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

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**

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 :

- Weight of solid in SI system.
- Upthrust on solid in SI system.
- Apparent weight of solid in alcohol.
- 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}]

**Answer:**

Density of solid = ρ= 2700 kgm^{3
}Volume of solid = V = 0.0015 m^{3
}Density of alcohol = ρ’ = 800 kgm^{3}

- 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 - Volume of alcohol displaced = Volume of solid

V = 0.0015 m^{3 }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 - Apparent weight of solid in alcohol

= Actual weight of solid – Upthrust

= 40.5 -12 = 28.5 N - 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 kgm^{3} is lying submerged in water of density 1000 kgm^{3}. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms^{2}]
**Answer:**

Density of stone =ρ = 3000 kgm^{3
}Density of water = ρ’ = 1000 kgm^{3
}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 m^{3
}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 m ^{2} and length 0.60 m is completely immersed in water of density 1000 kgm^{3}. Calculate :**

- Wt of solid in SI system
- Upthrust acting on the solid in SI system.
- Apparent weight of solid in water.
- Apparent weight of solid in brine solution of density 1050 kgm
^{3}.

[Take g = 10 N/kg; Density of solid = 7200 kgm^{3}]

**Answer:**

Area of cross-section of solid = A = 0.004 m^{2
}Length of the solid = l = 0.60 m

Density of water = p’ = 1000 kgm^{3
}Acceleration due to gravity = g = 10 ms^{-2
}Density of solid = p = 7200 kgm^{3
}**(1)** Volume of solid = V = A x l

V = 0.004 x 0.60 = 0.0024 m^{3
}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 m^{3
}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 =ρ_{b}= 1050 kgm^{3
}Volume of brine solution displaced = Volume of solid = V

V = 0.0024 m^{3
}Mass of brine solution displaced

= m_{b} = V x ρ_{b} = 0.0024 x 1050

m_{b} = 2.52 kg

Upthrust acting on solid in brine solution = Weight of brine solution displaced -m_{b}g

= 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 kgm^{3} is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm^{3}, find the apparent weight of solid in liquid.

**Answer:**

Weight of solid in air = 0.950 kgf

∴ Mass of solid in air = m = 0.950 kg

Density of solid =ρ = 7600 kgm^{3
}

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} m^{2} 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.

**Answer:**

Length of glass cylinder = l = 12 x 10^{-2} m

Area of cross-section = A = 5 x 10^{-4} m^{2
}Volume of glass cylinder = V = A x l

V= 5 x lo^{-4} x 12 x 10-^{2
}V= 0.00006 m^{3
}Acceleration due to gravity = g = 9.8 m/s^{2
}Density of glass cylinder = ρ = 2500 kgm^{3
}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 kgm^{3}]
**Answer:**

Weight of solid in air = 0.08 kgf

Weight of solid in water = 0.065 kgf

Density of water = 1000 kgm^{3
}**(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.

**Answer:**

Relative density of solid = R.D. = 2.5

Weight of solid in water = W’ = 0.120 kgf

Weight of solid in air = W = ?

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