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$1 \mathrm{Kcal}$ of heat flowing through a rod of iron. When the rod is cut down to 4 pieces then what will be the heat flowing through each piece having same differential temperature?
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Verified Answer
The correct answer is:
$1 \mathrm{Kcal}$
We know that the heat flowing through a conductor is given by
$$
\begin{aligned}
\frac{Q}{t} & =-K A\left(\frac{d \theta}{d x}\right) \\
\Rightarrow Q & =-K A\left(\frac{d \theta}{d x}\right) t
\end{aligned}
$$
Where
$$
\begin{aligned}
K & =\text { thermal conductivity of the material } \\
A & =\text { area of cross-section } \\
\frac{d \theta}{d x} & =\text { temperature gradient } \\
t & =\text { time }
\end{aligned}
$$
Now if we cut the rod into 4 pieces, $A$ remains same $K$ remains same, $\frac{d \theta}{d x}$ is said to be same, so for same time $t, Q$ is going to be same as earlier.
$$
\begin{aligned}
\frac{Q}{t} & =-K A\left(\frac{d \theta}{d x}\right) \\
\Rightarrow Q & =-K A\left(\frac{d \theta}{d x}\right) t
\end{aligned}
$$
Where
$$
\begin{aligned}
K & =\text { thermal conductivity of the material } \\
A & =\text { area of cross-section } \\
\frac{d \theta}{d x} & =\text { temperature gradient } \\
t & =\text { time }
\end{aligned}
$$
Now if we cut the rod into 4 pieces, $A$ remains same $K$ remains same, $\frac{d \theta}{d x}$ is said to be same, so for same time $t, Q$ is going to be same as earlier.
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