Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5.  We Provide Step by Step Answer of Exercise, MCQs, Numericals Practice Problem Questions of Exercises, Subjective Upthrust and Archimedes Class-9, Visit official Website CISCE  for detail information about ICSE Board Class-9 Physics.

## Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5

-: Select Exercise :-

Practice Problems I

Practice Problems II

Objective Questions

Subjective Questions

### Practice Problems I

Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5

Practice Problem 1:

Question 1.
A solid of density 2700 kgm’ and of volume 0.0015 m3 ¡s completely immersed in alcohol of density 800 kgm’.Calculate :

1. Weight of solid in SI system.
2. Upthrust on solid in SI system.
3. Apparent weight of solid in alcohol.
4. Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, [g = 10 ms-2]

Density of solid = ρ= 2700 kgm3
Volume of solid = V = 0.0015 m3
Density of alcohol = ρ’ = 800 kgm3

1. Mass of solid = m = V x p
m = 0.0015 x 2700 = 4.05 kg
Weight of solid = mg = 4.05 x 10 = 40.5 N
2. Volume of alcohol displaced = Volume of solid
V = 0.0015 m3
Mass of alcohol displaced = m’ = V x p’
m’ = 0.0015 x 800 = m’ = 1.2 kg
Upthrust = Weight of alcohol displaced
= m’g= 1.2 x 10= 12N
3. Apparent weight of solid in alcohol
= Actual weight of solid – Upthrust
= 40.5 -12 = 28.5 N
4. When a solid is immersed completely in brine solution, then upthrust acts on it in upward direction, as a result, its apparent weight of solid will be less than actual weight of solid will be less than actual weight of solid.

Question 2.
A stone of density 3000 kgm3 is lying submerged in water of density 1000 kgm3. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms2]
Density of stone =ρ = 3000 kgm3
Density of water = ρ’ = 1000 kgm3
Mass of stone = m = 150 kg
Acceleration due to gravity = g = 10 ms-2

Actual weight of stone = mg = 150 x 10 = 1500 N
Volume of water displaced = Volume of stone
V = 0.05 m3
Mass of water displaced = m’ = V x p’
m’ = 0.05 x 1000 = 50 kg
Upthrust = m’g = 50 x 10 = 500 N
Force required to lift the stone
= Actual weight of stone – upthrust
= 1500 -500 = 1000 N

Question 3.
A solid of area of cross-section 0.004 m2 and length 0.60 m is completely immersed in water of density 1000 kgm3. Calculate :

1. Wt of solid in SI system
2. Upthrust acting on the solid in SI system.
3. Apparent weight of solid in water.
4. Apparent weight of solid in brine solution of density 1050 kgm3.
[Take g = 10 N/kg; Density of solid = 7200 kgm3]

Area of cross-section of solid = A = 0.004 m2
Length of the solid = l = 0.60 m
Density of water = p’ = 1000 kgm3
Acceleration due to gravity = g = 10 ms-2
Density of solid = p = 7200 kgm3
(1) Volume of solid = V = A x l
V = 0.004 x 0.60 = 0.0024 m3
Mass of solid = m = V x p
m = 0.0024 x 7200 = 17.28 kg
Weight of the solid = mg = 17.28 x 10 = 172.8 N
(2) Volume of water displaced = Volume of solid
= V = 0.0024 m3
Mass of water displaced = m’ = V x p’
m’ = 0.0024 x 1000 = 2.4 kg
Upthrust = Weight of water displaced
= m’g = 2.4 x 10 = 24 N
(3) Apparent weight of solid=Actual weight of solid – upthrust
= 172.8-24= 148.8 N
(4) Density of brine solution =ρb= 1050 kgm3
Volume of brine solution displaced = Volume of solid = V
V = 0.0024 m3
Mass of brine solution displaced
= mb = V x ρb = 0.0024 x 1050
mb = 2.52 kg
Upthrust acting on solid in brine solution = Weight of brine solution displaced -mbg
= 2.52 x 10 = 25.2 N
Apparent weight of solid in brine solution
= Actual weight – Upthurst
= 172.8-25.2= 147.6 N

Practice Problem 2:     (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)

Question 1.
A solid of density 7600 kgm3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm3, find the apparent weight of solid in liquid.
Weight of solid in air = 0.950 kgf
∴ Mass of solid in air = m = 0.950 kg
Density of solid =ρ = 7600 kgm3

Apparent weight of solid in liquid
= Actual weight – Upthrust
= 0.950-0.09 = 0.860 kgf

Question 2.
A glass cylinder of length 12 x 10-2 m and area of cross­section 5 x 10-4 m2 has a density of 2500 kgm-3. It is immersed in a liquid of density 1500 kgm-3, such that 3/8. of its length is above liquid. Find the apparent weight of glass cylinder in newtons.
Length of glass cylinder = l = 12 x 10-2 m
Area of cross-section = A = 5 x 10-4 m2
Volume of glass cylinder = V = A x l
V= 5 x lo-4 x 12 x 10-2
V= 0.00006 m3
Acceleration due to gravity = g = 9.8 m/s2
Density of glass cylinder = ρ = 2500 kgm3

Mass of glass cylinder = m = V x ρ
m = 0.00006 x 2500 m
= 0.15 kg
Weight of glass cylinder = mg = 0.15 x 10=1.5 N
Mass of liquid displaced by glass cylinder = V’ x ρ’
m’= 0.0000375 x 1500
m’ = 0.05625 kg
Upthrust = Weight of liquid displaced by the glass cylinder
= m’g = 0.05625 x 10 = 0.5625 N
Apparent weight of glass cylinder in liquid = Actual weight of glass cylinder – Upthrust
= 1.5 – 0.5625 = 0.9375 N

Practice Problems 3:

Question 1.
A solid weighs 0.08 kgf in air and 0.065 kgf in water.Find
(1) R.D. of solid
(2) Density of solid in SI system. [Density of water = 1000 kgm3]
Weight of solid in air = 0.08 kgf
Weight of solid in water = 0.065 kgf
Density of water = 1000 kgm3
(1) Relative density (RD) of solid
= Weight of solid in air
wt. of solid in air-wt. of solid in water

Question 2.
A solid of R.D. = 2.5 is found to weigh 0.120 kgf in water. Find the wt. of solid in air.
Relative density of solid = R.D. = 2.5
Weight of solid in water = W’ = 0.120 kgf
Weight of solid in air = W = ?

Question 3.
A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
Relative density of solid = R.D. = 4.2
Weight of solid in air = W = 0.200 kgf

Practice Problems 4:

Question 1.
A sinker is found to weigh 56.7 gf in water. When the sinker is tied to a cork of weight 6 gf, the combination is found to weigh 40.5 gf in water. Calculate R.D. of cork.
Weight of sinker in water = 56.7 gf
Weight of cork = 6 gf
Wt. of sinker in water + Wt. of cork in air
= 56.7 + 6 = 62.7 gf                                                 …(1)
Wt. of cork in water + Wt. of sinker in water = 40.5 gf …(2)
Subtract eq. (1) from eq. (2)
Wt. of cork in air – Wt. of cork in water = 62.7 – 40.5 = 22.2 gf

Question 2.      (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A solid lighter than water is found to weigh 7.5 gf in air. When tied to a sinker the combination is found to weigh .If the sinker alone weighs 72.5 gf in water, find R.D. of solid.
Weight of solid in air = 7.5 gf
Weight of sinker in water = 72.5 gf
Wt. of sinker in water + Wt. of solid in air
= 72.5 + 7.5 = 80.0 gf                                              …(1)
Wt. of solid in water + Wt. of sinker in water = 62.5 gf …(2)
Subtract eq. (1) from eq. (2)
Wt. of solid in air – Wt. of solid in water = 80- 62.5 = 17.5 gf.

Practice Problems 5:

Question 1.
An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80. Find the tension in thread.
Side of an aluminium cube = l = 5 cm
Volume of aluminium cube = V = Z3 = (5)3 =125 cm3
Relative density of aluminium = R.D. = 2.7
Relative density of alcohol = R.D. = 0.80
Density of water = 1 g cm-3

Density of aluminium =ρ = 2.7 g cm-3
Mass of aluminium = V x ρ
m = 125 x 2.7 = 337.5 g
Wt. of aluminium cube acting downwards = 337.5 gf
Volume of alcohol displaced = Volume of cube = V = 125 cm3
Upthrust due to alcohol = V x ρalcohol x g

Question 2.
A cube of lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.
Length of side of cube = l = 8 cm
Volume of cube =β= (8)3 = 512 cm3
V = 512 cm3
Relative density of lead cube = R.D. = 10.6 Relative density of sugar solution = R.D. 1.4
Density of water = 1 g cm-3

Practice Problems II

Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5

Practice Problems 1:

Question 1.
A hollow cylinder of copper of length 25 cm and area of cross-section 15 cm2, floats in water with 3/5 of its length inside water. Calculate :
(1) apparent density of hollow copper cylinder.
(2) wt. of cylinder.
(3) extra force required to completely submerge it in water.
(1)
Length of hollow cylinder of copper = hcu= l= 25 m
Length of hollow cylinder of copper inside water

Question 2.
A cork cut in the form of a cylinder floats in alcohol of density 0.8 gcm-3, such that 3/4 of its length is outside alcohol. If the total length of cylinder is 35 cm and area of cross-section 25 cm2, calculate :
(1) density of cork
(2) wt. of cork
(3) extra force required to submerge it in alcohol

Practice Problems 2:

Question 1.
A cylinder made of copper and aluminium floats in mercury of density 13.6 gem-3, such that 0.26th part of it is below mercury. Find the density of solid.
Density of mercury = ρHg =13.6 g.cm-3
Density of solid cylinder = ρsolid = ?
0.26th part of the cylinder is below mercury
Let Vsolid = Volume of solid cylinder
Volume of mercury displaced by immersed part of the solid cylinder

Question 2.
An iceberg floats in sea water of density 1.17 g cm 3, such that 2/9 of its volume is above sea water. Find the density of iceberg.
Density of sea water =ρw = 1.17 g cm-3
Density of solid ice berg =ρi = ?

Practice Problems 3:     (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)

Question 1.
A wooden block floats in alcohol with 3/8 of its length above alcohol. If it is made to float in water, what fraction of its length is above water? Density of alcohol is 0.80 g cm-3.
Let length of wooden block = x

Question 2.
A hollow metal cylinder of length 10 cm floats in alcohol of density 0.80 g cm-3, with 1 cm of its length above it. What length of cylinder will be above copper sulphate solution of density 1.25 g cm-3?
Length of hollow metal cylinder = x= 10 cm
1 cm length of cylinder is above the alcohol
∴Length of the cylinder below alcohol = (10-1) = 9cm
Density of alcohol =ρalcohol = 0.80 g cm-3
Density of copper sulphate solution = ρCuS04 = 1.25 g cm-3
When block floats in alcohol By the law of floatation :

Length of metal block below copper sulphate solution = 5.76 cm
So, length of metal block above copper sulphate solution
= 10 – 5.76 = 4.24 cm

Practice Problems 4:
Question 1.
What fraction of an iceberg of density 910 kgm-3 will be above the surface of sea water of density 1170 kgm-3 ?
Ans.
Let volume of iceberg = Vi = x
Volume of iceberg inside sea water = Vw
Density of iceberg = ρi = 910 kmg-3
Density of sea waer = ρw = 1170 kgm-3
By law of floatation:
Weight of icebeig = Weight of sea water displaced by the iceberg

Question 2.
What fraction of metal of density 3400 kgm-3 will be above the surface of mercury of
density 13600 kgm-3, while floating in mercury?
Density of metal =ρm = 3400 kgm-3
Density of mercury = pHg = 13600 kgm-3
Let volume of metal = x
and volume of metal inside mercury = y
By law of floatation:
Weight of mercury displaced by metal = wt. of metal

Practice Problems 5:

Question 1.
A balloon of volume 1000 m3 is filled with a mixture of hydrogen and helium of density 0.32 kgm-3. If the fabric of balloon weighs 40 kgf and the density of cold air is 1.32 kgm-3, find the tension in the tope, which is holding the balloon to ground.
Volume of balloon = V = 1000 m3
Density of mixture of hydrogen and helium =ρ = 0.32 kgm3
Density weight of empty balloon = 40 kgf
Density of cold air = p’ = 1.32 kgm3
Volume of balloon=Volume of mixture of hydrogen and helium gas
= Volume of cold air displaced by balloon
= V= 1000 m3
Weight of mixture of hydrogen and helium gas in balloon = Vρg
= 1000 x 0.32 x g = 320 kgf
Down thrust = Weight of empty balloon + Weight of mixture of hydrogen and helium gas
= 40+ 320 = 360 kgf
Upthrust = Weight of cold air displaced by balloon = Vρ’g
= 1000 x 1.32 x g= 1320kgf
Tension in the rope = Upthrust – downthrust
= 1320-360 = 960 kgf

Question 2.
A balloon of volume 800 cm3 is filled with hydrogen gas of density 9 x 10-5 gem-3. If the empty balloon weighs 0.3 gf and density of air is 1.3 x 10-3 gem-3, calculate the lifting power of balloon.
Volume of balloon = V = 800 cm3
Density of hydrogen gas = pH = 9 x 10-5 g cm-3
Weight of empty balloon = 0.3 gf
Density of air = ρa = 1.3 x 10-3 g cm-3
Weight of hydrogen gas in balloon = VρHg
= 800 x 9 x 10-5 x g
= 72 x 10-3gf
= 0.072 gf
Volume of balloon = Volume of hydrogen gas in the balloon = Volume of air displaced by balloon.
Downthrust = Wt. of empty balloon + wt. of hydrogen gas in balloon
= 0.3 + 0.072 = 0.372 gf
Upthrust = Wt. of air displaced by balloon
= Vρag
= 800 x 1.3 x 1o-3 x  g
= 8 x 1.3 x 1o-1 x g
= 10.4 x 10-1 x g = 1.04 gf
Lifting power of balloon = Upthrust – Down thrust
=1.04-0.372 = 0.668 gf

Question 3.
A balloon of volume 120 m3 is filled with hot air, of density 38 kg-3. If the fabric of balloon weighs 12 kg, such that an additional equipment of wt. x is attached to it, calculate the magnitude of Density of cold air is 1.30 kgm-3.
Volume of balloon = V = 120 m3
Density of hot air = ρhot air = 0.38 kgm-3
Mass of empty balloon = 12 kg
Weight of the empty balloon = 12 kgf
Weight of the additional equipment attached with the balloon
=x kgf
Density of cold air = ρcoldair = 1.30 Kgm3
Volume of balloon=Volume of hot air inside the balloon=Volume
of cold air displaced by balloon = V = 120 mWeight of hot air = Vphotair g
= 120 x 0.38 x g = 45.6kgf
Weight of empty balloon + Weight of hot air inside the balloon +
Weight of equipment = Downthrust
12 + 45.6 + x = Downthrust
Downthrust = 57.6 + x
Upthrust = Weight of cold air diplaced by balloon
=Vρcoldair g
= 120 x 1.30 xg= 156kgf
By law of floatation :
Downthrust = Upthrust
57.6+x= 156
x= 156-57.6
x =98.4 kgf

Practice Problems 6:   (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)

Question 1.
A test tube weighing 17 gf, floats in alcohol to the level P. When the test tube is made to float in water to the level P, 3 gf of the lead shots are added in it. find the R.D. of alcohol.
When tube floats in alcohol :
Weight of test tube = 17 gf
By law of floatation :

Weight of alcohol displaced by test tube = Weight of test tube = 17 gf
When tube floats in water :
When test tube is made to float in water to the same level, as in alcohol then 3g lead stones are added in it.
.’. Weight of test tube = 17 gf + 3 gf = 20 gf
Weight of water displaced by test tube = 20 gf
Volume of alcohol displaced = Volume of water displaced

Question 2.
A test tube loaded with lead shots, weighs 150 gf and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that 27 gf of lead shots have to be removed, so as to float it to level X. Find R.D. of alcohol.
When tube floats in water :
Weight of test tube = 150 gf
By law of floatation:
Weight of water displaced = Weight of test tube = 150 gf
When test tube floats in alcohol :
When test tube is made to float in alcohol, then 27 gf of lead shots have to removed, so that it can float upto the same level as in water.
∴Weight of test tube in alcohol = 150 – 27 = 123 gf
By law of floatation:
Weight of alcohol displaced by test tube = Weight of test tube in alcohol = 123 gf
As volume of alcohol displaced = Volume of water displaced
∴ R.D. of alcohol

(A) Objective Questions

Multiple Choice Questions.

### Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5

Select the correct option.

1.The force experienced by a body when partially or fully immersed in water is called :
(a) aparent weight
(b) upthrust
(c) down thrust
(d) none of these

2. When a body is floating in a liquid :
(a) The weight of the body is less than the upthrust due to immersed part of the body
(b) The weight of body is more than the upthrust due to the immersed part of the body
(c) The weight of body is equal to the upthrust due to the immersed part of the body
(d) none of the above

3. With the increase in the density of the fluid, the upthrust experienced by a body immersed in it :
(a) decreases
(b) increases
(c) remains same
(d) none of these

4. The apparent weight of a body in a fluid is :
(a) equal to weight of fluid displaced
(b) volume of fluid displaced
(c) difference between its weight in air and weight of fluid displaced
(d) none of the above

5. The phenomenon due to which a solid experiences upward force when immersed in water is called :
(a) floatation
(b)    buoyancy
(c)    density
(d)    none of these

6. When an object sinks in a liquid, its :
(a) buoyant force is more than the weight of object
(b) buoyant force is less than the weight of object
(c) buoyant force is equal to the weight of the object
(d) none of the above

7. The SI unit of density is :
(a)   gem-3
(b)    kgem-3
(c)    kgm-3
(d)   gm-3

8. When a body is wholly or partially immersed in a liquid, it experiences a buoyant force which is equal to :
(a) volume of liquid displaced by it
(b) weight of liquid displaced by it
(c) both (a) and (b)
(d) none of the above

9. The ratio between the mass of a substance and the mass of an equal volume of water at 4°C is called :
(a)   relative density

(b)   density
(c)    weight
(d)   pressure

10. A body has density 9.6 gcm -3. Its density in SI system is :
(a) 96 kgm-3
(b) 960 kgm-3
(c)  9600 kgm-3
(d) 96,000 kgm-3
Ans:
Explanation :
Density = 9.6 g cm-3

(B) Subjective Questions

### Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5

Question 1.
A wooden block floats in water with two third of its volume submerged.
(1) Calculate density of wood.
(2) When the same block is placed in oil, three quarter of its volume is immersed in oil. Calculate the density of oil.

Question 2.
A.metal cube of 5cm edge and relative density 9 is suspended by a thread so as to be completely immersed in a liquid of relative density 1.2. Find the tension in the thread.
Volume of metal cube = (side)3 = 53 = 125cm3
Density of cube = 9 g cm-3
Weight of cube acting downward = mg = v × d
F↓= 125 x 9 = 1125 gf
Density of liquid dL– 1.2 g cm-3
.’. Upthrust due to liquid in the upward direction.
F2↑ = v dL= 125 x 1.2 = 150.0 gf Tension in the string = Net downward force = F – F2
= 1125- 150 = 975 gf= 9.75 N

Question 3.
A weather forecasting plastic balloon of volume 15 mcontains hydrogen of density 0.09 kgm-3. The volume of equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon is 7.15 . kg. The balloon is floating in air of density 1.3 kgm-3.
(1) Calculate the mass of hydrogen in balloon.
(2)    Calculate the mass of hydrogen and the balloon.
(3) If the mass of equipment is x kg, write down the total mass of hydrogen, the balloon and the equipment,
(4) Calculate the mass of air displaced by balloon.
(5) Using the law of floatation, calculate the mass of equipment.
Volume of Hydrogen V = 15 m3
Density of hydrogen = d= 0.09 kg m-3
(1)  Mass of hydrogen in balloon = Vd= 15 × 0.09
= 1.35 kg
The mass of empty balloon alone = 7.15 kg
(2) The mass of hydrogen and balloon = 1.35 + 7.15
= 8.50 kg
Mass of equipment = x kg
(3) Total mass of hydrogen + Balloon + Equipment
= (8.50 + :c) kg
Density of air = 1.3 kg m-3
(4) Mass of air displaced by balloon = v x d=15 x 1.3
= 19.5 kg
(5) According to law of floatation
Total downward wt. = UPTHRUST
8.5+x= 19.5
Mass of equipment x = 11 kg

Question 4.
(a) State the principle of floatation.
(b) The mass of a block made of certain material is 1.35 kg and its volume is 1.5 x 10-3 m3.

1. Find the density of block.
2. Will this block float or sink? Give reasons for your answer.

(a) PRINCIPLE OF FLOATATION : “When a solid is floating in a fluid, the weight of whole solid acting vertically downward at its CENTRE OF GRAVITY, is equal to the weight of fluid displaced by the IMMERSED part of solid acting upward, at its CENTRE OF BUOYANCY or at the centre of the BULK OF LIQUID displaced.”

OR

“The weight of a floating body is equal to the weight of the liquid displaced by its SUBMERGED part.”

(b) Mass of block = m = 1.35 kg
Volume of block = V = 1.5 x 10-3 m3
(1)
(2)   Density of block (900 kgm-3) is less than density of water (1000 kgm-3)
.’. Block will float in water

Question 5.
(a) State Archimedes’ Principle.
(b) A block of mass 7 kg and volume 0.07 m3 floats in a liquid of density 140 kg/m3. Calculate :

1. Volume of block above the surface of liquid.
2. Density of block.

(a) ARCHIMEDES’ PRINCIPLE : “Whenever a body is immersed in a liquid (fluid), wholly or partially, it loses weight equal to the weight of liquid displaced by it.”
(b) Mass of block = m = 7 kg
Volume of block = V = 0.07 m3
Density of liquid =ρl = 140 kgm-3
Let V’ = Volume of block immersed in the liquid
By law of floatation:
Weight of block = Weight of liquid displaced by the immersed pa

Question 6.          (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
(a) A body whose volume is 100 cm3 weighs 1 kgf in air. Find its weight in water.
(b) Why is it easier to swim in sea water than in river water?
(a) Volume of body = V = 100 cm3 = 10-4 m3
Weight of body in air = 1 kgf
Density of water = pw = 1000 kgm3
We know volume of water displaced = Volume of body
= V = 10-4 m3
Upthrust = Weight of water displaced by body = Vpwg
= 10-4 x 1000 x g .
= 10-1 kgf =0.1 kgf
Weight of body in water=Weight of body in air – Upthrust
= 1-0.1 =0.90 kgf
(b) With smaller portion of man’s body submerged in sea water, the wt. of sea water displaced is equal to the total weight of body.
While to displace the same weight of river water, a larger portion of the body will have to be submerged in water. It is easier for man to swim in sea water.

Question 7.
Why does a ship made of iron not sink in water, while an iron nail sinks in it?
Density of iron is more than density of water, therefore weight of iron nail is more than wt. of water displaced by it and nail SINKS. While shape of iron ship is made in such a way that it displaces MORE WEIGHT OF WATER than its own weight. Secondly the ship is HOLLOW and THE EMPTY SPACE contains AIR which makes the AVERAGE DENSITY OF SHIP LESS THAN THAT OF WATER and hence ship floats on water.

Question 8.
A solid of density 5000 kgm-3 weighs 0.5 kgf in air. It is completely immersed in a liquid of density 800 kgm-3. Calculate the apparent weight of the solid in liquid.
Density of solid, ds = 5000 kg m-3
Weight of body in air = 0.5 kgf
mg = 0.5 kgf
∴  m – 0.5 kg

(1) Apparent weight of the solid in water = 0.5 – 0.08
= 0.42 kgf
(2) Apparent weight of body in liquid of density 800 kg m-3 is zero.
Density of solid is less than density of liquid i.e. upthrust is more than weight of body.

Question 9.
(a) A body dipped in a liquid experiences an upthrust. State the factors on which the upthrust depends
(b) While floating, is the weight of body greater than, equal to or less than upthrust?
Ans.
(a) Factors on which upthrust depends are :

1. Volume of body immersed in fluid.
Upthrust is maximum when body completely immersed in the fluid.
2. Density of the fluid.
Upthrust α density of fluid
Larger the density of the fluid, large will be the upthrust acting on the body.

(b) When the body floats then weight of the body is equal to the upthrust acting on the body.

Question 10.
A sinker is first weighed alone under water. It is then tied to a cork and again weighed under water. In which of the two cases weight under water is less and why?
Weight of sinker, when tied to a cork, under water is less than that when it is alone weighed under water. Because cork displaces more water than its own weight and hence large upthrust acts on the sinker.

Question 11.         (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A solid weighs 105 kgf in air. When completely immersed in water, it displaces 30,000 cm3 of water, calculate relative density of solid.
Weight of solid in air =105 kgf
Volume of solid = Volume of water displaced = 30000 cm3
= 30000 x 10-6 m3 = 0.03 m3
pw = Density of water = 1000 kgm-3
Wt. of water displaced by solid .

Question 12.       (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A test tube loaded with lead shots weighs 25 gf and floats upto the mark X in water. When the test tube is made to float in brine solution, it needs 5 gf more of lead shots to float upto level X. Find the relative density of brine solution.
When test tube floats is water :
Weight of test tube = 25 gf
By law of floatation
Weight of water displaced = Weight of test tube = 25 gf
When test tube floats in brine solution, it needs 5 gf more of lead shots to float upto same level as in water.
Weight of test tube = 25 + 5 = 30 gf
By law of floatation :
Weight of brine solution displaced = Weight of test tube = 30 gf As, volume of brine solution displaced = Volume of water displaced

Question 13.
A wooden block is weighed with iron, such that combination just floats in water at room temperature. State your observations when :
(1) water is heated above room temperature
(2) water is cooled below 4°C. Give reasons to your answers in (1) and (2).
(1) We know density of water decreases with rise in temperature and hence upthrust decreases.
(2) Density of water is maximum at 4°C. When water cooled below 4°C, then its density and hence upthrust acting on it decreases. So, wooden block weighed with iron, sinks more than earlier.

Question 14
A rubber ball floats in water with 2/7 of its volume above the surface of water. Calculate the average relative density of rubber ball.
Let volume of rubber ball = V

Question 15.          (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A cube of ice whose side is 4.0 cm is allowed to melt. The volume of water formed is found to be 58.24 cm3. Find the density of ice.
Side of ice cube = l = 4 cm
Volume of ice cube = V = β = (4)3 = 64 cm3
Volume of water = V = 5824 cm3
Density of ice cube = ρi= ?
Density of water = ρw = 1 gem-3
By law of floatation:
Volume of ice cube x Density of ice=Volume of water x Density of water
64 x ρi = 58.24 x 1

Question 16.
A jeweller claims to make ornaments of pure gold of relative density 19.3. A customer buys from him a bangle of weight 25.25 gf. The customer then weighs the bangle under water and finds its weight 23.075 g with the help of suitable calculations explain whether the bangle is of pure gold or not.
[R.D. of gold is 11.61 and bangle is not of pure gold] Ans. R.D of pure gold = 19.3 …given

Question 17.
(a) When a piece of ice floating in water melts, the level of water inside the glass remains same. Explain.
(b) An inflated balloon is placed inside a big glass jar which is connected to an evacuating pump. What will you observe when the evacuating pump starts working? Give a reason for your answer.
(a) A piece of ice displaces an amount of water equal to its own weight. Volume of water displaced is equal to volume of submerged part of ice cube. When ice cube melts, its volume decreases and gets occupied in that volume of water which is displaced by it. As a result, level of water inside the glass remains same when piece of ice (ice cube) melts.
(b) When evacuating pump starts working, pressure inside the glass jar reduces. As the pressure inside the balloon is more than pressure outside the balloon inside the glass jar, so balloon with burst.

Question 18.
(a) A trawler is fully loaded in sea water to maximum capacity. What will happen to this trawler, if moved to river water? Explain your answer.
(b) A body of mass 50 g is floting in water. What is the apparent weight of body in water? Explain your answer.
(a) Density of sea water is more than the density of river water.So river water offers less upthrust to the trawler as compared to sea water.So, when a trawler is fully loaded sea water to maximum capacity, is moved to river water, it will sink.
(b) Mass of body = 50 g
Apparent weight of body = Weight of body in air – Weight of water displaced by body.
But when a body floats, then weight of body in air is equal to the weight of water displaced by the body.
=> Apparent weight of the body = 0

Question 19.
A body of mass ‘m’ is floating in a liquid of density ‘p’
(1) what is the apparent weight of body?
(2) what is the loss of weight of body?
Mass of body = m
Density of liquid = ρ
(1) Apparent weight of body = Weight of body in air – Weight of liquid displaces by body.
When a body floats in the liquid, then weight of the body in a liquid is equal to weight of liquid displaced by the body.
=> Apparent weight of body = 0
(2) Loss in weight of body is equal to the weight of liquid displaced by the body.

Question 20.   (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A block of wood of volume 25 cm3 floats in water with 20 cm3 of its volume immersed in water.
Calculate :
(1) density of wood
(2) the weight of block of wood.
Volume of wooden block = V = 25 cm3
Volume of wooden block immersed in water = 20 cm3
Volume of water displaced by wooden block=volume of wooden
block immersed in water = 20 cm3
Density of water = ρwater = 1 gcm-3
Density of wooden block =ρwood = ?
By law of floatation:
Volume of wooden block x Density of wood=Volume of water displaced x Density of water

Question 21.
A solid body weighs 2.10 N. in air. Its relative density is 8.4. How much will the body weigh if placed
(1) in water,
(2) in liquid of relative density 1.2?
Weight of solid body in air = 2.10 N
R.D. of solid = 8.4

Weight of body in water = Weight of body in air – Weight of water displaced by body
= 2.10 -0.25 = 1.85 N

(2) Upthrust due to water = Weight of water displaced by body
= 0.25 N
Upthrust due to liquid = Upthrust due to water ×R.D. of liquid
= 0.25 x 1.2 = 0.30 N
Weight of body in liquid = Weight of body in air – Upthrust due to liquid
= 2.10-0.30= 1.80 N

-: End of Goyal Brothers : Upthrust and Archimedes’ Principle  Class-9 ICSE Physics Ch-5:-

Thanks

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