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Standard XII

Physics

Ropes

Two unequal m...

Question

Two unequal masses (M $_{1}$ and M $_{2}$ ) are connected by a string which passes over a frictional pulley. If M $_{1}$ , M $_{2}$ and the table are frictionless, the acceleration of the masses would be

\frac{M_{1} g}{M_{1} + M_{2}}

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\frac{M_{1} + M_{2}}{M_{1} g}

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\frac{M_{2} g}{M_{1} + M_{2}}

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none of these

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Solution

The correct option is A $\frac{M_{1} g}{M_{1} + M_{2}}$
Let tension developed in the string be $T$ .
Equations of motion are as follows-
$T = M_{2} a$
$M_{1} g - T = M_{1} a$
which gives, $a = g \frac{M_{1}}{M_{1} + M_{2}}$

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Two unequal masses (M1 and M2) are connected by a string which passes over a frictional pulley. If M1, M2 and the table are frictionless, the acceleration of the masses would be

The correct option is A M1gM1+M2Let tension developed in the string be T.Equations of motion are as follows-T=M2aM1g−T=M1awhich gives, a=gM1M1+M2

Two unequal masses (M $_{1}$ and M $_{2}$ ) are connected by a string which passes over a frictional pulley. If M $_{1}$ , M $_{2}$ and the table are frictionless, the acceleration of the masses would be

The correct option is A $\frac{M_{1} g}{M_{1} + M_{2}}$
Let tension developed in the string be $T$ .
Equations of motion are as follows-
$T = M_{2} a$
$M_{1} g - T = M_{1} a$
which gives, $a = g \frac{M_{1}}{M_{1} + M_{2}}$