NCERT Physics Class 11 Exemplar Ch 9 Mechanical Properties of Solids Part 7

Doorsteptutor material for IEO Class-7 is prepared by world's top subject experts: fully solved questions with step-by-step explanation- practice your way to success.

Q. 28 In nature, the failure of structural members usually results from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this

bending about the central axis of the tree is given by . Y is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Answer: When the tree is about to buckle

If then the centre of gravity is at a height from the



If 0 w is the weight/volume

Stone tied to string

Stone Tied to String

Loading image

Q.29 A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.

(a) Find the distance y from the top when the mass comes to rest for an instant, for the first time.

Answer: Till the stone drops through a length L it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at y.

The loss in P.E. of the stone is the P.E. stored in the stretched string



PE and KE of string

PE and KE of String

Loading image

Retain the positive sign.

(b) What is the maximum velocity attained by the stone in this drop?

Answer: The maximum velocity is attained when the body passes, through the “equilibrium, position” i.e. when the instantaneous acceleration

is zero. That is mg where x is the extension from L:

Let the velocity be Then


(c) What shall be the nature of the motion after the stone has reached its lowest point?

Answer: Consider the particle at an instantaneous position y. Then

Make a transformation of variables:


Thus the stone performs SHM with angular frequency about the point