physicscatalyst.com logo







Elasticity




Question 1. A block of gelatin is 60 mm by 60 mm by 20 mm when unstressed. A force of .245 N is applied tangentially to the upper surface causing a 5 mm displacement relative to the lower surface. The block is placed such that 60X60 comes on the lower and upper surface. Find the shearing stress, shearing strain and shear modulus
(a) (68.1 N/m2 , .25 , 272.4 N/m2)
(b) (68 N/m2 , .25 , 272 N/m2)
(c) (67 N/m2 , .26 , 270.4 N/m2)
(d) (68.5 N/m2 , .27 , 272.4 N/m2)

Solution:

Shear stress$=\frac{F}{A}=\frac{.245}{36 \times 10^{-4}} =68.1 N/m^{2}$

Shear strain$= tan \theta = \frac{d}{h}=\frac{5}{20}=.25$

Shear modulus (S) $\frac{\emph{shear stress}}{\emph{shear strain}}=272.4 N/m$

Question 2.A steel wire of diameter 4 mm has a breaking strength of 4X105N. The breaking strength of similar steel wire of diameter 2 mm is
(a)1X105N.
(b) 4X105N.
(c) 16X105N.
(d) none of the these
Solution 2.
Breaking strength is proportional to square of diameter,Since diameter becomes half,Breaking strength reduced by $\frac{1}{4}$ Hence A is correct.


Question 3.What is the SI unit of modulus of elasticity of a substance?
(a) Nm-1
(b) Nm-2
(c) Jm-1
(d) Unit less quantity

Solution 3
Answer is b

Question 4A thick uniform rubber rope of density 1.5 gcm-3 and Young Modulus 5X10106 Nm-2 has a length 8 m. when hung from the ceiling of the room, the increase in length due to its own weight would be ? (a) .86m
(b) .2m
(c) .1m
(d) .096m
Solution 4 The weight of the rope can be assumed to act at its mid point. Now the extension x is proportional to the original length L. if the weight of the rope acts at its mid point, the extension will be that produced by the half of the rope.So replacing L by $\frac{L}{2}$ in the expression for Young 's Modulus ,we have $Y=\frac{FL}{2Al}
or
$l=\frac{FL}{2AY}$
Since
$F=mg=\rho V = \rho AL$
Therefore
$l=\frac{gL^{2}\rho}{2Y}
Substituting the values, we get
$l=.096m$

Go Back to Class 11 Maths Home page Go Back to Class 11 Physics Home page



link to us