Question

Speed of transverse wave on a straight wire (mass $6·0\mathrm{gm}$, length $60\mathrm{cm}$ and area of cross-section $1·0{\mathrm{mm}}^2$) is $90\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$. If the Young’s modulus of wire is $16\times {10}^{11}{\mathrm{Nm}}^{-2}$, the extension of wire over its natural length is

• A. $0·03\mathrm{mm}$
• B. $0·04\mathrm{mm}$
• C. $0·02\mathrm{mm}$
• D. $0·01\mathrm{mm}$

Answer 1

The correct option is A

$0·03\mathrm{mm}$

Step 1: Given data:

Mass of wire $\left(m\right)$=$6·0\mathrm{gm}=6·0\times {10}^{-3}\mathrm{kg}$

Length of wire$\left(l\right)$=$60\mathrm{cm}=60\times {10}^{-2}\mathrm{m}$

Cross-section area$\left(A\right)$=$1{\mathrm{mm}}^2=1·0\times {10}^{-6}{\mathrm{m}}^2$

Young's Modulus of wire$\left(Y\right)$=$16\times {10}^{11}{\mathrm{Nm}}^{-2}$

The speed of the transverse wave$\left(v\right)=90\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step2: Formula used:

From the relation of the speed of wave and mass per length

$Speedofwave\left(v\right)=\sqrt{\frac{Tension\left(T\right)}{Massperlength\left(\mu \right)}}$

Mass per unit length is the linear density given as-

$\mathrm{mass}\mathrm{per}\mathrm{length}\left(\mathrm{\mu }\right)=\frac{\mathrm{mass}\left(\mathrm{m}\right)}{\mathrm{length}\left(\mathrm{l}\right)}$

Young's modulus will be calculated by-

$\frac{stress}{\text{strain}}=Y$

Step 3: Compute the tension in the wire

From the formula,

$\mathrm{Speed}\mathrm{of}\mathrm{wave}\left(\mathrm{v}\right)=\sqrt{\frac{\mathrm{Tension}\left(\mathrm{T}\right)}{\mathrm{Mass}\mathrm{per}\mathrm{length}\left(\mathrm{\mu }\right)}}$

$$\begin{gathered} Speedofwave\left(v\right)=\sqrt{\frac{Tension\left(T\right)}{{\displaystyle \frac{Massofwire\left(m\right)}{Lengthofwire\left(l\right)}}}} \\ 90=\sqrt{\frac{T}{{\displaystyle \frac{6·0\times {10}^{-3}}{60\times {10}^{-2}}}}} \\ 90=\sqrt{\frac{T}{1\times {10}^{-2}}} \\ 8100=\frac{T}{1\times {10}^{-2}} \\ T=81\mathrm{N} \end{gathered}$$

Thus, the tension will be $81\mathrm{N}$.

Step4: Compute extension in the wire

We know that,

$\frac{stress}{\text{strain}}=Y$

Strain will be-$\frac{∆l}{l}$

Stress will be-$\frac{\mathrm{force}}{\mathrm{area}}=\frac{\mathrm{Tension}}{\mathrm{area}}$

On putting the given values,

$$\begin{gathered} \frac{\text{stress}}{Y}=\frac{∆l}{l} \\ \frac{{\displaystyle \frac{T}{A}}}{Y}=\frac{∆l}{l} \\ ∆l=\frac{T\times l}{A\times Y} \\ ∆l=\frac{81\times 60\times {10}^{-2}}{\left(1\times {10}^{-6}\right)\left(16\times {10}^{11}\right)} \\ ∆l=3.0375\times {10}^{-5}\mathrm{m} \\ ∆l=0·03\mathrm{mm} \end{gathered}$$

Thus, the extension of wire over its natural length is $0.03\mathrm{mm}$.

Hence, option A is the correct answer.