Q: Question 1(iv)For a floating body of volume V, the relation between the volume of its submerged part v, the densities of liquid (ρL) and the body (ρS) is :vV\dfrac{\text v}{\text V}Vv = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}ρLρsVv\dfrac{\text V}{\text v}vV = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}ρLρsv×ρs=V×ρL\text v \times \text ρ_\text s = \text V \times \text ρ_\text Lv×ρs=V×ρLnone of these

vV\dfrac{\text v}{\text V}Vv = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}ρLρsReason —Given,Volume of body = VVolume of body submerged in liquid = vDensity of body = ρsDensity of liquid = ρLLet weight of the body be W.W = volume of the body x density of the body x g = V ρs gWeight of liquid displaced by the body will be equal to upthrust. Let it be FB.FB = volume of the liquid displaced x density of the liquid x g = v ρL gFrom principle of floatation,W = FB⇒ V ρs g = v ρL g⇒ vV\dfrac{\text v}{\text V}Vv = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}ρLρsHence proved.

Question 1

Question 1With the use of Archimedes' principle, state how you will find relative density of a solid denser than water and insoluble in it. How will you modify your experiment if the solid is soluble in water?

Accepted Answer

Relative density of a solid denser than water and insoluble in itProcedure —(i) Suspend a piece of the given solid with a thread from hook of the left pan of a physical balance and find it's weight W1.(ii) Now balance a wooden bridge over the left pan of balance and place a beaker nearly two-third filled with water on the bridge. Take care that the bridge and beaker do not touch the pan of balance,(iii) Immerse the solid completely in water such that it does not touch the walls and bottom of beaker and find the weight W2 of solid in water.Observation —Weight of solid in air = W1 gfWeight of solid in water= W2 gfCalculation —Loss in weight of solid when immersed in water = (W1 - W2) gfR.D.= Weight of solid in airLoss in weight of solid in water	ext{R.D.} = \dfrac{	ext{ Weight of solid in air}}{	ext{Loss in weight of solid in water}}R.D.=Loss in weight of solid in water Weight of solid in air​orR.D.=W1W1−W2	ext{R.D.} = \dfrac{	ext W_{1}}{	ext W_1 - 	ext W_2}R.D.=W1​−W2​W1​​Relative density of a solid denser than water and soluble in it —Procedure —If solid is soluble in water, instead of water, we take a liquid of known relative density in which solid in insoluble and it sinks in that liquid. Then the process described above is repeated. NowR.D.=Wt of solid in airLoss in wt of solid in liq×R.D. of liq	ext{R.D.} = \dfrac{	ext{Wt of solid in air}}{	ext{Loss in wt of solid in liq}} 	imes 	ext{R.D. of liq}R.D.=Loss in wt of solid in liqWt of solid in air​×R.D. of liq

Question 2

Question 3A body weighs W1 gf in air and when immersed in a liquid, it weighs W2 gf, while it weighs W3 gf on immersing it in water. Find: (i) volume of the body (ii) upthrust due to liquid (iii) relative density of the solid and (iv) relative density of the liquid.

Accepted Answer

Given,Weight of the body in air = W1 gfWeight of the body in liquid = W2 gfWeight of the body in water = W3 gf(i) Let V be the volume of the body.Upthrust due to water = loss in weight when immersed in water = (W1 - W3) gf ...[Eq 1]Weight of water displaced = Volume of water displaced x density of water x gVolume of water displaced = Volume of the body = VDensity of water = 1 g cm-3∴ Weight of water displaced = V x 1 x g = V gf ...[Eq 2]But weight of water displaced is equal to upthrust due to water∴ From eqns 1 & 2,V = (W1 - W3) cm3∴ Volume of the body = V = (W1 - W3) cm3(ii) Upthrust due to liquid = loss in weight when immersed in liquid = (W1 - W2) gf(iii) Relative density of the solidR.D. of solid=(Wt of body)air(Wt of body)air−(Wt of body)water=W1W1−W3\text{R.D. of solid} \[0.5em] = \dfrac{(\text{Wt of body})\text{air}}{(\text{Wt of body})\text{air}-(\text{Wt of body})_\text{water}} \[0.5em] = \dfrac{\text W_1}{\text W_1 - \text W_3}R.D. of solid=(Wt of body)air−(Wt of body)water(Wt of body)air=W1−W3W1(iv) Relative density of the liquidR.D. of liquid=(Wt of body)air−(Wt of body)liquid(Wt of body)air−(Wt of body)water=W1−W2W1−W3\text{R.D. of liquid} = \dfrac{(\text{Wt of body})\text{air}-(\text{Wt of body})\text{liquid}}{(\text{Wt of body})\text{air}-(\text{Wt of body})\text{water}} \[0.5em] = \dfrac{\text W_1 - \text W_2}{\text W_1 - \text W_3}R.D. of liquid=(Wt of body)air−(Wt of body)water(Wt of body)air−(Wt of body)liquid=W1−W3W1−W2

Question 3

Question 8A body of volume 100 cm3 weighs 1 kgf in air. Find (i) it's weight in water and (ii) it's relative density.

Accepted Answer

Given,Volume of body = 100 cm3Weight of the body in air (W1) = 1 kgf = 1000 gfLet weight of the body in water be W2Density=MassVolume=1000100=10 g cm−3\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \[0.5em] = \dfrac{1000}{100} \[0.5em] = 10 \text { g cm}^{-3}Density=VolumeMass​=1001000​=10 g cm−3Hence,Density of body = 10 g cm-3R.D. of body=Density of body in g cm−31.0 g cm−3=101=10\text{R.D. of body} = \dfrac{\text {Density of body in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \[0.5em] = \dfrac{10}{1} \[0.5em] = 10R.D. of body=1.0 g cm−3Density of body in g cm−3​=110​=10Therefore,Relative Density of body = 10R.D. of body=W1W1−W2⇒10=10001000−W2⇒10×(1000−W2)=1000⇒10000−10W2=1000⇒10W2=9000⇒W2=900010⇒W2=900 gf\text{R.D. of body} = \dfrac{\text W_{1}}{\text W_{1} − \text W_{2}} \[1 em] \Rightarrow 10 = \dfrac{1000}{1000 − \text W_{2}} \[1 em] \Rightarrow 10 \times (1000 − \text W_{2}) = 1000 \[1 em] \Rightarrow 10000 - 10\text W_{2} = 1000 \[1 em] \Rightarrow 10\text W_{2} = 9000 \[1 em] \Rightarrow \text W_{2} = \dfrac{9000}{10} \[1 em] \Rightarrow \text W_{2} = 900 \text { gf}R.D. of body=W1​−W2​W1​​⇒10=1000−W2​1000​⇒10×(1000−W2​)=1000⇒10000−10W2​=1000⇒10W2​=9000⇒W2​=109000​⇒W2​=900 gfHence,(i) Weight in water = 900 gf(ii) Relative density of body = 10

Question 4

Question 9A body of mass 70 kg, when completely immersed in water, displaces 20,000 cm3 of water. Find (i) the weight of body in water and (ii) the relative density of material of body.

Accepted Answer

Given,Mass of body = 70 kgVolume of water displaced = 20,000 cm3Converting cm3 to m3, we get,100 cm = 1m100 cm x 100 cm x 100 cm = 1 m3Therefore, 20,000 cm3 = 110,00,000\dfrac{1}{10,00,000}10,00,0001​ x 20,000 = 0.02 m3Hence, volume of water displaced = 0.02 m3(i) Mass of body immersed in water = mass of the water displaced = volume of water displaced x density of water= 0.02 x 1000= 20 kgWeight of the body = mg = 70 x g = 70 kgfWeight of water displaced = mg = 20 x g = 20 kgfWeight of body in water = weight of body in air - upthrust due to liquid= 70 kgf - 20 kgf    [∵ upthrust is equal to weight of water displaced]= 50 kgf(ii) Formula for density is:Density=MassVolume=700.02=3500 Kg m−3\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \[1 em] = \dfrac{70}{0.02} \[1 em] = 3500 \text { Kg m}^{-3}Density=VolumeMass​=0.0270​=3500 Kg m−3R.D. of body=Density of body in kg m−31000 kg cm−3=35001000=3.5\text{R.D. of body} = \dfrac{\text {Density of body in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \[1 em] = \dfrac{3500}{1000} \[1 em] = 3.5R.D. of body=1000 kg cm−3Density of body in kg m−3​=10003500​=3.5Therefore,Relative density of body = 3.5

Question 5

Question 10A solid weighs 120 gf in air and 105 gf when it is completely immersed in water. Calculate the relative density of solid.

Accepted Answer

Given,Weight of the solid in air (W1) = 120 gfWeight of the solid when completely immersed in water (W2) = 105 gfRelative density of solid=(W1W1−W2)=(120120−105)=8\text{Relative density of solid} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \[0.5em] = \big(\dfrac{120}{120 - 105}\big) \[0.5em] = 8Relative density of solid=(W1​−W2​W1​​)=(120−105120​)=8∴ Relative density of solid = 8

Question 6

Question 11A solid weighs 32 gf in air and 28.8 gf in water. Find (i) the volume of solid, (ii) R.D. of solid, and (iii) the weight of solid in a liquid of density 0.9 g cm-3.

Accepted Answer

Given,Weight of the solid in air (W1) = 32 gfWeight of the solid immersed completely in water (W2) = 28.8 gfRelative density of solid=(W1W1−W2)=(3232−28.8)=(323.2)=10But, R.D. of solid=Density of solid in g m−31 g cm−3⇒10=Density of solid in g m−31 g cm−3⇒Density of solid=10 g m−3Density=MassVolume∴Volume=MassDensity=3210=3.2 cm3\text{Relative density of solid} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \[1 em] = \big(\dfrac{32}{32 - 28.8}\big) \[1 em] = \big(\dfrac{32}{3.2}\big) \[1 em] = 10 \[1.5em] \text{But, R.D. of solid} = \dfrac{\text {Density of solid in g m}^{-3}}{\text{1 g cm}^{-3}} \[1 em] \Rightarrow 10 = \dfrac{\text {Density of solid in g m}^{-3}}{\text{1 g cm}^{-3}} \[1 em] \Rightarrow \text {Density of solid} = 10 \text{ g m}^{-3} \[1 em] \text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \[1em] \therefore \text{Volume} = \dfrac{\text{Mass}}{\text{Density}} \[1 em] = \dfrac{32}{10} \[1 em] = 3.2 \text { cm}^{3}Relative density of solid=(W1​−W2​W1​​)=(32−28.832​)=(3.232​)=10But, R.D. of solid=1 g cm−3Density of solid in g m−3​⇒10=1 g cm−3Density of solid in g m−3​⇒Density of solid=10 g m−3Density=VolumeMass​∴Volume=DensityMass​=1032​=3.2 cm3Hence, volume of solid = 3.2 cm3Let weight of solid in liquid of density 0.9 g cm-3 be W.R.D. of solid=(W1W1−W2)×R.D. of liquidR.D. of liquid=Density of liquid in g cm−31.0 g cm−3=0.91=0.9\text{R.D. of solid} = \Big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\Big) \times \text{R.D. of liquid} \[1em] \text{R.D. of liquid} = \dfrac{\text {Density of liquid in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \[1 em] = \dfrac{0.9}{1} \[1 em] = 0.9R.D. of solid=(W1​−W2​W1​​)×R.D. of liquidR.D. of liquid=1.0 g cm−3Density of liquid in g cm−3​=10.9​=0.9Substituting the values in the formula for relative density of solid :10=(3232−W)×0.910=28.832−W10×(32−W)=28.8320−10W=28.810W=320−28.8W=291.210W=29.12 gf10 = \Big(\dfrac{32}{32 − \text W}\Big) \times 0.9 \[1 em] 10 = \dfrac{28.8}{32 − \text W} \[1 em] 10 \times (32 - \text W) = 28.8 \[1 em] 320 - 10\text W = 28.8 \[1 em] 10\text W = 320 - 28.8 \[1 em] \text W = \dfrac{291.2}{10} \[1 em] \text W = 29.12 \text{ gf}10=(32−W32​)×0.910=32−W28.8​10×(32−W)=28.8320−10W=28.810W=320−28.8W=10291.2​W=29.12 gfSummarizing the answers:(i) Volume of solid = 3.2 cm3(ii) Relative density of solid = 10(iii) Weight of solid in a liquid of density 0.9 g cm-3 = 29.12 gf

Question 7

Question 12A body weighs 20 gf in air and 18 gf in water. Calculate relative density of the material of the body.

Accepted Answer

Given,Weight of the solid in air (W1) = 20 gfWeight of the solid in water (W2) = 18 gfRelative density of body=(W1W1−W2)=(2020−18)=(202)=10\text{Relative density of body} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \[0.5em] = \big(\dfrac{20}{20 - 18}\big) \[0.5em] = \big(\dfrac{20}{2}\big) \[0.5em] = 10Relative density of body=(W1​−W2​W1​​)=(20−1820​)=(220​)=10

Question 8

Question 13A solid weighs 1.5 kgf in air and 0.9 kgf in a liquid of density 1.2 x 103 kg m-3. Calculate R.D. of solid.

Accepted Answer

Given,Weight of the solid in air (W1) = 1.5 kgfWeight of the solid in liquid (W2) = 0.9 kgfDensity of the liquid = 1.2 x 103 kg m-3R.D. of liquid=Density of liquid in kg m−31000 kg cm−3=1.2×1031000=1.2\text{R.D. of liquid} = \dfrac{\text {Density of liquid in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \[0.5em] = \dfrac{1.2 \times 10^3}{1000} \[0.5em] = 1.2R.D. of liquid=1000 kg cm−3Density of liquid in kg m−3​=10001.2×103​=1.2R.D. of solid=(W1W1−W2)×R.D. of liquid=(1.51.5−0.9)×1.2=(1.50.6)×1.2=3.0\text{R.D. of solid} = \Big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\Big) \times \text{R.D. of liquid} \[0.75em] = \Big(\dfrac{1.5}{1.5 - 0.9}\Big) \times 1.2 \[0.75em] = \Big(\dfrac{1.5}{0.6}\Big) \times 1.2 \[0.75em] = 3.0R.D. of solid=(W1​−W2​W1​​)×R.D. of liquid=(1.5−0.91.5​)×1.2=(0.61.5​)×1.2=3.0

Question 9

Question 14A jeweller claims that he makes ornament of pure gold of relative density 19.3. He sells a bangle weighing 25.25 gf to a person. The clever customer weighs the bangle when immersed in water and finds that it weighs 23.075 gf in water. With the help of suitable calculations find out whether the ornament is made of pure gold or not.[Hint — Calculate R.D of material of bangle which comes out to be 11.6]

Accepted Answer

Given,Relative density of pure gold = 19.3Weight of the bangle in air (W1) = 25.25 gfWeight of the bangle in water (W2) = 23.075 gfRelative density of bangle=(W1W1−W2)=(25.2525.25−23.075)=(25.252.175)=11.6\text{Relative density of bangle} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \[1 em] = \big(\dfrac{25.25}{25.25 - 23.075}\big) \[1 em] = \big(\dfrac{25.25}{2.175}\big) \[1 em] = 11.6Relative density of bangle=(W1​−W2​W1​​)=(25.25−23.07525.25​)=(2.17525.25​)=11.6It is given that relative density of gold is 19.3 where as relative density of bangle is 11.6.Hence, we can deduce that the bangle is not made of pure gold.

Question 10

Question 16A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate —(a) the weight of the piece of stone in air,(b) the volume of the piece of stone,(c) the relative density of stone,(d) the relative density of the liquid.

Accepted Answer

Given,Mass of the stone = 15.1 g(a) Weight = mass x acceleration due to gravity= 15.1 x g= 15.1 gfHence,Weight of the piece of stone in air (W1) = 15.1 gf(b) W1 = 15.1 gfWeight of the stone when immersed in water(W3) = 9.7 gfUpthrust on the stone = loss in weight when immersed in water= Weight in air (W1) - Weight in water (W3)= 15.1 – 9.7= 5.4 gfLet volume of the piece of stone be V.From the relation, Upthrust on stone = volume of stone x density of water x acceleration due to gravity5.4 x g = V x 1 x g⇒ V = 5.4 cm3∴ Volume of piece of stone = 5.4 cm3(c) From the relation,Relative density of stone=(W1W1−W3)=(15.115.1−9.7)=(15.15.4)=2.79≈2.8\text{Relative density of stone} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{3}}\big) \[1 em] = \big(\dfrac{15.1}{15.1 - 9.7}\big) \[1 em] = \big(\dfrac{15.1}{5.4}\big) \[1 em] = 2.79 \approx 2.8Relative density of stone=(W1​−W3​W1​​)=(15.1−9.715.1​)=(5.415.1​)=2.79≈2.8Hence,Relative density of stone = 2.8(d) From the relation,Relative density of liquid=W1−W2W1−W3\text{Relative density of liquid} = \dfrac{\text W_{1} - \text W_{2} }{\text W_{1} − \text W_{3}}Relative density of liquid=W1​−W3​W1​−W2​​where,W1 is the weight of piece of stone in air,W2 is the weight of piece of stone in liquid,W3 is the weight of piece of stone in waterSubstituting the values in the formula above we get,Relative density of liquid=15.1−10.915.1−9.7=4.25.4=0.777≈0.78\text{Relative density of liquid} = \dfrac{15.1 - 10.9}{15.1 - 9.7} \[0.5em] = \dfrac{4.2}{5.4} \[0.5em] = 0.777 \approx 0.78\[0.5em]Relative density of liquid=15.1−9.715.1−10.9​=5.44.2​=0.777≈0.78Hence,the relative density of the liquid = 0.777 = 0.78

Question 11

Question 1(iv)For a floating body of volume V, the relation between the volume of its submerged part v, the densities of liquid (ρL) and the body (ρS) is :vV\dfrac{	ext v}{	ext V}Vv​ = ρsρL\dfrac{	ext ρ_	ext s}{	ext ρ_	ext L}ρL​ρs​​Vv\dfrac{	ext V}{	ext v}vV​ = ρsρL\dfrac{	ext ρ_	ext s}{	ext ρ_	ext L}ρL​ρs​​v×ρs=V×ρL	ext v 	imes 	ext ρ_	ext s = 	ext V 	imes 	ext ρ_	ext Lv×ρs​=V×ρL​none of these

Accepted Answer

vV\dfrac{	ext v}{	ext V}Vv​ = ρsρL\dfrac{	ext ρ_	ext s}{	ext ρ_	ext L}ρL​ρs​​Reason —Given,Volume of body = VVolume of body submerged in liquid = vDensity of body = ρsDensity of liquid = ρLLet weight of the body be W.W = volume of the body x density of the body x g = V ρs gWeight of liquid displaced by the body will be equal to upthrust. Let it be FB.FB = volume of the liquid displaced x density of the liquid x g = v ρL gFrom principle of floatation,W = FB⇒ V ρs g = v ρL g⇒ vV\dfrac{	ext v}{	ext V}Vv​ = ρsρL\dfrac{	ext ρ_	ext s}{	ext ρ_	ext L}ρL​ρs​​Hence proved.

Question 12

Question 1(xv)A student is given three liquids A, B and C of densities ρ1, ρ2 and ρ3. He has to arrange these liquids in order of their increasing densities. To do so, he takes an object and dips it in each liquid and observes that the object floats with 1/9, 2/11 and 3/7 parts of its volume outside the surface of liquid A, B and C respectively. The correct order is :ρ1 > ρ2 > ρ3ρ2 > ρ3 > ρ1ρ3 > ρ2 > ρ1ρ3 > ρ1 > ρ2

Accepted Answer

ρ3 > ρ2 > ρ1ReasonGiven,An object is floating in three liquids A, B, and C and the fraction of volume outside the liquid is :A : 19\dfrac{1}{9}91 outside ⇒ 89\dfrac{8}{9}98 submergedB : 211\dfrac{2}{11}112 outside ⇒ 911\dfrac{9}{11}119 submergedC : 37\dfrac{3}{7}73 outside ⇒ 47\dfrac{4}{7}74 submergedMaking the denominator common for all the submerged fractions:LCM of 9, 11, 7 = 693A : 89\dfrac{8}{9}98 = 8×779×77\dfrac{8 \times 77}{9 \times 77}9×778×77 = 616693\dfrac{616}{693}693616B : 911\dfrac{9}{11}119 = 9×6311×63\dfrac{9 \times 63}{11 \times 63}11×639×63 = 567693\dfrac{567}{693}693567C : 47\dfrac{4}{7}74 = 4×997×99\dfrac{4 \times 99}{7 \times 99}7×994×99 = 396693\dfrac{396}{693}693396For a floating object :Fraction submerged=Volume submergedTotal volume=Density of objectDensity of liquid\text {Fraction submerged} =\dfrac {\text {Volume submerged}}{\text {Total volume}} \[1em] =\dfrac{\text {Density of object}}{\text {Density of liquid}}Fraction submerged=Total volumeVolume submerged=Density of liquidDensity of object Since the object is the same in all three cases then submerged fraction is inversely proportional to density of liquid i.e., less submerged volume → greater liquid density and vice versa.On comparing submerged fractions:C : 396693\dfrac{396}{693}693396 < B : 567693\dfrac{567}{693}693567 < A : 616693\dfrac{616}{693}693616So, the order of density of liquids is : ρ3 > ρ2 > ρ1

Question 13

Question 2A block of wood of mass 24 kg floats on water. The volume of wood is 0.032 m3.Find —(a) the volume of block below the surface of water,(b) the density of wood.(Density of water = 1000 kg m-3)

Accepted Answer

(a) Given,Mass = 24 kgVolume = 0.032 m3Upthrust = Volume of block below the water x density of liquid x acceleration due to gravityHence,24kgf=v×1000×gv=241000⇒v=0.024 m324 \text {kgf} = \text v \times 1000 \times \text g \[0.5em] \text v = \dfrac{24}{1000} \[0.5em] \Rightarrow \text v = 0.024 \text{ m}^{3} \[0.5em]24kgf=v×1000×gv=100024​⇒v=0.024 m3Hence, volume of block below the surface of water = 0.024 m3(b) By the principle of floatation,Volume of immersed part Total volume=Density of woodDensity of water\dfrac{\text {Volume of immersed part }}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}}Total volumeVolume of immersed part ​=Density of waterDensity of wood​andDensity of water = 1000 kg m-3Substituting the values in the formula we get,0.0240.032=Density of wood1000⇒Density of wood=0.0240.032×1000=7.5×102 kg m−3\dfrac{0.024}{0.032} = \dfrac{\text {Density of wood}}{1000} \[0.5em] \Rightarrow \text{Density of wood} = \dfrac{0.024}{0.032} \times 1000 \[0.5em] = 7.5 \times 10^{2} \text{ kg m}^{-3} \[0.5em]0.0320.024​=1000Density of wood​⇒Density of wood=0.0320.024​×1000=7.5×102 kg m−3

Question 14

Question 5If the density of ice is 0.9 g cm-3, what portion of an iceberg will remain below the surface of water in a sea? (Density of sea water = 1.1 g cm-3)

Accepted Answer

Given,Density of ice = 0.9 g cm-3Density of sea water = 1.1 g cm-3By the principle of floatation,Volume of immersed partTotal volume=Density of iceDensity of sea water∴Volume of immersed partTotal volume=0.91.1=911=0.818\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of ice}}{\text {Density of sea water}} \[1em] \therefore \dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{0.9}{1.1} \[1 em] = \dfrac{9}{11} = 0.818Total volumeVolume of immersed part​=Density of sea waterDensity of ice​∴Total volumeVolume of immersed part​=1.10.9​=119​=0.818Hence, the volume of iceberg that will remain below the surface of water in the sea = 911\dfrac{9}{11}119​th of total volume (or 0.818th) part.

Question 15

Question 6A piece of wood of uniform cross section and height 15 cm floats vertically with it's height 10 cm in water and 12 cm in spirit. Find the density of (i) wood and (ii) spirit.

Accepted Answer

Given,Height of the wood = 15 cmHeight of the wood in water = 10 cmHeight of the wood in spirit = 12 cmAs the block of wood is of uniform cross-sectional area, the height of the block is proportional to the volumeBy the principle of floatation,Volume of immersed partTotal volume=Density of woodDensity of water∴1015=Density of wood1⇒Density of wood=1015=0.667g cm−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}} \[1em] \therefore \dfrac{10}{15} = \dfrac{\text {Density of wood}}{1} \[1 em] \Rightarrow \text{Density of wood} = \dfrac{10}{15} = 0.667 \text {g cm}^{-3}Total volumeVolume of immersed part​=Density of waterDensity of wood​∴1510​=1Density of wood​⇒Density of wood=1510​=0.667g cm−3In the case of spirit, substituting the values in the formula above, we get,Volume of immersed partTotal volume=Density of woodDensity of spirit1215=0.667Density of spiritDensity of spirit=1512×0.667=0.833 g cm−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of spirit}} \[0.5em] \dfrac{12}{15} = \dfrac{0.667}{\text{Density of spirit}} \[0.5em] \text{Density of spirit} = \dfrac{15}{12} \times 0.667 = 0.833 \text { g cm}^{-3} \[0.5em]Total volumeVolume of immersed part​=Density of spiritDensity of wood​1512​=Density of spirit0.667​Density of spirit=1215​×0.667=0.833 g cm−3

Question 16

Question 7A wooden block floats in water with two-third of it's volume submerged. (a) Calculate the density of wood. (b) When the same block is placed in oil, three-quarter of it's volume is immersed in oil. Calculate the density of oil.

Accepted Answer

(a) Let volume of wooden block be V.Volume of block submerged in water = 23V\dfrac{2}{3}\text V32​VDensity of water = 1000 kg m-3By the principle of floatation,Volume of immersed partTotal volume=Density of woodDensity of water∴23VV=Density of wood1000⇒Density of wood=23×1000=666.6666 kg m−3≈667 kg m−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}} \[1em] \therefore \dfrac{\dfrac{2}{3}\text V}{\text V} = \dfrac{\text {Density of wood}}{1000} \[1 em] \Rightarrow \text{Density of wood} = \dfrac{2}{3} \times 1000 = 666.6666 \text { kg m}^{-3} \approx 667 \text { kg m}^{-3}Total volumeVolume of immersed part​=Density of waterDensity of wood​∴V32​V​=1000Density of wood​⇒Density of wood=32​×1000=666.6666 kg m−3≈667 kg m−3(b) Volume of block submerged in oil = 34V\dfrac{3}{4}\text V43​VBy the principle of floatation,Volume of immersed partTotal volume=Density of woodDensity of oil∴34VV=667Density of oil⇒Density of oil=43×667=889.33 kg m−3≈889 kg m−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of oil}} \[1em] \therefore \dfrac{\dfrac{3}{4}\text V}{\text V} = \dfrac{\text {667}}{\text{Density of oil}} \[0.5em] \Rightarrow \text{Density of oil} = \dfrac{4}{3} \times 667 = 889.33 \text { kg m}^{-3} \approx 889 \text { kg m}^{-3}Total volumeVolume of immersed part​=Density of oilDensity of wood​∴V43​V​=Density of oil667​⇒Density of oil=34​×667=889.33 kg m−3≈889 kg m−3

Upthrust in Fluids, Archimedes' Principle and Floatation

Exercise 5B Long Answer Type

Exercise 5B Numericals

Exercise 5C Multiple Choice Type

Exercise 5C Numericals