Do we need the heights and to solve the problem?

No, because the ratio of velocities is already directly given as 4. The relation means , which confirms the same ratio without extra information.

JEE MainPhysics

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Work, Energy and Power: JEE Main Physics Question with Solution

A body is dropped on ground from a height '

\text{h}_1

' and after hitting the ground, it rebounds to a height '

\text{h}_2

' If the ratio of velocities of the body just before and after hitting ground is

4

, then percentage loss in kinetic energy of the body is

\frac{\text{x}}{4}

. The value of

\text{x}

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Step-by-step solutionView

Correct answer

The value of x is 375.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Ratio of velocity just before impact to velocity just after impact is $\frac{v_1}{v_2} = 4$ . Percentage loss in kinetic energy is given as $\frac{x}{4}\%$ .

02find

Find the value of $x$ .

03visualise

A body falls freely, attains velocity $v_1$ just before collision, and rebounds upward with velocity $v_2 = \frac{v_1}{4}$ . Kinetic energy is lost during the inelastic collision with the ground.

04strategise

Express initial kinetic energy as $K_i = \frac{1}{2}mv_1^2$ and final kinetic energy as $K_f = \frac{1}{2}mv_2^2 = \frac{1}{2}m\left(\frac{v_1}{4}\right)^2 = \frac{K_i}{16}$ . Compute percentage loss: $\text{Loss}\% = \frac{K_i - K_f}{K_i} \times 100\% = \left(1 - \frac{1}{16}\right) \times 100\% = \frac{15}{16} \times 100\%$ . Equate to $\frac{x}{4}$ .

05execute

$\text{Percentage loss} = \frac{15}{16} \times 100 = \frac{15 \times 25}{4} = \frac{375}{4}\%$ . Comparing with $\frac{x}{4}\%$ , we get $x = 375$ .

✓verify

Fractional loss is $\frac{15}{16} = 93.75\%$ . $\frac{375}{4}\% = 93.75\%$ . The value $x = 375$ is consistent and correct.

Hints that build this answer step by step

If the ratio of the velocity before collision to after collision is $\frac{v_1}{v_2} = 4$ , what is the ratio of final kinetic energy to initial kinetic energy, $\frac{K_2}{K_1}$ ?
$\frac{K_2}{K_1} = \frac{1}{16}$
Using the ratio $\frac{K_2}{K_1} = \frac{1}{16}$ , what is the percentage loss in kinetic energy?
$\frac{1500}{16}\% = \frac{375}{4}\%$
Equating the percentage loss $\frac{375}{4}\%$ to $\frac{x}{4}\%$ , what is the value of $x$ ?
$x = 375$

✓ Source and academic review

Question type: Numerical
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

Do we need the heights

h_1

and

h_2

to solve the problem?

No, because the ratio of velocities $v_1/v_2$ is already directly given as 4. The relation $v = \sqrt{2gh}$ means $h_1/h_2 = (v_1/v_2)^2 = 16$ , which confirms the same ratio without extra information.

Answer

The value of x is 375.

Why each option works or fails

Step-by-step solution

given: Ratio of velocity just before impact to velocity just after impact is $\frac{v_1}{v_2} = 4$ . Percentage loss in kinetic energy is given as $\frac{x}{4}\%$ .
find: Find the value of $x$ .
visualise: A body falls freely, attains velocity $v_1$ just before collision, and rebounds upward with velocity $v_2 = \frac{v_1}{4}$ . Kinetic energy is lost during the inelastic collision with the ground.
strategise: Express initial kinetic energy as $K_i = \frac{1}{2}mv_1^2$ and final kinetic energy as $K_f = \frac{1}{2}mv_2^2 = \frac{1}{2}m\left(\frac{v_1}{4}\right)^2 = \frac{K_i}{16}$ . Compute percentage loss: $\text{Loss}\% = \frac{K_i - K_f}{K_i} \times 100\% = \left(1 - \frac{1}{16}\right) \times 100\% = \frac{15}{16} \times 100\%$ . Equate to $\frac{x}{4}$ .
execute: $\text{Percentage loss} = \frac{15}{16} \times 100 = \frac{15 \times 25}{4} = \frac{375}{4}\%$ . Comparing with $\frac{x}{4}\%$ , we get $x = 375$ .
verify: Fractional loss is $\frac{15}{16} = 93.75\%$ . $\frac{375}{4}\% = 93.75\%$ . The value $x = 375$ is consistent and correct.