SATHEE: Rotational Motion Question 203

Rotational Motion Question 203

Question: From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is

Options:

A) $\frac{4 M R^{2}}{9 \sqrt{3 π}}$

B) $\frac{4 M R^{2}}{3 \sqrt{3 π}}$

C) $\frac{M R^{2}}{32 \sqrt{2 π}}$

D) $\frac{M R^{2}}{16 \sqrt{2 π}}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Here

$a = \frac{2}{\sqrt{3}} R$

Now, $\frac{M}{M^{'}} = \frac{\frac{4}{3} π R^{3}}{a^{3}}$ $= \frac{\frac{4}{3} π R^{3}}{{(\frac{2}{\sqrt{3}} R)}^{3}} = \frac{\sqrt{3}}{2} π .$

$M^{'} = \frac{2 M}{\sqrt{3} π}$

Moment of inertia of the cube about the given axis,

$I = \frac{M^{'} a^{2}}{6}$

$= \frac{\frac{2 M}{\sqrt{3} π} \times {(\frac{2}{\sqrt{3}} R)}^{2}}{6} = \frac{4 M R^{2}}{9 \sqrt{3} π}$

Rotational Motion Question 203

Question: From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

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Sample Mock Test

Mentorship

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Rotational Motion Question 203

Question: From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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