Chapter #6 :Gravitation : Short Q/A / C.R.Q's



Q. How would the weight of the body vary as it is taken from earth to moon? What will be the effect on mass?
Ans. As we know that the weight depends on the mass of body and the acceleration due to gravity there. As the mass being the constant quantity it will not vary on the surface of moon. But the value of g on moon is one sixth the value on the surface of earth so the weight of the body will decreases 1/6 times.

Q. Gravitational force is directly proportional to the mass of the body then why all bodies fall with same velocity?
Ans. A/c to the law of gravitation
                        F=G m1 m2/r2
Or        mg= G Me m/ Re 2
Then    g=G Me/ Re 2
            So the acceleration of any body in free fall motion is independent of mass.

Q. Suppose you are falling from a hill with a box in your hand so the weight of box during falling  experienced is grater or lesser?
Ans. As we know that in moving downward the apparent weight is less than the actual weight .i.e.
                                    Apparent weight = (mg) – ma
                                    In free fall motion a=g
            Then                Apparent weight =mg– mg
                                    Apparent weight =0
So the weight of the box experienced during falling is equal to zero.

Q. The Moon revolves around Earth due to gravitational force then why the apple on the earth doesn’t?
Ans. For the circular motion the force is required which attracts the body toward center of the circle and a force that pulls the body out of the circle (i.e. centripetal and centrifugal force).As the apple does not has any velocity for the centrifugal force so it is in rest whereas the earth has its velocity or the centrifugal force that’s why it moves around sun.

Q. What is the effect on the weight at the center of earth?
Ans.As we know that the weight of any body depends upon the value of g and g decreases with the depth.
                                    g’=g (1-d/ Re 2)
            at the center d=Re
then  g’=g (1- Re 2/ Re 2)
                        or         g’=g (1-1)
                        or         g’=0
Now if we are at the center of the earth the value of g becomes zero and also the weight will become zero.

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