QJEEM22Ph8f7b: JEE Main 2022 Physics - Class 11 JEEM22Ph8f7b Solution: A mass of 10 kg is suspended vertically by a rope of length 5 m …

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:

$\therefore x = 3$