Q: Q2. Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative veloc

From the given list, the two scalar quantities are:\n\n1. **Work**\n2. **Current**\n\n**Explanation:**\n* **Work:** It is defined as the dot product of force and displacement (W = F ⋅ d), which results in a scalar quantity. It only has magnitude.\n* **Current:** Although current has a direction (flow of charge), it does not obey the laws of vector addition (e.g., triangle law or parallelogram law). Therefore, it is considered a scalar quantity. It only has magnitude.

Q: Q3. Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, c

From the given list, the only vector quantity is:\n\n**Impulse**\n\n**Explanation:**\n* **Impulse:** Impulse is defined as the product of force and the time interval for which the force acts (J = F ⋅ Δt). Since force is a vector quantity and the time interval is a scalar, their product (impulse) is a vector quantity, having both magnitude and the direction of the force. It represents the change in momentum of an object.\n\nAll other quantities listed (Temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, charge) are scalar quantities, possessing only magnitude.

Q: Q7. Read each statement below carefully and state, with reasons, if it is true or false:\n(a) The net acceleration of a particle in circular motion is always along the radius of the ci

(a) False.\n(b) False.\n(c) True.

Question 1

Q1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displ

Accepted Answer

Here's the classification of the given physical quantities as scalar or vector:

1.  **Volume:** Scalar
2.  **Mass:** Scalar
3.  **Speed:** Scalar
4.  **Acceleration:** Vector
5.  **Density:** Scalar
6.  **Number of moles:** Scalar
7.  **Velocity:** Vector
8.  **Angular frequency:** Scalar
9.  **Displacement:** Vector
10. **Angular velocity:** Vector

Question 2

Q2. Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative veloc

Accepted Answer

From the given list, the two scalar quantities are:

1.  **Work**
2.  **Current**

**Explanation:**
*   **Work:** It is defined as the dot product of force and displacement (W = F ⋅ d), which results in a scalar quantity. It only has magnitude.
*   **Current:** Although current has a direction (flow of charge), it does not obey the laws of vector addition (e.g., triangle law or parallelogram law). Therefore, it is considered a scalar quantity. It only has magnitude.

Question 3

Q3. Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, c

Accepted Answer

From the given list, the only vector quantity is:

**Impulse**

**Explanation:**
*   **Impulse:** Impulse is defined as the product of force and the time interval for which the force acts (J = F ⋅ Δt). Since force is a vector quantity and the time interval is a scalar, their product (impulse) is a vector quantity, having both magnitude and the direction of the force. It represents the change in momentum of an object.

All other quantities listed (Temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, charge) are scalar quantities, possessing only magnitude.

Question 4

Q4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars
(b) adding a scalar to a vect

Accepted Answer

Let's analyze each algebraic operation with reasons:

### (a) Adding any two scalars
**Meaningful.**

**Reason:** Scalars are quantities that have only magnitude. They can be added algebraically using ordinary rules of arithmetic, provided they represent the same physical quantity (i.e., have the same dimensions). For example, adding 5 kg of mass to 3 kg of mass gives 8 kg of mass. Adding 10 J of energy to 5 J of energy gives 15 J of energy. It is meaningless to add a scalar of mass to a scalar of time, as they represent different physical quantities.

### (b) Adding a scalar to a vector of the same dimensions
**Not meaningful.**

**Reason:** A scalar quantity has only magnitude, while a vector quantity has both magnitude and direction. These are fundamentally different types of physical quantities. You cannot add a quantity that has only magnitude to a quantity that has both magnitude and direction, even if they have the same dimensions. For example, you cannot add 5 m (a scalar distance) to a displacement of 5 m East (a vector). Vector addition follows specific rules (triangle law, parallelogram law) that scalars do not obey.

### (c) Multiplying any vector by any scalar
**Meaningful.**

**Reason:** When a vector is multiplied by a scalar, the result is a new vector. The magnitude of the new vector is the product of the magnitude of the original vector and the scalar. The direction of the new vector is the same as the original vector if the scalar is positive, and opposite if the scalar is negative. This operation changes the scale (magnitude) of the vector. For example, if a vector **A** is multiplied by a scalar 'k', the result is k**A**. If **A** represents velocity, and 'k' represents mass, then k**A** represents momentum (mass × velocity), which is a vector.

### (d) Multiplying any two scalars
**Meaningful.**

**Reason:** Scalars can be multiplied using ordinary rules of arithmetic. The product of two scalars is always a scalar. For example, multiplying speed (scalar) by time (scalar) gives distance (scalar). Multiplying mass (scalar) by density (scalar) is not a standard physical operation, but mathematically, their product is a scalar. The dimensions of the resulting scalar will be the product of the dimensions of the individual scalars.

### (e) Adding any two vectors
**Meaningful.**

**Reason:** Vectors can be added, provided they represent the same physical quantity (i.e., have the same dimensions). Vector addition follows specific rules like the triangle law or parallelogram law of vector addition, which account for both magnitude and direction. For example, adding two displacement vectors gives a resultant displacement vector. Adding two force vectors gives a resultant force vector. It is meaningless to add a force vector to a velocity vector, as they are different physical quantities.

### (f) Adding a component of a vector to the same vector
**Meaningful.**

**Reason:** A component of a vector is itself a vector (e.g., the x-component of a force vector is a force vector along the x-axis). Since both the original vector and its component are vectors of the same physical quantity (and thus same dimensions), they can be added using vector addition rules. For example, if **A** is a vector and **A**x is its x-component, then **A** + **A**x is a meaningful vector sum.

Question 5

Q5. State with reasons, whether the following statements are true or false:
(a) The magnitude of a vector is always a scalar.
(b) Each component of a vector is always a scalar.
(c)

Accepted Answer

Let's evaluate each statement with reasons:

### (a) The magnitude of a vector is always a scalar.
**True.**

**Reason:** A vector quantity is defined by both its magnitude and direction. The magnitude of a vector is a numerical value that represents its 'size' or 'strength' and is always a positive real number. It does not have any associated direction. By definition, a quantity that has only magnitude and no direction is a scalar. For example, if a velocity vector is 5 m/s East, its magnitude is 5 m/s, which is a scalar quantity.

### (b) Each component of a vector is always a scalar.
**False.**

**Reason:** A component of a vector is itself a vector. For example, if a force vector **F** has components **F**x and **F**y along the x and y axes respectively, then **F**x and **F**y are also vectors. They have both magnitude (e.g., Fx) and direction (along the x-axis or y-axis). The *magnitude* of a component (e.g., Fx, Fy) is a scalar, but the component itself (e.g., Fx**i**, Fy**j**) is a vector.

### (c) The total path length is always equal to the magnitude of the displacement vector of a particle.
**False.**

**Reason:**
*   **Total path length (or distance):** This is the actual length of the path covered by a particle, irrespective of its direction. It is a scalar quantity and is always positive or zero.
*   **Magnitude of displacement:** This is the shortest straight-line distance between the initial and final positions of the particle. It is the magnitude of the displacement vector.

The total path length is equal to the magnitude of the displacement only when the particle moves in a straight line without changing its direction. In all other cases (e.g., curved path, path with change in direction, or returning to the starting point), the total path length will be greater than the magnitude of the displacement. If the particle returns to its starting point, the displacement is zero, but the path length is non-zero.

### (d) The average speed of a particle is always greater than or equal to the magnitude of the average velocity of the particle.
**True.**

**Reason:**
*   **Average speed:** Defined as (Total path length) / (Total time taken).
*   **Magnitude of average velocity:** Defined as (Magnitude of displacement) / (Total time taken).

From statement (c), we know that (Total path length) ≥ (Magnitude of displacement). Since the total time taken is always positive and the same for both, it follows that:
(Total path length / Total time taken) ≥ (Magnitude of displacement / Total time taken)
Therefore, Average speed ≥ Magnitude of average velocity.
Equality holds only when the particle moves in a straight line without changing direction.

### (e) Three vectors not lying in a plane can never add up to give a null vector.
**False.**

**Reason:** For three vectors to add up to a null vector (zero resultant vector), they must form a closed triangle (if coplanar) or a closed polygon in space. If three vectors **A**, **B**, and **C** are such that **A** + **B** + **C** = **0**, then **A** + **B** = -**C**. This means that the resultant of **A** and **B** must be equal in magnitude and opposite in direction to **C**. The resultant of two vectors (**A** + **B**) always lies in the plane containing **A** and **B**. If **C** does not lie in the plane of **A** and **B**, then -**C** also does not lie in that plane. Therefore, it is impossible for **A** + **B** (which is in the plane of **A** and **B**) to be equal to -**C** (which is not in that plane). Hence, three non-coplanar vectors can never add up to give a null vector. The statement says they *can never* add up to give a null vector, which is true. So the statement is true. My initial thought process was incorrect. Let me re-evaluate this point carefully.

**Re-evaluation for (e):**
If three vectors **A**, **B**, and **C** are non-coplanar, it means they do not lie in the same plane. For their sum to be a null vector (**A** + **B** + **C** = **0**), it implies that **A** + **B** = -**C**. The vector **A** + **B** is the resultant of **A** and **B**, and it must lie in the plane formed by **A** and **B**. For **A** + **B** to be equal to -**C**, the vector -**C** must also lie in the same plane as **A** and **B**. However, if **C** is non-coplanar with **A** and **B**, then -**C** will also be non-coplanar with **A** and **B**. Therefore, it is impossible for **A** + **B** (which is coplanar with **A** and **B**) to be equal to -**C** (which is non-coplanar with **A** and **B**). Hence, three non-coplanar vectors can never add up to give a null vector. The statement is **True**.

My apologies for the momentary confusion. The statement is indeed true. Three vectors not lying in a plane can never add up to give a null vector. This is a fundamental property of non-coplanar vectors.

Question 6

Q6. A particle moves in a circle of radius 10 cm with a constant speed of 10 m/s. What is the magnitude of the acceleration of the particle?

Accepted Answer

The magnitude of the acceleration of the particle is 1000 m/s².

Question 7

Q7. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the ci

Accepted Answer

(a) False.
(b) False.
(c) True.

Question 8

Q8. The position of a particle is given by $\mathbf{r} = 3.0t\hat{\mathbf{i}} - 2.0t^2\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}$ m, where $t$ is in seconds and the coefficients have the p

Accepted Answer

The position vector of the particle is given by $\mathbf{r} = 3.0t\hat{\mathbf{i}} - 2.0t^2\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}$ m.

**(a) Velocity and Acceleration:**
To find the velocity, we differentiate the position vector with respect to time:
$\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(3.0t\hat{\mathbf{i}} - 2.0t^2\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}})$ 
$\mathbf{v} = 3.0\frac{d}{dt}(t)\hat{\mathbf{i}} - 2.0\frac{d}{dt}(t^2)\hat{\mathbf{j}} + 4.0\frac{d}{dt}(1)\hat{\mathbf{k}}$ 
$\mathbf{v} = 3.0(1)\hat{\mathbf{i}} - 2.0(2t)\hat{\mathbf{j}} + 0\hat{\mathbf{k}}$ 
$\mathbf{v} = (3.0\hat{\mathbf{i}} - 4.0t\hat{\mathbf{j}})$ m/s

To find the acceleration, we differentiate the velocity vector with respect to time:
$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3.0\hat{\mathbf{i}} - 4.0t\hat{\mathbf{j}})$ 
$\mathbf{a} = 3.0\frac{d}{dt}(1)\hat{\mathbf{i}} - 4.0\frac{d}{dt}(t)\hat{\mathbf{j}}$ 
$\mathbf{a} = 0\hat{\mathbf{i}} - 4.0(1)\hat{\mathbf{j}}$ 
$\mathbf{a} = (-4.0\hat{\mathbf{j}})$ m/s$^2$

**(b) Magnitude and Direction of Velocity at $t = 2.0$ s:**
First, substitute $t = 2.0$ s into the velocity expression:
$\mathbf{v}(t=2.0	ext{ s}) = (3.0\hat{\mathbf{i}} - 4.0(2.0)\hat{\mathbf{j}})$ 
$\mathbf{v}(t=2.0	ext{ s}) = (3.0\hat{\mathbf{i}} - 8.0\hat{\mathbf{j}})$ m/s

Magnitude of velocity:
$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$ 
$|\mathbf{v}| = \sqrt{(3.0)^2 + (-8.0)^2}$ 
$|\mathbf{v}| = \sqrt{9.0 + 64.0}$ 
$|\mathbf{v}| = \sqrt{73.0}$ 
$|\mathbf{v}| \approx 8.54$ m/s

Direction of velocity:
Let $	heta$ be the angle the velocity vector makes with the positive $x$-axis.
$	an	heta = \frac{v_y}{v_x} = \frac{-8.0}{3.0} = -2.667$
Since $v_x$ is positive and $v_y$ is negative, the vector lies in the fourth quadrant.
$	heta = \arctan(-2.667) \approx -69.44^\circ$
So, the velocity vector makes an angle of approximately $69.44^\circ$ below the positive $x$-axis.

Question 9

Q9. A particle starts from the origin at $t = 0$ with a velocity of $10.0\hat{\mathbf{j}}$ m/s and moves in the $xy$-plane with a constant acceleration of $(8.0\hat{\mathbf{i}} + 2.0\h

Accepted Answer

Given:
Initial position $\mathbf{r}_0 = 0\hat{\mathbf{i}} + 0\hat{\mathbf{j}}$ (starts from origin)
Initial velocity $\mathbf{u} = 10.0\hat{\mathbf{j}}$ m/s
Constant acceleration $\mathbf{a} = (8.0\hat{\mathbf{i}} + 2.0\hat{\mathbf{j}})$ m/s$^2$

We use the kinematic equation for position: $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$
Since $\mathbf{r}_0 = 0$, we have $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$

Substitute the given values:
$\mathbf{r} = (10.0\hat{\mathbf{j}})t + \frac{1}{2}(8.0\hat{\mathbf{i}} + 2.0\hat{\mathbf{j}})t^2$
$\mathbf{r} = 10.0t\hat{\mathbf{j}} + (4.0t^2\hat{\mathbf{i}} + 1.0t^2\hat{\mathbf{j}})$ 
$\mathbf{r} = (4.0t^2)\hat{\mathbf{i}} + (10.0t + 1.0t^2)\hat{\mathbf{j}}$

From this, the $x$-coordinate is $x(t) = 4.0t^2$ and the $y$-coordinate is $y(t) = 10.0t + 1.0t^2$.

**(a) Time when $x$-coordinate is 16 m:**
We are given $x(t) = 16$ m.
$16 = 4.0t^2$
$t^2 = \frac{16}{4.0} = 4.0$
$t = \sqrt{4.0} = 2.0$ s (time cannot be negative)

**(b) $y$-coordinate at that time:**
Substitute $t = 2.0$ s into the expression for $y(t)$:
$y(2.0) = 10.0(2.0) + 1.0(2.0)^2$
$y(2.0) = 20.0 + 1.0(4.0)$
$y(2.0) = 20.0 + 4.0$
$y(2.0) = 24.0$ m

**(c) Speed of the particle at that time:**
First, find the velocity vector using the kinematic equation: $\mathbf{v} = \mathbf{u} + \mathbf{a}t$
$\mathbf{v} = (10.0\hat{\mathbf{j}}) + (8.0\hat{\mathbf{i}} + 2.0\hat{\mathbf{j}})t$
$\mathbf{v} = (8.0t)\hat{\mathbf{i}} + (10.0 + 2.0t)\hat{\mathbf{j}}$

Now, substitute $t = 2.0$ s into the velocity expression:
$\mathbf{v}(2.0) = (8.0(2.0))\hat{\mathbf{i}} + (10.0 + 2.0(2.0))\hat{\mathbf{j}}$
$\mathbf{v}(2.0) = (16.0)\hat{\mathbf{i}} + (10.0 + 4.0)\hat{\mathbf{j}}$
$\mathbf{v}(2.0) = (16.0\hat{\mathbf{i}} + 14.0\hat{\mathbf{j}})$ m/s

The speed is the magnitude of the velocity vector:
Speed $= |\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$
Speed $= \sqrt{(16.0)^2 + (14.0)^2}$
Speed $= \sqrt{256.0 + 196.0}$
Speed $= \sqrt{452.0}$
Speed $\approx 21.26$ m/s

Question 10

Q10. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s e

Accepted Answer

The net acceleration of the cyclist is the vector sum of the tangential acceleration and the centripetal acceleration. The tangential acceleration is given as 0.50 m/s² (opposite to velocity, as speed is reducing). The centripetal acceleration is calculated using the instantaneous speed. The magnitude of the net acceleration is 0.86 m/s² and its direction is at an angle of approximately 35.5° with the radius vector (towards the center of the circle) or 54.5° with the tangent (opposite to the direction of motion).

Question 11

Q11. A particle is projected in the XY-plane with a speed of 𝑣₀ at an angle of 𝜃 with the X-axis. Show that its trajectory is a parabola. (You can ignore air resistance).

Accepted Answer

To show that the trajectory of a projectile is a parabola, we derive the equations of motion for its horizontal and vertical components and then eliminate time to obtain an equation relating y and x. This equation will be in the form of a parabola.

Question 12

Q12. A particle is projected with a velocity 𝑣₀ such that its range R is maximum. If the maximum height attained by the particle is H, show that R = 4H.

Accepted Answer

For the range R of a projectile to be maximum, the angle of projection (𝜃) must be 45°. Using this condition, we can derive the expressions for maximum range (R_max) and maximum height (H) and then show that R_max = 4H.

Question 13

Q13. For a projectile, the range R and maximum height H are related by R = 4H tan θ, where θ is the angle of projection. Show that the angle of projection for which the horizontal range

Accepted Answer

Given the relation R = 4H tan 𝜃, we need to find the angle of projection 𝜃 for which R = H. By substituting R = H into the given relation, we can directly solve for tan 𝜃.

Question 14

Q14. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and directi

Accepted Answer

The magnitude of the acceleration of the stone is approximately 9.91 m/s², and its direction is always towards the center of the horizontal circle.

Question 15

Q15. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Accepted Answer

The centripetal acceleration of the aircraft is approximately 62.5 m/s². When compared with the acceleration due to gravity (g = 9.8 m/s²), the centripetal acceleration is about 6.38 times 'g'.

Question 16

Q16. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the ci

Accepted Answer

Here are the evaluations for each statement with reasons:

**(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.**
**False.**

**Reason:** This statement is only true for uniform circular motion, where the speed of the particle is constant. In uniform circular motion, there is only centripetal acceleration (a_c), which is directed towards the center along the radius. However, if the circular motion is non-uniform (i.e., the speed of the particle is changing), there is also a tangential acceleration (a_t) component, which is tangent to the circle and either in the direction of velocity (if speed is increasing) or opposite to it (if speed is decreasing). In non-uniform circular motion, the net acceleration is the vector sum of the centripetal and tangential accelerations (a_net = a_c + a_t). Therefore, the net acceleration is generally not directed along the radius towards the center, but rather at an angle to the radius.

**(b) The velocity vector of a particle at a point is always perpendicular to the acceleration vector at that point.**
**False.**

**Reason:** This statement is true only for uniform circular motion. In uniform circular motion, the velocity vector is tangential to the circle, and the acceleration vector (centripetal acceleration) is directed radially inwards, making them perpendicular to each other. However, in non-uniform circular motion, there is a tangential component of acceleration (a_t) in addition to the centripetal component (a_c). The tangential acceleration is either parallel or anti-parallel to the velocity vector. The net acceleration (a_net = a_c + a_t) will therefore have a component along the direction of velocity (a_t). This means the angle between the net acceleration vector and the velocity vector will not be 90 degrees (unless a_t = 0, which brings us back to uniform circular motion). For example, in projectile motion, at the highest point, the vertical component of velocity is zero, and acceleration (g) is vertically downwards, while velocity is horizontal, making them perpendicular. But at any other point, the velocity has both horizontal and vertical components, and the acceleration (g) is purely vertical, so they are not perpendicular.

**(c) The work done by a centripetal force is always zero.**
**True.**

**Reason:** Work done by a force is defined as W = F ⋅ d = F d cosθ, where F is the force, d is the displacement, and θ is the angle between the force and displacement vectors. In circular motion, the centripetal force (F_c) is always directed towards the center of the circle, i.e., along the radius. The instantaneous displacement (or velocity) vector of the particle is always tangential to the circle. Since the radius is always perpendicular to the tangent at any point on a circle, the angle (θ) between the centripetal force vector and the instantaneous displacement vector is always 90 degrees. As cos(90°) = 0, the work done by the centripetal force (W_c = F_c * d * cos 90°) is always zero. This is consistent with the fact that centripetal force only changes the direction of velocity, not its magnitude (in uniform circular motion), and thus does no work to change the kinetic energy of the particle.

Question 17

Q17. A stone is dropped from a height of 490 m and simultaneously another stone is projected horizontally with a velocity of 10 m/s from the same height. Find the time taken by each sto

Accepted Answer

Both stones will take 10 seconds to reach the ground. The distance of separation between the two stones when they reach the ground will be 100 meters.

Question 18

Q18. A cricket ball is thrown at a speed of 28 m s–1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same leve

Accepted Answer

Given:
Initial speed, u = 28 m/s
Angle of projection, θ = 30°
Acceleration due to gravity, g = 9.8 m/s²

(a) Maximum height (H):
The maximum height reached by a projectile is given by the formula:
H = (u² sin²θ) / (2g)

(b) Time taken by the ball to return to the same level (Time of flight, T):
The time of flight for a projectile is given by the formula:
T = (2u sinθ) / g

(c) Distance from the thrower to the point where the ball returns to the same level (Horizontal Range, R):
The horizontal range of a projectile is given by the formula:
R = (u² sin2θ) / g

Question 19

Q19. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and directi

Accepted Answer

Given:
Length of the string, r = 80 cm = 0.80 m
Number of revolutions, N = 14
Time taken, t = 25 s

First, we need to calculate the frequency (f) or the angular speed (ω) of the stone.
Frequency (f) = Number of revolutions / Time taken
f = N / t

Angular speed (ω) = 2πf

Then, we can calculate the linear speed (v) of the stone:
v = rω

Finally, the magnitude of the centripetal acceleration (a_c) is given by:
a_c = v²/r  or  a_c = rω²

The direction of the centripetal acceleration is always towards the center of the circular path.

Question 20

Q20. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Accepted Answer

Given:
Radius of the horizontal loop, r = 1.00 km
Speed of the aircraft, v = 900 km/h
Acceleration due to gravity, g = 9.8 m/s²

First, convert the given units to SI units (meters and seconds).
Then, calculate the centripetal acceleration (a_c) using the formula:
a_c = v²/r

Finally, compare this centripetal acceleration with the acceleration due to gravity (g) by finding the ratio a_c / g.

Question 21

Q21. An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of

Accepted Answer

Given:
Radius of the circular groove, r = 12 cm
Number of revolutions, N = 7
Time taken, t = 100 s

(a) To find angular speed (ω) and linear speed (v):
First, convert radius to meters. Calculate the frequency (f) or period (T).
Frequency, f = N / t
Angular speed, ω = 2πf
Linear speed, v = rω

(b) To determine if the acceleration vector is constant and find its magnitude:
In uniform circular motion, the magnitude of centripetal acceleration is constant, but its direction continuously changes. Therefore, the acceleration vector is not a constant vector.
The magnitude of acceleration (centripetal acceleration, a_c) is given by:
a_c = rω²

Question 22

Q22. A particle starts from the origin at t = 0 with a velocity of 5.0 i m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0 i + 2.0 j) m/s

Accepted Answer

Given:
Initial position, r₀ = 0 (origin)
Initial velocity, v₀ = 5.0 i m/s
Constant acceleration, a = (3.0 i + 2.0 j) m/s²

We can use the kinematic equation for position:
r = r₀ + v₀t + (1/2)at²
Since r₀ = 0, the position vector at time t is:
r(t) = v₀t + (1/2)at²

Separating into x and y components:
x(t) = v₀ₓt + (1/2)aₓt²
y(t) = v₀ᵧt + (1/2)aᵧt²

From the given initial velocity, v₀ₓ = 5.0 m/s and v₀ᵧ = 0 m/s.
From the given acceleration, aₓ = 3.0 m/s² and aᵧ = 2.0 m/s².

(a) Find the time (t) when x-coordinate is 84 m, then use this time to find the y-coordinate.
(b) Find the velocity vector v(t) at that time, then calculate its magnitude (speed). The kinematic equation for velocity is:
v = v₀ + at

Question 23

Q23. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and directi

Accepted Answer

The magnitude of the acceleration of the stone is 9.9 m/s², and its direction is always towards the center of the circular path.

Question 24

Q24. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Accepted Answer

The centripetal acceleration of the aircraft is 6.25 times the acceleration due to gravity.

Question 25

Q25. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the ci

Accepted Answer

Here's a careful analysis of each statement with reasons:

**(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.**

**False.**

**Reason:** This statement is only true for *uniform circular motion*. In uniform circular motion, the speed of the particle is constant, and thus there is no tangential acceleration. The only acceleration present is the centripetal acceleration, which is directed radially inwards towards the center of the circle.

However, in *non-uniform circular motion*, the speed of the particle changes. In this case, in addition to the centripetal acceleration (a_c) directed towards the center, there is also a tangential acceleration (a_t) directed along the tangent to the path, either in the direction of velocity (if speed is increasing) or opposite to it (if speed is decreasing). The net acceleration (a_net) is the vector sum of the centripetal and tangential accelerations (a_net = a_c + a_t). Therefore, the net acceleration is generally not directed along the radius towards the center; it makes an angle with the radius unless a_t is zero.

**(b) The velocity vector of a particle at a point is always perpendicular to the acceleration vector at that point.**

**False.**

**Reason:** This statement is true only for *uniform circular motion*. In uniform circular motion, the velocity vector is tangential to the circle, and the acceleration vector (centripetal acceleration) is directed radially inwards, making them perpendicular to each other at every point.

However, in *non-uniform circular motion*, as explained above, there is a tangential component of acceleration (a_t) in addition to the centripetal component (a_c). The velocity vector is always tangential. The net acceleration vector (a_net = a_c + a_t) will have a component along the tangent (a_t) and a component perpendicular to the tangent (a_c). Since a_t is parallel or anti-parallel to the velocity vector, the net acceleration vector will not be perpendicular to the velocity vector. The angle between the velocity vector and the net acceleration vector will be less than 90° if the speed is increasing (a_t in same direction as v) and greater than 90° if the speed is decreasing (a_t opposite to v).

**(c) The work done by a centripetal force in moving a body along a circular path is always zero.**

**True.**

**Reason:** Work done by a force is defined as W = F ⋅ d = Fd cosθ, where F is the force, d is the displacement, and θ is the angle between the force and displacement vectors.

In circular motion, the centripetal force (F_c) is always directed towards the center of the circle, i.e., along the radius. The instantaneous displacement (ds) of the particle is always along the tangent to the circular path. Since the radius is always perpendicular to the tangent at any point on a circle, the centripetal force is always perpendicular to the instantaneous displacement (θ = 90°).

Therefore, the work done by the centripetal force over any infinitesimal displacement ds is dW = F_c ⋅ ds = F_c ds cos(90°) = 0. Consequently, the total work done by the centripetal force over any finite path along the circle is also zero.

Question 26

Q26. A stone is dropped from a height h. Simultaneously, another stone is thrown horizontally from the same height h with a speed of √2gh. Which stone reaches the ground earlier?
(a) T

Accepted Answer

The correct option is (c) Both reach simultaneously.

Question 27

Q27. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and directi

Accepted Answer

The stone is undergoing uniform circular motion. The acceleration in uniform circular motion is centripetal acceleration, directed towards the center of the circle. We first calculate the frequency and angular velocity, then use these to find the magnitude of the centripetal acceleration.

Given:
Length of the string, r = 80 cm = 0.80 m
Number of revolutions = 14
Time taken, t = 25 s

Frequency (f) = Number of revolutions / Time taken
f = 14 revolutions / 25 s = 0.56 Hz

Angular velocity (ω) = 2πf
ω = 2π (0.56 Hz) = 1.12π rad/s ≈ 3.5186 rad/s

Centripetal acceleration (a_c) = ω²r
a_c = (1.12π rad/s)² × 0.80 m
a_c = (3.5186 rad/s)² × 0.80 m
a_c = 12.3805 rad²/s² × 0.80 m
a_c = 9.9044 m/s²

Alternatively, using a_c = v²/r:
First, calculate the speed (v).
Period (T) = 1/f = 25 s / 14 = 1.7857 s
v = 2πr / T = 2π (0.80 m) / 1.7857 s = 2.813 m/s
a_c = (2.813 m/s)² / 0.80 m = 7.913 m²/s² / 0.80 m = 9.891 m/s²

Rounding to two significant figures (as 80 cm has two), or three (as 25s has two, 14 is exact):
The magnitude of the acceleration of the stone is approximately 9.90 m/s².

The direction of the acceleration is always towards the center of the horizontal circle.

Question 28

Q28. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Accepted Answer

To compare the centripetal acceleration with the acceleration due to gravity (g), we first need to calculate the centripetal acceleration of the aircraft. Ensure all units are consistent (SI units).

Given:
Radius of the loop, R = 1.00 km = 1.00 × 10³ m
Speed of the aircraft, v = 900 km/h

Convert speed from km/h to m/s:
v = 900 km/h × (1000 m / 1 km) × (1 h / 3600 s)
v = 900 × (1000 / 3600) m/s
v = 900 × (5 / 18) m/s
v = 50 × 5 m/s = 250 m/s

Centripetal acceleration (a_c) is given by the formula:
a_c = v² / R
a_c = (250 m/s)² / (1.00 × 10³ m)
a_c = 62500 m²/s² / 1000 m
a_c = 62.5 m/s²

Now, compare this with the acceleration due to gravity, g ≈ 9.8 m/s².
Ratio = a_c / g
Ratio = 62.5 m/s² / 9.8 m/s²
Ratio ≈ 6.377

So, the centripetal acceleration of the aircraft is approximately 6.38 times the acceleration due to gravity.

Question 29

Q29. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the ci

Accepted Answer

Let's analyze each statement with reasons:

**(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.**
**False.**

**Reason:** This statement is only true for **uniform circular motion**, where the speed of the particle is constant. In uniform circular motion, there is only centripetal acceleration (a_c = v²/r), which is directed towards the center. 
However, in **non-uniform circular motion**, the speed of the particle changes. In this case, there is both a centripetal acceleration (a_c) directed towards the center and a tangential acceleration (a_t = dv/dt) directed along the tangent to the circle (either in the direction of velocity if speeding up, or opposite to velocity if slowing down). The net acceleration is the vector sum of these two components (a_net = a_c + a_t). Therefore, the net acceleration is generally not directed along the radius towards the center, but rather at an angle to the radius.

**(b) The velocity vector of a particle at a point is always perpendicular to the acceleration vector at that point.**
**False.**

**Reason:** This statement is true only for **uniform circular motion** where the velocity vector is tangential and the acceleration vector (centripetal acceleration) is radial (towards the center), making them perpendicular. It is also true at the highest point of a projectile's trajectory where vertical velocity is zero and acceleration (g) is vertical, while horizontal velocity is horizontal.
However, in general, this is not true. For example:
*   In **non-uniform circular motion**, the net acceleration has both radial (centripetal) and tangential components. The tangential component of acceleration is parallel or anti-parallel to the velocity vector. Thus, the net acceleration vector is not perpendicular to the velocity vector.
*   In **projectile motion** (except at the highest point), the velocity vector has both horizontal and vertical components, while the acceleration vector (g) is purely vertical. The angle between them is generally not 90 degrees.
*   When a particle is speeding up or slowing down in a straight line, the acceleration vector is parallel or anti-parallel to the velocity vector, respectively, not perpendicular.

**(c) The work done by a centripetal force in moving a body along a circular path is always zero.**
**True.**

**Reason:** Work done by a force is given by W = F ⋅ d = Fd cosθ, where θ is the angle between the force vector (F) and the displacement vector (d). In circular motion, the centripetal force (F_c) is always directed towards the center of the circle, i.e., along the radius. The instantaneous displacement (or velocity) vector of the particle is always tangential to the circle. Therefore, the centripetal force is always perpendicular to the instantaneous displacement (and velocity) vector (θ = 90°). Since cos(90°) = 0, the work done by the centripetal force is always zero. This is consistent with the fact that centripetal force only changes the direction of velocity, not its magnitude (in uniform circular motion), and thus does not change the kinetic energy of the particle.

Question 30

Q30. A stone is dropped from a height h. Simultaneously, another stone is projected horizontally with some velocity from the same height h. Which stone reaches the ground earlier?

Accepted Answer

Both stones reach the ground simultaneously.

**Reason:**
This is due to the principle of independence of motion. The horizontal motion of a projectile does not affect its vertical motion, and vice-versa. Both stones are subjected to the same acceleration due to gravity (g) in the vertical direction.

Let's consider the vertical motion for both stones:
*   **Stone 1 (dropped):** Initial vertical velocity (u_y) = 0. It undergoes free fall from height h.
*   **Stone 2 (projected horizontally):** Initial vertical velocity (u_y) = 0 (since it's projected *horizontally*). Its horizontal velocity does not influence its vertical motion.

For both stones, the vertical displacement is -h (taking upward as positive), initial vertical velocity is 0, and vertical acceleration is -g. Using the kinematic equation for vertical motion:
y = u_y t + (1/2)a_y t²
-h = 0 × t + (1/2)(-g)t²
h = (1/2)gt²
t = √(2h/g)

Since both stones start from the same height 'h' with zero initial vertical velocity and experience the same acceleration 'g' downwards, the time taken for them to reach the ground will be identical. The horizontal velocity of the second stone only determines how far it travels horizontally before hitting the ground, not when it hits the ground.

Question 31

Q31. Path of a projectile is described by y = ax – bx², where a and b are constants. What is the range of the projectile and the maximum height attained?

Accepted Answer

The given equation of the projectile's path is y = ax – bx².
This is a parabolic equation, characteristic of projectile motion.

### 1. Range of the Projectile
The range (R) is the horizontal distance covered by the projectile when it returns to its initial height (usually the ground, where y = 0).
So, to find the range, we set y = 0 in the given equation:
0 = ax – bx²
Factor out x:
0 = x(a – bx)

This equation gives two possible values for x:
1.  x = 0 (This corresponds to the starting point of the projectile).
2.  a – bx = 0
    a = bx
    x = a/b

This second value of x represents the horizontal distance when the projectile hits the ground again. Therefore, the range of the projectile is R = a/b.

### 2. Maximum Height Attained
The maximum height (H) is the highest vertical position reached by the projectile. At the maximum height, the vertical component of the velocity (v_y) becomes zero. This corresponds to the point where the slope of the trajectory (dy/dx) is zero.

First, differentiate the trajectory equation with respect to x to find the slope:
dy/dx = d/dx (ax – bx²)
dy/dx = a – 2bx

Set dy/dx = 0 to find the x-coordinate (x_H) at which the maximum height occurs:
a – 2bx_H = 0
a = 2bx_H
x_H = a / (2b)

Now, substitute this value of x_H back into the original trajectory equation to find the maximum height (H):
H = a(x_H) – b(x_H)²
H = a(a / (2b)) – b(a / (2b))²
H = a² / (2b) – b(a² / (4b²))
H = a² / (2b) – a² / (4b)

To subtract these terms, find a common denominator (4b):
H = (2a² / (4b)) – (a² / (4b))
H = a² / (4b)

Therefore, the maximum height attained by the projectile is H = a² / (4b).

Question 32

Q32. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direc

Accepted Answer

No, not everything that has magnitude and direction is necessarily a vector. For a physical quantity to be a vector, it must satisfy two conditions:
1.  It must possess both magnitude and direction.
2.  It must obey the laws of vector addition (e.g., triangle law of vector addition, parallelogram law of vector addition).

**Example:** Electric current has both magnitude (e.g., 5 Amperes) and direction (e.g., from positive to negative terminal), but it is a scalar quantity because it does not obey the laws of vector addition. If two wires carrying current meet at a junction, the total current is the algebraic sum (scalar sum) of the individual currents, not their vector sum.

Regarding rotation:
Small rotations can be treated as vector quantities. This is because infinitesimal (very small) rotations obey the commutative law of vector addition (i.e., the order of addition does not matter). For example, if a body undergoes a small rotation Δθ₁ about an axis and then another small rotation Δθ₂ about another axis, the final orientation is the same regardless of the order in which these rotations are performed.

However, large rotations are not vector quantities. This is because large rotations do not obey the commutative law of vector addition. The final orientation of a body after two large rotations depends on the order in which these rotations are performed. Since vector addition must be commutative, large rotations cannot be classified as vectors.

Question 33

Q33. Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere?

Accepted Answer

(a) **The length of a wire bent into a loop:**
No, a vector cannot be associated with the length of a wire bent into a loop. Length is a scalar quantity; it only has magnitude and no specific direction associated with it. Even if the wire forms a closed loop, its total length remains a scalar quantity.

(b) **A plane area:**
Yes, a vector can be associated with a plane area. This is known as an **area vector**. For a plane surface, the area vector is defined as a vector whose magnitude is equal to the area of the surface and whose direction is perpendicular to the plane of the surface. By convention, for an open surface, the direction can be chosen arbitrarily (e.g., upward or downward normal), but for a closed surface, the direction is taken to be along the outward normal to the surface. Area vectors are crucial in concepts like electric flux (Φ = E ⋅ A) and magnetic flux (Φ = B ⋅ A).

(c) **A sphere:**
No, a single vector cannot be associated with a sphere as a whole. A sphere is a three-dimensional object. Its volume is a scalar quantity. Its total surface area is also a scalar quantity. While an infinitesimal element of the surface area of a sphere can be represented by an area vector (pointing radially outward from the center), the entire sphere itself does not have a single, unique vector associated with it in the same way a plane area does.

Chapter 4: Motion in a Plane