99 Percentile Qs Bank for JEE MainPhysicsLaws of Motion
A block is between two surfaces as shown in the figure. Find the normal reaction at both surfaces. [Assume, g=10 ~m / s ^2 ]
Options
- AN₁=37.2 ~N and N₂=9.6 ~N
- BN₁=38.2 ~N and N₂=8.6 ~N
- CN₁=40 ~N and N₂=4 ~N
- DN₁=37.5 ~N and N₂=9.9 ~N
Correct answer
A. N₁=37.2 ~N and N₂=9.6 ~N
Step-by-step solution
The given situation is shown in the figure. Resolving applied force, we have following forces on block Given, = 3 4 , = 3 5 and = 4 5 So, horizontal component, F₁=12 =12 4 5 = 48 5 ~N and vertical component, F₂=12 =12 3 5 = 36 5 ~N So, total applied force on ground is 10+ 36 5 +2 10=37.2 ~N So, reaction N₁ of ground =37.2 ~N upwards Also, total force horizontally on wall is F₁= 48 5 =9.6 ~N So, reaction N₂ of wall =9.6 ~N towards left