The circuit depicted in the figure is employed to study Ohm's Law. Rather than employing a conventional resistor, a glass tube of length ‘l’ and cross-sectional area ‘a’, which is half-filled with mercury of resistivity ‘d’, is connected to the circuit via two electrodes, E₁ and E₂. These electrodes are linked to a battery with an emf of ‘E’ and negligible internal resistance. Provide the answer in terms of ‘a’, ‘l’, ‘d’, and ‘E’.

(a) Resistance of mercury in the tube.

(b) The ammeter reading.

(c) Voltmeter reading.

(d) Which of the measurements are altered when the tube is entirely filled with mercury?

Given,

Length of glass tube = l

Cross-sectional area of glass tube = a

Resistivity of mercury = d

Internal resistance of battery = 0

(a) When mercury is half filled in glass tube then cross sectional area of mercury = $\dfrac{\text{a}}{2}$

Resistance of mercury in the tube:

$\text{Resistance} = \dfrac{\text{resistivity} \times \text{length of mercury in tube}}{\text{area of cross section of mercury}} \\[1em] \text{R}=\dfrac{\text{d}\times \text{l}}{\dfrac{\text a}{2}} \\[1em] \text{R}=\dfrac{2\text{d}\text{l}}{\text a} \\[1em]$

(b) Reading of ammeter:

$\text I =\dfrac{\text E}{\text R} \\[1em] \text I =\dfrac{\text E}{\dfrac{2\text{d}\text{l}}{\text a}} \\[1em] \text I =\dfrac{\text {Ea}}{2\text{d}\text{l}} \\[1em]$

(c) Reading of voltmeter = E
(it does not depend on applied resistance)

(d) If mercury is entirely filled in tube then cross sectional area of mercury = a
Initially cross sectional area of mercury was $\dfrac{\text a}{2}$

We know that:

$\text{Resistance (R)} = \dfrac{\text{resistivity (d)} \times \text{length of mercury in tube(l)}}{\text{area of cross section of mercury(a)}}$

And

$\text I =\dfrac{\text E}{\text R}$

By changing cross sectional area, the value of resistance and the reading of ammeter will change.