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Standard XII

Physics

Calculating Friction Using Coeffcients

The maximum v...

Question

The maximum value of the block $m_{2}$ for which the system will remain in equilibrium (coefficient of friction between block $m_{1}$ and plane surface is $μ$ , pulleys are massless) is :

μ \frac{m_{1}}{2}

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\frac{m_{1}}{2}

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μ m_{1}

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2 μ m_{1}

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Solution

The correct option is A $μ \frac{m_{1}}{2}$

$2 T = μ m_{1} g$
$T = m_{2} g$
$2 m_{2} g = μ m_{1} g$
$m_{2} = \frac{μ m_{1}}{2}$ .

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The maximum value of the block m2 for which the system will remain in equilibrium (coefficient of friction between block m1 and plane surface is μ, pulleys are massless) is :

The correct option is A μm12 2T=μ m1g T=m2g 2m2 g=μ m1g m2=μ m12.

The maximum value of the block $m_{2}$ for which the system will remain in equilibrium (coefficient of friction between block $m_{1}$ and plane surface is $μ$ , pulleys are massless) is :

The correct option is A $μ \frac{m_{1}}{2}$

$2 T = μ m_{1} g$
$T = m_{2} g$
$2 m_{2} g = μ m_{1} g$
$m_{2} = \frac{μ m_{1}}{2}$ .