Question 61:
moderateA body weighs $200 \text{ N}$ on the surface of the earth. How much will it weigh half way down to the centre of the earth?
(2019)
The weight at depth $d$ is $W_d = W(1 - \frac{d}{R})$.nGiven $d = \frac{R}{2}$ (halfway to the center), we have $$W_d = 200(1 - \frac{1}{2})$.n$W_d = 200 \times \frac{1}{2} = 100 \text{ N}$$.
Question 62:
moderateThe height at which the weight of a body becomes $1/16^{\text{th}}$, its weight on the surface of earth (radius R), is:
(2012 Pre)
Weight at height $h$ is given by $W_h = \frac{W}{(1 + \frac{h}{R})^2}$. Given $W_h = \frac{W}{16}$, we equate: $\frac{1}{16} = \frac{1}{(1 + \frac{h}{R})^2}$. Taking the square root gives $$ 1 + \frac{h}{R} = 4 \Rightarrow \frac{h}{R} = 3 \Rightarrow h = 3R$$.
Question 63:
moderateA body of weight $72 \text{ N}$ moves from the surface of earth to a height half of the radius of the earth, then gravitational force exerted on it will be:
(2000)
Gravitational force (weight) at height $h$ is $F = \frac{W}{(1 + \frac{h}{R})^2}$.nSubstitute $h = \frac{R}{2}$ to get $F = \frac{72}{(1 + 0.5)^2}$.n$F = \frac{72}{2.25} = 32 \text{ N}$.