JEE Advanced · Physics · 5. Laws of Motion
A circular disc with a groove along its diameter is placed horizontally. A block of mass \(1 \mathrm{~kg}\) is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is \(\mu=2 / 5\). The disc has an acceleration of \(25 \mathrm{~m} / \mathrm{s}^2\). Find the acceleration of the block with respect to disc.

- A 15
- B 20
- C 10
- D 25
Answer & Solution
Correct Answer
(C) 10
Step-by-step Solution
Detailed explanation
Normal reaction in vertical direction \(N_1=m g\)
Normal reaction from side to the groove \(N_2=m a \sin 37^{\circ}\)
Therefore, acceleration of block with respect to discs
\(
a_r=\frac{m a \cos 37^{\circ}-\mu N_1-\mu N_2}{m}
\)
Substituting the values we get, \(\quad a_r=10 \mathrm{~m} / \mathrm{s}^2\)
Normal reaction from side to the groove \(N_2=m a \sin 37^{\circ}\)
Therefore, acceleration of block with respect to discs
\(
a_r=\frac{m a \cos 37^{\circ}-\mu N_1-\mu N_2}{m}
\)
Substituting the values we get, \(\quad a_r=10 \mathrm{~m} / \mathrm{s}^2\)
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