Question

2
02A
(4) 1A
If the radius of cross-section of the conductor is increased by 0.1% keeping volume constant, then percent
change in the resistance of the conductor is
(1)-0.2%
(2) -0.1%
(3) -0.3%
(4) -0.4%

NO SPAMMING
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Answer 1

Answer:

then percent  change in the resistance of the conductor is -0.4%

Explanation:

As we know that the volume of the wire is constant

So if the radius of the wire is increased by 0.1 %

in order to find the percentage increase in area

$A = \pi r^2$

$\frac{\Delta A}{A} = 2(\frac{\Delta r}{r}$

$\frac{\Delta A}{A} = 2(0.1)$%

so area is increased by 0.2%

now the volume remains constant

so percentage decrease in length must be 0.2%

now by the formula of resistance

$R = \frac{\rho L}{A}$

$\frac{\Delta R}{R} = \frac{\Delta L}{L} + \frac{\Delta A}{A}$

so we have

$\frac{\Delta R}{R} = -0.2 - 0.2$%

$\frac{\Delta R}{R} = -0.4$%

#Learn

Topic : Resistance of conductor

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