11.5) …….. (2)
Eq(2) – (1) gives 6 a = 30
Or a = 30/6 = 5m/s²
From (1), 45 = u + 5 × 5.5
u = 17.5m/s.
Distance travelled in t = 20s,
S = ut+ 1/2at2
= 17.5 × 20 + 1/2 × 5 × 20²
=1350m.

Question 8.
Draw a velocity-time graph of uniformly accelerated motion in one dimension. From the velocity-time graph of uniformly accelerated motion deduce the equations of motion in distance and time.
Answer:
A graph drawn by taking time along x-axis and velocity along y-axis is called velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle in t seconds is given by,
s = area under v-t graph
= area of the triangle ABC + area of the rectangle ABDO
= 1/2 BC × AB + OA × OD From the graph,
BC = CD – BD = v – u,
AB = OD = t and OA = u
∴ s=1/2 ( v-u) t + ut …….. (1)
But, a = Slope of the graph = \(\frac{v-u}{t}\)
∴ v – u = at
On substituting in (1) we have,
s = 1/2 at.t + ut
s = ut + 1/2 at².

Question 9.
Derive an equation for the distance covered by a uniformly accelerated body in the n^th second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.
Answer:
Consider a particle moving with uniform acceleration ‘a’.Let u be the initial velocity of the particle.
Distance travelled during nth second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds
s_n = un + 1/2 an² – [ u(n – 1) + 1/2 a(n – 1)²]
= un + 1/2 an² – [ un – u + 1/2 a(n² – 2n + 1)]
= un + 1/2 an² – [ un – u + 1/2 an² – an + 1/2 a]
= un + 1/2 an² – un + u – 1/2 an² + an – 1/2 a
= u + an – 1/2 a
s_n =u + a(n – 1/2)
s_n = 4 + a (n – 1/2)
Given u = 0, a = g
s = 1/2 g t² = 1/2 g n²
By the given condition 1/2 s = s_n
⇒ \(\frac{\mathrm{g} n^{2}}{4}\) = g(n – 1/2)
⇒ n² – 4 n + 2 = 0
⇒ n = \(\frac{4 \pm \sqrt{16-8}}{2}\)
⇒ n = 2 ± \(\sqrt{2}\)
∴ Time of the body’s fall = (2 ± \(\sqrt{2}\)) s

1st PUC Physics Motion in a Straight Line Numerical Problems Questions and Answers

Question 1.
A man runs from his home to office at a speed of 2 ms^-1 on a straight road and returns back to home at a speed of 4 ms^-1. Find

1. Average speed and
2. average velocity.

Solution:
1. To find the average speed. Let ‘s’ is the distance between the home and the office.
∴ Time taken to reach the office
t₁ \(\frac{\text { distance }}{\text { velocity }}\) = \(\frac{s}{2}\)
Similarly, time taken to walk back to the home from the office is t₂ = \(\frac{s}{4}\)
∴ Total time taken = t₁ + ₂ = \(\frac{s}{2}\) + \(\frac{s}{4}\) = \(\frac{3 s}{4}\)
Total distance moved = s + s = 2s
Hence. the average speed \(=\frac{\text { toal distance moved }}{\text { total time taken }}\)
= 2.67 ms^-1
2. To find the average velocity Since the man comes back to home, the final and initial position is same.
∴ Total displacement is zero.
Hence average velocity
\(=\frac{\text { toal displacement }}{\text { total time }}\) = 0

Question 2.
A car moving along a straight high-way with a speed of 35 m/s is brought to stop within a distance of 200 m. What Is the retardation, and how long does it take for the car to stop? Solution:
Initial speed of the car, u = 35 ms^-1
Final speed of the car, v = 0
Let a be the retardation of the car.
Then using the relation,
v² = u² + 2as, we have 0 = (35)² + 2a × 200
a = – \(\frac{35 \times 35}{2 \times 200}\) = – 3.06 ms^-2
Let t be the time taken by the car to come to stop. Then,
v = u + at ;
0 = 35 – 3.06 × t or
t = \(\frac{35}{3.06}\) = 11.43 s.

Question 3.
A car starts from rest and accelerates from rest uniformly for 10 s to a velocity of 36 kmhr^-1. It then runs at a constant velocity and is finally brought to rest in 50 m with uniform retardation. If the total distance covered by the car is 500 m, find

1. acceleration and
2. retardation.

Solution:
1. Initial velocity of the car u = 0;
Final velocity v = 36 kmhr^-1
= \(\frac{36 \times 1000}{3600}\) ms^-1
= 10 ms^-1
time taken t=10 s
∴ acceleration a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{\mathrm{10}-\mathrm{0}}{\mathrm{10}}\) 1ms^-2

2. To find the retardation
During the motion with retardation, Initial velocity of the car u = 10 ms^-1
Final velocity v = 0, distance travelled, s = 50 m
Using the relation v² = u² + 2as,
0 = 10² – 2a × 50
a = \(-\frac{100}{2 \times 50}\) = -1 ms^-2

Question 4.
A body travelling with a uniform acceleration travels a distance of 100m in the first 5 second and 200m in the next 5 second, calculate the initial velocity and acceleration of the body.
Solution:
If s₁ and s₂ are the distances travelled by a body in two successive intervals of t seconds, then the acceleration is given
by, a = \(\frac{s_{2}-s_{1}}{t^{2}}=\frac{200-100}{5^{2}}=\frac{100}{25}\) = 4ms^-2
we have, s = ut + 1/2 at²
Here, s = s₁ = 100m, a = 4ms^-2, t = 5 s
100 = u × 5 + 1/2 × 4 × 25 OR
5u = 50 ∴u= 10ms^-1.

Question 5.
A body moving with uniform acceleration along a straight line covers 60m in the 5^th second and 80m in the 8^th second of its motion. Calculate the initial velocity and uniform acceleration.
Solution:
The distance travelled during the n^th second is given by,
s_n = u + a/2(2n – 1), n = 5 and s_n = 60
∴ 60 = u+ a/2 (2 × 5 – 1)
60 = u + 9a/2 ………… (1)
Similarly, for n = 8, s_n =80
∴ 80 = u+ a/2 (2 × 8 – 1)
80 = u + 15a/2 …………….. (2)
(2) – (1), gives
20 = 6a/2 = 3a
∴ a = 20/3 ms^-2
Using this in equation (1), we get
60 = u + \(\frac{9 \times 20}{2 \times 3}\)
u = 60 – 30 = 30 ms^-1.

Question 6.
A stone is thrown vertically upwards with an initial velocity of 19.6 m^-1. After how long will the stone strike the ground? Take g = 9.8 ms^-2.
Solution:
Initial velocity u = +19.6 ms^-1. The stone after reaching the highest point comes back to the initial point
∴ The displacement = 0.
a = g = – 9.8 ms^-2
Let t be the time taken to reach the ground
From s = ut + 1/2at², we have
0=19.6t – 1/2 × 9.8 × t²
4.9t² – 19.6t = 0
Dividing throughout by 4.9t, we have t = 4s. Hence, the stone reaches the ground after 4s.

Question 7.
A stone is thrown vertically upwards with a velocity of 10ms^-1 from the top of a tower 40m tall and it finally falls to the ground,

1. Find the time taken by the stone to reach the ground
2. After how long will it pass through the point of projection
3. Calculate the velocity when it strikes the ground. Take g = 10 ms^-2

Solution:
Let the stone be thrown upward with a velocity u = 10ms^-1 from the tower AB of height 40 m. Let C be the highest point reached.

1. Let t be the total time taken by the stone to move from B to C and back to the ground.
∴ The displacement = BA = – 40 m (directed downward since displacement is measured from the initial to the final position)
a = -10 ms^-2.
From s = ut + 1/2at², we have
– 40= 10 × t – \(\frac{10 t^{2}}{2}\)
– 40 = 10t – 5t²
– 8 = 2t -t²
t² – 2t – 8 = 0
(t – 4) (t + 2) = 0.
OR
t = + 4s or – 2s
As time cannot be negative total time taken to reach the ground = 4s.

2. For the motion from B to C and back to B,
displacement = 0
u = 10ms^-1, a = g = -10ms^-2
Let t₁ be the time taken.
From s = ut + 1/2at², we have
0= 10t₁ – \(\frac{10 t_{1}^{2}}{2}\) or 10₁² – 5t₁² = 0
∴ t₁ = 2s.
Hence, the stone reaches the point of projection after 2 seconds.

3. Let ‘v’ be the velocity of the stone when it strikes the ground. For the motion from B to C and back to A, we have,
u = 10 ms^-1, a = g = -10 ms^-2, t = 4s
From, v = u + at, we have
v = 10 – 10 × 4 = – 30 ms^-1.
The -ve sign indicates that it is directed downward.

Question 8.
A stone is projected vertically upwards from the ground with a velocity of 49 ms^-1. At the same time, another stone is dropped from a height 98m to fall freely along the same path as the first. Find where and when the two stones meet each other.
Solution:

Let P be the stone dropped from B at a height 98m from the ground A. Q is the other stone thrown up vertically upwards with a velocity 49ms^-1 at the same instant. Let the two stones meet at C after a time t seconds.
For the stone P, we have
u = 0, a = + g, s = + h (=BC), t = t
From s = ut + 1/2 at², we have,
+h = 0 × t + 1/2 gt² or h = 1/2 gt² ………. (1)
For the stone Q,
u = 49ms^-1,
a = -g,
s = +AC = +(98 – h),
(98 – h) = 49t – 1/2 gt²
Using eqn. (1) in (2), we have,
(98 – h) = 49t – h or 49t =98
∴ t = 2s.
Using this in (1), we have
h = BC = 1/2 × 9.8 × 4 = 19.6 m
∴ AC =96 – 19.6 = 76.4 m
∴ Two stones meet at a height of 76.4 m from the ground, after 2 seconds.

Question 9.
A train is moving southwards with a speed of 30ms^-1. A man is running on the roof of the train with speed of 5ms^-1 with respect to the train. Find the velocity of the man as observed by the person on the ground if the man is running

1. southwards
2. northwards.

Solution:
If v_A and v_B are the velocity of the two bodies moving along the same direction with respect to the ground, relative velocity of A with respect to B is given by,
V_AB = V_A – V_B …………. (1)
Here, the velocity of the man with respect to the train,
v_AB = 5ms^-1.
velocity of the train with respect to the ground is,
v_B = 30 ms^-1
∴ velocity of the man with respect to the ground,
v_A = v_B + v_AB = 30 + 5 = 35 ms^-1 if the man moves southwards.
On the other hand, if the man moves northwards,
v_AB = – 5 ms^-1
∴ V_A = V_B – V_AB
= 25 ms^-1.

Question 10.
A police van moving on a highway with a speed of 10ms^-1 fires a bullet at a smuggler’s car speeding away in the same direction with a speed of 30ms^-1. If the muzzle velocity of the bullet is 140ms^-1, with what speed the bullet will hit the smuggler’s car?
Solution:
Speed of the police car v_p = 10 ms^-1.
Muzzle velocity of the bullet v_M=140 ms^-1
Net velocity of the bullet,
v_A = v_p + v_M = 10 + 140 = 150 ms^-1
(∵ gun is mounted on the van)
speed of the smuggler’s car v_B = 30 ms^-1
∴ Velocity of the bullet relative to smuggler’s car,
v_AB = V_A – V_B
= 150 – 30 = 120 ms^-1.

Question 11.
Two trains A and B are moving on two parallel tracks with uniform speed of 20ms^-1 in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms^-2. After 50 seconds, they are moving together, find the original distance of separation.
Solution:
Initially, both A and B are moving with the same velocity in the same direction,
∴ The initial relative velocity .
u_AB = u_A – u_B = 0
when t = 50 s,
relative acceleration = a_AB = 1 ms^-2
If s is the distance moved, then
From s = ut + 1/2at²
S = u_AB × t + 1/2 a_ABt²
S = 0 × t + 1/2 × 1 × 50²
= 2500/2 = 1250 m.
Hence, the Initial distance of separation is 1250m.

Question 12.
A body is starting from rest and is subjected to a uniform acceleration of 10ms^-2 determine.

1. Velocity of the body at the end of 5s
2. Displacement of the body In the first 3s
3. velocity of the body after a displacement of 20m.
4. Displacement of the body In the 4^th second of its motion (Tumkur 05)

Solutions:
1. Initial velocity, u = 0
acceleration, a = 10ms^-2
time, t = 5s,
v=?
From, the Equation, v = u + at we have,
v= 0 + 10 × 5
v= 50 ms^-1

2. Initial velocity, u = 0
acceleration a = 10 ms^-2
time, t = 36s
Distance travelled, s =?
From the equation s = ut+ 1/2 at² we have
= 0 + 1/2 × 10 × 9
s = 45 m

3. Initial velocity u= 0,
acceleration a = 10ms^-2
Displacement s = 20m
velocity v=?
From the equation v²= u² + 2as ,
= 0² + 2 × 10 × 20
= 400
v = \(\sqrt{400}\) = 20ms^-1

4. Initial velocity u=0
Acceleration a = 10ms^-1
n = 4 sec
Displacement S₄ = ?
From the Equation, S_n = u+ a/2 (2n-1)
S₄=0 + 10/2 (2 × 4 – 1) = 0 + 5(7)
S₄ = 35m.

Question 13.
A stone dropped from the top of a building travels 24.5 m in the last second of its fall. Find the height of the tower, (g = 9.8 m/s2)
Answer:
Distance travelled in the last second of its fall is S_n = 24.5m
Also, S_n = ut + a (n – 1/2)
But u = 0, a = g
∴ S_n = 0 + g (n – 1/2)

24.5 = 9.8n – 4.9
n =\(\frac{24.5+4.9}{9.8}\) = 3 seconds.
we know that S = ut + 1/2 at²
∴ Height of the tower;
h = 0 × 3+ 1/2 × 9.8(3)²
(∵ s = h, u = 0, a = g = 9.8m/s)
∴ h = 0 + 4.9(3)²
= 44.1m.

Question 14.
A body moving along a straight line with uniform acceleration covers 23m and 35m respectively In the 5^th and 8^th second of its motion calculate the distance travelled by the body in

1. 10^th second
2. complete 10 second and what is its velocity at the end of 6 seconds.

Answer:
S₅ = 23m, S₈ = 35m
W.k.that s_n = u + a(n – 1/2)
i.e., S₅ = u + a(5 – 1/2) or 23 = u + 4.5 a….(1)
S₈ = u + a(8 – 1/2) or 35 = u + a(7.5)……….(2)
Eq(2) – (1) gives 3a = 12
or a = 12/3 = 4 m/s²
From (1), 23 = u + 4.5 × 4
u = 5 m/s
1. Distance traveled in 10^th second
S₁₀=u + a(n – 1/2) = 5 + 4(10 – 1/2) = 43m

2. Distance traveled in 10s is
S= ut + 1/2at² = 5 × 10 + 1/2 × 4 × 10² = 250m
Velocity at the end of 6 seconds is v = u + at = 5 + 4 × 6 =29 m/s.

Question 15.
A point object is thrown vertically upwards at such a speed that it returns the thrown after 6 seconds. With what speed was it thrown up and how high did it rise? Plot speed-time graph for the object and use it to find the distance travelled by it in the last second of its journey.
Answer:
Time of rise t₁ = time of fall = t₂
t = 6 s
v = u – g t (upward direction is chosen as positive)
v = – u
– u=u-gt
⇒ – 2u = – 9.8 × 6
⇒ u = 29.4 ms^-1
v² = u² – 2 gh
At the top, v = 0
⇒ h = \(\frac{u^{2}}{2 g}\)
= \(\frac{(29.4)^{2}}{2 \times 9.8}\)
h = 44.1 m
Initial speed = 29.4 ms^-1
Height the object attained = 44.1 m

Distance covered in last second = area under speed – time graph.
= 1/2 × (6 – 5) × (19.6 + 29.4)
= 24.5 m

Question 16.
A race car is moving on a straight road with a speed of 180 km h^-1. If the driver stops the car in 25 s by applying the brakes, calculate the distance covered by the car during the time brakes are applied. Assume acceleration of the car is uniform throughout the retarding motion.
Answer:
Initial speed μ = 180 km h^-1
μ = 50 ms^-1
t = 25 s
v = u – at
v = 0
∴ 0 = 50 – a × 25
⇒ a = 2 ms^-2
v² = u² – 2as
⇒ s = μ²/2 a
\(\frac{50^{2}}{2 \times 2}\)
∴ Stopping distance = 625 m

Question 17.
A body covers 12 m in the 2^nd second and 20 m in the 4^th second. Find what distance the body will cover in the 4 seconds after the 5^th second.
Answer:
s_n= u + a/2 (2n -1)
s₂ = 12
⇒ 12 = u+ a/2 (2 × 2 – 1)
⇒ 12 = u + 3 a/2 ………. (1)
s₄ = 20
⇒ 20 = u+ a/2 (2 × 4 – 1)
⇒ 20 = u + 7a/2 ………. (2)
Subtract equation (1) from (2)
s = 4a/2
⇒ a = 4 ms^-2
12 = u + 3/2 × 4
⇒ u = 6 ms^-1
Distance covered in the 4 seconds after 5^th second,
= S₉ – S₅
= u (9) + 1/2 a (9)² – [u(5) + 1/2 a (5)² ]
= 4u + a/2 (81 – 25)
= 4 × 6 + 4/2 × 56 = 136 m.

Question 18.
A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a building of height 25 m from the ground.

1. How high will the ball reach?
2. How long will the ball tale to reach the ground ? (g = 10 ms^-2)

Answer:
u = 20 ms^-1
h = 25 m
g = 10ms^-2
1. v² = u² – 2gh₁
At the top, v = 0
⇒ h₁ = \(\frac{\mathrm{u}^{2}}{2 \mathrm{g}}\)
= \(\frac{20^{2}}{2 \times 10}\)
= 20 m
∴ Height the ball reaches = 20 + 25 = 45 m from the ground.
2. – h = u t – 1/2 g t²
– 25 = 20 t – 1/2 × 10 t²
⇒ 5 t² – 20 t – 25 = 0
⇒ t² – 4 t – 5 = 0
⇒ (t – 5 ) (t + 1 ) = 0
= t = 5, -1
∴ t = 5 s (Since t cannot be negative) Total time to reach the ground = 5 s

1st PUC Physics Motion in a Straight Line Hard Questions and Answers

Question 1.
The speed of a train increases at a constant rate a from zero to v and then remains constant for an interval, and finally decreases to zero at a constant rate p. If L be the total distance covered prove that the total time taken
is \(\frac{L}{v}+\frac{v}{2}\left(\frac{1}{\infty}+\frac{1}{\beta}\right)\)
Answer:
During time t₁, the train accelerates uniformly from 0 to V.
v = α t₁
Distance covered d₁ = 1/2 α t₁²
= 1/2 (∝ t₁) t₁
d₁ = 1/2 v t₁
⇒ \(t_{1}=\frac{2 d_{1}}{v} \alpha\)
During time t₂, the train travels at a uniform speed of v
Distance covered d₂ = vt₂
⇒ t₂ = \(\frac{\mathrm{d}_{2}}{\mathrm{v}}\)
During time t₃, the train decelerates uniformly from v to 0.
0 = v – β t₃
⇒ v = β t₃
Distance covered d₃ = vt₃ – 1/2 β t₃²
= vt₃ – 1/2 (β t₃) t₃
d₃ = vt₃ – 1/2 v t₃
⇒ d₃ = 1/2 v t₃
⇒ t₃ = \(\frac{2 \mathrm{d}_{3}}{\mathrm{v}}\)
Total time T = t₁ + t₂ + t₃

Using v² = u² + 2as
d₁ = \(\frac{v^{2}}{2 \alpha}\) and d₃ = \(\frac{v^{2}}{2 \beta}\)
∴ T = \(\frac{L}{v}+\frac{v}{2 \alpha}+\frac{v}{2 \beta}\)
T = \(\frac{L}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
Hence proved.

Question 2.
In a car race, car A takes ‘t’ seconds less than car B and passes the finish line with a velocity ‘V’ more than that of the car B. Of the cars start from rest and travel with constant acceleration a₁ and a₂ respectively, show that
V = \(t \sqrt{a_{1} a_{2}}\).
Answer:
Let the car A take t_A seconds to finish the race.
Car B takes t_B seconds to finish the race. Let the final velocities of the cars be v_A and v_B
By the given conditions,
t_B – t_A = t
v_A = a₁ t_A
v_B = a₂ t_B
v_A – v_B = v
Since the distance d covered is the same,

Now, v_A = a₁ t_A
⇒ v_B + v = a₁ (t_B – t)
⇒ a₂ t_B + v = a₁ (t_B – t)
⇒ v = (a₁ – a₂) t_B – a₁ t
Substitute for t from (1),

Substitute for t_B from (2),

Hence proved.

Question 3.
An object moves with a deceleration \(\alpha \sqrt{\mathbf{v}}\) in a straight line (a is a constant) At time t = 0, the velocity is v₀. What is the distance it traverses before coming to rest? What will be the total time taken?
Answer:
Deceleration = \(\alpha \sqrt{\mathbf{v}}\)
⇒ Acceleration a = – \(\alpha \sqrt{\mathbf{v}}\)
\(\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}\) = – αv^1/2
⇒ \(\frac{\mathrm{d} \mathrm{v}}{\mathrm{v}^{1 / 2}}\) = – αdt
Integrating,

2v^1/2 = – αt + c
At t = 0, v = v₀
⇒ 2v₀^1/2 = c
∴ 2v^1/2 = – αt + 2v₀^1/2 …………… (1)
When t = T, v = 0
∴ 2(0)^1/2 = – αT + 2v₀^1/2
⇒ T = \(\frac{2 \sqrt{v_{0}}}{\alpha}\)
∴ Total time taken = \(\frac{2 \sqrt{v_{0}}}{\alpha}\)
From (1), we get

You can Download Chapter 14 Oscillations Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 14 Oscillations

1st PUC Physics Oscillations Textbook Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(b) and (c)

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

1. the rotation of earth about its axis.
2. motion of an oscillating mercury column in a U-tube.
3. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
4. general vibrations of a poly-atomic molecule about Its equilibrium position.

Answer:

1. Periodic but not SHM (Simple Harmonic Motion)
2. SHM
3. SHM
4. Periodic but not SHM

Question 3.
Figure below depicts four x

 

 

 

-t plots for linear motion of a particle. Which of the piots represent periodic motion? What is the period of motion (in case of periodic motion)?


Answer:
(b) and (d) are in periodic motion with period of 2 sec.

Question 4.
Which of the following functions of time represent
(a) simple harmonic,
(b) periodic but not simple harmonic, and
(c) non-periodic motion?
Give period for each case of periodic motion (ω is any positive constant):
(a) sin ω t – cos ω t
(b) sin³ ω t
(c) 3 cos( π/4 – 2 ω t)
(d) cos ω t + cos 3 ω t + cos 5 ω t
(e) exp (- ω² t²)
(f) 1 + ω t+ ω² t²
Answer:
(a) sin ω t – cos ω t = \(\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t\right)\)
= \(\sqrt{2} \sin (\omega t-\pi / 4)\) ; SHM , period = \(\frac{2 \pi}{\omega}\)
(b) sin³ ω t = \(\frac{1}{4}\) (3 sin ωt – sin 3ωt) both terms are in SHM hence sin³ ωt is periodic.
period = \(\frac{2 \pi}{\omega}\)
(c) 3 cos (π/4 – 2 ω t) S H M with period = \(\frac{\pi}{\omega}\)
(d) cos ω t + cos 3 ω t + cos 5 ω t Superposition of 3 periodic motion,period = \(\frac{2 \pi}{\omega}\)
(e) exp (- ω² t²) non periodic motion
(f) 1 + ω t+ ω² t² non periodic motion

Question 5.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration, and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Answer:

Question 6.
Which of the following relationships between the acceleration α and the displacement x of a particle involve simple harmonic motion?

1. α = 0.7x
2. α = -200x²
3. α = -10x
4. α = 100x³

Answer:
α = -10 x (∵ acceleration is proportional and opposite to displacement)

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ω t + Φ). If the initial (t = 0) position of the particle is 1 cm and Its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ω t + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Given: x(t) = A cos(ω t + Φ) ……. (1)
at t = 0, x(0) = 1cm, ω = π s^-1 ,
substituting in (1) we get
1 cm = A cos (0 × π + Φ)
⇒ A cos Φ =1     …… (2)
Differentiate equation (1) w.r.t. t
⇒ \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) – A ω sin (ω t + Φ)    …… (3)
at t = 0, v = ω cm /s, ω = π s^-1
Substituting in 3
⇒ ω = – A ωsin Φ
⇒ 1 = – Asin Φ     ….. (4)
Divide equation (4) by (2)
⇒ 1 = \(\frac{-\sin \phi}{\cos \phi}\)
⇒ tan Φ = -1
⇒ Φ = \(\frac{-\pi}{4}\)
Φ = initial phase.
Substitute Φ value in equation (2) we get
1 = A cos \(\left(\frac{-\pi}{4}\right)\)
⇒ A = \(\sqrt{2} \mathrm{cm}\)
If x (t) = B sin (ω t + α)        ……. (5)
at t = 0,  x = 1 cm, ω = π s^-1
Substituting in (5) we get
1 = B sin(α)                 …… (6)
Differentiate (5) w.r.t. ‘ t’
⇒ v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\)= B ω cos (ω t + α) dt
at t = 0, v = ω = π
⇒ 1 = B cos(α)           …… (7)
Divide(7) by (6)
we get tan Φ = 1
⇒ Φ = π/4 is the initial phase
Substitute Φ in equation (6) we get
1 = B sin (π/4) ⇒ B = \(\sqrt{2} \mathrm{cm}\)

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Maximum length Y_max = 0.2m
Maximum weight M_max = 50 kg
We know that,
F = Mg = ky, at M = M_max, Y = Y_max

⇒ m = 22.36 kg
Weight of the body = 22.36 × 9.8 = 219.1 N

Question 9.
A spring having a spring constant. 1200 N m^-1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass Is then pulled sideways to a distance of 2.0 cm and released.

Determine

1. the frequency of oscillations,
2. maximum acceleration of the mass, and
3. the maximum speed of the mass.

Answer:
Given:
K = 1200 Nm^-1, m = 3 kg, a = 2 cm
1.  Frequency of Oscillation
We know that T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}=2 \pi \sqrt{\frac{3}{1200}}\).
⇒ T = 0.3144 s
⇒ f = \(\frac{1}{\mathrm{T}}\) = 3.18/s

2. Maximum acceleration of mass (α_max) maximum acceleration is obtained when y = a
⇒ m α_max = Ka
⇒ α_max = \(\frac{\mathrm{Ka}}{\mathrm{m}}=\frac{1200 \times 0.02}{3}\) = 8 ms^-2

3. Maximum speed of the mass
V_max = a ω = \(a \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}\)
V_max = 0.4 ms^-1

Question 10.
In Exercise 14.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t= 0), the mass is
1. at the mean position,
2. at the maximum stretched position, and
3. at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the Initial phase?
Answer:
Given a = 2 cm k = 1200Nm^-1 m = 3 kg
⇒ \(\omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=\sqrt{\frac{1200}{3}}=20 \mathrm{s}^{-1}\)
let equation of S H M ⇒ x = A sin ω t
1. time is measured from mean position.
x = A sin ω t
⇒ x = 2 sin 20 t

2.  at maximum structured position phase angle is π/2
⇒ x = a sin (ω t + π/2)
⇒ x = 2 cos (20t)

3. At maximum compressed position phase angle is 3 π/2
x= a sin (ω t + 3 π/2)
⇒ x = – a cos ω t

Question 11.
The figure below correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anticlockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:

(a) Let A be any point on the circle shown in fig (a). Draw AM ⊥ to x-axis. Point M refers to x – projection of the radius vector.
Now, ∠POA = θ = ωt, T = 2s OA = 3cm ⇒ ∠OAM =θ=ωt (∵ Alternate angles)

⇒ x = 3 sin \(\frac{2 \pi}{\mathrm{T}}\)t(cm)
⇒ x = – 3 sin π t (cm)

(b) let A be any point on the circles of fig. (b). From A draw AM ⊥ to x-axis
Now ∠MOA= θ = ωt

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the Initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise In every case: (x Is In cm and t is in s).

1. x = – 2 sin (3t + π/3)
2. x = cos(π/6 – t)
3. x = 3 sin (2 π t + π/4)
4. x = 2 cos π t

Answer:
1.


2.  x = cos(π/6 – t)
radius r = 1 cm
at t = 0, x = cos (π/6) = \(\frac{\sqrt{3}}{2} \mathrm{cm}\),
Φ = – π/6 = – 30°
ω = 1 rad / s

3.

4. x = 2 cos π t
r = 2 cm
ω = π rad / s
at t = 0 , x = 2 cm

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

1. What is the maximum extension of the spring in the two cases?
2. If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Answer:
1. Consider figure (a)
let y be the extension produced in the spring F = ky
consider fig (b) each mass acts as if it is fixed w.r.t the other
⇒ F = ky  ⇒ y = F/k

2. consider fig (a)
F = – ky
ma = – ky   a = \(\frac{-\mathrm{k}}{\mathrm{m}} \mathrm{y}\)
⇒ ω² = \(\frac{\mathrm{k}}{\mathrm{m}}\)
Therefore, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)
Consider fig (b)
let us assume (1) as centre of system and 2 springs each of length 1/2 attached to two masses. So k’ is the spring factor of each spring.
k’ = 2k

⇒ \(\mathrm{T}=\frac{2 \pi}{\omega}\)
\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}\) is the period of oscillation in the case of (b).

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/ min, what is its maximum speed?
Answer:
Stroke = 1 m
Amplitude = \(\frac{\text { stroke }}{2}\) = 1/2 m
ω = 200 rad / min
V_max = ωA
= 200 × 1/2
= 100 m / min
V_max = 1.67 m/s

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 m s^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s^-2)
Answer:
g_m =1.7 ms^-2  g_e = 9.8 ms^-2  T_e = 3.5 s

Question 16.
Answer the following questions :

1. Time period of a particle in SHM depends on the force constant k and mass m of the particle: T =\(2 \pi \sqrt{\frac{m}{k}}\) . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
2. The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than \(2 \pi \sqrt{\frac{1}{g}}\) . Think of a qualitative argument to appreciate this result.
3. A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
4. What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Answer:

1. In case of a simple pendulum, k is directly proportional to m. Hence ratio of m / k is a constant. Hence time period doesn’t depend on mass.

2. Restoring force that brings body of pendulum back to its mean position F = – mg sin θ
Form small θ, sin θ ≈ θ = \(\frac{\mathrm{y}}{1}\)

If approximation for sin θ = θ is not taken into account then Time period T > \(2 \pi \sqrt{\frac{1}{\mathrm{g}}}\) This happens when θ is not small.

3. Wristwatch work on the principle of spring action. Hence acceleration due to gravity plays no role in the functioning of the wristwatch hence it gives correct time during free fall.

4. During free-fall acceleration, due to gravity is zero hence the pendulum will not vibrate (i.e.: frequency of oscillation is zero).

Question 17.
A simple pendulum of length I and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations In a radial direction about its equilibrium position, what will be its time period?
Answer:
Body of the pendulum is under the action of two accelerations, acceleration due to gravity ‘g’ and centripetal acceleration \(\alpha=\frac{v^{2}}{R}\)
effective acceleration = \(\alpha^{\prime}=\sqrt{\alpha^{2}+g^{2}}\)

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density \(\rho_{\mathrm{t}}\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h} \rho}{\rho_{1} \mathrm{g}}}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Initially in equilibrium,
weight of cork = weight of water displaced

When cork is pushed in, then the restoring force acting on it is :
f = – weight of the portion dipped after pushing

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic, motion.
Answer:
Restoring force acting on the liquid when suction pump is removed is,
f = – mg  ⇒ f= -(A × 2y)ρ × g
where, A = Cross – section area of U tube
ρ = density of liquid
⇒ f = – 2Aρgy
acceleration produced in liquid column
a = f/mass of liquid

hence it is a SHM.

Question 20.
An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Answer:
Volume = V
mass of ball = m
cross-sectional Area = A
initial pressure on either side of the ball = atmospheric pressure = P
Let the charge in volume of air when ball is pressed be ∆V
∆V = Ay (y = displacement)
Now, Bulk modulus of elasticity


The above equation is of the form,

hence it is in SHM.

Question 21.
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of Its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of

1. the spring constant k and
2. the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Answer:
1. Total mass = 3000 kg
mass supported by each wheel = 750 kg
y = 0.15 m
We know that,
mg = kg

2.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the
same period.
Answer:
Let the particle executing SHM starts from mean position.
Displacement can be given by
x = Asin ω t
Velocity v = A ω cosωt
Kinetic energy over one complete cycle

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillators is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J= – α θ, where j is the restoring couple and θ the angle of twist).
Answer:
We know that,

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is

1. 5 cm
2. 3 cm
3. 0 cm.

Answer:
Given,
r = 5 cm
T = 0.2 s
\(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.2}=10 \pi \mathrm{rad} / \mathrm{s}\)
acceleration A = – ω²y
velocity V = \(\omega \sqrt{r^{2}-y^{2}}\)
1. y = 5 cm
A = – (10π)² × 0.05
⇒ A = 0.493 m / s²
V = \(10 \pi \sqrt{(0.05)^{2}-(0.05)^{2}}=0\)

2. y = 3 cm
A = (10π)² × 0.03
⇒ A = – 0.296 m / s²
V= \(10 \pi \sqrt{(0.05)^{2}-(0.03)^{2}}\)
V = 0.4 πm/s

3. y = 0 cm
A = 0
V= \(10 \pi \sqrt{(0.05)^{2}-0}\)
= 0.5 πm/s

Question 25.
A mass attached to a spring is freeto oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x₀ and v₀.
[Hint : Start with the equation x = a cos (ωt+θ) and note that the initial velocity is negative.]
Answer:
Let x = a cos (ωt + θ)
∴ v = \(\frac{d x}{d t}\) = – a ω sin (ωt + θ)
When t = 0, x = x₀ and \(\frac{d x}{d t}\) = – v₀
⇒ x_o = a cos θ ….. (1) and
– v₀ = – a ω sin θ
⇒ a sin θ = \(\frac{\mathrm{v}_{0}}{\omega}\) …..(2)
Squarring and adding (1) and (2),
a² = \(=x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}} \Rightarrow a=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}\)

1st PUC Physics Oscillations One Mark Questions and Answers

Question 1.
Mention the relation between period and frequency of periodic motion.
Answer:
Frequency f = \(\frac{1}{T}\)

Question 2.
A particle executes simple harmonic motion. At what point on its path is the acceleration maximum?
Answer:
Acceleration is maximum at a point of maximum displacement from the mean position.

Question 3.
At what position the KE of an oscillating simple pendulum Is maximum?
Answer:
Kinetic energy is maximum at the mean position.

Question 4.
What is the condition for motion of a particle to be SHM?
Answer:
Acceleration should be proportional to displacement and always directed towards mean position.

Question 5.
What is Oscillation?
Answer:
To and for motion in the same path is called Oscillation.

Question 6.
How will the period of a simple pendulum change when its length is doubled?
Answer:
T_new = \(\sqrt{2}\) T_old

Question 7.
Will a pendulum’s time period increases or decreases when taken to the top of the mountain?
Answer:
Increases, g decreases as high altitude and \(\mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{g}}}\)

Question 8.
Two simple pendulum of equal length cross each other at mean position. What is their phase difference?
Answer:
180° (π radians)

Question 9.
How many times in one vibration, KE and PE become maximum?
Answer:
Two.

Question 10.
When is the tension maximum in the spring of a simple pendulum?
Answer:
Mean position.

Question 11.
A spring of spring constant k Is cut Into two equal parts. What is the spring constant of each part?
Answer:
2 K

Question 12.
What Is the phase difference between the displacement and velocity in a SHM?
Answer:
90° (π/2 radians)

Question 13.
State force law for a SHM.
Answer:
Force, F = mω²x
m = mass
ω = angular speed
x = displacement

Question 14.
A pendulum is making one oscilla¬tion in every two seconds. What is the frequency of oscillation?
Answer:
f = \(\frac{1}{T}=\frac{1}{2} s^{-1}\)

Question 15.
What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer:
Frequency is 0 because the acceleration zero during free fall.

Question 16.
A simple pendulum Is inside a spacecraft. What should be its time period vibration?
Answer:
Pendulum does not oscillate.

Question 17.
What are isochronous vibrations?
Answer:
When the time period is independent of amplitude such an oscillation is called isochronous.

1st PUC Physics Oscillations Two Marks Questions and Answers

Question 1.
Define simple harmonic motion (SHM). Give an example.
Answer:
A particle is said to have SHM if the acceleration of the particle is directly proportional to its displacement from the mean position and directed towards the mean position.
E.g.:

1. Vertical oscillations of a loaded spring.
2. Oscillations of bob of a simple pendulum.
3. Vibrations of string of musical instruments.
4. Motion of air particles during the propagation of sound waves.
5. Vibration of a tuning fork.

Question 2.
Mention expression for velocity and acceleration of a particle executing SHM.
Answer:
Velocity of a particle executing SHM is v = \(\omega \sqrt{A^{2}-y^{2}}\)
Acceleration of a particle executing SHM is a = – ω²y
y is the displacement of a particle from its mean position in t seconds. A is the amplitude and ω is the angular frequency.

Question 3.
Mention expression for K.E, P.E and total energy of a particle executing SHM.
Answer:
Potential energy of particle at any instant t second is E_p = 1/2 m ω² y².
Kinetic energy of particle at any instant t second is, E_K = 1/2 m ω² (A² – y²).
Total energy of particle at any instant is constant E = E_K + E_p = 1/2 m ω² A².
A is the amplitude and w is the angular frequency and m is mass of the particle y is the displacement of a particle from its mean position in t seconds.

Question 4.
The acceleration of a particle executing SHM Is 20ms^-1 at a distance of 5 m from the mean position. Calculate its time period and frequency.
Answer:
a = 20m/s², y = 5m.
We know that a = | ω² y |
⇒ \(\omega=\sqrt{\frac{a}{y}}=\sqrt{\frac{20}{5}}\)= 2 rad/s or 2πf = 2
Frequency f = \(\frac{2}{2 \times 3.14}\) = 0.3185 Hz
Period T = \(\frac{1}{f}=\frac{1}{0.3185}\) = 3.14s

Question 5.
Differentiate between forced oscillations and resonance.
Answer:
1. Forced Oscillations:
A body oscillates with the help of external periodic force with a frequency different from natural frequency of body.

2. Resonance:
A body oscillating with its natural frequency with the help of external periodic force whose frequency is equal to natural frequency of body.

Question 6.
Two linear, simple harmonic motion of equal amplitudes and frequencies ω and 2ω are impressed on a particle along the axis of X and Y respectively. If the initial phase difference is π/2, then find the resultant path followed by particle
Answer:
Let
X = Asinωt …… (1)
Y= Acos2ωt …….(2)
(1) and (2) represent SHM with equal amplitude and phase difference π/2 with frequencies ω and 2ω.

Question 7.
Derive an expression for K.E and P.E of a particle SHM.
Answer:
We know that, Potential energy
U = 1/2 kx²
For a particle in SHM, k = mω² and x = Asin ωt
⇒ U = 1/2 mω² A² sin² ωt
Kinetic energy, k= 1/2 mv²
now, v = ωy = ωA cos ωt
⇒ K= 1/2 mω² A²cos²ωt

Question 8.
The amplitude of a oscillating simple pendulum Is doubled. What will be its effect on

1. periodic time
2. total energy
3. maximum velocity

Answer:

1. Periodic time does not change. T is independent of amplitude.
2. Total energy, T.E = 1/2 mω² A²
   A is doubled ⇒ T.E_new = 4 TE_old
3. Maximum velocity, V_max = ωA
   A is doubled ⇒ V_max is doubled.

Question 9.
The frequency of oscillations of a mass m suspended by a spring is V₁. If the length of spring is cut to one half, the same mass oscillates with frequency V₂. Calculate V₂/V₁.
Answer:
V₁ = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
where K is the spring constant and m is the mass
V₂ = \(\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}\)
∴ \(\frac{V_{2}}{V_{1}}=\sqrt{2}\)

Question 10.
Discuss some important characteristics of wave motion.
Answer:

1. Wave transports energy.
2. particles are not transported.
3. Elasticity and inertia determine the motion of particle distribution.

Question 11.
Differentiate between free oscillations and forced oscillations with the help of examples.
Answer:
Consider a pendulum in free space (without air) it oscillates freely. This is free oscillation. Consider another pendulum in a viscous liquid it oscillates only if there is external force this is forced oscillation.

Question 12.
List any two characteristics of SHM.
Answer:

1. SHM is always directed towards mean position.
2. Acceleration is directly proportioned to displacement but opposite in direction.

1st PUC Physics Oscillations Three Marks Questions and Answers

Question 1.
Define the terms

1. amplitude
2. period
3. frequency
4. phase related with a particle executes SHM.

Answer:
1. Amplitude:
Maximum displacement of the particle from the mean position is called amplitude.

2. Period :
Time taken by the particle to complete one oscillation is called time period. (T)

3. Frequency :
The number of oscillations completed by the particle in one second is called frequency (f)

4. Phase:
Phase of a particle is defined as the fraction of the time period that has elapsed since the particle last passed through its mean position in the positive direction.

Question 2.
For an oscillating pendulum, establish the relation \(\frac{d^{2} \theta}{d t^{2}}=-\omega^{2} \theta\) where \(\omega=\sqrt{\frac{g}{1}} \theta\) = small angular displacement.
Answer:

restoring force F_r = – mg sin θ
torque on pendulum \(\tau=\mathrm{I} \alpha=\mathrm{ml}^{2} \alpha\)
restoring torque = torque on pendulum

We know that = \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=\alpha\) and \(\omega=\sqrt{\frac{g}{1}}\)
⇒ \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-\omega^{2} \theta\)

Question 3.
A body of mass 1 kg is suspended from a weightless spring having force constant 600 Nm^-1. Another body of mass 0.5 kg moves vertically upwards lift the suspended body with a velocity of 3 ms^-1 and tets embedded in it. Find the frequency of oscillation and amplitude of motion.
Answer:
Total mass = (1 + 0.5)kg = 1.
k = 600 nm^-1
frequency of oscillations

Let V₁ be velocity of mass after collision
⇒ m₁ V₂ = (m₁ + m_2,) V₁
⇒ 0.5 × 3 = 1.5 V₁
⇒ V₁ = 1 m/s
According to law of conservation of energy
(PE)_max = (KE)_max
\(\frac{1}{2} m v_{1}^{2}=1/2^{K A^{2}}\)
\(\frac{1}{2} 1.5 \times 1^{2}=\frac{1}{2} 600 \mathrm{A}^{2}\)
⇒ A = 5 cm

Question 4.
The bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about this dip. Deduce an expression for the period of oscillation.
Answer:

Let the rikshaw of mass M be at any point P in the dip of radius R. Let O be the centre of this circular path. This case is similar to that of a simple pendulum and assume that e is small.
restoring force
F = -mg sinθ
F = – mg θ
displacement of rickshaw = R θ

Question 5.
Find an expression for body vibrating in SHM.
Answer:
Let the SHM equation be
x = Asin ω t

Potential energy is given by
= 1/2 kx² = 1/2 m ω ² k²
= 1/2 mω² A² sin² ω t
Kinetic energy = 1/2 mv²
= 1/2 m ω ² A² cos² ω t
Total energy    = PE + KE
= 1/2 mω² A² sin² ω t + 1/2 mω² A² cos² ω t
= 1/2 mω ² A²

Question 6.
Show that when a particle is moving in SHM its velocity at a distance \(\frac{\sqrt{3}}{2}\) its amplitude from the central position is half its velocity in central position.
Answer:
For a particle in SHM
\(\mathrm{V}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}\)

Question 7.
A body oscillates with SHM according to the equation.
x(t) = 5cos (2π t + π /4) Where x is in meters and t is in seconds calculate,

1. displacement att = 0
2. Angular frequency
3. V_max

Answer:
1. displacement at t = 0
x(0) = 5cos(2π 0 + π/4)
⇒ x = \(\frac{5}{\sqrt{2}} m\)

2. Angular frequency
ω = 2π radian/s

3. Vmax = aω = 2π × 5 = 10π m/s

Question 8.
What is spring factor? Find its value in case of two springs connected in

1. series
2. Parallel

Answer:
Spring factor (k):
Force acting for unit extension produced is called spring factor.
1. When two springs connected in series,

force experienced by both springs is same.
Total extension (x) is sum of individual extension.
⇒ x = x₁ + x₂
\(\frac{F}{-K_{e q}}=\frac{F}{-K_{1}}+\frac{F}{-K_{2}}\) ⇒ 1/K_eq = 1/K₁ + 1/K₂
K_eq = \(\frac{\mathrm{K}_{1} \mathrm{K}_{2}}{\mathrm{K}_{1}+\mathrm{K}_{2}}\)

2. When two springs are in parallel,

Net force experienced is the sum of force experienced in both springs. But the extension will be the same.
F_eq = F₁ + F₂
K_eqx =K₁ X + K₂X
⇒ K_eq = K₁ + K₂

Question 9.
Explain the relation in phase between displacement, velocity, and acceleration in SHM, graphically as well as theoretically.
Answer:
Let displacement x = Asinωt
Velocity v= ω Acosω t
acceleration a = – ω² Asin ω t = ω² Asin(ω t + π )
Displacement and velocity have phase difference of π/2 radians.
v and α has \(\frac{\pi}{2}\) radians of phase difference. α ans x has π radians of phase difference.
Assume ω >1

Question 10.
For a particle in SHM, the displacement x of the particle at a function of time t is given as x = Asin(2πt) …..(1) where x is in centimeters and t is in seconds. Let the time taken by the particle to travel from x=0 to x = A/2 beT₁ and the time taken to travel from x = A/2 to x = A be T₂. Find T₁/T₂
Answer:
ω =2π = \(\frac{2 \pi}{\mathrm{T}}\) ⇒ T= 1 s
at t = 0 , x = 0
now if x = A/2

now in a SHM time taken from 0 to A is \(\frac{T}{4}\)
⇒ time taken for x = A/2 to A is \(\frac{T}{4}\) -T/12

Question 11.
What is SHM? Show that in SHM acceleration is directly proportional to Its displacement at a given instant.
Answer:
Simple Harmonic Motion is a type of motion in which displacement is always directed towards mean position and acceleration is directly proportional to displacement and opposite in direction.
Let a SHM be represented by
x = Asin ωt
⇒ \(\frac{\mathrm{d} x}{\mathrm{dt}}\) = A ω cos ωt

acceleration is proportional to displacement and opposite in direction.

Question 12.
A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretched 7cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
Answer:
Mass added m = 0.02
Length stretched x= 7 cm = 0.07 m

Time period T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)

Question 13.
A particle executes SHM of time period 10 sec. The displacement of particle at any instant is x = 10 sinωt(cm) Find

1. Velocity of body 2 s after it passes through mean position.
2. Acceleration of body 2 s after it passes mean position.

Answer:
Given x = 10 sin ωt T = 10 sec
A = 10 cm   ω = \(\frac{2 \pi}{10}\) rad/s
Let at t = 0 body be at the mean position.
Now at t = 2
1.  V = Aω cos ωt
= 10ω cos(ω2)cm/s
⇒ \( V=10 \times \frac{2 \pi}{10} \cos \left(\frac{2 \pi}{10} 2\right)\)
⇒ V = 1.94 cm/s

2. Acceleration
a = – Aω² sinωt
\(= – 10 \times \frac{4 \pi^{2}}{100} \sin \left(\frac{4 \pi}{10}\right)\)
⇒ a = – 3.75 cm/s²

Question 14.
What is a simple pendulum? Show that the motion of the pendulum is SHM and hence deduce an expression for the time period of pendulum. Also, define Second’s pendulum.
Answer:
Simple pendulum is a point mass body suspended by a weightless thread or string from a rigid support about which it is free to oscillate.

Now restoring force acting on body can be given by
F = – mgsinθ (from figure)
If  θ is small
F= – mgθ    (∵ sin θ ≈ θ)
Let P be any point on the path of pendulum and makes ∠OMP = θ which is small and arc OP be x (displacement) then
\(\dot{\theta}=\frac{\mathrm{OP}}{1}=\frac{\mathrm{x}}{1}\)
⇒ F = \(\frac{-m g x}{1}\) ……(1)
⇒ force ∝ displacement and opposite in direction hence an SHM
Now this of form F = -kx ……(2)
⇒ k = mg/l (from (1) and (2)

Question 15.
A body of mass ‘m’ suspended from a spring executes SHM. Calculate the ratio of kinetic energy and potential energy of the body when it is at half the amplitude far from the mean position.
Answer:
KE of a mass suspend from a spring
KE = mω²(a² – y²)
Potential energy of a mass suspend from a spring
PE = 1/2 mω² y²
\(\frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\left(\mathrm{a}^{2}-\mathrm{y}^{2}\right) 2}{\mathrm{y}^{2}}\)
at y = a/2
⇒ \(\frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\left(\mathrm{a}^{2}-\mathrm{a}^{2} / 4\right)^{2}}{\mathrm{a}^{2} / 4}\)
⇒ \(\frac{\mathrm{KE}}{\mathrm{PE}}=3\)

Question 16.
If x = a cos ωt + b sin ωt, show that it represents SHM.
Answer:
We have,
x = a cos ωt + b sin ωt
Now, \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = – aω sinωt + bω cosωt
\(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – aω² cos ωt + bω² sin ωt
⇒ \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – ω² (a cos ωt+ b sin ωt)
⇒ \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – ω²x
⇒ α = – ω²x
Hence an SHM

Question 17.
Find the expression for the total energy of a particle executing SHM.
Answer:
For a SHM PE = 1/2 kx²
= 1/2 mω² x² = 1/2 mω A² sin² ωt
KE= 1/2 mv²
v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = – Aω cos ωt
⇒ KE = 1/2 mA²ω² cos² ωt
Total energy = KE + PE = 1/2 mω² A² cos² ωt + 1/2 mω² sin² ωt
Total energy = 1/2 mω² A²

1st PUC Physics Oscillations Five Marks Questions and Answers

Question 1.

1. Find the total energy of particle executing SHM?
2. Show graphically the variation of PE and KE with time In SHM.
3. What is the frequency of these energies w.r.t the frequency of the particle executing SHM?

Answer:
1. PE at any instant
PE = 1/2 kx²
= 1/2 mω² A² sin² ωt
(∵ k = mω² and x = A sin ωt)
⇒ PE = 1/2 mω² A² sin² ωt
KE at any instant KE = 1/2 mv²
KE = 1/2 mA² ω² cos² ωt
Total energy = KE + PE
⇒ Total energy = 1/2 mω² A²sin² ωt + 1/2 mω² A² cos² ωt
⇒ Total energy = 1/2 mω² A²

2. Graph

3. Frequency = \(\frac{2}{\text { period }}=\frac{2}{\mathrm{T}}\)

1st PUC Physics Oscillations Numerical Problems Questions and Answers

Question 1.
A spring compressed by 0.1m develops a restoring force of 10N. A body of mass 4 kg is placed on it. Deduce the

1. force constant of the spring
2. depression of the spring under the weight of the body
3. period of oscillation, if the body is disturbed.

Answer:
Given
Restoring force, F = 10N
Mass of the body, m = 4kg
displacement, x =0.1m

1. \(\mathrm{k}=\frac{\text { force }}{\text { displacement }}=\frac{10}{0.1}\) = 100 Nm^-1
2. Depression due to weight \(=\frac{\text { force }}{\mathbf{k}}=\frac{4 \times 10}{100}\) = 0.4 m
3. Period of oscillation T = \(2 \pi \sqrt{\frac{m}{k}}\) \(=2 \pi \sqrt{\frac{4}{100}}\) = 0.4 πs
   T = 0.4 πs

Question 2.
A vertical U- tube of uniform crosssection contains water up to a height to 20cm. Calculate the time period of the oscillation of water when it is disturbed.
Answer:
The length of liquid column
L = 2 × 20cm = 40 cm
Time period of oscillation
\(=2 \pi \sqrt{\frac{L}{2 g}}=2 \pi \sqrt{\frac{40}{2 \times 9.80}}\) = 0.9 s

Question 3.
A cylindrical piece of cork of base area ‘A’ and height ‘h’ floats in a liquid of density \(\rho_{\mathrm{e}}\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period of \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h} \rho}{\rho_{\mathrm{e}} \mathrm{g}}}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let x be the depression created.
Excess upthrust caused is, U_e= g\(\rho_{\mathrm{e}}\) Ax
Restoring force = U = gA\(\rho_{\mathrm{e}}\) x
mass = m = Ahρ
Now F = mα;  α = acceleration
Ahρα = -gA\(\rho_{\mathrm{e}}\) x

Question 4.
If this earth were a homogenous sphere and a straight hole bored in it through its centre. Show that if a body were dropped into the hole it would execute a SHM. Also, find its time period.
Answer:
Let mass of body dropped = m
mass of earth = M
radius of earth = R
mg = \(=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) ⇒  g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
⇒  \(\mathrm{g}=\frac{\mathrm{G}}{\mathrm{R}^{2}} \frac{4}{3} \pi \mathrm{R}^{2} \rho=\frac{4 \pi \mathrm{GR} \rho}{3}\)
where ρ is mean density of earth

Let R be the depth of the body fallen from centre then g at the point is
\(\mathrm{g}^{\prime}=\frac{\mathrm{Gm}^{\prime}}{(\mathrm{R}-\mathrm{d})^{2}}\)
where m’ is mass of sphere of radius (R – d)

Hence acceleration is proportional to displacement and in opposite direction go on SHM.

You can Download Chapter 2 Theory Base of Accounting Questions and Answers, Notes, 1st PUC Accountancy Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Accountancy Question Bank Chapter 2 Theory Base of Accounting

1st PUC Accountancy Theory Base of Accounting One Mark Questions and Answers

Question 1.
What is the need for theory base accounting?
Answer:
Theory base accountancy makes accounting information meaningful for internal and external uses. Such theory make information is reliable and comparable.

Question 2.
What is accounting concepts?
Answer:
Accounting concepts means assumptions upon which accounting is based and recorded.

Question 3.
What is accounting conventions?
Answer:
Accounting conventions refers to customs, traditions, usages or practices followed by accountants as guide for preparation of financial statements.

Question 4.
What is revenue said to be recognized?
Answer:
“Revenue is said to be recognised from sale of goods, or services only when revenue is actually realised”

Question 5.
Expand GAAP.
Answer:
Generally Accepted Accounting Principles: GAAP.

Question 6.
What is accrual concept of accounting?
Answer:
This concept distinguish between cash received and receivable, cash paid and payable on various income or expenees of business.

Question 7.
What do you mean by double entry system of Book Keeping?
Answer:
It is a method of book keeping. Double entry system means “The method of recording of two fold aspects of a transactions.

Question 8.
What is single entry system of book-keeping?
Answer:
It is a method of Book-keeping where both the aspects are not recorded, and for a few transactions none of the aspects is recorded.

Question 9.
Write any two Accounting standards, accepted under IAS.
Answer:
As 1. Disclosure of Accounting Policies As 2. Valuation of Inventories.

Question 10.
What is cost concept?
Answer:
An assets acquired by a concern is recorded in the books of accounts at cost called cost concept.

Question 11.
Write the accounting equation.
Answer:
Accounting equation is Assets = Liabilities + Capitals.

Question 12.
Find out the value of Liability, if capital is 50,000 and Assets is 70,000.
Answer:
Liability = Assets – Capital ⇒ Liability = 70,000 – 50,000
Liability = 20,000.

Question 13.
Find out capital, if liability is 70000, and Assets is 200000.
Answer:
Capital = Assets -Liability ⇒ Capital = 2,00,000 – 70,000
Capital = 1,30,000.

Question 14.
Find out Assets, if capital is 60000 and Liability is 90000.
Answer:
Assets = Capital + Liability ⇒ Assets = 60,000 + 9,0,000
Assets = 1,50,000.

Question 15.
What is Accounting cycles?
Answer:
It refers to the flow of accounting data, in the course of accounting during the period of accounting.

Question 16.
Write any two disadvantages of double entry system.
Answer:
The disadvantages of double entry system are :

1. It is a costly
2. It requires special knowledge and skills to maintain the accounts.

Question 17.
Write any two features of double entry system.
Answer:
The two features of double entry system are :

1. It maintain complete record of all transactions.
2. It is a costly system and requires a specialised skills to maintain.

Question 18.
State the two systems of book-keeping.
Answer:
The two systems of book-keeping are :

1. Single entry system of book-keeping.
2. Double.entry system of book-keeping.

Question 19.
Write the meaning of an account.
Answer:
An account refers to statement of business transactions relating to person, income, expenses related to a particular period.

Question 20.
Mention any one merit of single entry system of accounting.
Answer:
It is suitable for small business to recording business transaction is more helpful.

Question 21.
Complete the following work sheet:
(i) If a firm believes that some of its debtors may default?. It should act on this by making sure that all possible losses are recorded in the books. This is an example of the __________ concept.
(ii) The fact that a business is separate and distinguishable from its owner is best exemplified by the _________ concept.
(iii) Everything a firm owns, it also owns out to somebody. This’ co-incidence is explained _____________ by the concept.
(iv) The ____________ concept states that if straight line method of depreciation is used in one year, then it should also be used in the next year.
(v) A firm may hold stock which is heavily in demand. Consequently, the market value of this stock may be increased. Normal accounting procedure is to ignore this because of the ____________  .
(vi) If a firm receives an order for goods, it would not be included in the sales figure owing to the ____________
(vii) The management of a firm is remarkably incompetent, but the firms, accountants cannot take this into account while preparing book of accounts because of concept ____________.
Answer:
(i) The conservatism concept.
(ii) The business entity concept.
(iii) The dual aspect concept.
(iv) The consistency concept
(v) Conservatism concept
(vi) The revenue recognition concept
(vii) The money measurement concept.

1st PUC Accountancy Theory Base of Accounting Two Marks Questions and Answers

Question 1.
Why is necessary for accountancy to assume that business will remain a going concern?
Answer:
According to this the assumption is made that “Every business is carried on with a view to ’ continue it for an indefinite period of time in future and not to liquidate the affairs.

Question 2.
Write any four concepts of accounting.
Answer:

1. Money measurement concepts
2. Dual concepts
3. Business entity concept
4. Continuity concept.

Question 3.
Mention any four accounting conventions.
Answer:

1. Convention of materiality
2. Convention of conservatism
3. Convention of consistency
4. Full disclosure.

Question 4.
What is money measurement concept?
Answer:
In accounting, a record is made only of those transactions which can be expressed in terms of money called money measurement concept.

Question 5.
Write any two assumptions of double entry system.
Answer:
The assumptions of double entry system of book keeping are :

1. Every transactions affects the financial position in two ways.
2. The effect of change is in opposite directions.
3. The benefit measured in terms of money.

Question 6.
Write any two advantages of double entry system.
Answer:
The advantages of double entry systems are :

1. It maintains complete record of all transactions.
2. The correctness of records can be verified easily.
3. It helps to ascertain correct profit and loss of the business.

Question 7.
Why it is necessary for accountants to assume that business entity will remain a going concern?
Answer:
Going Concern Concept assumes that the business entity will continue its operation for an indefinite period of time. It is necessary to assume so, as it helps to bifurcate revenue expenditure (i.e. expenditure related to current year), and capital expenditure (i.e. expenditure whose benefits accrue over a period of time).

Question 8.
When should revenue be recognized? Are there exceptions to the general rule?
Answer:
Revenue should be recognized when sales take place either in cash or credit and /or right to receive income from any source is established. Revenue is not recognized, in case, if the income or payment is received in advance or the. payment is actually received from the debtors. In a nutshell, revenue will be recognized when the right to receive income is established.
The exceptions to this rule are given below.

• Hire purchase: When goods are sold on hire-purchase system, the amount received in installments is treated as revenue.
• Long term construction contract: The long term projects like construction of dams, highways, etc. have long gestation period. Income is recognized on proportionate basis of work certified and not on the completion of contract.

Question 9.
What is the basic accounting equation?
Answer:
The basic accounting equation is Assets = Liabilities + Capital
It means that, the monetary value of all assets of a firm is equal to the total claims, viz. owners and outsiders.

Question 10.
The realization concept determines when goods sent on credit to customers are to be included in the sales figure for the purpose of computing the profit or loss for the accounting period. This of the following tends to be used in practice to determine when to include a transaction in the sales figure for the period. When the goods have been:

a. Dispatched
b. Invoiced
c. Delivered
d. Paid for

Give reasons for your answer.
Answer:
According to the realization concept. Revenue is recognized when an obligation to receive the amount arises. When the goods are invoiced, it is treated as the transfer of ownership of goods from the seller to the buyer and hence the revenue is recognized.

Question 11.
What is the need for theory base accounting?
Answer:
Theory base accountancy makes accounting information meaningful for internal and external uses. Such theory make information is reliable and comparable.

Question 12.
What is accounting concepts?
Answer:
Accounting concepts means assumptions upon which accounting is based and recorded.

Question 13.
What is accounting conventions?
Answer:
Accounting conventions refers to customs, traditions, usages or practices followed by accountants as guide for preparation of financial statements.

Question 14.
Why is necessary for accountancy to assume that business will remain a going concern?
Answer:
According to this the assumption is made that “Every business is carried on with a view to continue it for an indefinite period of time in future and not to liquidate the affairs.

Question 15.
What is revenue said to be recognized? Are there exceptions to the general rule?
Answer:
“Revenue is said to be recognised from sale of goods, or services only when revenue is actually realsied”

Question 16.
Expand GAAP.
Answer:
Generally Accepted Accounting Principles.

Question 17.
Write any four concepts of accounting.
Answer:

1. Money measurement concepts
2. Dual concepts
3. Business entity concept
4. Continuity concept.

Question 18.
Mention any four accounting conventions.
Answer:

1. Convention of materiality
2. Convention of conservatism
3. Convention of consistancy
4. Full disclosure.

Question 19.
What is accrual concept of accounting?
Answer:
This concept distinguish between cash received and receivable, cash paid and payable on various income or expenees of business.

Question 20.
What do you mean by double entry system of Book Keeping?
Answer:
It is a method of book keeping. Double entry system means “The method of recording of two fold aspects of a transactions.

Question 21.
What is single entry system of book-keeping?
Answer:
It is a method of Book-keeping where both the aspects are not recorded, and for a few transactions none of the aspects is recorded.

Question 22.
Write any two Accounting standards, accepted under IAS.
Answer:
As 1. Disclosure of Accounting Policies
As 2. Valuation of Inventories.

Question 13.
What is money measurement concept?
Answer:
In accounting, a record is made only of those transactions which can be expressed in terms of money called money measurement concept.

Question 24.
What is cost concept?
Answer:
An assets acquired by a concern is recorded in the books of accounts at cost called cost concept.

Question 25.
Write the accounting equation.
Answer:
Accounting equation is: Assets = Liabilities + Capital.

Question 16.
Find out the value of Liability, if capital is 50,000 and Assets is 70,000.
Answer:
Liability = Assets – Capital
Liability = 70,000 – 50,000 Liability = 20,000.

Question 17.
Find out capital, if liability is 70000, and Assets is 200000.
Answer:
Capital = Assets – Liability
capital = 2,00,000 – 70,000
capital = 1,30,000.

Question 18.
Find out Assets, if capital is 60000 and Liability is 90000.
Answer:
Assets = Capital + Liability
Assets = 60,000 + 90,000
Assets = 1,50,000.

Question 19.
What is Accounting cycles?
Answer:
It refers to the flow of accounting data, in the course of accounting during the period of accounting.

Question 20.
Write any two assumptions of double entry system.
Answer:
The assumptions of double entry system of book keeping are :

1. Every transactions affects the financial position in two ways.
2. The effect of change is in opposite directions.
3. The benifit measured in terms of money.

Question 21.
Write any two advantages of double entry system.
Answer:
The advantages of double entry systems are :

1. It maintains complete record of all transactions.
2. The correctness of records can be verified easily.
3. It helps to ascertain correct profit and loss of the business.

Question 22.
Write any two disadvantages of double entry system.
Answer:
The disadvantages of double entry system are :

1.  It is a costly system
2. It requires special knowledge and skills to maintain the accounts.

Question 23.
Write any two features of double entry system.
Answer:
The two features of double entry system are :

1. It maintain complete record of all transactions.
2. It is a costly system and requires a specialised skills to maintain.

Question 24.
State the two systems of book-keeping.
Answer:
The two systems of book-keeping are :

1. Single entry system of book-keeping.
2. Double entry system of book-keeping.

Question 25.
Write the meaning of an account.
Answer:
An account refers to statement of business transactions relating to person, income, expenses related to a particular period.

1st PUC Accountancy Theory Base of Accounting One Mark Questions and Answers

Question 1.
Explain the different accounting concepts.
Answer:
Accounting concepts are basic assumptions and conditions on which the accounting is based, the different concept of accounting are:

(1) Business entity concept: According to this “the business is treated as a separate and distinct entity from the owner, who invests money or money’s worth”. If there is any branch or unit, is also treated as a distinct entity.

(2) Going concern concept: According to this the assumption is made that “Every business is carried on with a view to contiune it for an indefinite period of time in future and not to liquidate the affairs.

(3) Money measurement concept: According to this “the accounting entries made in the books are only of those transactions which can be measured and recorded in terms of money”.

(4) Cost concept: According to this “All the fixed assets which are acquired by a concern are recorded in the books of accounts at cost price”.

(5) Dual-aspect concept: According to this “Every business transactions has a two-fold- aspects (receiving and giving benifit) of same value”.

(6) Accounting period concept: As per the going concern concept every business is intended to be continued indefinitely for a long period in future. In that case the trading result cant be ascertained in the life time. For this “The convinient period of time is selected by dividing the estimated period of life of the business for ascertaining the net result of business during a given period as well as financial position of the business as on that date”.

(7) Realisation concept: According to the “revenue is said to be recognised from sale of goods, or services only when revenue is actually realised.

(8) Matching Concept: Earning prof it is the object of ever}’ business enterprise. It has been the duty of an accountant to calculate exact accurate prof it. The result of there efforts was the introduction of the principle of matching cost and Revenue. According to this principle income can be as curtained by matching revenue of the business with its costs.

(9) Accrual Concept: Accrual means recognition of revenue and costs as they are earned or incurred and not as money is received or paid. The accrual concept relates to measurement of income. Identifying assets is liabilities example: Recording salary payable to staff commission receivable etc.

Question 2.
What is matching concept?
Answer:
This concept mainly based on accounting period concept, according to this “The trading result of a business is calculated by matching the total revenues earned during the year with total amount of expenses incurred in the same year”. The difference between the two represents profit or loss.

Question 3.
Explain the different accounting conventions.
Answer:
Conventions are customs or traditions which are followed to maintain accounts and presentation of financial statements. The different accounting conventions are :

• Convention of conservatism: According to this “a safe policy is adopted in preparing the financial statements of a concern. Anticipate profits may be ignored not anticipated loss while financial statements are prepared.
• Convention of consistancy: According to this the accounting rules and practices should be continuously followed and applied in accounting”. Rules and practices once adopted should not be changed from year to year.
• Convention of full disclosure: According to this “all the important information should be fully disclosed in the financial statements of a concern”.
• Convention of materiality: According to this only the significant information which is material in nature, is disclosed in the financial statements.

Question 4.
Write a note on Generally Accepted Accounting Principles. (GAAP).
Answer:
Accounting is the systematic body of knowledge having cache and effect relationship. The subject has certain established concepts, conventions, standard language and terminology to enable the interested parties in the subject to understand it in the same sense as the accountant wants to communicate. There rules are usually called Generally Accepted Principles (GAAP). Accounting assumption rule of recording and reporting business transactions are also known by terms like concepts, principles, conventions, doctrines, axioms and postulates.

Question 5.
Write the difference between accounting and accountancy.
Answer:

Accounting	Accountancy
It is a process or activity	It is a profession or practice
It consist of principles, concepts, conventions and accounting standard	It involves application of accounting principles and conventions to practical problems
Accounting is a clerk	job It required public relations.
Without accountancy, accounting has no results or utility	Without accounting, accountancy has no conceptual foundation.
It is less scope compare to accountancy	It has wider scope than accounting.

Question 6.
Write the needs or importance of accounting.
Answer:
Accounting needs or importance are as follows-

• Book-keeping creates financial records in analytical and appropriate manner and also give reference in future.
• Accounting gives evidence in the eyes of law, it is accepted as evidence.
• Accounting provides relevant information to the management and helps in decision making.
• Accounting system develops reporting system, it helps to control the organisation.
• Accounting prevent fraud and errors, and also reduce the misappropriation of funds in business.
• As per legal requirement, some of the business should keep accounts compulsorily. It is statutory requirement.

Question 7.
Write a note on basis of accounting.
Answer:
The two basis of accounting are.
(a) Cash Basis of Accounting: It is a simple form of accounting and a payment is received for the sale of goods or services, a goods purchase and payment is recorded on same date. The payment or receipt recorded date wise and not post poring called cash basis accounting under this system accrual transactions are not considered.

(b) Accrual Basis of Accounting: Accrual basis of accounting matches revenue to the time period in which they are earned and matinees expenses to the time period in which they are incurred. It provides more information about business example: Commissioning on sales payable. Interest on fixed deposit receivable etc are recorded for the current period.

Question 8.
Mention the differences between cash basis accounting and accrual basis accounting.
Answer:

Cash basis	Accrual basis
Receivables are recessed on the date of receipt and not on the period which it belongs.	Receivables are recessed when they are earned and not on the day of received.
Expenses are recorded on the date of payment and not on the period which it belongs.	Expenses payables are recorded for the period which it belongs and not on the date of payment.
No receivables and payables and recorded.	Receivables by payables both are recorded.
Financial statements match only actual receipts and payments.	Financial statements match revenues to the expenses in cured in earning them and more accurately reflects the result and operations.
It reflects net cash profit for the year Accrual basis.	It reflects correct cash and for cash profit.

Question 9.
What do you mean by Accounting Standard? List out Indian Accounting Standard.
Answer:
According to Ghosh ‘Accounting standard are the policy documents issued by the recognised expert accountancy body relating to various aspects of measurements, treatment and disclosure of accounting transactions and events.’
The following are mandatory Accounting standards (AS) issued by institute of chartered accountants of India (ICAI).

AS 1 Disclosure of Accounting Policies.
AS 2 Valuation of Inventories.
AS 3 Cash Flow Statement
AS 4 Contingencies and events occuring after the balance sheet
AS 5 Net profit or loss for the period. Prior period items and changes in accounting policies AS 6 Depreciation Accounting.
AS 7 Constriction contracts (revised 2002) ‘
AS 8 Presently in in As 26 AS 9 Revenue Recognitions
AS 10 Accounting for Fixed Assets
AS 11 The effects of changes in foreign exchange rates (Revised 2003)
AS 12 Accounting for Government grants
AS 13 Accounting for Investments
AS 14 Accounting for Amalgamation
AS 15 Employes Benefits (revised 2005)
AS 16 Borrowing costs
AS 17 Segment Report
AS 18 Related party disclosure
AS 19 Leases
AS 20 Earning per share
AS 21 Consolidated financial statements.
AS 22 Accounting for Taxes and Incomes
AS 23 Accounting for investments in associates in Consolidated Financial statement.
AS 24 Discontinuing operation AS 25 Interim Financial Reporting
AS 26 Intangible Assets .
AS 27 Financial reporting of interest in joint venture.
AS 28 Impairment of Assets
AS 29 Provisions, contingent liabilities and contingent assets.
AS30 Financial Instruments: Recognition and Measurement.
AS31 Financial Instruments: Presentation
AS32 Financial Instruments: Disclosure.

1st PUC Accountancy Theory Base of Accounting Six Marks Questions and Answers

ESSAY TYPE / LONG ANSWERS

Question 1.
The accounting concepts and accounting standards are generally referred to as the essence of financial accounting. Comment.
Answer:
Financial accounting is concerned with the preparation of the financial statements and provides financial information to various accounting users. It is performed according to the basic accounting concepts like Business Entity, Money Measurements, consistency, Conservation, etc. These concepts allow various alternatives to treat the same transaction.

For example, there are a number of methods available for calculating stock and depreciation, which can be followed by various firms? This leads to wrong interpretation of financial results by external users due to the problem of inconsistency and incomparability of financial results among different business entities. In order to mitigate inconsistency and incomparability and to bring uniformity in preparation of the financial statements, accounting standards are being issued in India by the Institute of Chartered Accountant of India. Accounting standards help in removing ambiguities and inconsistencies.

Hence, accounting standards and accounting concepts are referred as the essence of financial accounting.

Question 2.
Why is it important to adopt a consistent basis for the preparation of financial statements? Explain.
Answer:
Financial Statements are drawn to provide information about growth or decline of business activities over a period of time or comparison of the results, i.e. intra-firm (comparison within the same organization) or inter-firm comparisons (comparison between different firms). Comparisons can be performed only when the accounting policies are uniform and consistent.

According to the Consistency Principle, accounting practices once selected should be continued over a period of time (i.e. years after years) and should not be changed very frequently. These help in a better understanding of the financial statements and thus make comparisons easy.

Although consistency does not prevent change in the accounting policies, but if change in the policies is essential for better presentation and better understanding of the financial results, then the firm must undertake change in its accounting policies and must fully disclose all the relevant information, reasons and effects of those changes in the financial statements.

Question 3.
What is matching concept? Why should a business concern follow this concept ? Discuss?
Answer:
Matching Concept states that all expenses incurred during the year, whether paid of not, and all revenues earned during the year, whether received or not, should be taken into account while determining the profit of that year. In other words, expenses incurred in a period should be set off against its revenues earned in the same accounting period for ascertaining profit or loss.

For example, insurance premium paid for a year is ₹ 1200 on July 01 and if accounts are closed on March 31, every year, then the insurance premium of the current year will be ’ ascertained for nine months (i.e. from July to March) and will be calculated as, ₹ 1200 – ₹ 900 = ₹ 300

Thus, according to the matching concept, the expense of ₹ 900 will be taken into account ₹ and not ₹ 1200 for determining profit, as the.benefit of only ₹ 900 is availed in the current accounting period.

The business entities follow this concept mainly to ascertain the true profit or loss during an accounting period. It is possible that in the same accounting period, the business may either pay or receive payments that may or may not belong to the same accounting period. This leads to either overcasting or under casting of the profit or loss, which may not reveal the truth, efficiency of the business and its activities in the concerned accounting period,

Question 4.
What is the money measurement concept? Which one factor can make it difficult to
compare the monetary values of one year with the monetary values of another year?
Answer:
Money Measurement Concept states that only those events that can be expressed in monetary terms are recorded in the books of accounts.

For example: 12 television sets of ₹ 10,000 each are purchased and this event js recorded in the books with a total amount of ₹ 1,20,000. Money acts a common denomination for all the transactions and helps in expressing different measurement units into a common unit, for example rupees. Thus, money measurement concept enables consistency in maintaining accounting records.

But on the other hand, the adherence to the money measurement concept makes it difficult to compare the monetary values of one period with that of another. It is because of the fact that the money measurement concept ignores the changes in the purchasing power of the money, i.e. only the nominal value of money is concerned with and not the real value. What ₹ 1 could buy 10 years back cannot buy today; hence, the nominal value of money makes comparison difficult. In fact, the real value of money would be a more appropriate measure as it considers the price level (inflation), which depicts the changes in profits, expenses, incomes, assets arid liabilities of the business.