Two wires `A` and `B` have equal lengths and are made of the same material , but diameter of wire `A` is twice that of wire `B`. Then, for a given load,

Options

A. The extension of `B` will be four times that of `A`
B. the extension of `A` and `B` will be equal
C. the strain in `B` is four times that in `A`
D. the strains in `A` and `B` will be equal

Answer

A, C

Explanation

Area of cross section
`A=(pid^(2))/4`
where `d` is the diameter of the wire. Therefore
`l=(4FL)/(pid^(2)Y)`
since `F, L` and `Y` are the same for wires `A` and `B`
`:. Lprop1/(d^(2))`
i.e., the extension is inversely proportional to the square of the diameter. Hence choice a is correct.The strain is
`l/L=(4\F)/(pid^(2)Y)`
Thus strain `prop1/(d^(2))`
i.e., the extension is inversely proportional material to the square of the diameter. Hence choice a is correct. The strain is
`l/L=(4F)/(pid^(2)Y)`
Thus, strain`prop1/(d^(2))`
Hence the correct choice are a and c.