# Gravitation

## Concept Of Physics

### H C Verma

1   Two spherical balls of mass $10$ kg each are placed $10$ cm apart. Find the gravitational force of attraction between them.

##### Solution :

Gravitational force of attraction $\\$ F = $\frac{GMm}{r^2}$ $\\$ $\frac{6.67\times 10^{-11} \times 10 \times 10}{(0.1)^2}$ = 6.67 $\times 10^{-7}$ N

2   Four particles having masses m, $2$ m, $3$ m and $4$ m are placed at the four corners of a square of edge a. Find the gravitational force acting on a particle of mass m placed at the centre.

##### Solution :

3   15. A pole of length $1'00 m$ stands half dipped in a swimming pool with water level $50'0 cm$ higher than the bed. The refractive index of water is $1'33$ and sunlight is coming at an angle of $45°$ with the vertical. Find the length of the shadow of the pole on the bed.

##### Solution :

Shadow length =$BA^1$ = $BD + A^1D$ = 0.5 + 0.5 tan r$\\$ Now. 1.33 = $\frac{ sin 45^0}{ sin r }$ $\Rightarrow$ sin r = 0.53.$\\$ $\Rightarrow$ cos r = $\sqrt{1-sin^2r}$ = $\sqrt{1-(0.53)^2}$ = 0.85 $\\$ So, tan r = 0.6235 $\\$ So, shadow length = (0.5) (1 + 0.6235) = 81.2 cm