Assume there are two identical simple pendulu | vedclass

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.The acceleration due t

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$3200$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R_E}{R_E + h} ight)^2$,which implies $\sqrt{\frac{g'}{g}} = \frac{R_E}{R_E + h}$.Given the time periods $T_1 = 4\,s$ and $T_2 = 6\,s$,we have the ratio:$\frac{T_1}{T_2} = \sqrt{\frac{g}{g'}} = \frac{R_E + h}{R_E}$.Substituting the values:$\frac{4}{6} = \frac{6400 + h}{6400}$.$\frac{2}{3} = 1 + \frac{h}{6400}$.$\frac{h}{6400} = \frac{2}{3} - 1 = -\frac{1}{3}$.Wait,re-evaluating the ratio: Since $T \propto \frac{1}{\sqrt{g}}$,then $\frac{T_2}{T_1} = \sqrt{\frac{g}{g'}} = \frac{R_E + h}{R_E}$.$\frac{6}{4} = \frac{6400 + h}{6400}$.$1.5 = 1 + \frac{h}{6400}$.$0.5 = \frac{h}{6400}$.$h = 0.5 	imes 6400 = 3200\,km$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

Similar Questions

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

$A$ body weighs $300 \ N$ on the surface of the earth. How much will it weigh at a distance $\frac{R}{2}$ below the surface of the earth (in $N$)? ($R$ is the radius of the earth.)

What will be the formula for the mass of the Earth in terms of $g$,$R$,and $G$?

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