In an experiment to determine the Young's mod | vedclass

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

Question

(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/L} = \frac{F \cdot L}{A \cdot \Delta l}$.Rearranging this,we get the rati

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/L} = \frac{F \cdot L}{A \cdot \Delta l}$.Rearranging this,we get the ratio of extension to load as $\frac{\Delta l}{F} = \frac{L}{YA}$.Here,the graph plots $\frac{\Delta l}{F}$ on the $y$-axis and $L$ on the $x$-axis.The slope of this graph is $m = \frac{\Delta l/F}{L} = \frac{1}{YA}$.From the graph,we can calculate the slope $m = \frac{0.25 	imes 10^{-5}}{1} = 0.25 	imes 10^{-5}\,m/N$.Given cross-sectional area $A = 2\,mm^2 = 2 	imes 10^{-6}\,m^2$.Substituting these values into the slope equation: $0.25 	imes 10^{-5} = \frac{1}{Y 	imes 2 	imes 10^{-6}}$.$Y = \frac{1}{0.25 	imes 10^{-5} 	imes 2 	imes 10^{-6}} = \frac{1}{0.5 	imes 10^{-11}} = 2 	imes 10^{11}\,N/m^2$.Comparing this with $x 	imes 10^{11}\,N/m^2$,we get $x = 2$.

Similar Questions

Young's modulus of a perfectly rigid body material is

Two wires $A$ and $B$ are of the same material. Their lengths are in the ratio $1: 2$ and diameters are in the ratio $2: 1$. When stretched by forces $F_{A}$ and $F_{B}$ respectively,they get equal increase in their lengths. Then the ratio $F_{A} / F_{B}$ is

The area of a cross-section of a steel wire is $0.1 \, cm^2$ and Young's modulus of steel is $2 \times 10^{11} \, N \, m^{-2}$. The force required to stretch it by $0.1 \%$ of its original length is ......... $N$.

To double the length of an iron wire having $0.5 \, cm^2$ area of cross-section,the required force will be $(Y = 10^{12} \, dyne/cm^2)$.

Young's modulus depends upon

Vedclass Test Series

Exam Paper Generator

Online Exam Module