Browse by Stream
Discuss Doubts
Home
Search for Exam, Articles, Questions
get app
Go To Your Study Dashboard
  1. Home
  2. Help me please, - Work Energy and Power - JEE Main
# Engineering

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface that its \left [ \frac{1}{n} \right ]^{th}part is hanging below the edge of the surface. To lift the hanging part of the cable upto the srface, the work done should be :

  • Option 1)

      \frac{MgL}{2n^{2}}

  • Option 2)

      \frac{MgL}{n^{2}}

  • Option 3)

     \frac{2MgL}{n^{2}}

  • Option 4)

     nMgL

 

Answers (1)
S solutionqc

Total length of rod=L

Length hanging=\frac{L}{n}

initial P.E = u_i=\left ( \frac{M}{n} \right )\left ( \frac{L}{n}\right )\frac{1}{2}g=\frac{MLg}{2n^{2}}

final P.E=u_f=\left ( \frac{M}{n} \right )g\left ( \frac{L}{n} \right )=\frac{MgL}{n^{2}}

               W=u_{f}-u_{i}

                   =\frac{MgL}{2n^{2}}

                


Option 1)

  \frac{MgL}{2n^{2}}

Option 2)

  \frac{MgL}{n^{2}}

Option 3)

 \frac{2MgL}{n^{2}}

Option 4)

 nMgL

Similar Questions

Preparation Products

Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions