The dimensions of four wires of the same mate | vedclass

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

Question

(D) Young's Modulus is given by $Y = \frac{F/A}{\Delta L/L} = \frac{FL}{\Delta L A}$.Since $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$,we have $\D

Vedclass · Accepted Answer

Length $50 \ cm$,Diameter $0.5 \ mm$

Vedclass · Answer

(D) Young's Modulus is given by $Y = \frac{F/A}{\Delta L/L} = \frac{FL}{\Delta L A}$.Since $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$,we have $\Delta L = \frac{FL}{YA} = \frac{4FL}{Y \pi D^2}$.For wires of the same material ($Y$ is constant) subjected to the same force ($F$ is constant),the increase in length $\Delta L \propto \frac{L}{D^2}$.$(a)$ $\frac{100}{1^2} = 100$$(b)$ $\frac{200}{2^2} = \frac{200}{4} = 50$$(c)$ $\frac{300}{3^2} = \frac{300}{9} \approx 33.33$$(d)$ $\frac{50}{0.5^2} = \frac{50}{0.25} = 200$Comparing the values,the increase in length is maximum for option $(d)$.

The dimensions of four wires of the same material are given below. The increase in length is maximum in the wire of

Similar Questions

The decrease in length of a metal bar of length $L$ and cross-sectional area $A$ when compressed with a load $F$ along its length is (where $Y$ is Young's modulus of the material of the metal bar).

The adjacent graph shows the extension $(\Delta l)$ of a wire of length $1\, m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-6}\, m^2$,calculate the Young's modulus of the material of the wire.

The force required to stretch a wire of cross-section $1 \ cm^{2}$ to double its length will be ........ $\times 10^{7} \ N$. (Given Young's modulus of the wire $= 2 \times 10^{11} \ N/m^{2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module