A wire of length 10 cm and diameter 0.5 mm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: Length $L = 10 \ cm = 0.1 \ m$,Diameter $d = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$,Temperature $T = 1727^{\circ} C = 1727 + 273 = 2000 \ K$,Po

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Given: Length $L = 10 \ cm = 0.1 \ m$,Diameter $d = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$,Temperature $T = 1727^{\circ} C = 1727 + 273 = 2000 \ K$,Power $P = 94.2 \ W$,Stefan-Boltzmann constant $\sigma = 6.0 	imes 10^{-8} \ W \ m^{-2} \ K^{-4}$.Using the Stefan-Boltzmann law for radiation: $P = \varepsilon \sigma A T^4$,where $A$ is the surface area of the wire.The surface area of the wire (cylindrical) is $A = \pi d L$.$A = 3.14 	imes (0.5 	imes 10^{-3} \ m) 	imes (0.1 \ m) = 1.57 	imes 10^{-4} \ m^2$.Substituting the values into the power equation:$94.2 = \varepsilon 	imes (6.0 	imes 10^{-8}) 	imes (1.57 	imes 10^{-4}) 	imes (2000)^4$.$94.2 = \varepsilon 	imes (6.0 	imes 10^{-8}) 	imes (1.57 	imes 10^{-4}) 	imes (16 	imes 10^{12})$.$94.2 = \varepsilon 	imes (6.0 	imes 1.57 	imes 16) 	imes 10^0$.$94.2 = \varepsilon 	imes (150.72)$.$\varepsilon = \frac{94.2}{150.72} = 0.625$.Since $\varepsilon = \frac{x}{8}$,then $0.625 = \frac{x}{8} \implies x = 0.625 	imes 8 = 5$.

Similar Questions

The adjoining diagram shows the spectral energy density distribution $E_\lambda$ of a black body at two different temperatures. If the areas under the curves are in the ratio $16 : 1$,the value of temperature $T$ is ......... $K$. (in $,000$)

If the temperature of a black body is increased from $27^{\circ}C$ to $327^{\circ}C$,the radiation emitted increases by a factor of:

The temperature at which a black body of unit area loses its energy at the rate of $1 \text{ J/s}$ is:

The unit of Stefan's constant is:

In an external environment of temperature $T$ Kelvin, a sphere at temperature $3T$ Kelvin has a cooling rate $R_1$. When the temperature of that sphere falls to $2T$ Kelvin, the cooling rate of the sphere will become:

Vedclass Test Series

Exam Paper Generator

Online Exam Module