By splitting a total mass $M$ in two masses $M_{1}$ and $M_{2}$ , the maximum value of tension that can be produced in the string $1$ as shown in the figure is:-

\frac{M g}{2}

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M g

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\frac{M g}{4}

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4 M g

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Solution

The correct option is A $\frac{M g}{2}$
Given,

$M = M_{1} + M_{2}$ . . . .(1)

Free body diagram,

Tension on string due to $M_{1}$

$T_{1} = M_{1} g$

Tension produce due to $M_{2}$

$T_{2} = M_{2} g$

The given pulley and string are ideal $T_{1} = T_{2} = T$ and both are produced in the upward direction.

The maximum tension produced in the string is

$2 T = M_{1} g + M_{2} g$

$T = \frac{(M_{1} + M_{2}) g}{2}$

$T = \frac{M g}{2}$ (from equation 1)

The correct option is A.

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By splitting a total mass M in two masses M1 and M2, the maximum value of tension that can be produced in the string 1 as shown in the figure is:-

By splitting a total mass $M$ in two masses $M_{1}$ and $M_{2}$ , the maximum value of tension that can be produced in the string $1$ as shown in the figure is:-