Four identical hollow cylindrical columns of | vedclass

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

Question

(D) Total mass $M = 50,000 \; kg$. Total force $F_{total} = Mg = 50,000 	imes 9.8 = 490,000 \; N$.Since there are $4$ columns,the force on each colum

Vedclass · Accepted Answer

$7.22 	imes 10^{-7}$

Vedclass · Answer

(D) Total mass $M = 50,000 \; kg$. Total force $F_{total} = Mg = 50,000 	imes 9.8 = 490,000 \; N$.Since there are $4$ columns,the force on each column is $F = \frac{490,000}{4} = 122,500 \; N$.The area of cross-section of each hollow column is $A = \pi(R^2 - r^2)$,where $R = 0.6 \; m$ and $r = 0.3 \; m$.$A = \pi(0.6^2 - 0.3^2) = \pi(0.36 - 0.09) = 0.27\pi \; m^2$.Stress on each column is $\sigma = \frac{F}{A} = \frac{122,500}{0.27\pi} \; Pa$.Using Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}}$,the strain is $\epsilon = \frac{\sigma}{Y}$.Given $Y = 2.0 	imes 10^{11} \; Pa$,we have $\epsilon = \frac{122,500}{0.27 	imes \pi 	imes 2.0 	imes 10^{11}}$.$\epsilon = \frac{122,500}{1.69646 	imes 10^{11}} \approx 7.22 	imes 10^{-7}$.

Similar Questions

One end of a uniform wire of length $L$ and of weight $W$ is attached rigidly to a point in the roof and a weight $W_1$ is suspended from its lower end. If $A$ is the area of cross-section of the wire,the stress in the wire at a height $3L/4$ from its lower end is

Young's modulus of rubber is $10^4 \ N/m^2$ and the area of cross-section is $2 \ cm^2$. If a force of $2 \times 10^5 \ dynes$ is applied along its length,then its initial length $L$ becomes:

$A$ load $W$ produces an extension of $1 \ mm$ in a thread of radius $r$. If the load is increased to $4W$ and the radius is increased to $2r$,while all other factors remain the same,what will be the new extension in $mm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module