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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2026 · 2 April · Morning Shift · Physics · Center of Mass, Momentum and Collision

The position of an object having mass 0.1 kg as a function of time...

The position of an object having mass 0.10.1 kg as a function of time tt is given as r⃗=(10t2i^+5t3j^)\vec{r} = \left(10t^2\hat{i} + 5t^3\hat{j}\right) m. At t=1t = 1 s, which of the following statements are correct?
A. The linear momentum p⃗=(2i^+1.5j^)\vec{p} = \left(2\hat{i} + 1.5\hat{j}\right) kg·m/s.
B. The force acting on the object F⃗=(2i^+3j^)\vec{F} = \left(2\hat{i} + 3\hat{j}\right) N.
C. The angular momentum of the object about its origin L⃗=15k^\vec{L} = 15\hat{k} J·s.
D. The torque acting on the object about its origin τ⃗=20k^\vec{\tau} = 20\hat{k} N·m.
Choose the correct answer from the options given below :

Options

  1. AA, B and C only
  2. BB, C and D only
  3. CA, C and D only
  4. DA, B and D only

Correct answer

D. A, B and D only

Step-by-step solution

Given the position vector r⃗=10t2i^+5t3j^\vec{r} = 10t^2\hat{i} + 5t^3\hat{j}

Velocity v⃗=dr⃗dt=20ti^+15t2j^\vec{v} = \dfrac{d\vec{r}}{dt} = 20t\hat{i} + 15t^2\hat{j}

Acceleration a⃗=dv⃗dt=20i^+30tj^\vec{a} = \dfrac{d\vec{v}}{dt} = 20\hat{i} + 30t\hat{j}

At t=1t = 1 s:
r⃗=10i^+5j^\vec{r} = 10\hat{i} + 5\hat{j} m
v⃗=20i^+15j^\vec{v} = 20\hat{i} + 15\hat{j} m/s
a⃗=20i^+30j^\vec{a} = 20\hat{i} + 30\hat{j} m/s2^2

Linear momentum p⃗=mv⃗=0.1(20i^+15j^)=2i^+1.5j^\vec{p} = m\vec{v} = 0.1(20\hat{i} + 15\hat{j}) = 2\hat{i} + 1.5\hat{j} kg⋅\cdotm/s (Statement A is correct)

Force F⃗=ma⃗=0.1(20i^+30j^)=2i^+3j^\vec{F} = m\vec{a} = 0.1(20\hat{i} + 30\hat{j}) = 2\hat{i} + 3\hat{j} N (Statement B is correct)

Angular momentum L⃗=r⃗×p⃗=(10i^+5j^)×(2i^+1.5j^)=(10×1.5−5×2)k^=5k^\vec{L} = \vec{r} \times \vec{p} = (10\hat{i} + 5\hat{j}) \times (2\hat{i} + 1.5\hat{j}) = (10 \times 1.5 - 5 \times 2)\hat{k} = 5\hat{k} J⋅\cdots (Statement C is incorrect)

Torque τ⃗=r⃗×F⃗=(10i^+5j^)×(2i^+3j^)=(10×3−5×2)k^=20k^\vec{\tau} = \vec{r} \times \vec{F} = (10\hat{i} + 5\hat{j}) \times (2\hat{i} + 3\hat{j}) = (10 \times 3 - 5 \times 2)\hat{k} = 20\hat{k} N⋅\cdotm (Statement D is correct)

Therefore, statements A, B, and D are correct.

Answer: A, B and D only
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