.. 9 = M × (s — S) and 7 = M′ × (s' – S)

.. M = 14

whence s and s'.

S

S- S

=

7

665.

Let M be the magnitude of the brass when weighed in vacuo, M + x when weighed in water; and let s, s' be the specific gravities of brass and gold, and S that of water; then the magnitude of the gold is

666.

of the fluids.

The altitudes are inversely as the specific gravities
Therefore

1:10: 140 inches the altitude required.

667.

If a be the side of the cube, the pressures upon

the

base and four faces of the cube are respectively

a

aa × a, and 4a" × 1

or as 1 : 2.

668.

Let M be the magnitude of the globe, s' the specific

gravity; then its absolute weight is

Ms'

and its weight in air is

WMs Ms M. (s's)

Hence if 2r be the diameter required, we have

669.

Let a be the length of the cylinder, r the radius of its base, s its specific gravity, and S that of the fluid; also let x be the depth required; then (Vince, Prop. XVI.)

If s be > S, as in the enunciation, the cylinder will not be at rest.

670.

Letr be the radius of the base of the cone, a its altitude, and the distance required; then since the area of the section is

671.

Let α, b be the vertical and horizontal sides of the rectangle, and suppose the required line to divide the base into the two parts x and b r; then the areas of the two segments of the rectangle are

672. If a denote the depth of the part immersed; then its volume or magnitude is (Vince's Fluxions)

s and S being the specific gravities of the body and the fluid.

673.

Let w be the given weight of the vessel, r its radius; then since the volume of any segment of a sphere whose axis is a, is

Hence if s be the specific gravity of the fluid, we have

Again, let x be additional weight required; then since by the question this, together with the vessel, is to have the effect of displacing

674.

Let a be the altitude of the cylinder, r the radius of

its base; then the pressure upon the base is measured by

πρα.

Again, let x be the breadth of the first annulus; then the pressure upon it is

Again, let ' be the breadth of the second annulus; then the pressure upon it is

Again, let a" be the breadth of the third annulus; then we

Now the height is

a = √ra × {1 + (√ 2 − 1) + (√3 − √ 2) + &c.

(√ n − √ n − 1) } = √ra × √n

and consequently the breadth of the pth. annulus becomes

675. Let S be the specific gravity of the fluid, s of the atmosphere, and s' that of the paraboloid; then if P be the part in air, and Q the part in the fluid, we have (Vince, Prop. XXI.) P: QS-s': 5′- s