Question

If radius of earth shrinks by $0.1\%$ without change in its mass, the percentage change in the duration of one day:

• A. decreases by $0.1\%$
• B. increases by $0.1\%$
• C. decreases by $0.2\%$
• D. increases by $0.2\%$

Answer 1

Answer: (C)

decreases by $0.2\%$

Let the initial radius be $r$ and final is $r_2=0.999r$

Now using the angular momentum conservation

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

We get $2m{r}^2/5\times {\omega }_1=2m{r}_2^2/5\times {\omega }_2$

So $\frac{{\omega }_1}{{\omega }_2}=\frac{(999{)}^2}{{1000}^2}=0.998$....(1)

$T_{initial}=2\pi /{\omega }_1$,

$T_{final}=2\pi /{\omega }_2$

So $\frac{T_{final}}{T_{initial}}=\frac{{\omega }_1}{{\omega }_2}$

from equation 1, decrease in time period is $(1-0.998)\times 100=0.2$%