A 3-4-5 inclined plane with inclination $θ = = \sin^{- 1} \frac{3}{5}$ is fixed to a rotating turn table as shown in figure. A block rests on the inclined plane and the co-efficient of static friction between the inclined plane and the block is $μ_{s} = 1 / 4$ . The block is to remain at a position 40 cm from the axes of the turn table. Then the minimum angular speed of turn table required to keep the block from sliding down the plane $(g = 9.8 m / s^{2})$

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$4.2 \frac{rad}{sec}$

$4 \frac{rad}{sec}$

$3.8 \frac{rad}{sec}$

$3.2 \frac{rad}{sec}$

answer is C.

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Detailed Solution

If block is observed from inclined plane then FBD is shown
$N = m g \cos θ + m w^{2} R \sin θ - (A) f_{r} + m w^{2} R \cos θ = m g \sin θ μ N = m g \sin θ - m w^{2} \sin θ - (B)$
Solving A&B for w
$w = \sqrt{\frac{g}{R} (\frac{\tan θ - μ}{1 + μ \tan θ})} = 3.2 \frac{r a d}{\sec}$