A tension of 22 ~N is applied to a copper wir | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Tension $F = 22 ~N$,Area $A = 0.02 ~cm^2 = 0.02 	imes 10^{-4} ~m^2$,Young's modulus $Y = 1.1 	imes 10^{11} ~N/m^2$,Poisson's ratio $\sigm

Vedclass · Accepted Answer

$1.28 	imes 10^{-6} ~cm^2$

Vedclass · Answer

(A) Given: Tension $F = 22 ~N$,Area $A = 0.02 ~cm^2 = 0.02 	imes 10^{-4} ~m^2$,Young's modulus $Y = 1.1 	imes 10^{11} ~N/m^2$,Poisson's ratio $\sigma = 0.32$.Longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{22}{0.02 	imes 10^{-4} 	imes 1.1 	imes 10^{11}} = \frac{22}{2.2 	imes 10^5} = 10^{-4}$.Poisson's ratio is defined as $\sigma = - \frac{\Delta r/r}{\Delta l/l}$. The lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l}$.The area $A = \pi r^2$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.Substituting the lateral strain,$\frac{\Delta A}{A} = -2 \sigma \frac{\Delta l}{l}$.The magnitude of decrease in area is $\Delta A = 2 \sigma A \frac{\Delta l}{l}$.$\Delta A = 2 	imes 0.32 	imes 0.02 ~cm^2 	imes 10^{-4} = 0.64 	imes 0.02 	imes 10^{-4} ~cm^2 = 0.0128 	imes 10^{-4} ~cm^2 = 1.28 	imes 10^{-6} ~cm^2$.

Column-$I$	Column-$II$
$A$. Tapetum	$i$. Degenerates during development
$B$. Intine	$ii$. Resistant organic material
$C$. Vegetative cell	$iii$. Pollen tube
$D$. Sporopollenin	$iv$. Provides nourishment to pollen grains

$A$ tension of $22 ~N$ is applied to a copper wire of cross-sectional area $0.02 ~cm^2$. The Young's modulus of copper is $1.1 \times 10^{11} ~N/m^2$ and the Poisson's ratio is $0.32$. The decrease in cross-sectional area will be:

Similar Questions

At $298 \ K$,the value of $-\frac{\Delta[Br^-]}{\Delta t}$ for the reaction $5 Br^- (aq) + BrO_3^- (aq) + 6 H^+ (aq) \rightarrow 3 Br_2 (aq) + 3 H_2 O (l)$ is $x \ mol \ L^{-1} \ min^{-1}$. What is the rate (in $mol \ L^{-1} \ min^{-1}$) of this reaction?

$540\, g$ of ice at $0\,^{\circ}C$ is mixed with $540\, g$ of water at $80\,^{\circ}C$. The final temperature of the mixture is:

$A$ vertical spring with force constant $K$ is fixed on a table. $A$ ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

Match the following columns correctly:

Column-$I$ Column-$II$

$A$. Tapetum $i$. Degenerates during development

$B$. Intine $ii$. Resistant organic material

$C$. Vegetative cell $iii$. Pollen tube

$D$. Sporopollenin $iv$. Provides nourishment to pollen grains

$\cos 36^{\circ} - \cos 72^{\circ}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module