- Introduction
- Stress
- Strain
- Hook's Law
- Elastic Modulus
- Poisson's Ratio
- Stress-Strain Digram
- Solved Examples

(a) (68.1 N/m

(b) (68 N/m

(c) (67 N/m

(d) (68.5 N/m

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) $\frac{\emph{shear stress}}{\emph{shear strain}}=272.4 N/m$

(a)1X10

(b) 4X10

(c) 16X10

(d) none of the these

Breaking strength is proportional to square of diameter,Since diameter becomes half,Breaking strength reduced by $\frac{1}{4}$ Hence A is correct.

(a) Nm

(b) Nm

(c) Jm

(d) Unit less quantity

Answer is b

(b) .2m

(c) .1m

(d) .096m

or

$l=\frac{FL}{2AY}$

Since

$F=mg=\rho V = \rho AL$

Therefore

$l=\frac{gL^{2}\rho}{2Y}

Substituting the values, we get

$l=.096m$

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