AP 9th Class Physical Science Chapter 7 Solution – Gravitation
Andhra Pradesh SCERT 9th Class Physical Science Chapter 7 Gravitation Solution for AP 9th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 9 Physical Science Textbook Solution or Study Material for AP 9th exam. Here you search will end! Here in this page we have provided for all question answer for AP SCERT Chapter 7 Gravitation.
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AP 9th Class Physics Chapter 7 Gravitation Solution
1) What path will the moon take when the gravitational interaction between the moon and earth disappears? (AS2)
Ans: – There is force which acting between the moon and the earth centre, this force is known as centripetal force. So when the gravitational interaction between the moon and earth will disappear which is not possible practically, the moon will move tangentially to its present position.
2) A car moves with constant speed of 10 m/s in a circular path of radius 10m. The mass of the car is 1000 kg. Who or what is providing the required centripetal force for the car? How much is it? (AS1)
Ans: – From the equation of centripetal force we know that F =mv^2/r, where m is mass of object, v is speed of that object and r is the radius of the circular path.
It given that m=1000kg, v= 10m/s, r=10m. So F=1000×10^2/10 = 10^4N.
So the centripetal force is 10^4N.
3) A small metal washer is placed on the top of a hemisphere of radius R. What minimum horizontal velocity should be imparted to the washer to detach it from the hemisphere at the initial point of motion? (See figure) (AS1 , AS4)
Ans: – Let the mass of the metal washer is m, velocity is v and radius of the hemisphere is R.
From the image if we calculate the centripetal force on the washer will be F=mv^2 / r……. (1).
Force by hemisphere on the washer is F= mg (g is gravitational constant)…… (2).
According to Newton third law these two force must be same,
Then, mv^2 /r = mg;v^2 = gr; v=√gr.
4) Explain why a long pole is more beneficial to the tight rope walker if the pole has slight bending. (AS1 , AS7)
Ans: –If a man carry a long pole to walk on the rope it helps to increase the rotational inertia by this the rope Walker made balance while walking.
5.) Why is it easier to carry the same amount of water in two buckets, one in each hand rather than in a single bucket? (AS7)
Ans: –When a man hold the two buckets in his two hand it is easier to walk rather than holding two buckets in a single hand. As centre of gravity remain same in case of holding two bucket in two hands it is easier to walk.
6) What is the speed of an apple dropped from a tree after 1.5 second? What distance will it cover during this time? Take g=10m/s2(AS1)
Ans: – According to Newton equation of motion, v = u + gt.
V= final velocity=? ; u=initial velocity=0, g=10. /s^2, t=1.5s.
So v= 15m/s.
Then speed of the apple dropped from a tree after 1.5 s is 15ms^-1.
As we know s = ut + 1/2at^2. Where s is distance covered, and a=g.
Then s = 0 + ½ × 10× 1.5^2 = 11.25m.
The distance cover during this time is 11.25m.
7) A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entire time of motion? What is the velocity at 5 seconds after the projection? Take g=10m/s2 (AS1)
Ans: – Let the maximum height be H
So according the formula v^2 = u^2 – 2gH (as it vertically upward)
Given that v=0 as maximum height reached, u= 40m/s, g=10 and H=? So 0 = 40^2 – 2×10×H; or H= 1600/20= 80 m.
The maximum height is 80m.
As T=2u/g; T= 2×40/ 10 = 8.
So entire time of motion is 8s.
The velocity after 5s is. V.
Here v= u + gt, here it starts from rest so u is zero; t=1s.
So v =10 m/s.
8) A boy is throwing balls into the air one by one in such a way that when the first ballthrown reaches maximum height he starts to throw the second ball. He repeats thisactivity. To what height do the balls rise if he throws twice in a second? (AS1 .AS7)
Ans: – As the boy throws two ball in one second so 1 ball throws at 1/2s. And we know time is equal to the ratio of u/g where u is the initial velocity and g is gravitational constant.
Then, 1/2=u/g; u = 5 (g=10).
The maximum height is equal to v^2/2g or 5^2/2×10 = 1.25m.
9) A man is standing against a wall such that his right shoulder and right leg are in contact with the surface of the wall along his height. Can he raise his left leg at this position without moving his body away from the wall? Why? Explain.(AS7)
Ans: – When a man is standing against a wall and his right shoulder and right leg are in contact with the surface of the wall along his height he can’t rise his left leg because the centre pf gravity is lies on the belly so he will move his body part to moves the centre of gravity.
10) A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hitting the ground, find the height from which it is dropped. Take g = 10m/ s2 (AS1)
Ans: –let the height be h where from the ball drop. H =s1 + s2.
S1 is the distance covered in 0.2s at last time so v=0.
According to Newton laws of equation, S= ut + ½ at^2.
Here S=s1=6m; t=0.2s; a=g=10, then s1=0.2u + ½×10×0.2^2.
Or 0.2u= 6 – 0.2; u = 29m/s.
This velocity is the final velocity of the first region.
V= u + gt, Here v=29m/s; t=?,g=10, u=0, so t=29/20 =2.9s.
So s2 is equal to s2=ut + 1/2gt^2, or s2= 0×2.9 + ½×10×2.9^2.
Or s2 = 42.50.
So the total height is (42.50 + 6) 48.50m, from which ball dropped.
11) A ball is dropped from a balloon going up at a speed of 5 m/s. If the balloon was at a height 60 m. at the time of dropping the ball, how long will the ball take to reach the ground?
Ans:-The balloon drop from a height h=60m , and its speed is 5m/s.
According to Newton laws of equation,S=ut + 1/2gt^2.
Here s= h=60m, u=-5m/s. T=?
So, 60 = -5×t + ½ × 10×t^2.
Or, 5(-t + t^2) = 60
Or, t = 4, -3.
So it take 4s to reach the ground.
12.) A ball is projected vertically up with a speed of 50 m/s. Find the maximum height ,the time to reach the maximum height, and the speed at the maximum height(g=10 m/ s2 ) (AS1)
Ans: – Speed of the ball is 50m/s which is initial velocity.
From the equation of projectile motion we know that, maximum height =u^2/2g = 50^2/2×10 (as g =10).
H or maximum height is 125m.
The time to reach the maximum height is u/g or 50/10 or 5s.
The ball when reached its maximum height speed will be zero.
13.) Two cars having masses m1 and m2move in circles of radii r1and r2 respectively. Ifthey complete the circle in equal time. What is the ratio of their speeds andcentripetal accelerations? (AS1)
Ans: –Let the time needed by car 1 is t1 and its speed and mass is v1 and m1. Then time t1= 2πr1/v1.
The time needed by car 2 is t2 and its speed and mass is v2 and m2. Then time t2=2πr2/v2.
According to the question t1= t2,2πr1/v1 = 2πr2/v2, or, v1/v2 =r1/r2.
So the ratio of the speed is r1/r2.
The centripetal acceleration of car 1 = v1^2 / r1.
And the centripetal acceleration of car 2 = v2^2 /r2.
So the ratio of their centripetal acceleration is (v1^2/r1)/(v2^2/r2)=(v1^2/v2^2)×(r2/r1)=(r1/r2)^2 ×(r2/r1)=r2/r1.
14) Two spherical balls of mass 10 kg each are placed with their centres 10 cm apart.Find the gravitational force of attraction between them. (AS1)
Ans: –From the gravitational force of attraction we came know that F=Gm1m2/r^2.
Here m1=m2=10kg,r=10cm=0.1m
Then F= 6.67×10^-11 × 10×10/o.1^2 = 6.67×10^-7N.
15) Find the free-fall acceleration of an object on the surface of the moon, if the radiusof the moon and its mass are 1740 km and 7.4 x 1022 kg respectively. Compare thisvalue with free fall acceleration of a body on the surface of the earth.
Ans: – The acceleration due to gravity of the free falling body is,
.g=GM/R^2; here M=7.4×10^22,R=1740km =1740×10^3m.
Then g=6.67×10^-11 × 7.4×10^22 / (1740×10^3)^2=1.63 m/s^2.
The gravity of earth is 9.8 m/s^2 which is (9.8/1.6=6) 6times more than the free falling body.
16) Can you think of two particles which do not exertgravitational force on each other?(AS2)
Ans: – There is no such particles which do not exert gravitational force on each other. According to Newton theory each particle exert gravitational force to other.
17.) An apple falls from a tree. An insect in the apple finds that the earth is fallingtowards it with an acceleration g. Who exerts the force needed to accelerate theearth with this acceleration? (AS7)
Ans:- When apple falls from a tree towards the earth it exert some acceleration to earth surface and the earth also react with forces according to third laws . As the mass of the earth very large compared to apple, apples force exert on earth is negligible.
18) Scooter weighing 150kg together with its rider moving at 36 km/hr is to take aturn of radius 30 m. What force on the scooter towards the centre is needed tomake the turn possible? Who or what provides this?
Ans: – The weight (m) of these two is 150kg, and the radius of the turning (r) is 30m. The velocity (v) of the scooter is 36km/h or 36×5/18=10m/s.
According to centripetal force, F =mv^2 /r = 150×10^2 /30 =500N.
These force are provided by the normal reaction of scooter and the road.
19) The bob of a simple pendulum of length 1 m has mass 100g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this moment.
Ans: –As the pendulum is moving from a point in asemi-circular path both centripetal and gravitational force will act to it.
So tension in the string = mv^2/r + mg
Here m=100gm or 0.1kg,v=1.4m/s,r=1m.
Tension = 0.1×1.4^2 /1 + 0.1×10 = 1.176N.
20.) How can you find the centre of gravity of an India map made of steel? Explain.(AS3)
Ans:- The steel sheath are make like the map and put three holes on it and from one holes place the map vertically with a thread . Then make a line along with the thread and repeat the process with the other three holes, then the point of intersection of these 3 line remark as a centre of gravity of the map.
21.) Explain some situations where the centre of gravity of man lies outside the body. (AS1)
Ans: –When we practice yoga then our centre of gravity lies in different position of the different posture of our body. And when we go gym in that time when we follow different exercise the centre of gravity changes.
22.) Where does the centre of gravity of the atmosphere of the earth lie? (AS2)
Ans: –The centre of gravity of the atmosphere of the earth lie at the centre of the earth.
23.) Where does the centre of gravity lie, when a boy is doing sit-ups? Does weightvector pass through the base or move away from the base? Explain. (AS7)
Ans:- The centre of gravity changes when a boy is doing sit ups as when the body move upward the centre of gravity lies in up and when body move downward the centre of gravity lies in down.
Although centre of gravity changes in different position but the weight vector of the body remain probably same as previousbut slightly movement seen.