A stone hangs in air from one end of a wire, which is stretched over a sonometer. The wire is in unison with a certain tuning fork when the bridges of sonometer are 45 cm apart. Now the stone hangs immersed in water at `4^(@)C` and the distance between the bridges has to be altered by 9cm to re-establish unison of the wire with the same fork. Calculate the density of the stone.

Explanation

Suppose M is mass of stone and `rho` is density of stone.
Initial tension in the wiere , `T_(1)=Mg`
On immersing the stone in water, it loses weight `=` weight of water displaced.
Taking density of water `= 1gram//c.c.,`
Final tension, `T_(2)=(Mg-Mg//rho)`
As tension has reduced, the resonating length would reduced.
As `L_(1)=45cm, L_(2)=(45-9)cm=36cm.` ltbr In both cases, wire resonates with the same tuning fork. Therefore, frequency of wire in both cases must be same.
`(1)/(2L)sqrt((T_(1))/(m))=(1)/(2L_(2))sqrt((T_(2))/(m))`
`sqrt((T_(1))/(T_(2)))=(L_(1))/(L_(2))=(45)/(36)=(5)/(4)`
`sqrt((Mg)/((Mg-Mg//rho)))=(5)/(4)`
`(1)/(1-1//rho)=((5)/(4))^(2)=(25)/(16)`
`25-(25)/(rho)=16`
`rho=(25)/(9)=2.8gm//cm^(3)`