JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2021
- Hard
- MCQ
A block of mass \(m\) slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is : Given \({m}=8 \,{kg}, {M}=16\, {kg}\) Assume all the surfaces shown in the figure to be frictionless.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{2}{3} g\)
Step-by-step Solution
Detailed explanation
Let acceleration of wedge is \(a_{1}\) and acceleration of block w.r.t. wedge is \(a_{2}\)
\({N} \cos 60^{\circ}={Ma}_{1}=16 {a}_{1}\)
\(\Rightarrow {N}=32 {a}_{1}\)
F.B.D. of block w.r.t wedge
\(\perp\) to incline
\({N}=8 {g} \cos 30^{\circ}-8 {a}_{1} \sin 30^{\circ} \Rightarrow 32 {a}_{1}=4 \sqrt{3} {g}-4 {a}_{1}\)
\(\Rightarrow {a}_{1}=\frac{\sqrt{3}}{9} {g}\)
Along incline
\(8 g \sin 30^{\circ}+8 a_{1} \cos 30^{\circ}=m a_{2}=8 a_{2}\)
\(a_{2}=g \times \frac{1}{2}+\frac{\sqrt{3}}{9} g \cdot \frac{\sqrt{3}}{2}=\frac{2 g}{3}\)
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