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.

Answer

2.78

Explanation

Let V be the volume and `rho` be the density of the stone. When the stone hangs in air, tension in the string
`T=V rhog`
Frequency of vibration,
`v=1/(2L)sqrt(T/m)=1/(2L) sqrt((Vrhog)/(m))" "...(i)`
When the stone is immersed in water, it loses in weight due to the upthrust of water. Tension in the string decreases to T.
T.= Apparent weight of stone
`= V rhog-Vsigmag = V(rho- sigma)g`
Where `sigma` is the density of water. The length of the wire has to be decreased to L. to bring it in unison with the tuning fork.
`:.v =1/(2L.)sqrt((T.)/m)=1/(2L.)sqrt((V(rho-sigma)g)/m)" "...(ii)`
From equations (i) and (ii), we get
`1/(2L)sqrt((Vrhog)/m)=1/(2L.)sqrt((V(rho-sigma)g)/m)`
or `(L.)/L=sqrt((rho-sigma)/rho)`
Now L = 45 cm, L. = 45 - 9 = 36 cm, `sigma =1gm^(-3)`
`:. 36/45=sqrt((rho-1)/rho)`
`4/5=sqrt((rho-1)/rho)`
or `(rho-1)/rho = 16/25` or `25 rho -25 =16 rho`
or `rho=25//9=2.778 ~~2.78gcm^(-3)`