Class 10 Physics Chapter 3 Machines

Key Concept: Class III Lever Applications

b) Knife
[Solution Description]
A Class III lever has the effort between the fulcrum and the load, resulting in gain in speed but no gain in force. Examples include knives, sugar tongs, and fishing rods. Among the options, the knife is explicitly mentioned as a Class III lever that provides gain in speed.

Key Concept: Direction of Effort, Practical Application

c) 1.83
[Solution Description]
To calculate the actual mechanical advantage ($M.A.$):
1. The arrangement uses a single movable pulley, so ideal $M.A. = 2$.
2. However, due to practical losses, the actual $M.A.$ is calculated as $\frac{\text{Load}}{\text{Effort}} = \frac{110}{60} = 1.833$.
The closest option is $1.83$.

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: 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, Class II Levers

c) 4.0
[Solution Description]
Since it's class II, $V.R. = \frac{\text{effort displacement}}{\text{load displacement}} = \frac{40}{10} = 4$. In ideal conditions, $M.A. = V.R. = 4$.

Key Concept: Class III Lever

b) Between the fulcrum and the load
[Solution Description]
In a Class III lever, the fulcrum and load are at the two ends of the lever, and the effort is applied somewhere between the fulcrum and the load.

Key Concept: Single Fixed Pulley

c) To change the direction of the effort
[Solution Description]
A single fixed pulley changes the direction of the applied effort but does not provide any mechanical advantage. It is commonly used to lift small loads (e.g., lifting a water bucket). The tension in the rope remains constant, and in equilibrium, the effort equals the load ($E = L$).

Key Concept: Practical Applications of Simple Machines

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that a single fixed pulley changes the direction of the applied effort, which is correct as it allows the user to pull downward to lift the bucket upward. However, the reason claims that the mechanical advantage is greater than 1, which is false for a single fixed pulley. The mechanical advantage of a single fixed pulley is 1 because the load and effort are equal in magnitude (assuming no friction). Therefore, while the assertion is true, the reason is false.

Key Concept: Mechanical Advantage, Efficiency

c) 800 N
[Solution Description]
First, calculate the Velocity Ratio (V.R.), which equals the number of pulleys:
$V.R. = 6$
Given efficiency ($\eta$) = 75% or 0.75.
Mechanical Advantage (M.A.) is calculated as:
$M.A. = \eta \times V.R. = 0.75 \times 6 = 4.5$
The effort ($E$) needed to lift a load ($L$) of 3600 N is:
$E = \frac{L}{M.A.} = \frac{3600}{4.5} = 800\,N$

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: 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: Class III levers, Mechanical Advantage, Velocity Ratio

c) 0.25
[Solution Description]
For a Class III lever, the velocity ratio (V.R.) is given by:
$V.R. = \frac{\text{Load Arm}}{\text{Effort Arm}}$
Given:
Effort Arm = 0.5 m
Load Arm = 2 m
Substituting the values:
$V.R. = \frac{2}{0.5} = 4$
However, for a Class III lever, the effort arm is always shorter than the load arm, but the calculation above seems to give V.R. > 1, which contradicts the concept that V.R. < 1 for Class III levers. Re-evaluating the positions: In Class III levers, the fulcrum is at one end, the load is at the other end, and the effort is in between. Therefore:
$V.R. = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}}$
But since effort arm < load arm, V.R. should be less than 1. The initial question setup might have swapped arms. Correcting this:
If Effort Arm (distance from fulcrum to effort) = 0.5 m
Load Arm (distance from fulcrum to load) = 2 m
Then:
$V.R. = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{0.5}{2} = 0.25$
Thus, the correct V.R. is 0.25.

Key Concept: Class III levers, Gain in Speed

c) 0.25
[Solution Description]
In Class III levers, the gain in speed is given by the ratio of the distance moved by the load to the distance moved by the effort. Here:
$\text{Gain in Speed} = \frac{\text{Distance moved by load}}{\text{Distance moved by effort}} = \frac{5}{20} = \frac{1}{4} = 0.25$
Since the distance moved by the load is less than the effort, there's a reduction in speed, not a gain. The correct interpretation is that the speed of the load is a fraction (0.25) of the speed of the effort.

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, 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: Mechanical Advantage, Velocity Ratio, Efficiency

c) 3
[Solution Description]
First, find the velocity ratio (V.R.) using the distances moved by effort and load:
$V.R. = \frac{d_E}{d_L} = \frac{5\, \text{m}}{1\, \text{m}} = 5$.
Given efficiency ($\eta$) is 60% or 0.6, use the relationship $\eta = \frac{M.A.}{V.R.}$ to find M.A.:
$M.A. = \eta \times V.R. = 0.6 \times 5 = 3$.

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: Mechanical Advantage in Block and Tackle System

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a block and tackle system with $n$ pulleys, the mechanical advantage ($M.A.$) is equal to the number of supporting strands ($n$) when friction is negligible. Therefore, for 4 pulleys, $M.A. = 4$, making Assertion (A) true.
The velocity ratio ($V.R.$) is also equal to the number of supporting strands ($n$) because the effort moves a distance $n$ times the distance moved by the load. Thus, Reason (R) correctly explains why $M.A. = n$ in such systems, as it directly relates to the underlying principle of pulley mechanics.
Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Mechanical Advantage Calculation

d) 4
[Solution Description]
Given:
Load (L) = 200 N
Effort (E) = 50 N
The mechanical advantage (M.A.) is calculated as:
$M.A. = \frac{L}{E} = \frac{200}{50} = 4$
Thus, the correct answer is option d).

Key Concept: Pulley Systems

b) $MA = 2$
[Solution Description]
The mechanical advantage of a single movable pulley is 2 because it divides the load equally between two segments of the rope, effectively doubling the effort applied.

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: Class II Levers, Force Multiplier

b) 3 cm
[Solution Description]
The mechanical advantage (MA) is:
$MA = \frac{\text{Load}}{\text{Effort}} = \frac{40}{10} = 4$
Using the formula for MA in levers:
$MA = \frac{\text{Effort arm}}{\text{Load arm}}$
Solving for the load arm:
$\text{Load arm} = \frac{\text{Effort arm}}{MA} = \frac{12}{4} = 3 \text{ cm}$

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: 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: 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: Combined Concepts - MA, VR, Efficiency

c) 84.9 cm
[Solution Description]
First find M.A.:
$M.A. = \frac{L}{E} = \frac{8000}{25} = 320$
Given efficiency is 60%, so:
$V.R. = \frac{M.A.}{\eta} = \frac{320}{0.6} = 533.33$
Now calculate lever length (l):
$V.R. = \frac{2\pi l}{p}$
$533.33 = \frac{2 \times 3.14 \times l}{0.01}$
$l = \frac{533.33 \times 0.01}{2 \times 3.14} = 0.849 \, \text{m} = 84.9 \, \text{cm}$

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: Mechanical Advantage

b) 62.5%
[Solution Description]
The relationship between Mechanical Advantage (M.A.), Velocity Ratio (V.R.), and Efficiency ($\eta$) is given by:
$\eta = \frac{\text{M.A.}}{\text{V.R.}}$
Given M.A. = 5 and V.R. = 8, substituting the values:
$\eta = \frac{5}{8} = 0.625$
To express as a percentage, multiply by 100:
$0.625 \times 100 = 62.5\%$

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: 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: 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: Mechanical Advantage and Efficiency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Efficiency ($\eta$) is defined as the ratio of Mechanical Advantage (M.A.) to Velocity Ratio (V.R.) for a machine: $\eta = \frac{\text{M.A.}}{\text{V.R.}}$. In an ideal machine where there are no energy losses, $\text{M.A.} = \text{V.R.}$ and $\eta = 1$ (or 100%). However, in practical machines, energy is lost due to factors like friction and weight of moving parts, which results in $\text{M.A.} < \text{V.R.}$. Therefore, $\eta < 1$ (or less than 100%).
Both Assertion and Reason are correct, and the Reason logically explains why the Assertion is true.

Key Concept: Equilibrium Condition for a Lever

b) Clockwise moment of load = Anticlockwise moment of effort
[Solution Description]
For an ideal lever, equilibrium is achieved when the clockwise moment of the load equals the anticlockwise moment of the effort. Mathematically, $L \times \text{Load Arm} = E \times \text{Effort Arm}$.

Key Concept: Pulleys, Work Input/Output

c) 1066.67 J
[Solution Description]
Ideal M.A. = $2^n = 2^3 = 8$.
Output work = $L × d_l = 400 × 2 = 800$ J.
Efficiency = $\frac{Output}{Input}$ ⇒ Input = $\frac{800}{0.75} ≈ 1066.67$ J.

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

c) Bicycle pedal system with M.A. = 0.7
[Solution Description]
A machine with mechanical advantage less than 1 acts as a speed multiplier. Among the options, only a bicycle pedal system has M.A. < 1 because it increases speed while reducing force.

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 III Lever, Human Body

b) 13.3 N
[Solution Description]
For Class III levers, M.A. = $\frac{\text{Effort arm}}{\text{Load arm}} = \frac{4}{30} \approx 0.133$.
Since M.A. = $\frac{\text{Load}}{\text{Effort}}$, Load = $100 \times 0.133 \approx 13.3$ N.

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: 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 a machine with $M.A. = 1$ can still be useful despite not multiplying force or speed. This is true because such machines are often used to change the direction of effort, as explained in the syllabus. The reason correctly explains why the assertion is valid, as changing the direction of effort can indeed be advantageous in practical scenarios, such as using a single fixed pulley to lift a load vertically by applying effort downward.

Key Concept: Class III Levers

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 a Class III lever is always less than 1, which is true because for Class III levers, M.A. $= \frac{\text{Effort Arm}}{\text{Load Arm}}$. Since the effort arm is always shorter than the load arm in Class III levers, M.A. will be less than 1. The reason correctly explains this by stating that the effort arm is always smaller than the load arm in such levers.

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: 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: 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: Principle of Moments

b) 10 kgf
[Solution Description]
Given: Radius of wheel, $R = 25$ cm; Radius of axle, $r = 5$ cm; Load, $L = 50$ kgf; Efficiency = 100%.
Since efficiency is 100%, Mechanical Advantage (M.A.) equals Velocity Ratio (V.R.).
First, calculate V.R.:
$V.R. = \frac{R}{r} = \frac{25}{5} = 5$
Thus, $M.A. = 5$.
Now, calculate Effort (E):
$E = \frac{L}{M.A.} = \frac{50}{5} = 10$ kgf
The minimum effort required is 10 kgf.

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: Purpose of using a combination of pulleys

b) To decrease the amount of force needed by increasing mechanical advantage
[Solution Description]
A combination of pulleys is used to increase the mechanical advantage beyond what a single movable pulley can provide, allowing heavy loads to be lifted or moved with less effort.

Key Concept: Normal Force on an Inclined Plane

b) $mg\cos\theta$
[Solution Description]
The normal force ($N$) acts perpendicular to the surface of the inclined plane and is calculated as:
$N = mg\cos\theta$
This is because the normal force counterbalances the perpendicular component of the gravitational force.

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: Mechanical Advantage

b) 1
[Solution Description]
In an ideal single fixed pulley, both the effort ($E$) and load ($L$) are equal ($E = L$). Mechanical Advantage ($M.A.$) is given by: $M.A. = \frac{L}{E}$. Substituting $E = L$, we get $M.A. = 1$.

Key Concept: Inclined Plane

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the acceleration of an object sliding down a frictionless inclined plane depends on the angle of inclination, which is true because the acceleration $a = g \sin\theta$, where $\theta$ is the angle of inclination. The reason states that the component of gravitational force along the inclined plane causes the acceleration, which is also true and correctly explains the assertion since $F = mg \sin\theta$ is the driving force for the motion.

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: 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: Mechanical Advantage and Efficiency

c) 50%
[Solution Description]
The efficiency of a machine is given by:
$\eta = \frac{\text{M.A.}}{\text{V.R.}} \times 100\%$
Substituting the given values:
$\eta = \frac{3}{6} \times 100\% = 50\%$

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 and Velocity Ratio

d) 80%
[Solution Description]
First, calculate the Mechanical Advantage (MA) using the formula:
$MA = \frac{L}{E} = \frac{400}{50} = 8$
Next, use the efficiency formula:
$\eta = \frac{MA}{VR} = \frac{8}{10} = 0.8 \, \text{(or 80\%)}$
Therefore, the efficiency is 80%.

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: Levers - Class I, II, III

c) Tongs
[Solution Description]
In a Class III lever, the effort is between the fulcrum and the load, resulting in the effort arm being shorter than the load arm. This makes the mechanical advantage (MA) always less than 1 because $MA = \frac{\text{Effort arm}}{\text{Load arm}}$.
Examples of Class III levers include tweezers, a fishing rod, or a human forearm where the elbow acts as the fulcrum, the biceps muscle provides the effort, and the load is held in the hand.

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: Mechanical Advantage, Load and Effort

d) 3.125
[Solution Description]
First, find the original effort ($E$) using the M.A. formula:
$\text{M.A.} = \frac{L}{E} \implies 2.5 = \frac{500}{E} \implies E = \frac{500}{2.5} = 200 \, \text{N}$.
Now, reduce the effort by 20% due to friction:
$E_{\text{new}} = 200 - (0.2 \times 200) = 160 \, \text{N}$.
Calculate the new M.A.:
$\text{M.A.}_{\text{new}} = \frac{L}{E_{\text{new}}} = \frac{500}{160} = 3.125$.

Key Concept: Lever Classes, Force Analysis

b) 20 N
[Solution Description]
First calculate theoretical M.A.: $M.A._{theo} = \frac{Effort Arm}{Load Arm} = \frac{15}{3} = 5$
Theoretical Load = $M.A._{theo} \times Effort = 5 \times 5 = 25$ N
With 80% efficiency: $Actual Load = 0.8 \times 25 = 20$ N

Key Concept: Mechanical Advantage, Efficiency

c) 4.8
[Solution Description]
First, calculate the ideal mechanical advantage (M.A.$_{\text{ideal}}$) for 6 pulleys:
$\text{M.A.}_{\text{ideal}} = n = 6$.
Next, use the efficiency formula:
$\eta = \frac{\text{M.A.}_{\text{actual}}}{\text{M.A.}_{\text{ideal}}}$
Given $\eta = 0.8$, rearrange to find M.A.$_{\text{actual}}$:
$\text{M.A.}_{\text{actual}} = \eta \times \text{M.A.}_{\text{ideal}} = 0.8 \times 6 = 4.8$.
The actual mechanical advantage is 4.8.

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) 1
[Solution Description]
A single fixed pulley does not provide any mechanical advantage in terms of force multiplication, but it helps in changing the direction of effort. Therefore, the mechanical advantage ($M.A.$) of a single fixed pulley is 1, as it does not multiply the effort applied.

Key Concept: Trade-off Between Force and Speed

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a Class III lever, the effort arm ($E$) is always shorter than the load arm ($L$). From the formula for mechanical advantage ($M.A.$), $M.A. = \frac{E}{L}$. Since $E < L$, $M.A. < 1$. Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains why the mechanical advantage is less than 1 in a Class III lever.

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: 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: Relationship Between M.A., V.R., and Efficiency in Pulley System

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To analyze the assertion and reason:
1. The formula for M.A. in a pulley system with weight $w$ is:
$M.A. = \frac{L}{E} = n - \frac{w}{E}$
As $E$ increases, $\frac{w}{E}$ decreases, so $M.A.$ increases.
2. Efficiency is given by:
$\eta = \frac{M.A.}{V.R.} = \frac{n - \frac{w}{E}}{n}$
Since $V.R. = n$ remains unchanged, $\eta$ increases as $M.A.$ increases (due to higher $E$).
Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

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: Principle of Wheel and Axle

c) 250 N
[Solution Description]
Given:
Radius of wheel ($R$) = 20 cm
Radius of axle ($r$) = 4 cm
Effort ($E$) = 50 N
Using the principle of moments for wheel and axle:
$E \times R = L \times r$
Substituting the given values:
$50 \times 20 = L \times 4$
Solving for $L$:
$L = \frac{50 \times 20}{4} = 250 \text{ N}$

Key Concept: Mechanical Advantage, Efficiency

b) 4.8
[Solution Description]
The relationship between mechanical advantage (MA), velocity ratio (VR), and efficiency ($\eta$) is given by:
$MA = VR \times \eta$
Given, $VR = 6$, $\eta = 80\% = 0.8$.
$MA = 6 \times 0.8 = 4.8$
Therefore, the mechanical advantage is 4.8.

Key Concept: Mechanical Advantage, Combined Systems

d) 7.6
[Solution Description]
First find theoretical M.A.: Each movable pulley gives M.A. of 2, so 3 movable pulleys would give $2^3 = 8$
Considering weight: Total force to lift = Load + Pulley weight = $350 + 30 = 380N$
Actual M.A. = $\frac{380}{50} = 7.6$
Note: The theoretical maximum without weight would be 8, but practical M.A. is less

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: 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: 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, 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 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: Efficiency of a Single Fixed Pulley

c) Because of friction at the axle or pulley bearings
[Solution Description]
In real scenarios, the efficiency $\eta$ of a single fixed pulley is less than 100% because:
$\eta = \frac{M.A.}{V.R.} \times 100\%$
Since $V.R. = 1$, but $M.A.$ decreases due to friction at the axle and pulley bearings, the efficiency becomes less than 100%.

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: 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: Wheel and Axle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a wheel and axle is a simple machine consisting of two co-axial cylinders of different diameters, which is true. The reason states that this arrangement amplifies the input force to lift heavy loads with less effort, which is also true and correctly explains why the wheel and axle is used as a simple machine.

Key Concept: Effort Calculation

b) 30 N
[Solution Description]
In an ideal single movable pulley, the effort $E$ is related to the load $L$ by the equation $E = \frac{L}{2}$. Substituting $L = 60\,N$, we get $E = \frac{60}{2} = 30\,N$.

Key Concept: Class I Levers and Mechanical Advantage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
- In a seesaw, if the fulcrum is exactly at the midpoint, the effort arm and load arm are equal in length.
- According to the law of levers, $M.A. = \frac{\text{Effort arm}}{\text{Load arm}}$. Since both arms are equal, $M.A. = 1$.
- The assertion correctly states that a seesaw with a midpoint fulcrum has a mechanical advantage of 1.
- The reason provides the correct formula for calculating the mechanical advantage.
- Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Rolling Motion, Torque

c) $23.52\,\text{rad/s}^2$
[Solution Description]
The torque due to gravity causes rotation. First, find the moment of inertia of a solid cylinder:
$I = \frac{1}{2}MR^2 = \frac{1}{2} \times 4 \times (0.5)^2 = 0.5\,\text{kg}\cdot\text{m}^2$
Compute the torque about the center of mass:
$\tau = MgR\sin\theta = 4 \times 9.8 \times 0.5 \times \sin(37^\circ) \approx 11.76\,\text{N}\cdot\text{m}$
Angular acceleration ($\alpha$) is given by:
$\alpha = \frac{\tau}{I} = \frac{11.76}{0.5} \approx 23.52\,\text{rad/s}^2$

Key Concept: Class III Levers, Mechanical Advantage

c) 4.0
[Solution Description]
For a class III lever, $V.R. = \frac{\text{load arm}}{\text{effort arm}} = \frac{80}{20} = 4$. However, since it's class III, $\text{actual } M.A. = \text{efficiency} \times V.R. = 0.75 \times 4 = 3$. But the question asks for $V.R.$ which remains 4 regardless of efficiency.

Key Concept: Tension in a System with Inclined Plane

c) $29.4 \, \text{N}$
[Solution Description]
For the $5 \, \text{kg}$ block on the incline, the downhill force is $5g \sin 37^\circ$ and uphill force is tension ($T$). For the hanging $3 \, \text{kg}$ block, downward force is $3g$ and upward force is $T$.
Setting up equations:
$T - 5g \sin 37^\circ = 5a \quad (\text{for } 5 \, \text{kg block})$
$3g - T = 3a \quad (\text{for } 3 \, \text{kg block})$
Adding both equations eliminates $T$:
$3g - 5g \sin 37^\circ = 8a$
Substituting $\sin 37^\circ \approx 0.6$ and $g = 9.8$:
$29.4 - 29.4 = 8a \implies a = 0 \, \text{m/s}^2$
Substituting back into the second equation:
$3g - T = 0 \implies T = 3g = 29.4 \, \text{N}$
Thus, the tension is $29.4 \, \text{N}$.

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:
$T = E = 60 \text{ N}$

Key Concept: Examples of Class III Levers

c) Wheelbarrow
[Solution Description]
Class III levers have the effort between the fulcrum and the load. Examples include sugar tongs, fishing rods, and fire tongs. A wheelbarrow, however, is a Class II lever (load is between the fulcrum and the effort). Thus, the correct answer is the wheelbarrow.

Key Concept: Class I Levers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A seesaw is indeed an example of a Class I lever because the fulcrum is positioned between the effort (the force applied by the person sitting on one end) and the load (the force exerted by the person sitting on the other end). The reason correctly explains why the seesaw is classified as a Class I lever.

Key Concept: Single Fixed Pulley, Velocity Ratio, Work Done

d) 400 J
[Solution Description]
In a single fixed pulley, the distance moved by the effort ($d_E$) is equal to the distance moved by the load ($d_L$). Here, $d_E = 4$ m.
Work done by the effort is calculated as:
$W = F \times d = 100 \, \text{N} \times 4 \, \text{m} = 400 \, \text{J}$
The same work is done on the load since $V.R. = 1$.

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: 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: Class I Levers - Speed Multiplier, Human Body Application

b) Spine is fulcrum, load is at the front, effort is at the rear
[Solution Description]
From the syllabus, in the action of nodding the head:
- The spine acts as the fulcrum (F).
- The front part of the head (e.g., the face) is the load (L).
- The rear part of the head (where muscles apply force) is the effort (E).
This configuration is a Class I lever with fulcrum between load and effort.

Key Concept: Class II Levers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the definition of Class II levers, the effort arm is always longer than the load arm ($\text{Effort arm} > \text{Load arm}$). This results in the mechanical advantage ($M.A. = \frac{\text{Effort arm}}{\text{Load arm}}$) being greater than 1. Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

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: 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

c) By measuring the distance moved by effort over the distance moved by load
[Solution Description]
The velocity ratio is the ratio of the distance moved by the effort ($d_e$) to the distance moved by the load ($d_l$) in the same time. Mathematically, $V.R. = \frac{d_e}{d_l}$ or $V.R. = \frac{V_e}{V_l}$, where $V_e$ and $V_l$ are the velocities of effort and load, respectively.

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: Changing Direction of Effort

c) It changes the direction of effort from upward to downward
[Solution Description]
According to the syllabus, without a pulley, effort has to be applied upwards to lift water. With a fixed pulley, the direction changes so that downward effort can be applied instead, allowing the person to use their own weight.

Key Concept: Definition of Mechanical Advantage

b) $M.A. = \frac{L}{E}$
[Solution Description]
The mechanical advantage (M.A.) is defined as the ratio of the load (L) to the effort (E). The formula is given by:
$M.A. = \frac{L}{E}$
This means option b) correctly represents the formula for mechanical advantage.

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: 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: 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: Block and Tackle System, Efficiency

c) 83.33%
[Solution Description]
Input force = $60 \times 10 = 600$ N, Output force = $300 \times 10 = 3000$ N.
$V.R. = \text{Number of pulleys} = 6$, $M.A. = \frac{3000}{600} = 5$.
Efficiency = $\frac{M.A.}{V.R.} \times 100 = \frac{5}{6} \times 100 = 83.33\%$.

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: Speed Multiplier

c) Using scissors to cut cloth
[Solution Description]
The syllabus states that scissors are an example where small movement of handles produces larger movement of blades, resulting in gain in speed. This matches the definition of speed multiplier.

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: 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: 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: 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: 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: Velocity Ratio, Work Done

b) 600 J
[Solution Description]
First find V.R.: $\frac{12m}{1.5m} = 8$
Theoretical effort needed: $E_{ideal} = \frac{L}{V.R.} = \frac{400}{8} = 50N$
Actual effort considering efficiency: $E_{actual} = \frac{E_{ideal}}{\eta} = \frac{50}{0.8} = 62.5N$
Work done by effort: $W_{effort} = F \times d = 62.5 \times 12 = 750J$
Work done on load: $W_{load} = 400 \times 1.5 = 600J$
Work done against gravity is equal to $W_{load} = 600J$

Key Concept: Examples of Class II Levers

c) A wheelbarrow
[Solution Description]
Examples of Class II levers include nut crackers, bottle openers, and wheel barrows, where the fulcrum and effort are at the ends and the load is in between.

Key Concept: Class I Lever, Mechanical Advantage

d) 800 N
[Solution Description]
For a Class I lever, mechanical advantage (M.A.) = $\frac{\text{Effort arm}}{\text{Load arm}} = \frac{3}{1.5} = 2$.
M.A. = $\frac{\text{Load}}{\text{Effort}}$, so Load = Effort × M.A. = $400 \times 2 = 800$ N.

Key Concept: Machines as force multipliers

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that a single fixed pulley is used to lift a bucket from a well because it acts as a force multiplier. However, this is incorrect because a single fixed pulley does not multiply force; instead, it only changes the direction of the applied effort. The reason states that in a single fixed pulley, the mechanical advantage is greater than 1. This is also incorrect because the mechanical advantage of a single fixed pulley is equal to 1. Therefore, the assertion is false, and the reason is also false.

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: Examples of Class III Levers

c) Fishing rod
[Solution Description]
Examples of Class III levers include sugar tongs, a forearm lifting a load, fire tongs, foot treadle, knife, spade used to lift coal, fishing rod, etc. These have the fulcrum and load at the ends and the effort in between.

Key Concept: Energy Considerations and Losses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the efficiency of a real machine can never exceed 100%, which is true because it would violate the principle of conservation of energy if output work exceeded input work. The Reason explains this by stating that energy losses (e.g., friction, non-ideal conditions) ensure output work is always less than input work, which is also true and correctly explains the Assertion. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Effort Arm Determination

c) 45 cm
[Solution Description]
Given M.A. = 3 and Load Arm = 15 cm, we use the formula M.A. = Effort Arm / Load Arm. Rearranging gives Effort Arm = M.A. $\times$ Load Arm = 3 $\times$ 15 cm = 45 cm.

Key Concept: Mechanical Advantage Calculation

c) 4
[Solution Description]
Mechanical Advantage (M.A.) for a lever is calculated as the ratio of the effort arm ($d_E$) to the load arm ($d_L$). Given $d_E = 8\,\text{cm}$ and $d_L = 2\,\text{cm}$,
$M.A. = \frac{d_E}{d_L} = \frac{8}{2} = 4.$
Thus, the mechanical advantage is 4, indicating a force multiplier effect.

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: Acceleration on a Frictionless Inclined Plane

b) $g\sin\theta$
[Solution Description]
On a frictionless inclined plane, the net force accelerating the object is the parallel component of gravity:
$F_{\text{parallel}} = mg\sin\theta$
Using Newton's second law ($F = ma$), we can find the acceleration ($a$):
$a = \frac{F_{\text{parallel}}}{m} = g\sin\theta$

Key Concept: Relationship Between M.A., V.R., and Efficiency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that reducing the weight of moving parts improves efficiency. This is correct because, as per the syllabus, the weight of moving parts (e.g., in pulley systems) contributes to energy loss through friction and additional effort needed to move them. The reason provided is also correct because reducing the weight of moving parts minimizes these energy losses, thereby increasing efficiency. Furthermore, the reason directly explains why the assertion is valid, as it elaborates on how reduced weight leads to higher efficiency.

Key Concept: Forces on an Inclined Plane

b) $mg\sin\theta$
[Solution Description]
The component of gravitational force acting parallel to the inclined plane is given by:
$F_{\text{parallel}} = mg\sin\theta$
This formula is derived from resolving the gravitational force ($mg$) into components along and perpendicular to the slope.

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

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Mechanical advantage (MA) of a wheel and axle is given by $\text{MA} = \frac{r_{\text{wheel}}}{r_{\text{axle}}}$, where $r_{\text{wheel}}$ is the wheel's radius and $r_{\text{axle}}$ is the axle's radius. Increasing the wheel radius ($r_{\text{wheel}}$) increases MA, enabling a smaller input force to lift a heavier load (output resistance). Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

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: Velocity Ratio

c) Both the handle length and the screw pitch
[Solution Description]
The velocity ratio (V.R.) of a screw jack is given by:
$V.R. = \frac{2\pi r}{p}$
Here, $r$ is the radius of the handle and $p$ is the pitch of the screw. Thus, it depends on both the handle length and the screw pitch.

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: 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: Mechanical Advantage in Block and Tackle System

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For an ideal block and tackle system with $n = 6$ pulleys, the mechanical advantage should be equal to the number of pulleys, i.e., $\text{M.A.} = 6$. However, due to friction and the weight of the pulley and string, the actual mechanical advantage is less than the ideal value. Therefore, the assertion that $\text{M.A.} = 5$ for a 6-pulley system is plausible under practical conditions. The reason correctly explains why the actual mechanical advantage is lower than the ideal value.

Key Concept: Work Done on an Inclined Plane

b) $68 \, \text{J}$
[Solution Description]
First, find the distance moved along the incline using $h = d \sin \theta$:
$d = \frac{h}{\sin \theta} = \frac{2}{\sin 30^\circ} = 4 \, \text{m}$
The normal force ($N$) is $mg \cos \theta$:
$N = 10 \times 9.8 \times \cos 30^\circ = 84.87 \, \text{N}$
Frictional force ($F_f$) is $\mu_k N$:
$F_f = 0.2 \times 84.87 = 16.97 \, \text{N}$
Work done against friction is $W = F_f \times d$:
$W = 16.97 \times 4 = 67.88 \, \text{J}$
Thus, the work done against friction is approximately $68 \, \text{J}$.

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: Class II Lever, Velocity Ratio

d) 3.125
[Solution Description]
For Class II levers, V.R. = $\frac{\text{Effort arm}}{\text{Load arm}}$ since M.A. = V.R. in ideal case. Here, V.R. = $\frac{25}{8} = 3.125$.

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: Functions and Uses of Simple Machines

d) Jack lifting a car
[Solution Description]
A force multiplier allows lifting a heavy load with less effort. Among the options, the jack lifting a car demonstrates this principle, as it reduces the effort needed to lift the heavy car.

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: Class III Levers

c) Class III
[Solution Description]
In a Class III lever, the fulcrum and load are at the two ends, and the effort is applied somewhere in between. A pair of tongs fits this description as the fulcrum is at one end, the load (object being gripped) is at the other end, and the effort is applied in the middle.

Key Concept: Mechanical Advantage, Velocity Ratio

c) 80%
[Solution Description]
Use the relation between M.A., V.R., and efficiency:
$\text{M.A.} = \text{V.R.} \times \eta$.
Given M.A. = 8 and V.R. = 10:
$8 = 10 \times \eta \implies \eta = \frac{8}{10} = 0.8$.
Convert to percentage:
$\eta = 0.8 \times 100 = 80\%$.

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: Levers, Principle of Moments

b) 25 N
[Solution Description]
Step 1: Understand the lever arrangement. The fulcrum divides the lever into two arms:
- Load Arm = 20 cm
- Effort Arm = 80 cm
Step 2: Apply the principle of moments for equilibrium:
$E \times \text{Effort Arm} = L \times \text{Load Arm}$
$E \times 80\,cm = 100\,N \times 20\,cm$
Step 3: Solve for $E$:
$E = \frac{100 \times 20}{80} = \frac{2000}{80} = 25\,N$
Thus, the effort needed is 25 N.

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: Single Fixed Pulley, Tension, Equilibrium

c) 30 N
[Solution Description]
In a single fixed pulley:
- The tension $T$ is the same throughout the string.
- The load $L$ is balanced by the tension acting upwards.
$T = L = 30 \, \text{N}$
Since the system is in equilibrium, the effort $E$ must also be equal to $T$.

Key Concept: Types of Levers

c) Tongs
[Solution Description]
A Class III lever has the effort applied between the fulcrum and the load. Examples include tongs and the human forearm when bending at the elbow.

Key Concept: Class II Lever Definition

c) The fulcrum and effort are at the two ends, with the load in between.
[Solution Description]
A Class II lever has the fulcrum and effort at the two ends, with the load in between. The effort arm is longer than the load arm, making the mechanical advantage always greater than 1. It acts as a force multiplier.

Key Concept: Effort Calculation

b) 50 N
[Solution Description]
In an ideal single movable pulley, the effort $E$ is related to the load $L$ by:
$E = \frac{L}{2}$
Given $L = 100$ N, the effort required is:
$E = \frac{100}{2} = 50 \text{ N}$
Therefore, the effort needed is 50 N.

Key Concept: Force Multiplier

b) Using a jack to lift a car
[Solution Description]
A force multiplier allows lifting a heavy load with less effort. The syllabus mentions that a jack is used to lift a car - this is an example where less effort is applied to lift a heavier load, making it a force multiplier.