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prashant3670 (0)

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Q. A square loop of side length L is to be consirdered. Point A & C are maintained at temperatute T1 & T2 respectively. The rate of heat flow was observed to be 30X. Now, the reservoir by which the temp at C was maintained is now shifted to point D. Find the new rate of heat flow, if its value is KX, then find the value of (K+5)*(K+5)-18?
    
magiclko (4210)

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we knw the rate of heat flow is given by I = KA(T2-T1)/L
here L/KA can be considered as resistance of the rods.... we'll denote it by R
so now when temperature is maintained across A and C, we have an equivalent resistance of value R, across A and C....
thrfore 30X= (T2-T1)/R
now when it is maintained between A and D, we have an equivalent resistance of 3R/4
thrfore KX =4(T2-T1)/3R
now u can easily calculate the value of K and find the value of the required expression.......


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bhupesh (723)

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dear
   The solution provided in the post is correct for further clarification let me know

Bhupesh.M
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