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A small block of mass m is projected on a larger block of mass $10 m$ and length l with a velocity v as shown in the figure. The coefficient of friction between the two blocks is $μ_{2}$ while that between the lower block and the ground is $μ_{1}$ . Given that $μ_{2} > 11 μ_{1}$ . Find the minimum value of $v$ , such that the mass $m$ falls off the block of mass $10 m$ .

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$\sqrt{\frac{52 (μ_{2} - μ_{1}) g l}{10}}$

$\sqrt{\frac{22 (μ_{2} - μ_{1}) g l}{10}}$

$\sqrt{\frac{12 (μ_{2} - μ_{1}) g l}{10}}$

$\sqrt{\frac{11 (μ_{2} - μ_{1}) g l}{10}}$

answer is B.

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Detailed Solution

Force of friction at different contacts are shown in figure.

Here,

$f_{1} = μ_{2} m g f_{2} = μ_{1} (11 m g) μ_{2} > 11 μ_{1} f_{1} > f_{2}$

Retardation of upper block

$a_{1} = \frac{f_{1}}{m} = μ_{2 g}$

Acceleration of lower block

$a_{2} = \frac{f_{1} - f_{2}}{m} = \frac{(μ_{2} - 11 μ_{1}) g}{10}$

Relative retardation of upper block

$a_{r} = a_{1} + a_{2} or a_{r} = \frac{11}{10} (μ_{2} - μ_{1}) g Now, 0 = v_{\min}^{2} - 2 a l ∴ v_{\min} = \sqrt{2 a l} = \sqrt{\frac{22 (μ_{2} - μ_{1}) g l}{10}}$

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