JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 13)
Two bodies of masses $$m_{1}=5 \mathrm{~kg}$$ and $$m_{2}=3 \mathrm{~kg}$$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass $$\mathrm{m}_{1}$$ will be : [Take $$\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$$
_29th_July_Evening_Shift_en_13_1.png)
30 N
40 N
50 N
60 N
Explanation
_29th_July_Evening_Shift_en_13_2.png)
Force on equilibrium, $m_{2} g=m_{1} g \sin \theta$
$\Rightarrow m_2=m_1 \sin \theta$
$$ \Rightarrow $$ $\sin \theta=\frac{m_2}{m_1}=\frac{3}{5}$
Normal force on m1, $R=m_{1} g \cos \theta$
$\Rightarrow \frac{R}{m_{2} g}=\cot \theta$
$\Rightarrow R=3 \times 10 \times \cot \theta=3 \times 10 \times \frac{4}{3} \quad\left(\because \sin \theta=\frac{3}{5}\right)$
$\Rightarrow R=40$ N
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