Question

density remaining constant, if earth contracts to half of its present radius duration of the day would be (in minutes)
1)45

2)80

3)100

4)120​

Answer 1

Duration of the day would be 3 Hour means 120 minute, here given option is wrong.  

Explanation:

1. Here density remain constant , means mass of earth is constant.

    $M_{1}=M_{2}$       ...1)

2. Relation between initial radius and final radius

    $R_{1}=2\times R_{2}$

   so $\frac{R_{2}}{R_{1}}= \frac{1}{2}$      ...2)

3.  Earth shape is almost like solid sphere

Mass moment of inertia of solid sphere (I) = $\frac{2}{5}\times M\times R^{3}$

4.  Relation between initial mass moment of inertia and final mass moment of inertia

    $\frac{I_{2}}{I_{1}}= (\frac{M_{2}}{M_{1}})\times (\frac{R_{2}}{R_{1}})^{3}$

now from equation 1 and equation 2

$\frac{I_{2}}{I_{1}}= \frac{1}{8}$        ...3)

5. Here on earth no torque is acting, so angular momentum(J=Iω) of earth is constant . Where ω is angular speed of earth.

angular speed $\omega = \frac{2\pi }{T}$      ...4)

  Relation between initial angular momentum and final angular momentum

  $J_{1}= J_{2}$        

  $I_{1}\omega _{1}= I_{2}\omega _{2}$

$I_{1}\times \frac{2\pi }{T_{1}}=I_{2}\times \frac{2\pi }{T_{2}}$   ...5)

6.  Equation 5 can also be written as

     $T_{2}= (\frac{I_{2}}{I_{1}})\times T_{1}$  

 So $T_{2}= (\frac{1}{8})\times T_{1}$    ...6)

    where $T_{1}$ = 24 hours

     So  $T_{2}$ = 3 hours = 180 minutes