the frequency of a sonometer wire is 10 Hz. When the weight producing th tensions are completely immersed in water the frequency becomes 80 Hz and on immersing the weight in a certain liquid the frequency becomes 60 Hz. The specific gravity of the liquid is

Options

A. `1.42`
B. `1.77`
C. `0.36`
D. `1.82`

Answer

b

Explanation

Case I `n=1/(2l) sqrt(T_(L)/(pir^(2)p))`
`100 = 1/(2l)sqrt((Vd_(1)g)/(pir^(2)p))` ...(i)
Case II `80=1/(2l) sqrt((Vg(d_(1)-d))/(pir^(2)p))` ...(ii)
Case III `60=1/(2l) sqrt((Vg(d_(1)-d_(2)))/(pir^(2)p))` ...(iii)
where, `d_(1)=` Density of weight
`d_(2)=` Density of water
`d_(2)=` Density of second liquid
From Eqs. (ii) and (i), we get ltbtgt `80/100 = sqrt((d_(1)-d)/d_(1))rArr 4/5 = sqrt(1-d/d_(1))`
`rArr 16/25 = 1 - d/d_(1) rArr therefore d_(1)/d=25/9` ...(iv)
From Eqs. (iii) and (iv), we get
`80/60= sqrt((d_(1)-d)/(d_(1)-d_(2)))rArr 4/3 = sqrt((d_(1)/d-1)/(d_(1)/d-d_(2)/d))`
`rArr 16/9 = (25/9-1)/(25/9-d_(2)/d)`
Putting `d_(1)/d-25/9` from Eqs. (iv)
`rArr 16/9=(16//9)/(25/9-d_(2)/d)rArr 25/9 - d_(2)/d=1 rArrd_(2)/d=16/9`
Thus, The specific gravity of the liquid, `d_(2)/d=16/9=1.77`