
If it takes $8$ minutes to boil a quantity of water electrically, how long will it take to boil the same quantity of the water using the same heating coil but with the current doubled:
A.$32$ minutes
B.$16$ minutes
C.$4$ minutes
D.$2$ minutes
Answer
536.1k+ views
Hint: In order to solve this question, we will use the concept of heat energy produced by a heating coil which helps us to boil a certain quantity of water electrically. And if the same heating coil is used then no other quantities other than current (as mentioned) would change.
Formula used:
$E={{I}^{2}}Rt$
Complete answer:
The heat energy using the heating coil will be used to boil the quantity of water. The energy of a heating coil is as follows:
$E={{I}^{2}}Rt$
Here $I$ is the amount of current flowing through the heating coil,
$R$ is the resistance of the heating coil,
And $t$ is the time taken to boil the quantity of water.
In the question it is stated that the same quantity of water is boiled electrically with varying currents, therefore the energy of the heating coil must remain same:
$I_{1}^{2}R{{t}_{1}}=I_{2}^{2}R{{t}_{2}}$
Since the same heating coil is used, resistance will remain the same.
$I_{1}^{2}{{t}_{1}}=I_{2}^{2}{{t}_{2}}$
It is given that the current is doubled, then ${{I}_{2}}=2{{I}_{1}}$. Substituting the value in the equation, we get:
$\begin{align}
& I_{1}^{2}{{t}_{1}}={{\left( 2{{I}_{1}} \right)}^{2}}{{t}_{2}} \\
& \Rightarrow I_{1}^{2}{{t}_{1}}=4I_{1}^{2}{{t}_{2}} \\
& \Rightarrow {{t}_{1}}=4{{t}_{2}} \\
& \Rightarrow {{t}_{2}}=\dfrac{{{t}_{1}}}{4} \\
\end{align}$
It is given that ${{t}_{1}}=8$ minutes, hence:
$\begin{align}
& {{t}_{2}}=\dfrac{8}{4} \\
& \therefore {{t}_{2}}=2\text{ min} \\
\end{align}$
Thus, the time taken to boil the same quantity of water using the same heating coil but with the current developed will be $2$ minutes
Hence option $D$ is correct.
Note:
If the current was doubled in the heating coil, then we could have definitely concluded that the heat energy produced by the heating coil would have increased and hence the heating coil would take less amount of time to boil the same quantity of water, but to find out the exact value we need to use the formula $E={{I}^{2}}Rt$.
Formula used:
$E={{I}^{2}}Rt$
Complete answer:
The heat energy using the heating coil will be used to boil the quantity of water. The energy of a heating coil is as follows:
$E={{I}^{2}}Rt$
Here $I$ is the amount of current flowing through the heating coil,
$R$ is the resistance of the heating coil,
And $t$ is the time taken to boil the quantity of water.
In the question it is stated that the same quantity of water is boiled electrically with varying currents, therefore the energy of the heating coil must remain same:
$I_{1}^{2}R{{t}_{1}}=I_{2}^{2}R{{t}_{2}}$
Since the same heating coil is used, resistance will remain the same.
$I_{1}^{2}{{t}_{1}}=I_{2}^{2}{{t}_{2}}$
It is given that the current is doubled, then ${{I}_{2}}=2{{I}_{1}}$. Substituting the value in the equation, we get:
$\begin{align}
& I_{1}^{2}{{t}_{1}}={{\left( 2{{I}_{1}} \right)}^{2}}{{t}_{2}} \\
& \Rightarrow I_{1}^{2}{{t}_{1}}=4I_{1}^{2}{{t}_{2}} \\
& \Rightarrow {{t}_{1}}=4{{t}_{2}} \\
& \Rightarrow {{t}_{2}}=\dfrac{{{t}_{1}}}{4} \\
\end{align}$
It is given that ${{t}_{1}}=8$ minutes, hence:
$\begin{align}
& {{t}_{2}}=\dfrac{8}{4} \\
& \therefore {{t}_{2}}=2\text{ min} \\
\end{align}$
Thus, the time taken to boil the same quantity of water using the same heating coil but with the current developed will be $2$ minutes
Hence option $D$ is correct.
Note:
If the current was doubled in the heating coil, then we could have definitely concluded that the heat energy produced by the heating coil would have increased and hence the heating coil would take less amount of time to boil the same quantity of water, but to find out the exact value we need to use the formula $E={{I}^{2}}Rt$.
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