A load suspended by a massless spring produces an extention of xcm in equilibrium.When it is cut into two unequal parts ,the same load produces an extention of 7.5cm when suspended by the larger part of length 60cm .When it is suspended by the smaller part,the extention is 0.5cm.Then

Options

A. x=12.5cm
B. x=3.0cm
C. the length of the original spring is 90cm
D. the length of the original spring is 80cm

Answer

A

Explanation

(a) Elongation of the wire `Deltal=(Fl)/(AY)` (can also be applied for a spring)
`therefore Deltal prop l`
`(7.5)/(5.0)=(60)/(l_(2))`
`therefore l_(2)=40 cm`
`therefore` Length of originl is (60+40)cm=100 cm
Now `(x)/(7.5)=(100)/(60)`
`therefore x=12.5 cm`