यदि एक तार को तीन हिस्सों में बांटे तो उनकी मूल आवृत्तियाँ `n_(1),n_(2)&n_(3)` है तथा पुरे तार की मूल आवृत्ति n है तो-

Options

A. `n=n_(1)+n_(2)+n_(3)`
B. `(1)/(n)=(1)/(n_(1))+(1)/(n_(2))+(1)/(n_(3))`
C. `(1)/(sqrtn)=(1)/(sqrt(n_(1)))+(1)/(sqrt(n_(2)))+(1)/(sqrt(n_(3)))`
D. `sqrt(n)=sqrt(n_(1))+sqrt(n_(2))+sqrt(n_(3))`.

Answer

B

Explanation

`n=(1)/(2l)sqrt((T)/(m))`
या `nprop(1)/(l)`
या `nl=`नियतांक, K
`therefore n_(1)l_(1)=K`,
`n_(2)l_(2)=K,n_(3)l_(3)=K`
Also, `l=l_(1)+l_(2)+l_(3)`
`(K)/(n)=(K)/(n_(1))+(K)/(n_(2))+(K)/(n_(3))`
`(1)/(n)=(1)/(n_(1))+(1)/(n_(2))+(1)/(n_(3))`