Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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)

best_answer

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

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question