Statement-1: If the Earth suddenly contracts to ${\frac{1}{n}}^{t h}$ of its present size without any change in its mass.
Statement-2:The duration of the new day will be $\frac{24}{n}$ hrs.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

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Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1

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Statement-1 is True, Statement-2 is False

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Statement-1 is False, Statement-2 is True

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Solution

The correct option is B Statement-1 is True, Statement-2 is False
Statement -1 is true.
$T \propto \frac{1}{\sqrt{r^{3}}}$
If earth contracts to 1/n th of current size, length of new day will become $\frac{24}{n^{2}}$ hrs.
So, statement -2 is wrong and option C is right.

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

I^{-} < B r^{-} < C l^{-} < F^{-}

I^{-}

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