Two conducting wires A and B are made of same material. If the length of B is twice that of A and the radius of circular cross-section of A is twice that of B, then their resistances RA and RB Eire in the ratio

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Q: 89 (NDA-I/2014)
Two conducting wires A and B are made of same material. If the length of B is twice that of A and the radius of circular cross-section of A is twice that of B, then their resistances RA and RB Eire in the ratio

question_subject: 

Maths

question_exam: 

NDA-I

stats: 

0,5,13,6,4,5,3

keywords: 

{'resistances ra': [0, 0, 0, 1], 'ratio': [1, 0, 1, 12], 'radius': [0, 0, 2, 2], 'length': [0, 0, 1, 0], 'same material': [0, 0, 2, 4], 'rb eire': [0, 0, 0, 1]}

In this scenario, we have two conducting wires A and B made of the same material. We are given that the length of wire B is twice that of wire A. Additionally, the radius of the circular cross-section of wire A is twice that of wire B.

Resistance is directly proportional to the length of the wire and inversely proportional to the cross-sectional area. Therefore, we can use the formula for resistance (R = ρ * L / A) to compare the resistances of wires A and B.

Let`s consider the length and radius of wire A as L and R respectively. So, the length of wire B will be 2L and the radius will be R/2.

Plugging the values into the formula, we get the resistance of wire A as RA = ρ * L / (π * R²) and the resistance of wire B as RB = ρ * 2L / (π * (R/2)²).

Simplifying the equations, we get RA = 8 * ρ * L / (π * R²) and RB = ρ * 4L / (π * R²).

Now, we can observe that the ratio of RA to RB is 8:4, which simplifies