What should be the diameter of a copper wire | vedclass

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

Question

(C) The formula for Young's modulus is $Y = \frac{F L}{A \Delta L}$, where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Rearranging for elongation:

Vedclass · Accepted Answer

$2.3$

Vedclass · Answer

(C) The formula for Young's modulus is $Y = \frac{F L}{A \Delta L}$, where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Rearranging for elongation: $\Delta L = \frac{4 F L}{\pi d^2 Y}$.Since $F$, $L$, and $\Delta L$ are the same for both wires, we have $\frac{1}{d_c^2 Y_c} = \frac{1}{d_a^2 Y_a}$, where $c$ denotes copper and $a$ denotes aluminium.Thus, $d_c^2 Y_c = d_a^2 Y_a$.Substituting the values: $d_c^2 (12 	imes 10^{10}) = (3 	ext{ mm})^2 (7 	imes 10^{10})$.$d_c^2 = \frac{9 	imes 7}{12} 	ext{ mm}^2 = \frac{63}{12} 	ext{ mm}^2 = 5.25 	ext{ mm}^2$.$d_c = \sqrt{5.25} 	ext{ mm} \approx 2.29 	ext{ mm} \approx 2.3 	ext{ mm}$.

Similar Questions

Explain the experimental determination of Young's modulus.

$A$ $5\, m$ long aluminium wire $(Y = 7 \times 10^{10}\, N/m^2)$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10}\, N/m^2)$ of the same length under the same weight,the diameter should now be,in $mm$.

$A$ ring of a thin wire of cross-sectional area $a$ and length $l$ is dipped into a liquid of surface tension $\sigma$ and taken out so that a film of liquid is formed in the ring. If Young's modulus of the material of the wire is $Y$,then the longitudinal strain developed in the ring will be:

$A$ uniform rod of mass $m$,length $L$,area of cross-section $A$ and Young's modulus $Y$ hangs from the ceiling. Its elongation under its own weight will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module