Class 10 Physics Chapter 1 Force

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

b) 5 m s$^{-2}$
[Solution Description]
Using Newton's second law of motion, $F = ma$, where $F$ is the force, $m$ is the mass, and $a$ is the acceleration.
Given: $F = 10 \text{ N}$, $m = 2 \text{ kg}$.
To find acceleration ($a$), rearrange the formula:
$a = \frac{F}{m} = \frac{10}{2} = 5 \text{ m s}^{-2}$.
The acceleration produced is $5 \text{ m s}^{-2}$.

Key Concept: Translational Motion

b) Pushing a ball lying on a floor.
[Solution Description]
Translational motion occurs when a force acts on a stationary rigid body free to move, causing it to move in a straight path in the direction of the force.
Among the given options:
- Pushing a ball lying on a floor causes it to roll in a straight line, demonstrating translational motion.
- Other options involve rotational motion or no motion at all.

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: Force on Rigid and Non-rigid Bodies

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a force applied to a rigid body can only cause motion, which is true according to the syllabus. The reason states that a rigid body cannot change its shape or size when a force is applied, which is also true. Moreover, the reason correctly explains why the assertion is true because the inability of a rigid body to change its shape or size means force can only cause motion in it.

Key Concept: Mathematical Definition of Force

b) Rate of change of momentum
[Solution Description]
If mass m is constant, then $F = \frac{d(mv)}{dt}$ simplifies to $F = m \frac{dv}{dt}$, which is $F = ma$ where a is acceleration.

Key Concept: Principle of Moments

b) 20 N
[Solution Description]
According to the Principle of Moments, for equilibrium:
$\text{Sum of clockwise moments} = \text{Sum of anticlockwise moments}$
Let $W$ be the unknown weight. The clockwise moment is $30 \, \text{N} \times 2 \, \text{m} = 60 \, \text{N m}$. The anticlockwise moment is $W \times 3 \, \text{m}$. For equilibrium:
$W \times 3 = 60$
$W = \frac{60}{3} = 20 \, \text{N}$

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

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Assertion (A) states that a wheel pivoted at its centre will rotate when a force is applied tangentially to its rim, which is true because a tangential force creates rotational motion. Reason (R) states that the application of a tangential force produces a moment of force about the pivot point, which is also true and correctly explains why the wheel rotates (as moment of force causes rotation). Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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: Moment of Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the moment of force is determined by the product of the force and the perpendicular distance from the axis of rotation, which is correct according to the syllabus. The reason explains that increasing this distance enhances the turning effect, which directly supports the assertion as it aligns with the definition of moment of force. Therefore, both are true, and the reason correctly explains the assertion.

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 $\text{Moment} = F \times d_{\perp}$, where $F$ is the force and $d_{\perp}$ is the perpendicular distance from the axis of rotation to the line of action of the force. When the force is applied perpendicular to the line joining the point of application and the axis, $d_{\perp}$ is maximum because the entire distance contributes to the moment. Thus, the Assertion is true. The Reason states that the perpendicular distance determines the magnitude of the moment, which is also correct because the formula explicitly depends on $d_{\perp}$. Moreover, the Reason correctly explains why the Assertion is true since maximizing $d_{\perp}$ maximizes the moment.

Key Concept: Static Equilibrium, Conditions for Translational Motion

c) 10 N, upward
[Solution Description]
The book is in static equilibrium, meaning the net force acting on it is zero. The weight of the book acts downward due to gravity, and the normal reaction force from the table acts upward to balance it.
Given:
Weight ($W$) = 10 N (downward)
For equilibrium:
Normal reaction force ($N$) = Weight ($W$) = 10 N (upward)

Key Concept: Equilibrium Condition

b) When the net force acting on it is zero
[Solution Description]
A body is in static equilibrium when the net force acting on it is zero ($\sum F = 0$), meaning all forces balance each other out and no translational motion occurs.

Key Concept: Relation to Force and Motion, Units of Force

b) 19.6 m/s$^2$
[Solution Description]
First, convert the force from kgf to N:
$10 \text{ kgf} = 10 \times 9.8 \text{ N} = 98 \text{ N}$
Using Newton's second law, $F = ma$, we can find the acceleration ($a$):
$a = \frac{F}{m} = \frac{98}{5} = 19.6 \text{ m/s}^2$
Thus, the acceleration of the block is 19.6 m/s$^2$.

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: Couple

c) Opening a door by applying force at the handle
[Solution Description]
A couple consists of two equal and opposite parallel forces acting on a body but not along the same line, causing rotation without translation. Examples include opening a door or turning a wrench.

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: 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) -24 Nm
[Solution Description]
Each force acts at a distance of 2 m from the pivot (half the length of the rod).
The downward force creates a clockwise moment, while the upward force also creates a clockwise moment (as both are acting in opposite directions but symmetrically).
Total moment = $(6 \times 2) + (6 \times 2) = 12 + 12 = 24 \, N \, m$ (clockwise).
Since clockwise moment is negative, the resultant moment is $-24 , Nm$.

Key Concept: Equilibrium, Couple

b) 10 N upward
[Solution Description]
To balance the beam, the net moment about the pivot must be zero.
- Moment due to 8 N force: Distance from pivot = 1 m (left side), causing a clockwise moment.
$\tau_1 = -8 \, \text{N} \times 1 \, \text{m} = -8 \, \text{Nm}$.
- Moment due to 12 N force: Distance from pivot = 1 m (right side), causing a clockwise moment.
$\tau_2 = -12 \, \text{N} \times 1 \, \text{m} = -12 \, \text{Nm}$.
Total unbalanced moment: $-8 \, \text{Nm} - 12 \, \text{Nm} = -20 \, \text{Nm}$.
To balance, apply a force $F$ at the right end (distance = 2 m) to create an equal anticlockwise moment.
$+F \times 2 \, \text{m} = 20 \, \text{Nm} \Rightarrow F = 10 \, \text{N}$ upward.
Hence, a 10 N upward force is needed at the right end.

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: Moment of Force, Perpendicular Distance

b) 7.071 N m
[Solution Description]
To find the effective moment, we need the perpendicular component of the force relative to the wrench handle:
$F_{\perp} = F \times \sin(\theta) = 50 \, \text{N} \times \sin(45^\circ) = 50 \times 0.7071 = 35.355 \, \text{N}$
Now, calculate the moment:
$M = F_{\perp} \times d = 35.355 \, \text{N} \times 0.2 \, \text{m} = 7.071 \, \text{N m}$

Key Concept: Factors Affecting Moment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that increasing the perpendicular distance from the axis of rotation increases the moment of force, which is true based on the formula for moment of force: $\text{Moment of Force} = \text{Force} \times \text{Perpendicular Distance}$. The reason explains that the moment of force is directly proportional to the perpendicular distance, which correctly justifies the assertion because it describes the mathematical relationship in the formula.

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: Principle of Moments, Conditions for Equilibrium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a uniform meter rule pivoted at its center, if equal weights are suspended from both ends at equal distances from the pivot, the clockwise and anticlockwise moments are equal and opposite, resulting in a net moment of zero. According to the principle of moments, this satisfies the condition for rotational equilibrium ($\sum M = 0$). Thus, both Assertion and Reason are true, and Reason correctly explains why the rule remains horizontal.

Key Concept: Conditions for Equilibrium

c) The sum of anticlockwise moments must equal the sum of clockwise moments
[Solution Description]
For a body to be in rotational equilibrium, the algebraic sum of moments about any point must be zero. This means the sum of clockwise moments must equal the sum of anticlockwise moments.

Key Concept: Moment of Couple, Rotational Equilibrium

c) 25 N$\cdot$m
[Solution Description]
The moment of a couple is given by $M = F \times d$, where $F$ is the magnitude of either force and $d$ is the perpendicular distance between the forces. Here, $F = 50\ \text{N}$ and $d = 0.5\ \text{m}$.
Calculation: $M = 50\ \text{N} \times 0.5\ \text{m} = 25\ \text{N}\cdot\text{m}$.

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 Arm, Non-Perpendicular Forces

c) 13.86 N$\cdot$m
[Solution Description]
First, find the perpendicular component of the force contributing to the rotation: $F_{\perp} = F \sin(\theta)$. Here, $\theta = 60^\circ$ and $F = 40\ \text{N}$. Thus, $F_{\perp} = 40 \sin(60^\circ) = 40 \times 0.866 = 34.64\ \text{N}$.
Next, calculate the perpendicular distance (couple arm): $d = 0.4\ \text{m}$.
Moment of the couple: $M = F_{\perp} \times d = 34.64 \times 0.4 = 13.86\ \text{N}\cdot\text{m}$.

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, rotational equilibrium

c) 22 Nm
[Solution Description]
The net moment is the sum of the moments of the two couples. First couple: $M_1 = 5 \times 2 = 10$ Nm. Second couple: $M_2 = 3 \times 4 = 12$ Nm. Since both couples act in the same direction, the net moment is $M_1 + M_2 = 10 + 12 = 22$ Nm.

Key Concept: Moment of couple, static equilibrium

c) 12 Nm
[Solution Description]
The moment of the couple is calculated as $M = F \times d$, where $F = 15$ N and $d = 0.8$ m (distance from the hinge). Note that the reaction force at the hinge does not contribute to the moment since its distance is zero. Thus, $M = 15 \times 0.8 = 12$ Nm.

Key Concept: Centre of Gravity, Composite Bodies

c) Remains unchanged
[Solution Description]
The original C.G. of the square lamina is at its geometric centre. After removing the circular hole symmetrically about the centre, the C.G. remains unchanged because the hole is symmetrical and centrally positioned. Thus, there is no shift in the C.G.

Key Concept: Centre of Gravity Position

d) At the intersection of its diagonals
[Solution Description]
For a uniform rectangular lamina, the centre of gravity lies at the intersection of its diagonals.

Key Concept: Centre of Gravity Calculation

c) $\frac{2h}{3}$
[Solution Description]
The centre of gravity of a hollow cone of height $h$ is located at a distance $\frac{h}{3}$ from its base or $\frac{2h}{3}$ from its vertex.

Key Concept: Centre of Gravity Position

c) At a distance $\frac{2h}{3}$ from the vertex.
[Solution Description]
For a hollow cone of height $h$, the centre of gravity is not at the centroid of a solid cone. Instead, it is located along the axis of symmetry but closer to the base due to the hollow nature. Specifically, for a hollow cone, the centre of gravity lies at a distance $\frac{h}{3}$ from the base or $\frac{2h}{3}$ from the vertex. This is derived from the distribution of mass in a hollow cone.

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 Experiment

c) By suspending it from two different points and intersecting the plumb lines.
[Solution Description]
To determine the centre of gravity of an irregular lamina, suspend it freely from a point and allow it to come to rest. Draw a vertical line (plumb line) below the suspension point. Repeat this process by suspending the lamina from another point and drawing another plumb line. The intersection point of these two lines gives the centre of gravity. If more accuracy is needed, repeat the process with a third suspension point to confirm the location.

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: Definition of Centre of Gravity

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that for a uniform rectangular lamina, the C.G. lies at the intersection of its diagonals. This is true as per the syllabus examples. The reason states that for regular shaped objects, the C.G. coincides with the geometrical centre. This is also true as per the syllabus. However, the reason does not specifically explain why the C.G. of a rectangular lamina is at the intersection of diagonals - it just gives a general property. Therefore, both are true but reason is not the correct explanation of assertion.

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: 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: 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: Centre of Gravity, Uniform Circular Motion

c) Remains unchanged at the center
[Solution Description]
The centre of gravity of a uniform circular disc is always at its geometric center, regardless of its rotational speed, because mass distribution remains symmetric.

Key Concept: Principle of C.G.

c) Vertically below the point of suspension
[Solution Description]
When the lamina is suspended freely, it comes to rest such that its centre of gravity lies vertically below the point of suspension. This principle is used to locate the centre of gravity.

Key Concept: Intersection Point Significance

b) The common intersection point represents the position of the centre of gravity
[Solution Description]
The common intersection point of the lines drawn from different suspension points indicates the position where the centre of gravity lies. This is because the centre of gravity must lie vertically below each point of suspension when the lamina is at rest. If the lines do not intersect at a single point, the experiment has errors in alignment or execution.

Key Concept: Plumb line method

b) To mark the vertical line when the lamina comes to rest
[Solution Description]
The plumb line helps to mark the vertical line when the lamina comes to rest after suspension. According to the principle, the centre of gravity lies vertically below the point of suspension. Hence, the intersection of these lines from different suspension points gives the position of the centre of gravity.

Key Concept: Dynamic Equilibrium

d) The net force on the raindrop is zero, resulting in constant velocity.
[Solution Description]
The raindrop is in dynamic equilibrium because the net force acting on it is zero. The gravitational force (weight) is balanced by the sum of the buoyant force and air resistance (viscous drag). Since there is no net force, the raindrop moves with a constant velocity.

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, Center of Gravity

b) 57.7 N
[Solution Description]
For static equilibrium, the net forces and moments must balance.
1. Draw a free-body diagram:
- Weight $W = 20 \times 10 = 200$ N acts downward at the midpoint (2.5 m from the bottom).
- Normal force $N$ from the wall is horizontal (since the wall is smooth).
- Frictional force $f$ and normal force $R$ act upward and horizontally from the ground.
2. Take moments about the base of the ladder to eliminate $f$ and $R$:
Moment due to weight $W$: $200 \times (2.5 \cos 60°) = 200 \times 1.25 = 250$ Nm (clockwise).
Moment due to normal force $N$: $N \times 5 \sin 60° = N \times 4.33$ Nm (anticlockwise).
For equilibrium:
$\sum \tau = 0 \Rightarrow 250 - 4.33 N = 0$.
Solving: $N = \dfrac{250}{4.33} = 57.74$ N ≈ 57.7 N.
Thus, the horizontal reaction force by the wall is approximately 57.7 N.

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: Static and Dynamic Equilibrium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a book lying on a table is in static equilibrium because the net force acting on it is zero. This is correct as the weight of the book (W) is balanced by the normal reaction (N) from the table, making the net force zero.
The reason states that in static equilibrium, the body remains at rest, and the algebraic sum of moments of all forces about any point is also zero. This is also correct because static equilibrium requires both translational equilibrium (net force = 0) and rotational equilibrium (net moment = 0).
Moreover, the reason correctly explains the assertion since the condition of net force being zero is part of the criteria for static equilibrium.

Key Concept: Static Equilibrium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Static equilibrium occurs when a body remains at rest under the influence of balanced forces. In this case, the book's weight ($W$) acts downward due to gravity, and the table exerts an equal and opposite normal reaction force ($N$) upward. Since these two forces are equal and opposite, their sum is zero ($\sum F = 0$). Therefore, the book does not accelerate and remains in static equilibrium. The reason correctly explains why the assertion is true.

Key Concept: Dynamic Equilibrium

c) The net force on the raindrop is zero.
[Solution Description]
For the raindrop to be in dynamic equilibrium, the net force acting on it must be zero. The weight of the raindrop acting downward is balanced by the sum of the buoyant force and the air resistance (frictional force) acting upward.

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: 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: Dynamic Equilibrium

d) The sum of buoyant force and drag force equals the gravitational force.
[Solution Description]
In dynamic equilibrium, the net force acting on the object is zero because it moves with constant velocity. Here, the downward gravitational force (weight) is balanced by the upward buoyant force and air resistance (drag force). Thus, these forces cancel each other out, resulting in no acceleration.

Key Concept: Types of Equilibrium

d) Static equilibrium
[Solution Description]
A book lying stationary on a table has no net force or torque acting on it. The weight of the book is balanced by the normal reaction from the table, resulting in static equilibrium where there is no motion.

Key Concept: Conditions for Equilibrium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for a body to be in equilibrium, the resultant of all forces acting on it must be zero. This is correct as per the first condition of equilibrium \[\sum F = 0\]. The reason given is that this ensures no translational acceleration of the body, which is also correct and directly explains why the resultant force must be zero for equilibrium. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Moment of force, Principle of moments

b) 10 g
[Solution Description]
First, convert masses to forces: $F_1 = 20\,g \times g = 0.02\,kg \times 9.8\,m/s^2 = 0.196\,N$,
$F_{rule} = 100\,g \times g = 0.98\,N$ acts at 50 cm mark.
Calculate moments about fulcrum (40 cm):
Clockwise moment due to $F_1 = 0.196\,N \times (40-10)\,cm = 0.196 \times 0.3 = 0.0588\,Nm$.
Anticlockwise moment due to $F_{rule} = 0.98\,N \times (50-40)\,cm = 0.98 \times 0.1 = 0.098\,Nm$.
Net unbalanced anticlockwise moment = $0.098 - 0.0588 = 0.0392\,Nm$.
Let $W$ be the unknown weight in kg. The required clockwise moment from $W$ must balance this:
$W \times 9.8 \times (80-40)\,cm = 0.0392$ → $W \times 9.8 \times 0.4 = 0.0392$ → $W = 0.01\,kg = 10\,g$.

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: 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 door is easier to open when the force is applied at the handle rather than near the hinges, which is true because the handle is placed farther from the pivot (hinges), increasing the moment arm. The reason explains that the moment of force increases with an increase in the perpendicular distance from the pivot point, which is also true and directly explains why applying force at the handle makes it easier to open the door. Hence, both Assertion and Reason are true, and Reason correctly explains the Assertion.

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: Basic Definition

c) A device that aids in performing work by modifying force
[Solution Description]
A machine is any device that helps in performing work by either changing the direction or magnitude of force applied. It makes tasks easier by utilizing mechanical advantage.

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: Types of Machines

c) Computer
[Solution Description]
Simple machines include lever, pulley, wheel and axle, inclined plane, wedge, and screw. A computer is not a simple machine; it is an electronic device.

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: 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: Moment of Force, Principle of Moments

b) 100 Nm
[Solution Description]
To find the moment of force ($\tau$), we use the formula:
$\tau = F \times r$
where $F = 200 \, \text{N}$ and $r = 50 \, \text{cm} = 0.5 \, \text{m}$. Substituting the values:
$\tau = 200 \times 0.5 = 100 \, \text{Nm}$

Key Concept: Concept: Spanner and Nut

c) By generating a large moment of force with a small effort.
[Solution Description]
A long handle increases the perpendicular distance from the nut (pivot), allowing a small applied force at the end of the handle to generate a large enough moment of force to loosen or tighten the nut. Turning the spanner anticlockwise loosens the nut.

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, Resultant Moment

c) $0\, \text{N m}$
[Solution Description]
The moment of a force is calculated as: $M = F \times d$, where $F$ is the force and $d$ is the perpendicular distance from the pivot. Here, both forces are $10\, \text{N}$.
- For the upward force at $A$ (left end), distance from $O$ is $2\, \text{m}$. Moment: $10 \times 2 = 20\, \text{N m}$ (anticlockwise).
- For the downward force at $B$ (right end), distance from $O$ is $2\, \text{m}$. Moment: $10 \times 2 = 20\, \text{N m}$ (clockwise).
Resultant moment: $20\, \text{N m}$ (clockwise) $- 20\, \text{N m}$ (anticlockwise) = $0\, \text{N m}$.

Key Concept: Resultant Moment of Forces on a Pivoted Rod

c) 8 Nm (anticlockwise)
[Solution Description]
The moment due to the first force ($8 \, \text{N}$):
$M_1 = 8 \, \text{N} \times 2 \, \text{m} = 16 \, \text{N m} \ (\text{clockwise})$
The moment due to the second force ($12 \, \text{N}$):
$M_2 = 12 \, \text{N} \times 2 \, \text{m} = 24 \, \text{N m} \ (\text{anticlockwise})$
Resultant moment:
$M_{\text{resultant}} = M_2 - M_1 = 24 \, \text{N m} - 16 \, \text{N m} = 8 \, \text{N m} \ (\text{anticlockwise})$

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]
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: Moment of force and turning effect

c) Far from the hinge
[Solution Description]
The moment of force depends on both the magnitude of the force and its perpendicular distance from the pivot. For a door, the handle is placed far from the hinge to maximize the perpendicular distance, allowing a smaller force to produce the same moment. Applying the force near the hinge makes it harder to open the door.

Key Concept: Direction of Rotation

b) Anticlockwise
[Solution Description]
An anticlockwise moment causes the body to rotate in the anticlockwise direction. This is because the convention defines anticlockwise rotation as positive.

Key Concept: Moment of Force, Applications in Daily Life

b) 16 Nm
[Solution Description]
The moment of force is given by the product of the force and the perpendicular distance from the pivot. Here, the force is 20 N and the distance is 0.8 m. So, the moment of force is:
$M = F \times d = 20 \text{ N} \times 0.8 \text{ m} = 16 \text{ Nm}$
This moment acts anticlockwise if the force is applied outward.

Key Concept: Moment of Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a longer spanner makes it easier to loosen a tight nut, which is true because a longer handle increases the perpendicular distance ($r$) from the pivot (nut). According to the formula for moment of force $\tau = F \times r$, a larger $r$ means less force ($F$) is required to achieve the same torque ($\tau$). The Reason correctly explains this by stating that the moment of force increases with the perpendicular distance from the pivot point. Thus, both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

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: 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: Wheel and Axle, Moment of Force

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 system amplifies force by reducing the required input force, which is true as it follows the principle of moments. The reason explains this by stating that the moment of force at the wheel balances the moment due to the load on the axle, which is also correct. Moreover, the reason accurately justifies the assertion by explaining how the system works based on the principle of moments.

Key Concept: Couple concept, steering wheel

b) To create pure rotational motion without side effects
[Solution Description]
Applying two equal and opposite forces creates a couple which produces:
1. Pure rotational effect without translation
2. Net force remains zero while creating maximum torque
3. More efficient control over steering compared to single force application
The torque is calculated as $\tau = F \times d$, where $d$ is diameter

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: 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: Direction of Torque

b) Negative
[Solution Description]
Conventionally, if the effect on the body is to turn it clockwise, the moment of force is called the clockwise moment and is taken as negative.

Key Concept: Principle of Moments

d) 20 N
[Solution Description]
According to the principle of moments, for equilibrium:
$W_1 \times l_1 = W_2 \times l_2$
Given:
$W_1 = 10 \, \text{N}$, $l_1 = 40 \, \text{cm}$,
$l_2 = 20 \, \text{cm}$.
We need to find $W_2$.
Rearranging the equation:
$W_2 = \frac{W_1 \times l_1}{l_2} = \frac{10 \times 40}{20} = 20 \, \text{N}$

Key Concept: Static Equilibrium

c) A book lying on a table
[Solution Description]
A book lying on a table is in static equilibrium because the weight of the book is balanced by the normal force from the table, resulting in no net force and no motion.

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: Dynamic Equilibrium

b) A raindrop falling with a constant velocity
[Solution Description]
A raindrop falling with a constant velocity is in dynamic equilibrium because the gravitational force is balanced by air resistance, resulting in no net force but continued motion at a constant speed.

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: Translational Equilibrium

c) 50 N
[Solution Description]
The box is in translational equilibrium, so the net force acting on it is zero. The forces acting are:
Weight of the box ($W$) downward: $W = m \times g = 5 \times 10 = 50 \, \text{N}$
Normal force ($N$) upward balances the weight:
$N = W = 50 \, \text{N}$

Key Concept: Translational Equilibrium, Dynamic Equilibrium

c) 0.44
[Solution Description]
Step 1: Since velocity is constant, net force must be zero (dynamic equilibrium).
Step 2: Break applied force into components: $F_x = 200\cos30° = 173.2N$, $F_y = 200\sin30° = 100N$
Step 3: For vertical equilibrium: $N + F_y = mg$ → $N = 50×9.81 - 100 = 390.5N$
Step 4: For horizontal equilibrium: $F_x = f_k$ → $173.2 = μ_k N$ → $μ_k = 173.2/390.5 = 0.44$

Key Concept: Centre of Gravity

b) (3, 1.33)
[Solution Description]
The centre of gravity (C.G.) of a triangle is at the intersection of its medians. For a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, the C.G. coordinates are:
$\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Substituting the given vertices:
$\left( \frac{0 + 6 + 3}{3}, \frac{0 + 0 + 4}{3} \right) = (3, 1.\overline{3})$
So, the C.G. is at approximately (3, 1.33).

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: Centripetal force, Circular motion

c) Frictional force
[Solution Description]
The centripetal force required for circular motion is provided by the frictional force between the tires of the car and the road. This force acts towards the center of the circular path, allowing the car to maintain its circular motion.

Key Concept: Centripetal and Centrifugal Force

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that centrifugal force is a real force, which is false because according to the syllabus, centrifugal force is a fictitious force, not a real one. The Reason states that centrifugal force is experienced by an observer moving in a circular path along with the body, which is true as it explains why centrifugal force is considered (to account for the apparent motion observed from a rotating frame of reference). Therefore, Assertion is false, but Reason is true.

Key Concept: Uniform Circular Motion

c) It becomes four times
[Solution Description]
The centripetal force required to keep the stone moving in a circular path is given by $F = \frac{mv^2}{r}$. If the speed ($v$) is doubled while the mass ($m$) and radius ($r$) remain the same, the centripetal force becomes four times the original force. Since the tension in the string provides this force, the tension must also increase to four times its original value.

Key Concept: Centripetal and Centrifugal Force

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that centrifugal force is a real force, which is false because centrifugal force is a fictitious force observed in a rotating frame of reference. The Reason claims it is the reaction force to the centripetal force, which is also false because action-reaction pairs act on different bodies, and centrifugal force does not arise from any interaction but from the inertia of the body in a non-inertial frame. Thus, both the Assertion and Reason are false.