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Q.

A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as T(t)=T0  1+βt1/4 where β  is a constant with appropriate dimensions while T0 is a constant with dimensions of temperature. The heat capacity of metal is

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a

4PTt−T03β4T04

b

4PTt−T0β4T02

c

4PTt−T02β4T03

d

4PTt−T04β4T05

answer is A.

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Detailed Solution

P=msdTdt=(H)ddtT01+βt1/4=HT0β4t−3/4…..(1)here H is heat capacity Given :T=T01+βt1/4⇒TT0−1=βt1/4⇒t−1/4=βT0T−T0⇒t−3/4=βT0T−T03…….. (2)  Substitute (2) in (1)P=HT0β4β3T03T−T03⇒H=4PT−T03β4T04
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A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as T(t)=T0  1+βt1/4 where β  is a constant with appropriate dimensions while T0 is a constant with dimensions of temperature. The heat capacity of metal is