Acceleration Due to Gravity and its variation - NEET Physics Questions
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Acceleration Due to Gravity and its variation

Practice Questions:
Question 11: moderate

The 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)

1. $5R$
2. $15R$
3. $3R$
4. $4R$
View Answer

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 $$ 1 + \frac{h}{R} = 4 \Rightarrow \frac{h}{R} = 3 \Rightarrow h = 3R$$.

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Question 12: moderate

A 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)

1. $36 \text{ N}$
2. $32 \text{ N}$
3. $144 \text{ N}$
4. $50 \text{ N}$
View Answer

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$F = \frac{72}{2.25} = 32 \text{ N}$.

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