JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 19)
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $$\theta$$ = tan$$-$$1 (x $$\times$$ 10$$-$$1). The value of x is ____________.
(Given, g = 10 m/s2)
Explanation
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The vertical component of the tension, $$T\cos\theta$$, balances the weight of the mass:
$$T\cos \theta = mg$$
Since $$ mg = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}$$:
$$T\cos \theta = 100\, \text{N}$$ ...... (i)
The horizontal component of the tension is given by:
$$T\sin \theta = 30\, \text{N}$$ ........ (ii)
Dividing equation (ii) by equation (i):
$$ \frac{T\sin \theta}{T\cos \theta} = \frac{30}{100}$$
Therefore:
$$ \tan \theta = \frac{3}{10}$$
Hence, the value of x is:
$$ \therefore x = 3$$
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