JEE MainPhysics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Gravitation: JEE Main Physics Question with Solution

Q: Why does change if only mass is explicitly doubled?

Because density is given as constant. If mass doubles and density is unchanged, volume must double, which forces to increase by a factor of .

A planet has double the mass of the earth. Its average density is equal to the that of the earth. An object weighing

\mathrm{W}

on earth will weigh on that planet :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

2^{2/3}\text{ W}

\text{W}

2^{1/3}\text{ W}

2\text{ W}

Step-by-step solutionView

Correct answer

Because mass scales with the cube of the radius for constant density, doubling the mass increases the radius by a factor of $2^{1/3}$ , which leads to a weight of $2^{1/3}\text{ W}$ .

Option analysis

Why each option works or fails

A ·

2^{2/3}\text{ W}

Inverting the power relationship when substituting $R \propto M^{1/3}$ into $g = GM/R^2$ , arriving at $M / M^{1/3} = M^{2/3}$ instead of $M / M^{2/3} = M^{1/3}$ . Compute $g \propto M / (M^{1/3})^2 = M / M^{2/3} = M^{1-2/3} = M^{1/3}$ carefully.

B ·

\text{W}

Believing that having the same density means the surface gravity remains completely unchanged regardless of total mass. Surface gravity depends on radius as well as density ( $g = \frac{4}{3}\pi G \rho R$ ); since mass doubled at constant density, radius must increase, altering surface gravity.

C ·

2^{1/3}\text{ W}

Correct option. Since $M = \frac{4}{3}\pi R^3 \rho$ , $R \propto M^{1/3}$ . Therefore, $g = \frac{4}{3}\pi G \rho R \propto M^{1/3}$ , yielding $W' = 2^{1/3}W$ .

D ·

2\text{ W}

Assuming weight is directly proportional to mass alone ( $W \propto M$ ) without accounting for the increased planetary radius that reduces the surface gravitational field. Remember that surface gravity depends on radius via $g = GM/R^2$ ; a larger mass at constant density occupies a larger volume and thus has a larger radius.

Reviewed route

Solution

StepWorking

01given

$M' = 2M$ , $\rho' = \rho$ , Weight on Earth $= W = mg = m \frac{GM}{R^2}$ .

02find

Find the weight $W'$ of the object on the new planet.

03visualise

A spherical planet with mass $M = \frac{4}{3}\pi R^3 \rho$ . For constant density $\rho$ , radius scales with mass as $R \propto M^{1/3}$ .

04strategise

Express surface gravity $g = \frac{GM}{R^2}$ purely in terms of $M$ by substituting $R \propto M^{1/3}$ , yielding $g \propto \frac{M}{(M^{1/3})^2} = M^{1/3}$ . Then weight $W \propto g \propto M^{1/3}$ .

05execute

Since $M' = 2M$ , the new weight is $W' = W \left(\frac{M'}{M}\right)^{1/3} = 2^{1/3} W$ .

✓verify

Double mass at constant density means larger volume and larger radius ( $R' = 2^{1/3}R$ ). Surface gravity increases by a factor of $2 / (2^{1/3})^2 = 2^{1/3}$ , so $W' = 2^{1/3}W$ , which is physically consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does

R

change if only mass is explicitly doubled?

Because density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ is given as constant. If mass doubles and density is unchanged, volume must double, which forces $R$ to increase by a factor of $2^{1/3}$ .

Answer

Because mass scales with the cube of the radius for constant density, doubling the mass increases the radius by a factor of $2^{1/3}$ , which leads to a weight of $2^{1/3}\text{ W}$ .

Why each option works or fails

A: $2^{2/3}\text{ W}$ - Inverting the power relationship when substituting $R \propto M^{1/3}$ into $g = GM/R^2$ , arriving at $M / M^{1/3} = M^{2/3}$ instead of $M / M^{2/3} = M^{1/3}$ . Compute $g \propto M / (M^{1/3})^2 = M / M^{2/3} = M^{1-2/3} = M^{1/3}$ carefully.
B: $\text{W}$ - Believing that having the same density means the surface gravity remains completely unchanged regardless of total mass. Surface gravity depends on radius as well as density ( $g = \frac{4}{3}\pi G \rho R$ ); since mass doubled at constant density, radius must increase, altering surface gravity.
C · correct: $2^{1/3}\text{ W}$ - Correct option. Since $M = \frac{4}{3}\pi R^3 \rho$ , $R \propto M^{1/3}$ . Therefore, $g = \frac{4}{3}\pi G \rho R \propto M^{1/3}$ , yielding $W' = 2^{1/3}W$ .
D: $2\text{ W}$ - Assuming weight is directly proportional to mass alone ( $W \propto M$ ) without accounting for the increased planetary radius that reduces the surface gravitational field. Remember that surface gravity depends on radius via $g = GM/R^2$ ; a larger mass at constant density occupies a larger volume and thus has a larger radius.

Step-by-step solution

given: $M' = 2M$ , $\rho' = \rho$ , Weight on Earth $= W = mg = m \frac{GM}{R^2}$ .
find: Find the weight $W'$ of the object on the new planet.
visualise: A spherical planet with mass $M = \frac{4}{3}\pi R^3 \rho$ . For constant density $\rho$ , radius scales with mass as $R \propto M^{1/3}$ .
strategise: Express surface gravity $g = \frac{GM}{R^2}$ purely in terms of $M$ by substituting $R \propto M^{1/3}$ , yielding $g \propto \frac{M}{(M^{1/3})^2} = M^{1/3}$ . Then weight $W \propto g \propto M^{1/3}$ .
execute: Since $M' = 2M$ , the new weight is $W' = W \left(\frac{M'}{M}\right)^{1/3} = 2^{1/3} W$ .
verify: Double mass at constant density means larger volume and larger radius ( $R' = 2^{1/3}R$ ). Surface gravity increases by a factor of $2 / (2^{1/3})^2 = 2^{1/3}$ , so $W' = 2^{1/3}W$ , which is physically consistent.

Shortcut:

When to use it: Use when density is given as constant, utilizing $g = \frac{4}{3}\pi G \rho R$ .

given: $M' = 2M$ , $\rho' = \rho$ .

strategise: Recall $g = \frac{4}{3}\pi G \rho R$ , so for constant $\rho$ , $g \propto R$ . Since $M = \frac{4}{3}\pi R^3 \rho \propto R^3$ , we have $R \propto M^{1/3}$ . Therefore, $g \propto M^{1/3}$ .

execute: $\frac{W'}{W} = \frac{g'}{g} = \frac{R'}{R} = \left(\frac{M'}{M}\right)^{1/3} = 2^{1/3}$ , hence $W' = 2^{1/3}W$ .

verify: Directly matches Option (2).

Gravitation: JEE Main Physics Question with Solution

What feels right?

Because mass scales with the cube of the radius for constant density, doubling the mass increases the radius by a factor of $2^{1/3}$ , which leads to a weight of $2^{1/3}\text{ W}$ .

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

We think you should solve this next

The resistivity $(\rho)$ of semiconductor varies with temperature. Which of the following…

The kinetic energy of an electron, $\alpha$ -particle and a proton are given as…

The energy levels of an hydrogen atom are shown below. The transition corresponding to em…

Students also ask

Gravitation: JEE Main Physics Question with Solution

Because mass scales with the cube of the radius for constant density, doubling the mass increases the radius by a factor of 21/32^{1/3}21/3, which leads to a weight of 21/3 W2^{1/3}\text{ W}21/3 W.

01given

02find

03visualise

04strategise

05execute

✓verify

The resistivity (ρ)(\rho)(ρ) of semiconductor varies with temperature. Which of the following…

The kinetic energy of an electron, α\alphaα-particle and a proton are given as…

The energy levels of an hydrogen atom are shown below. The transition corresponding to em…

Students also ask

Because mass scales with the cube of the radius for constant density, doubling the mass increases the radius by a factor of $2^{1/3}$ , which leads to a weight of $2^{1/3}\text{ W}$ .

The resistivity $(\rho)$ of semiconductor varies with temperature. Which of the following…

The kinetic energy of an electron, $\alpha$ -particle and a proton are given as…