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Rotational Motion question

2019 · 10 Apr · Shift 2 · Q50
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Rotational Motion question

2019 · 10 Apr · Shift 2 · Q50

JEE MainPhysicsRotational MotionMCQ+4 / −1
A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7M8{{7M} \over 8}87M​ and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio I1/I2 is given by :
  1. A
    65
  2. B
    140
  3. C
    185
  4. D
    285
View written solutionFree

Correct answer: B

  1. Masses of the two parts

The original solid sphere has mass MMM and radius RRR.

It is divided into two parts:

  • First part: m1=7M8m_1 = \dfrac{7M}{8}m1​=87M​
  • Second part: m2=M−7M8=M8m_2 = M - \dfrac{7M}{8} = \dfrac{M}{8}m2​=M−87M​=8M​
  1. Moment of inertia of the disc

The first part is converted into a uniform disc of radius 2R2R2R.

For a uniform disc about its central axis,

I1=12m1a2I_1 = \frac{1}{2} m_1 a^2I1​=21​m1​a2

where a=2Ra=2Ra=2R.

So,

I1=12⋅7M8⋅(2R)2I_1 = \frac{1}{2}\cdot \frac{7M}{8} \cdot (2R)^2I1​=21​⋅87M​⋅(2R)2 I1=12⋅7M8⋅4R2I_1 = \frac{1}{2}\cdot \frac{7M}{8} \cdot 4R^2I1​=21​⋅87M​⋅4R2 I1=7M4R2I_1 = \frac{7M}{4}R^2I1​=47M​R2
  1. Radius of the new solid sphere

The second part of mass m2=M8m_2=\dfrac{M}{8}m2​=8M​ is converted into a uniform solid sphere of the same material.

Since density remains same,

m2M=V2V\frac{m_2}{M} = \frac{V_2}{V}Mm2​​=VV2​​

For spheres, volume is proportional to cube of radius, so

m2M=(rR)3\frac{m_2}{M} = \left(\frac{r}{R}\right)^3Mm2​​=(Rr​)3

Thus,

18=(rR)3\frac{1}{8} = \left(\frac{r}{R}\right)^381​=(Rr​)3 r=R2r = \frac{R}{2}r=2R​
  1. Moment of inertia of the new sphere

For a solid sphere about its diameter,

I2=25m2r2I_2 = \frac{2}{5} m_2 r^2I2​=52​m2​r2

Substitute m2=M8m_2=\dfrac{M}{8}m2​=8M​ and r=R2r=\dfrac{R}{2}r=2R​:

I2=25⋅M8⋅(R2)2I_2 = \frac{2}{5}\cdot \frac{M}{8}\cdot \left(\frac{R}{2}\right)^2I2​=52​⋅8M​⋅(2R​)2 I2=25⋅M8⋅R24I_2 = \frac{2}{5}\cdot \frac{M}{8}\cdot \frac{R^2}{4}I2​=52​⋅8M​⋅4R2​ I2=MR280I_2 = \frac{MR^2}{80}I2​=80MR2​
  1. Compute the ratio I1/I2I_1/I_2I1​/I2​
I1I2=7MR24MR280\frac{I_1}{I_2} = \frac{\frac{7MR^2}{4}}{\frac{MR^2}{80}}I2​I1​​=80MR2​47MR2​​

Cancel MR2MR^2MR2:

I1I2=74⋅80=7⋅20=140\frac{I_1}{I_2} = \frac{7}{4}\cdot 80 = 7\cdot 20 = 140I2​I1​​=47​⋅80=7⋅20=140
  1. Check with options

The correct value is:

140\boxed{140}140​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.

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