1.

Two slabs are of the thicknesses d_(1) and d_(2). Their thermal conductivities are K_(1) and K_(2) respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures theta_(1) and theta_(2) . Assume theta_(1)gttheta_(2). The temperature theta of their common junction is :

Answer»

`(K_(1)theta_(1)+K_(2)theta_(2))/(theta_(1)+theta_(2))`
`(K_(1)theta_(1)d_(1)+K_(2)theta_(2)d_(2))/(K_(1)d_(2)+K_(2)d_(1))`
`(K_(1)theta_(1)d_(2)+K_(2)theta_(2)d_(1))/(K_(1)d_(2)+K_(2)d_(1))`
`(K_(1)theta_(1)+K_(2)theta_(2))/(K_(1)+K_(2))`

Solution :For first slab

HEAT current, `H_(1)=(K_(1)(theta_(1)-THETA)A)/(d_(1))`
For second slab,
Heat current, `H_(2)=(K_(2)(theta-theta_(2))A)/(d_(2))`
As slabs are in series
`H_(1)=H_(2)`
`:.(K_(1)(theta_(1)-theta)A)/(d_(1))=(K_(2)(theta-theta_(2))A)/(d_(2))`
`rArr""theta=(K_(1)theta_(1)d_(2)+K_(2)theta_(2)d_(1))/(K_(2)d_(1)+K_(1)d_(2))`
So, correct choice is (c ).


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