{"id":2877,"date":"2020-08-26T05:53:51","date_gmt":"2020-08-26T00:23:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2877"},"modified":"2021-07-02T15:04:28","modified_gmt":"2021-07-02T09:34:28","slug":"kseeb-solutions-class-9-science-chapter-10","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-science-chapter-10\/","title":{"rendered":"KSEEB Solutions for Class 9 Science Chapter 10 Gravitation"},"content":{"rendered":"

Karnataka Board Class 9 Science Chapter 10 Gravitation<\/h2>\n

KSEEB Solutions Class 9 Science Chapter 10 Intext Questions<\/h3>\n

Question 1.
\nState the universal law of gravitation.
\nAnswer:
\nEvery object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.
\n\"KSSEB<\/p>\n

Question 2.
\nWrite the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
\nAnswer:
\n\"KSSEB<\/p>\n

Question 3.
\nWhat do you mean by free fall?
\nAnswer:
\nWhenever objects fall towards the earth under the gravitational force alone, we say that the objects are in free fall.<\/p>\n

Question 4.
\nWhat do you mean by the acceleration due to gravity?
\nAnswer:
\nWhenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore this acceleration is called the acceleration due to the gravitational force of the earth.<\/p>\n

Question 5.
\nWhat are the differences between the mass of an object and its weight?
\nAnswer:<\/p>\n\n\n\n\n
Mass<\/strong><\/td>\nWeight<\/strong><\/td>\n<\/tr>\n
Mass of an object is the measure of the inertia.<\/td>\nThe weight of an object is the force with which it is attracted to the earth.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 6.
\nWhy is the weight of an object on the moon l\/6th its weight on the earth?
\nAnswer:
\nMass of moon is lesser than that of earth. Due to this the moon exerts lesser force of attraction on objects. Hence weight of an object on the moon l\/6th its weight on the earth.<\/p>\n

Question 7.
\nWhy is it difficult to hold a school bag having a strap made of a thin and strong string?
\nAnswer:
\nIt is difficult to hold a school bag having a starp made of a thin and strong string because the area and starp is small. Hence large pressure is exerted by the starp on the fingers.<\/p>\n

Question 8.
\nWhat do you mean by buoyancy?
\nAnswer:
\nThe upward force exerted by the water on the bottle is known as upthrust or buoyant force.<\/p>\n

Question 9.
\nWhy does an object float or sink when placed on the surface of the water?
\nAnswer:
\nUpward thrust of liquids, (different-in liquid densities) Sink of an object depends upon density. The density of an object is more than water then it sinks. Object density is less than that of water it floats.<\/p>\n

Question 10.
\nYou find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
\nAnswer:
\nOur mass is more than 42 kg.<\/p>\n

Question 11.
\nYou have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the others. Can you say which one is heavier and why?
\nAnswer:
\nIron bar is heavier than the cotton bag because for the mass the iron bar will have a lesser surface area and apply more force, so the iron bar is heavier than a cotton bag.<\/p>\n

KSEEB Solutions for Class 9 Science Chapter 10 Textbook Exercises<\/h3>\n

Question 1.
\nHow does the force of gravitation between two objects change when the distance between them is reduced to half?
\nAnswer:
\nForce of gravitation between two objects.
\n\"KSSEB
\ni.e. the force of gravitation becomes 4 times than the original value.<\/p>\n

Question 2.
\nGravitational force acts on all objects in proportion, to their masses. Why then a heavy object does not fall faster than a light object?
\nAnswer:
\nEach object falls towards the earth with the acceleration equal to the acceleration due to gravity, which is constant (9.8ms-2<\/sup>) and does not depend on the mass of the object. So heavy object does not fall faster than the light object.<\/p>\n

Question 3.
\nWhat is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 1023<\/sup> kg and radius of the earth
\n6.4 x 106<\/sup>m)
\nAnswer:
\nGravitational force between the earth and an F = \\(\\frac{\\mathrm{GM}_{\\mathrm{m}}}{\\mathrm{R}^{2}}\\)
\nWhere G = Gravitational constant = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup>kg-2<\/sup>.
\nM = Mass of earth = 6 \u00d7 1024<\/sup> kg.
\nR = Radius of earth = 6.4 \u00d7 106<\/sup>m
\nm = Mass of object = 1 kg.
\nHence F = \\(\\frac{6.67 \\times 10^{-11} \\times 6 \\times 10^{24} \\times 1}{\\left(6.4 \\times 10^{6}\\right)^{2}}\\)
\n= 9.8N.<\/p>\n

Question 4.
\nThe earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
\nAnswer:
\nWhen two objects attract each other then the gravitational force of attraction applied by first object on the second object is same as the force applied by the second object on the first object. So both earth and moon attract each other by the same gravitational force of attraction.<\/p>\n

Question 5.
\nIf the moon attracts the earth, why does the earth not move towards the moon?
\nAnswer:
\nThe earth does not move towards the moon inspite of the attraction by moon because the force with which moon attracts the earth is same by which earth attracts the moon. So net force on the system becomes zero.<\/p>\n

Question 6.
\nWhat happens to the force between two objects, if
\ni) the mass of one object is doubled?
\nii) the distance between the objects is doubled and tripled?
\niii) The masses of both objects are doubled?
\nAnswer:
\nForce of attraction between two objects.
\nF = \\(\\frac{\\mathrm{G} \\mathrm{m}_{1} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\)
\nwhere m1<\/sub>, m2<\/sub> = masses of the objects
\nr = Distance between the objects.
\nG = Gravitational constant.<\/p>\n

(i) If mass of one object is doubled, then the new force
\n\"KSSEB
\n\u2234 force becomes double.<\/p>\n

(ii) If the distance between the objects is tripled, then new force
\n\"KSSEB
\n\u2234 force becomes one-ninth.<\/p>\n

iii) If masses of both objects are doubled, then new force .
\n\"KSSEB
\ni.e. force becomes four times.<\/p>\n

Question 7.
\nWhat is the importance of universal law of gravitation?
\nAnswer:
\nThe importance of universal law of gravitation can be understood from the following points:<\/p>\n

    \n
  • All the planets revolve around the sun due to gravitational force between the sun and planets. Hence gravitational force is responsible for the existence of solar system.<\/li>\n
  • Tides in oceans are formed due to the gravitational force between the moon and water in oceans.<\/li>\n
  • The atmosphere of the earth is attached with it due to gravitational force of earth.<\/li>\n
  • Artificial satellites revolve around the earth due to gravitational force of earth.<\/li>\n
  • We can stand on the surface of earth due to gravitational force of earth.<\/li>\n<\/ul>\n

    Question 8.
    \nWhat is the acceleration of free fall?
    \nAnswer:
    \nThe acceleration produced in the motion of an object when it falls freely towards the earth is termed acceleration of free fall. It is also called acceleration due to gravity. Its value on the earth’s surface is 9.8 ms-2<\/sup>.<\/p>\n

    Question 9.
    \nWhat do we call the gravitational force between the earth and an object?
    \nAnswer:
    \nThe gravitational force between the earth and an object is called the force of gravity.<\/p>\n

    Question 10.
    \nAmit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator will the friend agree with the weight of Gold bought? If not, why? (Hint: The value of g is greater at the poles than at the equator)
    \nAnswer:
    \nWeight of an object, W = mg.
    \nwhere g = acceleration due to gravity.
    \nThe value of g is greater at the poles than at the equator. So the weight of the gold at the equator will be less than the weight of the gold at poles. So Amit’s friend will find the weight of the gold less than the weight told by Amit.<\/p>\n

    Question 11.
    \nWhy will a sheet of paper fall slower than one that is crumpled into a ball?
    \nAnswer:
    \nThe sheet of paper falls slower than that is compelled into a ball because in the first case the area of the sheet is more. So it experiences a large opposing force due to air. while the sheet crumpled into a ball experience less opposing force due to small area.<\/p>\n

    Question 12.
    \nThe gravitational force on the surface of the moon is only 1\/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?
    \nAnswer:
    \nMass of the object m = 10 kg.
    \nWeight on the earth W = mg.
    \n= 10 \u00d7 9.8 = 98 N
    \nWeight on the moon = \\(\\frac{1}{6}\\) of the weight on the earth.
    \n= 1 \u00d7 98 = 16.33 N
    \n= 16.33 N.<\/p>\n

    Question 13.
    \nA ball is thrown vertically upwards with a velocity of 49m\/s. Calculate.
    \ni) the maximum height to which it rises.
    \nii) the total time it takes to return to the surface of the earth.
    \nAnswer:
    \nInitial velocity u = 49 ms-1<\/sup>
    \n(i) At the maximum height, velocity becomes zero.
    \n\u2234 Final velocity v = 0
    \nFrom equation of upward motion
    \nv2<\/sup> = u2<\/sup> – 2gh
    \n0 = (49)2<\/sup> – 2 \u00d7 9.8 \u00d7 h
    \n\u2234 h = \\(\\frac{(49)^{2}}{2 \\times 9.8}\\) = 1225 m
    \nMaximum height attained = 122.5 m.<\/p>\n

    (ii) Time is taken by the ball to reach the maximum height.
    \nFrom equation of upward motion
    \nv = u – gt
    \n0 = 49 – 9.8 \u00d7 t
    \n\u2234 t = \\(\\frac{49}{9.8}\\) = 5s
    \nIn the motion against gravity, the time of descent is same as the time of ascent.
    \nSo time taken by the ball to fall from maximum height = 5s.
    \n\u2234 Total time taken by the ball to return to surface of the earth = 5 + 5 = 10s.<\/p>\n

    Question 14.
    \nA stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
    \nAnswer:
    \nh = 19.6 m.
    \nInitial velocity u = 0 (\u2235 It starts from rest)
    \nFrom equation of motion.
    \n\"KSSEB
    \nFinal velocity of stone before touching the ground = 19.6 ms-1<\/sup>.<\/p>\n

    Question 15.
    \nA stone is thrown vertically upward with an initial velocity of 40m\/s. Taking g = 10m\/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
    \nAnswer:
    \nInitial velocity u = 40 ms-1<\/sup>
    \nAt maximum height, final velocity becomes zero i.e., v = 0
    \nFrom equation of motion
    \nv2<\/sup> = u2<\/sup> – 2gh
    \n(0)2<\/sup> = (40)2<\/sup> – 2 \u00d7 10 \u00d7 h
    \n0 = 1600 – 20h
    \nTotal time taken by the ball to return to surface of the earth = 5 + 5 = 10s.
    \n\u2234 h = \\(\\frac{1600}{20}\\) = 80 m
    \nMaximum height reached by the stone = 80 m
    \nAfter reaching the maximum height, the stone will fall towards the earth and will reach the earth surface covering the same distance.
    \nSo distance covered by the stone
    \n= 80 + 80 = 160 m
    \nDisplacement of the stone = 0
    \n(because the stone starts from the earth surface and finally reaches the earth surface again i.e the initial and final position of the stone are same<\/p>\n

    Question 16.
    \nCalculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 \u00d7 1024<\/sup>kg and of the sun = 2 \u00d7 1030<\/sup> kg. The average distance between the two is 1.5 \u00d7 1011<\/sup>m)
    \nAnswer:
    \nThe force of attraction between the earth and the sun F = \\(\\frac{\\mathrm{GM}_{1} \\mathrm{M}_{2}}{\\mathrm{r}^{2}}\\)
    \nwhere G = 6.67 \u00d7 10-11<\/sup>Nm2<\/sup>kg-2<\/sup>
    \nGiven mass of the Sun M1<\/sub>= 2 \u00d7 1030<\/sup> kg
    \nMass of the earth M1<\/sub> = 6 \u00d7 1024<\/sup> kg.
    \nAverage distance between the earth and the sun = 1.5 \u00d7 1011<\/sup>M
    \n\"KSSEB<\/p>\n

    Question 17.
    \nA stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from
    \nthe ground with a velocity of 25m\/s. Calculate when and where the two stones will meet.
    \nAnswer:
    \nLet after time t both stone meet and s be the distance travelled by the stone dropped from the top of tower at which the stones will meet. Distance travelled by the stone dropped = S
    \n\u2234 Distance travelled by the stone projected upwards (100 -s) m
    \nFor the stone dropped fromt the tower.
    \n\"KSSEB
    \n(u = 0 because stone is dropped ie it starts from rest)
    \ns = 5t2<\/sup> …………. (i)
    \nFor the stone projected upward
    \ns = ut – \\(\\frac{1}{2}\\) gt2<\/sup>
    \n(Due to upward motion negative sign is taken)
    \n\"KSSEB
    \nThe stones will meet after; 4s
    \ns = 5t2<\/sup> = 5 \u00d7 (4)2<\/sup> = 80 m
    \nThe stones will be at a distance of 80 m from the top of tower or 20 mt (100 m – 80 m) from the base of the tower 20m (100m – 80m).<\/p>\n

    Question 18.
    \nA ball thrown up vertically returns to the thrower after 6s. Find
    \na) the velocity with which it was thrown up
    \nb) the maximum height it reaches, and
    \nc) its position after 4s.
    \nAnswer:
    \nTotal time taken = 6 s
    \nTime taken to reach the maximum height = \\(\\frac{6}{2}\\) = 3s
    \n(\u2235 Time of ascent = time)
    \na) From equation of motion v = u – gt
    \n(Negative sign is taken due to upward motion)
    \n0 = 4- 9.8 \u00d7 3 (\u2235 at maximum height v=0)
    \n0 = u – 9.8 \u00d7 3
    \n0 = u – 9.8 \u00d7 3
    \nu = 29.4 ms-1<\/sup>.
    \n(\u2235 at maximum height v = 0)
    \nu = 29.4<\/p>\n

    b) From equation of motion
    \nv2<\/sup> = u2<\/sup> – 2gh
    \n0 = (29.4)2<\/sup> – 2 \u00d7 9.8 \u00d7 h
    \n\u2234 h = \\(\\frac{(29.4)^{2}}{2 \\times 9.8}\\) = 44.1 m
    \nMaximum height attained by the ball is 44.1 m<\/p>\n

    c) In initial 35. the ball will rise then in next 35 it falls towards the earth
    \n\u2234 The position after 4s(3 + 1)
    \n= Distance covered in Is in the download motion.
    \nFrom the equation of motion
    \n\"KSSEB
    \n\u2234 ie the ball will be at 4.9m from the top of the tower.<\/p>\n

    Question 19.
    \nIn what direction does the buoyant force on an object immersed in a liquid act?
    \nAnswer:
    \nThe buoyant force on an object immersed in a liquid always act in upward direction.<\/p>\n

    Question 20.
    \nWhy does a block of plastic released under water come up to the surface of water.
    \nAnswer:
    \nThe buoyant force on an object immersed in a liquid always act in upward direction.<\/p>\n

    Question 21.
    \nThe volume of 50g of a substance is 20 cm3<\/sup>. If the density of water is 1gcm-3<\/sup> will the substance float or sink?
    \nAnswer:
    \nMass of a substance = 50g
    \nVolume of substance = 20 cm3<\/sup>
    \n\"KSSEB
    \n= \\(\\frac{50}{20}\\) = 2.5 g cm-3<\/sup>
    \nie the density of the substance is greater than the density of water. So it will sink in water.<\/p>\n

    Question 22.
    \nThe volume of 500g sealed packet is 350 cm-3<\/sup>, will the packet float or sink in water if the density of water is 1gmcm-3<\/sup>? What will be the mass of the water displaced by this packet?
    \nAnswer:
    \nMass of packet = 500g
    \nVolume of packet = 350 cm-3<\/sup>
    \n\"KSSEB
    \n\u2234 ie the density of packet is greater than density of water, so it will sink in water.
    \nMass of water displaced by the packet = Volume of packet \u00d7 Density of water = 350 \u00d7 1 = 350 gm.<\/p>\n

    KSEEB Solutions for Class 9 Science Chapter 10 Additional Questions and Answer<\/h3>\n

    Fill in the blanks with suitable words.
    \nQuestion 1.
    \nThe force that causes acceleration and keeps the body moving along the circular path acting towards the centre is called the _________
    \nAnswer:
    \nCentripetal force.<\/p>\n

    Question 2.
    \nForce is the product of and _________
    \nAnswer:
    \nMass, acceleration.<\/p>\n

    Question 3.
    \nThe ________ of an object is the force with which it is attracted towards the earth.
    \nAnswer:
    \nweight.<\/p>\n

    Question 4.
    \nThe force acting on an object perpendicular to the surface is called the ___________
    \nAnswer:
    \nthrust.<\/p>\n

    Question 5.
    \nThe upward force exerted by the water on objects is known as ________ force.
    \nAnswer:
    \nbuoyant.<\/p>\n

    Question 6.
    \nThe objects of density greater than that of a liquid _________ in the liquid.
    \nAnswer:
    \nsink.<\/p>\n

    Question 7.
    \nThe value of \u2018g\u2019 in the poles is _________ than the equator.
    \nAnswer:
    \ngreater.<\/p>\n

    Answer the following questions:<\/p>\n

    Question 1.
    \nWhy is Newton\u2019s law of Gravitation called universal law?
    \nAnswer :
    \nIt is applicable to all bodies either they ‘ are small or big. Whether they are celestial or terrestrial.<\/p>\n

    Question 2.
    \nMention two factors on which the buoyant force acting on an object depends?
    \nAnswer :<\/p>\n