Question

answer is 5 and 0.67... plzzz i will mark brainliest to genius....plzzz

Answer 1

Actual weight of the block in air, W = 22 N

When immersed in water, the apparent weight is, WW = 17 N

When immersed in another liquid, the apparent weight is, WL = 18 N

Let, the volume of the block be V. Density of block be db, density of water be dw and density of liquid be dl.

Now,

Weight of the block in air is, W = V(db)g = 22 => Vg = 22/db ………….(1)

When immersed in water, the buoyant force is, BW = V(dw)g

When immersed in liquid, the buoyant force is, BL = V(dl)g

Thus,

WW = W - BW

=> 17 = 22 – V(dw)g

=> (22/db)(dw) = 5

=> (db/dw) = 22/5 = 4.4

This is the relative density of the block.

Again, db = (4.4)(dw) = (4.4)(1000) kg/m3 = 4400 kg/m3

This is the density of the block.

Using (1),

Vg = 22/db

=> V = 22/(dbg) = 22/(4400 × 9.8) = 5.1 × 10-4 m3

This is the volume of the block.

WL = W - BL

=> 18 = 22 – V(dl)g

=> (22/db)(dl) = 4

=> dl = 4(db/22)

=> dl/dw = (4/22)(db/dw)

=> dl/dw = (4/22)(4.4) = 0.8

This is the relative density of the liquid.