JEE MAIN - Physics (2024 - 9th April Evening Shift - No. 19)
A $$1 \mathrm{~kg}$$ mass is suspended from the ceiling by a rope of length $$4 \mathrm{~m}$$. A horizontal force '$$F$$' is applied at the mid point of the rope so that the rope makes an angle of $$45^{\circ}$$ with respect to the vertical axis as shown in figure. The magnitude of $$F$$ is :
(Assume that the system is in equilibrium and $$g=10 \mathrm{~m} / \mathrm{s}^2$$)
_9th_April_Evening_Shift_en_19_1.png)
1 N
10 N
$$\frac{10}{\sqrt{2}} N$$
$$\frac{1}{10 \times \sqrt{2}} N$$
Explanation
$$\begin{aligned}
& T_1 \cos 45=F \\
& \text { and, } T_1 \sin 45=m g \\
& \therefore F=m g \\
& \Rightarrow F=10
\end{aligned}$$
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