A solid sphere rotates about a vertical axis on frictionless bear… - Vidyadip

MCQ

A solid sphere rotates about a vertical axis on frictionless bearing. A massless cord passes around the equator of sphere, then passes through over a solid cylinder and then is connected to block of mass $M$ as shown in figure. If the system is released from rest then the speed acquired by block after it has fallen through distance $h$ is

A
$\sqrt {\frac{{10gh}}{9}} $
✓
$\sqrt {\frac{{20gh}}{19}} $
C
$\sqrt {\frac{{15gh}}{139}} $
D
$\sqrt {\frac{{18gh}}{159}} $

✓

Answer

Correct option: B.

$\sqrt {\frac{{20gh}}{19}} $

b
$\mathrm{T}_{1} \mathrm{R}=\frac{2}{5} \mathrm{MR}^{2} \times \frac{\mathrm{a}}{\mathrm{R}}$

$\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \mathrm{R}=\frac{1}{2} \mathrm{MR}^{2} \frac{\mathrm{a}}{\mathrm{R}}$

$\mathrm{Mg}-\mathrm{T}_{2}=\mathrm{Ma}$

$\mathrm{Mg}=\mathrm{Ma}\left(\frac{2}{5}+\frac{1}{2}+1\right)$

$g=a \times \frac{19}{10}$

$a=\frac{10 g}{19}$

$v=\sqrt{2 \mathrm{as}}$

$v=\sqrt{2 \times \frac{10 g h}{19}}$

$v=\sqrt{\frac{20 g h}{19}}$

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