At t=0 , a body of mass 100 g starts moving under the influence of a force (5 i +10 j ) N. After 2 s its position is (2x i +5y j ) m. The ratio x:y is ______.

Correct answer: D. 5:4 Mass m = 100 g = 0.1 kg Force F = 5 i + 10 j N Acceleration a = F m = 5 i + 10 j 0.1 = 50 i + 100 j m/s^2 Since the body starts from rest, initial velocity u = 0 . Position after t = 2 s is given by r = u t + (1)/(2) a t^2 r = (1)/(2)(50 i + 100 j )(2)^2 = 100 i + 200 j m Comparing with the given position r = 2x i + 5y j : 2x = 100 x = 50 5y = 200 y = 40 The ratio x:y = 50:40 = 5:4 Answer: 5:4

JEE Main 20264 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 4 April
Shift: Evening Shift
Subject: Physics
Chapter: Laws of Motion

JEE Main 2026 · 4 April · Evening Shift · Physics · Laws of Motion

At t=0 , a body of mass 100 g starts moving under the influence of...

t=0

, a body of mass

100

g starts moving under the influence of a force

(5\hat{i}+10\hat{j})

N. After

2

s its position is

(2x\hat{i}+5y\hat{j})

m. The ratio

x:y

is ______.

Options

A $1:2$
B $2:5$
C $5:2$
D $5:4$

Correct answer

D. $5:4$

Step-by-step solution

Mass

m = 100 \text{ g} = 0.1 \text{ kg}

Force

\vec{F} = 5\hat{i} + 10\hat{j} \text{ N}

Acceleration

\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{5\hat{i} + 10\hat{j}}{0.1} = 50\hat{i} + 100\hat{j} \text{ m/s}^2

Since the body starts from rest, initial velocity

\vec{u} = 0

.

Position after

t = 2 \text{ s}

is given by

\vec{r} = \vec{u}t + \dfrac{1}{2}\vec{a}t^2

\vec{r} = \dfrac{1}{2}(50\hat{i} + 100\hat{j})(2)^2 = 100\hat{i} + 200\hat{j} \text{ m}

Comparing with the given position

\vec{r} = 2x\hat{i} + 5y\hat{j}

2x = 100 \Rightarrow x = 50

5y = 200 \Rightarrow y = 40

The ratio

x:y = 50:40 = 5:4

Answer:

5:4

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