(B) The length of the hanging part is $l = L/n$.Since the cable is uniform,the mass of the hanging part is $m = (M/L) \times l = M/n$.The center of ma

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

(B) The length of the hanging part is $l = L/n$.Since the cable is uniform,the mass of the hanging part is $m = (M/L) \times l = M/n$.The center of mass of the hanging part is at a distance $h_{COM} = l/2 = L/(2n)$ below the edge of the surface.To lift the hanging part onto the surface,we must raise its center of mass to the level of the surface.The work done $W$ is equal to the change in potential energy of the hanging part:$W = m \times g \times h_{COM}$$W = (M/n) \times g \times (L/(2n))$$W = \frac{MgL}{2n^2}$

Class 11 Physics Work, Energy and Power Potential Energy

Difficult

$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

JEE MAIN 2019 All JEE MAIN

A
$nMgL$
B
$\frac{MgL}{2n^2}$
C
$\frac{2MgL}{n^2}$
D
$\frac{MgL}{n^2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Work, Energy and Power

Subtopic

Potential Energy

Previous Year

JEE MAIN 2019 Paper All JEE MAIN years →

The potential energy between two atoms is given by the function $U(r) = a/r^{12} - b/r^{6}$. Find the equilibrium distance between them.

Difficult

View Solution

The potential energy function (in $J$) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$. Here $x, y$ and $z$ are in meters. The magnitude of the $x$-component of the force (in $N$) acting on the particle at point $P(1, 2, 3) \ m$ is:

DifficultJEE MAIN 2024

View Solution

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

Difficult

View Solution

In the equation $W = -mgh$,what does the negative sign indicate?

Medium

View Solution

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium 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

$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

Similar Questions

The potential energy between two atoms is given by the function $U(r) = a/r^{12} - b/r^{6}$. Find the equilibrium distance between them.

The potential energy function (in $J$) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$. Here $x, y$ and $z$ are in meters. The magnitude of the $x$-component of the force (in $N$) acting on the particle at point $P(1, 2, 3) \ m$ is:

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

In the equation $W = -mgh$,what does the negative sign indicate?

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module