The Geometry of Power Advanced Simple Machines Class 9 Question Answer Advanced Science Chapter 4

Students can use NCERT Class 9 Advanced Science Solutions Chapter 4 The Geometry of Power Advanced Simple Machines Question Answer to understand complex concepts with ease.

The Geometry of Power Advanced Simple Machines Class 9 Questions and Answers

The Geometry of Power Advanced Simple Machines Question Answer Class 9

Quick Check

In a mechanical watch, a single power source (a spring or motor) must move three different hands at three different speeds. This is achieved through a Gear Train, where the ‘output’ of one gear becomes the ‘input’ for the next. The seconds-to-minutes gear ratio is 60:1 and the minutes-to-hours ratio is 60:1;

Question 1.
If the seconds gear is 2 mm, how large would the hour gear be in metres?
Answer:
Given:
Seconds-to-minutes ratio = 60:1
Minutes-to-hours ratio = 60:1
Seconds gear size = 2 mm
Seconds to Minutes: To get a 60:1 ratio, the minute gear must be 60 times larger than the 2 mm seconds gear. So, it will be 2 × 60 = 120 mm.
Minutes to Hours: To get a 60:1 ratio, the hour gear must be 60 times larger than the minute gear. So, it will be 120 × 60 = 7200 mm

Question 2.
Which of the three hands gear should be directly connected to the motor? Why?
Answer:
In a mechanical watch, the seconds hand rotates at the highest speed compared to the minute and hour hands. Therefore, the gear directly connected to the motor is the seconds-hand gear.

This arrangement allows the gear train to function efficiently by reducing speed in stages.
At each stage, the speed is reduced according to the required time intervals:

• The minutes hand moves slower than the seconds hand (by a factor of 60).
• The hours hand moves even slower than the minutes hand (again by a factor of 60).

Let us Calculate

Question 1.
Tension and acceleration are produced when two unequal masses are connected over a pulley.
1. Do the setup of weights, string and a simple pulley as shown.

2. Since 0.55 kg > 0.5 kg, the 0.55 kg mass will move downward. The 0.5 kg mass will move upward. Both masses will move with the same acceleration because they are connected by the same string.
3. For 0.55 kg mass (moving downward):
a. Downward force = Weight = _________
b. Upward force = Tension (T)

4. Net force: 0.55 g – T = 0.55 a

5. For 0.5 kg mass (moving upward):
a. Downward force = Weight = _________
b. Upward force = Tension (T)

6. Net force: T – 0.5g = 0.5a
7. Add both equations: _________
8. Acceleration of the system: _________
9. Find Tension: _________
Answer:
3. For 0.55 kg mass (moving downward):
a. Downward force = Weight = 0.55 ×9.8 = 5.39 N
b. Upward force = Tension (T)

4. Net force: 0.55g – T = 0.55a

5. For 0.5 kg mass (moving upward):
a. Downward force = Weight = 0.5 × 9.8 = 4.9 N
b. Upward force = Tension (T)

6. Net force: T – 0.5g = 0.5a
7. Add both equations:
(0.55g – T) + (T – 0.5g) = 0.55a + 0.5a
0.55g – 0.5g = 1.05a
0.05g = 1.05a
a = \(\frac{0.05 \times 9.8}{1.05}\)
a = \(\frac{0.49}{1.05}\) ≈ 0.467 m/s²

8. Acceleration of the system: 0.467 m/s²

9. Substitute a back into the equation from step 6.
T – 0.5g = 0.5a
T = 0.5(g + a)
T = 0.5(9.8 + 0.467)
T = 0.5 kg (10.267 m/s²)
T ≈ 5.13 N

Check Your Understanding

Question 1.
Show the direction of weight and tension for both objects m₁ and m₂.

Answer:

Question 2.
An 8 kg mass hangs freely from a single fixed pulley. The system is at rest. Find the tension in the rope.
Answer:
As the system is at rest (equilibrium), the upward tension equals the downward weight.
T = W = m × g
T = 8 kg × 9.8 m/s²
T = 78.4 N

Question 3.
Observe the given diagram. Find out in which direction the rope will move? What will be the net downward force?

Answer:
The rope will move toward the heavier side, so it moves downward on the 20 kg side and upward on the 10 kg side.
Net downward force: This is the difference between the two weights.
F_net = (m₂ × g)- (m₁ × g)
F_net = (20 × 9.8) – (10 × 9.8)
= 196 – 98
F_net = 98 N

Question 4.
A 6 kg mass hangs freely from a single fixed pulley. The system is at rest. Find the tension in the rope.
Answer:
T = m × g
T = 6 kg × 9.8 m/s²
T = 58.8 N

Question 5.
Two objects having masses of 2 kg and 6 kg are connected over a frictionless pulley with the help of a rope. Find acceleration and tension in the rope.
Answer:
Let m₁ = 2 kg and m₂ = 6 kg.
Acceleration (a):
a = \(\frac{\left(m_2-m_1\right) g}{m_1+m_2}\)
a = \(\frac{(6-2) \times 9.8}{6+2}\)
= \(\frac{4 \times 9.8}{8}\)
= \(\frac{39.2}{8}\) = 4.9 m/s²

Tension:
T = \(\frac{2 \times m_1 \times m_2 \times g}{m_1+m_2}\)
T = \(\frac{2 \times 2 \times 6 \times 9.8}{2+6}\)
= \(\frac{235.2}{8}\) = 29.4 N

The Geometry of Power Advanced Simple Machines Class 9 Extra Questions and Answers

Short Answer Type Questions

Question 1.
Name a machine which can be used to:
(a) Multiply force
(b) Change the direction of force applied
Answer:
(a) Wheel and axle / Pulley.
(b) Fixed pulley.

Question 2.
A type of single pulley is very often used as a machine even though it does not give any gain in mechanical advantage.
(a) Name the type of pulley used.
(b) For what purpose is such a pulley used?
Answer:
(a) Fixed pulley
(b) It is used to change the direction of the applied force, making work more convenient.

Question 3.
A wheel and axle machine has a wheel radius of 40 cm and an axle radius of 5 cm. A load of 800 N is lifted using the machine.
(a) Find the Mechanical Advantage (MA).
(b) Calculate the effort required to lift the load.
Answer:
(a) For a wheel and axle:
MA = \(\frac{R_{\text {wheel }}}{R_{\text {axle }}}\)
MA = \(\frac{40 \mathrm{~cm}}{5 \mathrm{~cm}}\) = 8

(b) MA = \(\frac{\text { Load }}{\text { Effort }}\)
8 = \(\frac{800 \mathrm{~N}}{\text { Effort }}\)
Effort = \(\frac{800 \mathrm{~N}}{8}\) = 100 N

Question 4.
Two objects of masses 3 kg and 5 kg are connected by a light string passing over a frictionless pulley.
(a) In which direction will the system move?
(b) Calculate the net force acting on the system. (Take g = 10 m/s²)
Answer:
(a) The system will move toward the 5 kg mass because it is heavier.
(b) Net force:
F = (m₂ – m₁) g
F = (5 kg – 3 kg) × 10 m/s² = 20 N

Question 5.
Define mechanical advantage and velocity ratio. A pulley system has a velocity ratio of 4 and efficiency 60%. Calculate mechanical advantage.
Answer:
Mechanical Advantage (MA): Mechanical Advantage is the ratio of the load lifted by a machine to the effort applied.
MA = \(\frac{\text { Load }}{\text { Effort }}\)
Velocity Ratio (VR): Velocity Ratio is the ratio of the distance moved by the effort to the distance moved by the load.
VR = \(\frac{\text { Distance moved by effort }}{\text { Distance moved by load }}\)
Given, V.R. = 4
Efficiency = 60%
\(\frac{M A}{V R}\) = efficiency
\(\frac{M A}{4}=\frac{60}{100}\)
MA = \(\frac{4 \times 60}{100}\) = 2.4

Question 6.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
Answer:
Suppose m₁ and m₂ be the masses suspended at the ends of a light inextensible string passing over the pulley.

m₁ = 8 kg, m₂ = 12 kg T = tension in the string
a = common acceleration with which m₁ moves upward and m₂ moves downward.
The equations of motion of m1 and m2 are given by
T – m₁g = m₂a ………….. (1)
and
m₂g-T = m₂a Adding equations (i) and (ii),
(m₂ – m₁)g = (m₁ + m₂)a
a = \(\frac{\left(m_2-m_1\right) g}{m_1+m_2}\)
From equations (i) and (iii),
T = m₁g + m₁\(\frac{\left(m_2-m_1\right) g}{m_1+m_2}\)
T = \(\frac{m_1 g}{m_1+m_2}\) (m₁ + m₂ + m₃ + m₄)
T = \(\frac{2 m_1 m_2}{m_1+m_2} g\) ………. (iv)
Putting m₁ = 8 kg and m₂ = 12 kg and g = 10 ms^-2, in equation (iii) and (iv), we get
a = \(\frac{(12-8)}{(8+12)}\) × 10
= \(\frac{4}{20}\) × 10 = 2ms^-2
= 2 ms^-2
From eq. (i)
T = m₁a + m₁g
= 8 kg × 2 m/s² + 8 kg × 10 m/s²
= 96 N

Long Answer Type Questions

Question 1.
State the relationship between Mechanical Advantage (MA), Velocity Ratio (VR), and Efficiency (q) for a machine. Hence, mathematically derive the expression that connects these three terms. Why is the Mechanical Advantage of a real machine always less than its Velocity Ratio?
Answer:
Efficiency (η) is the ratio of Mechanical Advantage (MA) to the Velocity Ratio (VR). It is usually expressed as a percentage.
Efficiency (η) = \(\frac{\text { Mechanical Advantage (MA) }}{\text { Velocity Ratio (VR) }}\)
Efficiency (η) = \(\frac{\text { Work Output }}{\text { Work Input }}\)
Work Output = Load (L) × Distance moved by Load (d_L)
Work Input = Effort (E) × Distance moved by Effort (d_E)
So,
η = \(\frac{L \times d_L}{E \times d_E}=\left(\frac{L}{E}\right) \times\left(\frac{d_L}{d_E}\right)\)
η = MA × \(\frac{1}{\mathrm{VR}}\) ⇒ η = \(\frac{\text { MA }}{\text { VR }}\)
Explaining Real vs. Ideal Machines

• In any real machine, some input energy is always lost as heat to overcome friction between moving parts and the weight of the machine elements.
• Because of this energy loss, the real Mechanical Advantage drops (MA < VR), making the efficiency less than 100% (η < 1 ).
• The Velocity Ratio (VR) depends purely on the dimensions (like wheel radii or distance moved) and remains constant regardless of friction.

Question 2.
A maintenance engineer is inspecting a mechanical lift system used to transport materials between two floors. The lift operates as a simple vertical pulley system (Atwood machine). A heavy equipment crate of mass m2 = 60 kg is connected to a counterweight of mass m, = 40 kg by a light, inextensible steel cable passed over a frictionless, ideal pulley. The system is released from rest, allowing the crate to descend safely to the lower floor.
(Take acceleration due to gravity, g = 10 m/s².)
Derive the expressions for the acceleration (a) of the system and the tension (T) in the connecting cable by drawing the free-body diagrams (FBD) for both masses. Hence, calculate the numerical values of both the acceleration and tension for this factory lift system.
Answer:
Free Body Diagrams for Connected Masses (m₂; m₂)

Overall System:
m₁ = 40 kg, m₂ = 60 kg

Net Force is:
ΣF = m₂ × g – T = m₂ × a
For Mass m₁ (Moving Upward):
The forces acting on it are Tension (T) upward and weight (m₁g) downward.
T – m₁g = m₁a ….. (1)
For Mass m₂ (Moving Downward):
The forces acting on it are weight (m₂g) downward and Tension (T) upward.
m₂g – T = m₂a ……….. (2)
For Acceleration (a): Adding Equation 1 and Equation 2 eliminates T:

ΣF = T = m₁ × g = m₁ × a
m₂g – m₁g =(m₁ + m₂)a
⇒ a = \(\left(\frac{m_2-m_1}{m_1+m_2}\right) g\)
Substituting the value of a into Equation 1 gives:
T = \(\frac{2 m_1 m_2}{m_1+m_2} g\)
Substitute the given values (m₁ = 40 kg, m₂ = 60 kg, g = 10 m/s²):
a = \(\left(\frac{60-40}{60+40}\right)\) × 10
\(\left(\frac{20 \mathrm{~kg}}{100 \mathrm{~kg}}\right)\) × 10 m/s² = 2m/s²
Substitute the values into the Tension formula:
T = \(\frac{2 \times 40 \times 60}{40+60}\) × 10 = \(\frac{4800}{100}\) × 10 = 480 N

Case-Based MCQs

Question 1.
Rahul is helping his uncle move heavy items in a farm. First, he tries to lift a heavy bucket of water straight up out of a deep well, but it requires too much force. His uncle helps him to set up a simple wheel and axle system (a water well crank).
The large handle wheel has a radius of 30 cm, and the small axle where the rope winds up have a radius of 6 cm. Next, his uncle explains that because of friction in the rusty parts of the old machine, some energy will be lost as heat.

(i) What is the Mechanical Advantage (MA) of the wheel and axle system used on the well?
(A) 2
(B) 5
(C) 10
(D) 36
Answer:
Option (B) is correct.
Explanation: MA = \(\frac{\text { Radius of wheel }}{\text { Radius of axle }}=\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}}\) = 5

(ii) If the bucket of water weighs 200 Newtons, how much effort force does Rahul ideally need to apply to turn the handle?
(A) 40 N
(B) 100 N
(C)200 N
(D) 1200 N
Answer:
Option (A) is correct.
Explanation: Effort = \(\frac{\text { Load }}{M A}=\frac{200 N}{5}\) = 40 N

(iii) Rahul’s uncle explains that this wheel and axle machine works as a ‘force multiplier”. Why is it called that?
(A) It makes the bucket move much laster than the handle.
(B) It allows a small input force to lilt a much heavier load.
(C) It creates free energy out of nowhere.
(D) It reduces the total amount of work done.
Answer:
Option (B) is correct.
Explanation: It allows a small input force to lift a much heavier load.

(iv) Because the old well machine is a “real machine” with friction, what is true about its efficiency?
(A) Its efficiency is exactly 100%.
(B) Its efficiency is more than 100%.
(C) Its efficiency is less than 100%.
(D) It has zero friction losses.
Answer:
Option (C) is correct.
Explanation: Real machines always lose some energy to friction.

Case Based Subjective Questions

Question 1.
An old village well uses a manual mechanical device called a windlass (a wheel and axle system) to fetch water. The system consists of a large handle wheel with a radius of 45 cm rigidly connected to a smaller inner axle with a radius of 5 cm. A villager uses this machine to lift a heavy bucket of water that requires an upward pulling force (load) of 180 N. Due to rust and wear over time, friction is present inside the bearings of this real-world mechanism.
(i) State the relationship between the radius of the wheel and the mechanical advantage of a wheel and axle system.
(ii) Calculate the theoretical Mechanical Advantage (MA) of this wheel and axle system.
(iii) If the system behaves ideally without any energy losses, calculate the minimum effort force the villager needs to apply to lift the 180 N bucket.
Answer:
(i) Mechanical Advantage (MA) is directly proportional to the radius of the wheel (MA ∝ R). A larger wheel radius provides a greater mechanical advantage.

(ii) MA = \(\frac{\text {Radius~of~wheel~}\left(R_{\text {wheel }}\right)}{\text {Radius~of~axle}\left(R_{\text {axle }}\right)}\)
MA = \(\frac{45 \mathrm{~cm}}{5 \mathrm{~cm}}\) = 9

(iii) Effort (E) = \(\frac{\text { Load }}{M A}\)
E = \(\frac{180 \mathrm{~N}}{9}\) = 20 N

Question 2.
Two students, Amit and Priya, set up a physics experiment using a single fixed frictionless pulley. They hang two unequal masses over the pulley using a light, inextensible string. Mass m₁ is 2 kg and mass m₂ is 3 kg. When the system is released from rest, the weights become unbalanced. The heavier mass accelerates downward while pulling the lighter mass upward.
(Take acceleration due to gravity, g = 9.8 m/s²)
(i) Define the term ‘Tension’.
(ii) Explain why both masses move with the exact same acceleration magnitude.
(iii) Calculate the net downward force (F_net) acting on this unbalanced pulley system.
Answer:
(i) Tension is the pulling or stretching contact force transmitted through the string when pulled from opposite ends.

(ii) Both masses are connected tightly by the same inextensible string. Therefore, the distance moved by one mass is exactly equal to the distance moved by the other in the same interval of time.
(iii) F_net = (m₂ g) – (m₁ × g)
= (3 × 9.8) – (2 × 9.8)
= 29.4 – 19.6
= 9.8N

The Geometry of Power Advanced Simple Machines Class 9 MCQ

Question 1.
Tension in a rope always acts:
(A) Perpendicular to the rope
(B) Along the length of the rope
(C) Downward only
(D) Upward only
Answer:
Option (B) is correct.
Explanation: Tension is a pulling force transmitted through a rope or string. It always acts along the length of the rope because the rope can only pull, not push.

Question 2.
A moment of couple has a tendency to rotate the body in an anticlockwise direction. Then the moment of couple is taken as:
(A) Positive
(B) Negative
(C) Maximum
(D) Zero
Answer:
Option (A) is correct.
Explanation: A moment of couple consists of two parallel forces those are equal in magnitude but opposite in direction. If body rotates in anticlockwise direction, the moment is said to be positive. If force pair or couple rotates in clockwise direction, it is said to be negative.

Question 3.
Mechanical Advantage (MA) is the ratio of:
(A) Effort to Load
(B) Load to Effort
(C) Distance to Time
(D) Work to Power
Answer:
Option (B) is correct.
Explanation: Mechanical Advantage tells how much a machine multiplies force. It is calculated by dividing the load by the effort.
MA = \(\frac{\text { Load }}{\text { Effort }}\)

Question 4.
A 10 kg object hangs freely from a rope and remains at ‘ rest. What is the tension in the rope? (Take g = 10 m/s²)
(A) 10 N.
(B) 50 N
(C) 100 N
(D) 200 N
Answer:
Option (C) is correct.
Explanation: When the object is at rest, the upward tension equals the downward weight.
T = mg
T = 10 kg × 10 m/s² = 100 N

Question 5.
In a wheel and axle machine, the wheel radius is greater than the axle radius. This helps to:
(A) Reduce friction
(B) Increase mechanical advantage
(C) Decrease speed
(D) Reduce weight
Answer:
Option (B) is correct.
Explanation: In a wheel and axle machine, a larger wheel radius produces a greater turning effect (torque). This allows a small effort to lift or move a larger load easily. So, it increases the Mechanical Advantage (MA).

Question 6.
Why is the steering wheel of a truck made large?
(A) To increase weight
(B) To reduce speed
(C) To increase turning effect
(D) To save fuel
Answer:
Option (C) is correct.
Explanation: A truck steering wheel is made large to produce a greater turning effect (torque). A larger wheel radius helps the driver turn the vehicle with less effort.

Assertion-Reason Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): Efficiency of a machine can be greater than Mechanical Advantage.
Reason (R): Efficiency depends on work output and work input.
Answer:
Option (D) is correct.
Explanation: The assertion is false because efficiency cannot exceed 100% .and is not simply greater than Mechanical Advantage.

The reason is true because efficiency is calculated using work output and work input.

Question 2.
Assertion (A): Velocity Ratio depends on the design of the machine.
Reason (R): Velocity Ratio is independent of friction.
Answer:
Option (B) is correct.
Explanation: Both statements are true, but the reason does not explain the assertion.

Velocity Ratio depends on the machine’s design, while friction does not affect Velocity Ratio.

Question 3.
Assertion (A): The efficiency of a real machine is always less than 100%.
Reason (R): Some energy is lost due to friction.
Answer:
Option (A) is correct.
Explanation A real machine has less than 100% efficiency because some input energy is lost due to friction.

Question 4.
Assertion (A): In an ideal pulley system, the tension is the same throughout the rope.
Reason (R): An ideal rope is massless and friction in the pulley. 0
Answer:
Option (A) is correct.
Explanation: In an ideal pulley system, the rope has no mass and the pulley has no friction. Therefore, no force is lost while transmitting tension, so the tension remains equal throughout the rope.