Question

Current in a coil decreases down to zero from $l_0$ uniformly during a time interval $t_0$ If amount of heat generated in time $t=0tot=\frac{t_0}{2}is{H}_1$ and heat generated in the $t=\frac{t_0}{2}$ to $t=t_{0}is{H}_{2}then\frac{H_1}{H_2}is$

• A. $3:1$
• B. $5:1$
• C. $1:1$
• D. $7:1$

Answer 1

The correct option is D $7:1$

$l=t_0(1-\frac{t}{t_0})$

$H_1=\underset{0}{\overset{t_0/2}{\int }}{l}_0^{2}R(1+\frac{t^2}{t_0^2}-\frac{2t}{t_0})dt$

$=l_0^2{(t+\frac{t^3}{3t_0^2}-\frac{t^2}{t_0})}_0^{t_0/2}R$

$=l_0^{2}R(\frac{1}{2}+\frac{1}{24}-\frac{1}{4})t_0=\frac{7}{24}{l}_0^{2}R{t}_0$

$H_2=\underset{t_0/2}{\overset{t_0}{\int }}(1+\frac{t^2}{t_0^2}-\frac{2t}{t_0})l_0^{2}Rdt$

$=l_R^2{(t+\frac{t^3}{3t_0}-\frac{t^2}{t_0})}_{t_0/2}^{t_0}$

$=l_0^{2}R{t}_0[(1+\frac{1}{3}-1)-\left(\frac{7}{24}\right)]$

$H_2-l_0^{2}R{t}_0\left(\frac{1}{24}\right)$

$\frac{H_1}{H_2}=\frac{l_0^{2}R{T}_0\frac{7}{24}}{l_0^{2}R{t}_0\frac{1}{24}}=\frac{7}{1}$