Two wires have their lengths, diameters and resistivities, all in the ratio of 1 : 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is

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Q: 100 (IAS/2001)
Two wires have their lengths, diameters and resistivities, all in the ratio of 1 : 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is

question_subject: 

Science

question_exam: 

IAS

stats: 

0,13,23,10,13,11,2

keywords: 

{'thinner wire': [0, 0, 1, 0], 'thicker wire': [0, 0, 1, 0], 'resistance': [0, 0, 1, 2], 'wires': [0, 0, 1, 1], 'resistivities': [0, 0, 1, 0], 'ratio': [1, 0, 1, 12], 'diameters': [0, 0, 2, 0], 'lengths': [0, 0, 1, 0], 'ohms': [0, 0, 1, 0]}

The resistance of a wire is determined by its resistivity, length, and diameter or cross-sectional area. The formula is Resistance (R) = Resistivity (ρ) * Length (L) / Area (A).

In this problem, the lengths and resistivities are twice as large for the thicker wire, which would suggest that its resistance would be four times greater. However, because the diameter is also twice as large, the area (π*(d/2)^2) of the thicker wire is four times the area of the thinner wire. Therefore, the increase in resistance due to a longer length and higher resistivity is offset by the decrease in resistance due to the larger area or diameter.

So, the resistance remains the same as the thinner wire, which is 10 ohms. So, option 1 is the correct answer.

Alert - correct answer should be 10 ohms.

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