Question

5. The periods of a pendulum on two planets are
in the ratio 3 : 4. The acceleration due to
gravity on them are in the ratio
1) 9:16 2) 3:4 3) 4:3 4) 16:9​

Answer 1

Given:

The periods of a pendulum on two planets are

in the ratio 3 : 4.

To find:

Ratio of acceleration due to gravity on the two planets.

Calculation:

General formula for time period of simple pendulum on a planet is :

$\boxed{ \rm{T = 2\pi \sqrt{ \dfrac{l}{g} } }}$

On the 1st planet :

$\rm{T1 = 2\pi \sqrt{ \dfrac{l}{(g1)} } }$

On the 2nd planet:

$\rm{T2= 2\pi \sqrt{ \dfrac{l}{(g2)} } }$

So, required ratio is:

$\rm{ \therefore \: \dfrac{T1}{T2} = \sqrt{ \dfrac{g2}{g1} } }$

$\rm{ \therefore \: \dfrac{3}{4} = \sqrt{ \dfrac{g2}{g1} } }$

$\rm{ = > \: { \bigg(\dfrac{3}{4} \bigg )}^{2} = \dfrac{g2}{g1} }$

$\rm{ = > \dfrac{g2}{g1} = \dfrac{9}{16} }$

$\rm{ = > \dfrac{g1}{g2} = \dfrac{16}{9} }$

So, final answer is

$\boxed{ \rm{ \large{ g1 : g2 = 16 : 9 }}}$

Answer 2

$\bf{\red{\underline{\large{Question:-}}}}$

The time periods of a pendulum on two planets are

in the ratio 3 : 4. The acceleration due to

gravity on them are in the ratio

1) 9:16

2) 3:4

3) 4:3

4) 16:9

$\rule{400}{4}$

$\bf{\red{\underline{\large{Answer:-}}}}$

We know,

$\rm{\boxed{\large{T\:=\:2\pi\sqrt{\dfrac{l}{g}}}}}$

where T is time period, l is length of string of pendulum, and g is acceleration due to gravity on a particular planet

Wherever a pendulum is taken, it's length of string will be constant. And $\sf{\pi}$ will also remain constant.

Ratio of time period on two planets = 3 : 4

Suppose time period on first period is 3x and that on 2nd period is 4x

$\texttt{\underline{On first planet:-}}$

Suppose acceleration due to gravity is $\sf{{g}_{1}}$

Then,

$\sf{\large{{T}_{1}\:=\:2\pi\sqrt{\dfrac{l}{{g}_{1}}}}}$

And we know $\sf{{T}_{1}\:=\:3x}$

So,

$\sf{\large{3x\:=\:2\pi\sqrt{\dfrac{l}{{g}_{1}}}}}$--------(i)

$\texttt{\underline{On second planet:-}}$

Suppose acceleration due to gravity is $\sf{{g}_{2}}$

Then,

$\sf{\large{{T}_{2}\:=\:2\pi\sqrt{\dfrac{l}{{g}_{2}}}}}$

And we know $\sf{{T}_{2}\:=\:4x}$

So,

$\sf{\large{4x\:=\:2\pi\sqrt{\dfrac{l}{{g}_{2}}}}}$--------(ii)

$\mathtt{\underline{(i)\:\div\:(ii)}}$

$\sf{\dfrac{3x}{4x}}\:=\:\dfrac{\sf{\large{2\pi\sqrt{\dfrac{l}{{g}_{1}}}}}}{\sf{\large{2\pi\sqrt{\dfrac{l}{{g}_{2}}}}}}}$

$\implies\sf{\dfrac{3}{4}}\:=\:\sqrt{\dfrac{{g}_{2}}{{g}_{1}}}}$

$\implies\sf{\dfrac{9}{16}}\:=\:\dfrac{{g}_{2}}{{g}_{1}}}$

$\implies\sf{\dfrac{{g}_{1}}{{g}_{2}}\:=\:\dfrac{16}{9}}$

So, option 4 is correct