Answer

Verified

95.7k+ views

**Hint:**The heat produced during the flow of current through the conductor due to collision of electrons inside the conductor moving under the applied potential difference is absorbed by the conductor which leads to increases in its temperature.

**Formula used:**

$H = {I^2}Rt$ Where $H$ is heat produced $I$ is the current $R$ is the resistance and $t$ is time.

$H = ms\Delta T$ Where $m$ is the mass of the body $s$ is the body’s specific heat capacity and $\Delta T$ is the change in temperature

**Complete step by step solution:**

In the question the temperature of the conductor is increasing as current flows through it as heat is produced during the flow of current through the conductor due to collision of electrons inside the conductor moving under the applied potential difference, this heat is absorbed by the conductor which leads to a rise in its temperature.

We know that,

$ \Rightarrow H = {I^2}Rt$

$H$ is heat produced $I$ is the current $R$ is the resistance and $t$ is time.

Let the resistance of wire be $R$ initially current flowing through conductor be $I$and heat produced be${H_1}$and in the second case the current flowing through conductor be $2I$and heat produced be${H_2}$

Hence,

\[ \Rightarrow {H_1} = {I^2}Rt\]

$ \Rightarrow {H_2} = {(2I)^2}Rt$

As the heat is produced during the flow of current is the heat is absorbed by the conductor,

(We know that $H = ms\Delta T$ where$H$ is heat produced, $m$is the mass of the body, $s$ is body’s specific heat capacity, and $\Delta T$ is the change in temperature)

\[ \Rightarrow {H_1} = ms\Delta {T_1}\] ($\Delta {T_1}$is the initial change in temperature)

\[ \Rightarrow {H_2} = ms\Delta {T_2}\]($\Delta {T_2}$is the initial change in temperature)

From the above equations, we can assert that,

\[ \Rightarrow {I^2}Rt = ms\Delta {T_1}\]

\[ \Rightarrow {(2I)^2}Rt = ms\Delta {T_2}\]

Taking the ratio of the above two we get,

\[ \Rightarrow \dfrac{{{I^2}Rt}}{{{{(2I)}^2}Rt}} = \dfrac{{ms\Delta {T_1}}}{{ms\Delta {T_2}}}\]

\[ \Rightarrow \dfrac{1}{4} = \dfrac{{\Delta {T_1}}}{{\Delta {T_2}}}\]

$ \Rightarrow \Delta {T_2} = 4\Delta {T_1}$

It’s given in the question that $\Delta {T_1}$ is\[{5^0}C\].

$ \Rightarrow \Delta {T_2} = {20^0}C$

**Hence the correct answer is (C)\[{20^0}C\].**

**Note:**As current flows through the conductor its temperature increases and this also increases the resistance of the wire but this increase is very small thus we have neglected this change in the resistance of the wire in the calculation done above.

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

Other Pages

If the number of integral terms in the expansion of class 11 maths JEE_Main

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

When three capacitors of equal capacities are connected class 12 physics JEE_Main

A flask contains a mixture of compound A and B Both class 12 chemistry JEE_Main