Upthrust in Fluids Class 9 ICSE Important Questions with Answers

Answer: (b) weight of the fluid displaced.

By Archimedes' principle the buoyant force (upthrust) equals the weight of the fluid displaced by the immersed part of the body.

Answer: (c) no unit.

Relative density is a ratio of two densities (substance to water), so the units cancel and it is a pure number.

Answer: (c) about seven-eighths.

For a floating body, $\dfrac{\text{volume immersed}}{\text{total volume}}=\dfrac{\text{density of body}}{\text{density of liquid}}=\dfrac{0.9}{1.03}\approx0.87$(volume immersed)/(total volume)=(density of body)/(density of liquid)=0.9/1.03≈0.87, i.e. about $\dfrac78$7/8 of the iceberg is hidden below the surface.

Answer: (b) has an average density less than water.

The ship is hollow, so its average density (mass ÷ total volume including the enclosed air) is less than that of water; it displaces enough water to equal its weight and floats. The solid nail's density exceeds water's, so it sinks.

Answer: (a) Sea water has a higher density, so for the same submerged volume it provides a larger upthrust ($F_B=V\rho g$F_B=Vρ g), making swimming easier; R correctly explains A.

Archimedes' principle: When a body is wholly or partly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.

The apparent loss in weight of the submerged body equals the upthrust (buoyant force), which also equals the weight of the fluid displaced.

Relative density of a substance is the ratio of its density to the density of water at $4^{\circ}\text{C}$4°C; it has no unit.

$\text{R.D.}=\dfrac{\text{weight of solid in air}}{\text{apparent loss of weight of solid in water}}=\dfrac{W_{air}}{W_{air}-W_{water}}$R.D.=(weight of solid in air)/(apparent loss of weight of solid in water)=W_airW_air-W_water

Loss of weight in water $=60-52=8\ \text{gf}$=60-52=8 gf (weight of water displaced).

Loss of weight in the liquid $=60-50=10\ \text{gf}$=60-50=10 gf (weight of the same volume of liquid).

$\text{R.D. of liquid}=\frac{\text{loss of weight in liquid}}{\text{loss of weight in water}}=\frac{10}{8}=1.25$R.D. of liquid=(loss of weight in liquid)/(loss of weight in water)=10/8=1.25

Convert volume: $V=100\text{ cm}^3=100\times10^{-6}=1\times10^{-4}\text{ m}^3$V=100 cm³=100×10^-6=1×10^-4 m³.

Upthrust = weight of water displaced $=V\rho g$=Vρ g.
$F_B=1\times10^{-4}\times1000\times10=\boxed{1\text{ N}}$F_B=1×10^-4×1000×10=1 N

By the law of floatation, weight of block = weight of water displaced.

Let total volume $=V$=V, submerged volume $=V_s$=V_s. Then
$V\rho_{wood}g=V_s\rho_{water}g\Rightarrow \dfrac{V_s}{V}=\dfrac{\rho_{wood}}{\rho_{water}}=\text{R.D.}=0.8$Vρ_woodg=V_sρ_waterg⇒ V_s/V=ρ_woodρ_water=R.D.=0.8

So the fraction submerged $=0.8$=0.8, and the fraction above the surface
$=1-0.8=\boxed{0.2\ \left(\tfrac{1}{5},\ \text{i.e. } 20\%\right)}$=1-0.8=0.2 (1/5, i.e. 20%)

Fully immersed, the body displaces $100\text{ cm}^3$100 cm³ of water, so the upthrust is $F_B=100\text{ gf}$F_B=100 gf.

(i) Case I, $F_B<W$F_B<W ($100<150$100<150): the body sinks. Apparent weight $=150-100=50\text{ gf}$=150-100=50 gf.

(ii) Case II, $F_B=W$F_B=W ($100=100$100=100): the body floats fully submerged wherever it is placed. Apparent weight $=0$=0.

(iii) Case III, $F_B>W$F_B>W ($100>60$100>60): the body rises and floats with only part of it under water. At rest, the upthrust equals its weight, $60\text{ gf}$60 gf, so it displaces $60\text{ cm}^3$60 cm³ of water; $60\%$60% of its volume is immersed. Apparent weight $=0$=0. (This is Archimedes' principle applied to floating bodies: a floating body displaces its own weight of liquid.)

Law of floatation: A floating body displaces a weight of liquid equal to its own weight. When floating, its weight acting downward is balanced by the upthrust acting upward.

Conditions (comparing weight $W$W and upthrust $F_B$F_B):

• If $W>F_B$W>F_B (body's average density greater than the liquid's), the body sinks.
• If $W=F_B$W=F_B (density equal), the body floats fully submerged in equilibrium.
• If $W<F_B$W<F_B when fully immersed (density less than the liquid's), the body rises and floats partly submerged, displacing liquid whose weight equals its own weight.

Submarine: A submarine has large ballast tanks. To dive, the tanks are filled with sea water, increasing the submarine's weight beyond the upthrust so it sinks. To rise, compressed air pushes the water out, reducing its weight below the upthrust so it floats up.

Method: Hang the solid from a spring balance and note its weight in air, $W_1$W₁. Then lower the solid so that it is fully immersed in water (not touching the sides or bottom of the beaker) and note the new reading, its apparent weight in water, $W_2$W₂.

The apparent loss of weight $(W_1-W_2)$(W₁-W₂) equals the weight of water displaced (Archimedes' principle). Then
$\text{R.D.}=\dfrac{\text{weight in air}}{\text{apparent loss of weight in water}}=\dfrac{W_1}{W_1-W_2}.$R.D.=(weight in air)/(apparent loss of weight in water)=W₁/(W₁-W₂).

Numerical: $W_1=60\text{ gf}$W₁=60 gf, $W_2=52\text{ gf}$W₂=52 gf.
$\text{R.D.}=\dfrac{60}{60-52}=\dfrac{60}{8}=\boxed{7.5}$R.D.=60/(60-52)=60/8=7.5

Density $=\text{R.D.}\times\rho_{water}=7.5\times1000=\boxed{7500\text{ kg m}^{-3}}$=R.D.× ρ_water=7.5×1000=7500 kg m^-3 (i.e. $7.5\text{ g cm}^{-3}$7.5 g cm^-3).

$V=200\text{ cm}^3=2\times10^{-4}\text{ m}^3$V=200 cm³=2×10^-4 m³, $m=1.6\text{ kg}$m=1.6 kg.

(i) Weight in air $=mg=1.6\times10=\boxed{16\text{ N}}$=mg=1.6×10=16 N.

(ii) Upthrust $=V\rho_{water}g=2\times10^{-4}\times1000\times10=\boxed{2\text{ N}}$=Vρ_waterg=2×10^-4×1000×10=2 N.

(iii) Apparent weight $=$= weight in air $-$- upthrust $=16-2=\boxed{14\text{ N}}$=16-2=14 N.

(iv) It will sink. The upthrust ($2\text{ N}$2 N) is less than the weight ($16\text{ N}$16 N); equivalently the solid's density $=\dfrac{1.6}{2\times10^{-4}}=8000\text{ kg m}^{-3}$=1.62×10^-4=8000 kg m^-3 is greater than that of water.