Question

Moment of inertia of a big drop is I. If 8 droplets are formed from big drop, then moment of inertia of
small droplets is
​

Answer 1

Answer:

Moment of inertia is directly proportional to the mass so if 8 drops make a big drop and one drop has the moment of inertia as I so moment of inertia for small drops will be I/8

Answer 2

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

Explanation:

Given that,

Moment of inertia of a big drop = I

Number of small droplets = 8

We know that the moment of inertia of spherical drop

Using formula of moment of inertia

$I=\dfrac{2}{5}MR^2$....(I)

One big drops divided into 8 small droplets.

8m = M

We need to calculate the radius of small drop

Using formula of mass

$8\rho V=\rho V'$

$8\rho\times\dfrac{4}{3}\pi\times r^3=\rho\times\dfrac{4}{3}\pi\timesR^3$

$(2r)^3=R^3$

$2r=R$

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

We need to calculate the moment of inertia of small drops

Using formula of moment of inertia

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

Put the value of m and r into the formula

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

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

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

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