Question

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A silver wire has a diameter of 1 mm and a resistivity of 1.6 * 10^-6 ohm cm.

How much of this wire would be required to make a 10 ohm coil ?

How much does the resistance change if the diameter is doubled ?

Answer 1

Hey mate..!!

Here's your answer..

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We are given that----

$diameter = 1mm = 1 \times {10}^{ - 3} m \\ resitivity \: = 1.6 \times {10}^{ - 6} ohm \: cm = 1.6 \times {10}^{ - 8} m \\ required \: resistance = 10 \: ohm \\ \\ as \: resistance = resistivity \times length \div area \\ \\ so \: length = 10 \times \pi {r}^{2} \div 1.6 \times {10}^{ - 8} \\ \\ length = 10 \times 3.14 \times \: ({1 \times {10}^{ - 3} })^{2} \div 1.6 \times {10}^{ - 8} \times 4 \: m \\ \\ length = 490.625 \: m \\ \\ since = > \\ \\ resistance \: is \: inversely \: proportional \: to \: the \: square \: of \: diameter \: and \: radius.. \: so \: \\ \\ \\ when \: diameter \: is \: doubled \: resistance \: becomes \: 1 \div 4 \: times....$

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HOPE IT HELPS...

@Rêyaañ11

Answer 2

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