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

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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] Answer:
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] Answer:
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. 