Answer:
correct option is C) Â 2.8
Step-by-step explanation:
given data
string vibrates form = Â 8 loops
in water loop formed = Â 10 loops
solution
we consider  mass of stone = m
string length = l
frequency of tuning = f
volume = v
density of stone = [tex]\rho[/tex]
case (1) Â
when 8 loop form with 2 adjacent node is [tex]\frac{\lambda }{2}[/tex]
so here
[tex]l = \frac{8 \lambda _1}{2}[/tex] Â Â Â ..............1
[tex]l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}[/tex]
and we know velocity is express as
velocity = frequency × wavelength  .....................2
[tex]\sqrt{\frac{Tension}{mass\ per\ unit \length }}[/tex]  =  f × [tex]\lambda_1[/tex]
here tension = mg
so
[tex]\sqrt{\frac{mg}{\mu}}[/tex]  =  f × [tex]\lambda_1[/tex]   ..........................3
and
case (2) Â
when 8 loop form with 2 adjacent node is [tex]\frac{\lambda }{2}[/tex]
[tex]l = \frac{10 \lambda _1}{2}[/tex] Â Â Â ..............4
[tex]l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}[/tex]
when block is immersed
equilibrium  eq will be
Tenion + force of buoyancy = mg
T + v × [tex]\rho[/tex] × g = mg
and
T = v × [tex]\rho[/tex] - v × [tex]\rho[/tex] × g  Â
from equation 2
f × [tex]\lambda_2[/tex] = f  × [tex]\frac{1}{5}[/tex] Â
[tex]\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}[/tex] Â Â .......................5
now we divide eq 5 by the eq 3
[tex]\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}[/tex]
solve irt we get
[tex]1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}[/tex]
so
relative density [tex]\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}[/tex]
relative density = 2.78 ≈ 2.8
so correct option is C) Â 2.8
