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Byju's Answer

Standard XII

Physics

Velocity of COM

Three blocks ...

Question

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings as shown on a friction less table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 k g, m_{2} = 6 k g a n d m_{3} = 4 k g$ , the tension $T_{2}$ will be

20 N

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40 N

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10 N

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32 N

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Solution

The correct option is D
32 N

$T_{2} = (m_{1} + m_{2}) \times \frac{T_{3}}{m_{1} + m_{2} + m_{3}} = \frac{(10 + 6) \times 40}{20} = 32 N$

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Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be:

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Three blocks of masses m1,m2 and m3 are connected by massless strings as shown on a friction less table. They are pulled with a force T3=40 N. If m1=10 kg, m2=6 kg and m3=4 kg, the tension T2 will be

The correct option is D 32 N T2=(m1+m2)×T3m1+m2+m3=(10+6)×4020=32 N

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings as shown on a friction less table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 k g, m_{2} = 6 k g a n d m_{3} = 4 k g$ , the tension $T_{2}$ will be

The correct option is D
32 N

$T_{2} = (m_{1} + m_{2}) \times \frac{T_{3}}{m_{1} + m_{2} + m_{3}} = \frac{(10 + 6) \times 40}{20} = 32 N$