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A cylindrical wire AB consists of three equal parts AC, CD and DB, made of the same material but have different radii of cross-sections r, 2r and 3r, respectively. A battery is connected across the wire. Let V1, V2 and V3 be the potential differences across AC, CD and DB, respectively. Then:

A.
V1 ? V2 ? V3 = 36 ? 9 ? 4
B.
V1 ? V2 ? V3 = 1 ? 4 ? 9
C.
V1 = V2 = V3
D.
V1 ? V2 ? V3 = 1 ? 2 ? 3

Solution:

The correct answer is V1 ? V2 ? V3 = 36 ? 9 ? 4.Concept:Ohm's Law:It states that the potential difference between two points is directly proportional to the current passing through the conductor, provided the temperature remains the same. V = IR, where V = potential difference between two points, I = current, R = Resistance.Resistance:The property of a substance to resist the flow of current through it. R = ? L / A, where ? = resistivity, L = length of conductor, and A = area of cross section.Explanation:The current remains the same in series combination. Let the current flowing through the wire be I ampere. For Wire ACLength = L, Radius = r, Area of cross- section = ? r2Resistance, R1 = \(\frac{\rho L}{\pi r^2}\)Voltage, V1 = I R1 = \(\frac{I\rho L}{\pi r^2}\)For Wire CDLength = L, Radius = 2r, Area of cross- section = 4? r2Resistance, R1 = \(\frac{\rho L}{4\pi r^2}\)Voltage, V1 = I R1 = \(\frac{I\rho L}{4\pi r^2} \)For Wire BDLength = L, Radius = 3r, Area of cross- section = 9? r2Resistance, R1 = \(\frac{\rho L}{9\pi r^2}\)Voltage, V1 = I R1 = \(\frac{I\rho L}{9\pi r^2} \)The ratio V1: V2 : V3 = \(\frac{I\rho L}{\pi r^2}\) : \(\frac{I\rho L}{4\pi r^2} \) : \(\frac{I\rho L}{9\pi r^2} \) = 1 : 1/4 : 1/9 = 36 : 9 : 4. V1 ? V2 ? V3 = 36 ? 9 ? 4.

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