JEE Advanced · Physics · 5. Laws of Motion
A small block of mass of \(0.1 \mathrm{~kg}\) lies on a fixed inclined plane PQ which makes an angle \(\theta\) with the horizontal. A horizontal force of \(1 \mathrm{~N}\) acts on the block through its centre of mass as shown in the figure.
The block remains stationary if (take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )

- A \(\theta=45^{\circ}\)
- B \(\theta>45^{\circ}\) and a frictional force acts on the block towards P.
- C \(\theta>45^{\circ}\) and a frictional force acts on the block towards Q.
- D \(\theta < 45^{\circ}\) and a frictional force acts on the block towards Q.
Answer & Solution
Correct Answer
(C) \(\theta>45^{\circ}\) and a frictional force acts on the block towards Q.
Step-by-step Solution
Detailed explanation
The various forces acting on the block are as shown in the figure.

When \(\theta=45^{\circ}, \sin \theta=\cos \theta\)
The block will remain stationary and the frictional force is zero.
When \(\theta>45^{\circ}, \sin \theta>\cos \theta\)
Therefore a frictional force acts towards \(Q\).
When \(\theta < 45^{\circ}, \cos \theta>\sin \theta\)
Therefore a frictional force acts towards \(P\).

When \(\theta=45^{\circ}, \sin \theta=\cos \theta\)
The block will remain stationary and the frictional force is zero.
When \(\theta>45^{\circ}, \sin \theta>\cos \theta\)
Therefore a frictional force acts towards \(Q\).
When \(\theta < 45^{\circ}, \cos \theta>\sin \theta\)
Therefore a frictional force acts towards \(P\).
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