To solve this problem, we need to understand the relationship between the tension in the string and the speed of the bob.$\\$The tension $T$ in the string when the bob is moving in a circle is given by the centripetal force formula:$\\ T = m \cdot \omega^2 \cdot r \\$where:$\\$• $m$ is the mass of the bob,$\\$• $\omega$ is the angular velocity in radians per second,$\\$• $r$ is the radius of the circle.$\\$The angular velocity $\omega$ is related to the speed $v$ (in rpm) by:$\\ \omega = \frac{2\pi v}{60} \\$Initially, the tension $T$ is:$\\ T = m \cdot \left(\frac{2\pi v}{60}\right)^2 \cdot r \\$When the speed becomes $2v,$ the new angular velocity $\omega'$ is:$\\ \omega' = \frac{2\pi (2v)}{60} = \frac{4\pi v}{60} \\$The new tension $T'$ is:$\\ T' = m \cdot \left(\frac{4\pi v}{60}\right)^2 \cdot r \\$Simplifying the expression for $T': \\ T' = m \cdot \left(\frac{4\pi v}{60}\right)^2 \cdot r \\ = m \cdot \left(\frac{2 \cdot 2\pi v}{60}\right)^2 \cdot r \\ = m \cdot \left(2 \cdot \frac{2\pi v}{60}\right)^2 \cdot r \\ = 4 \cdot m \cdot \left(\frac{2\pi v}{60}\right)^2 \cdot r \\$Comparing with the initial tension $T: \\ T' = 4 \cdot T \\$Thus, the tension in the string becomes $4T$ when the speed is doubled.$\\$Therefore, the correct option is Option 4.