Two strings of the same material and the same area of cross-section are used in an wxperiment. One is loaded with 12 kg and the other with 3 kg The fundamental fequency of the first string is equal to the first overtone of the second string. If the length of the second string is 100 cm, then the length of the first string is

Options

A. 300 cm
B. 200 cm
C. 100 cm
D. 50 cm

Answer

c

Explanation

The frequency of the string
`n = 1/(2l) sqrt(T/m)=1/(2l) sqrt((Mg)/(pir^(2)d))(because T=Mg, m = pi r^(2)d)`
`rArr n_(1) =1/(2l_(1)) sqrt((Mg)/(pir^(2)d))and n_(2) =2/(2l_(2)) sqrt((M_(2)g)/(pir^(2)d))`
From the question`n_(1) n_(2) rArr l_(2)/l_(1)=2sqrt(M_(2)/M_(1)`
`rArr 100/l_(1) = 2sqrt(3/12)=2sqrt(1/4)=2xx1/2rArrl_(1)=100` cm