
Water boils in the electric kettle in \[15\] minutes after switching it on. If the length of the heating wire is decreased to \[\dfrac{2}{3}\] of its initial value, then the same amount of water will boil with the same supply voltage in
A. \[8\] minutes
B. \[10\] minutes
C. \[12\] minutes
D. \[15\] minutes
Answer
161.4k+ views
Hint: The amount of heat necessary to bring water to a boil is constant. The total length reduces as the number of turns decreases. This has an effect on the circuit's resistance. We can calculate the time taken by comparing the two examples, studying the difference in resistance value and the relationship between resistance and number of spins.
Complete step by step solution:
We have been given the question that water boils in the electric kettle in \[15\] minutes after switching on and the length of the heating wire is decreased to \[2/3\] of its initial value. We have been already known that the heat produced is given as,
\[{\rm{H}} = \dfrac{{{{\rm{V}}^2}}}{{\rm{R}}}\]
The same and equal amount of heat will produce, since supply voltage is the same.
\[\dfrac{{{R^1}}}{{{t^1}}} = \dfrac{{{R^2}}}{{{t^1}}}\]
The above expression can be written as,
\[\dfrac{{{R^1}}}{{{R^2}}} = \dfrac{{{t^1}}}{{{t^2}}} \ldots {\rm{ (1) }}\]
We have been already known that the length of the wire is termed as I,
\[{\rm{R}} = \dfrac{{\rho {\rm{I}}}}{{\rm{A}}}\]
This implies that,
\[ \Rightarrow {\rm{R}} \propto {\rm{I}}\]
Thus, we can write it as,
\[ \Rightarrow \dfrac{{{{\rm{R}}_1}}}{{{{\rm{R}}_2}}} = \dfrac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} \ldots {\rm{ (2)}}\]
Now, from the equation (1) and equation (2), we get
\[\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{t_1}}}{{{t_2}}}\]……… (3)
We have been given the final length,
\[{I_2} = \dfrac{2}{3}{I_1}\]
\[ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{3}{2}\]
On solving the above by equation (3), we get
\[\dfrac{3}{2} = \dfrac{{15}}{{{t_2}}}\]
\[ \Rightarrow {t_2} = 10\] Minutes.
Therefore, the same amount of water will boil with the same supply voltage in 10 minutes.
Hence, option B is correct.
Note:The energy required by the kettle is equal to the heat energy required to boil the water, which is given by the equation \[{\rm{H = ms}}\Delta {\rm{t}}\] where m signifies the mass of substance present, s is the specific heat of the liquid (water), and \[\Delta {\rm{t}}\] denotes the time required.
Complete step by step solution:
We have been given the question that water boils in the electric kettle in \[15\] minutes after switching on and the length of the heating wire is decreased to \[2/3\] of its initial value. We have been already known that the heat produced is given as,
\[{\rm{H}} = \dfrac{{{{\rm{V}}^2}}}{{\rm{R}}}\]
The same and equal amount of heat will produce, since supply voltage is the same.
\[\dfrac{{{R^1}}}{{{t^1}}} = \dfrac{{{R^2}}}{{{t^1}}}\]
The above expression can be written as,
\[\dfrac{{{R^1}}}{{{R^2}}} = \dfrac{{{t^1}}}{{{t^2}}} \ldots {\rm{ (1) }}\]
We have been already known that the length of the wire is termed as I,
\[{\rm{R}} = \dfrac{{\rho {\rm{I}}}}{{\rm{A}}}\]
This implies that,
\[ \Rightarrow {\rm{R}} \propto {\rm{I}}\]
Thus, we can write it as,
\[ \Rightarrow \dfrac{{{{\rm{R}}_1}}}{{{{\rm{R}}_2}}} = \dfrac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} \ldots {\rm{ (2)}}\]
Now, from the equation (1) and equation (2), we get
\[\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{t_1}}}{{{t_2}}}\]……… (3)
We have been given the final length,
\[{I_2} = \dfrac{2}{3}{I_1}\]
\[ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{3}{2}\]
On solving the above by equation (3), we get
\[\dfrac{3}{2} = \dfrac{{15}}{{{t_2}}}\]
\[ \Rightarrow {t_2} = 10\] Minutes.
Therefore, the same amount of water will boil with the same supply voltage in 10 minutes.
Hence, option B is correct.
Note:The energy required by the kettle is equal to the heat energy required to boil the water, which is given by the equation \[{\rm{H = ms}}\Delta {\rm{t}}\] where m signifies the mass of substance present, s is the specific heat of the liquid (water), and \[\Delta {\rm{t}}\] denotes the time required.
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