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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 radius of cross-sections 2r, 1.5r and r. 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 = 9 ? 16 ? 36
B.
V1 ? V2 ? V3 = 3 ? 4 ? 6
C.
V1 ? V2 ? V3 = 2 ? 1.5 ? 1
D.
V1 = V2 = V3

Solution:

The correct answer is V1 ? V2 ? V3 = 9 ? 16 ? 36.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 = 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 CDLength = L, Radius = 1.5 r, Area of cross- section = 2.25 ? r2Resistance, R2 = \(\frac{\rho L}{2.25\pi r^2}\)Voltage, V2 = I R2 = \(\frac{I\rho L}{2.25\pi r^2} \)For Wire BDLength = L, Radius = r, Area of cross- section = ? r2Resistance, R1 = \(\frac{\rho L}{\pi r^2}\)Voltage, V3 = I R3 = \(\frac{I\rho L}{\pi r^2} \)The ratio V1: V2 : V3 = \(\frac{I\rho L}{4\pi r^2}\) : \(\frac{I\rho L}{2.25\pi r^2} \) : \(\frac{I\rho L}{\pi r^2} \) = 1/4 : 1/2.25 : 1 = 1/4 : 100/225 : 1 = 1/4 : 4/9 : 1 = 9 : 16 : 36V1 ? V2 ? V3 = 9 ? 16 ? 36.

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