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 is : 229kg−m2s−1\; 22/9 {kg} - m² s^{-1}922kg−m2s−1

Correct Answer is : 229kg−m2s−1\; 22/9 {kg} - m² s^{-1}922kg−m2s−1

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BITSAT 2022 Paper

Section: Physics

Question : 21 of 130

Marks: +1, -0

A solid sphere of

80\ \text{kg}

15\ \text{m}

20\ \text{m}

1\ \text{h}

Solution: 👈: Video Solution

Given, mass of solid sphere

=80\ \text{kg}

radius of solid sphere,

R_s =15\ \text{m}

radius of circular disc,

R_c =20\ \text{m}

and time

= 1\ \text{hour} = 60\ \text{minutes} = 60\times 60\ \text{sec}

∴ Moment if inertia of solid sphere,

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

=\frac{2}{5}\times 80\times (15)^2

=7200\ \text{kg-m}^2

Similarly,

moment of inertia of the disc,

I_c = \frac{1}{2} M R_c^2

=\frac{1}{2}\times 80\times (20)^2

=16000\ \text{kg-m}^2

Rate of change of moment of Inertia

= \frac{I_c - I_s}{t}

= \frac{16000-7200}{60\times 60}

= \frac{22}{9}\ \text{kg-m}^2\text{s}^{-1}

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