- 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

$l=.096m$

Class 11 Maths Class 11 Physics Class 11 Chemistry

Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.

We really do hope that this resolve the issue. If you still hare facing problems then feel free to contact us using feedback button or contact us directly by sending is an email at **[email protected]**

We are aware that our users want answers to all the questions in the website. Since ours is more or less a one man army we are working towards providing answers to questions available at our website.