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Centre of Gravity

Two masses ...

Question

Two masses $m_{1}$ and $m_{2}$ are attached to a spring balance S as shown in Figure.if $m_{1}$ > $m_{2}$ then the reading of spring balance will be

(M_{1} - M_{2})

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(M_{1} {+ M}_{2})

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\frac{2 m_{1} m_{2}}{m_{1} + m_{2}}

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\frac{m_{1} m_{2}}{m_{1} + m_{2}}

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Solution

The correct option is D $\frac{2 m_{1} m_{2}}{m_{1} + m_{2}}$
Since the masses are unequal ( $m_{1} > m_{2}$ ), the whole system will attain a constant acceleration ' $a$ ' (say). At that moment, we calculate the reading of the spring. Since the spring and the pulleys are assumed massless, the tension in the whole system is same.
Writing , force equations for each body individually ,
$m_{1} g - T = m_{1} a$
$T - m_{2} g = m_{2} a$
Simplifying , we have $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}}$
The spring balance measures the reading by tension and hence , Reading is $\frac{2 m_{1} m_{2}}{m_{1} + m_{2}}$

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Two masses m1 and m2 are attached to a spring balance S as shown in Figure.if m1 > m2 then the reading of spring balance will be

Two masses $m_{1}$ and $m_{2}$ are attached to a spring balance S as shown in Figure.if $m_{1}$ > $m_{2}$ then the reading of spring balance will be