(C) Let $T$ be the tension in the string.The maximum vertical force on the clamp is given by $T \sin 30^{\circ} = 40 \,N$.Substituting the value of $\

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

(C) Let $T$ be the tension in the string.The maximum vertical force on the clamp is given by $T \sin 30^{\circ} = 40 \,N$.Substituting the value of $\sin 30^{\circ} = 1/2$:$T \cdot (1/2) = 40$$T = 80 \,N$Now,consider the free-body diagram of the monkey of mass $m = 5 \,kg$ climbing with acceleration $a$.The forces acting on the monkey are tension $T$ upwards and weight $mg$ downwards.According to Newton's second law: $T - mg = ma$$a = (T - mg) / m$Substituting the values $T = 80 \,N$,$m = 5 \,kg$,and $g = 10 \,ms^{-2}$:$a = (80 - 5 \times 10) / 5$$a = (80 - 50) / 5$$a = 30 / 5 = 6 \,ms^{-2}$Thus,the maximum acceleration is $6 \,ms^{-2}$.

Class 11 Physics Newton's Laws of Motion and Friction Tension Force and Pulley Block System

Difficult

One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of $30^{\circ}$ with the ground. The clamp can tolerate a vertical force of $40 \,N$. If a monkey of mass $5 \,kg$ were to climb up the rope,then the maximum acceleration in the upward direction with which it can climb safely is $\left(g=10 \,ms^{-2}\right)$ (in $\,ms^{-2}$)

AP EAMCET 2018 All AP EAMCET

A
$2$
B
$4$
C
$6$
D
$8$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

Subtopic

Tension Force and Pulley Block System

Previous Year

AP EAMCET 2018 Paper All AP EAMCET years →

$A$ bob of mass $m$ is hanging from a pulley inside a car by a string. The other end of the string is held by a person standing in the car. The car is moving with a constant horizontal acceleration '$a$' as shown in the figure. The other end of the string is pulled with a constant acceleration '$a$' vertically downward. The tension in the string is equal to-

Medium

View Solution

$A$ student calculates the acceleration of $m_1$ in the figure shown as $a_1 = \frac{(m_1 - m_2)g}{m_1 + m_2}$. Which assumption is not required to do this calculation?

Difficult

View Solution

$A$ block of mass $M$ is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force $F$. The kinetic energy of the block increases by $20\,J$ in $1\,s$.

Medium

View Solution

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium,the angle $\theta$ should be:

MediumMHT CET 2024

View Solution

$A$ light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (where $m_2 > m_1$). If the acceleration of the system is $\frac{g}{\sqrt{2}}$,then the ratio of the masses $\frac{m_1}{m_2}$ is:

DifficultJEE MAIN 2024

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

Similar Questions

$A$ student calculates the acceleration of $m_1$ in the figure shown as $a_1 = \frac{(m_1 - m_2)g}{m_1 + m_2}$. Which assumption is not required to do this calculation?

$A$ block of mass $M$ is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force $F$. The kinetic energy of the block increases by $20\,J$ in $1\,s$.

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium,the angle $\theta$ should be:

$A$ light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (where $m_2 > m_1$). If the acceleration of the system is $\frac{g}{\sqrt{2}}$,then the ratio of the masses $\frac{m_1}{m_2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module