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The rotation of the Earth (of radius $R$ ) about its axis speeds up to a value such that a man at latitude angle $45^{\circ}$ feels weightlessness. The duration of a day in such a case is
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Verified Answer
The correct answer is:
$\pi \sqrt{\frac{2 R}{g}}$
Given that, angle of latitude, $\lambda=45^{\circ}$
We know that, the expression of gravity at latitude angle is
$$
g_\lambda=g-\omega^2 R \cos ^2 \lambda
$$
The weight of man of mass $m$ at this point
$$
w=m g_\lambda=m\left(g-\omega^2 R \cos ^2 45^{\circ}\right)
$$
According to question, man feels weightlessness
$$
\therefore \quad \begin{aligned}
w & =0 \\
m\left(g-\omega^2 R \cos ^2 45^{\circ}\right) & =0
\end{aligned}
$$
But $m \neq 0$, then $g-\omega^2 R \cos ^2 45^{\circ}=0$
$$
\Rightarrow \quad \omega=\sqrt{\frac{2 g}{R}}
$$
Now, using time period, (duration of day)
$$
=T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{2 g}}=\pi \sqrt{\frac{2 R}{g}}
$$
We know that, the expression of gravity at latitude angle is
$$
g_\lambda=g-\omega^2 R \cos ^2 \lambda
$$
The weight of man of mass $m$ at this point
$$
w=m g_\lambda=m\left(g-\omega^2 R \cos ^2 45^{\circ}\right)
$$
According to question, man feels weightlessness
$$
\therefore \quad \begin{aligned}
w & =0 \\
m\left(g-\omega^2 R \cos ^2 45^{\circ}\right) & =0
\end{aligned}
$$
But $m \neq 0$, then $g-\omega^2 R \cos ^2 45^{\circ}=0$
$$
\Rightarrow \quad \omega=\sqrt{\frac{2 g}{R}}
$$
Now, using time period, (duration of day)
$$
=T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{2 g}}=\pi \sqrt{\frac{2 R}{g}}
$$
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