(2)

[IMAGE 172]

The two resistances are connected in parallel, so, they both have same potential difference as $V$.
Heat energy is given by:

           $H=\frac{V^2}{R}t$

So, $H_1=\frac{V^2}{100}t ; H_2=\frac{V^2}{200}t$

So, $\frac{H_1}{H_2}=\frac{200}{100}=2:1$

(1)

Fuse is an electrical safety device that operates to provide over current protection to an electrical circuit.

(3)

Given, $Q=at-b{t}^2$

$\therefore$   $I=\frac{dQ}{dt}=a-2bt$

At     $t=0,Q=0\Rightarrow I=0$

Also,  $I=0$ at $t=\frac{a}{2b}$

$\therefore$   Total heat produced in resistance $R,$

$H={\int }_0^{a/2b}{I}^{2}Rdt=R{\int }_0^{a/2b}{(a-2bt)}^{2}dt$

     $=R{\int }_0^{a/2b}(a^2+4b^2{t}^2-4abt)dt=R{[a^{2}t+4b^2\frac{{\displaystyle t^3}}{{\displaystyle 3}}-4ab\frac{{\displaystyle t^2}}{{\displaystyle 2}}]}_0^{a/2b}$

     $=R[a^2\times \frac{{\displaystyle a}}{{\displaystyle 2b}}+\frac{{\displaystyle 4b^2}}{{\displaystyle 3}}\times \frac{{\displaystyle a^3}}{{\displaystyle 8b^3}}-\frac{{\displaystyle 4ab}}{{\displaystyle 2}}\times \frac{{\displaystyle a^2}}{{\displaystyle 4b^2}}]$

      $=\frac{a^{3}R}{b}[\frac{{\displaystyle 1}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle 1}}{{\displaystyle 2}}]=\frac{a^{3}R}{6b}$

(2)

Here, 

Distance between two cities = 150 km

Resistance of the wire, $R=(0.5 \Omega {\text{km}}^{-1})(150 \text{km})=75\Omega$

Voltage drop across the wire, $V=(8 \text{V} {\text{km}}^{-1})(150 \text{km})=1200$$\text{V}$

Power loss in the wire is

$P=\frac{V^2}{R}=\frac{{(1200 \text{V})}^2}{75 \Omega }=19200 \text{W}=19.2$ $\text{kW}$