There rods of the same dimensions have therma | vedclass

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

Question

$\frac{{dQ}}{{dt}} = KA\frac{{\Delta T}}{L}$

$For\,the\,first\,rod,\left( {\frac{{dQ}}{{dt}}} ight) = \frac{{3KA}}{L}\left( {100 - 	heta } igh

Vedclass · Accepted Answer

$70$

Vedclass · Answer

$\frac{{dQ}}{{dt}} = KA\frac{{\Delta T}}{L}$

$For\,the\,first\,rod,\left( {\frac{{dQ}}{{dt}}} ight) = \frac{{3KA}}{L}\left( {100 - 	heta } ight)$

$Similarly,{\left( {\frac{{dQ}}{{dt}}} ight)_2} = 2K\frac{A}{L}\left( {	heta  - 50} ight)$

${\left( {\frac{{dQ}}{{dt}}} ight)_3} = K\frac{A}{L}\left( {	heta  - 20} ight)$

$Now,{\left( {\frac{{dQ}}{{dt}}} ight)_1} = {\left( {\frac{{dQ}}{{dt}}} ight)_2} + {\left( {\frac{{dQ}}{{dt}}} ight)_3}$

$ \Rightarrow \,\,3\left( {100 - 	heta } ight) = 2\left( {	heta  - 50} ight) + \left( {	heta  - 20} ight)$

$ \Rightarrow 	heta  = {70^ \circ }$

There rods of the same dimensions have thermal conductivities $3K, 2K$ and $K$. They are arranged as shown in fig. with their ends at $100\,^oC, 50\,^oC$ and $20\,^oC$. The temperature of their junction is....... $^oC$

Similar Questions

Assuming newton's law of cooling to be valid, body at temperature $50^o\ C$ in surrounding of temperature $20^o\ C$ , achieve steady state with help of $100\ W$ heater. If same body has temperature $35^o\ C$ in same surrounding, then power of heater required to maintain steady state ........ $W$

Two rods of crossectional area $A$ and $2A$ and equal length having thermal conductivities $2K$ and $3K$ are joined in parallel, then equivalent thermal conductivity of their combination will be-

Five wires each of cross-sectional area $A$ and length $l$ are combined as shown. The thermal conductivity of copper and steel are $k_1$ and $k_2$ respectively. The equivalent thermal resistance between $A$ and $C$ is

The power radiated by a black body is $P$ and it radiates maximum energy around the wavelength $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy around wavelength $\frac{3}{4}\lambda_0$, the power radiated by it will increase by a factor of

If the wavelengths of maximum intensity of radiations emitted by the sun and the moon are $0.5\times10^{-6}\, m$ and $10^{-4}\, m$ respectively. The ratio of their temperature is