Gujarati
Hindi
6.System of Particles and Rotational Motion
hard

Two vertical walls are separated by a distance of $2\  m$. Wall $A$ is smooth while wall $B$ is rough with a coefficient of friction $0. 5$. A uniform rod is placed between them as shown. The length of longest rod that can be placed between walls is equal to

A

$2\  m$

B

$2\sqrt 2\ m$

C

$\sqrt 5\ m$

D

$\frac{{\sqrt {17} }}{2}\ m$

Solution

$\tau_{\mathrm{com}}=0$

$\mathrm{N} \cdot \frac{\ell}{2} \sin \theta .2=\mu \mathrm{N} \frac{\ell}{2} \cos \theta$

$\Rightarrow \tan \theta=\frac{1}{4}$

$\cos \theta=\frac{2}{\ell}=\frac{-4}{\sqrt{17}}$

$\ell=\frac{\sqrt{17}}{2}$

Standard 11
Physics

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