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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 II. If 88 droplets are formed from the big drop, then the moment of inertia of the small droplet is.
(A) I32\dfrac{I}{{32}}
(B) I16\dfrac{I}{{16}}
(C) I8\dfrac{I}{8}
(D) I4\dfrac{I}{4}

Explanation

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 88 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=25MR2I = \dfrac{2}{5}M{R^2}
where MM stands for the mass of the sphere and RR 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=25MR2I = \dfrac{2}{5}M{R^2}.
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 rr be the radius of the small drops,
Then the volume of nn small drops will be, Vn=n43πr3{V_n} = n\dfrac{4}{3}\pi {r^3}
Let RR be the radius of the big drop,
Then the volume of the big drop will be V=43πR3V = \dfrac{4}{3}\pi {R^3}
The volume of the big drop will be equal to the volume of nn small drops.
n43πr3=43πR3n\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}
Canceling the common terms we get
nr3=R3n{r^3} = {R^3}
Taking the cube root of the above equation,
n13r=R{n^{\dfrac{1}{3}}}r = R
We know that n=8n = 8. Putting this value in the above equationMM
813r=R{8^{\dfrac{1}{3}}}r = R
From this, the radius of the small drop can be written as,
r=R2r = \dfrac{R}{2}
The mass of the big drop is MM.
Then the mass of one small drop will be M8\dfrac{M}{8}.
Therefore, the moment of inertia of the small droplet will be
Is=25[M8][R2]2{I_s} = \dfrac{2}{5}\left[ {\dfrac{M}{8}} \right]{\left[ {\dfrac{R}{2}} \right]^2}
Solving the above equation,
Is=132[25MR2]{I_s} = \dfrac{1}{{32}}\left[ {\dfrac{2}{5}M{R^2}} \right]
That is
Is=I32{I_s} = \dfrac{I}{{32}}

The answer is: Option (A): I32\dfrac{I}{{32}}

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.