A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle with a uniform angular velocity of 2 rad/s. The tension of the string is (in N)

The correct option is 4.

When a stone is whirled in a horizontal circle, the tension in the string acts as the centripetal force required to maintain the circular motion of the stone.

The formula for centripetal force (which is equal to the tension, $T$) in terms of angular velocity is:

$T=m{\omega }^{2}r$

Where:
• $m$ is the mass of the stone
• $\omega$ is the angular velocity
• $r$ is the radius of the circular path (equivalent to the length of the string)

From the given problem, we have the following values:
• Mass, $m=1\text{kg}$
• Length of the string (radius), $r=1\text{m}$
• Angular velocity, $\omega =2\text{rad/s}$

Substituting these values into the formula:

$T=1\times 2^2\times 1$

Calculating the value:

$T=1\times 4\times 1$

$T=4\text{N}$

Thus, the tension of the string is 4 N.