A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R…
Physics · JEE Advanced · NTA Exams — Properties of Solids and Liquids
A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density ρ where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is

- Mg
- Mg – Vrg
- Mg + pR2hrg
- rg (V + pR2h).
Answer
(D) r g ( V + p R 2 h ).
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- Water of volume 2 litre in a container is heated with a coil of 1 kW at 27°C. The lid of the container is…
- Water from a tap emerges vertically downwards with an initial speed of 1.0 m s – 1 . The cross-sectional area…
- Statement I : When a body floats such that its parts are immersed into two immiscible liquids then force…
- Figure shows the stress-strain graphs for materials A and B. From the graph it follows that
- The cylindrical tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If…
- A cubical block of wood of specific gravity 0.5 and a chunk of concrete of specific gravity 2.5 are fastened…
- Water is flowing continuously from a tap having an internal diameter 8 × 10 –3 m. The velocity as it leaves…
- Statement I : All the raindrops hit the surface of the earth with the same constant velocity. Statement II …
More Properties of Solids and Liquids questions · Browse all practice questions