Two weights $w_{1}$ and $w_{2}$ are suspended to the two ends of a string passing over a smooth pulley. If the pulley is pulled up with an acceleration $g$ , then the tension in the string is:

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

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

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Solution

The correct option is A $\frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$
When pulley is fixed,

$m_{1} a = m_{1} g - T . . . . (1)$

$m_{2} a = T - m_{2} g . . . . . (2)$

$(1) + (2), a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$

Using (1), we will get , $T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g$

When pulley will move upward with acceleration $g$ , then $T$
becomes, $T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} 2 g$

Now, $m_{1} = \frac{w_{1}}{g}, m_{2} = \frac{w_{2}}{g}$

$T = \frac{\frac{w_{1}}{g} \times \frac{w_{2}}{g}}{\frac{w_{1}}{g} + \frac{w_{2}}{g}} \times 4 g$

$∴ T = \frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$

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Two weights w1 and w2 are suspended to the two ends of a string passing over a smooth pulley. If the pulley is pulled up with an acceleration g, then the tension in the string is:

The correct option is A 4w1w2w1+w2When pulley is fixed, m1a=m1g−T....(1)m2a=T−m2g.....(2)(1)+(2),a=m1−m2m1+m2gUsing (1), we will get , T=2m1m2m1+m2gWhen pulley will move upward with acceleration g , then T becomes, T=2m1m2m1+m22gNow, m1=w1g,m2=w2gT=w1g×w2gw1g+w2g×4g∴T=4w1w2w1+w2

Two weights $w_{1}$ and $w_{2}$ are suspended to the two ends of a string passing over a smooth pulley. If the pulley is pulled up with an acceleration $g$ , then the tension in the string is: