At any instant the velocity of a particle of mass 500 g is $\left(2 t \hat{\mathrm{i}}+3 \mathrm{t}^{2} \hat{\mathrm{j}}\right) \mathrm{ms}^{-1}$. If the force acting on the particle at $\mathrm{t}=1 \mathrm{~s}$ is $(\hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}}) \mathrm{N}$. Then the value of x will be:

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Laws of Motion

At any instant the velocity of a particle of mass 500 g is

(2 t \hat{i} + 3 t^{2} \hat{j}) ms^{- 1}

. If the force acting on the particle at

t = 1 s

(\hat{i} + x \hat{j}) N

. Then the value of x will be:

(A)3✓ Correct
(B)4
(C)6
(D)2

Answer: (A)

Solution

Given velocity:

v = 2 t \hat{i} + 3 t^{2} \hat{j}

m/s

Acceleration:

a = \frac{d v}{d t} = 2 \hat{i} + 6 t \hat{j}

m/s²

At

t = 1

a = 2 \hat{i} + 6 \hat{j}

m/s²

Force:

F = m a = 0.5 \times (2 \hat{i} + 6 \hat{j}) = \hat{i} + 3 \hat{j}

N

Comparing with

F = \hat{i} + x \hat{j}

, we get

x = 3

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option

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