Answer
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Hint: We will first find the resistance when the heater is connected to a 220V supply using the formula \[R = \dfrac{{Supply\,voltage}}{{Supply\,current}}\] .
Then we use the formula of dependence of resistance with temperature i.e. \[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] as the resistance vary with the temperature.
After that we will calculate the temperature by putting all the values given in the question.
Complete step by step solution
It is given resistance of wire in a heater at room temperature i.e. \[{R_1} = 65\Omega \] and temperature is given i.e. room temperature \[{T_1} = 27\;^\circ C\] .
Now when the heater is connected to a 220V supply resistance \[{R_2} = \dfrac{{Supply\,voltage}}{{Supply\,current}}\] , now we put the values as given in question.
So \[{R_2} = \dfrac{{220V}}{{2.8A}} = 78.6\Omega \]
Now, using the relation
\[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] , where \[{R_2}\] is the resistance at temperature \[{T_2}\] , \[{R_1}\] is the resistance at temperature \[{T_1}\] and \[\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}}\] .
we can find \[{T_2} - {T_1}\] from this equation as we know the value of \[{R_2},{R_1}\] and \[\alpha \] .
So, we solve this equation:
\[{R_2} = {R_1} + {R_1}\alpha ({T_2} - {T_1})\]
\[{T_2} - {T_1} = \dfrac{{{R_2} - {R_1}}}{{{R_1}}} \times \dfrac{1}{\alpha }\] , now we put the values and find the value of \[{T_2} - {T_1}\]
\[{T_2} - {T_1} = \dfrac{{78.6 - 65}}{{65}} \times \dfrac{1}{{1.7 \times {{10}^{ - 4}}}}\] =1231
So \[{T_2} = 1231 + {T_1}\] and we know that \[{T_1} = 27\;^\circ C\]
So, after putting value of \[{T_1}\] in \[{T_2}\] we get:
\[{T_2} = 1258\;^\circ C\] .
So, option D is correct.
Note: Always remember that we will take the temperature of room \[27\;^\circ C\] and also the formula of resistance when a conductor is connected with supply voltage.
Also remember the general rule i.e. resistivity increases with increasing temperature in conductors and decreases with increasing temperature in insulators.
Then we use the formula of dependence of resistance with temperature i.e. \[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] as the resistance vary with the temperature.
After that we will calculate the temperature by putting all the values given in the question.
Complete step by step solution
It is given resistance of wire in a heater at room temperature i.e. \[{R_1} = 65\Omega \] and temperature is given i.e. room temperature \[{T_1} = 27\;^\circ C\] .
Now when the heater is connected to a 220V supply resistance \[{R_2} = \dfrac{{Supply\,voltage}}{{Supply\,current}}\] , now we put the values as given in question.
So \[{R_2} = \dfrac{{220V}}{{2.8A}} = 78.6\Omega \]
Now, using the relation
\[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] , where \[{R_2}\] is the resistance at temperature \[{T_2}\] , \[{R_1}\] is the resistance at temperature \[{T_1}\] and \[\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}}\] .
we can find \[{T_2} - {T_1}\] from this equation as we know the value of \[{R_2},{R_1}\] and \[\alpha \] .
So, we solve this equation:
\[{R_2} = {R_1} + {R_1}\alpha ({T_2} - {T_1})\]
\[{T_2} - {T_1} = \dfrac{{{R_2} - {R_1}}}{{{R_1}}} \times \dfrac{1}{\alpha }\] , now we put the values and find the value of \[{T_2} - {T_1}\]
\[{T_2} - {T_1} = \dfrac{{78.6 - 65}}{{65}} \times \dfrac{1}{{1.7 \times {{10}^{ - 4}}}}\] =1231
So \[{T_2} = 1231 + {T_1}\] and we know that \[{T_1} = 27\;^\circ C\]
So, after putting value of \[{T_1}\] in \[{T_2}\] we get:
\[{T_2} = 1258\;^\circ C\] .
So, option D is correct.
Note: Always remember that we will take the temperature of room \[27\;^\circ C\] and also the formula of resistance when a conductor is connected with supply voltage.
Also remember the general rule i.e. resistivity increases with increasing temperature in conductors and decreases with increasing temperature in insulators.
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