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A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fit inside the tube. The speed of block is 22 m/s, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The magnitude of the work done by the tube on the block is ____ J. (Given : \left.g=10 \mathrm{~m} / \mathrm{s}^2\right)
                                                             

Option: 1

245


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{R}_{\mathrm{arg}}=15 \mathrm{~cm}=.15 \mathrm{~m}
By WET
\begin{aligned} & \mathrm{W}_{\mathrm{f}}+\mathrm{W}_{\text {gravily }}=\Delta \mathrm{K}=\mathrm{Kf}-\mathrm{Ki} \\ & \mathrm{W}_{\mathrm{f}}+10 \times 3=0-\frac{1}{2} \times 1 \times(22)^2 \\ & \mathrm{~W}_{\mathrm{f}}=-3-\frac{484}{2}=3-242=-245 \end{aligned}
Work by friction = – 245
 

Posted by

Kuldeep Maurya

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