The weight of a body on the surface of the earth is 100 N. The gr… — Solution | JEE Physics

MCQEasyJEE 2023Variation with Altitude & Depth

The weight of a body on the surface of the earth is $100 \, \text{N}$ . The gravitational force on it when taken at a height, from the surface of earth, equal to $\frac{1}{4}$ the radius of the earth is:

A
$64 \, \text{N}$
B
$25 \, \text{N}$
C
$100 \, \text{N}$
D
$50 \, \text{N}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The weight on the surface of the earth is $100 \, \text{N}$ .

Find: The gravitational force at a height equal to $\frac{R_e}{4}$ above the earth's surface.

Using Newton's formula,

F = \frac{GMm}{r^2}

At the surface of the earth,

F = \frac{GMm}{R_e^2}

r = R_e + \frac{R_e}{4} = \frac{5R_e}{4}

the force becomes

F' = \frac{GMm}{\left(\frac{5R_e}{4}\right)^2} = \frac{16GMm}{25R_e^2}

Therefore,

F' = \frac{16}{25}F = \frac{16}{25} \times 100 = 64 \, \text{N}

Therefore, the gravitational force is $64 \, \text{N}$ .

Common mistakes

Using the height $\frac{R_e}{4}$ directly as the distance from the earth's centre is incorrect because Newton's law uses the distance from the centre. Use $r = R_e + \frac{R_e}{4} = \frac{5R_e}{4}$ instead.
Assuming gravitational force decreases linearly with height is wrong because $F \propto \frac{1}{r^2}$ , not $\frac{1}{r}$ . Always apply the inverse-square relation.
Confusing the given weight on the surface with the answer option mapping can lead to selecting B from the provided key even though the calculation gives $64 \, \text{N}$ . Verify the computed value against the options carefully.

Practice more Variation with Altitude & Depth questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Variation with Altitude & Depth

Answer

Step-by-step solution

Common mistakes

Practice more Variation with Altitude & Depth questions

Related questions