Two blocks m1 and m2 are resting on a rough inclined plane of inclination 37- as shown in figure- The contact force between the blocks is m1- -4 kg - m 2-2 kg - -1-0-8-2-0-5- g -10 m - s 2- sin 37-3-5A- 3-2 NB- 3-6 NC- 7-2 ND- Zero

The correct option is D ZeroFor a mass sliding down on a inclined plane , we know that ⇒a = g sinθ μg cosθ. Hence acceleration depends on the coefficient of ...

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Standard XII

Physics

Newton's Second Law

Two blocks m1...

Question

Two blocks $m_{1}$ and $m_{2}$ are resting on a rough inclined plane of inclination $37^{\circ}$ as shown in figure. The contact force between the blocks is $(m_{1} = 4 k g, m_{2} = 2 k g, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = \frac{3}{5})$

3.2 N

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3.6 N

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7.2 N

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Zero

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Solution

The correct option is D Zero
For a mass sliding down on a inclined plane , we know that
$\Rightarrow a = g sin θ - μ g cos θ$ .
Hence acceleration depends on the coefficient of friction ,
As $μ_{1} > μ_{2}$ $\Rightarrow a_{1} < a_{2}$ . (directly from the equation we can say)
Hence the acceleration of $m_{2}$ is greater than that of $m_{1}$ . Hence, there will not be any contact.
So, $N = 0$

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Similar questions

Q. Two blocks

m_{1}

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are resting on a rough inclined plane of inclination

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as shown in figure. The contact force between the blocks is

(m_{1} = 4 k g, m_{2} = 2 k g, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = \frac{3}{5})

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Two blocks m1 and m2 are resting on a rough inclined plane of inclination 37∘ as shown in figure. The contact force between the blocks is (m1=4 kg,m2=2 kg,μ1=0.8,μ2=0.5,g=10m/s2,sin37∘=35)

Two blocks $m_{1}$ and $m_{2}$ are resting on a rough inclined plane of inclination $37^{\circ}$ as shown in figure. The contact force between the blocks is $(m_{1} = 4 k g, m_{2} = 2 k g, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = \frac{3}{5})$