Goyal Brothers Pressure in Fluids Class-9 ICSE Physics Ch-4

Goyal Brothers Pressure in Fluids Class-9 ICSE Physics Ch-4.  We Provide Step by Step Answer of Exercise, MCQs, Numericals Practice Problem Questions of Exercises, Subjective Pressure in Fluids Class-9, Visit official Website CISCE  for detail information about ICSE Board Class-9 Physics.

Goyal Brothers Pressure in Fluids Class-9 ICSE Physics Ch-4


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Practice Problems 

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Objective Questions

 Subjective Questions


Practice Problems 

Goyal Brothers Pressure in Fluids Class-9 ICSE Physics Ch-4

Practice Problems 1.
Question 1.
Calculate pressure exerted by 0.8 m vertical length of alcohol of density 0.80 gcnr5 in SI units.
[Take g = 10 ms-2].
Answer:
Vertical length of alcohol column = h = 0.8 m
Density of alcohol = ρ = 0.80 g cm -3
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids
ρ = 0.80 x 1000 kgm-3 .
ρ = 800 kg m-3
Pressure = P = ?
p= hpg
P = 0.8 x 800 x 10
= 6400 Pa

Question 2.
What is the pressure exerted by 75 cm vertical column of mercury of density 13600 kgm-3 in SI units.
[Take g = 9.8 ms-2].
Answer:
Vertical length of mercury column = h = 75 cm = 0.75 m
Density of mercury = p = 13600 kg m-3
Acceleration due to gravity = g = 9.8 ms-2
Pressure = P = ?
p= hpg
P = 0.75 x 13600×9.8
P = 99960 Pa

Practice Problems 2.

Question 1.
66640 Pa pressure is exerted by 0.50 m vertical column of a liquid. If g = 9.8 Nkg-1, calculate density of the liquid.
Answer:
Pressure + P = 66640 Pa .
Vertical length of liquid column = h = 0.50 m
Acceleration due to gravity -g = 9.8 ms-2
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.011

Question 2.
What vertical height of water will exert pressure of 333200 Pa? Density of water is 1000 kgnr3 and g = 9.8 ms-2.
Answer:
Vertical height of water = h = ?
Pressure due to water column = P = 333200 Pa Acceleration due to gravity = g = 9.8 ms-2
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.012
Question 3.
Pressure at bottom of sea at some particular place is 8968960 Pa. If density of sea water is 1040 kgm3 calculate the depth of sea. Take g = 9.8 ms-2. Neglect the pressure of the atmosphere.
Answer:
Pressure at the bottom of the sea = P = 8968960 Pa
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.013

Practice Problems 3.

Question 1.
Atmospheric pressure at sea level is 76 cm of mercury. Calculate the vertical height of air column exerting the above pressure. Assume the density of air 1.29 kgm-3 and that of mercury is 13600 kgm-3. Why the height calculated by you is far less than actual height of atmosphere?
Answer:
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.014

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Question 2.
Calculate the equivalent height of mercury, which will exert as much pressure as 960 m of sea water of density 1040 kgm-3. Density of mercury is 13600 kgm-3.
Answer:
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.014

A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.015

Practice Problems 4.

Question 1.

The pressure of water on the ground floor, in a water pipe is 150000 Pa, whereas pressure on the fourth floor is 30000 Pa. Calculate height of fourth floor. Take g = 10 ms-2.
Answer:
Pressure of water at fourth floor = P2 = 3000 Pa Let h be the height of fourth floor
Difference in pressure of water at ground floor and fourth floor
= P1 – P2 = 150000 – 30000
= 120000 Pa
Pressure of water due to height (h) = hρg
=> P1 – P2 = h p
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.016
Question 2.        (Goyal Brothers Pressure in Fluids Class-9 ICSE Physics Ch-4)
The pressure of water on ground floor is 160000 Pa. Calculate the pressure at the fifth floor, at a height of 15 m.
Answer:
Pressure of water at ground floor = P, = 160000 Pa
Pressure of water at fifth floor = P2 = ?
Height of fifth floor = h = 15 m
Density of water = ρ = 1000 kgm-3
Difference in pressure of water at ground and fifth floor
= P1 – P2
Pressure of water due to height (h) = hρg
= P1 – P2 = hρg
160000 -P2= 15 x 1000 x 10
P2= 160000-150000
P2 = 10000 Pa

Practice Problems 5.

Question 1.
(a)
 The area of cross-sections of the pump plunger and press plunger of a hydraulic press are 0.02 m2 and 8 m2 respectively. If the hydraulic press overcomes a load of 800 kgf, calculate the force acting on pump plunger.
(b) If the mechanical advantage of the handle of pump plunger is 8, calculate the force applied at the end of the handle of pump plunger.
Answer:
(a) Load on the press plunger = L = 800 kgf
Let the effort acting on the pump plunger = E
Area of cross-section of pump plunger = A1 = 0.02 m2
Area of cross-section of press plunger = A2 = 8 m2
A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.017

(b)

A New Approach to ICSE Physics Part 1 Class 9 Solutions Pressure in Fluids.018

Question 2.
The radii of the press plunger and pump plunger are in ratio of 50 : 4. If an effort of 20 kgf acts on the pump plunger, calculate the maximum effort which the press plunger can over come.
Answer:

Effort acting on the pump plunger – E = 20 kgf
Load acting on the press plunger = L = ?
So let radius of pump plunger = 5 Ox = R and radius of press plunger = Ax = r
 radius of pump plunger

radius of press plunger


(A) Objective Questions

Multiple Choice Questions.

Goyal Brothers Pressure in Fluids Class-9 ICSE Physics Ch-4

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Select the correct option.

1. Unit of thrust in SI system is
(a) dynes
(b) joule
(c) N/m2
(d) newton

2. The unit Nmis the unit of
(a) force
(b) pressure
(c) thrust
(d) momentum

3. One Pascal is equal to:
(a) Nm2
(b) Nm-2
(c) Nm2
(d) Nm-1

4. Thrust acting perpendicularly on the unit surface area is called :
(a) pressure
(b) moment of force
(c) down thrust
(d) none of these

5. Pressure applied in liquids is transmitted with undiminished force:

(a) in downward direction
(b) upward direction only
(c) sides of containing vessel
(d) in all directions

6. As we move upwards, the atmospheric pressure :
(a) increases
(b) decreases
(c) remains same
(d) cannot be said

7. A dam for water reservoir is built thicker at the bottom than at the top because :
(a) pressure of water is very large at the bottom due to its large depth

(b) water is likely to have more density at the bottom due to its large depth
(c) quantity of water at the bottom is large
(d) variation in value of ‘g’

8. The pressure exerted by 50 kg (g = 10 m/s2) on an area of cross section of 2 m2 is :
(a) 50 Pa
(b) 200 Pa
(c) 250 Pa  
(d) 1000 Pa
Ans. (c) 250 Pa
Explanation : m = 50 kg; g = 10 ms-2
Area of cross-section = A = 2m2
Area

9. Pressure at a point inside a liquid does not depend on :
(a) The depth of the point below the surface of the liquid
(b) The nature of the liquid
(c) The acceleration due to gravity at that point
(d) The shape of the containing vessel

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