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# Engineering

A solid sphere  of mass M  and radius R is divided into two unequal  parts . The first part has a mass of 7M/8 and is converted into a uniform disc  of radius  2 R . The second part is converted into a uniform disc of radius 2R . 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 :

 

  

  • Option 1)

    185 

  • Option 2)

    140

  • Option 3)

    285

  • Option 4)

    65

Answers (1)
P prateek

I_{disc } = \frac{7 M }{8} \frac{(2R)^2}{2} = I_1 \\\\ solid sphere \\\\ \frac{M}{8 } =( \frac{4}{3} \pi r^3 ) ( \delta ) \\\\ \frac{\delta \left ( \frac{4}{3}\pi R^3 \right )}{8}= 4/3 \pi r ^ 3 \delta \\\\ r = R/2 = radius \: \: of \: \: of \: \: solid \: \: sphere \\\\ I_{2} = \left ( \frac{M}{8} r ^ 2 \right ) (2/5)

= \left ( \frac{M}{8} \right ) \left ( \frac{R}{2} \right )^2 2/5 = I_2 \\\\ I_1 / I_2 = \frac{\frac{7M}{8}\frac{(2 R )^2}{2}}{\frac{2}{5}\frac{M}{8}(\frac{R}{2})^2}= 140


Option 1)

185 

Option 2)

140

Option 3)

285

Option 4)

65

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