A triangular block of mass M with angles 30°, 60° and 90° rests with its 30°-90° side on a horizontal table. A cubical block of mass m rests on the 60°-30° . The acceleration which M must have relative to the table to keep m stationary relative to the triangular block assuming frictionless contact is

Question

A triangular block of mass M with angles 30°, 60° and 90° rests with its 30°-90° side on a horizontal table. A cubical block of mass m rests on the 60°-30° . The acceleration which M must have relative to the table to keep m stationary relative to the triangular block assuming frictionless contact is (A) g (B) (g/√2) (C) (g/√3) (D) (g/√5)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/228a23715f01fb9295ae1db1a249b517-.png" /> When M is accelerated with acceleration a, pseudo force m a acts on the block m. The ' m a ' can be resolved as m a cos 30° along the inclined plane and m a sin 30° normal to it. For 0 acceleration of block m w.r.t. M m g sin 30°-m a cos 30°=0 ⇒ a=g tan 30°=(g/√3)

$g$

$\frac{g}{2}$

$\frac{g}{3}$

$\frac{g}{5}$

g

2​g​

3​g​

5​g​

When M is accelerated with acceleration a, pseudo force ma acts on the block m. The ' ma ' can be resolved as ma cos30∘ along the inclined plane and masin30∘ normal to it. For 0 acceleration of block m w.r.t. M mgsin30∘−macos30∘=0⇒a=gtan30∘=3​g​

$g$

$\frac{g}{2}$

$\frac{g}{3}$

$\frac{g}{5}$