Answer:
Explanation:
Let the volume of bronze statue be V .
mass = vd where d is its density . If T₁ be the tension in cable when bronze statue hangs in air from it
T₁ = mg = vdg
when the statue is in water , tension in cable T₂
T₂ = v g ( d - 1 ) , density of water = 1 , here Tension reduces due to buoyant force acting on the statue .
If n be the frequency of cable when tension in it is T
n is proportional to √T .
[tex]\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2} }[/tex]
[tex]\frac{203}{187} = \sqrt{\frac{vdg}{vg(d-1)} }[/tex]
1.178 = d / (d-1 )
1.178 d - 1.178 = d
.178 d = 1.178
d = 6.6 gm / cc
