1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A hemispherical bowl just floats without sinking in a liquid of density 1.2×103 kg/m3. If the outer diameter and the density of the bowl are 1 m and 2×104 kg/m3 respectively, then the inner diameter of the bowl will be

A
0.92 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.94 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.96 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.98 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 0.98 m Weight of the bowl =mg =Vρg=23π[(D2)3−(d2)3]ρg Where D= Outer diameter d= Inner diameter, ρ= Density of bowl Now, weight of the liquid displaced by the bowl =Vσg=23π(D2)3σg where σ is the density of the liquid By law of floatation, 23π(D2)3σg=23π[(D2)3−(d2)3]ρg ⇒(12)3×1.2×103=[(12)3−(d2)3]2×104 By solving we get, d=0.98 m

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program