Two wires A and B are stretched by the same l | vedclass

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

Question

(B) Stress is defined as the force applied per unit area,given by the formula: $	ext{Stress} = \frac{F}{A}$.Given that both wires are stretched by th

Vedclass · Accepted Answer

Twice that on $A$

Vedclass · Answer

(B) Stress is defined as the force applied per unit area,given by the formula: $	ext{Stress} = \frac{F}{A}$.Given that both wires are stretched by the same load,the force $F$ is constant for both wires.Let $A_A$ be the area of wire $A$ and $A_B$ be the area of wire $B$. According to the problem,$A_A = 2 A_B$.The stress on wire $A$ is $\sigma_A = \frac{F}{A_A} = \frac{F}{2 A_B}$.The stress on wire $B$ is $\sigma_B = \frac{F}{A_B}$.Comparing the two,we get $\sigma_B = 2 	imes \left(\frac{F}{2 A_B}ight) = 2 \sigma_A$.Therefore,the stress on wire $B$ is twice that on wire $A$.

Two wires $A$ and $B$ are stretched by the same load. If the area of cross-section of wire $A$ is double that of wire $B$,then the stress on wire $B$ is

Similar Questions

The elasticity of Invar:

Hooke's law defines:

Explain the spring-ball model of the elastic behavior of solids.

Why do rail tracks have an '$I$-shaped' girder?

The elastic limit of brass is $3.5 \times 10^{10} \, N/m^2$. Find the maximum load that can be applied to a brass wire of $0.75 \, mm$ diameter without exceeding the elastic limit $....... \times 10^4 \, N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module