RBSE Solutions Class 11 Physics Chapter 2 Basic Mathematical Concepts

RBSE Class 11 Physics Chapter 2 Textbook Exercises with Solutions

RBSE Class 11 Physics Chapter 2 Very Short Answer Type Questions

Question 1. How many types of vectors are there?
Answer: Vectors are mainly of two types:
• Polar vectors
• Axial vectors
In simple words: Vectors are divided into two main categories: polar vectors and axial vectors.

🎯 Exam Tip: Remember to name both types of vectors when asked about their classification.

Question 2. What are equal vectors?
Answer: Equal vectors are vectors that have the same magnitude (length) and point in the same direction. This means they are identical in both size and orientation.
In simple words: Equal vectors are exactly alike in their strength and where they are pointing.

🎯 Exam Tip: For equal vectors, both magnitude and direction must match precisely. If either differs, they are not equal.

Question 4. What are the unit vectors corresponding to X, Y and Z-axes respectively?
Answer: The unit vectors corresponding to the X, Y, and Z-axes are \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) respectively. These vectors each have a magnitude of one and point along their specific axis.
In simple words: The special vectors that show direction along the X, Y, and Z-axes are called \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \).

🎯 Exam Tip: Always associate \( \hat{i} \) with the X-axis, \( \hat{j} \) with the Y-axis, and \( \hat{k} \) with the Z-axis in a right-handed coordinate system.

Question 5. What is a zero vector?
Answer: A zero vector is a vector that has a magnitude of zero. It indicates the direction at a particular point but has no length or displacement. It is also called a null vector.
In simple words: A zero vector is like a point; it has no size but can show a direction.

🎯 Exam Tip: A key property of a zero vector is its zero magnitude. Its direction is indeterminate or arbitrary, often used as a placeholder in vector equations.

Question 6. In how many types does the vectors be resolved?
Answer: Vectors can be resolved into components in two main ways:
• In two dimensions (e.g., into x and y components)
• In three dimensions (e.g., into x, y, and z components)
In simple words: We can break down vectors into parts, either in a flat space (two directions) or in a 3D space (three directions).

🎯 Exam Tip: When resolving vectors, remember that the components are usually perpendicular to each other, making calculations simpler using trigonometry.

Question 7. Can the magnitude of the resultant vector of two vectors be less than that of any of the two vectors?
Answer: Yes, the magnitude of the resultant vector of two vectors can be less than that of either of the individual vectors. This happens if the angle between the two vectors is greater than 90°. For example, if two vectors oppose each other, their resultant will be smaller than the larger individual vector.
In simple words: Yes, if two vectors push in different enough directions, their combined effect can be smaller than one of them alone. This happens when they are more than 90 degrees apart.

🎯 Exam Tip: This scenario is crucial for understanding vector subtraction and is common when vectors partially cancel each other out, such as forces acting at obtuse angles.

Question 8. Classify the scalar and vector quantities of the following: Force, torque, surface tension, momentum and temperature.
Answer: Here is the classification of the given quantities:
Scalar Quantities: Surface tension, Temperature
Vector Quantities: Force, Torque, Momentum
In simple words: Surface tension and temperature are scalar (just a number), while force, torque, and momentum are vector (number and direction).

🎯 Exam Tip: Always remember that scalar quantities only have magnitude, while vector quantities have both magnitude and direction.

Question 9. Can we add a scalar to a vector quantity?
Answer: No. You cannot directly add a scalar quantity to a vector quantity. This is because vectors have both magnitude and direction, while scalars only have magnitude. These are fundamentally different types of physical quantities.
In simple words: No, you cannot add a simple number to a quantity that also has a direction.

🎯 Exam Tip: Vector addition only works between two or more vectors. Scalars and vectors must be treated separately in equations.

Question 11. What is the magnitude of \( \vec{A} \cdot \vec{A} \)?
Answer: The magnitude of \( \vec{A} \cdot \vec{A} \) is \( A^2 \), which can also be written as \( |\vec{A}^2| \). This represents the square of the magnitude of vector \( \vec{A} \). The dot product of a vector with itself always gives the square of its magnitude.
In simple words: The result of dot-multiplying a vector by itself is simply the square of its length, or \( A^2 \).

🎯 Exam Tip: Remember that the dot product \( \vec{A} \cdot \vec{A} \) results in a scalar quantity, specifically \( A^2 \), not a vector.

Question 12. Is the vector product obey the commutative law?
Answer: No. The vector product (cross product) does not obey the commutative law. This means that \( \vec{A} \times \vec{B} \) is not equal to \( \vec{B} \times \vec{A} \). Instead, \( \vec{A} \times \vec{B} = -\vec{B} \times \vec{A} \), meaning the direction of the resultant vector is reversed.
In simple words: No, the order matters when you multiply vectors using the cross product; switching the order reverses the direction of the answer.

🎯 Exam Tip: Keep in mind that the cross product is anti-commutative, which is a crucial distinction from scalar multiplication and dot product.

Question 13. What is the direction of the resultant vector obtained from the vector product of two vectors?
Answer: The direction of the resultant vector obtained from the vector product (cross product) of two vectors is perpendicular to the plane containing both of the original vectors. This direction is determined by the right-hand screw rule.
In simple words: When two vectors are cross-multiplied, the new vector points straight out from the flat surface where the first two vectors lie. We find this direction using the right-hand screw rule.

🎯 Exam Tip: Always apply the right-hand screw rule or the right-hand thumb rule to correctly determine the direction of the cross product vector.

Question 14. What is the vector product of two parallel vectors?
Answer: The vector product (cross product) of two parallel vectors is zero. This is because the angle between parallel vectors is \( 0^\circ \), and \( \sin(0^\circ) = 0 \). Since the magnitude of the cross product involves \( \sin\theta \), the entire product becomes zero.
In simple words: If two vectors point in the same direction (parallel), their cross product is zero.

🎯 Exam Tip: The cross product being zero is a common test for determining if two vectors are parallel or anti-parallel.

RBSE Class 11 Physics Chapter 2 Short Answer Type Questions

Question 1. Differentiate between scalar quantities and vector quantities.
Answer: Physical quantities can be classified as either scalar or vector based on their properties:

Scalar Quantities: These are physical quantities that only have magnitude (a numerical value and a unit) but no specific direction. Examples include mass, length, distance, time, density, work, temperature, charge, specific heat, energy, and speed.

Vector Quantities: These are physical quantities that possess both magnitude and a specific direction. Examples include displacement, velocity, acceleration, force, momentum, electric field, impulse, and gravitational field.

It's important to note that vectors cannot be added, subtracted, multiplied, or divided in the same straightforward way as scalars. Vector operations follow specific rules of vector algebra because their direction must also be considered.
In simple words: Scalar quantities only have a size (like mass or temperature), while vector quantities have both a size and a direction (like force or velocity). You add and subtract them differently.

🎯 Exam Tip: When differentiating, clearly state the definition of each, provide relevant examples, and mention the fundamental difference in how they are manipulated mathematically.

Question 2. Write the law of triangle of vector addition.
Answer: The Triangle Law of Vector Addition states that: "If two vectors acting on a body are represented both in magnitude and direction by the two sides of a triangle, taken in the same order, then their resultant is represented by the third side of the triangle, taken in the opposite order, in both magnitude and direction."

This means that if you arrange two vectors such that the head of the first vector touches the tail of the second, then the vector connecting the tail of the first to the head of the second will be their resultant (sum).
In simple words: If you draw two vectors as two sides of a triangle, one after the other, then the third side of the triangle, drawn in the opposite direction, shows their total combined effect.

🎯 Exam Tip: To score full marks, precisely state the law and explain what 'same order' and 'opposite order' refer to in the context of vector arrangement. A simple diagram helps immensely.

Question 3. What is the law of parallelogram of vector addition?
Answer: The Law of Parallelogram of Vector Addition states that: "If two vectors acting simultaneously at a point are represented in both magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant vector is represented in both magnitude and direction by the diagonal of the parallelogram drawn from the same point."

This means that if you place two vectors with their tails at the same point, forming two sides of a parallelogram, their sum (resultant) will be the diagonal starting from that same tail point.
In simple words: If two vectors start from the same spot and make two sides of a parallelogram, then the diagonal starting from that same spot is their combined effect.

🎯 Exam Tip: Always specify that the vectors must be adjacent sides and the resultant diagonal must start from their common point of origin.

Question 4. Explain the resolution of vectors in two dimensions.
Answer: The resolution of a vector in two dimensions means breaking it down into two component vectors that are mutually perpendicular to each other. These components typically lie along the X-axis and Y-axis in a plane.

Consider a vector \( \vec{A} \) (represented as \( \overrightarrow{OP} \)) that needs to be resolved into two component vectors along the X-axis and Y-axis. Let \( \hat{i} \) and \( \hat{j} \) be the unit vectors along the X-axis and Y-axis respectively.

From a point P, we drop perpendiculars PM to the X-axis and PN to the Y-axis. According to the parallelogram law of vector addition, the vector \( \vec{A} \) can be written as:
\( \vec{A} = A_x \hat{i} + A_y \hat{j} \) ... (i)

In this equation, \( A_x \hat{i} \) is the x-component of \( \vec{A} \), and \( A_y \hat{j} \) is the y-component of \( \vec{A} \). \( A_x \) and \( A_y \) are the magnitudes of these component vectors.

If \( A \) is the magnitude of vector \( \vec{A} \) and \( \theta \) is its angle with the X-axis, then from the right-angled triangle OMP:
\( \cos\theta = \frac{OM}{OP} = \frac{A_x}{A} \implies A_x = A\cos\theta \)
\( \sin\theta = \frac{MP}{OA} = \frac{A_y}{A} \implies A_y = A\sin\theta \)

Using these components, the magnitude of \( \vec{A} \) can be found using the Pythagorean theorem:
\( A^2 = A_x^2 + A_y^2 \)
\( A = \sqrt{A_x^2 + A_y^2} \) ... (iv)

The direction of the vector \( \vec{A} \) can be found by dividing the y-component by the x-component:
\( \frac{A_y}{A_x} = \frac{A\sin\theta}{A\cos\theta} \implies \frac{A_y}{A_x} = \tan\theta \)
\( \implies \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \) ... (v)

In three dimensions, a vector \( \vec{A} \) can be resolved into three mutually perpendicular components along the X, Y, and Z axes, expressed as:
\( \vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k} \) ... (ii)
The magnitude in three dimensions is given by:
\( A = \sqrt{A_x^2+A_y^2+A_z^2} \) ... (iii)
In simple words: Resolving a vector means breaking it into parts. In two dimensions, we split a vector into two perpendicular parts, usually along the X and Y axes, using sine and cosine. These parts help us understand its effect in different directions. In three dimensions, we do this for X, Y, and Z axes.

🎯 Exam Tip: Clearly define vector resolution, explain the use of unit vectors \( \hat{i} \) and \( \hat{j} \), and derive the expressions for the components and the magnitude in terms of \( A \) and \( \theta \).

Question 5. Explain the scalar product of two vectors.
Answer: The scalar product, also known as the dot product, of two vectors is a way to multiply them such that the result is a scalar quantity (just a magnitude, no direction).

If two vectors, say \( \vec{a} \) and \( \vec{b} \), are inclined at an angle \( \theta \) to each other, their dot product is defined as the product of their magnitudes and the cosine of the angle between them:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \)

Here:
\( |\vec{a}| \) is the magnitude (length) of vector \( \vec{a} \)
\( |\vec{b}| \) is the magnitude (length) of vector \( \vec{b} \)
\( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \)

This operation makes sense because it effectively multiplies the length of one vector by the component of the other vector that points in the same direction. For example, \( |\vec{b}|\cos\theta \) is the component of \( \vec{b} \) along \( \vec{a} \). So, the dot product \( \vec{a} \cdot \vec{b} \) can be seen as the magnitude of \( \vec{a} \) multiplied by the component of \( \vec{b} \) along \( \vec{a} \). Conversely, it can also be the magnitude of \( \vec{b} \) multiplied by the component of \( \vec{a} \) along \( \vec{b} \).
In simple words: The scalar product, or dot product, multiplies two vectors to give a single number. You find it by multiplying their lengths and the cosine of the angle between them. It tells you how much one vector acts in the direction of the other.

🎯 Exam Tip: Remember the formula \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \), and emphasize that the result is always a scalar (a number), not another vector.

Question 6. Explain the right-hand screw rule for the vector product of two vectors.
Answer: The right-hand screw rule is used to determine the direction of the resultant vector obtained from the vector product (cross product) of two vectors. Imagine placing a right-handed screw at the common tail of the two vectors, say \( \vec{A} \) and \( \vec{B} \). Now, turn the screw from the first vector (\( \vec{A} \)) towards the second vector (\( \vec{B} \)) through the smaller angle between them. The direction in which the screw advances gives the direction of the resultant vector \( \vec{A} \times \vec{B} \). If the screw moves upwards, the cross product is upwards; if it moves downwards, the cross product is downwards.
In simple words: To find the direction of the cross product of two vectors, imagine turning a right-hand screw from the first vector to the second. The direction the screw moves (in or out) is the direction of the cross product.

🎯 Exam Tip: Clearly describe the process of turning the screw and how its advancement corresponds to the direction of the resultant vector for full clarity.

RBSE Class 11 Physics Chapter 2 Long Answer Type Questions

Question 1. Explain the application of Differential calculus in Physics.
Answer: Differential calculus is a powerful tool in Physics, used to study how physical quantities change with respect to one another. Many physical quantities depend on other variables, and calculus helps us analyze these relationships.

Here are some key applications:
1. **Rate of Change (Velocity and Acceleration):**
- **Velocity:** If displacement (\( \vec{r} \)) is a function of time (\( t \)), then velocity (\( \vec{v} \)) is the rate of change of displacement with respect to time. It's the first derivative of displacement:
\( \vec{v} = \frac{d\vec{r}}{dt} \)
- **Acceleration:** If velocity (\( \vec{v} \)) is a function of time (\( t \)), then acceleration (\( \vec{a} \)) is the rate of change of velocity with respect to time. It's the first derivative of velocity, or the second derivative of displacement:
\( \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac{d\vec{r}}{dt}\right) = \frac{d^2\vec{r}}{dt^2} \)

2. **Work and Power:**
- **Power:** If work done (W) by a force is a function of time (\( t \)), then power (P) is the rate of change of work done. This means power is the first derivative of work with respect to time:
\( P = \frac{dW}{dt} \)
- Similarly, if energy (E) is a function of time, then power can also be expressed as:
\( P = \frac{dE}{dt} \)

3. **Radioactive Decay:**
- The number of active atoms (N) in a radioactive substance changes over time. The rate of decay (A) is the negative rate of change of the number of active atoms with respect to time:
\( A = -\frac{dN}{dt} \)

These examples show how differential calculus allows physicists to calculate instantaneous rates of change, which are fundamental to understanding motion, energy transfer, and decay processes.
In simple words: Differential calculus helps us understand how things change in physics. For example, it tells us how fast an object is moving (velocity from displacement) or how quickly its speed changes (acceleration from velocity). It also helps calculate power from work and the rate of radioactive decay.

🎯 Exam Tip: When explaining applications, always provide the relevant physical quantity, its relationship to another quantity, and the corresponding derivative expression to illustrate the concept clearly.

RBSE Class 11 Physics Chapter 2 Numerical Questions

Question 1. The vertical component of a force is 200 N. The force makes an angle of 30° with the horizontal. Calculate the force.
Answer:
Given:
Angle with the horizontal, \( \theta = 30^\circ \)
Vertical component of force, \( F_v = 200 \mathrm{N} \)
We need to find the total force, \( F \).

We know that the vertical component of a force is given by \( F_v = F \sin\theta \).
So, to find \( F \), we can rearrange the formula:
\( F = \frac{F_v}{\sin\theta} \)

Substitute the given values:
\( F = \frac{200 \mathrm{N}}{\sin 30^\circ} \)
Since \( \sin 30^\circ = \frac{1}{2} \):
\( F = \frac{200 \mathrm{N}}{1/2} \)
\( F = 200 \times 2 \mathrm{N} \)
\( F = 400 \mathrm{N} \)
Therefore, the total force is 400 N.
In simple words: We know the upward part of a force is 200 N and it's at a 30-degree angle. Since sine of the angle gives the vertical part, we divide the vertical part by \( \sin(30^\circ) \) to find the full force, which is 400 N.

🎯 Exam Tip: Always draw a simple diagram to visualize the force and its components. Use the correct trigonometric function (sine for vertical component, cosine for horizontal component) related to the given angle.

Question 2. Determine the magnitude of the vector \( \vec{A}=3 \hat{i}+4 \hat{j} \).
Answer:
Given vector: \( \vec{A}=3 \hat{i}+4 \hat{j} \)

For a vector \( \vec{A}=x \hat{i}+y \hat{j}+z \hat{k} \), its magnitude is calculated using the formula:
\( |\vec{A}| = \sqrt{x^2 + y^2 + z^2} \)

In our case, \( x=3 \), \( y=4 \), and \( z=0 \) (since there is no \( \hat{k} \) component).
Substitute these values into the formula:
\( |\vec{A}| = \sqrt{(3)^2 + (4)^2 + (0)^2} \)
\( |\vec{A}| = \sqrt{9 + 16 + 0} \)
\( |\vec{A}| = \sqrt{25} \)
\( |\vec{A}| = 5 \)
Therefore, the magnitude of the vector \( \vec{A} \) is 5.
In simple words: To find the length (magnitude) of the vector \( 3 \hat{i}+4 \hat{j} \), we square each number (3 and 4), add them up, and then take the square root of the total. This gives us 5.

🎯 Exam Tip: Always remember the Pythagorean theorem for finding the magnitude of a vector from its perpendicular components. Don't forget to include all components (even if zero) in the calculation.

Question 3. Determine the value of the unit vector of vector \( \vec{A}=(4 \hat{i}+3 \hat{j}-5 \hat{k}) \).
Answer:
Given vector: \( \vec{A} = 4\hat{i} + 3\hat{j} - 5\hat{k} \)

To find the unit vector (\( \hat{n} \)) of \( \vec{A} \), we use the formula:
\( \hat{n} = \frac{\vec{A}}{|\vec{A}|} \)

First, calculate the magnitude of \( \vec{A} \):
\( |\vec{A}| = \sqrt{(4)^2 + (3)^2 + (-5)^2} \)
\( |\vec{A}| = \sqrt{16 + 9 + 25} \)
\( |\vec{A}| = \sqrt{50} \)
We can simplify \( \sqrt{50} \) as \( \sqrt{25 \times 2} = 5\sqrt{2} \).
So, \( |\vec{A}| = 5\sqrt{2} \).

Now, substitute \( \vec{A} \) and \( |\vec{A}| \) into the unit vector formula:
\( \hat{n} = \frac{4\hat{i} + 3\hat{j} - 5\hat{k}}{5\sqrt{2}} \)
This can also be written as:
\( \hat{n} = \frac{4}{5\sqrt{2}}\hat{i} + \frac{3}{5\sqrt{2}}\hat{j} - \frac{5}{5\sqrt{2}}\hat{k} \)
\( \hat{n} = \frac{4}{5\sqrt{2}}\hat{i} + \frac{3}{5\sqrt{2}}\hat{j} - \frac{1}{\sqrt{2}}\hat{k} \)
In simple words: To find the unit vector, first calculate the length of the vector \( 4\hat{i} + 3\hat{j} - 5\hat{k} \), which is \( 5\sqrt{2} \). Then, divide each part of the original vector by this length.

🎯 Exam Tip: Always remember to first calculate the magnitude of the vector accurately before dividing the vector by its magnitude to find the unit vector.

Question 4. Find out the value of \( \frac{d}{dx}\left(x^{7}\right) \).
Answer:
We need to find the derivative of \( x^7 \) with respect to \( x \).

The formula for differentiating a power of \( x \) is:
\( \frac{d}{dx} x^n = nx^{n-1} \)

In this problem, \( n=7 \).
Applying the formula:
\( \frac{d}{dx} (x^7) = 7x^{7-1} \)
\( \frac{d}{dx} (x^7) = 7x^6 \)
In simple words: To find the derivative of \( x \) to the power of 7, you bring the power (7) to the front and then subtract 1 from the power, making it \( 7x^6 \).

🎯 Exam Tip: Remember the power rule for differentiation: \( \frac{d}{dx} x^n = nx^{n-1} \). This is a fundamental rule in calculus.

Question 5. Find out the value of \( \frac{d}{d x}\left(x^{-3}\right) \).
Answer:
We need to find the derivative of \( x^{-3} \) with respect to \( x \).

Using the power rule for differentiation:
\( \frac{d}{dx} x^n = nx^{n-1} \)

In this problem, \( n = -3 \).
Applying the formula:
\( \frac{d}{dx} (x^{-3}) = (-3)x^{-3-1} \)
\( \frac{d}{dx} (x^{-3}) = -3x^{-4} \)
This can also be written as \( -\frac{3}{x^4} \).
In simple words: To differentiate \( x \) raised to the power of -3, bring the -3 to the front and subtract 1 from the exponent, giving us \( -3x^{-4} \).

🎯 Exam Tip: Be careful with negative exponents when applying the power rule; subtracting 1 from a negative number makes it more negative (e.g., -3 - 1 = -4).

Question 6. Find out the value of \( \int x^{4} d x \).
Answer:
We need to find the indefinite integral of \( x^4 \) with respect to \( x \).

The formula for integrating a power of \( x \) is:
\( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

In this problem, \( n=4 \).
Applying the formula:
\( \int x^4 dx = \frac{x^{4+1}}{4+1} + C \)
\( \int x^4 dx = \frac{x^5}{5} + C \)
In simple words: To integrate \( x \) to the power of 4, you add 1 to the power (making it 5) and then divide by this new power. Don't forget to add 'C' at the end for indefinite integrals.

🎯 Exam Tip: Always remember to add the constant of integration \( C \) for indefinite integrals. This is a common point where marks are lost.

Question 7. Find out the value of \( \int x^{-5} d x \)
Answer: To find the integral of \( x^{-5} \), we use the power rule for integration, which states that \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \). Here, \( n = -5 \).
\[ \int x^{-5} d x = \frac{x^{-5+1}}{-5+1} + C \] \[ = \frac{x^{-4}}{-4} + C \] \[ = -\frac{1}{4x^4} + C \] So, the value of the integral is \( -\frac{1}{4x^4} + C \).
In simple words: When you integrate \( x \) raised to a power, you add one to the power and divide by the new power. For \( x^{-5} \), this means \( x^{-4} \) divided by \( -4 \), plus a constant.

🎯 Exam Tip: Remember to always add the constant of integration \( C \) when performing indefinite integration, as it represents any constant value whose derivative is zero.

Question 8. What is the characteristic of the number 83256?
Answer: The characteristic of the logarithm of a number is found by subtracting 1 from the number of digits before the decimal point.
For the number 83256, there are 5 digits before the decimal point (8, 3, 2, 5, 6).
So, the characteristic is \( 5 - 1 = 4 \).
Alternatively, in scientific notation, 83256 can be written as \( 8.3256 \times 10^4 \). The characteristic is the power of 10, which is 4.
In simple words: Just count how many numbers are there before the decimal point and then take away 1. For 83256, it's 5 digits, so \( 5-1=4 \).

🎯 Exam Tip: When finding the characteristic of a whole number, count the digits to the left of the implied decimal point and subtract one. For numbers greater than or equal to 1, the characteristic is positive or zero.

Question 9. What is the characteristic of the number 0.00356.
Answer: The characteristic of a number less than 1 (a decimal number) is found by counting the number of zeros immediately after the decimal point and adding 1, then making the result negative (represented by a bar over the number).
For the number 0.00356:
There are 2 zeros immediately after the decimal point (0.00).
So, the characteristic is \( -(2+1) = -3 \), which is written as \( \overline{3} \).
In scientific notation, 0.00356 can be written as \( 3.56 \times 10^{-3} \). The characteristic is the power of 10, which is -3 or \( \overline{3} \).
In simple words: For numbers smaller than one, count the zeros right after the decimal point, add one, and make it negative. For 0.00356, there are two zeros, so the characteristic is \( \overline{3} \).

🎯 Exam Tip: For numbers between 0 and 1, the characteristic is negative. It is found by counting the number of zeros immediately following the decimal point and preceding the first non-zero digit, and then adding one to that count, with a negative sign.

Question 11. Two forces, \( \vec{F_1} = (2 \hat{i}-3 \hat{j}) \) N and \( \vec{F_2} = (-\hat{i}+3 \hat{j}) \) N acts at a point in X-Y plane. Calculate the resultant force.
Answer: To find the resultant force, we add the two given force vectors.
Given:
\( \vec{F_1} = (2 \hat{i}-3 \hat{j}) \, \text{N} \)
\( \vec{F_2} = (-\hat{i}+3 \hat{j}) \, \text{N} \)
The resultant force \( \vec{F} \) is:
\[ \vec{F} = \vec{F_1} + \vec{F_2} \] \[ \vec{F} = (2 \hat{i}-3 \hat{j}) + (-\hat{i}+3 \hat{j}) \] \[ \vec{F} = (2\hat{i} - \hat{i}) + (-3\hat{j} + 3\hat{j}) \] \[ \vec{F} = (2-1)\hat{i} + (-3+3)\hat{j} \] \[ \vec{F} = 1\hat{i} + 0\hat{j} \] \[ \vec{F} = \hat{i} \, \text{N} \] Therefore, the resultant force is \( \hat{i} \) N.
In simple words: To find the total force, we just add the individual forces. We combine the \( \hat{i} \) parts and the \( \hat{j} \) parts separately. The final answer shows a force of 1 N acting along the positive X-axis.

🎯 Exam Tip: To add vectors, always combine their respective components (x-components with x-components, y-components with y-components, etc.) separately.

Question 12. One of the two perpendicular components of velocity (10 m/s) of a particle makes an angle of 60° with the intended velocity. Then, find out the another component of velocity.
Answer: Let \( V \) be the resultant (intended) velocity. Let \( v_1 \) and \( v_2 \) be the two perpendicular components of velocity. Given that one component \( v_1 = 10 \, \text{m/s} \) makes an angle of \( 60^\circ \) with the resultant velocity \( V \).
When components are perpendicular, we can use trigonometry.
If \( v_1 \) is the component along the direction making \( 60^\circ \) with \( V \), and \( v_2 \) is the component perpendicular to \( v_1 \), then:
\( v_1 = V \cos 60^\circ \)
\( v_2 = V \sin 60^\circ \)
We are given \( v_1 = 10 \, \text{m/s} \).
From the first equation:
\( 10 = V \cos 60^\circ \)
\( 10 = V \times \frac{1}{2} \)
\( V = 20 \, \text{m/s} \)
Now, we can find the other component \( v_2 \):
\( v_2 = V \sin 60^\circ \)
\( v_2 = 20 \times \frac{\sqrt{3}}{2} \)
\( v_2 = 10\sqrt{3} \, \text{m/s} \)
So, the other component of velocity is \( 10\sqrt{3} \, \text{m/s} \).
In simple words: We know one part of the velocity (10 m/s) and the angle it makes with the total velocity. Using these, we first find the total velocity and then calculate the other perpendicular part of the velocity using sine.

🎯 Exam Tip: Always draw a vector diagram to visualize perpendicular components and their resultant. This helps in correctly applying sine and cosine functions.

Question 13. If an experiences a force of \( \vec{F}=(\hat{i}+\hat{j}-2 \hat{k}) \) N and displacement, \( d\vec{r}=(-\hat{i}+2 \hat{j}-\hat{k}) \) m. Then, determine the work and the angle between \( \vec{F} \) and \( d\vec{r} \).
Answer: To determine the work done and the angle between the force and displacement vectors, we use the dot product formula.
Given:
Force vector \( \vec{F} = (\hat{i}+\hat{j}-2 \hat{k}) \, \text{N} \)
Displacement vector \( d\vec{r} = (-\hat{i}+2 \hat{j}-\hat{k}) \, \text{m} \)

First, calculate the work done \( W \). Work done is given by the dot product of force and displacement:
\[ W = \vec{F} \cdot d\vec{r} \] \[ W = (1)(-1) + (1)(2) + (-2)(-1) \] \[ W = -1 + 2 + 2 \] \[ W = 3 \, \text{J} \]
Next, calculate the magnitudes of \( \vec{F} \) and \( d\vec{r} \).
Magnitude of force \( |\vec{F}| \):
\[ |\vec{F}| = \sqrt{(1)^2 + (1)^2 + (-2)^2} \] \[ |\vec{F}| = \sqrt{1 + 1 + 4} \] \[ |\vec{F}| = \sqrt{6} \]
Magnitude of displacement \( |d\vec{r}| \):
\[ |d\vec{r}| = \sqrt{(-1)^2 + (2)^2 + (-1)^2} \] \[ |d\vec{r}| = \sqrt{1 + 4 + 1} \] \[ |d\vec{r}| = \sqrt{6} \]
Now, calculate the angle \( \theta \) between \( \vec{F} \) and \( d\vec{r} \) using the dot product formula:
\[ \vec{F} \cdot d\vec{r} = |\vec{F}| |d\vec{r}| \cos \theta \] \[ \cos \theta = \frac{\vec{F} \cdot d\vec{r}}{|\vec{F}| |d\vec{r}|} \] \[ \cos \theta = \frac{3}{\sqrt{6} \times \sqrt{6}} \] \[ \cos \theta = \frac{3}{6} \] \[ \cos \theta = \frac{1}{2} \]
Therefore, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \] \[ \theta = 60^\circ \] The work done is 3 J and the angle between the force and displacement is \( 60^\circ \).
In simple words: Work is found by multiplying the matching parts of force and displacement vectors and adding them up. The angle is found by dividing the work by the product of the lengths of the force and displacement vectors, and then finding the angle that has that cosine value.

🎯 Exam Tip: Remember that work done is a scalar quantity (just a number), while force and displacement are vector quantities. The angle is always calculated using the dot product formula.

Question 14. If an electron moves with a velocity, \( \vec{v}=(2 \hat{i}-3 \hat{j}+\hat{k}) \) m/s in a magnetic field, \( \vec{B}=(2 \hat{i}+\hat{j}-\hat{k}) \) T. Then, calculate the Lorentz force.
Answer: The Lorentz force \( \vec{F}_L \) on a charged particle moving in a magnetic field is given by the formula \( \vec{F}_L = q (\vec{v} \times \vec{B}) \). For an electron, the charge \( q = -1.6 \times 10^{-19} \, \text{C} \).
Given:
Velocity \( \vec{v} = (2 \hat{i}-3 \hat{j}+\hat{k}) \, \text{m/s} \)
Magnetic field \( \vec{B} = (2 \hat{i}+\hat{j}-\hat{k}) \, \text{T} \)

First, calculate the cross product \( \vec{v} \times \vec{B} \):
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 2 & 1 & -1 \end{vmatrix} \] \[ = \hat{i}((-3)(-1) - (1)(1)) - \hat{j}((2)(-1) - (1)(2)) + \hat{k}((2)(1) - (-3)(2)) \] \[ = \hat{i}(3 - 1) - \hat{j}(-2 - 2) + \hat{k}(2 + 6) \] \[ = \hat{i}(2) - \hat{j}(-4) + \hat{k}(8) \] \[ = (2\hat{i} + 4\hat{j} + 8\hat{k}) \]
Now, substitute this into the Lorentz force formula:
\[ \vec{F}_L = q (\vec{v} \times \vec{B}) \] \[ \vec{F}_L = (-1.6 \times 10^{-19}) (2\hat{i} + 4\hat{j} + 8\hat{k}) \, \text{N} \] The Lorentz force acting on the electron is \( (-1.6 \times 10^{-19}) (2\hat{i} + 4\hat{j} + 8\hat{k}) \) N.
In simple words: To find the magnetic force on the electron, we first find the cross product of its velocity and the magnetic field. Then, we multiply this result by the electron's charge, which is a negative value.

🎯 Exam Tip: Remember the sign of the charge (negative for electrons) and correctly calculate the determinant for the cross product to avoid errors in the direction of the force.

Question 15. The potential energy of a particle is \( U = y^2 \sin y \). The force defined on this particle is \( F = -\frac{d U}{d y} \). Calculate the force
Answer: To calculate the force \( F \), we need to find the negative derivative of the potential energy \( U \) with respect to \( y \).
Given:
Potential energy \( U = y^2 \sin y \)
Force \( F = -\frac{d U}{d y} \)

First, find the derivative of \( U \) with respect to \( y \) using the product rule \( \frac{d}{dy}(uv) = u'v + uv' \), where \( u=y^2 \) and \( v=\sin y \).
\( \frac{du}{dy} = 2y \)
\( \frac{dv}{dy} = \cos y \)
So, the derivative of \( U \) is:
\[ \frac{d U}{d y} = \frac{d}{d y}(y^2 \sin y) \] \[ = (2y)(\sin y) + (y^2)(\cos y) \] \[ = 2y \sin y + y^2 \cos y \]
Now, substitute this into the force formula:
\[ F = -\left(2y \sin y + y^2 \cos y\right) \] The force acting on the particle is \( F = -(2y \sin y + y^2 \cos y) \).
In simple words: We find the force by taking the negative of the rate at which potential energy changes with position. This involves using a specific rule to differentiate (find the rate of change of) the given potential energy formula.

🎯 Exam Tip: When calculating force from potential energy, always apply the negative derivative. Ensure correct application of differentiation rules, especially the product rule for functions like \( y^2 \sin y \).

Question 16. The angular momentum and torque of a particle are related to each other as : \( \vec{\tau}=\frac{d \vec{J}}{d t} \). If \( \vec{\tau}=\left(t \hat{i}+t^{2} \hat{j}+t^{3} \hat{k}\right) \) and at instant \( t = 0 \), \( \vec{J}=(3 \hat{i}+4 \hat{j}+7 \hat{k}) \), then represent \( \vec{J} \) as a function of time.
Answer: We are given the relationship between torque \( \vec{\tau} \) and angular momentum \( \vec{J} \) as \( \vec{\tau}=\frac{d \vec{J}}{d t} \). This means we can find \( \vec{J} \) by integrating \( \vec{\tau} \) with respect to time \( t \).
Given:
\( \vec{\tau}=\left(t \hat{i}+t^{2} \hat{j}+t^{3} \hat{k}\right) \)
From the definition, \( d\vec{J} = \vec{\tau} dt \).
Integrate both sides:
\[ \int d\vec{J} = \int \vec{\tau} dt \] \[ \vec{J}(t) = \int \left(t \hat{i}+t^{2} \hat{j}+t^{3} \hat{k}\right) dt \] \[ \vec{J}(t) = \left( \int t \, dt \right) \hat{i} + \left( \int t^{2} \, dt \right) \hat{j} + \left( \int t^{3} \, dt \right) \hat{k} \] \[ \vec{J}(t) = \left( \frac{t^2}{2} \right) \hat{i} + \left( \frac{t^3}{3} \right) \hat{j} + \left( \frac{t^4}{4} \right) \hat{k} + \vec{C} \] Here, \( \vec{C} \) is the constant of integration, which is a vector.

We are given the initial condition: at \( t = 0 \), \( \vec{J}=(3 \hat{i}+4 \hat{j}+7 \hat{k}) \).
Substitute \( t=0 \) into the equation for \( \vec{J}(t) \):
\[ \vec{J}(0) = \left( \frac{0^2}{2} \right) \hat{i} + \left( \frac{0^3}{3} \right) \hat{j} + \left( \frac{0^4}{4} \right) \hat{k} + \vec{C} \] \[ (3 \hat{i}+4 \hat{j}+7 \hat{k}) = (0) \hat{i} + (0) \hat{j} + (0) \hat{k} + \vec{C} \] \[ \vec{C} = (3 \hat{i}+4 \hat{j}+7 \hat{k}) \]
Now substitute the value of \( \vec{C} \) back into the expression for \( \vec{J}(t) \):
\[ \vec{J}(t) = \left( \frac{t^2}{2} \right) \hat{i} + \left( \frac{t^3}{3} \right) \hat{j} + \left( \frac{t^4}{4} \right) \hat{k} + (3 \hat{i}+4 \hat{j}+7 \hat{k}) \] Combine the components:
\[ \vec{J}(t) = \left( \frac{t^2}{2} + 3 \right)\hat{i} + \left( \frac{t^3}{3} + 4 \right)\hat{j} + \left( \frac{t^4}{4} + 7 \right)\hat{k} \] This is the angular momentum \( \vec{J} \) as a function of time.
In simple words: To find how angular momentum changes over time, we integrate the torque expression. We then use the given starting angular momentum at time \( t=0 \) to find the constant part of the answer.

🎯 Exam Tip: When integrating vector quantities, integrate each component separately. Always use initial conditions to determine the constant of integration for definite solutions.

RBSE Solutions for Class 11 Physics