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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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