Class 10 Physics Chapter 3 Machines

Key Concept: Single Movable Pulley

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a single movable pulley, the load is supported by two segments of the rope, each carrying half the load. Therefore, the effort required to lift the load is $E = \frac{L}{2}$, giving a mechanical advantage of 2. The assertion correctly states that the mechanical advantage is 2. The reason explains why this happens: the tension in the rope is equally distributed across both segments, leading to the effort being halved. Both statements are true, and the Reason correctly explains the Assertion.

Key Concept: Block and Tackle System Efficiency

b) 50%
[Solution Description]
For a block and tackle system with $n = 3$ movable pulleys:
Velocity Ratio ($V.R.$) = $2^3 = 8$.
Mechanical Advantage ($M.A.$) = $\frac{L}{E} = \frac{800}{200} = 4$.
Efficiency ($\eta$) = $\frac{M.A.}{V.R.} \times 100\% = \frac{4}{8} \times 100\% = 50\%$.

Key Concept: Velocity Ratio

a) Radius of wheel to radius of axle
[Solution Description]
Velocity Ratio ($V.R.$) for a wheel and axle is the ratio of the distance moved by the effort (wheel) to the distance moved by the load (axle). This simplifies to the ratio of their radii:
$V.R. = \frac{R}{r}$
Hence, it compares the radius of the wheel ($R$) to the radius of the axle ($r$).

Key Concept: Velocity Ratio, Displacement Relationship

b) V.R. = 4; force multiplier
[Solution Description]
The velocity ratio (V.R.) is defined as the ratio of displacement of effort to displacement of load:
$V.R. = \frac{d_e}{d_l} = \frac{12}{3} = 4$.
Since $V.R. > 1$, the machine acts as a force multiplier (i.e., the effort required is less than the load lifted). The machine sacrifices speed for force multiplication.

Key Concept: Levers

b) 5
[Solution Description]
For a lever, the mechanical advantage (M.A.) is given by:
$\text{M.A.} = \frac{\text{Effort arm}}{\text{Load arm}}$
Substituting the given values (Effort arm = 10 cm, Load arm = 2 cm):
$\text{M.A.} = \frac{10}{2} = 5$

Key Concept: Block and Tackle System

c) 8
[Solution Description]
In a block and tackle system, the Velocity Ratio ($V.R.$) is calculated as $V.R. = 2^n$, where $n$ is the number of movable pulleys. For $n = 3$, $V.R. = 2^3 = 8$.

Key Concept: Examples of Class I Levers

a) Seesaw
[Solution Description]
From the syllabus, examples of Class I levers include seesaw, pair of scissors, crowbar, handle of water pump, etc. Among the options, the seesaw is explicitly mentioned as a Class I lever.

Key Concept: Single Movable Pulley

b) $E = \frac{L}{2}$
[Solution Description]
For an ideal single movable pulley, the tension $T$ in the string supports the load $L$ on both sides. Thus, $L = 2T$.
Since the effort $E$ is equal to the tension $T$, we get $E = \frac{L}{2}$.
Therefore, the effort required is half the load.

Key Concept: Block and Tackle System

c) 90%
[Solution Description]
For a 4-pulley block and tackle system, ideal V.R. = 4. The actual M.A. is:
$\text{M.A.} = \frac{540}{150} = 3.6$
Efficiency $(\eta)$ is calculated as:
$\eta = \left( \frac{\text{M.A.}}{\text{V.R.}} \right) \times 100 = \left( \frac{3.6}{4} \right) \times 100 = 90\%$

Key Concept: Mechanical Advantage, Efficiency

b) 640 N
[Solution Description]
For a system with $n$ movable pulleys, Mechanical Advantage ($M.A.$) is given by:
$M.A. = 2^n$
Here, $n = 4$:
$M.A. = 2^4 = 16$
Efficiency $(\eta)$ is given by:
$\eta = \left( \frac{M.A.}{V.R.} \right) \times 100\%$
For $n$ movable pulleys, $V.R. = 2^n = 16$. Given $\eta = 80\%$:
$80 = \left( \frac{M.A.}{16} \right) \times 100$
Solving for $M.A.$:
$M.A. = \frac{80 \times 16}{100} = 12.8$
Load $L$ can be calculated using $M.A. = \frac{L}{E}$ where $E = 50$ N:
$L = M.A. \times E = 12.8 \times 50 = 640 \text{ N}$

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

c) $2.\overline{6}\ \text{m}$
[Solution Description]
Given: Number of pulleys (n) = 4, Efficiency ($\eta$) = 75% = 0.75, Distance moved by effort ($d_e$) = 8 m.
First find Velocity Ratio (VR): $VR = n = 4$
We know that $\eta = \frac{MA}{VR}$, so $MA = \eta \times VR = 0.75 \times 4 = 3$
But also $MA = \frac{d_e}{d_l}$, so $3 = \frac{8}{d_l}$
Therefore, distance moved by load ($d_l$) = $\frac{8}{3}$ m = 2.666... m

Key Concept: Single Movable Pulley

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the mechanical advantage (M.A.) of an ideal single movable pulley is 2. This is true because in an ideal case (neglecting friction and weight of the pulley and string), the load $L$ is balanced by the tension $T$ in two segments of the string ($L = 2T$), and the effort $E = T$. Thus, M.A. = $\frac{L}{E} = 2$. The reason states that the velocity ratio (V.R.) is 2, which is also correct because when the effort moves a distance $2d$, the load moves a distance $d$, making V.R. = $\frac{2d}{d} = 2$. Moreover, the reason correctly explains why the mechanical advantage is 2, as the mechanical advantage is directly related to the velocity ratio in an ideal situation where efficiency is 100%.

Key Concept: Velocity Ratio, Efficiency, Mechanical Advantage

b) 240 N
[Solution Description]
First, calculate the mechanical advantage (M.A.) using the relation:
$\eta = \frac{M.A.}{V.R.} \times 100\%$.
Substituting the given values:
$80\% = \frac{M.A.}{6} \times 100\%$.
Simplifying: $M.A. = \frac{80 \times 6}{100} = 4.8$.
Now, since $M.A. = \frac{\text{Load}}{\text{Effort}}$, the load can be calculated as:
$\text{Load} = M.A. \times \text{Effort} = 4.8 \times 50 = 240 \text{ N}$.

Key Concept: Class III levers, Mechanical Advantage, Velocity Ratio

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in a Class III lever, the mechanical advantage ($M.A.$) is always less than 1 because the effort arm is shorter than the load arm. This is correct as per the syllabus, which explicitly mentions $M.A. < 1$ for Class III levers due to the effort arm being smaller than the load arm. The reason explains that for an ideal lever, $M.A.$ equals $V.R.$, and since $V.R. < 1$, $M.A.$ must also be less than 1. The reason correctly justifies the assertion by linking $M.A.$ and $V.R.$ under ideal conditions.

Key Concept: Velocity Ratio, Work Done

d) 2500 J
[Solution Description]
For a velocity ratio (V.R.) of 5, if the load moves up by 1 m, the effort moves:
$\text{Distance moved by effort} = V.R. \times \text{Distance moved by load} = 5 \times 1 = 5\,m$
Work done by the effort (input work):
$W_{\text{input}} = E \times d = 500\,N \times 5\,m = 2500\,J$

Key Concept: Examples of Class I Levers

c) Nutcracker
[Solution Description]
Among the given options, a nutcracker is a Class II lever (fulcrum at one end, load in the middle, and effort at the other end), whereas all others are Class I levers where the fulcrum lies between the effort and the load.

Key Concept: Class III levers, Mechanical Advantage, Efficiency

c) 0.2
[Solution Description]
First, calculate the ideal Mechanical Advantage (M.A.):
$M.A._{\text{ideal}} = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{10}{30} = \frac{1}{3} \approx 0.333$
Given efficiency $(\eta)$ = 60% = 0.6
Actual M.A. is calculated as:
$M.A._{\text{actual}} = \eta \times M.A._{\text{ideal}} = 0.6 \times 0.333 \approx 0.2$

Key Concept: Mechanical Advantage and Velocity Ratio

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
An ideal machine has no energy losses, so its efficiency ($\eta$) is 100%. The relationship between Mechanical Advantage (M.A.), Velocity Ratio (V.R.), and efficiency for any machine is given by:
$\eta = \frac{M.A.}{V.R.} \times 100\%$
For an ideal machine, since $\eta = 100\%$, we have $M.A. = V.R.$. Here, both the Assertion and Reason state this condition correctly, and the Reason explains why the Assertion is true. Thus, both are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Types of Machines Based on M.A.

c) Force multiplier
[Solution Description]
According to the syllabus, a machine with a mechanical advantage (M.A.) greater than 1 acts as a force multiplier. This means it reduces the effort needed to lift a load by multiplying the input force. Therefore, option c) correctly describes such a machine.

Key Concept: Block and Tackle System Efficiency

c) 75%
[Solution Description]
Given:
$V.R. = 4, \quad E = 20 \, \text{kgf}, \quad L = 60 \, \text{kgf}$
Mechanical Advantage (M.A.) is calculated as:
$M.A. = \frac{L}{E} = \frac{60}{20} = 3$
Efficiency ($\eta$) is given by:
$\eta = \frac{M.A.}{V.R.} \times 100\% = \frac{3}{4} \times 100\% = 75\%$
The efficiency of the system is 75%.

Key Concept: Efficiency Calculation

c) 80%
[Solution Description]
Efficiency $\eta$ is given by:
$\eta = \frac{M.A.}{V.R.}$
Given $M.A. = 4$ and $V.R. = 5$,
$\eta = \frac{4}{5} = 0.8 = 80\%.$
Thus, the efficiency is 80%.

Key Concept: Mechanical Advantage and Efficiency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given: Efficiency ($η$) = 80%, Velocity Ratio (V.R.) = 4.
The relationship between efficiency, Mechanical Advantage (M.A.), and V.R. is given by:
$η = \frac{M.A.}{V.R.} \times 100\%$
Substituting the given values:
$80\% = \frac{M.A.}{4} \times 100\%$
Simplify to find M.A.:
$M.A. = \frac{80\% \times 4}{100\%} = 3.2$
Since $M.A. = 3.2 > 1$, the assertion is true.
The reason states that a machine with $M.A. > 1$ acts as a force multiplier, which is correct as per the syllabus.
Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Class II Levers, Mechanical Advantage

c) 100 N
[Solution Description]
For class II levers, $M.A. = \frac{Effort Arm}{Load Arm}$. Given effort arm = 1.5 m, load arm = 0.5 m.
First calculate M.A: $M.A. = \frac{1.5}{0.5} = 3$
Then use $M.A. = \frac{Load}{Effort} \Rightarrow 3 = \frac{300}{Effort} \Rightarrow Effort = \frac{300}{3} = 100$ N

Key Concept: Mechanical Advantage

c) 4
[Solution Description]
The mechanical advantage ($M.A.$) is given by:
$M.A. = \frac{L}{E}$
Substituting values:
$M.A. = \frac{800}{200} = 4$
So, the mechanical advantage is 4.

Key Concept: Mechanical Advantage of a Lever

c) 3
[Solution Description]
The Mechanical Advantage (M.A.) is calculated by dividing the effort arm by the load arm.
Here, Effort Arm = 30 cm, Load Arm = 10 cm.
Therefore, M.A. = Effort Arm / Load Arm = 30 / 10 = 3.

Key Concept: Energy Losses in Machines

d) Gravity
[Solution Description]
The primary reasons for energy losses in machines are:
1. Friction between moving parts.
2. Imperfect elasticity of strings or belts.
3. Non-rigid parts causing deformation.
"Gravity" is not listed as a direct cause of energy loss in machines under the syllabus.

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

c) 76.43%
[Solution Description]
First calculate the velocity ratio (V.R.):
$V.R. = \frac{2\pi l}{p} = \frac{2 \times 3.14 \times 0.5}{0.008} = 392.5$
Then calculate mechanical advantage (M.A.):
$M.A. = \frac{L}{E} = \frac{12000}{40} = 300$
Now calculate efficiency ($\eta$):
$\eta = \frac{M.A.}{V.R.} \times 100\% = \frac{300}{392.5} \times 100\% = 76.43\%$

Key Concept: Class of Levers

b) Class II Lever
[Solution Description]
According to the principle of levers, Class II levers always have the effort arm greater than the load arm, resulting in M.A. > 1.

Key Concept: Pulley System, Velocity Ratio

c) 4
[Solution Description]
For an ideal pulley system, the velocity ratio (VR) is equal to the number of pulleys in the system. Given there are 4 pulleys:
$VR = 4$
Therefore, the velocity ratio is 4.

Key Concept: Velocity Ratio

b) 2 meters
[Solution Description]
For a single movable pulley, the velocity ratio ($V.R.$) is given by $V.R. = \frac{\text{distance moved by the effort}}{\text{distance moved by the load}} = 2$. Given that the effort moves 4 meters, let the distance moved by the load be $d$. Then, $2 = \frac{4}{d}$, solving for $d$ gives $d = \frac{4}{2} = 2$ meters.

Key Concept: Velocity Ratio, Block and Tackle System

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, for a block and tackle system with $n$ movable pulleys, the velocity ratio is given by $V.R. = 2n$. Here, $n = 3$, so $V.R. = 2 \times 3 = 6$. Thus, the Assertion is true. The Reason correctly explains the Assertion because it provides the formula from which the velocity ratio is derived. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Efficiency of a Machine

b) 80%
[Solution Description]
The formula for efficiency $(\eta)$ is given by:
$\eta = \left( \frac{\text{Work output}}{\text{Work input}} \right) \times 100$
Substituting the values:
$\eta = \left( \frac{400}{500} \right) \times 100 = 80\%$
Therefore, the efficiency of the machine is 80%.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For an ideal machine, friction is absent, so efficiency $(\eta)$ is 100% or 1. From the formula $M.A. = V.R. \times \eta$, if $\eta = 1$, then $M.A. = V.R.$
The reason correctly explains why the assertion is true because a 100% efficiency implies no energy loss, making M.A. equal to V.R. in an ideal scenario.

Key Concept: Mechanical Advantage and Velocity Ratio

b) 4
[Solution Description]
The mechanical advantage ($M.A.$) can be found using the relation between M.A., V.R., and efficiency: $M.A. = V.R. \times \eta$. Given that $V.R. = 5$ and $\eta = 0.8$, we calculate:
$M.A. = 5 \times 0.8 = 4$

Key Concept: Effect of Friction and Weight

c) 90%
[Solution Description]
Given, $M.A. = 4.5$ and $V.R. = n = 5$ for the block and tackle system. Efficiency $\eta$ is calculated as:
$\eta = \frac{M.A.}{V.R.} = \frac{4.5}{5} = 0.9 = 90\%$

Key Concept: Single Movable Pulley

c) 90.9%
[Solution Description]
For a single movable pulley, ideal M.A. = 2. Here, actual M.A. is calculated as:
$\text{M.A.} = \frac{\text{Load}}{\text{Effort}} = \frac{200}{110} \approx 1.818$
The Velocity Ratio (V.R.) for a single movable pulley is 2. Efficiency $(\eta)$ is given by:
$\eta = \left( \frac{\text{M.A.}}{\text{V.R.}} \right) \times 100 = \left( \frac{1.818}{2} \right) \times 100 = 90.9\%$

Key Concept: Basic Definition and Formula

b) $\text{M.A.} = \frac{\text{Load (L)}}{\text{Effort (E)}}$
[Solution Description]
The formula for Mechanical Advantage (M.A.) is given by the ratio of Load (L) to Effort (E). It is represented as:
$\text{M.A.} = \frac{\text{Load (L)}}{\text{Effort (E)}}$
Therefore, the correct answer is option b).

Key Concept: Load Lifted

c) 110 N
[Solution Description]
The load $L$ is given by:
$L = nE - w$
Here, $n = 3$, $E = 40$ N, and $w = 10$ N.
Substituting the values,
$L = 3 \times 40 - 10 = 120 - 10 = 110 \text{ N}.$
Therefore, the load lifted is 110 N.

Key Concept: Block and Tackle System

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a block and tackle system with 4 pulleys, the velocity ratio (V.R.) = number of pulleys = 4. Given: Load (L) = 800 N, Effort (E) = 200 N. Mechanical Advantage (M.A.) = $\frac{L}{E} = \frac{800}{200} = 4$. Efficiency ($\eta$) = $\frac{M.A.}{V.R.} \times 100 = \frac{4}{4} \times 100 = 100\%$. Thus, Assertion (A) is true. For Reason (R), M.A. equals the number of supporting strands, which in an ideal case equals the number of pulleys when effort is applied downward. So, Reason (R) is also true and correctly explains Assertion (A).

Key Concept: Mechanical Advantage

c) 30 N
[Solution Description]
For a block and tackle system with $n$ pulleys, the effort required to lift a load $L$ is given by:
$E = \frac{L}{n}$
Here, $n = 6$ and $L = 180\,\text{N}$. Substituting the values:
$E = \frac{180}{6} = 30\,\text{N}$
The effort required is 30 N.

Key Concept: Levers, Mechanical Advantage

d) 600 N
[Solution Description]
For a lever in equilibrium: $L \times \text{Load Arm} = E \times \text{Effort Arm}$.
Substituting values: $L \times 0.5 = 200 \times 1.5$.
Solving for $L$: $L = \frac{200 \times 1.5}{0.5} = 600 N$.

Key Concept: Ideal vs Actual Machine

c) The work output equals the work input.
[Solution Description]
An ideal machine has no energy loss, meaning the work output equals the work input (efficiency = 100%). It assumes no friction or other losses.

Key Concept: Tension in Strings, Load Distribution

b) 10 N
[Solution Description]
With one fixed and two movable pulleys, the tension $T$ in the string attached to the load is equal to the effort divided by the number of supporting strands. Here, there are 3 strands supporting the load (since $n = 2$ movable pulleys imply $2 + 1 = 3$ strands). Thus:
$T = \frac{E}{3} = \frac{30}{3} = 10 \text{ N}$

Key Concept: Mechanical Advantage, Efficiency

c) Assertion is true, but Reason is false.
[Solution Description]
For a block and tackle system with $n = 4$ pulleys, the ideal mechanical advantage ($M.A._{\text{ideal}}$) is equal to $n = 4$. Given the efficiency $\eta = 80\% = 0.8$, the actual mechanical advantage ($M.A._{\text{actual}}$) is calculated as:
$M.A._{\text{actual}} = M.A._{\text{ideal}} \times \eta = 4 \times 0.8 = 3.2$
However, the Reason incorrectly states that $M.A. = n \times \eta$ as a general formula, which is not accurate. The correct relationship is $M.A._{\text{actual}} = M.A._{\text{ideal}} \times \eta$, where $M.A._{\text{ideal}} = n$ for a block and tackle system. Thus, the Assertion is true, but the Reason is false.

Key Concept: Trade-off Between Force and Speed

c) Class I lever with gain in speed
[Solution Description]
Scissors with longer blades than handles are an example of a Class I lever where the effort arm (handle) is shorter than the load arm (blade). This configuration results in a velocity ratio less than 1 ($d_L > d_E$), providing a gain in speed. Thus, a small movement of the handles produces a larger displacement of the blades, useful for cutting cloth or paper efficiently.

Key Concept: Single Fixed Pulley, Mechanical Advantage, Efficiency

c) 62.5 kgf
[Solution Description]
In an ideal case (100% efficiency), the effort $E$ equals the load $L$, so $E = L = 50$ kgf.
However, efficiency $\eta = 80\%$ means the actual effort $E_{\text{actual}}$ is higher due to friction.
Efficiency is given by:
$\eta = \frac{M.A.}{V.R.} \times 100\%$
For a single fixed pulley, $M.A. = \frac{L}{E_{\text{actual}}}$ and $V.R. = 1$.
Substituting:
$80\% = \left(\frac{50}{E_{\text{actual}}}\right) \times 100\%$
Solving for $E_{\text{actual}}$:
$E_{\text{actual}} = \frac{50 \times 100}{80} = 62.5 \, \text{kgf}$

Key Concept: Single Movable Pulley Velocity Ratio

d) 6 m
[Solution Description]
The velocity ratio ($V.R.$) for a single movable pulley is given by:
$V.R. = \frac{\text{Distance moved by effort } (d_E)}{\text{Distance moved by load } (d_L)} = 2$
Given $d_L = 3 \, \text{m}$,
Therefore, $d_E = V.R. \times d_L = 2 \times 3 = 6 \, \text{m}$.

Key Concept: Direction of Effort

c) To change the direction of applied force
[Solution Description]
A single fixed pulley does not provide any mechanical advantage but changes the direction of the applied force. It allows the user to apply effort in a convenient downward direction to lift a load upwards.

Key Concept: Single Fixed Pulley

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion (A) states that a single fixed pulley is used to lift a bucket of water from a well. This is true because a single fixed pulley is commonly used for small loads like a water bucket. The Reason (R) states that it changes the direction of effort, making it easier to pull the load upwards. While this is also true, the reason does not fully explain why it is specifically used for lifting a bucket of water. The use is due to convenience in changing the direction of force rather than reducing the effort.

Key Concept: Pulley System - Single Fixed Pulley

b) 1
[Solution Description]
First, calculate the Mechanical Advantage (MA):
$MA = \frac{L}{E} = \frac{200}{220} = \frac{10}{11}$
Now, use the efficiency formula to find the Velocity Ratio (VR):
$\eta = \frac{MA}{VR} \implies VR = \frac{MA}{\eta} = \frac{\frac{10}{11}}{0.9} = \frac{10}{9.9} \approx 1.01$
However, for a single fixed pulley, the theoretical VR is 1 (since the distance moved by effort equals the distance moved by load). The slight discrepancy here is due to rounding errors or practical inefficiencies.

Key Concept: Mechanical Advantage in Block and Tackle System

c) 3.33
[Solution Description]
The mechanical advantage $M.A.$ can be calculated using the formula:
$M.A. = n - \frac{w}{E}$
Here, $n = 4$, $w = 20$ N, and $E = 30$ N.
Substituting the values,
$M.A. = 4 - \frac{20}{30} = 4 - 0.6667 = 3.3333.$
Therefore, the mechanical advantage is approximately 3.33.

Key Concept: Mechanical Advantage of Lever, Principle of Moments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To analyze the assertion and reason:
Step 1: The principle of moments states that for a lever in equilibrium, $L \times FB = E \times FA$, where $L$ is load, $FB$ is load arm, $E$ is effort, and $FA$ is effort arm.
Step 2: From this, the mechanical advantage (M.A.) is derived as $\frac{L}{E} = \frac{FA}{FB}$. Thus, M.A. > 1 implies $\frac{FA}{FB} > 1$, meaning the effort arm ($FA$) must indeed be longer than the load arm ($FB$).
Step 3: The Reason correctly defines M.A. as the ratio of the effort arm to the load arm, and when M.A. > 1, the lever indeed acts as a force multiplier (e.g., crowbar).
Conclusion: Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Combination of Pulleys

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a block and tackle system with four pulleys, the total number of strands supporting the load is equal to the number of pulleys in the movable block. Here, assuming two fixed and two movable pulleys, there are four strands supporting the load. According to the syllabus, the mechanical advantage (M.A.) is equal to the number of strands supporting the load. Hence, M.A. = 4.
The reason states that the mechanical advantage of a pulley system is equal to the number of strands supporting the load, which is correct as per the syllabus.
Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Class III Levers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A fishing rod is indeed a Class III lever where the fulcrum is at one end (the hand holding the rod), the load is at the other end (the fish being caught), and the effort is applied somewhere in between (the hand pulling the rod). For Class III levers, the effort arm is always shorter than the load arm, resulting in a mechanical advantage less than 1. This means the reason correctly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Mechanical Advantage and Efficiency

c) 32 kgf
[Solution Description]
Given: Radius of wheel, $R = 40$ cm; Radius of axle, $r = 8$ cm; Load, $L = 120$ kgf; Efficiency = 75%.
First, calculate Velocity Ratio (V.R.):
$V.R. = \frac{R}{r} = \frac{40}{8} = 5$
Next, calculate Mechanical Advantage (M.A.) using efficiency:
$M.A. = Efficiency \times V.R. = 0.75 \times 5 = 3.75$
Now, find the Effort (E):
$E = \frac{L}{M.A.} = \frac{120}{3.75} = 32$ kgf
Therefore, the effort required is 32 kgf.

Key Concept: Single Fixed Pulley

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For an ideal single fixed pulley:
1. Mechanical Advantage (M.A.) is given by $M.A. = \frac{L}{E}$. Since effort $E$ equals load $L$ in an ideal case, $M.A. = 1$.
2. Velocity Ratio (V.R.) is given by $V.R. = \frac{d_e}{d_l}$, where $d_e$ and $d_l$ are distances moved by effort and load respectively. Since both move the same distance, $V.R. = 1$.
Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

c) 90%
[Solution Description]
For a block and tackle system with $n$ movable pulleys, the ideal mechanical advantage ($M.A.$) is $n = 4$. The actual $M.A.$ is calculated as:
$M.A. = \frac{L}{E} = \frac{180}{50} = 3.6$
The velocity ratio ($V.R.$) equals the number of pulleys, which is $4$. Therefore, the efficiency $\eta$ is:
$\eta = \frac{M.A.}{V.R.} = \frac{3.6}{4} = 0.9 \text{ or } 90%$

Key Concept: Mechanical Advantage Calculation

b) 100 N
[Solution Description]
For a block and tackle system with $n$ pulleys, the mechanical advantage is equal to $n$. The formula for effort ($E$) in terms of load ($L$) and number of pulleys ($n$) is:
$E = \frac{L}{n}$
Given, $L = 600$ N and $n = 6$. Substituting these values:
$E = \frac{600}{6} = 100 \text{ N}$

Key Concept: Energy Loss in Machines

b) Friction between moving parts.
[Solution Description]
In an actual machine, energy is lost primarily due to friction between moving parts, which converts some useful energy into heat.

Key Concept: Single Movable Pulley

b) 2
[Solution Description]
A single movable pulley provides a mechanical advantage by reducing the effort needed to lift a load. The theoretical $M.A.$ is 2 because the load is supported by two parts of the rope, each carrying half the load ($L = 2E$).

Key Concept: Mechanical Advantage

c) 5
[Solution Description]
To find the mechanical advantage (M.A.), we use the formula:
$M.A. = \frac{L}{E}$
Here, Load $(L) = 500\,\text{N}$ and Effort $(E) = 100\,\text{N}$.
Substituting the values:
$M.A. = \frac{500}{100} = 5$.
Therefore, the mechanical advantage is 5.

Key Concept: Class II Levers

c) It is always greater than 1
[Solution Description]
In a Class II lever, the fulcrum and effort are at the two ends, and the load is somewhere in between. This means the effort arm is always longer than the load arm, resulting in a mechanical advantage always greater than 1.

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for a real pulley system, $MA < VR$. This is correct because in practical scenarios, no machine is ideal due to energy losses like friction and weight of moving parts. These losses result in efficiency ($\eta$) being less than 100% and hence $\eta = \frac{MA}{VR} < 1$, which implies $MA < VR$.
The reason explains that friction and the weight of moving parts reduce $\eta$ below 100%, leading to $MA < VR$. This is also correct and directly explains the assertion because $\eta < 1$ mathematically results in $MA < VR$ given $\eta = \frac{MA}{VR}$. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Class I Lever Definition

a) Between the effort and the load
[Solution Description]
A Class I lever has the fulcrum placed between the effort and the load. This is the defining characteristic of a Class I lever as per the syllabus.

Key Concept: Block and Tackle System

c) 6
[Solution Description]
In a block and tackle system with $n$ movable pulleys, the velocity ratio (V.R.) is given by $V.R. = 2n$. Here, $n = 3$, so the calculation is:
$V.R. = 2 \times 3 = 6$
Hence, the velocity ratio is 6.

Key Concept: Pulley System, Work Input/Output

b) 13333.33 J
[Solution Description]
Step 1: Calculate the work output ($W_{output}$):
$W_{output} = L \times d_l = 800\,N \times 10\,m = 8000\,J$
Step 2: Use efficiency to find work input ($W_{input}$):
$\eta = \frac{W_{output}}{W_{input}} \implies 0.6 = \frac{8000\,J}{W_{input}}$
$W_{input} = \frac{8000\,J}{0.6} \approx 13333.33\,J$
The work input required is approximately 13333.33 Joules.

Key Concept: Lever Classes, Mechanical Advantage

a) 0.33
[Solution Description]
Given: Third-class lever, Fulcrum to effort distance ($d_E$) = 10 cm, Fulcrum to load distance ($d_L$) = 30 cm, Load (L) = 120 N, Effort (E) = 80 N, Efficiency ($\eta$) = 60% = 0.6
For levers, VR is given by: $VR = \frac{d_E}{d_L} = \frac{10}{30} = \frac{1}{3}$
Actual MA = $\frac{L}{E} = \frac{120}{80} = 1.5$
We can verify using efficiency: $\eta = \frac{MA}{VR} \Rightarrow 0.6 = \frac{1.5}{VR} \Rightarrow VR = 2.5$
However, for levers the VR should be equal to $\frac{d_E}{d_L}$ which gives $\frac{1}{3}$. The discrepancy occurs because the MA calculated includes friction, while VR remains constant for a given machine design.

Key Concept: Calculation of efficiency

c) 90%
[Solution Description]
The efficiency of the pulley system is calculated using the formula:
$\eta = 1 - \frac{w}{nE} = 1 - \frac{20}{4 \times 50} = 1 - \frac{20}{200} = 1 - 0.1 = 0.9$
So, the efficiency is 90%.

Key Concept: Block and Tackle System - Mechanical Advantage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, for a block and tackle system with $n$ pulleys, the load $L$ is supported by $n$ segments of the string, each having tension $T$. Thus, $L = nT$. Since the effort $E$ equals the tension $T$ in each segment, we have $E = T$. The Mechanical Advantage ($M.A.$) is calculated as:
$M.A. = \frac{L}{E} = \frac{nT}{T} = n$
This shows that both Assertion (A) and Reason (R) are true, and the Reason correctly explains why the Mechanical Advantage is equal to the number of pulleys.

Key Concept: Single Movable Pulley - M.A., V.R., and Efficiency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In an ideal single movable pulley, the load $L$ is balanced by the tension $T$ in both segments of the string, giving $L = 2T$. The effort $E$ equals the tension $T$, so $E = T$. Thus, $M.A. = \frac{L}{E} = 2$.
For the velocity ratio, when the effort moves a distance $d_E = 2d$, the load moves $d_L = d$, so $V.R. = \frac{d_E}{d_L} = 2$.
Both Assertion and Reason are correct, and the Reason correctly explains the Assertion.

Key Concept: Velocity Ratio Concept

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the velocity ratio (V.R.) is always greater than 1 if the effort moves a longer distance than the load in the same time. This is true because, according to the definition, $V.R. = \frac{d_e}{d_l}$. If $d_e > d_l$, then $V.R. > 1$.
The reason defines velocity ratio correctly as the ratio of the distance moved by the effort to the distance moved by the load ($V.R. = \frac{d_e}{d_l}$). However, it does not explain why V.R. must be greater than 1 when $d_e > d_l$. Therefore, both statements are true, but the reason does not fully explain the assertion.

Key Concept: Mechanical Advantage, Velocity Ratio

d) 92%
[Solution Description]
To find the efficiency ($\eta$) of the pulley system:
1. Ideal Mechanical Advantage ($M.A._{ideal}$) for a single movable pulley is $2$.
2. Actual Mechanical Advantage ($M.A._{actual}$) = $\frac{\text{Load}}{\text{Effort}} = \frac{240}{130} \approx 1.846$.
3. Velocity Ratio ($V.R.$) = $2$ (for a single movable pulley).
4. Efficiency ($\eta$) = $\left(\frac{M.A._{actual}}{V.R.}\right) \times 100 = \left(\frac{1.846}{2}\right) \times 100 \approx 92.3\%$.
The closest option is $92\%$.

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

c) 1200 J
[Solution Description]
First, find the effort using Mechanical Advantage ($M.A.$):
$M.A. = \frac{L}{E} \Rightarrow 5 = \frac{500}{E} \Rightarrow E = \frac{500}{5} = 100 N$.
Next, calculate $W_{input}$ using $W_{input} = E \times d_e$.
To find $d_e$, use Velocity Ratio ($V.R.$): $V.R. = \frac{d_e}{d_l} \Rightarrow 6 = \frac{d_e}{2} \Rightarrow d_e = 12 m$.
Now, $W_{input} = 100 \times 12 = 1200 J$.

Key Concept: Efficiency

b) 2000 J
[Solution Description]
Efficiency ($\eta$) is given by the ratio of work output to work input: $\eta = \frac{W_{output}}{W_{input}}$. Rearranging for work input ($W_{input}$):
$W_{input} = \frac{W_{output}}{\eta}$
Substituting the given values, $W_{output} = 1200 \text{ J}$ and $\eta = 0.6$:
$W_{input} = \frac{1200 \text{ J}}{0.6} = 2000 \text{ J}$

Key Concept: Velocity Ratio for Block and Tackle System

b) 6
[Solution Description]
For a block and tackle system with $n$ movable pulleys, the velocity ratio (V.R.) is given by $V.R. = 2n$. Here, $n = 3$, so $V.R. = 2 \times 3 = 6$.

Key Concept: Efficiency Formula

d) $\eta = \frac{\text{M.A.}}{\text{V.R.}}$
[Solution Description]
The efficiency of a machine is given by the ratio of Mechanical Advantage (M.A.) to Velocity Ratio (V.R.). Therefore, the formula is: $\eta = \frac{\text{M.A.}}{\text{V.R.}}$.

Key Concept: Single Movable Pulley - Mechanical Advantage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In an ideal single movable pulley, the mechanical advantage ($M.A.$) is given by $M.A. = \frac{L}{E}$. Here, the load $L$ is supported by two segments of the string, each carrying a tension $T$. Therefore, $L = 2T$. The effort $E$ applied at the free end equals the tension $T$, so $E = T$. Substituting these values, we get $M.A. = \frac{2T}{T} = 2$. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Mechanical Advantage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a machine with a mechanical advantage of 5 requires less effort to lift a load compared to a machine with a mechanical advantage of 1. This is true because mechanical advantage ($\text{M.A.} = \frac{\text{Load}}{\text{Effort}}$). A higher M.A. means the same load can be lifted with less effort. The reason explains that mechanical advantage is the ratio of load to effort and a higher value indicates a greater reduction in effort needed. This is also correct and directly explains why the assertion is true.

Key Concept: Efficiency

c) 80%
[Solution Description]
The efficiency ($\eta$) of the machine can be calculated using:
$\eta = \frac{M.A.}{V.R.} \times 100\%$
Given $M.A. = 4$ and $V.R. = 5$:
$\eta = \frac{4}{5} \times 100\% = 80\%$
Therefore, the efficiency is 80%.

Key Concept: Applications

d) cutting metal sheets
[Solution Description]
Screw jacks are commonly used to lift heavy loads such as vehicles, adjust machinery height, or support structures during construction. They are not used for cutting materials, which is the work of tools like saws or drills.

Key Concept: Practical Uses of Pulleys

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a single movable pulley, the load is indeed supported by two segments of the string, leading to a mechanical advantage (M.A.) of 2. The velocity ratio (V.R.) is also 2, as the effort moves twice the distance compared to the load. In the ideal case (without friction and pulley weight), M.A. equals V.R. Hence, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Work Output

b) 600 J
[Solution Description]
Work output is calculated using the formula:
$W_{\text{output}} = L \times d_l$
Given Load ($L$) = 200 N and displacement of load ($d_l$) = 3 m, substituting the values:
$W_{\text{output}} = 200 \times 3 = 600 \, \text{J}$

Key Concept: Single Movable Pulley

b) 2
[Solution Description]
For a single movable pulley, the load $L$ is shared equally between the two segments of the string. Thus, $L = 2T$, where $T$ is the tension in each segment. The effort $E$ equals the tension $T$. Therefore, the mechanical advantage is:
$M.A. = \frac{L}{E} = \frac{2T}{T} = 2$

Key Concept: Block and Tackle System, Multiple Concepts

c) 200 N
[Solution Description]
For 3 movable pulleys:
Theoretical M.A. = $2^n = 2^3 = 8$
V.R. = Number of strands = 6 (for typical arrangement)
$\eta = \frac{M.A.}{V.R.} \times 100\%$
70% = $\frac{M.A.}{6} \times 100\%$ → M.A. = 4.2
$M.A. = \frac{L}{E}$
4.2 = $\frac{840}{E}$
$E = \frac{840}{4.2} = 200$ N

Key Concept: Velocity Ratio

c) 6
[Solution Description]
The velocity ratio ($V.R.$) in an ideal block and tackle system is equal to the number of pulleys ($n$). Here, $n = 6$.
$V.R. = n = 6$
Thus, the velocity ratio is 6.

Key Concept: Class II Levers

b) Class II lever
[Solution Description]
A nutcracker is a type of Class II lever. In a Class II lever, the fulcrum is at one end, the effort is applied at the other end, and the load is placed between them. The mechanical advantage is always greater than 1 because the effort arm is longer than the load arm. Examples include nutcrackers, bottle openers, and wheelbarrows.

Key Concept: Work Output

c) 600 J
[Solution Description]
The work output ($W_{\text{output}}$) is given by:
$W_{\text{output}} = \eta \times W_{\text{input}}$
Given $\eta = 60\% = 0.6$ and $W_{\text{input}} = 1000$ J:
$W_{\text{output}} = 0.6 \times 1000 = 600 \text{ J}$
Hence, the work output is 600 J.

Key Concept: Mechanical Advantage in Class III Levers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For Class III levers, the effort arm ($E$) is always shorter than the load arm ($L$), so mechanical advantage ($M.A.$) is given by $M.A. = \frac{E}{L} < 1$. For an ideal lever, velocity ratio ($V.R.$) equals $M.A.$, so $V.R. < 1$. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

d) Assertion is false, but Reason is true.
[Solution Description]
In an ideal block and tackle system, the mechanical advantage ($M.A.$) is given by the number of strands supporting the load. Here, with 4 movable pulleys and 1 fixed pulley, there are 5 strands supporting the load. However, only the movable pulleys contribute to the mechanical advantage, so $M.A. = 4$. The velocity ratio ($V.R.$) is also the number of strands supporting the load, which is 5 in this case. Therefore, the Assertion is false because $M.A. \neq V.R.$ (4 $\neq$ 5). The Reason is true because it correctly states that for an ideal system, both $M.A.$ and $V.R.$ are numerically equal to the number of strands supporting the load, but it does not consider the distinction between movable and fixed pulleys in the context of $M.A.$. Hence, while the Reason is true, it does not correctly explain the Assertion, and the Assertion itself is false.

Key Concept: Class II Levers - Efficiency

c) 4.0
[Solution Description]
The actual mechanical advantage ($M.A._{\text{actual}}$) is calculated using the efficiency ($\eta$) and the ideal mechanical advantage ($M.A._{\text{ideal}}$).
Calculation:
$M.A._{\text{actual}} = \eta \times M.A._{\text{ideal}} = 0.8 \times 5 = 4$
Therefore, the actual mechanical advantage is 4.

Key Concept: Inclined Plane: Mechanical Advantage and Efficiency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The mechanical advantage (M.A.) of an inclined plane is given by:
$\text{M.A.} = \frac{L}{E} = \frac{l}{h}$.
The velocity ratio (V.R.) is also $\text{V.R.} = \frac{l}{h}$.
For a frictionless inclined plane, the effort $E$ required is $L \cdot \sin \theta$, and since $\sin \theta = \frac{h}{l}$, substituting gives:
$\text{M.A.} = \frac{L}{L \cdot \sin \theta} = \frac{1}{\sin \theta} = \frac{l}{h} = \text{V.R.}$.
The efficiency ($\eta$) is defined as:
$\eta = \frac{\text{M.A.}}{\text{V.R.}} \times 100\%$.
Since M.A. equals V.R., efficiency becomes $100\%$.
Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Mechanical Advantage of Class I Lever

c) 3
[Solution Description]
The mechanical advantage ($M.A.$) of a lever is given by:
$M.A. = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{30\,\text{cm}}{10\,\text{cm}} = 3$
Hence, the mechanical advantage is 3.

Key Concept: Mechanical Advantage

c) Less than 1
[Solution Description]
The mechanical advantage (M.A.) of a Class III lever is always less than 1 because the effort arm is shorter than the load arm, resulting in M.A. < 1.

Key Concept: Velocity Ratio, Efficiency

b) 72 kgf
[Solution Description]
First, calculate the Mechanical Advantage (M.A.):
$M.A. = \text{Efficiency} \times V.R. = 0.6 \times 8 = 4.8$.
Now, find the load (L) that can be lifted:
$L = M.A. \times E = 4.8 \times 15 = 72 \, \text{kgf}$.

Key Concept: Mechanical Advantage

c) 5
[Solution Description]
Given:
Radius of wheel ($R$) = 25 cm
Radius of axle ($r$) = 5 cm
Mechanical Advantage ($M.A.$) is given by:
$M.A. = \frac{R}{r} = \frac{25}{5} = 5$

Key Concept: Class III Levers

c) They are used to gain speed rather than force.
[Solution Description]
Class III levers have the effort applied between the fulcrum and the load. The effort arm is always shorter than the load arm, resulting in a mechanical advantage less than 1, which means they act as speed multipliers. Examples include tongs, fishing rods, and human forearms.

Key Concept: Single Movable Pulley, Mechanical Advantage

a) 180 N
[Solution Description]
For single movable pulley:
Theoretical M.A. = 2
$\eta = \frac{M.A.}{V.R.} \times 100\%$
V.R. = 2 (for single movable pulley)
60% = $\frac{M.A.}{2} \times 100\%$ → M.A. = 1.2
$M.A. = \frac{L}{E}$
1.2 = $\frac{L}{150}$
$L = 1.2 \times 150 = 180$ N

Key Concept: Velocity Ratio

d) 5
[Solution Description]
Velocity ratio (V.R.) is calculated using the formula:
$V.R. = \frac{d_e}{d_l}$
Here, $d_e = 10\,\text{m}$ and $d_l = 2\,\text{m}$.
Substituting the values:
$V.R. = \frac{10}{2} = 5$.
Therefore, the velocity ratio of the machine is 5.

Key Concept: Mechanical Advantage, Velocity Ratio

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let us analyze the Assertion and Reason step by step:
1. Mechanical Advantage (M.A.) formula for wheel and axle is $M.A. = \frac{R}{r}$, where $R$ is the radius of the wheel and $r$ is the radius of the axle. If $R$ is doubled, the new $M.A. = \frac{2R}{r} = 2 \times \frac{R}{r}$. Thus, the M.A. doubles. So, the Assertion is true.
2. The velocity ratio (V.R.) is given by $V.R. = \frac{R}{r}$. This shows that V.R. is indeed directly proportional to the ratio $\frac{R}{r}$. Hence, the Reason is also true.
3. Although both statements are true, the Reason does not explain why doubling the wheel radius doubles the M.A. It only states the relation between V.R. and radii. Therefore, the Reason is NOT the correct explanation of the Assertion.

Key Concept: Class II Levers, Mechanical Advantage

b) 2.4
[Solution Description]
First, calculate the ideal mechanical advantage (MA) using the formula:
$MA_{ideal} = \frac{\text{Effort arm}}{\text{Load arm}} = \frac{15}{5} = 3$
Then, use the efficiency to find the actual mechanical advantage:
$MA_{actual} = \text{Efficiency} \times MA_{ideal} = 0.8 \times 3 = 2.4$

Key Concept: Mechanical Advantage of a Single Fixed Pulley

c) Equal to 1
[Solution Description]
For an ideal single fixed pulley, since $E = L$ (effort equals load when no friction or string mass is considered), the mechanical advantage $M.A.$ is given by:
$M.A. = \frac{L}{E} = 1$
Therefore, the mechanical advantage is 1.

Key Concept: Block and Tackle System

c) 4
[Solution Description]
To find the mechanical advantage ($M.A.$), we use the relationship between efficiency ($\eta$), mechanical advantage ($M.A.$), and velocity ratio ($V.R.$):
$\eta = \frac{M.A.}{V.R.} \times 100\%$
Given $\eta = 80\%$ and $V.R. = 5$, we can solve for $M.A.$:
$80 = \frac{M.A.}{5} \times 100$
Simplifying this equation gives:
$M.A. = \frac{80 \times 5}{100} = 4$
Therefore, the mechanical advantage is 4.

Key Concept: Class II Levers, Velocity Ratio

c) 3.0
[Solution Description]
The velocity ratio (VR) is given by:
$VR = \frac{\text{Effort arm}}{\text{Load arm}} = \frac{1.5}{0.5} = 3$
Since VR is independent of efficiency, it remains 3.

Key Concept: Mixed Concept: M.A., Equilibrium Condition

c) 400 N, M.A. = 2
[Solution Description]
For a Class I lever:
Effort Arm (FA) = 4 m, Load Arm (FB) = 2 m, Effort (E) = 200 N.
Using equilibrium condition: $L \times FB = E \times FA$.
Substituting values: $L \times 2 = 200 \times 4$.
Solving for L: $L = \frac{200 \times 4}{2} = 400$ N.
M.A. = $\frac{FA}{FB} = \frac{4}{2} = 2$ (>1), making it a force multiplier.

Key Concept: Practical Applications of Simple Machines

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion (A) is true because a single fixed pulley changes the direction of the applied effort, making it convenient to pull the bucket downward instead of lifting it upward. However, the reason (R) is false because the mechanical advantage of a single fixed pulley is 1 (it does not multiply the force). Thus, while the assertion is correct, the reason provided is incorrect and does not explain the assertion.

Key Concept: Mechanical Advantage, Efficiency, Weight of Pulleys

c) 83.33%
[Solution Description]
First calculate ideal M.A.: For 4 pulleys, $M.A._{ideal} = n = 4$.
Actual M.A. considering weight: $M.A. = \frac{L + w}{E} = \frac{180 + 20}{60} = \frac{200}{60} = 3.33$
V.R. remains unchanged at $n = 4$
Efficiency $\eta = \frac{M.A.}{V.R.} \times 100\% = \frac{3.33}{4} \times 100\% = 83.25\%$

Key Concept: Single Fixed Pulley Mechanical Advantage

b) 10 N
[Solution Description]
For a single fixed pulley, the mechanical advantage ($M.A.$) is given by:
$M.A. = \frac{L}{E} = 1$
Here, Load ($L$) = 10 N.
Therefore, Effort ($E$) = $\frac{L}{M.A.} = \frac{10}{1} = 10 \, \text{N}$.

Key Concept: Mechanical Advantage

c) 50
[Solution Description]
The mechanical advantage (M.A.) of a screw jack is given by the ratio of the load to the effort.
Given:
Load ($L$) = 1500 N
Effort ($E$) = 30 N
Using the formula for M.A.:
$M.A. = \frac{L}{E}$
Substituting the values:
$M.A. = \frac{1500}{30} = 50$
Therefore, the mechanical advantage is 50.

Key Concept: Velocity Ratio

c) $200\pi$
[Solution Description]
The velocity ratio (V.R.) of a screw jack is calculated as:
$V.R. = \frac{2\pi l}{p}$
Given:
Length of the handle ($l$) = 40 cm
Pitch of the screw ($p$) = 0.4 cm
Substituting the values:
$V.R. = \frac{2 \times \pi \times 40}{0.4}$
$V.R. = \frac{80\pi}{0.4} = 200\pi$
Therefore, the velocity ratio is $200\pi$.

Key Concept: Pulley System, Efficiency

c) 80%
[Solution Description]
In a block and tackle with 4 pulleys, $V.R. = 4$.
Calculate $M.A.$: $M.A. = \frac{L}{E} = \frac{800}{250} = 3.2$.
Efficiency $\eta = \frac{M.A.}{V.R.} = \frac{3.2}{4} = 0.8$ or 80%.

Key Concept: Combination of Pulleys, Multi-Concept

c) 4
[Solution Description]
For a block and tackle system with $n$ movable pulleys, the velocity ratio ($V.R.$) is given by $2n$.
1. Here, $n = 2$ (two movable pulleys), so $V.R. = 2 \times 2 = 4$.
2. The given data about the effort and load is not needed to compute $V.R.$, as it depends only on the pulley configuration.
Thus, the correct answer is $4$.

Key Concept: Work Input, Work Output, Efficiency

a) 500 J
[Solution Description]
The efficiency formula is $\eta = \frac{W_{output}}{W_{input}}$.
Given $\eta = 80\% = 0.8$ and $W_{output} = 400$ J, rearrange to find $W_{input}$:
$W_{input} = \frac{W_{output}}{\eta} = \frac{400\, \text{J}}{0.8} = 500$ J.

Key Concept: Effect of weight on M.A. and efficiency

c) Decreases both M.A. and efficiency
[Solution Description]
The weight of the lower block reduces the mechanical advantage and efficiency because part of the effort is used to lift the weight of the block itself rather than the load. The efficiency formula is given by $\eta = 1 - \frac{w}{nE}$ where $w$ is the weight of the lower block, $n$ is the number of pulleys, and $E$ is the effort applied.

Key Concept: Energy Losses, Power Calculation

b) 300 W and 25%
[Solution Description]
Given:
- Useful energy output = 6000 J
- Input energy = 8000 J
- Time = 20 s
Energy lost as heat:
$\text{Energy lost} = \text{Input energy} - \text{Useful output} = 8000 - 6000 = 2000 \text{ J}$
Percentage energy lost:
$\% \text{Energy lost} = \left(\frac{2000}{8000}\right) \times 100\% = 25\%$
Power output:
$P = \frac{\text{Useful energy output}}{\text{Time}} = \frac{6000}{20} = 300 \text{ W}$

Key Concept: Class II Levers - Mechanical Advantage

d) 4.0
[Solution Description]
The mechanical advantage ($M.A.$) for a class II lever is given by the ratio of the effort arm to the load arm. Here, the effort arm is 12 cm and the load arm is 3 cm.
Calculation:
$M.A. = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{12}{3} = 4$
Therefore, the mechanical advantage is 4.

Key Concept: Efficiency and Mechanical Advantage

b) 4
[Solution Description]
The relationship between mechanical advantage (M.A.), velocity ratio (V.R.), and efficiency ($\eta$) is given by:
$\text{M.A.} = \text{V.R.} \times \eta$
Given, V.R. $= 5$ and $\eta = 80% = 0.8$. Substituting these values:
$\text{M.A.} = 5 \times 0.8 = 4$

Key Concept: Force Decomposition, Friction

c) $3.20\,\text{m/s}^2$
[Solution Description]
First, resolve the gravitational force into components parallel and perpendicular to the incline:
$F_{\text{parallel}} = mg\sin\theta = 5 \times 9.8 \times \sin(30^\circ) = 24.5\,\text{N}$
$F_{\text{normal}} = mg\cos\theta = 5 \times 9.8 \times \cos(30^\circ) \approx 42.43\,\text{N}$
Next, calculate the frictional force opposing the motion:
$F_{\text{friction}} = \mu_k F_{\text{normal}} = 0.2 \times 42.43 \approx 8.49\,\text{N}$
The net force acting down the incline is:
$F_{\text{net}} = F_{\text{parallel}} - F_{\text{friction}} = 24.5 - 8.49 \approx 16.01\,\text{N}$
Finally, compute the acceleration using Newton's second law:
$a = \frac{F_{\text{net}}}{m} = \frac{16.01}{5} \approx 3.20\,\text{m/s}^2$

Key Concept: Velocity Ratio of a Single Fixed Pulley

b) 5 m
[Solution Description]
In an ideal single fixed pulley, the velocity ratio ($V.R.$) is:
$V.R. = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} = 1$
Thus, if the effort moves 5 m downward, the load moves the same distance upward:
$d_{\text{load}} = d_{\text{effort}} = 5 \, \text{m}$
The load moves 5 m upward.

Key Concept: Mechanical Advantage

b) 2
[Solution Description]
The mechanical advantage ($M.A.$) of a single movable pulley in an ideal case is given by $M.A. = \frac{L}{E}$. Since the effort $E$ required to lift the load $L$ is half of the load ($E = \frac{L}{2}$), substituting this into the formula gives $M.A. = \frac{L}{\frac{L}{2}} = 2$. Therefore, the mechanical advantage is 2.

Key Concept: Mechanical Advantage and Velocity Ratio

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a machine with mechanical advantage (M.A.) greater than 1 acts as a force multiplier, which is correct because M.A. = $\frac{L}{E}$ > 1 implies $L > E$. The reason is also correct as it explains why M.A. > 1: the effort ($E$) needed is less than the load ($L$). Therefore, the reason is the correct explanation for the assertion.

Key Concept: Velocity Ratio

b) $VR = \frac{d_E}{d_L}$
[Solution Description]
The velocity ratio (VR) is the ratio of the distance moved by the effort (d_E) to the distance moved by the load (d_L), or equivalently, the ratio of the velocity of effort (V_E) to the velocity of load (V_L). The correct formula is:
$VR = \frac{d_E}{d_L} = \frac{V_E}{V_L}$

Key Concept: Mechanical Advantage, Efficiency of Screw Jack

c) Assertion is true, but Reason is false.
[Solution Description]
The mechanical advantage (MA) of a screw jack is given by:
$MA = \frac{W}{P} = \frac{2\pi R}{p \tan(\alpha + \phi)}$
where $W$ is the load, $P$ is the effort, $R$ is the radius of the handle, $p$ is the pitch, $\alpha$ is the helix angle, and $\phi$ is the friction angle.
As $\alpha$ increases, $\tan(\alpha + \phi)$ increases, reducing the denominator, thus decreasing MA. The assertion is true.
The reason states that a larger helix angle reduces frictional resistance, which is incorrect because friction depends on the coefficient of friction and normal force, not just the helix angle. Hence, the reason is false.

Key Concept: Forces on an Inclined Plane, Friction, Newton’s Laws

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
An object moving with **constant velocity** has **zero acceleration**, which by **Newton’s second law ($F_{\text{net}} = ma$)** implies that the **net force** is zero. On an inclined plane, the gravitational force $mg$ can be resolved into two components:
- Parallel to the plane: $mg\sin\theta$ (accelerates object downward)
- Perpendicular to the plane: $mg\cos\theta$ (balanced by normal force).
If the object moves at **constant velocity**, the **frictional force ($f$)** must cancel the parallel component of gravity ($mg\sin\theta$) to make net force zero. Thus, the **Assertion** is correct.
The **Reason** correctly states that **frictional force balances the gravitational component** ($f = mg\sin\theta$) to allow constant velocity motion. Hence, **Reason explains Assertion**.
Both statements are **true**, and **Reason correctly explains Assertion**.

Key Concept: Class II Levers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a class II lever, the fulcrum and effort are at the two ends, while the load lies between them. The mechanical advantage ($M.A.$) is calculated as $M.A. = \frac{\text{Effort Arm}}{\text{Load Arm}}$. Since the effort arm is always longer than the load arm in a class II lever, the $M.A.$ is always greater than 1. Thus, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Velocity Ratio

d) 8 m
[Solution Description]
The velocity ratio ($V.R.$) for a block and tackle system is equal to the number of pulleys ($n$) because:
$V.R. = \frac{d_E}{d_L} = n$
Given $n = 4$ and $d_L = 2\,\text{m}$, the effort distance $d_E$ is:
$d_E = n \times d_L = 4 \times 2 = 8\,\text{m}$
The effort end moves 8 meters.

Key Concept: Single Fixed Pulley

b) 1
[Solution Description]
For a single fixed pulley, in equilibrium, neglecting the mass of the string and friction at the axle, we have:
$L = T$ and $E = T$, thus $E = L$.
The mechanical advantage ($M.A.$) is given by:
$M.A. = \frac{L}{E} = 1$.
Therefore, the mechanical advantage is 1.

Key Concept: Simple Machines - Levers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that for a Class I lever with equal effort and load arm lengths, $MA = 1$. This is correct because $MA$ is given by the ratio $\frac{\text{Effort arm}}{\text{Load arm}}$, which becomes 1 when both lengths are equal.
The Reason explains that in such levers, the moments of effort and load balance each other, leading to equilibrium. This is also true and correctly explains why $MA = 1$ in this case. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Effort Calculation

c) 200,N
[Solution Description]
In an ideal block and tackle system, the effort ($E$) is calculated as:
$E = \frac{L}{n}$
Given $L = 600\,N$ and $n = 3$:
$E = \frac{600}{3} = 200\,N$
So, the effort required is $200\,N$.

Key Concept: Principle of a Lever and M.A.

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the principle of a lever, the mechanical advantage (M.A.) is given by the ratio of the effort arm to the load arm ($M.A. = \frac{Effort\ arm}{Load\ arm}$). For Class II levers, the effort arm is always longer than the load arm, which means $M.A. > 1$. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Class I Levers, Mechanical Advantage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a Class I lever with an effort arm longer than the load arm will always have a mechanical advantage greater than 1. This is true because the mechanical advantage (M.A.) of a Class I lever is calculated as $M.A. = \frac{\text{Effort arm}}{\text{Load arm}}$. If the effort arm is longer than the load arm, the ratio will be greater than 1.
The reason states that the mechanical advantage of a Class I lever is given by the ratio of the effort arm to the load arm. This is also true and directly explains why the assertion is correct. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Load Arm Calculation

d) 30 cm
[Solution Description]
Using the formula M.A. = Effort Arm / Load Arm, and given M.A. = 2 and Effort Arm = 60 cm, we can find the Load Arm as Load Arm = Effort Arm / M.A. = 60 cm / 2 = 30 cm.

Key Concept: Mechanical Advantage, Efficiency

d) Assertion is false, but Reason is true.
[Solution Description]
For a block and tackle system, the mechanical advantage (M.A.) is given by $M.A. = \frac{L}{E}$. When there are $n$ pulleys in the movable block, the number of strands supporting the load is $n+1$ (including the strand pulled by effort). Thus, theoretically, $M.A. = n + 1$. Here, with 3 pulleys in the movable block, $M.A. = 4$, not 6. Hence, Assertion (A) is false. However, Reason (R) is true as the velocity ratio (V.R.) is indeed equal to the number of strands supporting the movable block.
Therefore, Assertion is false, but Reason is true.

Key Concept: Mechanical Advantage

b) $MA = \frac{L}{E}$
[Solution Description]
The mechanical advantage (MA) of a machine is defined as the ratio of the load (L) to the effort (E). Thus, the formula is given by:
$MA = \frac{L}{E}$
This represents how much the machine multiplies the input force.

Key Concept: Work input and output

c) 200 J
[Solution Description]
The work output is calculated using the formula $W_{output} = L \times d_l$, where $L$ is the load and $d_l$ is the displacement of the load. Given $L = 200 \text{ N}$ and $d_l = 1 \text{ m}$, we can compute the work output as follows:
$W_{output} = 200 \text{ N} \times 1 \text{ m} = 200 \text{ J}$

Key Concept: Velocity Ratio Calculation

c) 2
[Solution Description]
The velocity ratio (V.R.) is calculated as the ratio of the displacement of effort to the displacement of load. Given $d_e = 8$ cm and $d_l = 4$ cm, $V.R. = \frac{d_e}{d_l} = \frac{8}{4} = 2$.

Key Concept: Effect of Weight on Efficiency

c) Efficiency increases
[Solution Description]
The efficiency ($\eta$) of a machine is given by:
$\eta = \frac{M.A.}{V.R.}$
Here, M.A. = 4 and V.R. = 6, so:
$\eta = \frac{4}{6} = \frac{2}{3} \approx 66.67\%$
When the weight of the lower block decreases, the effective effort (E) required to lift a load (L) reduces, increasing the M.A. Since V.R. remains unchanged, efficiency ($\eta$) increases as per the equation $\eta = \frac{M.A.}{V.R.}$

Key Concept: Self-locking Condition, Friction Angle

d) α ≤ ϕ
[Solution Description]
The self-locking condition occurs when the friction force is sufficient to prevent the screw from turning back under load. This happens when:
α ≤ ϕ
The helix angle must be less than or equal to the angle of friction for the screw jack to be self-locking.

Key Concept: Mechanical Advantage, Velocity Ratio, Efficiency

c) 80%
[Solution Description]
To find the efficiency ($\eta$), we use the relationship $\eta = \frac{M.A.}{V.R.}$. Given M.A. = 4 and V.R. = 5, substituting the values gives $\eta = \frac{4}{5} = 0.8$, or 80%.

Key Concept: Class I Levers, Mechanical Advantage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a Class I lever, the mechanical advantage (M.A.) is given by M.A. = Effort Arm / Load Arm. If the effort arm is longer than the load arm, M.A. > 1. The velocity ratio (V.R.) is also equal to Effort Arm / Load Arm in an ideal lever. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion because the M.A. being greater than 1 directly results from the ratio of the effort arm to the load arm.

Key Concept: Functions of Machines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a single fixed pulley changes the direction of effort to a more convenient downward direction when lifting a bucket from a well. This is true because, as mentioned in the syllabus, a single fixed pulley allows the effort to be applied in the downward direction instead of upward, making it easier for the person to use their own weight as effort.
The Reason states that changing the direction of effort is one of the functions of machines. According to the syllabus, changing the direction of effort to a convenient direction is indeed one of the four primary functions of machines.
Therefore, both the Assertion and Reason are true, and the Reason correctly explains why the Assertion is valid in this context.

Key Concept: Levers

b) Wheelbarrow
[Solution Description]
A Class II lever has the load between the fulcrum and the effort. Among the options, a wheelbarrow fits this description since the load is in the middle and the effort is applied at the handles.

Key Concept: Velocity Ratio and Displacement

c) 3
[Solution Description]
The velocity ratio (V.R.) is given by the ratio of the displacement of effort ($d_e$) to the displacement of load ($d_l$). Given $d_e = 15$ cm and $d_l = 5$ cm, we substitute these values into the formula:
$V.R. = \frac{d_e}{d_l} = \frac{15}{5} = 3$
Therefore, the velocity ratio is 3.

Key Concept: Velocity Ratio

b) 2 meters
[Solution Description]
For a single movable pulley, the velocity ratio ($V.R.$) is 2, which means:
$V.R. = \frac{\text{distance moved by effort } (d_1)}{\text{distance moved by load } (d_2)} = 2$
Given that the effort moves $d_1 = 4$ meters, we can find the distance moved by the load as:
$d_2 = \frac{d_1}{V.R.} = \frac{4}{2} = 2 \text{ meters}$
Hence, the load is raised by 2 meters.

Key Concept: Velocity Ratio

b) 3
[Solution Description]
The relationship between mechanical advantage ($M.A.$), velocity ratio ($V.R.$), and efficiency ($\eta$) is given by $\eta = \frac{M.A.}{V.R.}$. Rearranged, this gives $M.A. = V.R. \times \eta$.
Given:
$V.R. = 4$ and $\eta = 75\% = 0.75$.
Step 1: Calculate $M.A.$ using the formula:
$M.A. = 4 \times 0.75 = 3$
The mechanical advantage of the machine is 3.

Key Concept: Tension, Single Movable Pulley

b) 60 N
[Solution Description]
For a single movable pulley, the tension $T$ in the string supports both the load $L$ and the pulley's weight (negligible here).
Thus, the effort $E$ required is:
$E = \frac{L}{2} = \frac{120}{2} = 60 \text{ N}$
Since the tension in the string equals the applied effort:
$T = E = 60 \text{ N}$