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A piece of wire of resistence 20 ohm drawn out to increase its length to twice as original length .calculate its resistence now.
When the length of the wire is doubled the area of cross section is reduced to half and resistance becomes four times the original value.

Thus,
Initial resistance R is 20 Ω for length L and area of cross section is A
We know,
R = ρL/A  ... (1)
When length is doubled L = 2L area of cross section becomes A =A/2
The new resistance becomes
R' = ρ(2L)/(A/2) .... (2) (resistivity 'ρ' is same for given substance)
= 4(ρL/A)
Thus,
R' = 4R ... (from (1) and (2))
R' = 4 × 20 = 80 Ω

Answered by Shiwani Sawant | 30 Mar, 2020, 02:11: PM

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