JEE PYQ: Properties Of Matter Question 14
Question 14 - 2019 (10 Apr 2019 Shift 2)
In an environment, brass and steel wires of length 1 m each with areas of cross section 1 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, $120 \times 10^9$ N/m$^2$ and $60 \times 10^9$ N/m$^2$]
(1) $1.2 \times 10^6$ N/m$^2$
(2) $4.0 \times 10^6$ N/m$^2$
(3) $1.8 \times 10^6$ N/m$^2$
(4) $8 \times 10^6$ N/m$^2$
Show Answer
Answer: (4)
Solution
$\Delta l_{net} = \sigma\left(\frac{l_1}{Y_1} + \frac{l_2}{Y_2}\right)$
$\sigma = \Delta l \cdot \frac{Y_1 Y_2}{Y_1 + Y_2} = 0.2 \times 10^{-3} \times \frac{120 \times 60}{180} \times 10^9 = 8 \times 10^6$ N/m$^2$
BETA
