Question

1)
Moment of inertia of a big drop is I. If 8 droplets are formed from big drop, then moment of inertia
small droplets is
B)
are The M of A about an a​

Answer 1

The moment of inertia  small droplets is $\frac{I}{32}$

Explanation:

As we know that the moment of inertia of one spherical drop is given as

$I = \frac{2}{5}MR^2$

now we know that one big drop is divided into 8 small drops

so we will have

$8m = M$

now we can also say

$8(\rho)\frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$

$r = \frac{R}{2}$

now the moment of inertia of each small drop is given as

$I_{small} = \frac{2}{5}(\frac{M}{8})(\frac{R}{2})^2$

$I_{small} = (\frac{2}{5}MR^2)(\frac{1}{32})$

$I_{small} = \frac{I}{32}$

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Answer 2

The moment of inertia of small droplets is $\dfrac{I}{32}$

Explanation:

Given that,

Moment of inertia of big drop is I.

We know that the moment of inertia of spherical drop

$I=\dfrac{2}{5}MR^2$

Now, A big drop divided into 8 small drops.

We need to calculate the radius of drop

Using formula of mass

$8m = M$

$8\times\rho\times V=\rho\times V'$

Where, $\rho$ = density of drop

V = volume of drop

$8\rho\times\dfrac{4}{3}\pi\times(r)^3=\rho\times\dfrac{4}{3}\pi\times(R)^3$

$r=\dfrac{R}{2}$

We need to calculate the moment of inertia of small droplets

Using formula of moment of inertia

$I'=\dfrac{2}{5}mr^2$

Put the value of m and r in to the formula of moment of inertia

$I'=\dfrac{2}{5}\times(\dfrac{M}{8})\times(\dfrac{R}{2})^2$

$I'=\dfrac{I}{32}$

Hence, The moment of inertia of small droplets is $\dfrac{I}{32}$

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Topic : moment of inertia

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