Question

Mass remaining constant. it the earth shrinks until its radius becomes $2/3^{rd}$ of the present value, then the length of the day will be:

• A. $32/3hr$
• B. $54hr$
• C. $16hr$
• D. $36hr$

Answer 1

The correct option is A $32/3hr$

The moment of inertia before and after contraction is given as,

$I_1{\omega }_1=I_2{\omega }_2$

$\left(\frac{2}{5}M{R_1}^2\right)\left(\frac{2\pi }{T_1}\right)=\left(\frac{2}{5}M{R_2}^2\right)\left(\frac{2\pi }{T_2}\right)$

$T_2={\left(\frac{R_2}{R_1}\right)}^2{T}_1$

$T_2={\left(\frac{\frac{2}{3}{R}_1}{R_1}\right)}^2\times 24$

$T_2=\frac{32}{3}hr$

Thus, the length of the day is $\frac{32}{3}hr$.