The area of cross-section of steel wire is `0.1 cm^(2)` and Young's modulus of steel is `2 times 10^(11)N" "m^(-1)`. The force required to stretched by 0.1% of its length is

Options

A. 1000 N
B. 2000 N
C. 5000 N
D. 4000 N

Answer

B

Explanation

`because` Young.s modulus of steel,
`" "Y=F/(A((Deltal)/l))`
`because` Given, `Y=2 times 10^(11)Nm^(-2)`,
`" "A=0.1 cm^(2)=10^(-5)m^(2)`
and `" "(Deltal)/l=0.1/100=1 times 10^(-3)`
Hence, the force required to stretched by 0.1% of it.s length,
`F=YA((Deltal)/l) rArr F=2 times 10^(11) times 10^(-5) times 1 times 10^(-3)N`
or `" "` F = 2000 N