Question

In specific resistance measurement of a wire using a meter bridge, the key k in the main circuit is kept open when we are not taking readings. The reason is
A. The emf of cell will decrease
B. The value of resistance will change due to the Joule heating effect.
C. The galvanometer will stop working.
D. None of these.

Answer 1

Hint: Express the specific resistance (resistivity) of the wire in terms of resistance and its length. Recall the effect of temperature on the resistance of the wire. Use Joule’s law of heating to answer this question.

Complete step by step solution:
We know that the specific resistance is the resistivity. The resistance of the wire of length l and cross-sectional area A is given as,
\[R = \rho \dfrac{l}{A}\]
\[ \Rightarrow \rho = \dfrac{{RA}}{l}\]
Here, \[\rho \] is the resistivity and R is the resistance.
Therefore, we can see that the resistivity depends upon the resistance R.
We have according to Joule’s heating effect, when the current passes through a conductor of resistance R for time t, the heat produced in the conductor is given as,
\[H = {I^2}Rt\]
Here, I am the current through the conductor.
Therefore, when we keep the key open for a longer time, the heat produced in the wire will be high. The heat generated in the wire increases the temperature. We know that the resistance changes with temperature by the expression,
\[R = {R_0}\left( {1 + \alpha \Delta T} \right)\]
Here, \[{R_0}\] is the original resistance, \[\alpha \] is the temperature coefficient of resistance, and \[\Delta T\] is the change in temperature.
If we open the key as soon as we take the readings, the heat will not generate in the short span of time. So, the specific resistance of the wire will not change.
So, the correct answer is option (B).
The emf of the cell does not decrease with continuous flow of supply. Therefore, the option (A) is incorrect.
The galvanometer does not stop working as long as current passing through it becomes zero. Therefore, the option (C) is incorrect.

So, the correct answer is “Option B”.

Note:
To answer this question, students must be familiar with the meter bridge experiment. If we know the current and resistance of the wire, we can express the power rating of the wire as, \[P = {I^2}R\]. Therefore, the heat generating in the wire will be expressed as, \[H = Pt\].