$A$ 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
- A$nMgL$
- B$\frac{MgL}{2n^2}$
- C$\frac{2MgL}{n^2}$
- D$\frac{MgL}{n^2}$
Explore More
Similar Questions
$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)
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$A$ particle is released from rest at the point $x = a$ and moves along the $x$ axis subject to the potential energy function $U(x)$ shown. The particle:
Medium
View SolutionWhat does a body at rest possess?
Easy
View Solution$A$ particle of mass $1 \ kg$ is free to move along the $x$-axis. Its potential energy is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the particle is $2 \ J$,find the maximum speed of the particle.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo