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ScienceClass 9CBSE

Law of gravitation gives the gravitational force between (a) the earth and a point mass only (b) the earth and Sun only (c) any two bodies having some mass two charged bodies only

Soln: Answer is (c) any two bodies having some mass

ScienceClass 9CBSE

In the relation F = G M m/d2, the quantity G (a) depends on the value of g at the place of observation (b) is used only when the earth is one of the two masses (c) is greatest at the surface of the earth is universal constant of nature

Soln: Answer is (d) is universal constant of nature Explanation: G is called as Newton’s constant. It is the force of gravity on a body. Value of G is 6.66x 10-11 Nm2kg-2

ScienceClass 9CBSE

An object is put one by one in three liquids having different densities. The object floats with 12 3 , and 9 11 7 parts of their volumes outside the liquid surface in liquids of densities d1, d2 and d3 respectively. Which of the following statement is correct? (a) d1> d2> d3 (b) d1> d2 d3 d1< d2< d3

Soln: Answer is (d) d1< d2< d3

ScienceClass 9CBSE

A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone (a) will continue to move in the circular path (b) will move along a straight line towards the centre of the circular path (c) will move along a straight line tangential to the circular path will move along a straight line perpendicular to the circular path away from the boy

Soln: Answer is (c) will move along a straight line tangential to the circular path Explanation: At any instance of time object in circular motion tend to be in rectilinear motion. Object keeps on moving due to centripetal force and it moves along a straight line tangential to the circular path when strings breaks.

ScienceClass 9CBSE

The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become (a) F/4 (b) F/2 (c) F (d) 2 F

Soln: Answer is (a) F/4

ScienceClass 9CBSE

The value of acceleration due to gravity (a) is same on equator and poles (b) is least on poles (c) is least on equator increases from pole to equator

Soln: Answer is (c) is least on equator Explanation: The value of acceleration due to gravity is least on equator because distance between surface of the earth and its centre is more on equator than in poles.

ScienceClass 9CBSE

Two objects of different masses falling freely near the surface of moon would (a) have same velocities at any instant (b) have different accelerations (c) experience forces of same magnitude undergo a change in their inertia

Soln: Answer is (a) have same velocities at any instant Explanation: Acceleration of an object depends on acceleration due to gravity irrespective of its mass. Hence object under free fall have the same velocities.

ScienceClass 9CBSE

Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 m s–2) (i) a butterfly of mass 1.0 g that flies upward at a rate of 0.5 m s–1. (ii) a 250 g squirrel climbing up on a tree at a rate of 0.5 m s–1.

Soln: Butterfly force= mg=1/100 power=f×s÷t. {s÷t=velocity} power=1÷100×.5 power=1÷200W squirrel force=mg=25 power=25×.5 power=12.5W

ScienceClass 9CBSE

Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s–1?

Soln: Power of an object which does work at 1 Joules/sec is called watt 1 watt= 𝑱𝒐𝒖𝒍𝒆 𝒔𝒆𝒄𝒐𝒏𝒅 1 kilowatt= 1000watts=1000J/sec mass of car = 150kg power for each kg=500x 150=7500w speed=20m/s power=Force x v &there4;force=power/v 7500 20 =3750N

ScienceClass 9CBSE

How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 m s–1 vertically? (g = 10 m s–2)

Soln: Power = work/time work=force x displacement force = mass x acceleration acceleration=velocity/time Therefore Power= velocity×mass× displacement time × time Here Power,P=100W velocity,v=1m/s since time,t=1s displacement, s=1m acceleration,a=10m/s From equation P=m×a×s t 100=m×10 m=10

ScienceClass 9CBSE

What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 m s–2)

Soln: Power is the rate of transfer of energy or rate of doing work. Watt is the unit of power and kilowatt is 1000 watts. h = 20 m, and mass = 2000 × 103 kg = 2 × 106 kg Power = m g h / t = 2 x 106 x 10 x 20 60 w = 4x107w = 6 2 x107w 3

ScienceClass 9CBSE

Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it. (a) How much work is done by the men in lifting the box? (b) How much work do they do in just holding it? Why do they get tired while holding it? (g = 10 m s–2)

Soln: (a) F= 250kg×g= 2500 N S= 1m W=F.s= 250Nm= 2500 J (b) While holding box there is no displacement hence workdone is zero c) While holding the box they apply muscular force which is equal and opposite to the gravitational force hence they feel tired while holding box.

ScienceClass 9CBSE

A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s– 1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?

(a) Initial velocity of the trolley,u = 4 m/s Final velocity of the trolleyv = 0 Mass of the trolleym = 5 kg Distance covered by the trolley before coming to rest,s = 16 m From Equation 2 as =v2-u2, a = v2-u22 S =0-(4)22x16 =0.5 m/s2 Force (frictional) acting on the trolley = ma = 40 (- 0.5) = - 20 N Work done on the trolley = Fs = (20 N) (16 m) = 320 J (b) Since the girl does not move w.r.t. the trolley (as she is sitting on it),work done by the girl = 0.

ScienceClass 9CBSE

An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.

Soln: m(A)= m(B) =1000 kg. v = 36 km/h =10 m/s Frictional force = 100 N car A moves with a uniform speed, which means engine of car applies a force equal to the frictional force Power = Force × distance / time = F . V = 100 N × 10 m/s = 1000 W After collision mA uA + mB uB = mA vA + mB vB. 1000 ×10 +1000 × 0 = 1000 × 0 + 1000 × vB vB = 10 m s–1

ScienceClass 9CBSE

A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

Soln: Kinetic energy equation: Ek= 1mv2 2 Ek= 1mv2 2 Momentum equation: p=mv p=mv Kinetic energy equation in terms of momentum: Ek= 𝑝2 2𝑚 Kinetic energy of two objects will be same if their mass and momentum are same. Among light and heavy bodies Heavy body has highest momentum. When the momentum is equal light body has more kinetic energy.

ScienceClass 9CBSE

If an electric iron of 1200 W is used for 30 minutes everyday, find electric energy consumed in the month of April.

Soln: Power of electric iron=1200W Usage per day =30min =30/60hrs =0.5hrs Number of days in the month of April=30days Electrical energy consumed, E=Pxt 1200x0.5x30 =18000WH =18KWH =18units Therefore, The Total Electricity consumed in april month is 18 units

ScienceClass 9CBSE

A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2).

Soln: mgh=m×10×10= 100m joules when 40% energy is reduced remaining energy will be 60mjoules Hence 60-m×10×h H= -6m

ScienceClass 9CBSE

Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example.

Force always acts in perpendicular to the displacement direction. If the object is moving in circular path then no work will be done despite of action of force.

ScienceClass 9CBSE

The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Soln: v2-u2 = 2as s= v2-u2 2a F= ma W=ma(v2-u2) 2a = 1mv2-1mu2 2 2 =(K.E)f-(K.E)t

ScienceClass 9CBSE

The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A?

Soln: Person on Planet A can jump a height of 0.8 m because weight of person on plat A is half the weight that of earth. Hence acceleration due to gravity on planet A will also be half of that on earth.

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