Question

A metal piece weighs 200 gf in air and 150 gf when completely immersed in water. (i) Calculate the relative density of the metal piece. (ii) How much will it weigh in a liquid of density 0.8 gcm-3?

Answer 1

Answer:(I). $R_{d} = 4$ , (II). $W = 1176\ N$

Explanation:

Given that,

Weight of metal piece in air $W = 200gf= 1.96\ N$

Weight of metal piece in water $W = 150gf= 1.47\ N$

We know that,

Mass = Volume X Density

Weight in air ,

$W = mg = V\rho_{solid} g$

$200gf = mg = V\rho_{solid}g$....(I)

Weight in water,

$W = mg - F_{up}$

$F_{up}$ = upthrust force

$150gf = 200gf -F_{up}$

$F_{up} = 50gf$

We know that,

Upthrust of water =  apparent loss of weight of solid = weight in air - weight in liquid

$F_{up} = 200gf - 150 gf$

$F_{up} = 50 gf$

We know that,

$F_{up} = V\rho_{water}g$

$V\rho_{water}g = 50gf = 1.47\ N$.....(II)

Now, volume of the solid

$V_{solid} = 0.15\ m^{3}$

Now, divided equation (I) by (II)

$\dfrac{V\rho_{solid}g}{V\rho_{water}g} = \dfrac{200 gf}{50gf}$

(I). The relative density of the metal piece is

$R_{d} = \dfrac{\rho_{solid}}{\rho_{water}} = 4$

$R_{d} = 4$

(II). The weight in liquid when density $\rho_{liquid} = 0.8gcm^{3}$

Weight in liquid $W = V\rho_{liquid}g$

$W = 1176\ N$

Hence, this is the required solution.

Answer 2

Explanation:

R.D. = 50 × 0.8

=40 ANS 1

Ans 2 = 200 - 40

=160