
It takes 12 minutes to boil 1 liter of water in an electric kettle. Due to some defect it becomes necessary to remove 20% turns of the heading coil of the kettle. After repair, how much time will it take to boil 1 liter of water?
A. 9.6 minutes
B. 14.4 minutes
C. 16.8 minutes
D. 18.2 minutes
Answer
163.5k+ views
Hint:In order to solve this problem we have to apply the concept of heating effect of current. As it is given in the question that after repairing the kettle 20% of the turns are removed which means its resistance gets reduced by 20%. By applying this idea we can obtain the correct solution.
Formula Used:
Joule's equation of electrical heating:
$H = {I^2}Rt$
Where H = heat produced, I = current flowing in the circuit, R = resistance of the circuit and t = time for which current flows.
Complete step by step solution:
Heat energy is produced in a conductor when current runs through it. An electric current's ability to heat things depends on three factors:
(1) the conductor's resistance (R). More heat is produced by higher resistance.
(2) The time (t) for which current flows. The amount of heat generated increases with time.
(3) The amount of current (I). The more current flowing, the more heat is produced.
As the voltage of the electric kettle is constant here. In this case the heat produced by the electric kettle is given by,
$H = {I^2}Rt = {\left( {\dfrac{V}{R}} \right)^2}Rt = \dfrac{{Vt}}{R}$.............(1)
20% of the heating coil is removed and the resistance of the electric kettle will decrease by 20%. The new resistance will be,
${R^*} = R - 0.2R = 0.8R$.
The heat produced after the reduced heating coil will be.
$H' = \dfrac{{Vt'}}{{{R^*}}} = \dfrac{{Vt'}}{{0.8R}}$............(2)
Equating the equations (1) and (2) and we will get.
$H = H' \\
\Rightarrow \dfrac{{Vt}}{R} = \dfrac{{Vt'}}{{0.8R}}$
$\Rightarrow t = \dfrac{{t'}}{{0.8}} \\
\Rightarrow t' = 0.8 \times t \\
\therefore t'= 0.8 \times 12 = 9.6m$
Hence, the correct option is option A.
Note: Most of the time students will make mistakes while calculating the resistance of the wire. As in this case some part of the coil is removed so its resistance will definitely change and doing appropriate calculations of it is necessary to find the accurate answer.
Formula Used:
Joule's equation of electrical heating:
$H = {I^2}Rt$
Where H = heat produced, I = current flowing in the circuit, R = resistance of the circuit and t = time for which current flows.
Complete step by step solution:
Heat energy is produced in a conductor when current runs through it. An electric current's ability to heat things depends on three factors:
(1) the conductor's resistance (R). More heat is produced by higher resistance.
(2) The time (t) for which current flows. The amount of heat generated increases with time.
(3) The amount of current (I). The more current flowing, the more heat is produced.
As the voltage of the electric kettle is constant here. In this case the heat produced by the electric kettle is given by,
$H = {I^2}Rt = {\left( {\dfrac{V}{R}} \right)^2}Rt = \dfrac{{Vt}}{R}$.............(1)
20% of the heating coil is removed and the resistance of the electric kettle will decrease by 20%. The new resistance will be,
${R^*} = R - 0.2R = 0.8R$.
The heat produced after the reduced heating coil will be.
$H' = \dfrac{{Vt'}}{{{R^*}}} = \dfrac{{Vt'}}{{0.8R}}$............(2)
Equating the equations (1) and (2) and we will get.
$H = H' \\
\Rightarrow \dfrac{{Vt}}{R} = \dfrac{{Vt'}}{{0.8R}}$
$\Rightarrow t = \dfrac{{t'}}{{0.8}} \\
\Rightarrow t' = 0.8 \times t \\
\therefore t'= 0.8 \times 12 = 9.6m$
Hence, the correct option is option A.
Note: Most of the time students will make mistakes while calculating the resistance of the wire. As in this case some part of the coil is removed so its resistance will definitely change and doing appropriate calculations of it is necessary to find the accurate answer.
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