Two wires of same length and thickness are joined end to end. Their Young's moduli are 15×1010 Paand 20×1010 Pa. If the combination is stretched by a certain load, the elongations of these wires will be in the ratio:

Y1=15×1010 Pa; Y2=20×1010 PaBy Hooke's LawFA=YeL ⇒eL=FYAAs the wires are subjected to same tensile force,e∝1Y; ⇒e1e2=Y2Y1=20×101015×1010=4:3

Two wires of same length and thickness are joined end to end. Their Young's moduli are 15×1010 Paand 20×1010 Pa. If the combination is stretched by a certain load, the elongations of these wires will be in the ratio:

Y1=15×1010 Pa; Y2=20×1010 PaBy Hooke's LawFA=YeL ⇒eL=FYAAs the wires are subjected to same tensile force,e∝1Y; ⇒e1e2=Y2Y1=20×101015×1010=4:3

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Two wires of same length and thickness are joined end to end. Their Young's moduli are $15 \times 10^{10} P a$ and $20 \times 10^{10} P a$ . If the combination is stretched by a certain load, the elongations of these wires will be in the ratio:

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1:1

3:4

1:2

4:3

answer is B.

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Detailed Solution

$Y_{1} = 15 \times 10^{10} Pa; Y_{2} = 20 \times 10^{10} Pa$

By Hooke's Law

$\frac{F}{A} = Y \frac{e}{L}$

$\Rightarrow \frac{e}{L} = \frac{F}{YA}$

As the wires are subjected to same tensile force,
$e \propto \frac{1}{Y}; \Rightarrow \frac{e_{1}}{e_{2}} = \frac{Y_{2}}{Y_{1}} = \frac{20 \times 10^{10}}{15 \times 10^{10}} = 4 : 3$

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Two wires of same length and thickness are joined end to end. Their Young's moduli are 15×1010 Paand 20×1010 Pa. If the combination is stretched by a certain load, the elongations of these wires will be in the ratio:

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Two wires of same length and thickness are joined end to end. Their Young's moduli are $15 \times 10^{10} P a$ and $20 \times 10^{10} P a$ . If the combination is stretched by a certain load, the elongations of these wires will be in the ratio: