Two weight $w_{1}$ and $w_{2}$ are connected by a light thread which passes over a light smooth pulley. If the pulley is raised upward with an acceleration equal to g then tension in the thread will be

\frac{2 w_{1} w_{2}}{w_{1} + w_{2}}

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\frac{w_{1} w_{2}}{w_{1} + w_{2}}

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\frac{4 w_{1} w_{2}}{w_{1} + w_{2}}

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\frac{4 w_{1} w_{2}}{w_{1} - w_{2}}

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Solution

The correct option is A $\frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$
According to diagram B

$2 w_{1} - T = m_{1} a^{'} . . . . (I)$

$T - 2 w_{2} = m_{2} a^{'} . . . . . (I I)$

After solving,

$m_{2} (2 w_{1} - T) = m_{1} (T - 2 w_{2})$

$m_{2} 2 w_{1} - m_{2} T = m_{1} T - 2 w_{2} m_{1}$
Therefore, the tension will be

$T = \frac{2 m_{2} w_{1} + 2 m_{1} w_{2}}{m_{1} + m_{2}}$

$T = \frac{2 \frac{w_{2}}{g} \times w_{1} + 2 \frac{w_{1}}{g} \times w_{2}}{\frac{w_{1}}{g} + \frac{w_{2}}{g}}$

$T = \frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$
Hence, the tension in thread is $\frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$

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Two weight w1 and w2 are connected by a light thread which passes over a light smooth pulley. If the pulley is raised upward with an acceleration equal to g then tension in the thread will be

The correct option is A 4w1w2w1+w2According to diagram B 2w1−T=m1a′....(I) T−2w2=m2a′.....(II) After solving, m2(2w1−T)=m1(T−2w2) m22w1−m2T=m1T−2w2m1 Therefore, the tension will be T=2m2w1+2m1w2m1+m2 T=2w2g×w1+2w1g×w2w1g+w2g T=4w1w2w1+w2 Hence, the tension in thread is 4w1w2w1+w2

Two weight $w_{1}$ and $w_{2}$ are connected by a light thread which passes over a light smooth pulley. If the pulley is raised upward with an acceleration equal to g then tension in the thread will be