In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m 3. The fundamental frequency of the wire is 260 Hz . If the suspended mass is completely submerged in water, the fundamental frequency will become (take `g = 10 ms^(-2)`) [

Options

A. 200 Hz
B. 220 Hz
C. 230 Hz
D. 240 Hz

Answer

B

Explanation

Initially, `260=(1)/(2L)sqrt((T_(1))/(mu))" "…(i)`
Tension `T_(1)=50.7g = 507 N`
When the mass is submerged,
Upthrust = `(0.0075m^(3))(10^(3)kgm^(-3))(10ms^(-2))=75N`
New tension, `T_(2)=(507-75)N=432N`
New fundamental frequency `upsilon.=(1)/(2L)sqrt((T_(2))/(mu))" "...(ii)`
Divide (ii) by (i), we get
`(upsilon.)/(260)=sqrt((T_(2))/(T_(1)))=sqrt((432)/(507))=sqrt((144)/(169))=(12)/(13)orupsilon.=240Hz`