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JEE MAIN 2019
09-04-2019 - S1
Question
A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its $\left(\frac{1}{n}\right)^{t h}$ part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:
Select the correct option:
A
$\frac{\mathrm{MgL}}{\mathrm{n}^2}$
B
nMgL
C
$\frac{M g L}{2 n^2}$
D
$\frac{2 \mathrm{MgL}}{\mathrm{n}^2}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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