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Standard XII

Physics

Energy Conservation Method

A uniform rod...

Question

A uniform rod of length L and weight W is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached a distance from the right end. The tension in the second rope is

W/2

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W/3

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W/4

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2W/3

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Solution

The correct option is D
2W/3

$T_{1} + T_{2} = m g \sum C_{A} = 0 T_{1} (0) - m g \frac{L}{2} + T_{2} \frac{3 L}{4} = 0 T_{2} \frac{3 L}{4} = m g \frac{L}{2} T_{2} = \frac{2 m g}{3} = \frac{2 w}{3}$

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A uniform rod of length L and weight W is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached a distance from the right end. The tension in the second rope is

The correct option is D 2W/3 T1+T2=mg∑CA=0T1(0)−mgL2+T23L4=0T23L4=mgL2T2=2mg3=2w3

The correct option is D
2W/3

$T_{1} + T_{2} = m g \sum C_{A} = 0 T_{1} (0) - m g \frac{L}{2} + T_{2} \frac{3 L}{4} = 0 T_{2} \frac{3 L}{4} = m g \frac{L}{2} T_{2} = \frac{2 m g}{3} = \frac{2 w}{3}$