Verified Solution • 2025–26 Session moderate

QJEEAdv19Ph2782: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph2782 Solution: A current carrying wire heats a metal rod. The wire provides a…

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+βt14)T(t) = {T_0}\left( {1 + \beta {t^{{1 \over 4}}}} \right) , where β\beta is a constant with appropriate dimension while T0 is a constant with dimension of temperature. The heat capacity of the metal is

Multiple Choice Options

  • A 4P(T(t)−T0)4β4T05{{4P{{(T(t) - {T_0})}^4}} \over {{\beta ^4}T_0^5}}
  • B 4P(T(t)−T0)3β4T04{{4P{{(T(t) - {T_0})}^3}} \over {{\beta ^4}T_0^4}}
  • C 4P(T(t)−T0)β4T02{{4P(T(t) - {T_0})} \over {{\beta ^4}T_0^2}}
  • D 4P(T(t)−T0)2β4T03{{4P{{(T(t) - {T_0})}^2}} \over {{\beta ^4}T_0^3}}

Step-by-Step Verified Solution

Heat capacity, dQdt=HdTdt{{dQ} \over {dt}} = H{{dT} \over {dt}}

Power of the rod, P=H.T0.β.14.t−3/4P = H.{T_0}.\beta .{1 \over 4}.{t^{ - 3/4}}

4PT0.β=t−3/4.H{{4P} \over {{T_0}.\beta }} = {t^{ - 3/4}}.H

⇒\Rightarrow H=(4PT0β)t3/4H = \left( {{{4P} \over {{T_0}\beta }}} \right){t^{3/4}} .... (i)

Now, T−T0=T0βt1/4T - {T_0} = {T_0}\beta {t^{1/4}}

So, t3/4=(T−T0T0β)3{t^{3/4}} = {\left( {{{T - {T_0}} \over {{T_0}\beta }}} \right)^3}

Substituting this value of t3/4 in equation (i) we get,

H=4P(T−T0)3T04β4H = {{4P{{(T - {T_0})}^3}} \over {T_0^4{\beta ^4}}}

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