In order to determine the Young's Modulus of | vedclass

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

Question

(D) The formula for Young's Modulus is $Y = \frac{FL}{A\ell} = \frac{mgL}{\pi R^2 \ell}$.The fractional error is given by $\frac{\Delta Y}{Y} = \frac{

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(D) The formula for Young's Modulus is $Y = \frac{FL}{A\ell} = \frac{mgL}{\pi R^2 \ell}$.The fractional error is given by $\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta R}{R} + \frac{\Delta \ell}{\ell}$.Given values:$m = 1 \, kg = 1000 \, g$,$\Delta m = 1 \, g$$L = 1 \, m = 1000 \, mm$,$\Delta L = 1 \, mm$$R = 0.2 \, cm$,$\Delta R = 0.001 \, cm$$\ell = 0.5 \, cm$,$\Delta \ell = 0.001 \, cm$Substituting these into the error formula:$\frac{\Delta Y}{Y} 	imes 100 = \left( \frac{1}{1000} + \frac{1}{1000} + 2 	imes \frac{0.001}{0.2} + \frac{0.001}{0.5} ight) 	imes 100$$= \left( 0.001 + 0.001 + 0.01 + 0.002 ight) 	imes 100$$= (0.001 + 0.001 + 0.01 + 0.002) 	imes 100 = 0.014 	imes 100 = 1.4 \%$.

Similar Questions

Two blocks of masses $1 \, kg$ and $2 \, kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $2 \times 10^9 \, N/m^2$. What should be the minimum radius of the wire used if it is not to break? Take $g = 10 \, m/s^2$.

Why is the maximum height of a mountain about $\sim 10 \ km$ on Earth?

$A$ force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2R$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module