A copper wire and an aluminium wire have lengths in the ratio 5:2 diameters in the ratio 4:3 and forces applied in the ratio 4:5. Find the ratio of increase in length of the copper wire to that aluminium wier. (Let `Y_(Cu)=1.1xx10^(11)Nm^(-2),Y_(Al)=0.7xx10^(11)Nm^(-2)`)

Options

A. `(178)/(63)`
B. `(63)/(88)`
C. `(189)/(11)`
D. `(33)/(89)`

Answer

B

Explanation

Gvien ratio of length of copper wire and aluminium wire `l_(1):l_(2)=5:2`
and diameter ratio of the wire `=d_(1):d_(2)=4:3`
`:.r_(1):r_(2)=2:(3)/(2)`
If `Deltal_(1)andDeltal_(2)` are increase in the length of copper and aluminium wire, respectively then for copper wire,
`Deltal_(1)=(F_(1)l_(1))/(A_(1)y_(1))" "......(i)`
For aluminium wire,
`Deltal_(2)=(F_(2)l_(2))/(A_(2)y_(2))" "......(ii)`
From Eqs. (i) and (ii) we get
` (Deltal_(1))/(Deltal_(2))=(l_(1))/(l_(2)).(A_(2))/(A_(1)).(y_(2))/(y_(1)).(F_(1))/(F_(2))" "[:'(y_(2))/(y_(1))=(y_(Al))/(y_(Cu))=(0.7)/(1.1)]`
`=(5)/(2).(pir_(2)^(2))/(pir_(1)^(2)).(0.7xx10^(11))/(1.1xx10^(11)).(4)/(5)" "[:'"Given", (F_(1))/(F_(2))=(4)/(5)]`
`=(5)/(2).((3)/(2))^(2)/((2)^(2))xx(7)/(11)xx(4)/(5)=(63)/(88)`
Hence the ratio of increase in length of the copper and aluminium wire will be `(63)/(88)`.