A small solid sphere of mass M0, of radius R0, and of uniform density 0 is place

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. 1) The new sphere has radius R = R0 and density > 0. 2) The new sphere has mass M = M0 and radius R > R0. 3) The new sphere has radius R < R0 and density > 0. 4) The new sphere has radius R < R0 and density = 0. 5) The new sphere has radius R < R0 and mass M = M0. 6) The new sphere has mass M > M0 and density = 0. A small The possible options for all answers are: 1.R 2.F 3.U 4.R or U 5. F or U 6. R or F or U

Explanation / Answer

initially , mass = M0

radius = R0

density = p0

density = mass / volume

mass = density *volume

M0 = p0* ( 4/3)*pi*R03   -------------------------------1

level L will rise if mass increases.

1) radius same , R = R0

density p increases

from eq 1 , mass is directly proportional to density, so mass will increase.

level = rise (R)

2) mass same = M0

mass unchanged , L unchanged (U).

3) radius dec, R <R0

densirty inc. p>p0

density increase, but size decrease ,unknown net result on mass.

so L = R or F or U.

4) radius dec. R < R0

density same p = p0

size decrease , from eq 1 mass decrease, Level fall

L = Fall

5) radius decrese , R < R0

mass same , M = M0

mass unchanged , Level unchanged ( U)

6) mass inc. M > M0

density same p = p0

mass increase , Level rise ( R)

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