To solve this problem, we need to understand the relationship between the tension in the string and the speed of the bob whirling in a horizontal circle. The centripetal force acting on the bob is provided by the tension in the string, and it can be given by the formula:

$ F = \frac{m v^2}{r} $

where:

• F is the centripetal force (or tension in the string)
• m is the mass of the bob
• v is the tangential speed of the bob
• r is the radius of the circle

According to the problem, the initial speed of the bob is $10 \mathrm{~rpm}$, and the radius is kept constant. Let’s denote the initial tension in the string as $T_1$ and the new tension as $T_2$. Given that the tension in the string is quadrupled, we have:

$ T_2 = 4 T_1 $

Also, the centripetal force can be written in terms of tension:

$$ T_1 = \frac{m v_1^2}{r} $$

$$ T_2 = \frac{m v_2^2}{r} $$

By substituting $T_2 = 4 T_1$ into the second equation, we get:

$$ 4 T_1 = \frac{m v_2^2}{r} $$

We know from the first equation that:

$$ T_1 = \frac{m v_1^2}{r} $$

Substituting this into our equation for $T_2$:

$$ 4 \left(\frac{m v_1^2}{r}\right) = \frac{m v_2^2}{r} $$

The masses and radii cancel out, leaving us with:

$ 4 v_1^2 = v_2^2 $

Taking the square root of both sides:

$ v_2 = 2 v_1 $

The initial speed $v_1$ is given as $10 \mathrm{~rpm}$. Therefore, the new speed $v_2$ is:

$$ v_2 = 2 \times 10 \mathrm{~rpm} = 20 \mathrm{~rpm} $$

Hence, the new speed of the bob is 20 rpm. The correct answer is Option A.