The length of a seconds pendulum on the surfa | vedclass

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.For a seconds pendulum,the time period $T$ is fixed at $2\, s$ on b

Vedclass · Accepted Answer

$\frac{1}{6}\, m$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.For a seconds pendulum,the time period $T$ is fixed at $2\, s$ on both the Earth and the Moon.Since $T = 2\pi \sqrt{\frac{l}{g}}$,we have $T^2 = 4\pi^2 \frac{l}{g}$,which implies $l = \frac{T^2 g}{4\pi^2}$.Since $T$ is constant,$l \propto g$.Therefore,$\frac{l_m}{l_e} = \frac{g_m}{g_e}$.Given $g_m = \frac{1}{6} g_e$ and $l_e = 1\, m$,we get $l_m = \frac{1}{6} 	imes 1\, m = \frac{1}{6}\, m$.

Column-$I$	Column-$II$
$(a)$ $T^2 \to l$	$(i)$ Linear
$(b)$ $T^2 \to g$	$(ii)$ Parabolic
$(c)$ $T \to l$	$(iii)$ Hyperbolic

The length of a seconds pendulum on the surface of the Earth is $1\, m$. The length of a seconds pendulum on the surface of the Moon,where $g$ is $1/6$th the value of $g$ on the surface of the Earth,is:

Similar Questions

The acceleration due to gravity at a place is $\pi^2 \, m/s^2$. Then the time period of a simple pendulum of length $1 \, m$ is:

In the following table, the relation of the graph in column-$I$ and the shape of the graph in column-$II$ is shown. Match them appropriately.
Column-$I$ Column-$II$
$(a)$ $T^2 \to l$ $(i)$ Linear
$(b)$ $T^2 \to g$ $(ii)$ Parabolic
$(c)$ $T \to l$ $(iii)$ Hyperbolic

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