Question

hint;above image

question...What would the length of a seconds pendulum on
the surface of the Earth be if the mass of the Earth
remains constant but its volume shrinks to 1/8 th. of its
original volume.
(Take original value of acceleration due to gravity as
9.8 m s-2)​

Answer 1

Answer:

Given:

Mass of Earth remains constant but the volume becomes ⅛th of original volume.

To find:

New length of a seconds pendulum on the Earth surface.

Concept:

Since volume of Earth changes , the radius correspondingly changes leading to change in Gravitational acceleration on the Earth surface.

So the length of a pendulum has to changed correspondingly in order to keep the time period same.

Calculation:

Let new volume be V2.

$V2 = \dfrac{1}{8}( V1)$

$= > \dfrac{4}{3}\pi {(r2)}^{3} = \dfrac{1}{8} \bigg \{\dfrac{4}{3}\pi {(r1)}^{3} \bigg \}$

$= > {(r2)}^{3} = \dfrac{1}{8} {(r1)}^{3}$

$= > r2 = \dfrac{(r1)}{2} \: ........(1)$

Initial gravitational acceleration be g1

$\red{ \therefore \: g1 = \dfrac{GM}{ {(r1)}^{2} } }$

Similarly for final gravitational acceleration :

$\red{ \therefore \: g2 = \dfrac{GM}{ {(r2)}^{2} } }$

$\red{ = > \: g2 = \dfrac{GM}{ { \{\dfrac{(r1)}{2} \}}^{2} } }$

$\red{ = > g2 = 4(g1)}$

Now the time period of the pendulum has to remain constant :

$2\pi \sqrt{ \dfrac{(l1)}{(g1)} } = 2\pi \sqrt{ \dfrac{(l2)}{(g2)} } = 2$

$= > 2\pi \sqrt{ \dfrac{(l1)}{(g1)} } = 2\pi \sqrt{ \dfrac{(l2)}{ \{4(g1) \}} } = 2$

Taking the 2nd equality :

$= > 2\pi \sqrt{ \dfrac{(l2)}{4 \times 9.8} } = 2$

$= > \dfrac{(l2)}{39.2 } = \dfrac{1 }{ {\pi}^{2} }$

$= > l2 = \dfrac{39.2}{ {\pi}^{2} }$

$= > l2 = 3.97 \: metres$

So final answer :

$\boxed{ \huge{ \sf{\blue{ \bold{L 2 = 3.97 \: m}}}}}$

Answer 2

$\huge\bold\red{Answer:-}$

The time period of a seconds pendulum= 2s

$= 2\pi \sqrt{l \div g}$

When the volume of Earth(v2) Shrinks to 1/8 th of its original volume(v1), then the radius of the Earth becomes half its orginal radius. 

V2 = (1/8) V1

=> R2 = R1/2

We know that,

g = GM/R^2

where 'g' is acceleration due to gravity and 'M' is the mass of the Earth.

$= > g1 = gm \div r \frac{2}{1}$

$g2 = gm \div \: r \ \frac{2}{2}$

Then,

$2 = 2\pi \sqrt{l1 \div g1}$

$= 2\pi \times (l2 \div g2)$

Substitute, the value of g2 and obtain the value of l2.

2π4×9.8(l2)=2

l2/39.2= 1/ π^2

l2= 3.97m

Lenth of seconds = 3.97m