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  • A copper wire with a cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at  an

A copper wire with a cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at 25^{\circ} \mathrm{C} and thermally insulated from the surroundings. If a current of 10 A is set up in this wire then the time in which the wire starts melting. Density of Cu = 9 \times 10^3 \mathrm{Kg} \mathrm{m}^{-3} specific heat of Cu = 9 \times 10^{-2} \mathrm{Cal} \mathrm{Kg}^{-10} \mathrm{C}^{-1}$, M.P. (Cu) $1075^{\circ} \mathrm{C} and specific resistance = 1.6 \times 10^{-8} \Omega \mathrm{m}

Option: 1

558s


Option: 2

508s


Option: 3

528s


Option: 4

458s


Answers (1)

best_answer

Mass of Cu = Volume \times density= 0.5 \times 10^{-6} \times 0.1 \times 9 \times 10^3=45 \times 10^{-5} \mathrm{Kg} 
Rise in temperature = \theta=1075-25=1050^{\circ} \mathrm{C}
Specific heat = 9 \times 10^{-2} \mathrm{Kg}^{-1}{ }^{\circ} \mathrm{C} \times 4.2 \mathrm{~J}
\begin{aligned} \Rightarrow \quad \mathrm{l}^2 \mathrm{Rt} & =\mathrm{mS} \theta \Rightarrow \mathrm{t}=\frac{\mathrm{mS} \theta}{\mathrm{I}^2 \cdot \mathrm{R}} \\ \text { but } \quad & R=\frac{\rho L}{A}=\frac{1.6 \times 10^{-8} \times 0.1}{0.5 \times 10^{-6}}=3.2 \times 10^{-3} \Omega \\ & \Rightarrow t=\frac{45 \times 4.2 \times 10^{-5} \times 1050 \times 0.09}{10 \times 10 \times 3.2 \times 10^{-3}}=558 \mathrm{~s}\end{aligned}

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vishal kumar

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