Discussion Forum

The heater of an electric kettle is made of a wire of length  L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d.The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40K?

Answer:

Explanation:

 Resistance of initially  given kettle3

   $R=\rho\frac{1}{A}=\rho$

$\frac{L}{\pi(d/2)^{2}}=\frac{4\rho L}{\pi d^{2}}$

Power , $P=\frac{V^{2}}{R}$

  or            $P\propto\frac{1}{R}$

   Resistance of two replaced kettles

             $R_{1}=\frac{\rho L}{\pi d^{2}}$

 and 

                 $R_{2}=\frac{\rho L}{\pi d^{2}}$

So    $R_{1}=R_{2}=\frac{R}{4}$

 if wires are in parallel then  equivalent resistanece

   $R_{p}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{R}{8}$

 Then power P_p=8p                   (as $P\propto\frac{1}{R}$]

 So, it will take 0.5 minute

   (as H=Pt  or      $t=\frac{H}{p}$      or  $t\propto\frac{1}{p}$ )

 if wires are in series then  equivalent resistanece

R_S=R₁+R₂=R/2

 tthen power P_s= 2P

 So, it will take 2 minutes.