A solid sphere of 80 kg and radius 15 m moving in a space becomes a circular disc of radius 20 m in 1 h . The rate of change of moment of Inertia in this process is

Correct answer: D. (22)/(9) kg^ -m^2 s^ -1 Given, mass of solid sphere =80 kg radius of solid sphere, R_s=15 m radius of circular disc, R_c=20 m and time =1 hour =60 minute =60 × 60 Moment of inertia of solid sphere, I_s=(2)/(5) M R^2 aligned & =(2)/(5) ×80 ×(15)^2 & =7200 kg-m^2 aligned Similarly, moment of inertia of the disc, aligned I_c & =(1)/(2) M R_c^2 & =(1)/(2) ×80 ×(20)^2 & =16000 kg-m^2 aligned aligned Rate of change of moment of Inertia & =(I_c-I_s)/(t) & =(16000-7200)/(60 × 60) & =(22)/(9) kg-m^2 s^ -1 aligned

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Physics
Chapter: Rotational Motion

BITSAT 2022 · Physics · Rotational Motion

A solid sphere of 80 kg and radius 15 m moving in a space becomes a...

A solid sphere of

80 \mathrm{ kg}

and radius

15 \mathrm{ m}

moving in a space becomes a circular disc of radius

20 \mathrm{ m}

1 \mathrm{ h}

. The rate of change of moment of Inertia in this process is

Options

A $\frac{30}{9} \mathrm{ kg}^2 \mathrm{ m}^2 \mathrm{ s}^{-1}$
B $\frac{25}{9} \mathrm{ kg}^{-\mathrm{m}^2 \mathrm{ s}^{-1}}$
C $\frac{10}{9} \mathrm{ kg}-\mathrm{m}^2 \mathrm{ s}^{-1}$
D $\frac{22}{9} \mathrm{ kg}^{-\mathrm{m}^2 \mathrm{ s}^{-1}}$

Correct answer

D. $\frac{22}{9} \mathrm{ kg}^{-\mathrm{m}^2 \mathrm{ s}^{-1}}$

Step-by-step solution

Given, mass of solid sphere

=80 \mathrm{ kg}

radius of solid sphere,

R_s=15 \mathrm{ m}

radius of circular disc,

R_c=20 \mathrm{ m}

and time

=1

hour

=60

minute

=60 \times 60 \mathrm\sec\therefore

Moment of inertia of solid sphere,

I_s=\frac{2}{5} M R^2

\begin{aligned}& =\frac{2}{5} \times80 \times(15)^2 \\ & =7200 \mathrm{ kg}-\mathrm{m}^2 \end{aligned}

Similarly, \text{ moment

inertia of the disc, }

\begin{aligned} I_c & =\frac{1}{2} M R_c^2 \\ & =\frac{1}{2} \times80 \times(20)^2 \\ & =16000 \mathrm{ kg}-\mathrm{m}^2 \end{aligned}

\begin{aligned}\text{ Rate of change of moment of Inertia }& =\frac{I_c-I_s}{t}\\ & =\frac{16000-7200}{60 \times 60}\\ & =\frac{22}{9} \mathrm{ kg}-\mathrm{m}^2 \mathrm{ s}^{-1} \end{aligned}

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