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6.System of Particles and Rotational Motion
hard
A heavy iron bar, of weight $\mathrm{W}$ is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle $\theta$ with the horizontal. The weight experienced by the person is:
A
$\frac{W}{2}$
B
$W$
C
$\mathrm{W} \cos \theta$
D
$\mathrm{W} \sin \theta$
(JEE MAIN-2024)
Solution

$R=$ net reaction force by shoulder
Balancing torque about pt of contact on ground:
$\mathrm{W}\left(\frac{\mathrm{L}}{2} \cos \theta\right)=\mathrm{R}(\mathrm{L} \cos \theta)$
$\Rightarrow \mathrm{R}=\frac{\mathrm{W}}{2}$
Standard 11
Physics