In an experiment, brass and steel wires|ELASTICITY VISCOSITY

JEE MAIN 2019

10-04-2019 S2

Question

In an experiment, brass and steel wires of length 1 m each with areas of cross section $1 \mathrm{~mm}^2$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young's Modulus for steel and brass are, respectively, $120 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ and $60 \times 109 \mathrm{~N} / \mathrm{m}^2$ ]

Select the correct option:

$1.8 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

$1.2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

$4.0 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

$0.2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Corresponding to the stress $(\sigma)$ Total elongation $\Delta \mathrm{I}_{\text {net }}=\frac{\sigma \mathrm{L}_1}{\mathrm{Y}_1}+\frac{\sigma \mathrm{L}_2}{\mathrm{Y}_2}$ $$ \begin{aligned} & \sigma=\Delta I\left(\frac{Y_1 Y_2}{Y_1+Y_2}\right)=0.2 \times 10^{-3} \times\left(\frac{120 \times 60}{180}\right) \times 10^9 \\ & =8 \times 10^6 \frac{\mathrm{~N}}{\mathrm{~m}^2} \text { (Answer is not matching) } \end{aligned} $$

Question Tags

JEE Main

Physics

Medium

Start Preparing for JEE with Competishun

Report Issue