Question

Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Young's modulus of steel is 2.0 × 10¹¹ and Young's modulus of Brass 0.91 × 10¹¹ Pa. The elongations of steel and brass wires are respectively

• A. 1.3 × 10^-2 m;
  1.5 × 10^-4 m
• B. 1.5 × 10^-4 m;
  1.3 × 10^-4 m
• C. 2.0 × 10^-5 m;
  1.8 × 10^-3 m
• D. 1.8 × 10^-3 m;
  2.1 × 10^-3 m

Answer 1

Answer: (B)

Given, diameter of wires (2r) = 0.25 cm
r = 0.125 cm = 1.25 × 10^-3 m
For steel wire,
Load (F₁) = (4 + 6) kgf = 10 × 9.8 N = 98 N
Length of steel wire (l₁) = 1.5 m
Young's modulus $(Y)=\frac{F_1\times l_1}{A_1\times \Delta l_1}$ Change in length $\left(\Delta l_1\right)=\frac{F_1\times l_1}{A_1\times Y_1}$
$\begin{array}{l}=\frac{F_1\times l_1}{\pi r_1^2\times Y_1}\\ =\frac{98\times 1.5}{3.14\times {\left(1.25\times 10^{-3}\right)}^2\times 2.0\times 10^{11}}\\ =1.5\times 10^{-4}m\end{array}$
For brass wire,
Load $\left(F_2\right)=6kgf=6\times 9.8N=58.8N$
Length of brass wire $\left(I_2\right)=1.0m$
Young's modulus $\left(Y_2\right)=0.91\times 10^{11}Pa$
Change in length $\left(l_2\right)=\frac{F_2\times l_2}{\pi r_2^2\times Y_2}$
$\begin{array}{l}=\frac{58.8\times 1.0}{3.14\times {\left(1.25\times 10^{-3}\right)}^2\times 0.91\times 10^{11}}\\ =1.3\times 10^{-4}m\end{array}$