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A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. then for the two wires to have the same resistance, the ratio lB/lA of their respective lengths must be
- a)1
- b)1/2
- c)1/4
- d)2
Correct answer is option 'D'. Can you explain this answer?
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A material 'B' has twice the specific resistance of 'A'...
To understand why the correct answer is option 'D', let's break down the problem step by step.
1. Specific Resistance:
Specific resistance is a property of a material that determines its ability to conduct electric current. It is denoted by the symbol "ρ" and is measured in ohm-meter (Ω·m). The specific resistance of material A is given as ρA, and the specific resistance of material B is given as ρB. We are given that ρB = 2ρA.
2. Diameter of the wires:
The wire made of material A has a certain diameter, denoted as dA. The wire made of material B has twice the diameter of the wire made of A, denoted as dB = 2dA.
3. Resistance of the wires:
The resistance of a wire is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A). Mathematically, resistance (R) is given by the formula R = ρ(l/A).
Let's calculate the resistance of the wire made of material A:
RA = ρA(lA/π(dA/2)^2) [Using the formula for resistance]
RA = ρA(lA/π(dA^2/4)) [Simplifying the denominator]
RA = (4ρAlA)/(πdA^2) [Multiplying numerator and denominator by 4]
RA = (4ρAlA)/(πdA × dA) [Simplifying the expression]
Similarly, let's calculate the resistance of the wire made of material B:
RB = ρB(lB/π(dB/2)^2) [Using the formula for resistance]
RB = ρB(lB/π((2dA)/2)^2) [Substituting the given value of dB]
RB = ρB(lB/πdA^2) [Simplifying the expression]
RB = (2ρAlB)/(πdA^2) [Substituting the value of ρB]
RB = (2 × 2ρAlB)/(πdA × dA) [Simplifying the expression]
4. Equating the resistances:
To make the resistances of the two wires equal, we need to equate RA and RB:
(4ρAlA)/(πdA × dA) = (2 × 2ρAlB)/(πdA × dA) [Equating RA and RB]
4ρAlA = 4ρAlB [Canceling out common terms]
Dividing both sides by 4ρAlA gives:
1 = lB/lA
Therefore, the ratio of their respective lengths must be 1, which corresponds to option 'D'.
1. Specific Resistance:
Specific resistance is a property of a material that determines its ability to conduct electric current. It is denoted by the symbol "ρ" and is measured in ohm-meter (Ω·m). The specific resistance of material A is given as ρA, and the specific resistance of material B is given as ρB. We are given that ρB = 2ρA.
2. Diameter of the wires:
The wire made of material A has a certain diameter, denoted as dA. The wire made of material B has twice the diameter of the wire made of A, denoted as dB = 2dA.
3. Resistance of the wires:
The resistance of a wire is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A). Mathematically, resistance (R) is given by the formula R = ρ(l/A).
Let's calculate the resistance of the wire made of material A:
RA = ρA(lA/π(dA/2)^2) [Using the formula for resistance]
RA = ρA(lA/π(dA^2/4)) [Simplifying the denominator]
RA = (4ρAlA)/(πdA^2) [Multiplying numerator and denominator by 4]
RA = (4ρAlA)/(πdA × dA) [Simplifying the expression]
Similarly, let's calculate the resistance of the wire made of material B:
RB = ρB(lB/π(dB/2)^2) [Using the formula for resistance]
RB = ρB(lB/π((2dA)/2)^2) [Substituting the given value of dB]
RB = ρB(lB/πdA^2) [Simplifying the expression]
RB = (2ρAlB)/(πdA^2) [Substituting the value of ρB]
RB = (2 × 2ρAlB)/(πdA × dA) [Simplifying the expression]
4. Equating the resistances:
To make the resistances of the two wires equal, we need to equate RA and RB:
(4ρAlA)/(πdA × dA) = (2 × 2ρAlB)/(πdA × dA) [Equating RA and RB]
4ρAlA = 4ρAlB [Canceling out common terms]
Dividing both sides by 4ρAlA gives:
1 = lB/lA
Therefore, the ratio of their respective lengths must be 1, which corresponds to option 'D'.
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Community Answer
A material 'B' has twice the specific resistance of 'A'...
ρB = 2ρA
dB = 2dA
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A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. then for the two wires to have the same resistance, the ratio lB/lA of their respective lengths must bea)1b)1/2c)1/4d)2Correct answer is option 'D'. Can you explain this answer?
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