(A) The formula for the magnitude of torque is given by $\tau = rF \sin \theta$,where $r$ is the position vector magnitude,$F$ is the force magnitude,

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The formula for the magnitude of torque is given by $\tau = rF \sin \theta$,where $r$ is the position vector magnitude,$F$ is the force magnitude,and $\theta$ is the angle between them.Given: $\tau = 2.5\,Nm$,$F = 1\,N$,and $r = 5\,m$.Substituting the values into the formula: $2.5 = 5 \times 1 \times \sin \theta$.This simplifies to $\sin \theta = \frac{2.5}{5} = 0.5$.Since $\sin \theta = 0.5$,the angle $\theta = \arcsin(0.5) = \frac{\pi}{6}$ radians.

Class 11 Physics System of Particles and Rotational Motion Torque and Couple

Medium

The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

JEE MAIN 2019 All JEE MAIN

A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{8}$
D
$\frac{\pi}{4}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

System of Particles and Rotational Motion

Subtopic

Torque and Couple

Previous Year

JEE MAIN 2019 Paper All JEE MAIN years →

$A$ couple produces:

Easy

View Solution

Find the torque of a force $\overrightarrow F = 2\hat i + \hat j + 4\hat k$ acting at the point $\overrightarrow r = 7\hat i + 3\hat j + \hat k$.

Easy

View Solution

$A$ force $\vec{F} = 4\hat{i} - 3\hat{j} + 4\hat{k} \, N$ acts on a particle at a position $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ from the origin. The torque acting on the particle is:

Medium

View Solution

If $\vec{F} = (4\hat{i} - 10\hat{j})$ and $\vec{r} = (5\hat{i} - 3\hat{j})$,then calculate the torque $\vec{\tau} = \vec{r} \times \vec{F}$. (in $hat{k}$)

Easy

View Solution

$A$ force $\vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k}$ is applied at a point with position vector $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The torque about the point with position vector $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$ is:

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

Similar Questions

$A$ couple produces:

Find the torque of a force $\overrightarrow F = 2\hat i + \hat j + 4\hat k$ acting at the point $\overrightarrow r = 7\hat i + 3\hat j + \hat k$.

$A$ force $\vec{F} = 4\hat{i} - 3\hat{j} + 4\hat{k} \, N$ acts on a particle at a position $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ from the origin. The torque acting on the particle is:

If $\vec{F} = (4\hat{i} - 10\hat{j})$ and $\vec{r} = (5\hat{i} - 3\hat{j})$,then calculate the torque $\vec{\tau} = \vec{r} \times \vec{F}$. (in $hat{k}$)

$A$ force $\vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k}$ is applied at a point with position vector $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The torque about the point with position vector $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module