SATHEE: Rotation 1 Question 2

Rotation 1 Question 2

2. A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7 M}{8}$ and is converted into a uniform disc of radius $2 R$ . The second part is converted into a uniform solid sphere. Let $I_{1}$ be the moment of inertia of the disc about its axis and $I_{2}$ be the moment of inertia of the new sphere about its axis.

The ratio $I_{1} / I_{2}$ is given by

(2019 Main, 10 Apri II)

(a) 285

(b) 185

(d) 140

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Answer:

Correct Answer: 2. (d)

Solution:

The given situation is shown in the figure given below

Density of given sphere of radius $R$ is

$ρ = \frac{Mass}{Volume} = \frac{M}{\frac{4}{3} π R^{3}}$

Let radius of sphere formed from second part is $r$ , then mass of second part $=$ volume $\times$ density

$\begin{aligned} \frac{1}{8} M & = \frac{4}{3} π r^{3} \times \frac{M}{\frac{4}{3} π R^{3}} \\ ∴ r^{3} & = \frac{R^{3}}{8} \Rightarrow r = \frac{R}{2} \end{aligned}$

Now, $I_{1} =$ moment of inertia of disc (radius $2 R$ and mass $\frac{7}{8} M$ ) about its axis

$= \frac{Mass \times (Radius)^{2}}{2} = \frac{\frac{7}{8} M \times (2 R)^{2}}{2} = \frac{7}{4} M R^{2}$

and $I_{2} =$ moment of inertia of sphere

$(radius \frac{R}{2} and mass \frac{1}{8} M) about its axis$

$= \frac{2}{5} \times$ Mass $\times (Radius)^{2} = \frac{2}{5} \times \frac{1}{8} M \times \frac{R}{2}^{2} = \frac{M R^{2}}{80}$

$∴$ Ratio $\frac{I_{1}}{I_{2}} = \frac{\frac{7}{4} M R^{2}}{\frac{1}{80} M R^{2}} = 140$

Rotation 1 Question 2

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Rotation 1 Question 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Other Resources

Syllabus

Test Platform

Sample Mock Test

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