Question

The temperature of a well stirred liquid kept open to a cold surrounding is plotted against time. (Assume conditions favourable to Newton's law of cooling)
The value of $se{c}^2{\theta }_1$ is

• A. $1+9ta{n}^2{\theta }_2$
• B. $1+ta{n}^2{\theta }_2$
• C. $1+3ta{n}^2{\theta }_2$
• D. $3+ta{n}^2{\theta }_2$

Answer 1

The correct option is A $1+9ta{n}^2{\theta }_2$

According to Newton's law of cooling $\frac{dT}{dt}\propto \Delta T\Rightarrow \frac{tan{\theta }_1}{tan{\theta }_2}=\frac{35-20}{25-20}=\frac{15}{5}=\frac{3}{1}$

$\Rightarrow ta{n}^2{\theta }_1=9ta{n}^2{\theta }_2\because se{c}^2{\theta }_1=1+ta{n}^2{\theta }_1\Rightarrow se{c}^2{\theta }_1=1+9ta{n}^2{\theta }_2$