Solve this following

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Question:
A uniform cylinder of mass $M$ and radius $R$ is to be pulled over a step of height a $(a

$\mathrm{Mg} \sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{R}^{2}}}$
$\operatorname{Mg} \sqrt{\left(\frac{R}{R-a}\right)^{2}-1}$
$\mathrm{Mg} \frac{\mathrm{a}}{\mathrm{R}}$
$\operatorname{Mg} \sqrt{1-\left(\frac{\mathrm{R}-\mathrm{a}}{\mathrm{R}}\right)^{2}}$

Correct Option: , 4

Solution:

$(\tau)_{\mathrm{P}}=0$

F.R. $-\mathrm{mgx}=0$

$x=\sqrt{R^{2}-(R-a)^{2}}$

$F=m g \frac{X}{R}$

$\mathrm{F}=\mathrm{mg} \sqrt{1-\left(\frac{\mathrm{R}-\mathrm{a}}{\mathrm{R}}\right)^{2}}$

$=$ minimum value of force to pull