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B point of the rod shown in figure is ma...

B point of the rod shown in figure is maintained at `200^@C`. At left end A, there is water at `100^@C` and at right end C there is ice at `0^@C`. Heat currents `H_1 and H_2` will flow on both sides. Due to `H_1`, water will convert into steam and due to `H_2` ice will be melted. If latent heat of vaporization is `540 cal//g` and latent heat of fusion is `80 cal//g` then neglecting the radiation losses find `l_1/l_2` so that rate of melting of ice is two times the rate of conversion of water into stream.
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Text Solution

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The correct Answer is:
A, B, D
Using the relation of `(dm)/(dt)` derived in above example,
`(dm)/(dt) = (TD)/(RL) = ((TD)KA)/(lL)` `(as R= l/(KA))`
Given that,
`((dm)/(dt))_(RHS) = 2((dm)/(dt))_(LHS)`
or ` [((TD)KA)/(lL)]_(RHS) = 2[((TD)KA)/(lL)]_(LHS)`
K and A are same on both sides. Hence,
`((TD)/(lL))_(RHS) = 2((TD)/(lL))_(LHS)`
Substituting the proper values, we have
`((200)/(l_2xx80))=2[(100)/(l_1xx540)]`
`l_1/l_2 = 80/540= 4/27`
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