A uniformly tapering conical wire is made fro | vedclass

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

Question

(C) Let the upper end be at $x=0$ and the lower end at $x=L$. The radius $r(x)$ at a distance $x$ from the top is given by $r(x) = R + \frac{3R-R}{L}x

Vedclass · Accepted Answer

$L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$

Vedclass · Answer

(C) Let the upper end be at $x=0$ and the lower end at $x=L$. The radius $r(x)$ at a distance $x$ from the top is given by $r(x) = R + \frac{3R-R}{L}x = R(1 + \frac{2x}{L})$.The extension $d\Delta L$ of a small element of length $dx$ is given by $d\Delta L = \frac{Mg dx}{Y A(x)}$,where $A(x) = \pi r(x)^2 = \pi R^2 (1 + \frac{2x}{L})^2$.Integrating from $x=0$ to $x=L$:$\Delta L = \int_0^L \frac{Mg dx}{\pi Y R^2 (1 + \frac{2x}{L})^2} = \frac{Mg}{\pi Y R^2} \int_0^L (1 + \frac{2x}{L})^{-2} dx$.Let $u = 1 + \frac{2x}{L}$,then $du = \frac{2}{L} dx$,or $dx = \frac{L}{2} du$.When $x=0, u=1$; when $x=L, u=3$.$\Delta L = \frac{Mg}{\pi Y R^2} \cdot \frac{L}{2} \int_1^3 u^{-2} du = \frac{MgL}{2 \pi Y R^2} \left[ -\frac{1}{u} \right]_1^3 = \frac{MgL}{2 \pi Y R^2} \left( 1 - \frac{1}{3} \right) = \frac{MgL}{2 \pi Y R^2} \cdot \frac{2}{3} = \frac{MgL}{3 \pi Y R^2}$.The total extended length is $L + \Delta L = L + \frac{MgL}{3 \pi Y R^2} = L \left( 1 + \frac{1}{3} \frac{Mg}{\pi Y R^2} \right)$.

Similar Questions

The force required to stretch a steel wire of $1\,cm^2$ cross-section to $1.1$ times its original length is $(Y = 2 \times 10^{11}\,N/m^2)$.

The ratio of Young's modulus of three wires is $2 : 2 : 1$ and the ratio of their cross-sectional areas is $1 : 2 : 3$. If the same force is applied to each,what is the ratio of the increase in their lengths?

Young's modulus of a perfectly rigid body material is

Vedclass Test Series

Exam Paper Generator

Online Exam Module