Question

Two wires of equal cross section one made up of aluminium and other of brass are joined end to end. When the combination of wires is kept under tension the elongation in wires are found to be equal. Find the ratio of lengths of two wires. ($Y_{AL}$ = 7 X 10¹⁰ N / m² and $Y_{brass}$ = 9.1 x 10¹⁰ N / m²) (Ans : 0.7693 : 1)

Answer 1

we know, Young's modulus , Y = $\frac{FL}{A\Delta{L}}$
Here, F is tension acts on wire,
L is length of wire ,
A is cross sectional area of wire.

According to question,
F, A and ∆L remain constant.
So, Young's modulus is directly proportional to length of wire,
Hence, $\frac{Y_1}{Y_2}=\frac{L_1}{L_2}$

$\frac{L_1}{L_2}=\frac{Y_1}{Y_2}$

Here, $Y_1$ = 7 × 10^10 N/m²
$Y_2$ = 9.1 × 10^10 N/m²

So, $\frac{L_1}{L_2}=\frac{7\times10^{10}}{9.1\times10^{10}}=\frac{0.7693}{1}$

Hence, answer is 0.7693 : 1

Answer 2

Answer:

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