Question
Question: The moment of inertia of a big drop is \(I\). If \(8\) droplets are formed from the big drop, then t...
The moment of inertia of a big drop is I. If 8 droplets are formed from the big drop, then the moment of inertia of the small droplet is.
(A) 32I
(B) 16I
(C) 8I
(D) 4I
Solution
The moment of inertia of a particle about an axis of rotation is the product of the mass of the particle and the square of the distance of the particle from the axis. Here the moment of inertia of a big drop is given. If we combine 8 such drops to form a big drop then the moment of inertia of one drop should be found.
Formula used:
The moment of inertia of a sphere,
I=52MR2
where M stands for the mass of the sphere and R stands for the radius of curvature of the sphere.
Complete step by step solution:
Let us assume the drop to be spherical.
The moment of inertia of a sphere is given by, I=52MR2.
A big drop is formed by combining small drops. The volume will remain the same after the formation of the big drop from the small drops.
Let r be the radius of the small drops,
Then the volume of n small drops will be, Vn=n34πr3
Let R be the radius of the big drop,
Then the volume of the big drop will be V=34πR3
The volume of the big drop will be equal to the volume of n small drops.
n34πr3=34πR3
Canceling the common terms we get
nr3=R3
Taking the cube root of the above equation,
n31r=R
We know that n=8. Putting this value in the above equationM
831r=R
From this, the radius of the small drop can be written as,
r=2R
The mass of the big drop is M.
Then the mass of one small drop will be 8M.
Therefore, the moment of inertia of the small droplet will be
Is=52[8M][2R]2
Solving the above equation,
Is=321[52MR2]
That is
Is=32I
The answer is: Option (A): 32I
Note:
Rotational inertia is the inability of a body at rest to rotate by itself, and a body in uniform rotational motion to stop by itself is called the rotational inertia of a body. The moment of inertia is a measure of rotational inertia. Mass is a measure of inertia.