Consider a force $\vec{F}=-x\hat{i}+y\hat{j}$ . The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is : (all quantities are in SI units)![JEE Main 2020 Physics, Work Power Energy — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Work_Power_Energy/648b5a69417cc3fb48d66d7b/question_1__q_648b5a69417cc3fb48d66d7b__cdn-question-pool.getmarks.app__37e2ddc8-042c-4a98-8294-be89e21fd9a3-image__fdc08e43cd_final_ppt_sync.png)

Consider a force F=-x i+y j .… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Mechanics

2020mcqmedium

Consider a force $\vec{F}=-x\hat{i}+y\hat{j}$ . The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is : (all quantities are in SI units) JEE Main 2020 Physics, Work Power Energy — question figure

Official previous-year question

Held on 9 Jan 2020 · Verified 6 Jul 2026.

Options

A
$2$
B
$\frac{1}{2}$
C
$1$
D
$\frac{3}{2}$

Did you get this right?

Solution

It is given that a force $\vec{F}=-x\hat{i}+y\hat{j}$ acts on a particle. The particle is moved from point $A(1,0)$ to $B(0,1)$ along the line segment.

Work done by a variable force on the particle,

$W=\int \vec{F}\cdot \vec{dr}=\int \vec{F}\cdot (dx\hat{i}+dy\hat{j})$

(In two dimension, $\vec{dr}=dx\hat{i}+dy\hat{j}$

and it is given $\vec{F}=-x\hat{i}+y\hat{j}$ )

$W=\int (-x\hat{i}+y\hat{j})\cdot (dx\hat{i}+dy\hat{j})$

$\displaystyle =\int -xdx+ydy=\int -xdx+\int ydy$ .

As the particle is displaced from $A(1,0)$ to $B(0,1)$ , $x$ varies from $1$ to $0$ and $y$ varies from $0$ to $1$ .

So, work will be,

$W=-{\int }_{1}^{0}xdx+{\int }_{0}^{1}ydy=-{[\frac{{x}^{2}}{2}]}_{1}^{0}+{[\frac{{y}^{2}}{2}]}_{0}^{1}$

$=\frac{1}{2}(0+{(1)}^{2}+{(1)}^{2}-0)=1J$ .

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

Solution

It is given that a force $\vec{F}=-x\hat{i}+y\hat{j}$ acts on a particle. The particle is moved from point $A(1,0)$ to $B(0,1)$ along the line segment.

Work done by a variable force on the particle,

$W=\int \vec{F}\cdot \vec{dr}=\int \vec{F}\cdot (dx\hat{i}+dy\hat{j})$

(In two dimension, $\vec{dr}=dx\hat{i}+dy\hat{j}$

and it is given $\vec{F}=-x\hat{i}+y\hat{j}$ )

$W=\int (-x\hat{i}+y\hat{j})\cdot (dx\hat{i}+dy\hat{j})$

$\displaystyle =\int -xdx+ydy=\int -xdx+\int ydy$ .

As the particle is displaced from $A(1,0)$ to $B(0,1)$ , $x$ varies from $1$ to $0$ and $y$ varies from $0$ to $1$ .

So, work will be,

$W=-{\int }_{1}^{0}xdx+{\int }_{0}^{1}ydy=-{[\frac{{x}^{2}}{2}]}_{1}^{0}+{[\frac{{y}^{2}}{2}]}_{0}^{1}$

$=\frac{1}{2}(0+{(1)}^{2}+{(1)}^{2}-0)=1J$ .

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics