**Pressure in Fluids and Atmospheric Pressure Exe-4A** Pressure in Fluids and it’s Transmission **Numericals Answer** Type for Class-9 ICSE Concise Physics. There is the solutions of **Numericals ****Answer** type Questions of your latest textbook which is applicable in 2023-24 academic session**. **Visit official Website CISCE for detail information about ICSE Board Class-9.

**Pressure in Fluids and it’s Transmission Exe-4A Numericals Answer **

**(ICSE Class – 9 Physics Concise Selina Publishers)**

Board | ICSE |

Class | 9th |

Subject | Physics |

Writer / Publication | Concise selina Publishers |

Chapter-4 | Pressure in Fluids and Atmospheric Pressure |

Exe – 4A | Pressure in Fluids and it’s Transmission |

Topics | Solution of Exe-4(A) Numericals Answer Type |

Academic Session | 2023-2024 |

**Exe-4A Pressure in Fluids and it’s Transmission Numericals Answer Type**

**Ch-4 Pressure in Fluids and Atmospheric Pressure Physics Class-9 ICSE Concise**

**Page 103**

**Question 1. **A hammer exerts a force of 1.5 N on each of the two nails A and B. The area of cross section of tip of nail A is 2 mm^{2} while that of nail B is 6 mm^{2}. Calculate pressure on each nail in pascal.

**Answer:**

Given:

Force acting on nail A is 1.5N, area = 2mm^{2}

Expressing 2mm^{2}in metre

1mm=0.001m,

∴ 1mm^{2}= 1mm x 1mm = 0.001m x 0.001m = 1 x 10^{-6} m^{2}

Pressure on B = Force/area

= 1.5/( 2 x 1 x 10^{-6})

= 7.5 x 10^{5} Pa

Pressure on B

= 1.5/(6 x 1 x 10^{-6})

= 2.5 x 10^{5} Pa

**Question 2. **A block of iron of mass 7.5 kg and of dimensions 12 cm × 8 cm × 10 cm is kept on a table top on its base of side 12 cm × 8 cm.

Calculate :

(a) Thrust and

(b) Pressure exerted on the table top

Take 1 kgf = 10 N.

**Answer:**

(a) To calculate thrust

Force = mass x acceleration due to gravity

= 7.5 x 10

= 75N

Area of the base = 12 x 8 = 96cm^{2}

or 0.0096m^{2}

(b) To calculate pressure exerted

Pressure = thrust/area

= 75/0.0096

= 7182.5 Pa

**Question 3. **A vessel contains water up to a height of 1.5 m. Taking the density of water 10^{3} kg m^{-3}, acceleration due to gravity 9.8 m s^{-2} and area of base of vessel 100 cm^{2}, calculate: (a) the pressure and (b) the thrust at the base of vessel.

**Answer:**

(a) To calculate pressure:

Given: h=1.5m, ρ = 1000, g=9.8m/s^{2}

P = h ρ g

= 1.5 x 1000 x 9.8

= 14700 Pa

**(b) To calculate the thrust at the base of the vessel:**

Pressure = Force/area

Thrust = force = P x a

14700 x 100 x 10^{-4}

147 N

**Question 4. **The area of base of a cylindrical vessel is 300 cm^{2}. Water (density= 1000 kg m^{-3}) is poured into it up to a depth of 6 cm. Calculate: (a) the pressure and (b) the thrust of water on the base. (g = 10m s^{-2}.

**Answer:**

**(i) To calculate pressure**

Given: density of water, ρ = 1000kg/m^{3}

g=10m/s^{2}

h = 6cm or 0.06m

We know that:

P = h ρ g

= 1000 x 0.06 x 10

= 600 Pa

**(ii) Thrust of water on the base**

Pressure = thrust / area

-> Thrust = force = P x a

= 6 x 10^{-2 }x 1000 x 10 x area

= 6 x 10^{-2 }x 1000 x 10 x 300 cm^{2}

= 6 x 10^{-2 }x 1000 x 10 x 300 x 10^{-4} m^{2}

= 18N

**Pressure in Fluids and Atmospheric Pressure Exe-4A Numericals Class-9 ICSE**

**Ch-4 Pressure in Fluids and Atmospheric Pressure Physics Class-9 ICSE Concise**

**Page 104**

**Question 5.**

(a) Calculate the height of a water column which will exert on its base the same pressure as the 70 cm column of mercury. Density of mercury is 13-6 g cm^{-3}.

(b) Will the height of the water column in part (a) change if the cross section of the water column is made wider?

**Answer:**

**(a) We know that the pressure exerted by the water column, P = h ρ g**

Density of water = 1

As the pressure of water and mercury is same,

hw ρw g = hm ρm g

hw x1 x g = hm ρm g

hw = hm ρm

= (70/100) x 13.6

Height of the water column= 9.52m

**(b) If the water column is made wider, the height of the water column will be unaffected**

**Question 6. **The pressure of water on the ground floor is 40,000 Pa and on the first floor is 10,000 Pa. Find the height of the first floor.

(Take : density of water = 1000 kg m^{-3}, g = 10 m s^{-2})

**Answer:**

**To find the height of the first floor:**

**Given: Pw on ground floor Pwg= 40,000Pa**

Pw on first floor Pwf = 10,000Pa

From the formula of pressure,

P = h ρ g

In order to know the height of the first floor, let us calculate the difference in pressure

P = Pwg – Pwf

= 40000 – 10000

= 30000Pa

Substituting this value in the formula for pressure to calculate height;

P = h ρ g

30000 = 1000 x 10 x h

h = 3m

**The height of the floor is 3m**

**Question 7. **A simple U tube contains mercury to the same level in both of its arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in the other arm?

Given : density of mercury = 13.6 x 10^{3} kg m^{-3 }and density of water = 10^{3} kg m^{-3}.

**Answer:**

Rise of water in the other side of the u-tube when water is added from one end depends on the

density of water and mercury.

Given: Water poured to the height 13.6cm or 0.136m in one arm.

To find the rise at the other end of the u-tube:

Since it is a u-tube, pressure on both the arms is the same, hence:

Difference in pressure in the water column = difference in pressure in the mercury column

hw ρw g = hm ρm g

hm = hw ρw / ρm

= 13.6 x 10^{3 }/ 13.6 x 10^{3}

= 1cm

∴ The other end of the u-tube will see a rise of 1cm in the mercury level.

**Question 8. **In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm^{2}. What force is obtained on its piston of area of cross section 100 cm^{2} ?

**Answer:**

As per Pascal’s law, When pressure increases, it uniformly increases through all the points when

any force is exerted.

Pressure = force/ area

(2N x 10^{-4})/10 = (F x 10^{-4})/ 100

2N = F/10

F = 20N

**(Pressure in Fluids and Atmospheric Pressure Exe-4A Numericals Class-9 ICSE)**

**Question 9. **What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of its brake shoe by exerting a force of 0.5 N on the pedal ?

**Answer:**

**Question 10. **The areas of pistons in a hydraulic machine are 5 cm^{2} and 625 cm^{2}. What force on the smaller piston will support a load of 1250 N on the larger piston? State any assumption which you make in your calculation.

**Answer:**

Given:

Area of narrow piston = 5cm^{2} = A1, let force applied be F1

Area of wider piston = 625cm^{2} = A2, let force applied be F2 = 1250N

We know from the hydraulic machine,

= F1/A1 = F2/A2

= F1/5 = 1250/625

F1 = 10N

**Question 11.**

(a) The diameter of neck and bottom of a bottle are 2 cm and 10 cm, respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf, what force is exerted on the bottom of the bottle?

(b) Name the law/principle you have used to find the force in part (i).

**Answer:**

**Question 12. **A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.

**Answer:**

Comparing the diameter of narrow piston and broader piston = 5/25 or 5:25

= 25 : 625

Force exerted on narrow piston, F1 = 50kgf

Consider F2 to be the force exerted on the broader piston

We know from the principle of Hydraulic machine,

Pressure on narrow piston = pressure on broader piston

=>F1/A1 = F2/A2

=> F1/ F2 = A1/A2

**50/F2 = 25/625**

**F2 = 1250 kgf**

**Question 13. **Two cylindrical vessels fitted with pistons A and B of area of cross section 8 cm^{2} and 320 cm^{2} respectively, are joined at their bottom by a tube and they are completely filled with water. When a mass of 4 kg is placed on piston A, Find :

(i) the pressure on piston A,

(ii) the pressure on piston B, and

(iii) the thrust on piston B.

**Answer:**

Given that the force applied to the smaller piston A is 4 kg

Area of cross-section of piston A = 8 cm²

Area of cross-section of piston B = 320 cm²

(i) Pressure acting on piston A in the downward direction = Thrust /Area

=4kg / 8cm2

∴ Pressure acting on piston A = 0.5 kg cm^{-2}

(ii) According to Pascal’s Law

Pressure acting on piston B = Pressure acting on Piston A = 0.5 kg cm^{-2}

(iii) Thrust acting on piston B in the upward direction

= Pressure × Area of B

= 4 kg × 320cm² / 28cm²

∴ Thrust acting on piston B in the upward direction = 160 kg.

**(Pressure in Fluids and Atmospheric Pressure Exe-4A Numericals Class-9 ICSE)**

**Question 14. **What force is applied on a piston of area of cross section 2 cm^{2} to obtain a force 150 N on the piston of area of cross section 12 cm^{2} in a hydraulic machine?

**Answer:**

We know that pressure on smaller piston = pressure on wider piston in a hydraulic machine

∴ P1 = P2

-> F1/A1 = F2 A2

F1/(2 𝑥 10^{-4}) = 150/(12 𝑥 10^{-4})

**= F1 = 25N**

In a hydraulic machine

Pressure on narrow piston = Pressure on wider piston

— : End of Pressure in Fluids and it’s Transmission Exe-4A **Numericals** Type Solutions :–

Return to **Concise Selina Physics ICSE Class-9 Solutions**

Thanks

Please share with your friends