Question

PLEASE ANSWER THIS FAST
The apparent weight of a piece of solid in water is 80 gm-wt and that in kerosene oil is 100 gm-wt (density of kerosene =0•8 gm/cm3)
A. Calculate the weight of the piece of solid in air
B. Calculate its volume
C. What will be the apparent weight of the piece of solid in a liquid of density 1•4 gm/cm3?
NO SPAMS PLEASE

Answer 1

Answer:

(A) 160 gm-wt

(B) 80 cc

(C) 48 gm-wt

Explanation:

Let the weight of the piece of solid in air is W

Weight of piece of solid in water

= Weight of solid in air - bouyant force in water

= Weight of solid in air - weight of water displaced by the solid

Let the volume of the solid is V

Then the weight of water displaced = V × density of water × g

                                                            = V × 1 × g = Vg

Thus

80g= W - Vg                           .............. (1)

Similarly for kerosene we can write the equation

100g = W - V × 0.8 × g

or, 100g = W - 0.8Vg                    ........ (2)

Multiplying eq (1) by 0.8 and subtracting it from eq (2)

100g - 64g = 0.2W

or, 0.2W = 32g

or, W = 32g/0.2 = 160g

Therefore, the weight of the piece of solid in air is 160 gm-wt

(B) Putting the value of W in eq (1)

80g = 160g - Vg

or, V = 160 - 80 = 80 cm³

(C) Weight of the liquid displaced = V × density of the liquid × g

                                                        = 80 × 1.4 × g = 112g

Therefore, weight of the piece of solid in the liquid

= Weight of the solid - Weight of the liquid displaced

= 160g - 112g

= 48g

Thus, weight of the solid = 48 gm-wt.

Hope this answer is helpful.