NIOS Class 12 Physics Chapter 9 Properties of Fluids

NIOS Class 12 Physics Chapter 9 Properties of Fluids Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 9 Properties of Fluids Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.

NIOS Class 12 Physics Chapter 9 Properties of Fluids

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.

Properties of Fluids

Chapter: 9

Module – II: Mechanics of Solids and Fluids

INTEXT QUESTIONS 9.1

1. Why are the shoes used for skiing on snow made big in size?

Ans: Skiing shoes are made big in size to reduce pressure on the snow surface. 

According to the pressure formula:

Pressure = Force/Area

When the contact area is larger, the pressure exerted by the skier’s weight is distributed over a greater area, resulting in lower pressure per unit area. This prevents the skier from sinking deep into the snow and allows them to glide smoothly over the surface. If normal-sized shoes were used, the high pressure would cause the skier to sink into the snow, making skiing impossible.

2. Calculate the pressure at the bottom of an ocean at a depth of 1500 m. Take the density of sea water 1.024 × 10³ kg m^–3, atmospheric pressure=1.01 × 10⁵ Pa and g = 9.80 ms^–2.

Ans: Given: Depth, h = 1500 m

Density of sea water, ρ = 1.024 × 10³ kg m⁻³

Atmospheric pressure, P₀ = 1.01 × 10⁵ Pa

g = 9.80 m/s²

The total pressure at depth h in a liquid is given by:

P = P₀ + ρgh

Where:

P₀ = atmospheric pressure

ρghd column

P = 1.01 × 10⁵ + (1.024 × 10³)(9.80)(1500)

P = 1.01 × 10⁵ + 1.505 × 10⁷

P = 0.101 × 10⁷ + 1.505 × 10⁷

P = 1.606 × 10⁷ Pa

Pressure at the bottom = 1.606 × 10⁷ Pa

3. An elephant of weight 5000 kg f is standing on the bigger piston of area 10 m² of a hydraulic lift. Can a boy of 25 kg wt standing on the smaller piston of area 0.05m² balance or lift the elephant?

Ans: Given: Weight of elephant, W₁ = 5000 kgf

Area of bigger piston, A₁ = 10 m²

Weight of boy, W₂ = 25 kgf

Area of smaller piston, A₂ = 0.05 m²

According to Pascal’s law, pressure is transmitted equally in all directions in a confined liquid.

Pressure on bigger piston = W₁/A₁

= 5000/10

= 500 kgf/m²

Pressure on smaller piston = W₂/A₂ 

= 25/0.05 

= 500 kgf/m²

Since both pressures are equal, the boy can balance the elephant but cannot lift it. To lift the elephant, the boy would need to apply slightly more force to create higher pressure on the smaller piston.

Yes, the boy can balance the elephant but cannot lift it.

4. If a pointed needle is pressed against your skin, you are hurt but if the same force is applied by a rod on your skin nothing may happen. Why?

Ans: This happens due to the difference in pressure exerted by the needle and rod.

For the needle:

(a) Contact area is very small.

(b) Pressure = Force/Small area = High pressure.

(c) High pressure causes pain and can pierce the skin.

For the rod:

(a) Contact area is much larger (flat end).

(b) Pressure = Same force/Large area.

= Low pressure.

(a) Low pressure is distributed over a larger area, causing no harm

Since pressure is inversely proportional to area for the same applied force, the needle creates much higher pressure than the rod, explaining why it hurts more.

5. A body of 50 kg f is put on the smaller piston of area 0.1m² of a big hydraulic lift. Calculate the maximum weight that can be balanced on the bigger piston of area 10m² of this hydraulic lift.

Ans: Given: 

(a) Weight on smaller piston, F₁ = 50 kgf

(b) Area of smaller piston, A₁ = 0.1 m²

(c) Area of bigger piston, A₂ = 10 m²

(d) Weight on bigger piston, F₂ = ?

Using Pascal’s law:

P₁ = P₂

F₁/A₁ = F₂/A₂

F₂ = F₁ × (A₂/A₁)

F₂ = 50 × (10/0.1)

F₂ = 50 × 100

F₂ = 5000 kgf

Maximum weight that can be balanced = 5000 kgf