Class 10 Physics Chapter 1 Force

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: 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: Types of Motion Produced by Force

c) Translational Motion
[Solution Description]
When a force acts on a stationary rigid body free to move, the body begins moving in a straight path in the direction of the force. This type of motion is called translational motion.

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: Centripetal Force

c) 24 N
[Solution Description]
Centripetal force is given by:
$F = \frac{mv^2}{r}$
Given:
m = 2 kg,
v = 6 m/s,
r = 3 m
Substituting the values:
$F = \frac{2 \times 6^2}{3} = \frac{2 \times 36}{3} = \frac{72}{3} = 24 \text{ N}$

Key Concept: Rotational Motion

c) Rotational motion
[Solution Description]
If a body is pivoted at a point and a force is applied tangentially, it causes rotational motion around the pivot point. In this case, the wheel rotates about its center.

Key Concept: Rotational Motion, Moment of Force

b) 25 N m (anticlockwise)
[Solution Description]
First, determine the direction of each force's moment. The 10 N force at 0.5 m causes a clockwise rotation (negative moment). The 20 N force at 1.5 m causes an anticlockwise rotation (positive moment). Calculate each moment separately:
$\text{Moment due to 10 N} = -10 \, \text{N} \times 0.5 \, \text{m} = -5 \, \text{N m}$
$\text{Moment due to 20 N} = 20 \, \text{N} \times 1.5 \, \text{m} = 30 \, \text{N m}$
Net moment = sum of individual moments = $-5 + 30 = 25 \, \text{N m}$ (anticlockwise).

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 Calculation

b) 10 N m
[Solution Description]
The moment of force (torque) is calculated using the formula:
$\text{Moment} = \text{Force} \times \text{Perpendicular distance}$
Given:
$\text{Force} = 20 \, \text{N}, \quad \text{Distance} = 0.5 \, \text{m}$
Substituting the values:
$\text{Moment} = 20 \, \text{N} \times 0.5 \, \text{m} = 10 \, \text{N m}$

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

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: 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: Definition of Translational Motion

c) The motion of a body along a straight path due to an unbalanced force
[Solution Description]
Translational motion occurs when a body is free to move under the influence of an unbalanced force, resulting in movement along a straight path in the direction of the force.

Key Concept: Force and Linear Motion

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a book does not move when pushed gently due to the balance between the applied force and friction. This is true, as static friction prevents motion until the applied force exceeds the maximum static friction. The reason claims that translational motion requires a non-zero net force, which is correct based on Newton's first law of motion. Since the reason correctly explains why the book does not move (net force is zero), both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Principle of Moments, Rotational Motion

c) 60 kg
[Solution Description]
By the principle of moments, the clockwise moment equals the anticlockwise moment for equilibrium. The boy creates a moment of $40 \times 3 = 120$ kgm (perpendicular distance is half the seesaw length). Let the girl's mass be $m$. Her moment is $m \times 2$. So, $m \times 2 = 120$, which gives $m = 60$ kg.

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: Torque Calculation

c) 4.5 N m
[Solution Description]
The torque (\tau) is calculated using the formula:
$\tau = F \times d$
where:
$F = 15 \, \text{N}$,
$d = 0.3 \, \text{m}$.
$\tau = 15 \times 0.3 = 4.5 \, \text{N m}$
The torque produced is 4.5 N m.

Key Concept: Clockwise and Anticlockwise Moments

b) 20 Nm
[Solution Description]
To find the resultant moment of force:
Moment of force = Force $\times$ perpendicular distance ($F \times d$).
Here, $F = 10 \, N$, and $d = 2 \, m$ (radius).
The direction of rotation is anticlockwise since the force is applied tangentially.
Therefore, moment of force = $10 \times 2 = 20 \, N \, m$ (positive, as it is anticlockwise).

Key Concept: Moment of Force, Couple

b) 10 Nm clockwise
[Solution Description]
The net moment is calculated by summing the clockwise and anticlockwise moments.
- Moment due to force at 0.5 m: Since it is on one side of the pivot, let's assume it creates an anticlockwise moment (positive).
$\tau_1 = F \times d = 10 \, \text{N} \times 0.5 \, \text{m} = +5 \, \text{Nm}$.
- Moment due to force at 1.5 m: Applied in the opposite direction, so it creates a clockwise moment (negative).
$\tau_2 = -F \times d = -10 \, \text{N} \times 1.5 \, \text{m} = -15 \, \text{Nm}$.
Net moment: $\tau_{\text{net}} = \tau_1 + \tau_2 = 5 \, \text{Nm} - 15 \, \text{Nm} = -10 \, \text{Nm}$.
Thus, the net moment is 10 Nm in the clockwise direction.

Key Concept: Clockwise and Anticlockwise Moments

b) 18 Nm
[Solution Description]
The moment of force is calculated as $F \times d$.
Here, $F = 15 \, N$, and $d = 1.2 \, m$.
The direction depends on the side of application, but conventionally, opening the door creates an anticlockwise moment (positive).
Moment of force = $15 \times 1.2 = 18 \, N \, m$ (anticlockwise).

Key Concept: Magnitude of Force

c) 10 Nm
[Solution Description]
The formula for moment of force is given by $\text{Moment} = \text{Force} \times \text{Perpendicular distance}$. Here, Force = 5 N and Distance = 2 m. So, $\text{Moment} = 5 \, \text{N} \times 2 \, \text{m} = 10 \, \text{Nm}$.

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: Balancing moments

c) 30 cm
[Solution Description]
Let the distance from the pivot for the 10 N weight be $x$. To balance the moments: Clockwise moment = Anticlockwise moment. The 5 N weight creates a clockwise moment: $5 \, \text{N} \times (60 \, \text{cm} - 40 \, \text{cm}) = 5 \, \text{N} \times 20 \, \text{cm} = 100 \, \text{N\,cm}$. The 10 N weight must create an equal anticlockwise moment: $10 \, \text{N} \times (40 \, \text{cm} - x) = 100 \, \text{N\,cm}$. Solving for $x$: $40 \, \text{cm} - x = 10 \, \text{cm}$, so $x = 30 \, \text{cm}$. Thus, the 10 N weight should be placed at the 30 cm mark.

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, Torque Calculation

c) 60 N
[Solution Description]
The rod is balanced, so the sum of moments about the pivot is zero. Let the weight of the rod be $W$ acting at its center (2 m from either end). The pivot is 1.5 m from one end, so the distance of the rod's weight from the pivot is $2,m - 1.5,m = 0.5,m$. The 20 N weight is 1.5 m from the pivot. Applying the principle of moments:
$20,N \times 1.5,m = W \times 0.5,m \implies W = \frac{30}{0.5}\,N = 60,N$

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 metre rule remains balanced when the sum of clockwise moments equals the sum of anticlockwise moments. This is true because it aligns with the condition for rotational equilibrium as per the principle of moments. The reason correctly explains the assertion by stating the principle of moments, which requires the algebraic sum of moments to be zero for rotational equilibrium. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

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: Concept: Definition of Couple

c) Two equal and opposite parallel forces not acting along the same line
[Solution Description]
A couple consists of two equal and opposite parallel forces not acting along the same line, producing only rotational motion without any translational motion. Examples include turning a steering wheel or opening a door.

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: Moment of Couple

b) 3 N$\cdot$m
[Solution Description]
The moment of the couple is given by $\text{Moment} = F \times d$. Here, $F = 12 \, \text{N}$ and $d = 0.25 \, \text{m}$. Therefore, $\text{Moment} = 12 \times 0.25 = 3 \, \text{N} \cdot \text{m}$.

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 Couple, Perpendicular Distance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[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 two forces. This means it depends only on these two quantities. The reason states that the moment of a couple is independent of the point about which it is calculated because the resultant force is zero. This is true since a couple does not produce any translational motion and its rotational effect (moment) remains the same regardless of the reference point. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Centre of Gravity for Regular Objects

c) At a distance of $h/4$ from the base
[Solution Description]
For a solid cone, the centre of gravity is located along its axis. It is known that the C.G. of a solid cone lies at a height of $h/4$ from its base, where $h$ is the total height of the cone. This is derived from integrating the moments due to the weight distribution of the cone.

Key Concept: Centre of Gravity for Irregular Lamina

d) To accurately locate the single centre of gravity by intersecting plumb lines
[Solution Description]
Suspending the lamina from different points helps trace multiple vertical lines (plumb lines) on the lamina. The intersection point of these lines gives the centre of gravity because, in each case, the C.G. must lie vertically below the point of suspension. Using multiple points ensures accuracy in determining the exact location of the C.G.

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 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: Definition of Centre of Gravity

c) The point where the sum of moments due to weights of all particles is zero
[Solution Description]
The centre of gravity of a body is the point where the algebraic sum of moments due to weights of all particles about that point is zero. It can also be considered as the point where the entire weight of the body appears to act.

Key Concept: Definition of Centre of Gravity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The centre of gravity (C.G.) of a body is the point through which the resultant weight of the body acts. For a uniform circular ring, the mass is symmetrically distributed about its centre, making the C.G. coincide with its geometric centre. This property holds true for all symmetric bodies with uniform mass distribution. Therefore, both the Assertion and Reason are correct, and the Reason correctly explains the Assertion.

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, Deformation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The position of the centre of gravity depends on the distribution of mass in the body. For a uniform rod, the centre of gravity is at its midpoint. When the rod is bent into a semi-circular shape, the mass is redistributed along the arc of the semicircle. This redistribution causes the centre of gravity to shift from the original midpoint to a new position along the vertical axis of symmetry of the semicircle.
The reason correctly explains why the assertion is true because altering the shape of the body directly affects how mass is distributed, leading to a change in the position of the centre of gravity.

Key Concept: Center of gravity calculation

b) At R/2 distance from flat surface
[Solution Description]
For a hollow hemisphere, the center of gravity does not lie at the geometric center. It is calculated using the formula for centroid of a hemispherical shell which gives the distance from the flat surface as R/2. This shows the center of gravity lies halfway along the radius perpendicular to the flat surface, remaining within the hollow space of the hemisphere.

Key Concept: Centre of Gravity for Regular-Shaped Objects

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a circular disc, the mass is uniformly distributed, which means every part of the disc has equal mass per unit area. Due to this symmetry, the centre of gravity coincides with the geometric centre of the disc. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: C.G. of Regular-Shaped Objects

c) At the point of intersection of its diagonals
[Solution Description]
The C.G. of a rectangular lamina lies at the point of intersection of its diagonals.

Key Concept: C.G. of solid cone vs hollow cone

b) Solid cone's C.G. is higher than hollow cone
[Solution Description]
For a solid cone, C.G. is at $\frac{2h}{3}$ from vertex while for hollow cone it's at $\frac{h}{3}$ from vertex. This is because in hollow cone, more mass is concentrated near base making C.G. lower compared to solid cone where mass is evenly distributed.

Key Concept: Experimental determination of C.G. using a plumb line

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion describes the experimental method used to locate the centre of gravity (C.G.) of an irregular lamina by suspending it from different points and marking the plumb lines. The reason explains why this method works: when a body is freely suspended, it balances in such a way that its C.G. lies directly below the suspension point. Thus, the plumb line method relies on this principle for accurate determination of C.G.

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: Plumb Line Method

b) The marked line will not be truly vertical, causing incorrect C.G. determination
[Solution Description]
The plumb line method requires the lamina to come to rest so that the plumb line aligns vertically downward due to gravity. If the lamina is still moving, the line marked will not be truly vertical, leading to incorrect determination of the centre of gravity.

Key Concept: Static Equilibrium

b) A book lying on a table
[Solution Description]
Static equilibrium occurs when a body remains at rest under the influence of balanced forces. In the given examples, a book lying on a table shows static equilibrium because the downward force (weight of the book) is balanced by the upward normal force exerted by the table.

Key Concept: Dynamic Equilibrium

c) A raindrop falling with constant velocity
[Solution Description]
Dynamic equilibrium happens when a body moves with constant velocity (translational or rotational) under balanced forces. A rain drop falling with constant velocity is in dynamic equilibrium as the weight is balanced by air resistance and buoyant force.

Key Concept: Principle of Moments

c) Clockwise moments must equal anticlockwise moments
[Solution Description]
The principle of moments states that for a body to be in rotational equilibrium, the sum of clockwise moments about any pivot point must equal the sum of anticlockwise moments about the same point. Mathematically, $W_1 \times l_1 = W_2 \times l_2$.

Key Concept: Centre of Gravity

c) At the intersection of its diagonals.
[Solution Description]
The centre of gravity of a uniform body coincides with its geometric centre because the mass distribution is symmetrical. For a uniform rectangular block, the centre of gravity is at the intersection of its diagonals.

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: 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 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: 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: 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: 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: 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: Forces in equilibrium

c) The sum of clockwise moments equals the sum of anticlockwise moments.
[Solution Description]
For a body to be in rotational equilibrium, the sum of clockwise moments acting on it must be equal to the sum of anticlockwise moments ($\sum \text{Clockwise moments} = \sum \text{Anticlockwise moments}$). This ensures there is no net torque acting on the body.

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: Centre of gravity

c) The point where the entire weight of the body is assumed to act.
[Solution Description]
The centre of gravity is the point on an object where the entire weight of the body is assumed to act. It is the average location of the weight distribution.

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: Functions of Machines

c) To convert input energy into useful output work
[Solution Description]
One of the main functions of machines is to convert various forms of input energy (e.g., mechanical, electrical, thermal) into useful output energy or work to accomplish specific tasks efficiently.

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: 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: Function of Machine

b) Reduce workload by modifying force
[Solution Description]
Machines are designed to make work easier by altering the direction or magnitude of force applied. They help in reducing effort while increasing efficiency.

Key Concept: Couple in Rotation, Moment of Force

c) 200 N
[Solution Description]
Using the moment of force formula rearranged to solve for $F$:
$\tau = F \times r \implies F = \frac{\tau}{r}$
Here, $\tau = 60 \, \text{Nm}$ and $r = 0.3 \, \text{m}$. Thus:
$F = \frac{60}{0.3} = 200 \, \text{N}$

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: 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: 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: Moment of Force Calculation

c) 40 N m
[Solution Description]
Moment of force = Force × Perpendicular distance.
Given: $F = 8\, \text{N}$, $d = 5\, \text{m}$.
Calculation: $8 \times 5 = 40\, \text{N m}$ (anticlockwise).

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: 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: Couple and its application

b) It decreases the amount of force needed
[Solution Description]
A couple consists of two equal and opposite parallel forces that produce rotation. For a spanner, the torque $\tau$ is given by $\tau = F \times d$, where $F$ is the force applied and $d$ is the perpendicular distance (handle length). Increasing $d$ increases the torque for the same force, making the spanner more effective.

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: 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: 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: Steering Wheel

b) Decrease the moment of force needed to turn the wheel
[Solution Description]
A larger steering wheel increases the perpendicular distance from the axis of rotation, enabling the driver to turn the wheel with minimal effort using the formula $Moment = F \times r$.

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: Principle of Moments

b) 2 Nm
[Solution Description]
The radius of the steering wheel is half the diameter:
$r = \frac{40}{2} = 20 \text{ cm} = 0.2 \text{ m}$
Torque is calculated as:
$\text{Torque} = F \times r$
Substituting $F = 10$ N and $r = 0.2$ m:
$\text{Torque} = 10 \times 0.2 = 2 \text{ Nm}$
Hence, the torque produced is 2 Nm.

Key Concept: Moment of Force Calculation

b) 3 N$\cdot$m
[Solution Description]
To calculate the moment of force, we use the formula:
$\text{Moment of force} = F \times r$
Given:
$F = 15 \, \text{N}$,
$r = 20 \, \text{cm} = 0.2 \, \text{m}$.
Substituting the values:
$\text{Moment of force} = 15 \, \text{N} \times 0.2 \, \text{m} = 3 \, \text{N} \cdot \text{m}$

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: Torque, Vector Quantity

b) 16 Nm
[Solution Description]
Torque is given by the cross product of $\vec{r}$ and $\vec{F}$. Compute $\vec{\tau} = \vec{r} \times \vec{F} = (3\hat{i} + 4\hat{j}) \times (5\hat{i} + 12\hat{j})$. The cross product simplifies to $(3 \times 12) - (4 \times 5) = 36 - 20 = 16$ Nm in the $\hat{k}$ direction.

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]
To solve this problem, we apply the Principle of Moments, which states that for a system in equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments.
Step 1: Identify the given values.
- Clockwise moment (right side): Force = 50 N, Distance = 0.5 m
- Anticlockwise moment (left side): Force = W, Distance = 1 m
Step 2: Calculate the clockwise moment:
$\text{Clockwise moment} = 50 \, \text{N} \times 0.5 \, \text{m} = 25 \, \text{N}\cdot\text{m}$
Step 3: Equate the clockwise and anticlockwise moments to find W:
$25 \, \text{N}\cdot\text{m} = W \times 1 \, \text{m}$
Step 4: Solve for W:
$W = 25 \, \text{N}$
Conclusion: The system is in equilibrium as the condition $\text{Clockwise moment} = \text{Anticlockwise moment}$ is satisfied. The Principle of Moments correctly explains why the system remains balanced when the products of forces and their respective distances are equal on both sides of the pivot.

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: 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: Principle of Moments

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

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: 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: 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: Static Equilibrium

d) The net force on the book is zero.
[Solution Description]
The book is in static equilibrium because the weight of the book ($W$) acting downward is balanced by the normal reaction force ($N$) exerted by the table upward, i.e., $W = N$. Therefore, the resultant force on the book is zero.

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, Centrifugal Force, Dynamic Equilibrium

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that a satellite in a circular orbit experiences no net force due to the balance between centripetal and centrifugal forces. This is incorrect because the satellite is only under the influence of the gravitational force (centripetal force), and there is no real centrifugal force acting on it. The Reason states that the centrifugal force balances the centripetal force in a rotating frame, which is true, but the Assertion is false since the centrifugal force is fictitious and not a real force. Thus, the correct answer is that the Assertion is false, but the Reason is true.

Key Concept: Centripetal force, Circular motion, Acceleration

c) It becomes four times
[Solution Description]
The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{r}$. If the speed $v$ is doubled, the new acceleration becomes $a_c' = \frac{(2v)^2}{r} = \frac{4v^2}{r} = 4a_c$. Thus, the centripetal acceleration becomes four times the original value.

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 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.