A stretched wire emits a fundamental note of frequency 256Hz. Keeping the stretching force constant and reducing the length of the wire by 10cm, frequency becomes 320Hz. Calculate original length of the wire.

Options

A. 30 cm
B. 40 cm
C. 50 cm
D. 60 cm

Answer

C

Explanation

`n = 1/(2L) sqrt(T/m) :. 256 = 1/(2L) sqrt(T/m) ` .....(1)
and ` 320 = 1/(2(L - 10)) sqrt(T/m) ` ...(2)
Dividing (2) by (1) , we get
`(932)/(256) = (2l)/(2(L - 10)) :. 5/4 = L/(L - 10) `
` :. 5L - 50 = 4L :. L = 50 ` cm