A 5 m long aluminium wire (Y = 7 × 10^10 N/m2) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire….

Q: A 5 m long aluminium wire (Y = 7 × 1010 N/m2) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire (Y = 12 × 1010 N/m2) of the same length under the same weight, the diameter should now be, in mm.

(a) 1.75

(b) 1.5

(c) 2.3

(d) 5.0

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Ans: (c)
Sol: $\Delta l_1 = \Delta l_2$

$\displaystyle \Delta \frac{F L}{A_1 Y_1} = \frac{F L}{A_2 Y_2} $

$\displaystyle A_1 Y_1 = A_2 Y_2 $

$\displaystyle \frac{\pi D_1^2}{4} Y_1 = \frac{\pi D_2^2}{4} Y_2 $

$\displaystyle D_2^2 = D_1^2 \times \frac{Y_1}{Y_2}$

$\displaystyle D_2^2 = (3 \times 10^{-3})^2 \times \frac{7 \times 10^{10}}{12 \times 10^{10}}$

$\displaystyle D_2 = 2.3 mm $