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Home » Two wires of diameter , one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is and that of brass wire is . Compute the elongations of the steel and the brass wires. [Young’s modulus of steel is
Two wires of diameter 0.25 \mathrm{~cm}, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 \mathrm{~m} and that of brass wire is \mathbf{1 . 0} \mathrm{m}. Compute the elongations of the steel and the brass wires. [Young’s modulus of steel is 2.0 \mathrm{x} \left.10^{11} \mathrm{~Pa} .\left(1 \mathrm{~Pa}=1 \mathrm{~N} \mathrm{~m}^{2}\right)\right]
CBSE | Class 11 | Mechanical Properties of Solids | NCERT | Physics
Two wires of diameter 0.25 \mathrm{~cm}, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 \mathrm{~m} and that of brass wire is \mathbf{1 . 0} \mathrm{m}. Compute the elongations of the steel and the brass wires. [Young’s modulus of steel is 2.0 \mathrm{x} \left.10^{11} \mathrm{~Pa} .\left(1 \mathrm{~Pa}=1 \mathrm{~N} \mathrm{~m}^{2}\right)\right]

Diameter of the two wires is given as d=0.25 \mathrm{~m}

Radius of the wires is given as r=d / 2=0.125 \mathrm{~cm}

Unloaded length of the steel wire is given as l_{1}=1.5 \mathrm{~m}

Unloaded length of the brass wire is given as l_{2}=1.0 \mathrm{~m}

Force exerted on the steel wire will be,

\mathrm{F}_{1}=(4+6) \mathrm{g}=10 \times 9.8=98 \mathrm{~N}

Cross-section area of the steel wire is a_{1}=\pi r_{1}^{2}

Change in length of the steel wire is =\Delta I_{1}

Young’s modulus for steel is given as 2.0 \times 10^{11} \mathrm{~Pa}

Y_{1}=\frac{F_{1}}{a_{1}} \frac{l_{1}}{\Delta l_{1}}

Y_{1}=\frac{F_{1}}{\pi r_{1}^{2}} \frac{l_{1}}{\Delta l_{1}}

\Delta l_{1}=\frac{F_{1}}{\pi r_{1}^{2}} \frac{l_{1}}{Y_{1}}

\Delta l_{1}=\frac{9.8}{\pi\left(0.125 \times 10^{-2}\right)^{2}} \frac{1.5}{2.0 \times 10^{11}}

=1.49 \times 10^{-4} \mathrm{~m}

Force of the brass wire will be F_{2}=6 \times 9.8=58.8 \mathrm{~N}

Cross-section area of the brass wire is a_{2}=\pi r_{2}^{2}

Change in length of the brass wire is =\Delta \mathrm{l}_{2}

Young’s modulus of the brass wire is given as 0.91 \times 10^{11} \mathrm{~Pa}

Y_{2}=\frac{F_{2}}{a_{2}} \frac{l_{2}}{\Delta l_{2}}

\Delta l_{2}=\frac{F_{2}}{\pi r_{2}^{2}} \frac{l_{2}}{Y_{2}}

\Delta l_{2}=\frac{58.8}{\pi\left(0.125 \times 10^{-2}\right)^{2}} \frac{1}{0.91 \times 10^{11}}

=1.3 \times 10^{-4} \mathrm{~m}

As a result, Elongation of the steel wire is 1.49 \times 10^{-4} \mathrm{~m} and that of brass is 1.3 \times 10^{-4} \mathrm{~m}.

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