In the above problem, given that $M_{2} = 2 M_{1}$ and the tension in the string is T.If the positions of the masses are reversed, the tension in the string will be

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Solution

The correct option is C
Tension in the string $T = \frac{m_{1} + m_{2} (1 + s i n θ)}{m_{1} + m_{2}} g$
Even if the masses are reversed the tensin in the string remains same.

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In the above problem, given that M2=2M1 and the tension in the string is T.If the positions of the masses are reversed, the tension in the string will be

The correct option is C Tension in the string T=m1+m2(1+sinθ)m1+m2g Even if the masses are reversed the tensin in the string remains same.

In the above problem, given that $M_{2} = 2 M_{1}$ and the tension in the string is T.If the positions of the masses are reversed, the tension in the string will be

The correct option is C
Tension in the string $T = \frac{m_{1} + m_{2} (1 + s i n θ)}{m_{1} + m_{2}} g$
Even if the masses are reversed the tensin in the string remains same.