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A wire of 4 ohm is doubled to its length and half of diameter. What will be the new resistance |
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Answer» R1= 4ohmL1=LA1=rP1=not givenIt is of new wire, Length of wire is doubled it means 2L.It\'s diameter is half means A2=r=d/4=r/2.To find R2=? P1=not givenSo, R1/R2=p1/p2×L1/L2×A2/A14/R2=p1/p2×L/2L×(r)²/(r)²×(2)²P1 and p2,L and L2 are cancel out. 4/R2=1/2×r²/r²/4r²and r²are canceled. 4/R2=1/2×1/44/R2=1/8R2=32ohmHence, new resistance is 32ohm. We consider wire is doubled on it means to fold the length of wire. It means its length will get half and area of cross section will get doubled.∴ Let the resistance of the wire originally \'R\' of length \'L\' and area of cross-section \'A\' with resistivity of material is \'p\',ThenR = p l/A = 4ΩNow, for new arrangement,p\' = pl\' = l / 2A\' = 2AThus,R\' = p\' l\'/A\' = p (l /2) ÷ 2A = !/4 (p l / A) = 1/4R = 1/4 × 4 = 1Ω.\xa0 |
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