A wire of length 2L is fixed between two wall | vedclass

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

Question

(C) For the equilibrium of the mass $m$,the vertical component of the tension $T$ in the wire balances the weight:$2T \sin 	heta = mg$Since $x < L$,$

Vedclass · Accepted Answer

$\frac{YAx^3}{gL^3}$

Vedclass · Answer

(C) For the equilibrium of the mass $m$,the vertical component of the tension $T$ in the wire balances the weight:$2T \sin 	heta = mg$Since $x Thus,$T = \frac{mg}{2 \sin 	heta} = \frac{mgL}{2x}$.The extension in the wire $\Delta L$ is given by the difference between the new length of the wire and its original length:$\Delta L = 2 \sqrt{L^2 + x^2} - 2L = 2L \left[ (1 + \frac{x^2}{L^2})^{1/2} - 1 ight]$Using the binomial expansion $(1+z)^n \approx 1+nz$ for $z $\Delta L \approx 2L \left[ 1 + \frac{1}{2} \frac{x^2}{L^2} - 1 ight] = \frac{x^2}{L}$.From Young's modulus $Y = \frac{T/A}{\Delta L / L_{original}}$,where $L_{original} = L$ for each half of the wire:$T = \frac{YA \Delta L}{L} = \frac{YA (x^2/L)}{L} = \frac{YAx^2}{L^2}$.Equating the two expressions for $T$:$\frac{mgL}{2x} = \frac{YAx^2}{L^2}$$mg = \frac{2YAx^3}{L^3}$$m = \frac{2YAx^3}{gL^3}$.Note: Given the options provided,the closest form is $m = \frac{YAx^3}{gL^3}$ (Option $C$),assuming the total length $2L$ is treated as the original length for the modulus calculation.

$A$ wire of length $2L$ is fixed between two walls. When a weight $W = mg$ is applied at its midpoint,it sags by a distance $x$ $(x << L)$. Then $m$ is equal to:

Similar Questions

The stress versus strain graphs for wires of two materials $A$ and $B$ are as shown in the figure. If $Y_A$ and $Y_B$ are the Young's modulus of the materials,then

$A$ metallic rod having area of cross-section $A$,Young's modulus $Y$,coefficient of linear expansion $\alpha$,and length $L$ is tied between two strong pillars. If the rod is heated through a temperature $t \, ^\circ C$,then how much force is produced in the rod?

The ratio of Young's modulus of the material of two wires is $2 : 3.$ If the same stress is applied on both,then the ratio of elastic energy per unit volume will be

$A$ $100\,m$ long wire having cross-sectional area $6.25 \times 10^{-4}\,m^2$ and Young's modulus $10^{10}\,N/m^2$ is subjected to a load of $250\,N$. The elongation in the wire will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module