In an experiment,brass and steel wires of len | vedclass

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

Question

(D) Given: Length of each wire $\ell = 1\,m$,Area $A = 1\,mm^2 = 10^{-6}\,m^2$,Total elongation $\Delta \ell = 0.2\,mm = 0.2 	imes 10^{-3}\,m$. Young

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given: Length of each wire $\ell = 1\,m$,Area $A = 1\,mm^2 = 10^{-6}\,m^2$,Total elongation $\Delta \ell = 0.2\,mm = 0.2 	imes 10^{-3}\,m$. Young's Modulus for steel $Y_s = 120 	imes 10^9\,N/m^2$ and for brass $Y_b = 60 	imes 10^9\,N/m^2$. Since the wires are in series,the force $F$ applied is the same for both. The total elongation is $\Delta \ell = \Delta \ell_s + \Delta \ell_b = \frac{F \ell}{A Y_s} + \frac{F \ell}{A Y_b} = \frac{F \ell}{A} \left( \frac{1}{Y_s} + \frac{1}{Y_b} ight)$. Stress $\sigma = \frac{F}{A} = \frac{\Delta \ell}{\ell \left( \frac{1}{Y_s} + \frac{1}{Y_b} ight)}$. Substituting the values: $\sigma = \frac{0.2 	imes 10^{-3}}{1 \left( \frac{1}{120 	imes 10^9} + \frac{1}{60 	imes 10^9} ight)} = \frac{0.2 	imes 10^{-3} 	imes 120 	imes 10^9}{1 + 2} = \frac{0.2 	imes 120 	imes 10^6}{3} = 8 	imes 10^6\,N/m^2$. Since $8 	imes 10^6\,N/m^2$ is not among the given options,the correct choice is $D$.

Similar Questions

Two steel wires of same length $L$ but radii $r$ and $2r$ are connected together end to end and tied to a wall as shown. The force $F$ stretches the combination by $10 \ mm$. How far does the junction point $A$ move (in $mm$)?

If the temperature of a wire of length $2 \,m$ and area of cross-section $1 \,cm^2$ is increased from $0^{\circ}C$ to $80^{\circ}C$ and is not allowed to increase in length,then the force required for it is ............$N$ $\{Y=10^{10} \,N/m^2, \alpha=10^{-6}/^{\circ}C\}$

The Young's modulus of a wire of length $L$ and radius $r$ is $Y \ N/m^2$. If the length and radius are reduced to $L/2$ and $r/2,$ then its Young's modulus will be

In steel,the Young's modulus and the strain at the breaking point are $2 \times 10^{11} \, N/m^2$ and $0.15$ respectively. The stress at the breaking point for steel is therefore:

Vedclass Test Series

Exam Paper Generator

Online Exam Module