Question

A spherical shell and a solid cylinder of same radius
rolls down an inclined plane. The ratio of their
accelerations will be:-
(1715 : 14
(2) 9 : 10
(3) 2
:3
(4) 3:5​

Answer 1

Dear Student,

◆ Answer -

(1) 15 : 14

● Explaination -

Acceleration of the object rolling down inclined plane is -

a = mgR²sinθ / (I+mR²)

For spherical shell, I = 2mR²/5

as = mgR²sinθ / (2mR²/5 + mR²)

as = mgR²sinθ / (7mR²/5)

as = 5gsinθ/7

For solid cylinder, I = mR²/2

ac = mgR²sinθ / (mR²/2 + mR²)

ac = mgR²sinθ / (3mR²/2)

ac = 2gsinθ/3

Ratio acceleration of spherical shell to solid cylinder is called -

as/ac = (5gsinθ/7) / (2gsinθ/3)

as/ac = 15/14

Therefore, ratio of accelerations of spherical shell to solid cylinder is 15 : 14.

Thanks dear...

Answer 2

Answer:

1) 15:14

Explanation:

Acceleration of the object rolling down inclined plane is -

a = mgR²sinθ / (I+mR²)

In this question we have to deal with the moment of Inertia of both the spherical shell and a solid cylinder

Moment of Inertia is a quantity expressing a body's tendency to resist angular acceleration.

For spherical shell, the moment of inertia is given by I = 2mR²/5

as = mgR²sinθ / (2mR²/5 + mR²)

as = mgR²sinθ / (7mR²/5)

as = 5gsinθ/7

For solid cylinder, the momrnt of Inertia is given by I = mR²/2

ac = mgR²sinθ / (mR²/2 + mR²)

ac = mgR²sinθ / (3mR²/2)

ac = 2gsinθ/3

Now , Ratio acceleration of spherical shell to solid cylinder ;

as/ac = (5gsinθ/7) / (2gsinθ/3)

as/ac = 15/14

Therefore, ratio of accelerations of spherical shell to solid cylinder is 15 : 14.