Skip to main content

A body of mass 5 ~kg under the action of constant force F=F_x i+F_y j has velocity at t=0 ~s as overrightarrow v=(6 i-2 j m / s) and at t=10 ~s as…

JEE Main 2014 — Physics Mechanics

2014mcqmedium

A body of mass 5 kg5 \mathrm{~kg}5 kg under the action of constant force F⃗=Fxi^+Fyj^\vec{F}=F_x \hat{i}+F_y \hat{j}F=Fx​i^+Fy​j^​ has velocity at t=0 s\mathrm{t}=0 \mathrm{~s}t=0 s as v→=(6i^−2j^m/s)\overrightarrow{\mathrm{v}}=(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s})v=(6i^−2j^​m/s) and at t=10 s\mathrm{t}=10 \mathrm{~s}t=10 s as v→=+6j^m/s\overrightarrow{\mathrm{v}}=+6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}v=+6j^​m/s. The force F→\overrightarrow{\mathrm{F}}F is:

Official previous-year question

Held on 11 Apr 2014 · Verified 6 Jul 2026.

Options

  1. A

    (−3i^+4j^)N(-3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}(−3i^+4j^​)N

  2. B

    (−35i^+45j^)N\left(-\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right) N(−53​i^+54​j^​)N

  3. C

    (3i^−4j^)N(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \mathrm{N}(3i^−4j^​)N

  4. D

    (35i^−45j^)N\left(\frac{3}{5} \hat{\mathrm{i}}-\frac{4}{5} \hat{\mathrm{j}}\right) \mathrm{N}(53​i^−54​j^​)N

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Mechanics in other years

More JEE Main Mechanics questions

If $\epsilon, E$ and $t$ represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\epsilon E}{t}$ will be :

2026easy

When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\%$.

2026medium

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is _______ m/s$^2$.

2026hard

Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$. ![JEE Main 2026 Physics, Rotational Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/6983c19b8bfdbb24356396d6/question_1__q_6983c19b8bfdbb24356396d6__cdn-question-pool.getmarks.app__23S2_q_48_1__7376098484_final_ppt_sync.png)

2026hard

A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is $\_\_\_\_$. ($\pi=22 / 7$)

2026medium

Other Physics topics for JEE Main

Keep practicing Mechanics

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

Browse all Physics topics