Question

plzz give me......solution of 20,21......plzz koi answer de do...plz don,t sparm amyone itz my humble request ...​

Answer 1

Solution :

• Weight on the body on the surface of the earth = 300 N

• Radius of the earth = R

• Depth of the body below the surface of earth = d = R / 3

We know that ,

⇒ W = mg

⇒ 300 = m x 10

⇒ m = 30 kg

The acceleration due to gravity at a depth d below the surface of the earth is given by ,

⇒ g ' = (R - d / R ) g

⇒ g ' = (R - R /3 /R ) g

⇒ g ' =( 2R /3 R )g

⇒ g ' = 2g / 3

Hence ,

⇒ New weight of the body = m g' = 30 x 20 / 3 = 200 N

The new weight of the body is 200 N

Answer 2

$\begin{lgathered}\begin{lgathered}\begin{lgathered}\tt {\pink{Given}}\begin{cases} \sf{\green{Weight (W) = 300N}}\\ \sf{\blue{Radius = R}}\\ \sf{\orange{Depth (d) =\dfrac{R}{3}}}\end{cases}\end{lgathered} \:\end{lgathered}\end{lgathered}$

By using formula

$\large{\underline{\boxed{W = mg}}}$

→ 300 = m × 10

→ m = 300/10

→ m = 30 kg

Now,

$\implies{g' = \dfrac{R -d}{R} × g}$

$\implies{g' = \dfrac{R - \frac{R}{3}}{R} × g}$

$\implies{g' = \dfrac{2R}{3R} × g}$

$\implies{g' = \dfrac{2g}{R}}$

Hence,

★ New weight of the body = m × g'

→ 30 × $\dfrac{20}{3}$

→ 200 N