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Question: A 1m long metallic wire is broken into two unequal parts A and B. The part A is uniformly extended i...

A 1m long metallic wire is broken into two unequal parts A and B. The part A is uniformly extended into another wire C. The length of C is twice the length of A and resistance of C is equal to that of B. The ratio of resistances of parts A and C is

A

4

B

14\frac{1}{4}

C

2

D

12\frac{1}{2}

Answer

14\frac{1}{4}

Explanation

Solution

let LA and LB be length of parts A and B

Then RARB\frac{R_{A}}{R_{B}} = LALB\frac{L_{A}}{L_{B}} [as cross-section is same]

Now Lc = 2 LA and (volume)c = (volume)P

i.e. Lc × Ac = 2 LA × Ac = LA × AA

where Ac = AA are cross-sectional area of part C and A.

\ Ac = AA/2

\ RARC\frac{R_{A}}{R_{C}} = = LALC\frac{L_{A}}{L_{C}} × ACAA\frac{A_{C}}{A_{A}}

= LA2LA\frac{L_{A}}{2L_{A}} × AA/2AA\frac{A_{A}/2}{A_{A}} = 14\frac{1}{4}