ExamBro
ExamBro
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.
Verified Solution

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

\(8 g \sin 30^{\circ}+8 a_{1} \cos 30^{\circ}=m a_{2}=8 a_{2}\)

\(a_{2}=g \times \frac{1}{2}+\frac{\sqrt{3}}{9} g \cdot \frac{\sqrt{3}}{2}=\frac{2 g}{3}\)

Same subject
Explore more questions on app