(C) The resultant downward force along the incline is given by $F_{net} = mg(\sin \theta - \mu \cos \theta)$.The normal reaction force is $N = mg \cos

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

(C) The resultant downward force along the incline is given by $F_{net} = mg(\sin \theta - \mu \cos \theta)$.The normal reaction force is $N = mg \cos \theta$.According to the problem,the normal reaction is twice the resultant downward force:$mg \cos \theta = 2 \times mg(\sin \theta - \mu \cos \theta)$.Dividing both sides by $mg$:$\cos \theta = 2(\sin \theta - 0.5 \cos \theta)$.Expanding the equation:$\cos \theta = 2 \sin \theta - 1.0 \cos \theta$.Rearranging the terms:$2 \cos \theta = 2 \sin \theta$.Dividing by $2 \cos \theta$:$1 = \tan \theta$.Therefore,$\theta = \tan^{-1}(1) = 45^o$.

Class 11 Physics Newton's Laws of Motion and Friction Motion (or rest) on Rough Inclined Surface

Medium

$A$ body is sliding down an inclined plane having a coefficient of friction $0.5$. If the normal reaction is twice that of the resultant downward force along the incline,the angle between the inclined plane and the horizontal is ....... $^o$

A
$15$
B
$30$
C
$45$
D
$60$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

Subtopic

Motion (or rest) on Rough Inclined Surface

$A$ block of mass $m$ is at rest with respect to a lift when placed on an inclined plane of inclination $\theta$ inside the lift. If the lift moves upward at a constant velocity $v$,then the work done by the friction force on the block in time $t$ is:

Medium

View Solution

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

Difficult

View Solution

$A$ homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is:

Medium

View Solution

$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

Medium

View Solution

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

MediumTS EAMCET 2001

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

$A$ body is sliding down an inclined plane having a coefficient of friction $0.5$. If the normal reaction is twice that of the resultant downward force along the incline,the angle between the inclined plane and the horizontal is ....... $^o$

Similar Questions

$A$ block of mass $m$ is at rest with respect to a lift when placed on an inclined plane of inclination $\theta$ inside the lift. If the lift moves upward at a constant velocity $v$,then the work done by the friction force on the block in time $t$ is:

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

$A$ homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is:

$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

Vedclass Test Series

Exam Paper Generator

Online Exam Module