A stone of mass $1 ext{ kg}$ tied to a light inextensible string of length $L = \frac{10}{3} ext{ m}$ is w

Question

A stone of mass $1 	ext{ kg}$ tied to a light inextensible string of length $L = \frac{10}{3} 	ext{ m}$ is whirling in a circular path of radius $L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if $g$ is taken to be $10 	ext{ m/s}^2$, the speed of the stone at the highest point of the circle is:

TopperSquare Expert · Accepted Answer

The correct answer is Option (D): $10 	ext{ m/s}$. Let $v_1$ be the speed at the lowest point and $v_2$ be the speed at the highest point. 
By conservation of mechanical energy from the highest to the lowest point:
$$\frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2 + mg(2L)$$
$$v_1^2 = v_2^2 + 4gL$$

The maximum tension ($T_{max}$) occurs at the lowest point of the circular path:
$$T_{max} = \frac{mv_1^2}{L} + mg$$
Substitute $v_1^2 = v_2^2 + 4gL$ into the equation:
$$T_{max} = \frac{m(v_2^2 + 4gL)}{L} + mg = \frac{mv_2^2}{L} + 4mg + mg = \frac{mv_2^2}{L} + 5mg$$

The minimum tension ($T_{min}$) occurs at the highest point:
$$T_{min} = \frac{mv_2^2}{L} - mg$$

Given the ratio of maximum to minimum tension is 4:
$$\frac{T_{max}}{T_{min}} = 4$$
$$\frac{\frac{mv_2^2}{L} + 5mg}{\frac{mv_2^2}{L} - mg} = 4$$
$$mv_2^2 + 5mgL = 4(mv_2^2 - mgL)$$
$$mv_2^2 + 5mgL = 4mv_2^2 - 4mgL$$
$$3mv_2^2 = 9mgL$$
$$v_2^2 = 3gL$$

Given $g = 10 	ext{ m/s}^2$ and $L = \frac{10}{3} 	ext{ m}$, substitute these values:
$$v_2 = \sqrt{3 	imes 10 	imes \frac{10}{3}}$$
$$v_2 = \sqrt{100} = 10 	ext{ m/s}$$

Thus, the speed of the stone at the highest point is $10 	ext{ m/s}$.

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