An ordinary body cools from ' $4 \theta$ ' to ' $3 \theta$ ' in ' t ' minutes. The temperature of that body…
An ordinary body cools from ' $4 \theta$ ' to ' $3 \theta$ ' in ' t ' minutes. The temperature of that body after next 't' minutes is (Assume Newton's law of cooling and room temperature is $\theta$ )
$\frac{9 \theta}{4}$
$\frac{2 \theta}{5}$
$\frac{5 \theta}{3}$
$\frac{7 \theta}{3}$
Solution
According to Newton's law of cooling,
$\frac{\theta_1-\theta_2}{\mathrm{t}}=K\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right]$
where, $\theta_0=$ temperature of surrounding
$\begin{aligned}
\therefore \quad & \frac{4 \theta-3 \theta}{t}=K\left[\frac{4 \theta+3 \theta}{2}-\theta\right] \\
& \frac{\theta}{t}=K \times \frac{5 \theta}{2}...(i) \\
& K=\frac{2}{5 t}
\end{aligned}$
After another t min, let the temperature be x.
$\begin{array}{ll}
\therefore & \frac{3 \theta-x}{t}=\frac{2}{5 t}\left[\frac{3 \theta+x}{2}-\theta\right] \ldots[\text { using (i) }] \\
\therefore & 3 \theta-x=\frac{3 \theta+x-2 \theta}{5} \\
\therefore & 15 \theta-5 x=x+\theta \\
\therefore & 6 x=14 \theta \\
\therefore & x=\frac{7 \theta}{3}
\end{array}$