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Q.

A mass $m = 0.4 k g$ is attached to a vertical rod by mass of two strings of equal length $I = 2 m .$ The system rotates about the axis of the rod with $ω = 4 r a d / \sec,$ as shown. If $T_{1}$ is the tension in the upper string and $T_{2}$ is the tension in the lower string then ( $g = 10 m / s^{2}$ )

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a

$T_{1} = 3 N, T_{2} = 10 N$

b

$T_{1} = T_{2} + 4 N$

c

$T_{1} = T_{2}$

d

$T_{1} = 10, 4 N, T_{2} = 2.4 N$

answer is B.

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Detailed Solution

$\begin{array}{l} (T_{1} + T_{2}) \cos 30 = m r ω^{2} = m l \cos 30 ω^{2} \\ T_{1} + T_{2} = m l ω^{2} ......... (1) \\ (T_{1} - T_{2}) \sin 30 = m g \\ T_{1} - T_{2} = 2 m g .......... (2) \\ s o l v i n g T_{1} = 10.4 \\ T_{2} = 2.4 \end{array}$

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A mass m=0.4kg is attached to a vertical rod by mass of two strings of equal length I=2m. The system rotates about the axis of the rod with ω=4rad/sec, as shown. If T1 is the tension in the upper string and T2 is the tension in the lower string then ( g=10m/s2)