  Question

# A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3. The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0o. Then at θ=30o, the block will start sliding down the plane the block will remain at rest on the plane up to certain θ and then it will topple at θ=60o the block will start sliding down the plane and continue to do so at higher angles at θ=60o, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ

Solution

## The correct option is B the block will remain at rest on the plane up to certain θ and then it will topple The maximum angle when the block will not slide should be equal to the angle of repose.  ϕ=tan−1μ=tan−1√3=60o If the block does not topple, the line of action of weight should pass within the base of the block And for toppling, maximum angle of inclination,  θ=tan−1b/2h/2 θ=tan−1(10/2)(15/2)=tan−123 θ=34o<60o For 60o, mg is already outside the base.  ∴ As the angle θ is increased first it will topple before it starts sliding.  Suggest corrections   