JEE MainPhysics

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+4 marks−1 if incorrectNumericalPrevious-year question

Relating Loss of Kinetic Energy to Work Done by Retarding Force

Q: Why is the integral of 2x dx equal to x^2 rather than 2x^2?

Because integral of 2x dx = 2 ((x^2)/2) = x^2. The factor of 2 cancels with the denominator.

A particle of mass 10g moves in a straight line with retardation 2x, where x is the displacement in SI units. Its loss of kinetic energy for above displacement is (10/x)^(-n)J. The value of n will be ________.

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Step-by-step solutionView

Correct answer

The value of n is 2.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Mass $m = 10 g = 10^{- 2} kg$ , retardation $a_{r} = 2 x {m/s}^{2}$ , displacement $= x$ . Loss of kinetic energy $= {(\frac{10}{x})}^{- n} J$ .

02Find

Find the value of the exponent $n$ .

03Visualise

A particle moves along the x-axis subject to a retarding force opposing its motion, $F_{retard} = m a_{r} = 2 m x$ . As it travels from $0$ to $x$ , this retarding force extracts kinetic energy equal to the work done against it.

04Strategise

By the work-energy theorem, $Δ K = W_{net} = - \int_{0}^{x} F_{retard} d x$ . Therefore, $Loss of K.E. = - Δ K = \int_{0}^{x} m (2 x) d x = m x^{2}$ . Substitute $m = 10^{- 2} kg$ and rewrite in the form ${(\frac{10}{x})}^{- n}$ .

05Execute

$Loss of K.E. = 10^{- 2} x^{2} = {(\frac{x}{10})}^{2} = {(\frac{10}{x})}^{- 2} J$ . Comparing with ${(\frac{10}{x})}^{- n}$ , we get $n = 2$ .

✓Verify

Dimensionally, $10^{- 2} x^{2}$ has units of $kg \cdot m^{2} / s^{0}$ , which with the retardation constant $2 s^{- 2}$ yields Joules. The power of $x$ is 2, matching $n = 2$ .

Hints that build this answer step by step

How is the loss of kinetic energy related to the retarding force acting on the particle?
Loss of $K E = \int F_{retarding} d x = \int m a_{retarding} d x$
Given $m = 10 g = 10^{- 2} kg$ and retardation $a = 2 x$ , what is the integral for the loss of kinetic energy from $0$ to $x$ ?
$\int_{0}^{x} (10^{- 2}) (2 x) d x = 10^{- 2} x^{2}$
Equating $10^{- 2} x^{2}$ to ${(\frac{10}{x})}^{- n}$ , what is the value of $n$ ?
$n = 2$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is the integral of

2 x d x

equal to

x^{2}

rather than

2 x^{2}

Because $\int 2 x d x = 2 (\frac{x^{2}}{2}) = x^{2}$ . The factor of 2 cancels with the denominator.

Answer

Accepted answer: 2

The value of n is 2.

Why each option works or fails

Step-by-step solution

Given: Mass $m = 10 g = 10^{- 2} kg$ , retardation $a_{r} = 2 x {m/s}^{2}$ , displacement $= x$ . Loss of kinetic energy $= {(\frac{10}{x})}^{- n} J$ .
Find: Find the value of the exponent $n$ .
Visualise: A particle moves along the x-axis subject to a retarding force opposing its motion, $F_{retard} = m a_{r} = 2 m x$ . As it travels from $0$ to $x$ , this retarding force extracts kinetic energy equal to the work done against it.
Strategise: By the work-energy theorem, $Δ K = W_{net} = - \int_{0}^{x} F_{retard} d x$ . Therefore, $Loss of K.E. = - Δ K = \int_{0}^{x} m (2 x) d x = m x^{2}$ . Substitute $m = 10^{- 2} kg$ and rewrite in the form ${(\frac{10}{x})}^{- n}$ .
Execute: $Loss of K.E. = 10^{- 2} x^{2} = {(\frac{x}{10})}^{2} = {(\frac{10}{x})}^{- 2} J$ . Comparing with ${(\frac{10}{x})}^{- n}$ , we get $n = 2$ .
Verify: Dimensionally, $10^{- 2} x^{2}$ has units of $kg \cdot m^{2} / s^{0}$ , which with the retardation constant $2 s^{- 2}$ yields Joules. The power of $x$ is 2, matching $n = 2$ .

Relating Loss of Kinetic Energy to Work Done by Retarding Force

What feels right?

The value of n is 2.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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01Given

02Find

03Visualise

04Strategise

05Execute

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