Question

A wire of length $1.0$ m and radius ${10}^{-3}$ m is carrying a heavy current and is assumed to radiate as a black body. At equilibrium its temperature is $900K$ while that of the surrounding is $300K$. The resistivity of the material of wire at $300K$ is ${\pi }^2\times {10}^{-8}\Omega .m.$ Its temperature coefficient of resistance is $7.8\times {10}^{-3}{/}^{0}C.$ Then current in wire will be (Assume resistivity is linear function of temp.)

• A. 12 A
• B. 24 A
• C. 36 A
• D. 48 A

Answer 1

The correct option is C 36 A

At equilibrium,

$P_{Heatgenerated}=P_{Radiated}$

Given:

${\rho }_{300}={\pi }^2\times {10}^{-8}\Omega$m $L=1$m

Area $A=\pi \times r^2=\pi \times ({10}^{-3}{)}^2$ $m^2$

Surface area of the cylindrical wire$A_{surface}=(2\pi rL+2\pi r^2)$

$\alpha =7.8\times {10}^{-3}$

$I^{2}R=\sigma \times A_{surface}\times (T^4-T_s^4)$

$T={900}^{0}K$ $T_s={300}^{0}K$

$I^2\times \left(\frac{{\rho }_{900}\times L}{A}\right)=\sigma \times A_{surface}\times (T^4-T_s^4)$

$I^2\times \left(\frac{{\rho }_{300}(1+\alpha \times (T-T_s\left)\right)\times L}{A}\right)=\sigma \times A_{surface}\times (T^4-T_s^4)$ ....(1)

Substituting the values in eqn (1) and simplifying, we get

$I^2={36}^2$ $\therefore I=36Amp$