Question

It takes 16 min to boil some water in an electric kettle. Due to some defect it becomes necessary to remove 10% turns of the heating coil of the kettle. After repairs, how much time will it take to boil the same mass of water?
A. 17.7 min
B. 14.4 min
C. 20.9 min
D. 13.7 min

Answer 1

Hint: The amount of heat required to boil does not vary i.e. it remains constant. As the number of turns decrease the total length decreases. This affects the resistance of the circuit. Removing the turns by 10% does not affect the area of cross section and resistivity of the wire. By comparing the two cases, analyzing their difference in the value of resistance and the relation between resistance and number of turns we can find the time taken.

Formula used:
\[R = \dfrac{{\rho \times l}}{A}\]Where the resistance of the wire is$R$ , $\rho $ is the resistivity of the wire, $l$ is the length of the wire and $A$ is the area of cross section of the wire.
$P = V \times I$ ,$P$is the power of circuit,$V$ denotes the voltage and $I$ denotes the current flowing through the circuit.
$I = \dfrac{V}{R}$ (Ohm’s Law)
Here $V$ denotes the voltage, $I$ denotes the current flowing through the circuit and $R$denotes the resistance.
$E = P \times t$
$E$ Denotes the energy/heat required, $P$ is the power of the circuit and $t$ shows the time required in seconds.

Complete step by step answer:
The electrical energy is the capacity to do work. This is obtained by the product of power and time. In both the cases the energy/heat remains the same. We know, $P = V \times I$, where $P$ is the power of circuit, $V$ denotes the voltage and $I$ denotes the current flowing through the circuit. Substituting this in the equation energy we get $E = V \times I \times t$.
Using Ohm’s Law ($I = \dfrac{V}{R}$) in the equation of energy /heat $E = V \times I \times t$ , we get
$E = \dfrac{{{V^2}}}{R} \times t$
We know that resistance can be replaced in terms of resistivity$\rho $, area of cross section $A$ and length of the wire $L$.
Replacing \[R = \dfrac{{\rho \times l}}{A}\] in the formula of energy $E = \dfrac{{{V^2}}}{R} \times t$.
We get $E = \dfrac{{A{V^2}}}{{\rho \times L}} \times t$
If each turn has a radius of $r$ the length of one turn would be its perimeter $2\pi r$. If there are $n$ such turns in the initial case then the total length would be $L = n \times (2\pi r)$.
Whereas, in the second case 10% of the number of turns has been reduced. This implies that if $n$ was the initial number of turns then now it has become $0.9n$. Remember in both the cases energy needed is the same. So,
$E = \dfrac{{A{V^2}}}{{\rho \times n \times (2\pi r)}} \times {t_1} = \dfrac{{A{V^2}}}{{\rho \times 0.9n \times (2\pi r)}} \times {t_2}$
This implies $\dfrac{{{t_1}}}{n} = \dfrac{{{t_2}}}{{0.9n}}$
Given that ${t_1}$=16min Therefore $16\min = \dfrac{{{t_2}}}{{0.9}}$
${t_2}$=14.4 min
The correct option is B.

Note: Here the energy required by the kettle will be equal to the heat energy needed to boil the water, the amount of heat energy required is given by the equation $H = ms\Delta t$, $m$ denotes the mass of substance present, $s$ is the specific heat of the liquid(water) and $\Delta t$ shows the time required. Here this will be equal to the energy i.e. $H = ms\Delta t = \dfrac{{A{V^2}}}{{\rho \times n \times (2\pi r)}} \times {t_1} = \dfrac{{A{V^2}}}{{\rho \times 0.9n \times (2\pi r)}} \times {t_2}$.