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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

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 temp (T) in the metal rod changes with time (t) as

T(t)=T_0\left(1+\beta t^{1 / 3}\right)

The heat capacity of the metal is

Option: 1

\frac{4 P}{\beta^4 T_0^4}\left(T(t)-T_0\right)^3


Option: 2

\frac{4 p}{\beta^3 T_0{ }^3}\left(T(t)-T_0\right)^2


Option: 3

\frac{4 P}{\beta^2 T_0^2}\left(T(t)-T_0\right)^2


Option: 4

\frac{4 P\left(T(t)-T_0\right)^3}{\beta^4 \cdot T_0^2}


Answers (1)

best_answer

At equilibrium

c \frac{d t}{d t}=P \\

\Rightarrow \frac{dp}{d t}=\frac{T_0 B}{4} t^{-2 / 3}

So heat capacity C=\frac{4 P}{\beta T_0} \cdot t^{\beta / 4}

so \frac{T(t)-T(0)}{\beta T_0}=t^{1 / 3} \\

\Rightarrow t^{2 / 3}=\frac{[T(t)-T(0)]^2}{\beta^2 T_0^2}=x=\frac{4 p}{\beta^3 T_0^3}\left(T(t)-T_0\right)^2

Posted by

Ritika Jonwal

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