Class 10 Physics Chapter 1 Force

Key Concept: Force, Moment of Force, Rotational Motion

b) 30 N m
[Solution Description]
The moment of force (torque) is calculated using $\tau = F \times d$, where $F$ is the force and $d$ is the perpendicular distance from the pivot. Here, $F = 20 \text{ N}$, $d = 1.5 \text{ m}$. Therefore, $\tau = 20 \times 1.5 = 30 \text{ N m}$.

Key Concept: Moment of Force

b) 20 N m
[Solution Description]
Torque (\tau) is calculated using the formula $\tau = F \times d$, where $F$ is the force and $d$ is the perpendicular distance from the pivot. Plugging in the values:
$\tau = 10 \, \text{N} \times 2 \, \text{m} = 20 \, \text{N m}$
Thus, the torque produced is 20 N m.

Key Concept: Translational Motion, Force, Vector Quantity

c) 4 s
[Solution Description]
First, find the deceleration using $a = \frac{F}{m} = \frac{5000}{1000} = 5 \text{ m/s}^2$. Then, use $v = u + at$ where final velocity $v = 0$, initial velocity $u = 20 \text{ m/s}$, and $a = -5 \text{ m/s}^2$. Solving for time $t$, we get $0 = 20 - 5t \implies t = 4 \text{ s}$.

Key Concept: Definition of Force

c) Newton (N)
[Solution Description]
The S.I. unit of force is newton (N) or kilogram-force (kgf), as mentioned in the syllabus. So, the correct answer is newton (N).

Key Concept: Translational and Rotational Motion

b) The body must be pivoted at a point
[Solution Description]
For rotational motion, the body must be pivoted at a point, and the force should be applied at a suitable point such that it creates a turning effect (torque). The body cannot be free to move entirely; otherwise, it would result in translational motion.

Key Concept: Principle of Moments, Equilibrium

b) 1.5 m
[Solution Description]
For equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments:
$30 \, \text{kg} \times g \times 2 \, \text{m} = 40 \, \text{kg} \times g \times d$
Simplifying (canceling $g$):
$60 = 40d \implies d = \frac{60}{40} = 1.5 \, \text{m}$

Key Concept: Moment of Force

c) 40 N m
[Solution Description]
The moment of force (torque) is calculated using the formula:
$\text{Moment of force} = \text{Force} \times \text{Perpendicular distance}$
Given: Force = 20 N, Distance = 2 m.
$\text{Moment of force} = 20 \, \text{N} \times 2 \, \text{m} = 40 \, \text{N m}$

Key Concept: Effects of Force, Rotational Motion, Principle of Moments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a uniform metre rule pivoted at its center remains in equilibrium when equal forces are applied at both ends. This is true because when equal forces are applied at equal distances from the pivot, the clockwise and anticlockwise moments become equal, satisfying the condition for rotational equilibrium.
The reason states the principle of moments, which is correct as it defines the condition for rotational equilibrium where the sum of clockwise moments equals the sum of anticlockwise moments.
Here, the reason correctly explains the assertion since the equilibrium condition described in the assertion is directly based on the principle of moments mentioned in the reason.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Translational Motion

c) The body moves in a straight path in the direction of the force
[Solution Description]
When a force acts on a stationary rigid body that is free to move, the body starts moving in a straight path in the direction of the force. This motion is called translational or linear motion.

Key Concept: Moment of Force, Direction of Torque

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the direction of torque depends only on the point of application of the force and not on its direction. However, according to the syllabus, the direction of torque depends on both the point of application of the force and the direction of the force. The reason correctly states that the moment of force is a vector quantity whose direction is determined by the right-hand rule, which takes into account both the point of application and the direction of the force. Thus, the assertion is false because it ignores the role of the force's direction, while the reason is true as it accurately describes the vector nature of torque.

Key Concept: Units of Moment of Force

c) Newton meter (N m)
[Solution Description]
The SI unit of the moment of force is newton meter (\$N m\$), as derived from the formula Force $\times$ Distance. This is explicitly mentioned in the syllabus under "Units of moment of force."

Key Concept: Moment of force, perpendicular distance

c) 15 Nm anticlockwise
[Solution Description]
To find the net moment:
1. First calculate moment due to 10 N force: $M_1 = F_1 \times d_1 = 10 \times 0.5 = 5$ Nm (clockwise)
2. Then calculate moment due to 20 N force: $M_2 = F_2 \times d_2 = 20 \times 1 = 20$ Nm (anticlockwise)
3. Net moment = Anticlockwise moment - Clockwise moment = $20 - 5 = 15$ Nm anticlockwise

Key Concept: Unit Conversion

b) 19.6 N
[Solution Description]
Given $1\, \text{kgf} = g\, \text{N}$. Here, $g = 9.8\, \text{m/s}^2$.
So, $2\, \text{kgf} = 2 \times 9.8 = 19.6\, \text{N}$.

Key Concept: Translational Motion, Newton's Second Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a rigid body moving with constant velocity implies no net force acts on it. This is true because, from Newton's first law, if the velocity is constant ($v = \text{constant}$), then acceleration $a = 0$. Using $F = ma$, we get $F = 0$ when $a = 0$, meaning no net force. The Reason mentions Newton's second law, which states that a net force causes acceleration ($F = ma$). Since constant velocity means no acceleration, no net force exists. Hence, the Reason correctly explains the Assertion by linking the absence of acceleration to the absence of net force, as per Newton's laws.

Key Concept: Equilibrium in Translational Motion

c) The normal force equals the weight of the book.
[Solution Description]
For the book to remain stationary, the net force must be zero (translational equilibrium).
Weight acts downward ($W = 20\, \text{N}$). The normal force from the table must balance this weight upward ($N = 20\, \text{N}$).
Therefore, the net force $\sum F = N - W = 20 - 20 = 0$.

Key Concept: Moment of Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The moment of force (torque) is given by $\tau = F \times d$, where $F$ is the force applied and $d$ is the perpendicular distance from the pivot point. From this formula, it is clear that torque is directly proportional to the perpendicular distance $d$. Therefore, as $d$ increases, $\tau$ also increases. The assertion states that the moment of force increases with an increase in perpendicular distance, which is true. The reason correctly explains the assertion using the relation $\tau = F \times d$.

Key Concept: Equilibrium Condition

d) Sum of moments must be zero
[Solution Description]
For rotational equilibrium, the algebraic sum of all moments acting on the body about any point must be zero:
$\sum \tau = 0$
This means that the clockwise and anticlockwise moments must balance each other out.

Key Concept: Principle of Moments

a) 0 N m
[Solution Description]
The resultant moment is the sum of individual moments. Since the forces act in opposite directions, one creates a clockwise moment and the other an anticlockwise moment. Each has a lever arm of 2 m (half of 4 m).
$\text{Moment}_1 = 10\, \text{N} \times 2\, \text{m} = 20\, \text{N m (clockwise)}$
$\text{Moment}_2 = 10\, \text{N} \times 2\, \text{m} = 20\, \text{N m (anticlockwise)}$
The resultant moment is:
$20\, \text{N m} - 20\, \text{N m} = 0\, \text{N m}$

Key Concept: Clockwise and Anticlockwise Moments

c) Applying force at the point farthest from the pivot
[Solution Description]
According to the syllabus, the maximum turning effect is produced when the perpendicular distance of the line of action of the force from the axis of rotation is maximum.

Key Concept: Clockwise and Anticlockwise Moments

b) Inwards
[Solution Description]
The syllabus states that the direction of a clockwise moment is along the axis of rotation inwards.

Key Concept: Clockwise and Anticlockwise Moments

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to the convention mentioned in the syllabus, anticlockwise moments are considered positive. The right-hand rule is a common method to determine the direction of torque but is not explicitly mentioned as the reason for this convention in the provided syllabus. Thus, the Assertion is true, but the Reason is not directly supported by the syllabus.

Key Concept: Moment of force calculation

b) 5 N,m
[Solution Description]
The moment of force (torque) is calculated using the formula: $\text{Moment} = \text{Force} \times \text{Perpendicular distance}$. Here, Force $F = 10 \, \text{N}$ and perpendicular distance $d = 0.5 \, \text{m}$. Therefore, the moment of force is: $10 \, \text{N} \times 0.5 \, \text{m} = 5 \, \text{N\,m}$.

Key Concept: Direction of Moment, Net Moment

b) 2.5 N m anticlockwise
[Solution Description]
Calculate the moment due to the 15 N force (clockwise):
$M_1 = 15 \, \text{N} \times 0.5 \, \text{m} = 7.5 \, \text{N m} \, (\text{clockwise})$
Calculate the moment due to the 10 N force (anticlockwise):
$M_2 = 10 \, \text{N} \times 1 \, \text{m} = 10 \, \text{N m} \, (\text{anticlockwise})$
The net moment is the difference between the two moments:
$M_{\text{net}} = M_2 - M_1 = 10 \, \text{N m} - 7.5 \, \text{N m} = 2.5 \, \text{N m} \, (\text{anticlockwise})$

Key Concept: Moment of Force, Equilibrium Condition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that for a pivoted body to be in rotational equilibrium, the net torque acting on it must be zero, which is correct because equilibrium implies no rotation. The Reason states that when the sum of clockwise moments equals the sum of anticlockwise moments, the body does not rotate, which is also correct and directly explains the Assertion. Therefore, both are true, and the Reason correctly explains the Assertion.

Key Concept: Principle of Moments, Equilibrium Conditions

d) Assertion is false, but Reason is true.
[Solution Description]
First, calculate the moments about the fulcrum (50 cm mark). The anticlockwise moment due to the 20 g weight is $20 \times (50 - 30) = 400 \text{ g cm}$. For equilibrium, the clockwise moment due to the 10 g weight must be equal to this, so $10 \times (70 - 50) = 200 \text{ g cm}$. This is not equal to the anticlockwise moment (400 g cm), so the Assertion is false. However, the Reason correctly states the principle of moments and is true.

Key Concept: Principle of Moments

b) 24 N m anticlockwise
[Solution Description]
The net moment is calculated by taking the algebraic sum of clockwise and anticlockwise moments. The 6 N force is at 1 m from the pivot, creating a clockwise moment: $6 \, \text{N} \times 1 \, \text{m} = 6 \, \text{N m}$ (clockwise). The 10 N force is at 3 m from the pivot, creating an anticlockwise moment: $10 \, \text{N} \times 3 \, \text{m} = 30 \, \text{N m}$ (anticlockwise). Net moment = Anticlockwise moment - Clockwise moment = $30 \, \text{N m} - 6 \, \text{N m} = 24 \, \text{N m}$ (anticlockwise).

Key Concept: Principle of Moments, Multiple Forces

b) 47.5 cm
[Solution Description]
Calculate the moments due to existing forces:
$\text{Anticlockwise moment} = 10\,N \times (40\,cm - 10\,cm) = 10\,N \times 30\,cm = 300\,N\,cm$
$\text{Clockwise moment} = 5\,N \times (70\,cm - 40\,cm) = 5\,N \times 30\,cm = 150\,N\,cm$
Net unbalanced moment is $300\,N\,cm - 150\,N\,cm = 150\,N\,cm$ anticlockwise. To balance this, the 20 N weight must provide an equal clockwise moment. Let its distance from the pivot be $x$:
$20\,N \times x = 150\,N\,cm \implies x = \frac{150}{20}\,cm = 7.5\,cm$
Since it must be on the opposite side, its position is $40\,cm + 7.5\,cm = 47.5\,cm$.

Key Concept: Practical Application

b) The greater perpendicular distance increases the turning effect

[Solution Description]

Opening a door involves creating a rotational effect (moment). The moment depends on both the force applied and the perpendicular distance from the pivot (hinges). Applying force at the handle increases the perpendicular distance (couple arm), thus increasing the moment for the same amount of force. This makes it easier to rotate/open the door.
The correct answer is the greater perpendicular distance increases the turning effect.

Key Concept: Couple, Moment of Couple

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To evaluate the assertion and reason:
1. **Assertion Analysis**: The moment of a couple is indeed independent of the point about which it is calculated because it depends solely on the product of either force and the perpendicular distance between them (couple arm). This is a fundamental property of couples.
2. **Reason Analysis**: The reason correctly states that a couple consists of two equal and opposite parallel forces with non-coinciding lines of action. This is the definition of a couple.
3. **Connection**: While the assertion is true and the reason is also true, the reason does not explain why the moment is independent of the point. It is a property derived from the definition but not directly explained by it. Thus, the reason is not the correct explanation of the assertion.
Hence, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Moment of Couple

b) 5 Nm
[Solution Description]
To find the moment of couple:
Moment of couple = Force $\times$ perpendicular distance between forces.
Given Force F = 10 N, distance d = 0.5 m.
Moment = $F \times d = 10 \times 0.5 = 5$ Nm.
The correct answer is 5 Nm.

Key Concept: Moment of Couple

c) 20 N$\cdot$m
[Solution Description]
The moment of a couple is given by the formula: $\text{Moment} = F \times d$. Here, $F = 10 \, \text{N}$ and $d = 2 \, \text{m}$. Substituting the values, we get: $\text{Moment} = 10 \times 2 = 20 \, \text{N} \cdot \text{m}$.

Key Concept: Moment of Couple

b) 22.5 N$\cdot$m
[Solution Description]
Since the forces are equal, opposite, and parallel (angle = 180 degrees), they form a couple. The moment is calculated as $\text{Moment} = F \times d$. Here, $F = 15 \, \text{N}$ and $d = 1.5 \, \text{m}$. Thus, $\text{Moment} = 15 \times 1.5 = 22.5 \, \text{N} \cdot \text{m}$.

Key Concept: Moment of a Couple

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the moment of a couple is independent of the point about which it is calculated. This is true because a couple produces a pure moment (rotational effect) without any translational motion. The reason explains that a couple consists of two equal and opposite forces separated by a perpendicular distance, which indeed results in a pure rotational effect. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Centre of Gravity, Distribution of Mass

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the centre of gravity of a hollow hemisphere lies outside its material, which is true because for a hollow hemisphere, the C.G. is located at a distance of $\frac{3r}{8}$ from the flat surface, where $r$ is the radius, and this point does not lie within the material of the hemisphere. The reason states that the geometric centre does not coincide with any part of its material, which is also true since the geometric centre of a hollow hemisphere is at $\frac{r}{2}$ from the flat surface, but there is no material at that point. However, while both statements are true, the reason does not directly explain why the C.G. lies outside the material; it merely describes the geometric property. Therefore, the correct answer is (b).

Key Concept: Centre of Gravity and Balancing

c) Vertically below the point of suspension[Solution Description]When a body is freely suspended, its centre of gravity always comes to rest vertically below the point of suspension. For a uniform rectangular lamina, the C.G. is at the intersection of its diagonals. Thus, when suspended from a corner, the C.G. will lie directly below that corner in equilibrium.

Key Concept: Centre of Gravity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the centre of gravity of a uniform circular ring lies at its geometric centre, which is correct because for symmetrical bodies with uniform mass distribution, the C.G. coincides with the geometric centre. The reason given is also true as it correctly explains why the C.G. is at the centre - due to symmetric mass distribution. Hence, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Centre of Gravity Definition

c) It shifts to the geometric centre of the semicircle.
[Solution Description]
The centre of gravity of a uniform rod is initially at its midpoint since mass is uniformly distributed. When the rod is bent into a semicircle, the mass distribution changes symmetrically about the diameter. The centre of gravity shifts to the geometric centre of the semicircle, which lies along the perpendicular bisector of the diameter at a height $h = \frac{2r}{\pi}$ from the diameter. Thus, the centre of gravity moves from the initial midpoint.

Key Concept: Centre of Gravity of Regular Objects

c) Lies at the geometric centre of the ring
[Solution Description]
The centre of gravity of a circular ring lies at its geometric centre, even though there is no material at that point. This is because the mass is uniformly distributed along the circumference of the ring.

Key Concept: Centre of Gravity, Distribution of Mass

b) At a distance of $\frac{2R}{\pi}$ from the base
[Solution Description]
For a uniform wire bent into a semicircular arc, the centre of gravity is not at the midpoint but along the axis of symmetry. The formula to find the vertical distance from the base is given by $\frac{2R}{\pi}$.
Calculation: Since it's a uniform mass distribution, the y-coordinate of C.G. is calculated using integral calculus for a semicircular arc. The result shows that the C.G. lies $\frac{2R}{\pi}$ units above the diameter.
Solution Steps:
1. Identify the symmetry axis (vertical).
2. Integrate the mass elements over the semicircle.
3. Compute the average height using the integral formula.
Final Answer: $\frac{2R}{\pi}$ from the base.

Key Concept: Position of C.G. in Regular Objects

b) At the geometric centre of the disc.
[Solution Description]

For a uniform circular disc, the centre of gravity lies at the geometric centre of the disc because the mass is uniformly distributed.

Key Concept: C.G. Outside Material

c) Hollow sphere.
[Solution Description]

A hollow sphere has its centre of gravity at its geometric centre, which is a point not occupied by any material of the sphere.

Key Concept: Centre of Gravity Definition

c) The point where the entire weight of the body acts vertically downward.
[Solution Description]

The centre of gravity of a body is the point where the entire weight of the body can be considered to act vertically downwards. It depends on the distribution of mass in the body.

Key Concept: Centre of Gravity, Regular-shaped objects

d) At the centroid, which is two-thirds down the median from the vertex
[Solution Description]
The centre of gravity (C.G.) of a uniform triangular lamina lies at the intersection of its medians. When suspended from a vertex, the C.G. will position itself directly below the point of suspension in equilibrium.

Key Concept: C.G. of composite shapes

b) Remains at the same position
[Solution Description]
Originally C.G. of square was at its geometric center since it's uniform. After cutting the circular hole exactly at center, the mass distribution remains symmetric about both axes. Since symmetry is maintained, the C.G. remains at the original position.

Key Concept: C.G. of irregular lamina and balancing

c) Assertion is true, but Reason is false.
[Solution Description]
To determine the C.G. of an irregular lamina experimentally, we suspend it from three non-collinear points (not just two) and draw plumb lines. The intersection of these lines gives the C.G., which may or may not coincide with the geometrical centre. The Reason is incorrect because the C.G. of an irregular lamina does not necessarily lie at its geometrical centre unless the mass distribution is symmetric. Thus, Assertion is partially true but incomplete, while Reason is false.

Key Concept: Center of gravity, Plumb line method

c) Ensure the plumb line is perfectly vertical when marking the third line
[Solution Description]
For the plumb lines to intersect at the center of gravity (G):
1. Ensure the lamina is freely suspended without any external force or friction.
2. Verify the plumb line is perfectly vertical while marking the lines.
3. Re-suspend the lamina carefully from the third hole and check alignment.
The most crucial step is ensuring the plumb line is absolutely vertical when marking the third line, as slight deviations can cause misalignment.

Key Concept: Experimental determination of C.G., Plumb line method

b) The plumb line was not perfectly vertical during marking
[Solution Description]
The correct intersection point of the plumb lines should give the center of gravity (G). If they do not intersect at a single point, possible reasons include:
1. The lamina was not free to oscillate properly due to friction or obstruction.
2. The plumb line was not perfectly vertical while marking the lines.
3. The holes were not drilled accurately near the edge, causing misalignment.
The most probable reason is the inaccuracy in ensuring the plumb line is vertical when marking the lines, leading to non-intersecting lines.

Key Concept: Experimental determination of C.G.

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because suspending the lamina from three points and marking vertical lines with a plumb line helps locate the intersection point, which is the C.G. The reason is also true as it states the principle behind why the method works. Additionally, the reason correctly explains the assertion since the vertical alignment of C.G. below the suspension point is fundamental to this experimental method.

Key Concept: Principle of Moments

a) The sum of the clockwise moments equals the sum of the anticlockwise moments.
[Solution Description]
For the metre rule to be in equilibrium, the algebraic sum of moments about the pivot must be zero. The moment due to the 5 N weight (clockwise) is $5 \times 30 = 150 \text{ N cm}$. The moment due to the 10 N weight (anticlockwise) is $10 \times 20 = 200 \text{ N cm}$. Since $150 \neq 200$, the rule is not in equilibrium unless an additional moment balances the difference.

Key Concept: Dynamic Equilibrium, Centripetal Force

d) 8000 N
[Solution Description]
In dynamic equilibrium, the net force provides the centripetal force required for circular motion.
Centripetal force formula: $F_c = \dfrac{m v^2}{r}$.
Given:
Mass $m = 1000$ kg,
Speed $v = 20$ m/s,
Radius $r = 50$ m.
Substituting values:
$F_c = \dfrac{1000 \times (20)^2}{50} = \dfrac{1000 \times 400}{50} = 8000$ N.
Since the car is in dynamic equilibrium, this is also the net force acting on it.

Key Concept: Static Equilibrium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The book is an example of static equilibrium where the downward weight of the book is balanced by the upward normal reaction from the table. This satisfies the first condition of equilibrium where the resultant force is zero (Assertion). The Reason correctly explains the Assertion as static equilibrium requires the vector sum of all forces to be zero. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Static Equilibrium, Principle of Moments

b) 10 N
[Solution Description]
For equilibrium, the sum of moments about the pivot must be zero. The distance from the pivot to each end is 1 m (half of 2 m).
Moment due to the 10 N weight = $10 \times 1 = 10$ Nm (clockwise).
Moment due to the unknown weight $W$ = $W \times 1$ Nm (anticlockwise).
For equilibrium: $\sum \tau = 0$
So, $W \times 1 - 10 \times 1 = 0$
Simplifying: $W = 10$ N.
Therefore, the unknown weight $W$ must be 10 N for equilibrium.

Key Concept: Static Equilibrium

d) The weight of the book and the normal reaction are equal and opposite.
[Solution Description]
The book is in static equilibrium when the net force acting on it is zero. The weight of the book acts vertically downward, and the normal reaction from the table acts vertically upward. These forces are equal and opposite, balancing each other.

Key Concept: Principle of Moments

d) $W_1 \times l_1 = W_2 \times l_2$
[Solution Description]
The principle of moments states that for the meter rule to be in equilibrium, the sum of clockwise moments about point O must equal the sum of anticlockwise moments. Mathematically, this can be expressed as $W_1 \times l_1 = W_2 \times l_2$.

Key Concept: Conditions for equilibrium, Static equilibrium

b) $45^\circ$
[Solution Description]
For the ladder to be in static equilibrium, both net force and net torque must be zero. Let’s analyze forces:
- Vertical forces: Normal force $N$ from ground balances weight $W = mg = 200 \, \text{N}$. So $N = 200 \, \text{N}$.
- Horizontal forces: Friction $f$ balances normal force from wall $N_w$. So $f = N_w$.
- Torque about the base: Taking moments about the base, $N_w \times 5 \sin\theta = W \times 2.5 \cos\theta$. Simplifying, $N_w = 100 \cot\theta$.
For no slipping, $f \leq \mu N$ => $N_w \leq 0.5 \times 200$ => $100 \cot\theta \leq 100$ => $\cot\theta \leq 1$ => $\tan\theta \geq 1$ => $\theta \geq 45^\circ$.
Thus, the minimum angle is $45^\circ$.

Key Concept: Dynamic equilibrium, Principle of moments

b) The car is in dynamic equilibrium because net force is zero and it is moving
[Solution Description]
The car is moving with a constant velocity (constant speed and direction). This means the net force acting on the car is zero, satisfying the condition for equilibrium. Since the car is in motion, it is in dynamic equilibrium. No rotation is involved here, so torque conditions are irrelevant.

Key Concept: Dynamic Equilibrium, Net Force Balance

c) $v_t = \sqrt{\frac{mg}{k}}$
[Solution Description]
In dynamic equilibrium, net force is zero ($\sum F = 0$):
$mg = F_d$
Given $F_d = kv_t^2$, substitute:
$mg = kv_t^2$
Solve for $v_t$:
$v_t = \sqrt{\frac{mg}{k}}$

Key Concept: Principle of Moments

b) 20 kg
[Solution Description]
The principle of moments states that the sum of clockwise and anticlockwise moments must be equal for equilibrium. First, convert the mass of the first child to weight (force):
$W_1 = 40 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 392 \, \text{N}$
The anticlockwise moment is:
$\tau_{\text{anticlockwise}} = 392 \, \text{N} \times 1.5 \, \text{m} = 588 \, \text{Nm}$
Let $m$ be the mass of the second child. Its weight is $W_2 = m \times 9.8 \, \text{m/s}^2$. The clockwise moment is:
$\tau_{\text{clockwise}} = W_2 \times 3 \, \text{m} = (m \times 9.8) \times 3 \, \text{m}$
Setting the moments equal:
$588 \, \text{Nm} = (m \times 9.8) \times 3 \, \text{m} \implies m = \frac{588}{29.4} = 20 \, \text{kg}$
Hence, the mass of the second child is 20 kg.

Key Concept: Rotational and Translational Equilibrium, Principle of Moments

b) $\frac{Wx}{L} - \frac{mg}{2}$
[Solution Description]
Let $R_1$ be the reaction at the left pivot and $R_2$ at the right pivot.
Step 1: Apply translational equilibrium condition ($\sum F = 0$).
$R_1 + R_2 = W + mg$
Step 2: Apply rotational equilibrium about the left pivot ($\sum \tau = 0$).
Clockwise moments:
$W \times x$
Anticlockwise moments:
$R_2 \times L + mg \times \frac{L}{2}$
For equilibrium:
$Wx = R_2 L + \frac{mgL}{2}$
Solve for $R_2$:
$R_2 = \frac{Wx - \frac{mgL}{2}}{L}$
Simplifying:
$R_2 = \frac{Wx}{L} - \frac{mg}{2}$

Key Concept: Principle of Moments

d) $W_1 \times l_1 = W_2 \times l_2$
[Solution Description]
According to the principle of moments, for a balanced system, the clockwise moment must equal the anticlockwise moment. Mathematically, this is expressed as:
$W_1 \times l_1 = W_2 \times l_2$
This ensures rotational equilibrium.

Key Concept: Equilibrium, Centre of gravity

b) 3.06 kg
[Solution Description]
For equilibrium, sum of moments about P must be zero. Let $m$ be mass in kg.
Clockwise moment by object's weight = $m \times 9.8 \times 0.3\,m$.
Anticlockwise moment by applied force = $15\,N \times 0.6\,m = 9\,Nm$.
Balancing both: $m \times 9.8 \times 0.3 = 9$ → $m \times 2.94 = 9$ → $m = 9/2.94 ≈ 3.06\,kg$.

Key Concept: Moment of a force

c) Newton-metre
[Solution Description]
The moment of force is calculated as $F \times d$, where $F$ is the force and $d$ is the perpendicular distance from the pivot point. The S.I. unit of moment of force is newton-metre (N m).

Key Concept: Principle of Moments

c) At the pivot point
[Solution Description]
For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. Let the distance for the 5 N weight be $x$ cm from the pivot.
$4 \times 30 = 6 \times 20 + 5 \times x$
Solving for $x$:
$120 = 120 + 5x \\ \\ 5x = 0 \\ \\ x = 0 \, \text{cm}$
This implies the 5 N weight must be placed at the pivot point to maintain balance.

Key Concept: Couple, Moment of force

b) 5 Nm initially, 50% increase
[Solution Description]
Moment of couple = Force × perpendicular distance between forces.
Here, distance = spanner length = 20 cm = 0.2 m.
Initial moment = $25\,N \times 0.2\,m = 5\,Nm$.
When spanner length increases to 30 cm (0.3 m), new moment = $25 \times 0.3 = 7.5\,Nm$.
Percentage increase = $\frac{7.5-5}{5} \times 100 = 50\%$.

Key Concept: Definition of a Machine

c) A system that applies mechanical power and has components performing specific tasks to achieve a purpose
[Solution Description]
A machine is a device that uses mechanical power and has multiple parts, each with a definite function, to perform a particular task. It typically converts input energy into useful output energy.

Key Concept: Assertion-Reason

d) Assertion is false, but Reason is true.
[Solution Description]
Let us analyze the assertion and reason step by step:
- **Assertion (A):** This statement defines a machine in terms of energy transformation. While many machines do transform energy, this definition is incomplete because machines can also perform tasks like transferring or modifying motion without necessarily transforming energy (e.g., levers or pulleys).
- **Reason (R):** This statement correctly captures the practical purpose of most machines, which is to reduce human effort by either amplifying force or changing its direction.
Therefore, Assertion (A) is false because it provides an overly restrictive definition, while Reason (R) is true as it accurately describes the common purpose of machines.

Key Concept: Definition of Machine

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a bicycle can be considered as a machine. This is true as a bicycle meets the basic definition of a machine being a mechanical device that transmits and modifies human energy (pedaling) to produce motion. The reason correctly explains why a bicycle qualifies as a machine by providing the fundamental definition of what constitutes a machine - a device that transmits or modifies energy to perform work. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Work and Energy, Machine Efficiency

b) Output Work = 800 J, Energy Lost = 200 J
[Solution Description]
Output Work = Efficiency × Input Work = 0.8 × 1000 J = 800 J.
Energy Lost = Input Work - Output Work = 1000 J - 800 J = 200 J.

Key Concept: Principle of Moments, Equilibrium

b) 200 N
[Solution Description]
According to the principle of moments, for equilibrium:
$W_1 \times l_1 = W_2 \times l_2$
Here, $W_1 = 300 \, \text{N}$, $l_1 = 1 \, \text{m}$, $l_2 = 1.5 \, \text{m}$. Solving for $W_2$:
$300 \times 1 = W_2 \times 1.5 \implies W_2 = \frac{300}{1.5} = 200 \, \text{N}$

Key Concept: Moment of Force and Torque

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the handle of a spanner is made long to reduce the force required, which is true because a longer handle increases the perpendicular distance ($r$) from the axis of rotation (nut), thereby increasing the moment of force ($F \times r$). The reason states that the moment of force is directly proportional to the perpendicular distance, which is also true and correctly explains why a longer handle reduces the required force. Hence, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Concept: Steering Wheel Mechanism

c) The wheel rotates anticlockwise.
[Solution Description]
When a tangential force is applied at point A on the steering wheel's rim, it creates a moment that causes the wheel to rotate anticlockwise. Changing the point of application (e.g., to point B) while keeping the force direction the same reverses the rotation (clockwise).

Key Concept: Direction of moment of force

b) It reverses the direction of rotation
[Solution Description]
The direction of rotation of a steering wheel changes based on the point of application of the tangential force. If the force is applied at point A (one side of the rim), the wheel rotates anticlockwise, while applying the same force at point B (opposite side of the rim) causes clockwise rotation. This happens because the line of action of the force changes relative to the axis of rotation, altering the resultant torque direction.
Torque direction follows the right-hand rule.

Key Concept: Moment of Force, Principle of Moments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, we calculate the moment due to the 20 N force. Since it acts at a distance of 0.5 m (half the length of the rod) from the pivot and produces a clockwise rotation:
$\text{Moment}_{20\,N} = 20 \times 0.5 = 10\,Nm \, (\text{clockwise}).$
Next, we determine the unknown force $F$ that acts at 0.25 m from the pivot and produces an anticlockwise rotation:
$\text{Moment}_F = F \times 0.25 \, (\text{anticlockwise}).$
According to the principle of moments (rotational equilibrium), the sum of clockwise moments must equal the sum of anticlockwise moments:
$10 = F \times 0.25.$
Solving for $F$, we get:
$F = \frac{10}{0.25} = 40\,N.$
Thus, both the Assertion (rod equilibrium condition) and the Reason (principle of moments) are true, and the Reason correctly explains why the rod remains in equilibrium.

Key Concept: Balancing a Metre Rule with Weights

c) 25 cm from the pivot
[Solution Description]
Let $x$ be the distance from the pivot where the $40 \, \text{gf}$ weight should be placed. Using the principle of moments:
$20 \, \text{gf} \times 50 \, \text{cm} = 40 \, \text{gf} \times x$
Solving for $x$:
$x = \frac{20 \times 50}{40} = 25 \, \text{cm}$
The $40 \, \text{gf}$ weight should be placed $25 \, \text{cm}$ from the pivot.

Key Concept: Principle of Moments, Equilibrium

b) $20\, \text{cm}$ right of $O$
[Solution Description]
For equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments. Let $x$ be the distance from $O$ for the $80\, \text{gf}$ weight.
- Anticlockwise moment due to $40\, \text{gf}$: $40 \times 40 = 1600\, \text{gf cm}$.
- Clockwise moment due to $80\, \text{gf}$: $80 \times x$.
Setting them equal: $80x = 1600 \implies x = 20\, \text{cm}$.
So, the $80\, \text{gf}$ weight must be placed $20\, \text{cm}$ to the right of $O$.

Key Concept: Moment of Force Calculation

c) 20 Nm
[Solution Description]
The moment of force is calculated as:
$\text{Moment} = \text{Force} \times \text{Perpendicular Distance} = 10 \, \text{N} \times 2 \, \text{m} = 20 \, \text{N m}$

Key Concept: Moment of Force, Principle of Moments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a longer spanner handle reduces the force needed to loosen a nut. This is correct because the moment of force ($M$) is the product of the force ($F$) and the perpendicular distance ($r$) from the pivot point. Increasing $r$ (handle length) decreases the required $F$ for the same $M$, making it easier to turn the nut. The reason accurately describes the relationship between force, distance, and moment, correctly explaining the assertion.

Key Concept: Direction of rotation due to moment of force

c) The wheel rotates anticlockwise
[Solution Description]
The direction of rotation depends on the point of application of the force relative to the pivot. If the same force is applied at a diametrically opposite point (B), the moment arm reverses its direction, causing the wheel to rotate anticlockwise instead of clockwise.

Key Concept: Moment of Force, Couple

b) 3 Nm
[Solution Description]
The two equal and opposite forces form a couple. The moment of a couple is calculated as the product of one of the forces and the perpendicular distance between them. Here, the distance between the forces is equal to the length of the handle, 30 cm or 0.3 m. So, the net moment of force is:
$M = F \times d = 10 \text{ N} \times 0.3 \text{ m} = 3 \text{ Nm}$
The direction depends on the rotation produced, but the magnitude is 3 Nm.

Key Concept: Moment of Force

b) 10 N m
[Solution Description]
The moment of force (torque) is calculated as:
$\text{Moment} = \text{Force} \times \text{Perpendicular Distance} = 20 \, \text{N} \times 0.5 \, \text{m} = 10 \, \text{N m}.$

Key Concept: Moments and torque, spanner application

c) 600 N
[Solution Description]
To find the required force, we use the torque formula $\tau = F \times r$.
Given:
$\tau$ = 150 N$\cdot$m
$r$ = 25 cm = 0.25 m
Rearranging for $F$:
$F = \tau / r = 150 / 0.25 = 600$ N
Therefore, the minimum force needed is 600 N.

Key Concept: Principle of Moments

b) 20 Nm
[Solution Description]
The moment of force is calculated as:
$\text{Moment} = F \times r$
Here, $F = 25$ N and $r = 0.8$ m.
Substituting the values:
$\text{Moment} = 25 \times 0.8 = 20 \text{ Nm}$
Thus, the moment about the hinges is 20 Nm.

Key Concept: Door Handle

d) To maximize the moment of force for easier operation
[Solution Description]
Placing the handle far from the hinges maximizes the perpendicular distance ($d$) from the pivot point, allowing a small force ($F$) to produce a sufficient moment ($Moment = F \times d$) to open or close the door.

Key Concept: Principle of Moments

b) 10 Nm
[Solution Description]
The moment of force is calculated using the formula:
$\text{Moment} = F \times r$
where $F = 50$ N and $r = 0.2$ m.
Substituting the values:
$\text{Moment} = 50 \times 0.2 = 10 \text{ Nm}$
Therefore, the moment of force is 10 Nm.

Key Concept: Moment of Force, Equilibrium

b) 22.5 kg
[Solution Description]
For equilibrium, the sum of clockwise and anticlockwise moments must be equal. The first child's weight is $30 \times 10 = 300$ N, creating a moment of $300 \times 1.5 = 450$ Nm (clockwise). Let the second child's mass be $m$, so its weight is $10m$ N, creating a moment of $10m \times 2 = 20m$ Nm (anticlockwise). Set $20m = 450$ to get $m = 22.5$ kg.

Key Concept: Principle of Moments, Torque

c) 20 N
[Solution Description]
First, calculate the moment due to the 10 N force: $10 \times 1 = 10$ Nm (clockwise). Then, calculate the moment due to the 20 N force: $20 \times 1 = 20$ Nm (anticlockwise). The net unbalanced moment is $20 - 10 = 10$ Nm (anticlockwise). To balance this, we need an additional clockwise moment of 10 Nm. Let the required force be $F$. So, $F \times 0.5 = 10$, giving $F = 20$ N.

Key Concept: Basic Calculation of Moment of Force

b) 10 N m
[Solution Description]
Using the formula for moment of force:
$\text{Moment} = \text{Force} \times \text{Perpendicular distance}$
Given, Force = 20 N, Distance = 0.5 m
Moment = $20 \, \text{N} \times 0.5 \, \text{m} = 10 \, \text{N m}$

Key Concept: Unit Conversion

b) 9.8 N m
[Solution Description]
Given that $1 \, \text{kgf m} = 9.8 \, \text{N m}$.
Therefore, 1 kgf m is equal to 9.8 N m.

Key Concept: Static equilibrium, Principle of moments

d) Cannot be balanced within the ruler's length
[Solution Description]
For the meter rule to be in equilibrium, the sum of anticlockwise moments must equal the sum of clockwise moments about the pivot. The weight of the ruler acts at its center of mass (50 cm mark). Let the distance from the pivot to the 50 g weight be $x$ cm.
Anticlockwise moment due to the ruler: $100 \text{g} \times (50 - 40) \text{cm} = 1000 \text{g cm}$.
Clockwise moment due to 200 g weight: $200 \text{g} \times (40 - 10) \text{cm} = 6000 \text{g cm}$.
Total anticlockwise moment needed for equilibrium: $6000 \text{g cm}$.
Moment required from the 50 g weight: $6000 - 1000 = 5000 \text{g cm}$.
Thus, $50 \text{g} \times x = 5000 \text{g cm}$, so $x = 100 \text{cm}$.
Since the pivot is at 40 cm, the 50 g weight should be placed at $40 + 100 = 140$ cm, but since the ruler is only 100 cm long, this implies an error. Therefore, reconsider the direction:
Alternatively, place the 50 g weight on the other side: $(6000 + 1000) / 50 = 140$ cm from the pivot, which is again beyond the ruler's length. This suggests the initial setup may not be feasible, but recalculating:
Correct approach: The total anticlockwise moment is $1000 \text{g cm}$ (ruler) + $50 \text{g} \times d$ (additional weight). This must balance the clockwise moment of $6000 \text{g cm}$. So, $1000 + 50d = 6000$, giving $d = 100$ cm. Thus, the 50 g weight must be placed 100 cm from the pivot on the opposite side of the ruler, i.e., at the $40 + 100 = 140$ cm mark, which is not possible. Hence, the correct answer is that the 50 g weight cannot balance the system within the ruler's length.
However, assuming the question implies placing it on the other side, the correct position would be at the 90 cm mark (50 cm from pivot on the other side): $50 \times 50 = 2500$, total anticlockwise moment $1000 + 2500 = 3500$, which does not match. Therefore, revisiting:
Final correct calculation: To balance, $1000 + 50d = 6000$ ⇒ $d = 100$ cm. Since this exceeds the ruler's length, the correct answer is that it cannot be balanced with the given weights and positions.

Key Concept: Static and Dynamic Equilibrium

d) Assertion is false, but Reason is true.
[Solution Description]
The book lying on the table is not moving; hence, it is in static equilibrium. Therefore, the Assertion (A) is false. However, the Reason (R) is true because for a body to be in dynamic equilibrium, the net force acting on it must indeed be zero. Thus, Assertion is false but Reason is true.

Key Concept: Static Equilibrium

d) The weight of the book and the normal force are equal in magnitude and opposite in direction.
[Solution Description]
The book is in static equilibrium because the weight of the book (downward force) is balanced by the normal force (upward force) exerted by the table. There is no net force acting on the book, so it remains at rest.

Key Concept: Moment of couple, Rotational equilibrium

b) 6 N m
[Solution Description]
The moment of a couple is given by the product of one of the forces and the perpendicular distance between them.
Radius of the wheel: $r = \frac{30}{2} = 15 \text{cm} = 0.15 \text{m}$.
Distance between forces (diameter): $d = 0.3 \text{m}$.
Moment of couple: $M = F \times d = 20 \times 0.3 = 6 \text{N m}$.

Key Concept: Center of Gravity, Static Equilibrium

c) 45°
[Solution Description]
Step 1: Find center of gravity coordinates (x̄,ȳ):
Step 2: Area A1=1m² at (1,0.25), A2=0.75m² at (0.25,1.25)
Step 3: $x̄ = (1×1+0.75×0.25)/1.75 = 0.68m$
Step 4: $ȳ = (1×0.25+0.75×1.25)/1.75 = 0.68m$
Step 5: Toppling occurs when CG projects outside base support (2m wide)
Step 6: Critical angle θ when $x̄tanθ = ȳ$ → $0.68tanθ = 0.68$ → $θ = 45°$

Key Concept: Rotational Equilibrium, Principle of Moments

d) 294.3 N
[Solution Description]
Step 1: Total clockwise moment about hinge = $(20×9.81)×2 + (10×9.81)×2 = 588.6Nm$
Step 2: Anticlockwise moment due to rope = $T×4×\sin30° = 2T$
Step 3: Applying principle of moments: $2T = 588.6$ → $T = 294.3N$

Key Concept: Equilibrium Conditions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The book lying on the table remains at rest because the downward gravitational force (weight) is balanced by the upward normal force from the table, resulting in zero net force (translational equilibrium). Additionally, there is no rotational motion, so the net moment about any point is also zero (rotational equilibrium). Thus, both conditions for static equilibrium are satisfied, and the Reason correctly explains the Assertion.

Key Concept: Principle of Moments

c) 90 cm
[Solution Description]
For the metre rule to be in equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments. The pivot is at the 50 cm mark.
Let $x$ be the distance from the pivot where the 10 N weight should be placed.
Moment due to 20 N weight (clockwise):
$20 \times (50 - 30) = 20 \times 20 = 400 \, \text{Ncm}$
Moment due to 10 N weight (anticlockwise):
$10 \times x$
For equilibrium:
$10 \times x = 400$
$x = 40 \, \text{cm}$
Since it's measured from the pivot (50 cm mark), the 10 N weight should be placed at:
$50 + 40 = 90 \, \text{cm}$ mark.

Key Concept: Centrifugal Force

b) It is a fictitious force acting away from the center
[Solution Description]
Centrifugal force is not a real force but a fictitious force assumed by an observer in a rotating frame of reference. It acts away from the center of the circular path and has the same magnitude as the centripetal force.

Key Concept: Centripetal Force

b) Electrostatic force
[Solution Description]
The centripetal force required for an electron to move in a circular path around the nucleus is provided by the electrostatic force of attraction between the negatively charged electron and the positively charged nucleus.

Key Concept: Centrifugal force, Fictitious force

c) Centrifugal force
[Solution Description]
In the rotating frame of reference (merry-go-round), the person observes the object moving away from the center due to the centrifugal force, which is a fictitious force. This force appears to act outward because the observer is in a non-inertial frame.

Key Concept: Uniform Circular Motion

b) Because the direction of velocity continuously changes
[Solution Description]
Uniform circular motion is considered accelerated motion because although the speed remains constant, the direction of the velocity continuously changes. This change in direction implies a continuous change in velocity, which constitutes acceleration.