Consider a Ushaped frame with a sliding wire of length l and mass 'm' on its arm. It is dipped in a soap solution, taken out and placed in vertical position as shown in figure.Choose maximum value of m so that wire does not descend : (Surface tension of soap solution is S)

\frac{2 S l}{g}

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\frac{S l}{g}

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\frac{S l}{2 g}

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\frac{S l}{4 g}

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Solution

The correct option is A $\frac{2 S l}{g}$
For wire to be in equilibrium,
$m g = 2 S l$
$m = \frac{2 S l}{g}$
For mass less than $\frac{2 S l}{g}$ it will move upward
So the maximum mass should be $\frac{2 S l}{g}$

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Consider a Ushaped frame with a sliding wire of length l and mass 'm' on its arm. It is dipped in a soap solution, taken out and placed in vertical position as shown in figure.Choose maximum value of m so that wire does not descend : (Surface tension of soap solution is S)

The correct option is A 2SlgFor wire to be in equilibrium, mg=2Slm=2SlgFor mass less than 2Slg it will move upwardSo the maximum mass should be 2Slg

The correct option is A $\frac{2 S l}{g}$
For wire to be in equilibrium,
$m g = 2 S l$
$m = \frac{2 S l}{g}$
For mass less than $\frac{2 S l}{g}$ it will move upward
So the maximum mass should be $\frac{2 S l}{g}$