Question

A rigid bar of mass $15kg$ is supported symmetrically by three wire each of $2m$ long. These at each end are of copper and middle one is of steel. Young's modulus of elasticity for copper and steel are $110\times 10^{9}N/m^2$ and $190\times 10^{9}N/m^2$ respectively. If each wire is to have same tension, ratio of their diameters will be

• A. $\sqrt{\frac{11}{19}}$
• B. $\sqrt{\frac{19}{11}}$
• C. $\sqrt{\frac{30}{11}}$
• D. $\sqrt{\frac{11}{30}}$

Answer 1

Answer: (B)

Tension is same (given)
From free body diagram
$3T=150N$
Since the bar has to be supported symmetrically
Therefore extension in each wire will be same
We know $\Delta x=\frac{FL}{AY}$
Compare 1 copper wire with another steel wire
$\frac{FL}{A_C{Y}_C}=\frac{FL}{A_S{Y}_S}$
$\Rightarrow \frac{A_S}{A_C}=\frac{Y_C}{Y_S}$
where
$A_c$- Area of copper wire
$Y_c$ - Young's modulus copper
$A_s$ - Area of steel wire
$Y_s$ - Young's modulus steel
Substituing value of $Y_c$ and $Y_s$
$\frac{\pi d_S^2}{4\times \frac{\pi }{4}\times d_C^2}=\frac{110\times 10^9}{190\times 10^9}$
$\frac{d_S}{d_C}=\sqrt{\frac{11}{19}}$
$\{d_S-$ diameter of steel wire
$\{d_C-$ diameter of copper wire
$\frac{d_C}{d_s}=\sqrt{\frac{19}{11}}$