A steel wire of length l and cross sectional area A is stretched by 1 cm under a given load. When the same load is applied to another steel wire of double its length and half of its cross section area, the amount of stretching (extension) is

Options

A. 0.5
B. 2 cm
C. 4 cm
D. 1.5 cm

Answer

C

Explanation

(c) We know that , `Deltal=(FL)/(AY)`
Here, `Deltal prop (L)/(A)`
According to question
`implies (Deltal_(1))/(Deltal_(2))=(L_(1))/(L_(2))xx(A_(2))/(A_(1))`
`implies (Deltal_(1))/(Deltal_(2))=(1)/(2)xx(1)/(2)implies(Deltal_(1))/(Deltal_(2))=(1)/(4)`
`Deltal_(2)=4Deltal_(1)=4xx1, Deltal_(2)=4 cm`