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)`) [

Answer

240

Explanation

`c=v lambda and c=sqrt(T//m) therefore sqrt((T)/(m))=vlambda`
Where T= tension in the string and m= mass per unit length of wire When 50.7 kg mass is suspended for fundamental mode `lambda= 21`
`V_1xx21=sqrt((50.7xxg)/(m))`…(i)
When mass is submerged in water, new tension
`T_2=weight-upthrust =50.7g-0.0075xx1000xxg=432g`
`therefore V_2xx21 =sqrt((43.2g)/(m))`.....(ii)
On dividing Eqs (i) and (ii), we get :
`(V_2)/(V_1)=sqrt((43.2)/(50.7))implies v_2=v_1sqrt((43.2)/(50.07))=240 Hz`