MCQ
If radius of earth becomes $n$ times its present value, without change in mass, then duration of day becomes
- A$\frac{24}{n^2}$
- ✓$24 n^2$
- C$24\left(1-\frac{1}{n^2}\right)$
- D$24\left(1-n^2\right)$
✓
Answer
Correct option: B.
$24 n^2$
b
(b)
(b)
By conservation of angular momentum,
$\Rightarrow I _1 w = I _2 w _2$
$\frac{2}{5} MR ^2 \times \frac{2 \pi}{24 \text { hour }}=\frac{2}{5} M ( nR )^2 \frac{2 \pi}{ T }$
$\Rightarrow T = n ^2 24 h$
Hence, the answer is $n ^2 24 h$.
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