The correct option is A 10^12N/m^2 Y = F #160;lA #160;l l = FlAY So, l α1 ∴ #160; l_1 #160;L_2 = A_2A_1 = 3 l_13 = l_2 ...

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Standard XII

Physics

Work Energy Theorem Application

An elongation...

Question

An elongation of $0.1$ % in a wire of cross-section $10^{- 6} m^{2}$ causes a tension of $100 N$ . $Y$ for the wire is

10^{12} N / m^{2}

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10^{11} N / m^{2}

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10^{10} N / m^{2}

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100 N / m^{2}

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Solution

The correct option is A $10^{12} N / m^{2}$
$Y = \frac{F l}{A △ l}$
$△ l = \frac{F l}{A Y}$
$S o, △ l α \frac{1}{△}$
$∴ \frac{△ l_{1}}{△ L_{2}} = \frac{A_{2}}{A_{1}} = 3$
$\frac{△ l_{1}}{3} = △ l_{2}$
$△ l_{2} = \frac{0.1}{3} = 0.033 m m$

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Similar questions

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A block of weight $100$ N is suspended by copper and steel wires of same cross sectional area $0.5 c m^{2}$ and length $\sqrt{3} m$ and $1 m$ respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $30^{o}$ and $60^{o}$ respectively. If elongation in copper wire is $Δ l_{C}$ and elongation in steel wire is $Δ l_{S}$ , then the ratio $\frac{Δ l_{C}}{Δ l_{S}}$ is

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An elongation of 0.1% in a wire of cross-section 10−6m2 causes a tension of 100N. Y for the wire is

The correct option is A 1012N/m2Y=FlA△l△l=FlAYSo,△lα1△∴△l1△L2=A2A1=3△l13=△l2△l2=0.13=0.033mm

An elongation of $0.1$ % in a wire of cross-section $10^{- 6} m^{2}$ causes a tension of $100 N$ . $Y$ for the wire is

The correct option is A $10^{12} N / m^{2}$
$Y = \frac{F l}{A △ l}$
$△ l = \frac{F l}{A Y}$
$S o, △ l α \frac{1}{△}$
$∴ \frac{△ l_{1}}{△ L_{2}} = \frac{A_{2}}{A_{1}} = 3$
$\frac{△ l_{1}}{3} = △ l_{2}$
$△ l_{2} = \frac{0.1}{3} = 0.033 m m$