A small solid sphere of mass M 0 , of radius R 0 , and of uniform density 0 is p
ID: 1460788 • Letter: A
Question
A small solid sphere of mass M0, of radius R0, and of uniform density 0 is placed in a large bowl containing water. It floats and the level of the water in the dish is L. Given the information below, determine the possible effects on the water level L, (R-Rises, F-Falls, U-Unchanged), when that sphere is replaced by a new solid sphere of uniform density.
Read it to me
R F U R or U F or U R or F or U The new sphere has density = 0 and mass M > M0
R F U R or U F or U R or F or U The new sphere has density < 0 and radius R > R0
R F U R or U F or U R or F or U The new sphere has radius R > R0 and mass M = M0
R F U R or U F or U R or F or U The new sphere has density = 0 and mass M < M0
R F U R or U F or U R or F or U The new sphere has radius R = R0 and density < 0
R F U R or U F or U R or F or U The new sphere has mass M = M0 and radius R < R0
Explanation / Answer
Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
A floating object displaces an amount of water of the same mass. Assuming all objects float, the water will rise when the new object has higher mass.
"R" water level rise because the density is same so the surface area of M will propoionl to Mo and hence water level will raise propotionally .
density < 0 and radius R > R0
R or F or U { density decrease but radius increase but net mass remain unknown }
3 R > R0 and mass M = M0
3. U { mass is same so displace same mass of water and hence wtaer level remain unchanged }
4. = 0 and mass M < M0
4.F water levell falls because it displace less water as compare to Mo sphere.
5.R = R0 and density < 0
5. F water levell falls because it displace less water as compare to Mo sphere.
6. M = M0 and radius R < R0
U { mass remains same so level remains same }
.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.