JEE Advance - Physics (2019 - Paper 1 Offline - No. 2)
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 $$T(t) = {T_0}\left( {1 + \beta {t^{{1 \over 4}}}} \right)$$, where $$\beta $$ is a constant with appropriate dimension while T0 is a constant with dimension of temperature. The heat capacity of the metal is
$${{4P{{(T(t) - {T_0})}^4}} \over {{\beta ^4}T_0^5}}$$
$${{4P{{(T(t) - {T_0})}^3}} \over {{\beta ^4}T_0^4}}$$
$${{4P(T(t) - {T_0})} \over {{\beta ^4}T_0^2}}$$
$${{4P{{(T(t) - {T_0})}^2}} \over {{\beta ^4}T_0^3}}$$
Paaiškinimas
Heat capacity, $${{dQ} \over {dt}} = H{{dT} \over {dt}}$$
Power of the rod, $$P = H.{T_0}.\beta .{1 \over 4}.{t^{ - 3/4}}$$
$${{4P} \over {{T_0}.\beta }} = {t^{ - 3/4}}.H$$
$$ \Rightarrow $$ $$H = \left( {{{4P} \over {{T_0}\beta }}} \right){t^{3/4}}$$ .... (i)
Now, $$T - {T_0} = {T_0}\beta {t^{1/4}}$$
So, $${t^{3/4}} = {\left( {{{T - {T_0}} \over {{T_0}\beta }}} \right)^3}$$
Substituting this value of t3/4 in equation (i) we get,
$$H = {{4P{{(T - {T_0})}^3}} \over {T_0^4{\beta ^4}}}$$
Power of the rod, $$P = H.{T_0}.\beta .{1 \over 4}.{t^{ - 3/4}}$$
$${{4P} \over {{T_0}.\beta }} = {t^{ - 3/4}}.H$$
$$ \Rightarrow $$ $$H = \left( {{{4P} \over {{T_0}\beta }}} \right){t^{3/4}}$$ .... (i)
Now, $$T - {T_0} = {T_0}\beta {t^{1/4}}$$
So, $${t^{3/4}} = {\left( {{{T - {T_0}} \over {{T_0}\beta }}} \right)^3}$$
Substituting this value of t3/4 in equation (i) we get,
$$H = {{4P{{(T - {T_0})}^3}} \over {T_0^4{\beta ^4}}}$$