Karnataka SSLC Time Table 2020 (Released) | Check KSEEB 10th Time Table @ kseeb.kar.nic.in

Karnataka SSLC Time Table 2020: The officials of Karnataka Secondary Education Board (KSEEB) have released the Karanataka SSLC Time Table for the students of Class 10.  According to the Karnakata SSLC Time Table For Class 10th, the exams will commence from 27th March 2020 to 9th April 2020.

Karnataka SSLC Board Exam Time Table 2020 for Class 10

Before getting into the details of Karnataka SSLC Time Table for Class 10, let’s have an overview of the examination:

Description Details
Name of the Exam Karnataka SSLC Examinations
Conducting Body Karnataka Secondary Education Examination Board
Exam Mode Offline
Exam Start Date 27th March 2020
Exam End Date 9th April 2020
Category Karnataka SSLC Time Table For Class 10
Official Website kseeb.kar.nic.in

Karnataka SSLC Time Table for Class 10

The Karnataka SSLC Time Table for Class 10 is tabulated below:

Date and Day Subject Name
27 Mar 2020
Fri
First Language
Kannada (01)
Telugu (04)
Hindi (06)
Marathi (08)
Tamil (10)
Urdu (12)
English (14)
Sanskrit (16)
30 Mar 2020
Mon
Core Subjects
Science (83)
Political Science (97)
Karnataka Music/ Hindustani Music (98)
01 Apr 2020
Wed
English (31)
Kannada (33)
03 Apr 2020
Fri
Third Language
Hindi (61)
Kannada (62)
English (63)
Arabic (64)
Persian (65)
Urdu (66)
Sanskrit (67)
Konkani (68)
Tulu (69)
NSQF Exam Subjects
Information Technology (86)
Retail (87)
Automobile (88)
Health Care (89)
Brauty and Wellness (90)
04 Apr 2020
Sat
Elements of Electrical and Mechanical Engineering (71)
Engineering Graphics (72)
Elements of Electronic Engineering (73)
Elements of Comp[uter Science (74)
Economics (96)
07 Apr 2020
Tue
Mathematics (81)
Sociology (95)
09 Apr 2020
Thu
Social Science (85)

Karnataka SSLC Time Table 2020 in English

Karnataka SSLC Time Table in English is given below:
Karnataka SSLC Time Table for Class 10 in English
Karnataka SSLC Time Table for Class 10 in English 1

Karnataka SSLC Time Table 2020 in Kannada

Karnataka SSLC Time Table in Kannada is given below:
Karnataka SSLC Time Table for Class 10 in Kannada
Karnataka SSLC Time Table for Class 10 in Kannada 1

KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Students can Download Chapter 14 Practical Geometry Ex 14.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 1.
Draw a circle of radius 3.2 cm.
Solution:
KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 1
The required circle can be drawn as follows.
Step 1: First, open the compasses for the required radius 3.2 cm.
Step 2 : Mark a point ‘O’ where we want the centre of the circle to be.
Step 3 : Place the pointer of compasses on O.
Step 4 : Turn the compasses slowly to draw the circle.

KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 2.
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Solution:
KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 2
The required circle can be drawn as follows.
Step 1 : First, open the compasses for the required radius 4 cm.
Step 2 : Mark a point ‘O’ where we want the centre of the circle to be.
Step 3 : Place the pointer of compasses on O.
Step 4 : Turn the compasses slowly to draw the circle.
Step 5 : Now, open the compasses for 2.5 cm.
Step 6 : Again put the pointer of the compasses on point ‘O’ and turn the compasses slowly to draw the circle.

KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 3.
Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
Solution:
A circle can be drawn of any convenient radius, also having its centre as 0. Let AB and CD be two diameters of this circle. When we join the ends of these diameters, a quadrilateral ACBD is formed.
KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 20
As we know that the diameters of a circle are equal in length, therefore, the quadrilateral so formed will have its diagonals of equal length.
Also, OA = OB = OC = OD = radius r and if a quadrilateral has its diagonals of same length which are bisecting each other, then it will be a rectangle.
Let DE and FG be two diameters of this circle such that these are perpendicular to each other. A quadrilateral is formed by joining the ends of these diameters.
KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 22
Here, OD = OE = OF = OG = radius r
In this quadrilateral DFEQ the diagonals are equal and perpendicular to each other. Also, since these are bisecting each other, it will be a square.
The length of the sides of the quadrilateral so formed can be measured to check our answers.

KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 4.
Draw any circle and mark points A, B and C such that
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
Solution:
KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 23

Question 5.
Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D. Examine whether \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{C D}}\) are at right angles.
Solution:
Let us draw two circles of same radius which are passing through the centres of the other circle.
KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 24
Here, point A and B are the centres of these circles and these circles are intersecting each other at point C and D.
In quadrilateral ADBC,
AD = AC (Radius of circle centered at A)
BC = BD (Radius of circle centered at B)
As radius of both circles are equal, therefore, AD = AC = BC = BD
Hence, ADBC is a rhombus and in a rhombus, the diagonals bisect each other at 90°.
Hence, \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{C D}}\) are at right angles.

2nd PUC Physics Model Question Paper 2 with Answers

Students can Download 2nd PUC Physics Model Question Paper 2 with Answers, Karnataka 2nd PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Physics Model Question Paper 2 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 70

General Instructions:

  1. All parts are compulsory.
  2. Answers without relevant diagram/figure/circuit wherever necessary will not cany any marks
  3. Direct answers to the Numerical problems without detailed solutions will not carry any marks.

Part – A

I. Answer all the following questions ( 10 × 1 = 10 )

Question 1.
Write the SI unit of Electric field.
Answer:
Newton per coulomb (N/C)

Question 2.
When will the magnetic force on a movi¬ng charge be maximum in a mag notified?
Answer:
F = qVBsinθ , If θ = 90° (Fm – qυB)
The magnetic force is maximum only when moving charge is ⊥r to field.

Question 3.
Where on the Earth’s surface is the magnetic dip zero?
Answer:
At equator, dip is zero

Question 4.
State Curie’s law in magnetism.
Answer:
The magnetic susceptability of a para magnetic substance varies inversely to its absolute temperature(T).
i.e x α \(\frac{\mathrm{C}}{\mathrm{T}}\) (x = \(\frac{\mathrm{C}}{\mathrm{T}}\))

KSEEB Solutions

Question 5.
What is the significance of Lenz’s law?
Answer:
Law of conservation of energy.

Question 6.
Write the formula for Malus law.
Answer:
i = i0cos2θ
Where I → is intensity of the light transmitted by the analyser.
I0 → is intensity of the light incident on the analyser.
θ → is angle between the pass axes of the analyser & polariser.

Question 7.
What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:3?
Answer:
2nd PUC Physics Model Question Paper 2 with Answers 1
Nuclear density is independent of mass number & is approximately constant.
Hence \(\frac{P_{1}}{P_{2}}\) = 1

Question 8.
Define current amplification factor in a common – emitter mode of transistor.
Answer:
It is defined as the ratio of change in collector current ( ΔI)to the change in base current (ΔIb) at constant collector emitter voltage (VCE) when the transistor is in active state
2nd PUC Physics Model Question Paper 2 with Answers 2

Question 9.
Write the truth table of NAND gate.
Truth table of NAND gate
2nd PUC Physics Model Question Paper 2 with Answers 3

Question 10.
Why sky wave propagation is not possi¬ble for waves having frequency more than 30 MHz?
Answer:
The sky wave range is short & it is used in short wave broad cast service.

Part – B

II. Answer anyfive ofthe following questions. ( 5 × 3 = 15 )

Question 11.
Sketch the electric lines of force due to a point charge q. If i) q<0 and ii) q>0
Answer:
2nd PUC Physics Model Question Paper 2 with Answers 4

Question 12.
A galvanometer having a coil of resistance 12 Ω gives full scale deflection for a current of 4 mA. How can it be converted into a voltmeter of range 0 to 24V?
Answer:
Given G = 12 Ω, Ig = 4 x 10-3A. V = 24V
We have,
\(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{B}}}-\mathrm{G}\)
2nd PUC Physics Model Question Paper 2 with Answers 5
R = 5988Ω
A resistance of 5988 Ω must be connected in series with galvanometer.

KSEEB Solutions

Question 13.
Distinguish between paramagnetic and ferromagnetic substances.
Answer:

Paramagnetic substances  Ferromagnetic Substances
1. Paramagnetics are feebly attracted by magnets. 1. They are strongly attracted by magnets.
2. Magnetic permeability is slightly greater than one i.e. Mr>1 2. Relative permeability is greater than 1000
3. Magnetic susceptability is low & positive. 3. Magnetic susceptability is +ve & large.

Question 14.
What is meant by Self Inductance and Mutual Inductance?
Answer:

  1. The phenomena in which an emf is induced in a coil due to change of current through the same coil is known as self-induction.
  2. The phenomena in which an emf is induced in one coil due to change of current is the neighbouring coil is called mutual induction

Question 15.
What are electromagnetic waves? Write the expression for the velocity of electro magnetic waves in terms of permittivity and magnetic permeability of free space.
Answer:
The waves in which there are a sinusoidal variation of electric & magnetic field vectors at right angles to each other & also right angle to the direction of propagation of the wave is called e.m waves.
∴ The expression for velocity of e.m. wave is
\(C=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\) Where µ0 → is permeability of free
space, ε0 → is the permittivity of free space.

Question 16.
Write the relation between the path difference and wavelength of light wave used for constructive and destructive interference of light.
Answer:
a) For constructive interference
∴ Path difference = 2n\(\frac{\lambda}{2}\) = nλ
Wheve λ → is wave length of light used
n → 0, 1, 2,
b) For destructive interference :
∴ Path difference = (2n+l)\(\frac{\lambda}{2}\)
Wheve λ → is wave length of light used n → 0, 1, 2,………

Question 17.
Define :
i) photoelectric work function
ii) electron volt (ev)
i) Photoelectric work fraction : the minimum energy required to remove an electron from the metal surface is called work function, i.e. w = hV0
Where h → is planck’s constant
V0 → is threshold frequency
ii) Electron volt is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
i.e. lev = 1.602 × 10-19J.

Question 18.
Draw the block diagram of a AM receiver.
Answer:
2nd PUC Physics Model Question Paper 2 with Answers 6
x → Intermediate frequency stage
y → power amplifier

Part – C

III. Answer any five of the following questions. ( 2 × 5 = 10 )

Question 19.
Derive an expression for potential energy of a system of two charges in the absence of external electric field.
2nd PUC Physics Model Question Paper 2 with Answers 7
Consider 2 point charges q1 & q2 are separated by a distance ‘r’ are as sho in the fig.
∴ The change q1 is bringing from ∞ to given point A, No work done i.e. w1= 0 |||ly the charge q2 is bringing from ∞ to given point ‘B’ against us field q2
∴work is done it is given by
i.e. w2 = vq2
2nd PUC Physics Model Question Paper 2 with Answers 8

Question 20.
Arrive at an expression for drift velocity.
2nd PUC Physics Model Question Paper 2 with Answers 9
Controller a metalic conductor is connected to a battery
Let \(\overrightarrow{\mathrm{E}} \rightarrow\) is E.f. setup inside the conductor
m → is mass of electom
e → charge of free elector
Vd → is drift velovity
τ → is ralaxation time
the force experienced by an electron in the field is given by \(\overrightarrow{\mathrm{F}}=-\mathrm{e} \overrightarrow{\mathrm{E}}\)
-Ve sign shows that divertion of \(\overrightarrow{\mathrm{E}}\) & \(\overrightarrow{\mathrm{F}}\) are in apposite each other
2nd PUC Physics Model Question Paper 2 with Answers 10

KSEEB Solutions

Question 21.
State and explain Gauss law in magnetism.
Answer:
The net magnetic flux through any closed surface is always zero
\(\sum \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{As}}=0\)
2nd PUC Physics Model Question Paper 2 with Answers 11
consider a closed surface
S in a uniform
\(\mathrm{M} . \mathrm{F} \cdot \overrightarrow{\mathrm{B}}\) Let \(\overrightarrow{\Delta \mathrm{S}}\)
be a small area element of this surface with \(\hat{n}\)
along its normal
2nd PUC Physics Model Question Paper 2 with Answers 12
Question 22.
Derive the expression for motional emf induced in a conductor moving in a uniform magnetic field.
Answer:
Consider a straight metallic rod PQ of length ‘t’ placed in a uniform M.F. \(\overrightarrow{\mathrm{B}}\). The rod is moved with a velocity \(\overrightarrow{\mathrm{v}}\) is a direction ⊥r to
\(\overrightarrow{\mathrm{B}}\) Let the rod moved through a distance ‘x’ in time‘t’ them the area covered by the rod is A= 1 xx The magnetic flux linked with the rod is
Φ = B.A Φ = B1x
∴ The included emf in the rod is
2nd PUC Physics Model Question Paper 2 with Answers 13
Motional emf (e – Blv)
∴ \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\)

Question 23.
With a diagram, explain the working of a transformer.
2nd PUC Physics Model Question Paper 2 with Answers 14
Answer:
AC voltage is applied to the primary, it creates a varying M.F. & hence a changing magnetic flux in the core. Since the secondary coil is magnetically coupled, due to the mutual induction the changing flux causes induced emf in it. Thus, the power is transferred from primary to secondary.

Question 24.
What is total internal reflection? Mention two applications of optical fibres.
Answer:
The phenomena of complete reflection of light at the interface of two optical media when a ray of light travelling in denser medium is called total internal reflection.

  • Optical fibre are used in telecommunication
  • It is used to measure rate of flow of blood.

Question 25.
What are the matter waves? Write the expression for De – Broglie wavelength of a particle and explain the terms.
Answer:
The waves associated with material particles in motion is called matter waves. The expression for de Broglie wavelength is \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\)
where
h → is Planck’s constant
m → is mass of moving particle
v → is the velocity of moving particle.

Question 26.
Write three differences between n-type and p-type semiconductors.
Answer:
P – type S.C.

  1. It is a semiconductor doped with trivalent impurities.
  2. Majority charge carriers are holes
  3. The impurit atom is called acceptor impurity.

n-type S.c

  1. It is a semiconductor doped with pentavalent impurities.
  2. Majority charge carriers are electron
  3. The impurity atom is called Sonar impurity

Part – D

IV. Answer any two of the following questions  ( 2 × 5 = 10 )

Question 27.
Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss law.
Answer:
Consider an infinitely long thin straight wive with uniform linear charge q density λ. Let P be a point at ⊥r distance r from the wire.
2nd PUC Physics Model Question Paper 2 with Answers 15
To calculate the E.F \(\overrightarrow{\mathrm{E}}\) at P, imagine a cylindrical Gaussian surface.
∴ The surface area of the] curved part S = 2πrl
Total charge enclosed by the Gaussian surface q = λl
Electric fliix through the end Surfaces of the cylinder is Φ = 0
Electric flux through the curved Surfaces of the cylinder is Φ2 = Ecosθ.s
Φ2 = EX1X2πrl
The total electric flux Φ = Φ1 + Φ2
Φ = 0 + E2πrl, Φ2= 2πrlE …………… 1
A/C to Gauss law,
2nd PUC Physics Model Question Paper 2 with Answers 16
from (a) and (2)
2nd PUC Physics Model Question Paper 2 with Answers 17

Question 28.
Obtain the bridge balanced condition of Wheatstone’s bridge network by applying Kirchhoff’s rules.
Answer:
Wheatstone’s bridge is a device used to determine unknown resistance.
2nd PUC Physics Model Question Paper 2 with Answers 18

It resistances R1, R2, R3 & R4 a galvanometer of resistance G & a battery with key. Let I is the math current & splits into branch current I1 & I2 respectively
i.e. I = I1 +I2
At B, I1 = I3+ Ig………….(1)
A + D, I2 = 4 + Ig ………. (2)
Apply KVL to the mesh ABDA
I1R1 + igG – I2R2 = o ………….. (3)
to the mesh BCDB
I3R3 – I4R4– IgG = 0 ……………… (4)

The wheatstones bridge is said to be balanced
when no current flowing throught the galvano
meter i.e. Ig 0 ∴ eqn (1) is I1 = I3
eqn(2)is I2 = I4
eqn (3) is I1R1 – I2R2 = 0
∴ I1R1 = I2R2 …………. (5)
eqn (4) is I3R3 – I2R2
∴I3R3 = I4R4 …………………. (6)
2nd PUC Physics Model Question Paper 2 with Answers 36

KSEEB Solutions

Question 29.
Two straight parallel conductors are placed at certain distance in free space. The direction of current in both the conductors is same. Find the magnitude and direction of the force between them. Hence define ampere.
Answer:
2nd PUC Physics Model Question Paper 2 with Answers 19
Consider two infinitely long straight parallel condu ctors x & y carrying a current I1 & I2 respectively. Let d is the ⊥r distance between them.
Let I1 is the current flowing through x conductor, it produces a M.F. (B1)
2nd PUC Physics Model Question Paper 2 with Answers 20
Now the conductor Y carrying current I2 in the M.F.B, it experiences a mehanical force of length 1 is
2nd PUC Physics Model Question Paper 2 with Answers 21
The direction of force can be obtained using ampere’s left-hand rule.
∴ The force bn. two ||le conductors carrying currents in the same direction is attractive. ||ly the force bn two ||le conductors carrying current in opposite direction is repulsive.
Definition of ampere If I1 = I2 = IA, d=lm them F = 2 × 10-7N.

Ampere is the steady current which when flowing through each of 2 infinitely long straight ||le conductors placed in the air at a distance of lm produces a force of 2 × 10-7 N/m.

V. Answer any two of the following question ( 2 × 5 = 10 )

Question 30.
Derive Lens Maker’s formula for a con vex lens.
Answer:
2nd PUC Physics Model Question Paper 2 with Answers 22
Consider a thin convex lens of focal length f & R.I (n) placed in air as shown in fig let R1 & R2 → are the of curvatur of the surfaces ABC & ADC of the respectively.
o → Luminous point object on the principle axis.
A ray op invident at p, after refraction, emerges along QI & meet at I on the principal axis.
Image formution takes place in two stages. (1) Refraction at the surface ABC In the observe of ADC, the refracted ray is meet at I1, then
2nd PUC Physics Model Question Paper 2 with Answers 23
(2) Refraction at the surface ADC the image I1 acts as a virtual object to form a real image at a distance V, then
2nd PUC Physics Model Question Paper 2 with Answers 24

Question 31.
Assuming the expression for radius of the orbit, derive an expression for total energy of an electron in hydrogen atom.
Answer:
Consider an electron of mass m, charge, e revoking around the nucleus of radius of r. The charge on the nucleus + Te.
∴ T.E = KE + P.E
T. E = K + U ………… (1)
For circulation
Centripetal force = Electrostate force bn nucleus & electron.
2nd PUC Physics Model Question Paper 2 with Answers 25
2nd PUC Physics Model Question Paper 2 with Answers 26
The Potential energy of the electron is the field o nucleus is
2nd PUC Physics Model Question Paper 2 with Answers 27

KSEEB Solutions

Question 32.
With the help of circuit diagram, explain the working of NPN transistor as a common emitter amplifier.
Answer:
The circuit diagram of a CE amplifier using NPN transistor is as shown m fit. The input circuit is forward biased & the output circuit is reverse biased when the ac input signal to be amplified is fed to the base-emitter circuit. The output voltage V0 varies in accordance with the relation, V0=VCE =Vcc – IcRc, These variations is the collector voltage VCE uppers as amplified output.

During the -t-ve half cycle of ac input signal the forward base of emitter as junction increase Due to this base current IB increase & hence collector current IC increases. As a result of this ICRC increases output voltage VO this indicates that the +ve half cycle of input ac signal voltage is amplified through -ve half cycle.

During the -ve half cycle of ac input signal, the forward bias of emitter-base junction decreases. Due to this base current IB decreases & hence collector current IC decreases. As a result of this ICRC decreases the. output voltage V0 is +ve. This indicates that the -ve half cycle of input ac signal voltage is amplified through +ve half cycle.

Thus, the weak input signal is amplified & output signal is out of phase with the input signal by 180°
2nd PUC Physics Model Question Paper 2 with Answers 28

VI. Answer any three of the following questions ( 3 × 5 = 15 )

Question 33.
Charges 2 µ C, 4 µ C and 6 µ C are placed at the three corners A, B and C respectively of a square ABCD of side x metre. Find, what charge must be placed at the fourth corner so that the total potential at the center of the square is 0
Answer:
2nd PUC Physics Model Question Paper 2 with Answers 29
In the Δle ABC
AC2 = AB2 + BC2
AC2 = x2 + x2
AC2 = 2x2
∴ AC = √2 x,.m
2nd PUC Physics Model Question Paper 2 with Answers 30
2nd PUC Physics Model Question Paper 2 with Answers 31

Question 34.
A wire having length 2.0m, diameter 1.0 mm and resistivity 1.963 × 10-8 Ω m is connected in series with a battery of emf 3 V and internal resistance 1 Calculate the resistance of the wire and current in the circuit.
Answer:
Given,
length (l) = 2m
Diameter (D) = 1mm = 1 × 10-3 m
∴ radius (r) = \(\frac{\mathrm{D}}{2}\) = 0.5 × 10-3 m
resistivity (f) = 1.963 × 10-8Ωm
E = 3V Internal senstarce (r) = 1 Ω
R=? I = ?
We have,
2nd PUC Physics Model Question Paper 2 with Answers 32

KSEEB Solutions

Question 35.
An inductor and a bulb are connected in series to an AC source of 220 V, 50 Hz. A current of 11A flows in the circuit and phase angle between voltage and current is \(\frac{\pi}{4}\) radians. Calculate the impedance and inductance of the circuit.
Answer:
Given,
Vrms = 220v, f = 50 Hz
Irms = 11A
Φ = \(\frac{\pi}{4}\) radian, Z=?, L =?
2nd PUC Physics Model Question Paper 2 with Answers 33

Question 36.
In Young’s double slit experiment while using a source of light of wavelength
4500 Å, the fringe width is 5mm. 1f the distance between the screen and the plane of the slits ¡s reduced to half, what should be the wavelength of light to get fringe 4 mm?
Answer:
Given,
wavelength (λ) 45OOA°= 45OO × 10-10 m
fringe width (w) = 5mm = 5 × 10-3 m
D = \(\frac { D }{ 2 }\)
wavelength (λ1) = ?
fringe width (w1) = 4mm = 4 × 10-3 m
we have, fringe width (w) = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
5 × 10-3 =4500 × 1010 × \(\frac { D }{ d }\)
2nd PUC Physics Model Question Paper 2 with Answers 34

KSEEB Solutions

Question 37.
The activity of a radioactive substance is 4700 per minute. Five minutes later the activity is 2700 per minute. Find
a) decay constant and
b) half – life of the radioactive substance.
Answer:
Given A0=4700 per minute
A = 2700 per minute
t = 5 minute λ = ? Ty2 = ?
we have,
2nd PUC Physics Model Question Paper 2 with Answers 35

KSEEB Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

Students can Download Chapter 6 Integers Ex 6.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 6 Integers Ex 6.3

Question 1.
Find
Solution:
a) 35 – (20)
35 – (20) = 15

b) 72 – (90)
72 – (90) = -18

c) (-15) – (-18)
(-15) – (-18) = -15 + 18 = 3

d) (-20) – (-13)
(-20) – (-13) = -33

e) 23 – (-12)
23 – (-12) = 23 + 12 = 35

f) (-32) – (-40)
(-32) – (-40) = -32 + 40 = 8

Question 2.
Fill in the blanks with >,< or = sign
a) (-3) + (-6) ___ (-3) – (6)
(-3) + (-6) = -9
(-3) – (-6) = -3 + 6 = 3 – 9 < 3
Hence (3) +(-6) < (-3)-(-6) b) (-21) – (-10) ___ (-31) + (-11) -31 + (-11) = -42 – 11 > -42
(9 – 21) – (- 10) ≥ (-31) +(-11)

KSEEB Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

c) 45 – (-11) ___ 57 + (-4)
57 + (-4) = 57 – 4 = 53
56 > 53

d) (-25) – (-42) ___ (-42) – (-25)
-42 – (-25) = -42 + 25 = -17
117 > -17
(-25) – (-42) ≥ (-42) – (-25)

Question 3.
Fill in the blanks
a) (-8) + 8 = 0
b) 13 + (-13) = 0
c) 12 + (- 12) = 0
d) (-4) + (-8) =-12
e) 5 – 15 = -10

KSEEB Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

Question 4.
Find
Solution:
a) (-7) -8 – (-25)
(-7) – 8 – (-25) = (-7) – 8 + 25 = -15 + 25 = 10

b) (-13) + 32 – 8 – 1
(-13) + 32 – 8 – 1 = -13 + 32 – 8 – 1 = 32 – 22 = 10

c) (-7) + (-8) +(-90)
(-7) + (-8) + (-90) = -7 – 8 – 90 = -105

d) 50 – (-40) – (-2)
50 – (-40) – (-2) = 50 + 40 + 2 = 92

2nd PUC Hindi Previous Year Question Paper June 2017

Students can Download 2nd PUC Hindi Previous Year Question Paper June 2017, Karnataka 2nd PUC Hindi Model Question Papers with Answers help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Hindi Previous Year Question Paper June 2017

समय : 3 घंटे 15 मिनट
कुल अंक : 100

I. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (6 × 1 = 6)

प्रश्न 1.
जो मनुष्य सत्य बोलता है, वह किससे दूर भागता है?

प्रश्न 2.
किसका व्यापारीकरण हो रहा है?

प्रश्न 3.
मन्नू भंडारी को प्रभावित करनेवाली हिन्दी प्राध्यापिका का नाम लिखिए।

प्रश्न 4.
चीफ़ की दावत किसके घर पर थी?

KSEEB Solutions

प्रश्न 5.
स्वर्ग या नरक में निवास स्थान अलॉट करनेवाले कौन हैं?

प्रश्न 6.
‘नारा’. का प्रसिद्ध मंदिर कौनसा है?

आ) निम्नलिखित प्रश्नों में से किन्हीं तीन प्रश्नों के उत्तर लिखिएः (3 × 3 = 9)

प्रश्न 7.
घर में सुजान भगत का अनादर कैसे हुआ?

प्रश्न 8.
झूठ की उत्पत्ति और उसके कई रूपों के बारे में लिखिए।

प्रश्न 9.
गंगा मैया का कुर्सी से क्या अभिप्राय है?

प्रश्न 10.
विश्वेश्वरय्या की प्रसिद्धि तथा पदोन्नति देख कुछ इंजीनियर क्यों जलते थे?

प्रश्न 11.
भोलाराम का परिचय दीजिए।

II. अ) निम्नलिखित वाक्य किसने किससे कहे? (4 × 1 = 4)

प्रश्न 12.
“दिन भर एक न एक खुचड़ निकालते रहते हैं।”

प्रश्न 13.
“यू हैव मिस्ड समथिंग।”

KSEEB Solutions

प्रश्न 14.
“वह जरूर बना देंगी। आप उसे देखकर खुश होंगे।”

प्रश्न 15.
“महाराज, रिकार्ड सब ठीक है।”

आ) निम्नलिखित में से किन्हीं दो का ससंदर्भ स्पष्टीकरण कीजिए: (2 × 3 = 6)

प्रश्न 16.
“भगवान की इच्छा होगी, तो फिर रुपये हो जायेंगे। उनके यहाँ किस बात की कमी है?”

प्रश्न 17.
“वे बोलते जा रहे थे और पिताजी के चेहरे का संतोष धीरे-धीरे गर्व में बदला जा रहा था।”

प्रश्न 18.
“मेरी माँ गाँव की रहनेवाली हैं। उमर-भर गाँव में रही हैं।”

प्रश्न 19.
“यहाँ तो डेंटिस्ट मक्खी मारते होंगे।”

III. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (6 × 1 = 6)

प्रश्न 20.
श्रीकृष्ण के अनुसार किसने सब माखन खा लिया?

प्रश्न 21.
बाँसुरी किस रंग की है?

प्रश्न 22.
बेटी सजने-धजने से क्या महसूस करती है?

KSEEB Solutions

प्रश्न 23.
देश को किससे बचाना है?

प्रश्न 24.
भारतीयता कहाँ बहु रूप में सँवरती है?

प्रश्न 25.
दीवार किसकी तरह हिलने लगी?

आ) निम्नलिखित प्रश्नों में से किन्ही दो प्रश्नों के उत्तर लिखिए: (2 × 3 = 6)

प्रश्न 26.
श्रीकृष्ण के रूप सौन्दर्य का वर्णन कीजिए।

प्रश्न 27.
‘गहने’ कविता के द्वारा कवि ने क्या आशय व्यक्त किया है?

प्रश्न 28.
पर्यावरण के संरक्षण के संबंध में कवि कुँवर नारायण के विचार लिखिए।

प्रश्न 29.
दक्षिण प्रदेश की महत्ता को अपने शब्दों में लिखिए।

इ) ससंदर्भ भाव स्पष्ट कीजिए: (2 × 4 = 8)

प्रश्न 30.
अब कैसे छूटै राम रट लागी।
प्रभु जी तुम चंदन हम पानी,
जाकी अंग-अंग बास समानी।
अथवा
धनि रहीम जल पंक को, लघु जिय पिअत अघाय।
उदधि बड़ाई कौन है, जगत पिआसो जाय॥

प्रश्न 31.
ऐसा तेरा लोक, वेदना
नहीं, नहीं जिसमें अवसाद,
जलना जाना नहीं, नहीं
जिसने जाना मिटने का स्वाद!
अथवा
युद्धं देहि कहे जब पामर
दे न दुहाई पीठ फेर कर;
या तो जीत प्रीति के बल पर
या तेरा पद चूमे तस्कर।

IV. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (5 × 1 = 5)

प्रश्न 32.
मिश्रानी कितने वर्षों से मूलराज के परिवार में काम कर रही थी?

प्रश्न 33.
छोटी बहु के मन में किसकी मात्रा जरूरत से ज्यादा है?

KSEEB Solutions

प्रश्न 34.
कवि किस पर शासन करता है?

प्रश्न 35.
शास्त्रार्थों में पंडितों को किसने पराजित किया?

प्रश्न 36.
पितृ-हत्या का दण्ड क्या नहीं है?

आ) निम्नलिखित प्रश्नों में से किन्हीं दो प्रश्नों के उत्तर लिखिए: (2 × 5 = 10)

प्रश्न 37.
इन्दु को अपनी भाभी बेला पर क्यों क्रोध आया?
अथवा
बेला की चारित्रिक विशेषताओं पर संक्षेप में प्रकाश डालिए।

प्रश्न 38.
भारवि अपने पिता से क्यों बदला लेना चाहता था?
अथवा
प्रायश्चित को लेकर पिता और पुत्र के बीच हुए संवाद को लिखिए।

V. अ) वाक्य शुद्ध कीजिएः (4 × 1 = 4)

प्रश्न 39.
i) मेरा तो प्राण निकल गया।
ii) संदीप को पूछो।
iii) यह एक इतिहासिक घटना है।
iv) कोयल डाली में बैठी है।
उत्तरः
i) मेरे तो प्राण निकल गये।
ii) संदीप से पूछो
iii) यह एक ऐतिहासिक घटना है।
iv) कोयल डाली पर बैठी है।

आ) कोष्टक में दिये गए उचित शब्दों से रिक्त स्थान भरिए: (4 × 1 = 4)
(पावन, भला, समाज, समय)

प्रश्न 40.
i) आप …………… तो जग भला।
ii) ……… परिवर्तनशील है।
iii) वह सरस्वती देवी का …………… मंदिर है।
iv) साहित्य …………… का दर्पण है।
उत्तरः
i) भला
ii) समय
iii) पावन
iv) समाज।

इ) निम्नलिखित वाक्यों को सूचनानुसार बदलिए: (3 × 1 = 3)

प्रश्न 41.
i) वसुंधरा गाना गाती थी। (वर्तमानकाल में बदलिए)
ii) वह आग चिता पर रखेगा। (भूतकाल में बदलिए)
iii) पृथ्वीराज ने देश की सेवा की। (भविष्यत्काल में बदलिए)
उत्तरः
i) वसुंधरा गाना गाती है।
ii) उसने आग चिता पर रख दी।
iii) पृथ्वीराज देश की सेवा करेगा।

ई) निम्नलिखित मुहावरों को अर्थ के साथ जोड़कर लिखिए: (4 × 1 = 4)

KSEEB Solutions

प्रश्न 42.
i) खिल्ली उड़ाना a) रूठ जाना
ii) मुँह फुलाना b) सफल न होना
ii) टोपी उछालना c) हँसी उड़ाना
iv) दाल न गलना d) अपमानित करना
उत्तरः
i – c, ii – a, iii – d, iv – b.

उ) अन्य लिंग रूप लिखिए: (3 × 1 = 3)

प्रश्न 43.
i) श्रीमती
ii) गाय
iii) पुत्रवान।
उत्तरः
i) श्रीमान
ii) बैल
iii) पुत्रवती।

ऊ) अनेक शब्दों के लिए एक शब्द बनाइए : (3 × 1 = 3)

प्रश्न 44.
i) नीचे लिखा हुआ।
ii) जो पुत्र गोद लिया हो।
iii) जो छिपाने योग्य हो।
उत्तरः
i) निम्नलिखित
ii) दत्तक
iii) गुप्त/रहस्य।

ए) निम्नलिखित शब्दों के साथ उपसर्ग जोड़कर नए शब्दों का निर्माण कीजिए : (2 × 1 = 2)

प्रश्न 45.
i) परिवार
ii) शासन।
उत्तर:
i) परिवार = स + परिवार = सपरिवार।
ii) शासन = अनु + शासन = अनुशासनं।

KSEEB Solutions

ऐ) निम्नलिखित शब्दों में से प्रत्यय अलग कर लिखिए: (2 × 1 = 2)

प्रश्न 46.
i) बलवान
ii) घुमाव।
उत्तरः
i) बलवान = बल + वान।
ii) घुमाव = घुम + आव।

VI. अ) किसी एक विषय पर निबंध लिखिए : (1 × 5 = 5)

प्रश्न 47.
i) a) इंटरनेट की दुनिया।
b) खेल जगत में क्रिकेट का स्थान।
c) राष्ट्रीय एकता।
अथवा
ii) चरित्र प्रमाण-पत्र प्राप्त करने हेतु अपने कॉलेज के प्राचार्य को आवेदन-पत्र लिखिए।

आ) निम्नलिखित अनुच्छेद पढ़कर उस पर आधारित प्रश्नों के उत्तर लिखिए: (5 × 1 = 5)

प्रश्न 48.
यह कहावत सत्य है कि समय बलवान है, इस पर किसी का वश नहीं चलता। आनेवाले समय में अच्छा-बुरा क्या घट जाए, कोई नहीं जानता। ऐसे अनिश्चित समय के लिए यदि उसके पास कुछ संचित धन है तो उसके काम आ सकता है। कोई भी व्यक्ति दूसरों के सहारे न तो कल रह सका है, न आज रह पा रहा है, न ही भविष्य में रह सकता है। सत्य है। मनष्य का आज का जीवन कई प्रकार की आकस्मिकताओं वाला बन चुका है। अतः उन आकस्मिकताओं का ठीक प्रकार से सामना करने के लिए, प्रत्येक व्यक्ति को चाहिए कि वह प्रतिदिन जितनी भी हो अधिक-से-अधिक बचत करता रहे। इसी में उसकी भलाई है। आज की गई एक-एक पैसे की बचत कल का अनंत सुख सिद्ध हो सकती हैं। धन की कमी से सुखपूर्वक तो क्या सामान्य जीवन कतई संभव नहीं है। मनुष्य के जीवन में हमेशा से धन की आवश्यकता बनी रही है। उसे पूरा करने के लिए बचत करना नितांत आवश्यक है। अतः व्यक्ति अपने सभी तरह के स्रोतों से आज और कल में संतुलन बनाए रखकर ही सुख-चैन से जीवन जी सकता है।
प्रश्नः
i) कौन सी कहावत सत्य है?
ii) कटु सत्य क्या है?
iii) मनुष्य की भलाई किस में है?
iv) धन की कमी से क्या संभव नहीं है?
v) व्यक्ति सुख-चैन से कब जी सकता है?

KSEEB Solutions

इ) हिन्दी में अनुवाद कीजिए: (5 × 1 = 5)

प्रश्न 49.
2nd PUC Hindi Previous Year Question Paper June 2017

KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा

Students can Download Hindi Lesson 10 मेरा, हमारा, तेरा, तुम्हारा Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 6 Hindi helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा

मेरा, हमारा, तेरा, तुम्हारा Questions and Answers, Summary, Notes

KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा 1
KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा 2
KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा 3

KSEEB Solutions

अभ्यास

1. उचित शब्द चुनकर खाली :

  1. मोहन …………… भाई हैं (मेरी / मेरा)
  2. ……………घर बड़ा हैं (हमारा / हमारी)
  3. …………. नाम क्या है ? ( तेरी / तेरा )
  4. ………. गाँव कहाँ है ? (तुम्हारी / तुम्हारा)

उत्तर:

  1. मेरा
  2. हमारा
  3. तेरा
  4. तुम्हारा

2. मिलान करो :

  1. सतीश – a. गाँव के पास हैं
  2. कुत्ता  – b. सविता का बड़ा भाई हैं
  3. खरगोश – c. वफादार होता हैं
  4. विद्यालय – d. प्यारा हैं

उत्तर:

  1. b सविता का बड़ा भाई हैं
  2. c वफादार होता हैं
  3. d प्यारा हैं
  4. a गाँव के पास हैं

KSEEB Solutions

KSEEB Solutions for Class 5 English Prose Chapter 8 The Dinner Party

Students can Download English Lesson 8 The Dinner Party Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 English Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Board Class 5 English Prose Chapter 8 The Dinner Party

The Dinner Party Questions and Answers, Summary, Notes

Comprehension:

I. Answer the following questions

Question 1.
Who are the hosts and guests in the story?
Answer:
A colonial official and his wife, army officers and government official and their wives.

Question 2.
Describe the place where the dinner is hosted
Answer:
Dining room, which has a bare marble floor, open rafters and wide glass doors opening onto a veranda.

Question 3.
What is the young girl discussing with the colonel?
Answer:
A woman’s unfailing reaction in any crisis and a man’s nerve control.

Question 4.
According to the colonel Who has greater self-control, the man or the woman?
Answer:
Man.

KSEEB Solutions

Question 5.
Why do you think that the American did not join the discussion?
Answer:
The American is a naturalist, who just observes others and studies their reactions. So he did not join the discussion.

Question 6.
What did the hostess want her servants to do? Why?
Answer:
The hostess wants her servant to place a bowl of milk on the Veranda Just outside the open doors. Because a cobra had just crawled over her feet.

Question 7.
Do you think this action was a routine one?
Answer:
No.

KSEEB Solutions

Question 8.
What made the naturalist think that there must be a cobra in the room?
Answer:
In India milk in a bowl means only one thing bait for a snake. He realizes there must be cobra in the room.

Question 9.
Where did the American search for the cobra?
Answer:
He looked up at the rafters – the likeliest place, but they were bare. The three of the room were empty, and in the fourth, the servants were waiting to serve the next course. So he deduced that the cobra might be under the table.

Question 10.
What did the American do to test the self-control of the people at the dinner party?
Answer:
When the American Naturalist concluded that the cobra was under the table his first impulse was to jump back and warn others. But he did not want to frighten others and the cobra into striking one of them. In a sobering tone, he announced that he just wanted to know what self – control everyone at the table had.

He challenged them that he will start counting to three hundred – that’s five minutes and asked everyone to not move a muscle, and those who moved will lose fifty rupees. All the twenty guests sit like stone images while he counted. When he was counting two hundred and eight – the cobra emerged from under the table and went to the veranda where the bowl of milk was placed. When the guests saw the cobra, they were shocked and started, screaming. But the American quickly jumped up and slammed the veranda doors safely shut.

KSEEB Solutions

Question 11.
Who shows greater self-control in the stray The American guest or the hostess?
Answer:
The Hostess.

Question 12.
How does the American figure out that there is a cobra under the table?
Answer:
The American naturalist, while watching the hostess observes a strange expression come over the face of the hostess. She was staring straight ahead, he muscles contracting slightly.

She gestured to the servant standing behind her chain and whispered something to him. The servant’s eyes widened and he quickly left the room. The American then noticed that the servant placed a bowl of milk on the veranda just outside the open doors. In India, Milk in a bowl means – bait for a snake.

He looked up at the rafters – the likeliest place, but they were bare. The three of the room were empty, and in the fourth, the servants were waiting to serve the next course. So he deduced that the cobra might be under the table.

KSEEB Solutions

Question 13.
The colonel believes that women behave very differently from men in a crisis. Do you agree with the colonet’s belief.
Answer:
Yes, I agree that women behave very differently from men in a crisis. Women have more patience though the cobra was crawling across Mrs. Wynnes’s feet she kept her self – control.

Question 14.
How do you describe the followings:
a) The American Naturalist:
He is the one who kept control of the difficult situation in the lesson.

b) Mrs. Wynnes:
She realised that a cobra was in the room. Because it was crawling across her foot. But she kept her self – control.

c) The colonel:
The colonel is a male-centric person. He believed that men have more nerve control than woman.

d) The servants:
The servants were obedient and discrete.

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Language Exercise

I. Use the following words in sentences of your own.

  • Forfeit: The student cheated in the exams, so he had to for diet an year of schooling.
  • Sobar: The children were playing daily in the garden, but they all turned sober when they saw a cobra, crawling about.
  • Gesture: The teacher gestured at Nagaraj and called him to solve the Maths problem on the blackboard.
  • Stare: It is not right to stare at physically handicapped people.

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Grammar:

In one of the earlier lessons you have learnt the functions of adjectives. In this lesson lets learn how the three different forms of degrees of adjective are formed.

The three different forms/degree of adjectives are

  1. The Positive degree
  2. The Comparative degree
  3. The Superlative degree

Usually, the comparative degree is formed by the addition of ‘er’ to the positive degree form ex: tall-taller short – shorter
The superlative degree is usually formed by the addition of ‘est’.

ex: strong stronger strongest broad broader broadest near nearer nearest Now, form the comparative and superlative degrees of the adjectives given below

Positive Comparative Superlative

  1. fine – finer – finest
  2. sweet – sweeter – sweetest
  3. dear – dearer – dearest
  4. high – higher – highest
  5. fair – fairer – fairest
  6. long – longer – longest
  7. weak – weaker – weakest
  8. fair fairer fairest
  9. cool – coller – coolest
  10. low – lower – lowest
  11. bold – bolder – boldest
  12. few – fewer – fewest

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II. Match the adjectives given in column ‘A’ with the words given in B’

A B
1. Brilliant i) sea
2. Sharp ii) light
3. loud iii) shower
4. juicy iv) knife
5. heavy v) fruit
6. busy vi) noise
7. rough vii) cut
viii) street

Answer:

A B
1. Brilliant ii) light
2. Sharp iv) knife
3. loud vi) noise
4. juicy v) fruit
5. heavy iii) shower
6. busy viii) street
7. rough i) sea

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III. Complete the phrases below using appropriate adjectives

  1. a delicious meal
  2. a rough sea
  3. a cute child
  4. a gold ring
  5. a terrible accident
  6. a barking dog
  7. an interesting story
  8. a tough task
  9. a good friend
  10. a beautiful hat

IV. Write five sentences of your own using any of the completed phrases above.

  1. Nazir read a thrilling story about pirates.
  2. My uncle has a harking dog.
  3. Francis is a good friend.
  4. Arabian sea is a rough sea.
  5. My mother prepared a delicious meal.

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The Dinner Party Summary In English

The Dinner Party Summary In English 1

The lesson ‘The Dinner Party” written by Mona Gardner explains about the importance of self-control. A colonial official and his wife are giving a large dinner party for the army officers and government officials and their wives, and a visiting American naturalist. A spirited discussion springs up between a young girl and a woman.

The American notices the servant place a bowl of milk on the veranda just outside the open doors. He realizes there must be cobra in the room. The American told all that he would count three hundred, that was five minutes and not one of you is to move a muscle.

The Dinner Party Summary In English 2

Those who move will forfeit fifty rupees. The twenty people sat like stone images by the ti pie the cobra emerged and went to the varanda where the bowl of milk kept. The colonel appreciated the American who has just shown us an example of perfect control.

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The Dinner Party Summary In Kannada

The Dinner Party Summary In Kannada 1
The Dinner Party Summary In Kannada 2
The Dinner Party Summary In Kannada 3

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2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company

You can Download Chapter 3 Financial Statements of a Company Questions and Answers, Notes, 2nd PUC Accountancy Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company

2nd PUC Accountancy Financial Statements of a Company NCERT Textbook Questions and Answers

2nd PUC Accountancy Financial Statements of a Company Short Answer Type Questions and Answers

Question 1.
What is public company?
Answer:
A public company is defined as a company that offers a part of its ownership in the form of shares, debentures, bonds, securities to the general public through stock market:

Question 2.
What is private limited company?
Answer:
As defined by the Section 3. (1) (iii) of Companies Act 1956, private limited company is defined by the following characteristics:

  • It restricts the right to transfer its shares.
  • There must be at least two and a maximum of 50 members (excluding current and former employees) to form a private company.
  • It ca not invite application from the general public to subscribe its shares, or debentures.
  • It cannot invite or accept deposits from persons other than its members, Directors and their relatives.

Question 3.
Define Government Company?
Answer:
As per the Section 617 of Company Act of 1956, a Government Company means any company in which not less than 51% of the paid up share capital is held by the Central Government, or by any State Government or Governments, or partly the Central Government and partly by one or more State Governments and includes a company which is a subsidiary of a Government Company as thus defined.

Question 4.
What do you mean by a listed company?
Answer:
Those public companies whose shares are listed and can be traded in a recognised stock exchange for public trading like, Tata Moto. s, Reliance, etc are called Listed Company. These companies are also called Quota Cornpanies.

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Question 5.
What are the uses of securities premium?
Answer:
As per the Section 78 of the Companies Act of 1956, the amount of securities premium can be used by the company for the following activities:

  • For paying up unissued shares of the company to be issued to members of the company as fully paid bonus share.
  • For writing off the preliminary expenses of the company.
  • For writing off the expenses of, or the commission paid or discount allowed on, any issue of shares or debentures of the company.
  • For paying up the premium that is to be payable on redemption of preference shares or debentures of the company.
  • Further, as per the Section 77A, the securities premium amount can also be utilised by the company to Buy-back its own shares.

Question 6.
What is buy-back of shares?
Answer:
Buy-back of shares means repurchasing of its own shares by a company from the market for reducing the number of shares in the open market.

Question 7.
Write a brief note on ‘Minimum Subscription’.
Answer:
When shares are issued to the general public, the minimum amount that must be subscribed by the public so that the company can allot shares to the applicants is termed as Minimum Subscription. As per the Company Act of 1956, the Minimum Subscription of share cannot be less than 90% of the issued amount. If the Minimum Subscription is not received, the company cannot allot shares to its applicants and it shall immediately refund the entire application amount received to the public.

2nd PUC Accountancy Financial Statements of a Company Long Answer Type Questions and Answers

Question 1.
Explain the nature of the financial statements.
Answer:
The nature of the financial statements depends upon the following aspects;
1. Recorded facts: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Conventions: The preparation of financial statements is based on some accounting conventions like, Prudence Convention, Materiality Convention, Matching Concept, etc. The adherence to such accounting conventions makes financial statements easy to understand, comparable and reflects the true and fair financial position of the company.

3. Accounting Assumptions: These basic accounting assumptions like Going Concern Concept, Money Measurement Concept, Realisation Concept, etc are called as postulates. While preparing financial statements, certain postulates are adhered to. The nature of these postulates is reflected in the nature of the financial statements.

4. Personal Judgments: Personal value judgments play an important role in deciding the nature of the financial statements. Different judgments are attached to different practices of recording.transactions in the financial statements. Thus, personal judgments determine the nature of the financial statements to a great extent.

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Question 2.
Explain in detail about the significance of the financial statements.
Answer:
The importance of financial statements is mentioned below:
1. Provides Information: Financial statements provide information to various accounting users both internal as well as external users. It acts as a basic platform for different accounting users to derive information according to varying needs. For example, the financial statements on one hand help the shareholders and investors in assessing the viability and return on their investments, while on the other hand, the financial statements help the tax authorities in calculating the amount of tax liability of the company.

2. Cash Flow: Financial statements provide information about the cash flows of the company. The financial statements help the creditors and other investors.in determining solvency of company.

3. Effectiveness of Management: The comparability feature of the financial statements enables management to undertake comparisons like inter-firm and intra-firm comparisons. This not only helps in assessing the viability and performance of the business but also helps in’ designing policies and drafting policies. The financial statements enhance the effectiveness and efficacy of the management.

4. Disclosure of Accounting Policies: Financial statements provide information about the various policies, important changes in the methods, practices and process of accounting by the company. The disclosure of the accounting policies makes financial statements simple, true and enables different accounting users to understand without any ambiguity.

5. Policy Formation by Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policy measures and to address various economic problems like employment, poverty etc.

6. Attracts Investors and Potential Investors: They invest or plan to invest in the business. Hence, in order to assess the’viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

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Question 3.
Explain the limitations of financial statements.
Answer:
The following are the limitations of financial statements.
1. Historical Data: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Ignorance of Qualitative Aspect: A financial statement does not reveal the qualitative aspects of a transaction. The qualitative aspects like colour, size and brand position in the market, employee’s qualities and capabilities are not disclosed by the financial statements.

3. Biased: Financial statements are based on the personal judgments regarding the use of methods of recording. For example, the choice of practice in the valuation of inventory, method of depreciation, amount of provisions, etc. are based on the personal value judgments and may differ from person to person. Thus, the financial statements reflect the personal value judgments of the concerned accountants and clerks.

4. Inter-firm Comparisons: Usually, it is difficult to compare the financial statements of two companies because of the difference in the methods and practices followed by their respective accountants.

5. Window dressing: The possibility of window dressing is probable. This might be because of the motive of the company to overstate or understate the assets and liabilities to attract more investors or to reduce taxable profit. For example, Satyam showed high fixed deposits- in the Assets side of its Balance Sheet for better liquidity that gave false and misleading signals to the investors.

6. Difficulty in Forecasting: Since the financial statements are based on historical data, so they fail to reflect the effect of inflation. This drawback makes forecasting difficult. Prepare the format of income statement and explain its elements
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 1
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 2

Prepare the format of balance sheet and explain the various elements of balance sheet.
Vertical form of balance sheet
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 3
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 4
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 5

1. Share Capital: It is the first item on the Liabilities side. It consists of the following items:

  • Authorised Capital
  • Issued Capital: Equity share and preference share.
  • Subscribed Capital less Call in Arrears add Forfeited Shares

2. Reserve and Surplus: As per the Schedule VI,  it consists of the following items:

  • Capital Reserve
  • Capital Redemption Reserve
  • Security Premium
  • Other Reserve less Debit balance of P & L A/c
  • Credit balance of P & L A/c
  • Proposed Additions.
  • Sinking Fund

3 . Secured Loans

  1. Debentures
  2. Loan and advances from bank etc.

4. Unsecured Loans

  • Fixed Deposits
  • Loan & Advances from subsidiaries

5. Fixed Assets: These are those assets that are used for more than one year, like:

  • Goodwill
  • Land
  • Building
  • Plant ec Machinery
  • Patents, Trade Marks
  • Livestock
  • Vehicles, etc.

6. Current Assets: Assets that can be easily converted into cash or cash equivalents are termed as current assets. These are required to run day to day business activities; for example, cash, debtors, stock, etc.

7. Current Liabilities: Those liabilities that are incurred with an intention to be paid or are payable within a year; for example, bank overdraft creditors, bills payable, outstanding wages, short-term loans, etc are called current liabilities.

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Question 4.
Explain how financial statements are useful to the various parties who are interested in the affairs of an undertaking?
Answer:
The various parties that are directly or indirectly interested in the financial statements of a company can be categorized into the following two categories:
Internal Parties: The following are the various internal accounting users who are directly related to the company.

a) Owner: The owner/s is/’are interested in the profit earned or loss incurred during an accounting period. They are interested in assessing the profitability and viability of the capital invested by them in the business.

b) Management: The financial statements help the management in drafting various policies measures, facilitating planning and decision making process. The financial statements also enable management to exercise various cost controlling measures and to remove inefficiencies.

c) Employees and workers: They are interested in the timely payment of wages and salaries, bonus and appropriate increment in their wages and salaries. With the help of the financial statements they can know the amount of profit earned by the company and can demand reasonable hike in their wages and salaries.

External Parties: There are various external users of accounting who need accounting information for decision making, investment planning and to assess the financial position of the business. The various external users are given below.

a) Banks and other financial institutions: Banks provide finance in the form of loans and advances to various businesses. Thus, they need information regarding liquidity, creditworthiness, solvency and profitability to advance loans.

b) Creditors: These are those individuals and organisations to whom a business owes money on account of credit purchases of goods and receiving services; hence, the creditors require information about credit worthiness of the business.

c) Investors and potential investors: They invest or plan to invest in the business. Hence, in order to assess the viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

d) Tax “authorities: They need information about sales, revenues, profit and taxable income in order to determine the levy various types of tax on the business.

e) Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policies measures and to address various economic problems like employment, poverty etc.

f) Researchers: Various research institutes like NGOs and other independent research institutions like CRISIL, stock exchanges, etc. undertake various research projects and the accounting information facilitates their research work.

g) Consumers: Every business tries to build up reputation in the eyes of consumers, which can be created by the supply of better quality products and post-sale services at reasonable and alfordable prices. Business that has transparent financial records, assists the customers to know the correct cost of production and accordingly assess the degree of reasonability of the price charged by the business for its products and, thus, helps in repo building of the business.

h) Public: Public is keenly interested to know the proportion of the profit that the business spends on various public welfare schemes; for example, charitable hospitals, funding schools, etc. This information is also revealed by the profit and loss account and balance sheet of the business.

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Question 5.
Financial statements reflect a combination of recorded facts, accounting conventions and personal judgments’ discuss.
Answer:
The financial statements are the end-products of the accounting process. The financial statements not only reveal the true financial position of the company but also help various accounting users in decision making and policy designing process. The nature of the financial statements depends upon the following aspects like recorded facts, conventions, concepts, and personal judgment.

1. Recorded facts: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Conventions: The preparation of financial statements is based on some accounting conventions like,: Prudence Convention, Materiality Convention, Matching Concept, etc. The adherence to such accounting conventions makes financial statements easy to understand, comparable and reflects the true and fair financial position of the company.

3. Accounting Assumptions: These basic accounting assumptions like Going Concern Concept, Money Measurement Concept, Realisation Concept, etc are. called as postulates. While preparing financial statements, certain postulates are adhered to, The nature of these postulates is reflected in the nature of the financial statements.

4. Personal Judgments: Personal value judgments play an important role in deciding the nature of the financial statements. Different judgments are attached to different practices of recording transactions in the financial statements. Thus, personal judgments determine the nature of the financial statements to a great extent.

Question 6.
Explain the process of preparing income statement and balance sheet.
Answer:
The process of preparing Horizontal Form of Income Statement is explained below in a chronological order:

  1. Prepare a Trial Balance on the basis of the balances of various accounts in the ledger.
  2. Record Opening Stock, Purchases, Manufacturing Expenses and other direct expenses on the debit side of Trading Account
  3. Record Sales and Closing Stock on the credit side of the Trading Account.
  4. Ascertain the balancing figure by totalling both the sides of the Trading Account If the credit side exceeds the debit side, then the balancing figure is termed as Gross Profit, but if the debit side exceeds the credit side, then the balancing figure is termed as Gross Loss.
  5. Carry forward the Gross Profit (Gross Loss) to the credit (debit) side of the Profit and Loss Account.
  6. Record all current year’s operating and non-operating revenue expenditures with their relevant adjustments on the debit side of the Profit and Loss Account.
  7. Record ail current year’s operating and non-operating revenue incomes with their relevant adjustments on the credit side of the Profit and Loss Account.
  8. Ascertain the balancing figure by totalling both the sides of the Profit and Loss Account. If the credit exceeds the debit side, then the balancing figure is termed as Net Profit, but if the debit side exceeds the credit side, then the balancing figure is termed as Net Loss.

The process of preparing Horizontal Form of Balance Sheet is explained below in a chronological order:

  1. Prepare a Trial Balance on the basis of the balances of various accounts in the ledger.
  2. Record all the debit balances of Real and Personal Accounts on the left hand side (i.e. Assets side) of the Balance Sheet after making all adjustments for provision and other related items.
  3. Record all the credit balances of Real and Personal Accounts on the right hand side (i.e. Liabilities side) of the Balance Sheet after making all adjustments for interest and outstanding items.
  4. Add Net Profit to the Opening Capital and deduct Net Loss, if any from the Opening Capital
  5. Ascertain the total of two sides, jvhich must be equal.

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2nd PUC Accountancy Financial Statements of a Company Numerical Questions and Answers 

Question 1.
Show the following items in the balance sheet as per the provisions of the companies Act, 1956 in (Revised) Schedule VI:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 6
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 7

Question 2.
On 1stApril, 2013, Jumbo Ltd., issued 10,000; 12% debentures of₹ 100 each a discount of 20%; redeemable after 5 years. The company decided to write-off discount on issue of such debentures over the life time of the Debentures. Show the items of the company immediately after the issue of these debentures.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 8

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Question 3.
From the following information prepare the balance sheet of Gitanjali Ltd. Invntories ₹ 14,00,000; Equity Share Capital ₹ 20,00,000; Plant and Machinery ₹ 10,00, 000; Preference Share Capital ₹ 12,00,000; Debenture Redemption Reserve ₹ 6,00,000; Outstanding Expenses ₹ .3,00,000; Proposed-Dividend ₹ 5,00,000; Land and Building ₹ 20,00,000; Current Investment ₹ 8,00,000; Cash. Equivalent ₹ 10,00,000; Short term loan from Zaveri Ltd. (A Subsidiary Company of Twilight Ltd.) ₹ 4,00,000; Public Deposits ₹ 12,00,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 9
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 10

Question 4.
From the following information prepare the balance sheet of Jam Ltd. Inventories ₹ 7,00,000; Equity Share Capital ₹ 16,00,000; Plant and Machinery ₹ 8,00,000; Preference Share Capital ₹ 6,00,000; General Reserves ₹ 6,00,000; Bills payable Rs 1,50,000; Provision for taxation ₹ 2,50,000; Land and Building k 16,00,000; Non-current Investments ₹ 10,00,000; Cash at Bank ₹ 5,00,000; Creditors ₹ 2,00,000; 12% Debentures ₹ 12,00,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 11
Note to account
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 12
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 13

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Question 5.
Prepare the balance sheet of Jyoti Ltd., as at March 31, 2018 from the following information. Building ₹ 10,00,000; Investments in the shares of Metro Tyers Ltd. ₹ 3,00,000; Stores & Spares ₹ 1,00,000; Discount on issue of 10% debentures of ₹ 10,000; Statement of Profit and Loss (Pr.). ₹ 90,000; 5,00,000 Equity Shares of ₹ 20 each fully paid-up; Capital Redemption Reserve ₹ 1,00,000; 10% Debentures ₹ 3,00,000; Unpaid dividends ₹ 90,000; Share options outstanding account ₹ 10,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 14
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 15

Question 6.
Brinda Ltd., has furnished the following information:
(a) 25,000,10% debentures of ₹ 100 each;
(b) Bank Loan of ₹ 10,00,000 repayable after 5 years;
(c) Interest on debentures is yet to be paid.
Show the above items in the balance sheet of the company as at March 31, 2018.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 16

Question 7.
Prepare a Balance sheet of Black Swan Ltd., as at March 31, 2013 from the following information:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 17
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 18

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2nd PUC Chemistry Question Bank Chapter 2 Solutions

You can Download Chapter 2 Solutions Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions

2nd PUC Chemistry Solutions NCERT Textbook Questions and Answers

Question 1
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:
A true solution is a homogenous mixture of two or more substances. The constituent . particle which is in larger amount’’ is called a solvent and that in smaller quantity is called a solute.

Since the solvent ans solute may be either . gaseous, liquid and solid, the number of possible types of binary solutions than can be prepared are given below.
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 0

Question 2.
Suppose a solid solution is formed are very small. What kind of solid solution is this likely to be?
Answer:
Solid in solid type. E.g: Copper in gold. This type of solutions are called alloys.

Question 3.
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer:
(i) Mole fraction (X): The mole fraction of any component in a solution is the ratio of the number of moles of that component to the
sum of the number of moles of all the components present in the solution.
For a binary solution containing A and B, Mole fraction of A,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 1
where nA and nB are the numbers of moles of components A and B respectively.

(ii) Molality (m): Molality is the numberof moles of the solute dissolved in 1000 gms (1 kg) of the solvent. It B denoted by ‘m’ mathematically.
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 2

(iii) Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm3) of solution. It is denoted by ‘M’ mathematically.
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 3

(iv) Mass fraction multiplied by 100 gives mass percentage. E.g.: mass percentage of A
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 4

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Question 4.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mLr-1?
Answer:
68% of nitric acid by mass means that
Mass of nitric acid = 68g
Mass of solution = 100g
Molar mass of HN03 = 63g mol-1?
∴ 68 g HNO3 = \(\frac{68}{63}\) mole = 1.079 mole
Density of solution = 1.504g mL-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 5

Question 5.
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL-1, then what shall be the molarity of the solution?
Answer:
10 g glucose is present in 100 g solution, i.e., 90 g of water = 0.090 kg of H2O
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 6

Question 6.
How many mL of 0.1 M HCl are required to react completely with 1 g mixture Of Na2CO3 and NaHC03 containing equimolar amounts of both?
Answer:
Step 1: To calculate the number of moles of the components in mixture
suppose Na2CO3 present in the mixture = X g ,
NaHCO3 present in the mixture = (1 – x) g
Molar mass of Na2CO3 = 2 x 23+12+3 x 16=106g mol-1
Molar mass of NaHCO3.
= 23 + 1 + 12 + 3 × 16 = 84g mol-1
∴  Moles of Na2CO3 in xg = \(\frac{ x }{ 106}\)
Moles of NaHCO3 in (1-x) g = \(\frac{1-x}{84}\)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 7

Step 2: To calculate the moles of HC1 required.
Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
1 mole of Na2CO3 required HCl = 2 moles
0.00526 mole of Na2CO3 requires HCl
= 0.00526 × 2 moles = 0.01052
1 mole of NaHCO3 required HCl = 1 mole
0.00526 mole of NaHCO3 required HCl
= 0.00526 mole
∴Total HCl required = 0.01052 + 0.00526
=0.01578 moles

Step3: To calculate volume of 0.1M HCl
0.1 mole of 0.1 M HCl are present in 1000 mL of HCl
0.01578 mole of 0.1 M HCl will be present in HCl = \(\frac{1000}{0.1}\) x 0.01578
= 157.8 ml.

Question 7.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
300 g of 25% solutions contains solute = 75g
400 g of 40% solution contains solute = 160 g.
Total solute = 160 + 75 = 235 g
Total solution=300 + 400 = 700 g
% of solute in final solution = \(\frac{235}{700}\) × 100 = 33.5%
% of water in the final solution = 100 – 33.5 = 66.5%

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Question 8.
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?
Answer:
Mass of the solute, C2H4(OH)2 = 222.6g
Molar mass of C2H4 (OH)2 = 62 g mol-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 8
Mass of the solvent = 200 g = 0.200 kg

2nd PUC Chemistry Question Bank Chapter 2 Solutions - 9
Total mass of the solution = 422.6g
Volume of the solutions =

2nd PUC Chemistry Question Bank Chapter 2 Solutions - 10

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Answer:
15 ppm means 15 parts in million (106) parts by mass in solution
∴ % by mass = \(\frac{15}{10^{6}} \times\) × 100=15 × 10-4
Taking 15 g chloroform in 106g
Molar mass of CHCl3 = 12 + 1 + 3 × 35.5 = 119.5 g mol-1
Molality = \(\frac{15 / 119.5}{10^{6}}\) × 1000= 1.25 × 10-4 m

Question 10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
There is strong hydrogen bonding in alcohol molecules as well as water molecules. On mixing the molecular interactions are weekend. Hence, they show positive deviations from ideal behaviour. AS a result, the solution will have higher vapour pressure and lower boiling point than that of water and alcohol.

Question 11.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Dissolution of gas in liquid in an exothermic process (Gas + solvent ⇌ solution + Heat). As the temperature is increased, equilibrium shifts backward.

Question 12.
State Henry’s law and mention some important applications?
Answer:
Henry’s law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Applications: It is used

  • in the production of carbonated beverages
  • in the deep sea diving
  • in the function of lungs.

KSEEB Solutions

Question 13.
The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial pressure of the gas?
Answer:
Applying the relationship m = KH x p
In the first case, 6.56 x 102 g bar-1
In the second case, 5.00 x 10-2 g = (6.56 × 10-2 gbar-1) × p
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 11

Question 14.
What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔmixH related to positive and negative deviations from Raoult’s law?
Answer:
In +ve deviation, A-B interactions are weaker than those between A-A or B-B. In such molecules, A or B will find it easier to escape than in pure state. This increases vapour .pressure. In case of -ve deviation, A-B interaction = A-A or B-B. This leads to decrease in vapour pressure.

In +ve deviation, ΔmixH is + ve
In -ve deviation, ΔmixH is – ve

Question 15.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer:
Vapour pressure of pure water at boiling
point (P°) = 1 atm = 1.013 bar
vapour pressure of solution (ps) = 1.004 bar
Mass of solute = (w2) = 2g
Mass of solution = 100 g
Mass of solvent = 98g
Applying Roault’s law for dilute solution (being 2%)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 12

Question 16.
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer:
Molar mass of heptane (C7 H16) = 100 g mol-1
Molar mass of octane (C8H18) = 114 g mol-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 13
x (octane) = 1 – 0.456 = 0.544
p (heptane) = 0.456 x 105.2 kPa = 47.97 kPa
p (octane) = 0.544 x 46.8 kPa = 25.46 kPa
p total = 47.97 + 25.46 = 73.43 kPa

Question 17.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
1 molal solution means 1 mol of the solute in 1 kg of solvent (water)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 14
or ps = 12.08 kPa

Question 18.
Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer:
Ps = 80% of p° = 0.80 p° solute = \(\frac{w}{40} r\) mol solvent (octane) =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 15
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 16

Question 19.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.
Answer:
Suppose the molar mass of the solute = Mg mol-1
n2 (solute) = \(\frac{30}{M}\) moles
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 17
After adding 18 g of water,
n(H2O)i.e, n1 = 6 moles
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 18
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 19
Dividing equation (i) by equation (ii), we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 20

(ii) putting M = 23 in equation (i), we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 21

Question 20.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 A. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Answer:
Molality of sugar solution =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 22
∆Tf for sugar solution = 273.15 – 271 = 2.150
∆Tf = Kf × m ∴ Kf = 2.15/0.146
Molality of glucose solution =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 23
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 k

KSEEB Solutions

Question 21.
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. Tin molar depression – constant for benzene is 5.1 K kg mol-1. Calculate atomic masses of A and B.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 24
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 25
Suppose atomic masses of A and B are ‘a’ and ‘b’ respectively. Then
Molar mass of AB2 = a + 2b = 110.87 g mol-1
Molar mass of AB4 = a + 4b =196.15 g mol-1
Equation (ii) – Equation (i) gives
2b = 85.28 orb = 42.64
substituting in equation (i) we get
a + 2 × 42.64 = 110.87 or a = 25.59
Thus atomic mass A = 25.59 u
Atomic mass of B = 4.64 u.

Question 22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
π = CRT
∴ In the first case,
4.98 = \(\frac{36}{180}\) × R × 300 = 60R
In the second case, 1.52 = C × R × 300
Dividing (ii) by (i), we get C = 0.061 M.

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).
Answer:
(i) Both are non-polar. Hence, inter molecular interactions in them will be London/ dispersion forces (discussed in class XI)
(ii) Both are non-polar. Hence, inter molecular interactions in them will be London/ dispersion forces (discussed in class XI)
(iii) NaClO4 gives Na+ and ClO4 ions in the solution while water is polar molecule. Hence, intermolecular interactions in them will be ion – dipole interactions.
(iv) Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.
(v) Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer:
(i) Cyclohexane and n-octane both are non-polar. Hence they mix completely in all proportions.
(ii) KCl is an ionic compound while n-octane is nonpolar. Hence, KCl will not dissolve at all in n-octane.
(iii) CH3OH and CH3CN both are polar but CH3CN is less polar than CH3OH. As the solvent is non-polar, CH3CN will dissolve more than CH3OH is n-octane.
Thus the order of solubility will be KCl< CH3OH < CH3CN < Cyclohexane.

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) Partially soluble (because phenol has
polar-OH group but aromatic phenyl, C6H5 – group)
(ii) Insoluble because toluene is non polar while water is polar
(iii) Highly soluble because formic acid can form hydrogen bonds with water.
(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water
(v) Insoluble chloroform is an organic liquid
(vi) Partially soluble because-OH group is polar but the large hydrocarbon part (C5H11) is nonpolar.

KSEEB Solutions

Question 26.
If the density of some lake water is 1.25g mL-1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
Answer:
Number of moles in 92 g of Na+ ions
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 26
As these are present in 1 kg of water, by definition, molality = 4m.

Question 27
If the solubility product of CuS is 6 × 10-16, calculate the maximum molarity of CuS in aqueous solution.
Answer:
Maximum molarity of CuS in aqueous solution = solubility of CuS in mol L-1
If S in the solubility of CuS in mol L-1 , then
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 27

Question 28.
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 28

Question 29.
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10-3 m aqueous solution required for the above dose.
Answer:
1.5 × 10-3 m solution means that 1.5 × 10-3 mole of nalorphene is dissolved in I kg of water.
Molar mass of C19H21NO3 = 19 × 12 + 21 + 14 + 48
= 3119 mol-1
∴ 1.5 × 10-3 mole of C19H21NO3;
= 1.5 × 10-3 × 3119 = 0.467 g = 467 mg
∴ Mass of solution = 1000 g + 0.467 g ‘
=1000.467g
Thus, for 467 mg of nalorphene, solution required =1000.467 g for 1.5 mg of nalorphene, solution required =\(\frac{1000.467}{467} \times 1.5 = 3.219\)

Question 30.
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer:
0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L
i.e. 1000 mL of the solution.
Molar mass of benzoic acid (C6H5COOH) = 72 + 5 + 12 + 32 + 1 =122 g mol-1
∴ 0.15 mole of benzoic acid = 0.15 × 122g= 18.39
Thus, 1000 mL of the solution contain benzoic acid = 18.39
∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250= 4.575g

KSEEB Solutions

Question 31.
The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:
The depression in freezing points are in the order:
acetic acid < trichloroacetic acid
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 29
Fluorine, being most electronegative, has the highest electron withdrawing inductive effect. Consequently, trifluoroacetic acid is the Strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to minimum extent to give ions.in their.solutions . in water. Greater the ions produced, greater is the depression in freezing point. Hence the depression in freezing point is maximum for fluoroacetic acid and minimum for acetic acid.

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH3CH2 CHCICOOH is added to 250 g of water. Ka = 1.4 × 10-3, Kf= 1.86 K kg mol-1.
Answer:
Molar mass of CH3CH2CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 30
If α is the degree of dissociation of CH3CH2CHCICOOH, then CH3CH2CHCLCOOH □ CH3CHXHCICOO + H +
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 31
To calculate vant’s Hoff factor:
CH3CH2CHCICOOH □ CH3CH2HCICOO + H+
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 32

Question 33.
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid. Kf of water is 1.86 K kg mol-1.
Answer:
Here, w2 = 19.5 g, w1, = 500g, Kf = 1.86 K kg mol-1, (ATf) obs = 1.0°
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 33
M2 (Calculated) for CH2 FCOOH = 14+19+45 = 78 g mol-1
Vant Hoff factor (i) =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 34
calculation of dissociation constant. Suppose degree of dissociation at the given concentration is a.
Then
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 35
Taking volume of the solutions as 500ml,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 35(i)

KSEEB Solutions

Question 34.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:
Here, P° = 17.535 mm, w2 = 25g,
w1 = 450g
For solute (glucose, C6H12O6, M2 = 180 g mol-1
For solvent (H2O), M1 = 18g mol-1
Applying Raoult’s law,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 36
substituting the given values, we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 37
substituting the given values, we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 38

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10s mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer:
Here, KH = 4.27 × 105 mm
p = 760 mm
Applying Henry’s law
P = KH x
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 39
i.e, mole fraction of methane in benzene = 1.78 x 10-3

Question 36.
100 g of liquid A (molar mass 140 g mol-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer:
Number of moles of a liquid
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 40
Number of moles of a liquid B (Solvent)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 41
Mole fraction of B in the solution (xA)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 42
Mole fraction of B in the solution (xB)
= 1-0.114 = 0.886
Also, given p0B = 500 Torr
Applying Raoult’s law,
PA = xAA = 0.114 × P°A
PB= xBB = 0.886 × 500 = 443 Torr
p Total = pA + pB
475 = 0.114 P°A +443 or
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 43
Substituting this value in equation (i), we get pA = 0.114x 280.7 Torr = 32 Torr

Question 37.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50. 71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer:
Molar mass of benzene (C6H6 = 78 g mol-1
Molar mass of toluene (C6H5CH3) = 92 g mol-1
∴ Number of moles in 80 g of benzene
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 44
Number of moles in 100 g of toluene
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 45
In the solution, mole fraction of benzene
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 46
=0.486
mole fraction of toluene = 1-0,486 = 0.514
p° Benzene = 50.71 mm, p° Toluene = 32.06mm
Applying Raoult’s law,
p Benzene= xTolucno × p°Tolucne =0.514 × 32.06 mm = 16.48mm
Mole fraction of benzene in vapour phase
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 47

Question 38.
The air is a mixture of a number of gases. The major components are oxygen and, nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
Answer:
Total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volumes
partial pressure of oxygen :
(P ) = \(\frac{20}{100}\) x 10 atm = 2atm
= 2 x 760 mm = 1520 mm
partial pressure of nitrogen \(P_{N_{2}}\)
= \(\frac{79}{100}\) × 10 atm = 7.9 atm = 7.9 x 760 mm
= 6004 mm

KH(O2) = 3.30 x 107mm,KH (N2) = 6.51 × 107 mm
Applying Henry’s law
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 48

Question 39.
Determine the amount of CaCl (i= 2.47) dissolved in 2.51itre of water such that its osmotic pressure is 0.75 atm at 27° C.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 49
Molar mass of CaCl2= 40 + 2x 35.5 = 111 gmol-1
Amount dissolved = 0.0308 xlllg = 3.42g

Question 40.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4in 2 litre of water at 25° C, assuming that it is completely dissociated.
Answer:
K2SO4 dissolved = 0.025g
volume of solution = 2L
T = 250°C = 298 K
Molar mass of K2SO4 =
2×39 + 32 + 4 × 16= 174gmol-1
As, K2SO4 dissociates completely as K2SO4 → 2K+ SO42-
i.e., ions produced = 3 ∴ i =3
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 50

KSEEB Solutions

2nd PUC Chemistry Solutions Additional Questions and Answers

Question 1.
How does molarity of a solution change with temperature?
Answer:
Molarity decreases with increase in temperature because volume of solution increases with increase in temperature.

Question 2.
State Raoult’s law.
Ans:
It states that for a solution of volatile liquids its vapour pressure of each component is directly proportional to their mole fraction in the solution.
The relative lowering of vapour pressure is equal to mole fraction of solute in case of non volatile solute.

Question 3.
Why is liquid ammonia bottle first cooled in ice before opening it?
Answer:
At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out.

Question 4.
Sodium chloride and calcium chloride are used clear show from the roads. Why?
Answer:
Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature.

Question 5.
Two liquids A and B boil at 145°C and 190°C
respectively. Which of them has a higher vapour pressure of 80°C? (CBSE 2006)
Answer:
A being more volatile will have higher vapour pressure at 80°C.

KSEEB Solutions

Question 6.
Calculate the molality of sulphuric acid solution in which the mole fraction is 0.85.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 51

Question 7.
The molar freezing point depression constant for benzene is 4.90 kg mol’1. Selenium exists as polymer. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower for pure benzene. Decide the molecular formula of selenium (At Wt. of selenium is 78.8 g mol-1). (CBSE 2002)
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 52

Question 8.
CCI4 and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. (CBSE 2003,2001)
Answer:
CCI4 is a non-polar covalent compound.
Water is a polar compound. CCI4 can neither form H bonds with water molecules nor can it break H bonds between water molecules. Therefore, it is insoluble in water.

Ethanol is a polar compound and can form H’ bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions.

Question 9.
The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm3 )
Answer:
Let the solution be 1 litre or 1000 cm3
∴ Number of moles of H2SO4= 1.35
Wt. of solution = 1000 x 1.02 = 1020 g
Wt. of sulphuric acid = 1.35 x 98= 132.3g :
Wt. of water = 1020 – 132.3 = 887.79
Molality of H2SO4= ppp x 1000 = 1.52 m

From (i) M, = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15- 110.82 = B2 85.33 = B2 B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82=Atomic mass of A+85.33 Atomic mass of A = 110.82 – 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669

Question 10.
Two elements A and B form compounds having molecular formula AB2 and AB4. When dissolved 20 g of C6H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB„4 lower it by 1.3 K. The molar depression constant for benzene is 5.1 kg mol*1. Calculate the mass of A and B. (CBSE2004)
Answer:
Let the molar mass of AB2 and AB4 be M1and M2
Then, for AB2,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 53
From (I) M1 = 110.82, from (ii) M1 = 196.15
AB4– AB2= B2
196.15 – 110.82 =B2
85.33 = B2
B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B
110.82 = Atomic mass of A+ 85.33
∴ Atomic mass of A = 110.82-85.33 = 25.49
Atomic mass of A = 25 .499
Atomic mass of B = 42.669

KSEEB Solutions

KSEEB Solutions for Class 9 English Poem Chapter 10 Photograph

By understanding the academic needs we have prepared the Karnataka State Board English Class 9 Solutions Chapter Wise. Our aim is to help the students by providing the question and answers chapter wise and help them to gain a good score in the exams. Before you start your preparation go through the chapters covered in this academic. So go through them and Download KSEEB Solutions for Class 9 English Poem Chapter 10 Photograph Question and Answers Pdf for free.

Karnataka State Board Class 9 English Poem Chapter 10 Photograph

The topics covered in Karnataka Secondary Education Examination Board Class 9 Solutions for English Chapter 10 Photograph. The KSEEB Solutions Class 9 English Solutions Chapter 10 Photograph Question and Answers are prepared according to the latest edition.The Chapterwise page will help the students to revise the syllabus during the exams.

Photograph Questions and Answers, Summary, Notes

Comprehension:

C1. Answer the following in a sentence or two each:

Question 1.
How many people are there in the photograph?
Answer:
There are three people in the photograph.

Question 2.
How is the poet related to the people in the photograph?
Answer:
The people in the photograph are the poet’s mother and her cousins, Betty and Dolly.

KSEEB Solutions

Question 3.
Who was taking the snapshot?
Answer:
An uncle was taking the snapshot.

Question 4.
Is the mother described in the photo alive?
Answer:
No, she is dead.

Question 5.
Which aspect of the mother does the poet like very much?
Answer:
The poet has sweet memories of the smiling and sweet face of the mother which she likes. This nature of the mother is captured in the photo.

C2. Answer the following:

Question 1.
Why does the writer say
“And of this circumstance
There is nothing to say at all.
Its silence silences”?
Answer:
The poet says that her mother had been dead and now she finds herself in a situation in which there is nothing to be said but only emptiness. The silence of this situation silences her. In other words, she is left speechless. Fate has killed all the feelings in her.

By the phrase ‘of this circumstance’, the poet means the circumstance of the death of her mother. The poet puns upon the. word ‘silence’. Since there are no details, what she is left with is silence and this silence silences her.

There are different possible meanings of the use of the word ‘silence’ as verb: she cannot say anything as she does not know anything about the circumstance of the death; or she does not have words to express her feelings; or the loss of her mother has stifled her voice. If we go one step further, we can even wonder whether the silence is owing to the unnatural nature of the death of the mother.

Question 2.
Does the poet notice any change in the mother after the poet was bom? What do you think could have made the change in the mother’s face?
Answer:
Yes, the poet notices the change in the mother’s face after she was born. This could have been the outcome of sorrowful incidents or hardships in life. Age and ill health also might have made the mother lose the sweetness of her face and smile.

The poet is looking at her mother’s photograph which is indeed an old one. With it she can see how her mother looked when she was a little girl of twelve. The photo shows her on a beach with her two girl cousins who are younger than her, holding her hand.

It might have been windy at that time as their hair was flying on their faces when the uncle took the photograph. All the three smile through their flying hair. Looking at the photograph, the poet says that her mother had a sweet face, but it was a time before the poet was born.

The sea was washing their feet. The poet says that the sea has changed only a little but change has come about the ones whose feet it was washing. After 30 or 40 years, the mother would take out the photograph and take a look at it.

By that time, she was married and had a daughter. She would laugh a little and say “Look at Betty and Dolly, see how they have dressed for the beach”. By now, she can only remember those days. A huge change has come about her and she is no longer that small, innocent girl of twelve. After some years, when the poet’s mother dies, for the poet, her mother’s laughter becomes a thing of the past.

That’s why she says “the sea holiday was her past and mine is her laughter”. In the same way as the mother remembers her old days, the poet can remember her mother. The poem also shows that in due course of time, the two of them learned to live with their losses though the loss had made a permanent impression on their wry faces.

KSEEB Solutions

Question 3.
Why are the feet described as ‘transient feet’?
Answer:
The word ‘transient’ means fleeting/ passing or temporary. It is the opposite of permanent. The ‘feet’ are described with an adjective ‘transient’ to drive home the truth that the impression they make on the sand is transitory. They get washed off by the waves.

Everything about life is transient. Single incidents, as well as whole life itself, are transient. Only nature remains as a permanent feature around us. Thus, the poet presents a contrast between human life and the sea.

The poem lends itself to multiple interpretations with regard to the transitory nature of life. If the passage of time is one aspect of the transitory nature of life, sudden unexpected occurrences can also indicate how ephemeral everything is.

We can have a different reading of the poem here. At one point the children are posing for the photograph holding their aunt’s hand. At another, the sea washes their terribly transient feet. The word ‘terribly’ makes us wonder whether the children are swept away by the waves of the sea.

C3. Answer the following questions on your own.

Question 1.
What is the mood of the poet?
Answer:
Melancholy marks the utterances of the poet. The poet has a deep sense of loss on losing her mother and the tone of sadness is all-pervasive.

Question 2.
Which line in the poem do you like the most? Why?
Answer:
I like the line ‘All three stood still to smile through their hair’. On the one hand, it creates a powerful mental picture of the three human figures against the breeze on the seashore posing for the photograph. The golden time of childhood when children are filled with mirth and are free from all worries and complexities of life is powerfully pictured here.

Smiling through the hair is typical of people who pose on the seashore. The description captures the innocuous and blithe spirit of the children which is contrasted against the care and concern of old age. It shows the poet’s power of observation and description.

The line gets connected to the title also. A photograph becomes a metaphor for all that life captures and also loses. Life is transitory and we are likely to lose many things which we will remember only through photographs.

KSEEB Solutions

Question 3.
Is there any change in the life of the poet’s mother over the years? What kind of a person, you think, she was? Describe the mother in the poem in your own words.
Answer:
The three stanzas of the poem depict the three stages in the life of the mother – as a child with her cousins, as a mother looking at the old photograph, and as a memory for the daughter after her death. She had a smiling and sweet face in the photograph when she posed for it, holding the hands of her cousins.

However, the poet cannot remember witnessing the same cheerfulness on the mother’s face in her recollections of her mother after she was born. What could have been the reason for the change? Apparently, the poet’s mother had posed for the photograph with her cousins when she was young and was not yet bogged down by the responsibilities and hardships of life.

As people age, along with inevitable physical changes, they also experience a change in their mental make-up because of the challenges in life. The line,’… how they dressed us for the beach,’ indicates that the mother herself was quite young and followed the directions given by others when she accompanied her cousins to the beach.

Additional Questions:

Question 1.
What does the word ‘cardboard’ denote in the poem? Why has this word been used?
Answer:
The word ‘cardboard’ means a very stiff and thick paper. Here the cardboard is a part of the frame that keeps the photograph intact. Its use in the poem is ironical. It keeps the photograph of that twelve-year-old girl safe who herself was ‘terribly transient’. She had died years ago.

Question 2.
What has not changed over the years? What does this suggest?
Answer:
The sea has not changed over the years. It brings out the ‘transient’ nature of man when compared to nature and its objects. Time spares none. The pretty faces and the feet of the three girls are ‘terribly transient’ or mortal when compared to the ageless and the unchangeable sea.

Question 3.
The poet’s mother laughed at the snapshot. What did this laugh indicate?
Answer:
The poet’s mother would laugh at the snapshot as it would revive her memories of the old happy days on the sea beach and the strange way in which they were dressed up for the beach. The laugh indicates her youthful spirit.

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Question 4.
What is the meaning of the line “Both wry with the laboured ease of loss”?
Answer:
Both the poet’s mother and the poet suffer a sense of loss. The mother has lost her childlike innocence and joyful spirit that the photograph captured years ago. For the poet the smile of her mother has become a thing of the past. Ironically, both labour to bear this loss with ease.

Question 5.
What does ‘this circumstance’ refer to?
Answer:
‘This circumstance’ refers to the death of the poet’s mother. The photograph of the dead mother brings sad nostalgic feelings in the poet. But the poet has nothing to say at all about this circumstance. The silence of the poet makes the pall of silence prevailing there still deeper.

Question 6.
The three stanzas depict three different phases. What are they?
Answer:
In the first stanza, the poet’s mother is shown as a twelve-year-old girl with a pretty and smiling face. She went paddling with her two cousins. This phase is before the poet’s birth. The second phase describes the middle-aged mother laughing at her own snapshot. The third phase describes the chilling pall of silence that the death of the mother has left in the life of the poet.

Question 7.
“The sea, which appears to have changed less, washed their terribly transient feet.” – How does the poet contrast the girls’ ‘terribly transient feet’ with the sea?
Answer:
All the girls standing on the beach have ‘terribly transient’ existence. They are mortal and suffer physical changes with the passage of time. The mother’s sweet face and her smile has disappeared years later. But the vast sea remains unchanged of seems ‘to have changed less’ in their comparison.

Question 8.
‘Both wry with the laboured ease of loss’ – What is the loss? Describe the irony in the situation.
Answer:
Actually, both the poet and her mother suffer a sense of loss. The mother loses her carefree childhood. She can’t have those moments of enjoyment again that she once experienced at the beach. She can’t be a sweet smiling girl of twelve again.

This is also the poet’s loss. Perhaps she will never see that smiling face and experience her laughter again in life. The irony of the situation is that both of them struggle to bear the loss with tolerable ease.

Multiple Choice Questions:

Question 1.
The poet looks at the photograph of her mother taken when she was
A) ten years old
B) twelve years old
C) an old woman
D) a young lady
Answer:
B) twelve years old

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Question 2.
The mother had gone on a sea-holiday
A) with her father
B) with her mother
C) with her cousins
D) with her daughter
Answer:
C) with her cousins

Question 3.
The photograph was taken by her
A) father
B) cousin
C) sister
D) uncle
Answer:
D) uncle

Question 4.
The Phrase ‘Their terribly transient feet’ suggests that
A) their feet were moving
B) they were padding
C) life is not permanent
D) life is everlasting
Answer:
C) life is not permanent

Question 5.
The poet’s mother is dead and the poet has nothing to say about that situation because
A) she does not want to say anything
B) the silence of death has silenced her
C) she cannot relive the past
D) she knows her mother would come back.
Answer:
B) the silence of death has silenced her

Question 6.
The mother saw the photograph after …………. years.
A) twenty or thirty
B) twelve or thirteen
C) four or five
D) one or two.
Answer:
A) twenty or thirty

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Question 7.
The names of the cousins are
A) Bertha and Pinky
B) Betty and Dolly
C) Rosa and Mary
D) Laila and Pinky.
Answer:
B) Betty and Dolly

Question 8.
The cardboard shows the pictures of
A) two schoolgirls
B) two real sisters
C) two neighbours
D) narrator’s mother and her two cousins.
Answer:
D) narrator’s mother and her two cousins.

Question 9.
The ‘big girl’ here means
A) the eldest cousin of the mother
B) the tallest of the girls
C) narrator’s mother
D) the heaviest of the girls.
Answer:
C) narrator’s mother

Question 10.
The sweet face the photograph showed was that of the
A) narrator’s cousin
B) narrator’s father
C) narrator’s mother
D) narrator’s brother.
Answer:
C) narrator’s mother

Question 11.
The photograph was taken
A) when the narrator was about twelve
B) about twelve years ago
C) when the narrator was a child
D) when the narrator was not even born.
Answer:
D) when the narrator was not even born.

Question 12.
Who were ‘Betty’ and ‘Dolly’?
A) narrator’s cousins
B) her sisters
C) her mother’s cousins
D) her neighbours.
Answer:
C) her mother’s cousins

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Question 13.
‘Its silence silences’ means
A) death’s silence
B) silence only brings out deeper silence
C) poet’s silence
D) silence caused by the mother’s death gives birth to a pall of silence.
Answer:
D) silence caused by the mother’s death gives birth to a pall of silence.

Photograph by Shirley Toulson About The Poet:

Shirley Toulson was born in 1924 in Henley-on-Thames, England. She had a huge passion for writing and was greatly influenced by her father who was a writer too. She secured a B.A. in Literature from Brockenhurst College in London in the year 1953. Shortly, she took writing as a career but meantime also served as editor for many magazines.

Celtic Christianity, influenced her so greatly that most of her major works such as ‘Celtic Alternative’ in 1987 and ‘Celtic Year’ in 1993 were on that topic. But a whole lot of other works, including collections of poetry, essays on education and other publications on diverse topics brought her even more fame.

Photograph Summary in English

The poem ‘Photograph’ is an interesting piece describing how a photograph can set in motion recollections of different kinds. The photograph is also an indicator to the fact that everything in life is transitory. What is captured years ago on the photograph is a contrast to what people turn out to be years later.

The photograph can also be illusory in nature as it might not show exactly how the people being featured in the photograph felt at that particular point of time. The photograph can also refer to our desperate attempts to defy the passage of time and the inevitable changes that take place along with the passage of time.

The speaker begins the poem by saying, ‘A photograph: The cardboard shows me how it was.’ The photograph is of the speaker’s mother with her cousins. The photograph features the two girls holding the hands of the speaker’s mother and the speaker knows that the photograph was taken by an uncle. The speaker is struck by the realisation that her mother was sweet in her looks at the time the photo was taken.

This was even before the speaker was 4}orn. While referring to the sea washing the terribly transient feet of the people who posed for the photograph, the speaker introduces a contrast between the sea and the people.

People change with the passage of time, but the sea seems to have changed very little. This is an indicator to the permanence of nature and transience of man. Unfortunately, the change in human beings, both physically and emotionally, is for the worse.

The smiling, sweet face of the young girl is replaced by the melancholic face of the mother. Just as physical changes are inescapable, even mental deterioration also seems to be inevitable.

Twenty or thirty years passed since that photograph had been taken. Whenever mother looked at the photograph, she would laugh. She would point out to Dolly and Betty how they were dressed up (by their parents) for the sea-holiday.

That sea-holiday had become a thing of the past for the mother. And that sweet laughter of the mother which the photograph had captured had become the poet’s own past. Both of them suffered from a sense of loss. Ironically, both of them were labouring to ease the loss.

The poet says that her mother has been dead for years. She has been dead nearly as many years as she had lived. She (the poet) has nothing to say about the circumstances of her death or about the situation captured in the photo. Silence only brings out deeper silence and makes things mysterious.

The photograph describes three stages. In the first stage, the photograph shows the poet’s mother standing on the beach enjoying her holiday with her two girl cousins. She was 12 or so at that time.

The second stage takes us twenty or thirty years later. The mother would laugh at the way she and her cousins Betty and Dolly were dressed up for the sea-holidays. In the third stage, the poet remembers the mother with a heavy heart as the mother is already dead. The photograph revives a nostalgic feeling in the poet.

Thus, through an everyday, commonplace incident of a photo being clicked on the seashore, the poet is able to drive home philosophical reflections on the very nature of life.

Glossary:

paddling: moving like rowing.
transient: momentary; not lasting for long
wry: distorted.

We hope the information prevailed in this article is helpful for all the students of Class 9. The Karnataka State Board Solutions for Class 9 English Chapter 10 Photograph Question and Answers pdf enhance your skills and score good marks in the exams. Stay tuned to get the latest information about the KSEEB Solutions Class 9 English Solutions.