CBSE Class 12 Physics Short Answer Question Bank Worksheet Set 02

CBSE Class 12 Physics Short Answer Question Bank (2).The Short Answer questions in the worksheets have been specifically designed by best Physics teachers so that the students can practice them to clear their Short Answer concepts and get better marks in class 12 Physics tests and examinations. Students can free download these Short Answer Question Bank worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Short Answer questions chapter and other subjects too. Use them for better understanding of the subjects.

Section A ( Very Short Answer Type Questions )

1. Name the physical quantity whose S.I. unit is J/C. Is it a scalar or a vector quantity?

2. In a certain arrangement, a proton does not get deflected while passing through a magnetic field region. State the condition under which it is possible.

3. State Lenz’s law in electromagnetic induction.

4. What is the ratio of velocities of light rays of wavelengths 4000 Å and 8000 Å in vacuum?

5. Out of speed, frequency and wavelength which physical quantity remains unchanged with the refraction of wave?

6. Two nuclei have mass numbers in the ratio 1:8.What is the ratio of their nuclear radii?

7. Among alpha, beta and gamma radiations, which does not get affected by electric field?

8. Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities?

Section A ( Very Short Answer Type Questions )

Question 1. Name the physical quantity whose S.I. unit is J/C. Is it a scalar or a vector quantity?
Answer: The physical quantity whose SI unit is Joule per Coulomb (\( \text{J/C} \)) is electric potential (or potential difference). It is a scalar quantity.
In simple words: Electric potential is measured in Joules per Coulomb, which is also called a Volt. It only has a value and no direction, so it is a scalar quantity.

Exam Tip: Always state both the name of the quantity and its algebraic nature (scalar/vector) clearly as separate points to ensure full marks.

Question 2. In a certain arrangement, a proton does not get deflected while passing through a magnetic field region. State the condition under which it is possible.
Answer: This scenario occurs when the velocity vector \( \vec{v} \) of the proton is aligned parallel or anti-parallel to the magnetic field vector \( \vec{B} \). Under this condition, the angle \( \theta \) between \( \vec{v} \) and \( \vec{B} \) is either \( 0^\circ \) or \( 180^\circ \). Since the magnetic force is given by \( F = q v B \sin\theta \), the net deflecting force acting on the proton becomes zero.
In simple words: If a proton travels in the exact same direction as the magnetic field lines (or directly opposite to them), the magnetic field will not push it, allowing it to move straight.

Exam Tip: Mention the mathematical formula \( F = qvB\sin\theta \) and show that \( \theta = 0^\circ \) or \( 180^\circ \) leads to \( F = 0 \) to make your answer highly rigorous.

Question 3. State Lenz’s law in electromagnetic induction.
Answer: Lenz's law states that the direction of the induced electromotive force (emf) or induced current in a closed circuit is always such that it opposes the change in magnetic flux that produces it.
In simple words: When a magnetic field changes near a wire loop, it creates a current. This new current creates its own magnetic field to fight the original change.

Exam Tip: Lenz's law is a direct consequence of the law of conservation of energy - mentioning this connection is highly favored by examiners.

Question 4. What is the ratio of velocities of light rays of wavelengths 4000 Å and 8000 Å in vacuum?
Answer: In a vacuum, the speed of all electromagnetic waves is independent of their wavelength or frequency and is equal to \( c \approx 3 \times 10^8 \text{ m/s} \). Therefore, the ratio of their velocities is \( 1 : 1 \).
In simple words: All light travels at the exact same speed in empty space, no matter what color or wavelength it is.

Exam Tip: Clearly state that the speed of electromagnetic waves in a vacuum is a universal constant, which makes the ratio independent of wavelength.

Question 5. Out of speed, frequency and wavelength which physical quantity remains unchanged with the refraction of wave?
Answer: The frequency of a wave remains unchanged during refraction. This is because frequency is an intrinsic characteristic determined solely by the source of the wave.
In simple words: When a wave passes from one medium to another, its speed and wavelength change, but its frequency stays exactly the same.

Exam Tip: Remember that while velocity and wavelength depend on the medium, frequency is determined only by the wave source.

Question 6. Two nuclei have mass numbers in the ratio 1:8. What is the ratio of their nuclear radii?
Answer: The radius \( R \) of a nucleus is related to its mass number \( A \) by the formula:
\( R = R_0 A^{1/3} \)
Given the ratio of mass numbers is \( \frac{A_1}{A_2} = \frac{1}{8} \), the ratio of their radii is:
\( \frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} \)
Therefore, the ratio of their nuclear radii is \( 1 : 2 \).
In simple words: Since a nucleus is a sphere, its radius grows with the cube root of its mass. A mass ratio of 1 to 8 results in a radius ratio of 1 to 2.

Exam Tip: Always write down the core formula \( R = R_0 A^{1/3} \) before plugging in the values to secure partial marking in case of calculation errors.

Question 7. Among alpha, beta and gamma radiations, which does not get affected by electric field?
Answer: Gamma (\( \gamma \)) radiation is unaffected by an electric field because it consists of uncharged, high-frequency electromagnetic photons. Unlike alpha and beta particles, gamma rays carry no net electric charge.
In simple words: Gamma rays are pure energy and have no electric charge, so an electric field cannot pull or push them.

Exam Tip: Contrast the neutral nature of gamma rays with the positive charge of alpha particles and the negative charge of beta particles to write a complete answer.

Question 8. Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities?
Answer: The density of nuclear matter is a constant value of approximately \( 2.3 \times 10^{17} \text{ kg/m}^3 \), which is independent of the mass number \( A \) or the size of the nucleus. Thus, the ratio of their nuclear densities is \( 1 : 1 \).
In simple words: All nuclei are packed with nucleons at the exact same tightness, so their density is always identical regardless of their mass.

Exam Tip: Be ready to explain why density is independent of \( A \) by showing that both nuclear mass and volume are proportional to \( A \), which cancels out.

Section B ( Short Answer Type Questions Type 1)

Question 9. An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.
Answer:
(i) Let an electric dipole consisting of charges \( -q \) and \( +q \), separated by distance \( 2a \), be placed in a uniform electric field \( \vec{E} \).
The force acting on the positive charge is \( \vec{F}_+ = +q\vec{E} \) along the direction of the field.
The force acting on the negative charge is \( \vec{F}_- = -q\vec{E} \) opposite to the direction of the field.
The net force on the dipole is:
\( \vec{F}_{\text{net}} = \vec{F}_+ + \vec{F}_- = q\vec{E} - q\vec{E} = 0 \).

(ii) The work done in rotating a dipole in an electric field from an initial angle \( \theta_1 \) to a final angle \( \theta_2 \) is given by:
\( W = pE(\cos\theta_1 - \cos\theta_2) \)
When the dipole is aligned parallel to the field, \( \theta_1 = 0^\circ \). Rotating it by \( 180^\circ \) means \( \theta_2 = 180^\circ \).
\( W = pE(\cos 0^\circ - \cos 180^\circ) \)
\( \implies W = pE(1 - (-1)) = 2pE \).
In simple words: (i) The electric field pulls one side of the dipole forward and the other side backward with the same force, so the forces cancel out. (ii) Turning the dipole completely around against the field requires an amount of work equal to \( 2pE \).

Exam Tip: Clearly show the vector addition of forces for part (i) and the substitution of cosine values for part (ii) to achieve a perfect score.

Question 10. A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and original coil.
Answer: Let \( m_1 \) and \( m_2 \) be the magnetic dipole moments of the original and new coils respectively. The magnetic moment is given by \( m = N I A \).
For the original coil of radius \( R \) and \( N \) turns:
\( m_1 = N I (\pi R^2) \)
Since the total length of the wire remains constant during rewinding:
\( L = N(2\pi R) = N'\left(2\pi \frac{R}{2}\right) \implies N' = 2N \), where \( N' \) is the number of turns in the new coil.
For the new coil of radius \( \frac{R}{2} \):
\( m_2 = N' I A_2 = (2N) I \left[\pi \left(\frac{R}{2}\right)^2\right] = 2N I \pi \frac{R^2}{4} = \frac{1}{2} N I \pi R^2 \)
Now, calculating the ratio of their magnetic moments:
\( \frac{m_2}{m_1} = \frac{\frac{1}{2} N I \pi R^2}{N I \pi R^2} = \frac{1}{2} \).
Thus, the ratio of the new magnetic moment to the original magnetic moment is \( 1 : 2 \).
In simple words: Making the coil half as wide doubles the number of loops, but the area of each loop shrinks to a quarter. As a result, the total magnetic strength is cut in half.

Exam Tip: The key to this problem is using the conservation of wire length to find the relationship between the number of turns \( N \) and \( N' \).

Question 11. (i) Write two characteristics of a material used for making permanent magnets.
(ii) Why is the core of an electromagnet made of ferromagnetic materials?
Answer:
(i) A material suitable for making permanent magnets must possess: (a) High retentivity, so that it retains a strong magnetic field even when the magnetizing field is removed. (b) High coercivity, so that its magnetic alignment is not easily disrupted by external fields or temperature changes.
(ii) The core of an electromagnet is made of ferromagnetic materials (such as soft iron) because they have exceptionally high magnetic permeability and low retentivity. This ensures that the core magnetizes strongly when current flows and demagnetizes almost instantly when the current is turned off.
In simple words: (i) A permanent magnet needs to hold its magnetic power strongly (retentivity) and not lose it easily (coercivity). (ii) Electromagnets need to turn on and off instantly, so they are made of materials that magnetize strongly but lose their magnetism as soon as the power is cut.

Exam Tip: Be sure to clearly distinguish between the material requirements for a permanent magnet (high coercivity) and an electromagnet (low coercivity).

Question 12. Draw magnetic field line when a (i) diamagnetic, (ii) paramagnetic substance is placed in an external magnetic field. Which magnetic property distinguishes this behaviour of the field line due to the substances?
Answer:
(i) When a diamagnetic material is placed in an external magnetic field, the field lines tend to bend away from the material, resulting in a weaker field inside:
(ii) When a paramagnetic material is placed in an external magnetic field, the field lines are pulled into the material, concentrating inside it:
The magnetic property that distinguishes this behavior is **magnetic susceptibility** (\( \chi_m \)). For diamagnetic substances, \( \chi_m \) is negative, meaning they weakly oppose the field. For paramagnetic substances, \( \chi_m \) is positive, meaning they weakly assist the field.
In simple words: Diamagnetic materials push magnetic field lines out, while paramagnetic materials pull them in. This difference is defined by their magnetic susceptibility.

Exam Tip: Be prepared to state the range of magnetic susceptibility for both materials: \( -1 \le \chi_m < 0 \) for diamagnetic, and \( 0 < \chi_m < \varepsilon \) for paramagnetic.

Question 13. Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces.
Answer: Here is the plot of potential energy \( U \) (in MeV) of a pair of nucleons versus their separation \( r \) (in fm):
The potential energy is minimum at \( r_0 \approx 0.8 \text{ fm} \). For separations \( r > r_0 \), the force is attractive; for \( r < r_0 \), it is strongly repulsive.

Two characteristic features of nuclear forces:
1. **Short-range forces:** They operate only over extremely small distances (on the order of a few femtometers) and drop rapidly to zero beyond that.
2. **Charge independence:** The nuclear force between proton-proton, neutron-neutron, and proton-neutron is approximately equal.
In simple words: When nucleons are extremely close, they repel each other. When they are slightly further apart (around 0.8 to 2 fm), they attract each other. Nuclear forces do not depend on the electric charge of the particles.

Exam Tip: Clearly label \( r_0 \approx 0.8\text{ fm} \) on your graph, as this specific value is critical for scoring full marks.

Question 14. State the law of radioactive decay. Establish a mathematical relation between half-life period and disintegration constant of a radioactive nucleus.
Answer: **Law of Radioactive Decay:** The rate of disintegration of a radioactive substance at any instant is directly proportional to the total number of active nuclei present in the sample at that instant.
Mathematically, \( \frac{dN}{dt} = -\lambda N \), where \( \lambda \) is the decay constant.

Derivation of relation with Half-life (\( T_{1/2} \)):
Integrating the decay equation yields the exponential decay law:
\( N(t) = N_0 e^{-\lambda t} \)
By definition, at \( t = T_{1/2} \), the number of active nuclei reduces to half of its initial value, i.e., \( N = \frac{N_0}{2} \).
Substituting these values:
\( \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \)
\( \implies \frac{1}{2} = e^{-\lambda T_{1/2}} \)
\( \implies e^{\lambda T_{1/2}} = 2 \)
Taking natural logarithm (\( \ln \)) on both sides:
\( \lambda T_{1/2} = \ln(2) \)
\( \lambda T_{1/2} \approx 0.693 \)
\( \implies T_{1/2} = \frac{0.693}{\lambda} \).
This is the required relationship.
In simple words: The speed at which a radioactive material decays depends on how many atoms are left. The half-life is the time it takes for half the material to decay, and it is equal to 0.693 divided by the decay constant.

Exam Tip: Always state the final expression \( T_{1/2} = \frac{0.693}{\lambda} \) in a box to make it stand out to the examiner.

Question 15. Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5.
Answer: According to the Lens Maker's Formula:
\( \frac{1}{f} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
For a plano-convex lens:
The first surface is convex, so \( R_1 = R \).
The second surface is flat, so \( R_2 = -\infty \) (or \( \infty \)).
Given \( f = 0.3 \text{ m} \) and \( \mu = 1.5 \). Substituting these values:
\( \frac{1}{0.3} = (1.5 - 1)\left( \frac{1}{R} - \frac{1}{\infty} \right) \)
\( \implies \frac{1}{0.3} = 0.5 \left( \frac{1}{R} \right) \)
\( \implies R = 0.3 \times 0.5 = 0.15 \text{ m} \).
Converting to centimeters:
\( R = 15 \text{ cm} \).
Thus, the radius of curvature of the convex surface is \( 15 \text{ cm} \).
In simple words: Using the formula that links a lens's shape to how it bends light, we calculate that a plano-convex lens with a 0.3-meter focal point must have a curved side with a radius of 15 centimeters.

Exam Tip: Be careful with sign conventions. Remember that the flat side of a plano-lens always has a radius of curvature of infinity.

Question 16. A charge ‘q’ is placed at the centre of a cube of side l. Using Gauss law find electric flux passing through each face of the cube?
Answer: According to Gauss's law, the total electric flux \( \Phi \) passing through any closed surface enclosing a charge \( q \) is:
\( \Phi = \frac{q}{\varepsilon_0} \)
A cube is a highly symmetrical closed surface with 6 identical faces. By symmetry, the electric flux is distributed equally among all 6 faces.
Therefore, the electric flux \( \phi \) passing through each individual face of the cube is:
\( \phi = \frac{\Phi}{6} = \frac{q}{6\varepsilon_0} \).
In simple words: The total electric field lines coming out of the charge inside the cube must divide equally among its 6 sides. So, each side gets exactly one-sixth of the total flux.

Exam Tip: Mention the term "symmetry of the cube" to justify why the total flux is divided equally by 6.

Question 17. Two lines, A and B, in the plot given below show the variation of de Broglie wavelength, λ versus \( \frac{1}{\sqrt{V}} \), where V is the accelerating potential difference, for two particles carrying the same charge. Which one of two represents a particle of smaller mass?
Answer: The de Broglie wavelength \( \lambda \) of a charged particle accelerated through a potential \( V \) is given by:
\( \lambda = \frac{h}{\sqrt{2m q V}} \)
\( \implies \lambda = \left( \frac{h}{\sqrt{2mq}} \right) \cdot \frac{1}{\sqrt{V}} \)
This represents a straight line equation passing through the origin (\( y = m x \)), where the slope is:
\( \text{Slope} = \frac{h}{\sqrt{2mq}} \)
Since both particles carry the same charge \( q \), the slope is inversely proportional to the square root of their mass:
\( \text{Slope} \propto \frac{1}{\sqrt{m}} \)
From the plot, line B has a greater slope than line A.
\( \implies \text{Slope}_B > \text{Slope}_A \)
\( \implies \frac{1}{\sqrt{m_B}} > \frac{1}{\sqrt{m_A}} \implies m_B < m_A \).
Therefore, **line B** represents the particle of smaller mass.
In simple words: The slope of the line on this graph is steeper for lighter particles. Since line B is steeper than line A, the particle represented by line B has a smaller mass.

Exam Tip: Clearly write down the slope dependency relation \( \text{slope} \propto \frac{1}{\sqrt{m}} \) to establish a strong logical proof.

Question 18. Draw a circuit diagram of n-p-n transistor amplifier in CE configuration. Under what condition does the transistor act as an amplifier?
Answer: The circuit diagram of an n-p-n transistor amplifier in Common Emitter (CE) configuration is shown below:
The transistor acts as an amplifier when operating in the **active region**. This is achieved under the following biasing conditions:
1. The emitter-base (input) junction must be **forward biased** using a low-voltage supply \( V_{BB} \).
2. The collector-base (output) junction must be **reverse biased** using a high-voltage supply \( V_{CC} \).
In simple words: A transistor works as an amplifier when its input side is connected to turn "on" (forward biased) and its output side is connected to resist flow (reverse biased). This allows small input changes to create big output changes.

Exam Tip: Be sure to label key elements such as the base resistor \( R_B \), collector resistor \( R_C \), and the input/output coupling capacitors in your schematic.

Question 19. In the given block diagram of a receiver, identify the boxes labeled as X and Y and write their functions.
Answer: By analyzing the block diagram of the receiver:
- **Box X represents the IF (Intermediate Frequency) Stage.**
- **Box Y represents the Amplifier (or Power Amplifier).**

Functions:
- **IF Stage (X):** It changes the high carrier frequency of the received radio signal to a lower, fixed intermediate frequency. This process makes it much easier to filter and process the signal efficiently.
- **Amplifier (Y):** It increases the strength (amplitude) of the detected audio/data signal to make it strong enough to drive the output device (like a speaker).
In simple words: X shifts the signal's frequency to a lower, manageable level. Y boosts the volume or power of the signal before it goes to the speaker.

Exam Tip: Memorize the standard sequence of blocks in a receiver (Antenna - Amplifier - IF Stage - Detector - Amplifier - Output) as it is a common exam question.

Question 20. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table.
Answer: The given logic circuit consists of three NAND gates.
- The first two NAND gates have their inputs tied together, which means they act as NOT gates. Thus, their outputs are \( A' = \bar{A} \) and \( B' = \bar{B} \).
- These outputs are fed into the third NAND gate. The final output \( Y \) is:
\( Y = \overline{A' \cdot B'} = \overline{\bar{A} \cdot \bar{B}} \)
Applying De Morgan's Law:
\( Y = \bar{\bar{A}} + \bar{\bar{B}} = A + B \)
This is the Boolean expression for an **OR gate**.

Truth Table:

In simple words: This combination of gates behaves like an OR gate. The output is "1" if either input A or input B (or both) is "1", and "0" only when both are "0".

Exam Tip: Always show intermediate steps like \( \bar{A} \) and \( \bar{B} \) in your truth table to secure full step marks.

Section C ( Short Answer Type Question Type 2)

Question 21. Two point charges \( q_1 = 10 \times 10^{-8} \text{ C} \) and \( q_2 = -2 \times 10^{-8} \text{ C} \) are separated by a distance of 60 cm in air.
(i) Find at what distance from the 1st charge, \( q_1 \), would the electric potential be zero.
(ii) Also calculate the electrostatic potential energy of the system.
Answer: Given:
\( q_1 = 10 \times 10^{-8} \text{ C} \), \( q_2 = -2 \times 10^{-8} \text{ C} \)
Distance \( d = 60 \text{ cm} = 0.6 \text{ m} \).

(i) Position where electric potential is zero:
Let the potential be zero at a point P at a distance \( x \) from the first charge \( q_1 \).
- **Case A: Point P lies between the two charges.**
Distance of P from \( q_2 \) is \( (0.6 - x) \).
\( V = \frac{k q_1}{x} + \frac{k q_2}{0.6 - x} = 0 \)
\( \implies \frac{10 \times 10^{-8}}{x} + \frac{-2 \times 10^{-8}}{0.6 - x} = 0 \)
\( \implies \frac{10}{x} = \frac{2}{0.6 - x} \)
\( \implies 10(0.6 - x) = 2x \)
\( \implies 6 - 10x = 2x \implies 12x = 6 \implies x = 0.5 \text{ m} = 50 \text{ cm} \).
- **Case B: Point P lies outside, closer to the smaller charge.**
Distance of P from \( q_1 \) is \( x' \) and from \( q_2 \) is \( (x' - 0.6) \).
\( \frac{10}{x'} = \frac{2}{x' - 0.6} \)
\( \implies 10(x' - 0.6) = 2x' \)
\( \implies 10x' - 6 = 2x' \implies 8x' = 6 \implies x' = 0.75 \text{ m} = 75 \text{ cm} \).

(ii) Electrostatic potential energy of the system:
\( U = \frac{k q_1 q_2}{d} \)
\( U = \frac{9 \times 10^9 \times (10 \times 10^{-8}) \times (-2 \times 10^{-8})}{0.6} \)
\( U = \frac{-18 \times 10^{-6}}{0.6} = -3 \times 10^{-5} \text{ J} \).
In simple words: (i) The electric potential becomes zero at a point 50 cm away from the first charge (between them) and at 75 cm away (on the outer side of the smaller charge). (ii) The total energy stored in the pair is negative because they attract each other, equal to \( -3 \times 10^{-5} \text{ Joules} \).

Exam Tip: Remember that potential is a scalar quantity, so we simply add the potentials algebraically. Always check for both internal and external points where the potential is zero.

Question 22. Two point charges 4Q, Q are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero? Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, \( Q = 2 \times 10^{-7} \text{ C} \).
Answer: Let \( x \) be the distance from the \( 4Q \) charge where the net electric field is zero. The distance from the \( Q \) charge is \( (1 - x) \).
At this point, the magnitudes of the electric fields due to both charges are equal:
\( \frac{k (4Q)}{x^2} = \frac{k Q}{(1 - x)^2} \)
\( \implies \frac{4}{x^2} = \frac{1}{(1 - x)^2} \)
Taking square root on both sides:
\( \frac{2}{x} = \pm \frac{1}{1 - x} \)
- **Using positive sign:**
\( \frac{2}{x} = \frac{1}{1 - x} \implies 2(1 - x) = x \implies 2 - 2x = x \implies 3x = 2 \implies x = \frac{2}{3} \text{ m} \).
- **Using negative sign:**
\( \frac{2}{x} = -\frac{1}{1 - x} \implies 2 - 2x = -x \implies x = 2 \text{ m} \).
Since the field cannot be zero outside the region between two like charges, \( x = 2\text{ m} \) is neglected. Thus, the field is zero at a distance of \( \frac{2}{3}\text{ m} \) from the \( 4Q \) charge.

Potential Energy of the System:
\( U = \frac{k (4Q)(Q)}{d} = \frac{4 k Q^2}{1} \)
Substitute \( Q = 2 \times 10^{-7} \text{ C} \):
\( U = 4 \times 9 \times 10^9 \times (2 \times 10^{-7})^2 \)
\( U = 36 \times 10^9 \times 4 \times 10^{-14} = 1.44 \times 10^{-3} \text{ J} \).
In simple words: The electric field cancels out at a point \( \frac{2}{3} \) of a meter away from the larger charge. The potential energy stored in this system is positive because the charges repel each other, equal to \( 1.44 \times 10^{-3} \text{ Joules} \).

Exam Tip: Unlike potential, electric field is a vector. For like charges, the field can only be zero at a point between them, never on the outside.

Question 23. Identify the following electromagnetic radiations as per the wavelengths given below. Write one application of each. (a) \( 10^{-3} \text{ nm} \) (b) \( 10^{-3} \text{ m} \) (c) \( 1\text{ nm} \)
Answer:
(a) **Wavelength \( 10^{-3} \text{ nm} = 10^{-12} \text{ m} \):**
- **Type:** Gamma (\( \gamma \)) rays (or high-energy X-rays).
- **Application:** Used in radiation therapy to target and destroy cancer cells.

(b) **Wavelength \( 10^{-3} \text{ m} = 1\text{ mm} \):**
- **Type:** Microwaves.
- **Application:** Widely used in radar systems for aircraft navigation and speed detection.

(c) **Wavelength \( 1\text{ nm} = 10^{-9} \text{ m} \):**
- **Type:** X-rays.
- **Application:** Used in medical diagnostic imaging to detect bone fractures.
In simple words: (a) Extremely short waves are Gamma rays, used to treat cancer. (b) 1-millimeter waves are Microwaves, used in radar. (c) 1-nanometer waves are X-rays, used to take pictures of bones.

Exam Tip: Be sure to write both the correct wave type and a specific, well-known practical application to gain full marks.

Question 24. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E?
Answer: Looking at the circuit, we have two batteries of \( 8\text{ V} \) and \( 3\text{ V} \) opposing each other, along with resistors in series.
Applying Kirchhoff’s Voltage Law (KVL) around the single loop:
\( I(5) + I(R) + I(15) = V_{\text{net}} \)
\( \Rightarrow 5(0.2) + R(0.2) + 15(0.2) = 8 - 3 \)
\( \Rightarrow 1.0 + 0.2R + 3.0 = 5 \)
\( \Rightarrow 4.0 + 0.2R = 5 \)
\( \Rightarrow 0.2R = 1 \implies R = 5 \, \Omega \).

Potential Difference between B and E:
The branch between B and E has the resistor of \( 5 \, \Omega \) carrying a current of \( 0.2\text{ A} \).
\( V_{BE} = I \times 5 = 0.2 \times 5 = 1\text{ V} \).
In simple words: To get a current of 0.2 Amps in this circuit, the mystery resistor R must be 5 Ohms. The voltage drop between points B and E is exactly 1 Volt.

Exam Tip: Pay close attention to the polarity of the batteries. Since they oppose each other, the net emf is found by subtracting them (\( 8 - 3 = 5\text{ V} \)).

Section C ( Short Answer Type Question Type 2 )

Question 25. A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms\(^{-2}\)]
Answer: Let \( m \) be the mass per unit length of the wire CD. For the wire CD to remain suspended in mid-air, the downward gravitational force acting per unit length must be perfectly balanced by the upward magnetic force per unit length exerted by wire AB.
The upward magnetic force per unit length between two parallel current-carrying wires is given by:
\[ f = \frac{\mu_0}{2\pi}\frac{I_1 I_2}{r} \]
At equilibrium, this force balances the weight per unit length of the wire CD:
\[ \frac{\mu_0 I_1 I_2}{2\pi r} = m g \]
Given:
\( I_1 = 12\text{ A} \)
\( I_2 = 5\text{ A} \)
\( r = 1\text{ mm} = 1 \times 10^{-3}\text{ m} \)
\( g = 10\text{ ms}^{-2} \)
\( \mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A} \)

Substituting these values into the equilibrium equation:
\[ \frac{(4\pi \times 10^{-7}) \times 12 \times 5}{2\pi \times (1 \times 10^{-3})} = m \times 10 \]
\[ \Rightarrow \frac{2 \times 10^{-7} \times 60}{1 \times 10^{-3}} = 10 m \]
\[ \Rightarrow m = 1.2 \times 10^{-3}\text{ kg m}^{-1} \]

Direction of Current:
To produce an upward magnetic force on wire CD, the force between the two wires must be repulsive. Since parallel currents flowing in opposite directions repel each other, the current in wire CD must flow in the opposite direction to the current in wire AB.
In simple words: The upward magnetic push from the bottom wire needs to equal the downward pull of gravity on the top wire. For this push to happen, the currents in both wires must run in opposite directions.

Exam Tip: Always state the condition for equilibrium first (\( f = mg \)) and explicitly mention that opposite currents are required to generate the repulsive force needed to oppose gravity.

Question 26. Using Biot-Savart law, deduce an expression for the magnetic field on the axis of a circular current loop. Hence obtain the expression for the magnetic field at the centre of the loop.
Answer: Let us consider a circular loop of radius \( a \) carrying a steady current \( I \). Let \( P \) be a point on the axis of the loop at a distance \( r \) from its center \( C \).
Consider a small current element \( Idl \) at the top of the loop. The distance from this element to point \( P \) is \( s = \sqrt{a^2 + r^2} \).
According to the Biot-Savart law, the magnitude of the magnetic field \( dB \) at point \( P \) due to this current element is:
\[ dB = \frac{\mu_0}{4\pi}\frac{I dl \sin\theta}{s^2} \]
Since the angle \( \theta \) between the current element \( d\vec{l} \) and the position vector \( \vec{s} \) is \( 90^\circ \):
\[ dB = \frac{\mu_0}{4\pi}\frac{I dl}{s^2} \]
Resolving \( d\vec{B} \) into two perpendicular components:
1. \( dB\cos\phi \), perpendicular to the axis of the loop.
2. \( dB\sin\phi \), along the axis of the loop.
Due to symmetry, for any two diametrically opposite current elements, the components perpendicular to the axis are equal and opposite, and thus cancel each other out. The axial components, however, point in the same direction and add up.
Therefore, the total magnetic field \( B \) at \( P \) is the sum of all axial components:
\[ B = \int dB \sin\phi \]
From the triangle, \( \sin\phi = \frac{a}{s} \). Substituting the expressions for \( dB \) and \( \sin\phi \):
\[ B = \int \left( \frac{\mu_0}{4\pi}\frac{I dl}{s^2} \right) \frac{a}{s} = \frac{\mu_0 I a}{4\pi s^3} \int dl \]
Since the total integration of the element over the circular path \( \int dl = 2\pi a \):
\[ B = \frac{\mu_0 I a}{4\pi s^3} (2\pi a) = \frac{\mu_0 I a^2}{2s^3} \]
Substituting \( s = (a^2 + r^2)^{1/2} \):
\[ B = \frac{\mu_0 I a^2}{2(a^2 + r^2)^{3/2}} \]
If the loop has \( N \) turns, the total magnetic field is:
\[ B = \frac{\mu_0 N I a^2}{2(a^2 + r^2)^{3/2}} \]

Magnetic Field at the Centre:
At the center of the loop, \( r = 0 \). Substituting this into the formula:
\[ B = \frac{\mu_0 N I a^2}{2(a^2)^{3/2}} = \frac{\mu_0 N I}{2a} \].
In simple words: Off-axis magnetic components from opposite sides of the circle cancel each other out, while the components pointing straight along the axis add up. When you are at the exact center of the loop, the formula simplifies because the distance along the axis is zero.

Exam Tip: Be sure to explicitly state that the perpendicular components cancel out due to symmetry. This is a key conceptual point that evaluators check for.

Question 27. Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it. Write two important limitations of a refracting telescope over a reflecting type telescope.
Answer: The labelled ray diagram of a refracting telescope forming the final image at infinity is shown below:
Magnifying Power:
The magnifying power (\( m \)) of an astronomical telescope is defined as the ratio of the angle subtended by the final image at the eye (\( \beta \)) to the angle subtended by the object at the objective lens or eye (\( \alpha \)).
The expression for its magnifying power when the final image is formed at infinity is:
\[ m = -\frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece.

Two limitations of a refracting telescope over a reflecting type telescope:
1. **Chromatic Aberration:** Different colors of light focus at different points after passing through a lens, which causes a colored, blurred border around the observed image. This issue does not occur with mirrors.
2. **Spherical Aberration:** Spherical lenses do not focus all rays at a single point, causing spherical distortion. Parabolic mirrors can easily eliminate this defect.
In simple words: The magnifying power tells us how much larger the image appears compared to the real object. Refracting telescopes use lenses, which create colored halos (chromatic aberration) and blurriness (spherical aberration) that reflecting telescopes (which use mirrors) do not have.

Exam Tip: When drawing the ray diagram, make sure the objective lens is drawn significantly larger than the eyepiece, and the intermediate image is clearly labeled at the common focal plane.

Question 28. Write the principle of working of a potentiometer. Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a given cell.
Answer: **Principle of Potentiometer:**
When a steady current flows through a wire of uniform cross-sectional area and uniform composition, the potential drop across any portion of the wire is directly proportional to the length of that portion.
\[ V \propto l \implies V = \Phi l \]
where \( \Phi \) is the potential gradient (potential drop per unit length of the wire).

Determination of Internal Resistance:
The circuit diagram for measuring the internal resistance of a cell is shown below:
Let \( \Phi \) be the potential gradient of the potentiometer wire.
1. **When Key \( K_2 \) is open:** No current is drawn from the cell, and the potentiometer measures its electromotive force (emf) \( \varepsilon \). If the balancing length is \( l_1 \):
\[ \varepsilon = \Phi l_1 \] --- (1)
2. **When Key \( K_2 \) is closed:** Current is drawn from the cell through the shunt resistance box \( R \), and the potentiometer measures the terminal potential difference \( V \). If the new balancing length is \( l_2 \):
\[ V = \Phi l_2 \] --- (2)
Dividing equation (1) by (2):
\[ \frac{\varepsilon}{V} = \frac{l_1}{l_2} \] --- (3)
We know the relation between internal resistance \( r \), external resistance \( R \), emf \( \varepsilon \), and terminal voltage \( V \) is:
\[ r = R\left(\frac{\varepsilon}{V} - 1\right) \]
Substituting the value from equation (3) into this expression:
\[ r = R\left(\frac{l_1}{l_2} - 1\right) \]
By plugging in the values of \( R, l_1, \) and \( l_2 \), the internal resistance of the cell can be determined.
In simple words: The potentiometer wire drops voltage steadily along its length. By measuring the balancing lengths when the cell is idle (open circuit) and when it is working (closed circuit), we can calculate how much resistance exists inside the cell itself.

Exam Tip: Do not forget to include the resistance box \( R \) and the keys \( K_1, K_2 \) in your circuit diagram, as omitting them will lead to deduction of marks.

Question 31. Describe briefly, with the help of labelled diagram, working of a step-up transformer. A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.
Answer: **Step-Up Transformer:**
A step-up transformer consists of a laminated soft iron core with two coils wound around it: a primary coil (with \( N_p \) turns) and a secondary coil (with \( N_s \) turns, where \( N_s > N_p \)).

Working:**
When an alternating voltage is applied to the primary coil, an alternating magnetic flux is set up in the laminated core. This changing flux passes through the secondary coil, inducing an alternating emf across its terminals due to mutual induction. The induced emf is given by:
\[ V_s = -N_s \frac{d\Phi}{dt} \quad \text{and} \quad V_p = -N_p \frac{d\Phi}{dt} \]
Dividing these equations gives:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Since \( N_s > N_p \), we have \( V_s > V_p \), meaning the output voltage is stepped up.

Conservation of Energy:
No, this does not violate the principle of conservation of energy. For an ideal transformer (100% efficient), the input power is equal to the output power:
\[ \text{Power}_{\text{input}} = \text{Power}_{\text{output}} \implies I_p V_p = I_s V_s \]
\[ \Rightarrow \frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s} \]
Since \( V_s > V_p \), we must have \( I_s < I_p \). This means that when the transformer steps up the voltage, the current decreases proportionally to ensure that the total electric power remains constant. Therefore, energy is conserved.
In simple words: A step-up transformer increases the voltage, but at the same time, it lowers the current. Since power is voltage multiplied by current, the total energy output never exceeds the energy input.

Exam Tip: Underline the relation \( I_s V_s = I_p V_p \) when explaining the conservation of energy to make your explanation clear and concise.

Question 32. Write the expression for the force acting on a charged particle of charge q moving with velocity \( \vec{v} \) in the presence of magnetic field \( \vec{B} \). Show that in the presence of this force (i) the kinetic energy of the particle does not change (ii) its instantaneous power is zero.
Answer: The magnetic Lorentz force \( \vec{F} \) acting on a charged particle \( q \) moving with a velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by the vector cross product:
\[ \vec{F} = q(\vec{v} \times \vec{B}) \]

(i) Kinetic Energy does not change:
According to the properties of the vector cross product, the magnetic force \( \vec{F} \) is always perpendicular to the velocity \( \vec{v} \) of the particle:
\[ \vec{F} \cdot \vec{v} = 0 \]
The rate of change of kinetic energy is given by the work-energy theorem:
\[ \frac{d(K.E.)}{dt} = \vec{F} \cdot \vec{v} = 0 \]
Since the derivative of kinetic energy with respect to time is zero, the kinetic energy of the particle remains constant.

(ii) Instantaneous Power is zero:
The instantaneous power \( P \) delivered by a force \( \vec{F} \) to a particle moving with velocity \( \vec{v} \) is:
\[ P = \vec{F} \cdot \vec{v} \]
As shown above, because the magnetic force \( \vec{F} \) is perpendicular to the velocity \( \vec{v} \), we have:
\[ P = F v \cos 90^\circ = 0 \ ]
Thus, the instantaneous power delivered by the magnetic force is always zero.
In simple words: The magnetic force always pushes a moving charge sideways (at a 90-degree angle to its motion). Because the push is perpendicular, it only changes the direction of the particle, not its speed. Therefore, no work is done, no power is delivered, and the kinetic energy remains the same.

Exam Tip: Be sure to write the vector dot product \( \vec{F} \cdot \vec{v} = 0 \) to prove both parts mathematically.

Question 33. (i) Out of blue and red light which is deviated more by a prism? Give reason. (ii) Give the formula that can be used to determine refractive index of materials of a prism in minimum deviation condition.
Answer:
(i) **Blue light** is deviated more than red light when passing through a prism.
Reason: According to Cauchy's formula, the refractive index \( \mu \) of a material is inversely proportional to the wavelength \( \lambda \) of light (\( \mu \propto \frac{1}{\lambda} \)). Since the wavelength of blue light is shorter than that of red light (\( \lambda_{\text{blue}} < \lambda_{\text{red}} \)), the glass of the prism has a higher refractive index for blue light than for red light (\( \mu_{\text{blue}} > \mu_{\text{red}} \)).
The angle of deviation \( \delta \) for a thin prism is given by \( \delta = (\mu - 1)A \). Since \( \mu_{\text{blue}} > \mu_{\text{red}} \), the angle of deviation is larger for blue light.

(ii) **Prism Formula:**
The refractive index \( \mu \) of the material of a prism in the minimum deviation condition is given by:
\[ \mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
where \( A \) is the angle of the prism and \( D_m \) is the angle of minimum deviation.
In simple words: Blue light has shorter waves than red light, which makes the glass slow it down and bend it more. The refractive index of the prism can be calculated using the sine of the angles of the prism and the minimum deviation.

Exam Tip: In part (i), state the relationship between wavelength and refractive index (\( \mu \propto 1/\lambda \)) explicitly to make your explanation mathematically sound.

Question 34. Using Gauss’s law in electrostatic, drive an expression for the electric field due to an infinitely long straight wire of linear charge density \( \lambda \).
Answer: Consider an infinitely long straight wire with a uniform linear charge density \( \lambda \). To find the electric field at a distance \( r \) from the wire, we construct a coaxial cylindrical Gaussian surface of radius \( r \) and length \( l \).
The electric field \( \vec{E} \) is directed radially outward, perpendicular to the wire at all points.
The cylindrical Gaussian surface has three parts:
1. **Two flat end caps (A and B):** The area vector \( d\vec{A} \) is perpendicular to the electric field \( \vec{E} \) (\( \theta = 90^\circ \)). Thus, the flux through these flat ends is zero:
\[ \Phi_1 = \Phi_2 = \int E \, dA \cos 90^\circ = 0 \]
2. **The curved surface (C):** The area vector \( d\vec{A} \) is parallel to the electric field \( \vec{E} \) (\( \theta = 0^\circ \)). The electric field magnitude \( E \) is constant over this surface.
\[ \Phi_3 = \int E \, dA \cos 0^\circ = E \int dA = E (2\pi r l) \]
The total electric flux \( \Phi \) through the Gaussian surface is:
\[ \Phi = \Phi_1 + \Phi_2 + \Phi_3 = 0 + 0 + E (2\pi r l) = E (2\pi r l) \] --- (1)
According to Gauss's Law, the total flux is related to the enclosed charge \( q \):
\[ \Phi = \frac{q}{\varepsilon_0} \] --- (2)
Since the linear charge density is \( \lambda = \frac{q}{l} \), the enclosed charge is \( q = \lambda l \).
Comparing equations (1) and (2):
\[ E (2\pi r l) = \frac{\lambda l}{\varepsilon_0} \]
\[ \Rightarrow E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
This is the expression for the electric field.
In simple words: We draw an imaginary cylinder around the charged wire. The electric field only pushes outward through the curved side of the cylinder, not the flat ends. Equating this outward push to the charge inside gives the electric field formula.

Exam Tip: Be sure to write down the integration steps for both the flat surfaces and the curved surface separately to show a complete, rigorous proof.

Question 35. Using Bohr's postulates of the atomic model, derive the expression for radius of \( n^{\text{th}} \) electron orbit. Hence obtain the expression for Bohr's radius.
Answer: Let us consider an electron of mass \( m \) and charge \( e \) revolving around a nucleus of charge \( Ze \) in a circular orbit of radius \( r \) with a velocity \( v \).
1. **Electrostatic attraction provides centripetal force:**
\[ \frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{Z e^2}{r^2} \implies m v^2 = \frac{Z e^2}{4\pi\varepsilon_0 r} \] --- (1)
2. **Bohr's quantization condition:**
The angular momentum is an integral multiple of \( \frac{h}{2\pi} \):
\[ m v r = \frac{n h}{2\pi} \implies v = \frac{n h}{2\pi m r} \] --- (2)
Substituting the value of \( v \) from (2) into (1):
\[ m \left( \frac{n h}{2\pi m r} \right)^2 = \frac{Z e^2}{4\pi\varepsilon_0 r} \]
\[ \Rightarrow \frac{m n^2 h^2}{4\pi^2 m^2 r^2} = \frac{Z e^2}{4\pi\varepsilon_0 r} \]
Simplifying this expression for the radius \( r \):
\[ r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2} \]
This is the expression for the radius of the \( n^{\text{th}} \) orbit.

Bohr's Radius:
Bohr's radius is defined as the radius of the innermost orbit (\( n = 1 \)) of a hydrogen atom (\( Z = 1 \)):
\[ r_1 = \frac{h^2 \varepsilon_0}{\pi m e^2} \]
This value is a constant, approximately equal to \( 0.529\text{ \AA} \).
In simple words: By balancing the electric pull of the nucleus with the circular spin of the electron, and using the rule that the electron's spin momentum comes in fixed packets, we can calculate the exact size of the electron's orbit.

Exam Tip: Remember to state the conditions \( n=1 \) and \( Z=1 \) explicitly when defining Bohr's radius to avoid losing partial marks.

Section E ( long Answer Type Questions )

Question 41. Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies? A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = \( 6.4 \times 10^6 \text{ m} \))
Answer: **Type of Waves:** Space waves (or ultra-high frequency radio waves/microwaves) are used for line-of-sight (LOS) communication.
**Frequency Range:** The range of frequencies is \( 40\text{ MHz} \) and above.

Calculation of Maximum Distance (\( d_m \)):
Given:
Height of transmitting antenna, \( h_T = 20\text{ m} \)
Height of receiving antenna, \( h_R = 45\text{ m} \)
Radius of the Earth, \( R = 6.4 \times 10^6\text{ m} \)
The maximum line-of-sight distance \( d_m \) between the two antennas is given by:
\[ d_m = \sqrt{2 R h_T} + \sqrt{2 R h_R} \]
Substitute the values:
\[ d_m = \sqrt{2 \times (6.4 \times 10^6) \times 20} + \sqrt{2 \times (6.4 \times 10^6) \times 45} \]
\[ \Rightarrow d_m = \sqrt{256 \times 10^6} + \sqrt{576 \times 10^6} \]
\[ \Rightarrow d_m = 16 \times 10^3 + 24 \times 10^3 = 40 \times 10^3\text{ m} \]
\[ \Rightarrow d_m = 40\text{ km} \].
Thus, the maximum communication distance between the two antennas is \( 40\text{ km} \).
In simple words: Space waves are used for straight-line communication. Because the Earth is curved, the height of both towers determines how far they can "see" each other. For heights of 20m and 45m, the maximum distance is exactly 40 kilometers.

Exam Tip: Write down each term of the formula \( d_m = \sqrt{2 R h_T} + \sqrt{2 R h_R} \) clearly and show step-by-step simplification of the square roots to ensure full marks.

Question 42. (a) Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'.
(b) How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant K?
Answer: **(a) Derivation of Electrostatic Energy Stored:**
Let us consider a capacitor of capacitance \( C \). Suppose at any instant during the charging process, the charge on the plates is \( q \) and the potential difference is \( V \):
\[ V = \frac{q}{C} \]
If an additional small charge \( dq \) is transferred from one plate to the other, the small work done \( dW \) is:
\[ dW = V \, dq = \frac{q}{C} \, dq \]
The total work done \( W \) in charging the capacitor from an uncharged state (\( q = 0 \)) to a final charge \( Q \) is found by integration:
\[ W = \int_0^Q \frac{q}{C} \, dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q = \frac{Q^2}{2C} \]
This work is stored as electrostatic potential energy \( U \) in the electric field of the capacitor:
\[ U = \frac{Q^2}{2C} = \frac{1}{2} C V^2 = \frac{1}{2} Q V \]

**(b) Effect of introducing a Dielectric Material (assuming the charging battery is disconnected):**
When a dielectric slab of dielectric constant \( K \) is introduced, the charge \( Q \) on the plates remains constant (\( Q = Q_0 \)).
- **(i) Energy Stored:**
The new capacitance is \( C' = K C \). The new energy \( U' \) is:
\[ U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(KC)} = \frac{U}{K} \]
Thus, the stored energy decreases by a factor of \( K \).
- **(ii) Electric Field:**
The electric field \( E \) decreases by a factor of \( K \):
\[ E = \frac{E_0}{K} \]
This reduction occurs because of the opposing electric field induced inside the polarized dielectric material.
In simple words: (a) The energy stored in a capacitor is the total work done to push charge onto its plates. (b) If you fill a disconnected capacitor with a dielectric, the electric field and the stored energy both drop because the dielectric polarizes and fights the original electric field.

Exam Tip: Always state whether the charging battery remains connected or is disconnected when a dielectric is introduced, as this determines whether the charge or the voltage remains constant.

Question 44. (a) In Young's double slit experiment, derive the expression for position of points having condition for (i) constructive interference and (ii) destructive interference on the screen.
(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
Answer: **(a) Derivation of Fringe Positions:**
Let \( S_1 \) and \( S_2 \) be two coherent sources separated by a small distance \( d \). Let a screen be placed at a distance \( D \) from the slits. Consider a point \( P \) on the screen at a distance \( x \) from the center \( O \).
The path difference between the waves reaching \( P \) from \( S_1 \) and \( S_2 \) is:
\[ \Delta p = S_2P - S_1P \]
From the geometry of the setup, when \( d \ll D \), the path difference can be approximated as:
\[ \Delta p = \frac{x d}{D} \]
- **(i) For Constructive Interference (Bright Fringes):**
The path difference must be an integral multiple of wavelength \( \lambda \):
\[ \frac{x d}{D} = n\lambda \implies x_n = \frac{n\lambda D}{d} \quad (\text{where } n = 0, 1, 2 \dots) \]
- **(ii) For Destructive Interference (Dark Fringes):**
The path difference must be an odd multiple of half-wavelength:
\[ \frac{x d}{D} = (2n + 1)\frac{\lambda}{2} \implies x_n = (2n + 1)\frac{\lambda D}{2d} \quad (\text{where } n = 0, 1, 2 \dots) \]

**(b) Calculation of Coinciding Fringes:**
Given:
\( \lambda_1 = 800\text{ nm} = 800 \times 10^{-9}\text{ m} \)
\( \lambda_2 = 600\text{ nm} = 600 \times 10^{-9}\text{ m} \)
\( D = 1.4\text{ m} \)
\( d = 0.28\text{ mm} = 0.28 \times 10^{-3}\text{ m} \)
Let the \( n_1^{\text{th}} \) bright fringe of \( \lambda_1 \) coincide with the \( n_2^{\text{th}} \) bright fringe of \( \lambda_2 \):
\[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2 \]
\[ \Rightarrow \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{600}{800} = \frac{3}{4} \]
The minimum integral values are \( n_1 = 3 \) and \( n_2 = 4 \).
The least distance \( y_{\text{min}} \) from the central maximum is:
\[ y_{\text{min}} = \frac{n_1 \lambda_1 D}{d} = \frac{3 \times (800 \times 10^{-9}) \times 1.4}{0.28 \times 10^{-3}} \]
\[ y_{\text{min}} = \frac{3.36 \times 10^{-6}}{0.28 \times 10^{-3}} = 12 \times 10^{-3}\text{ m} = 12\text{ mm} \].
Thus, the least distance where the bright fringes of both wavelengths coincide is \( 12\text{ mm} \).
In simple words: (a) Waves add up to make bright lines when the path difference is a whole wavelength, and cancel out to make dark lines when it is a half-wavelength. (b) The third bright line of the 800nm light lands exactly on top of the fourth bright line of the 600nm light at a distance of 12mm from the center.

Exam Tip: In part (b), write down the ratio \( \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \) first, as establishing this correct ratio is often a key step in the marking scheme.

Question 45. Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point. (a) In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope.
Answer: The ray diagram showing the working of a compound microscope is shown below:
Objective Expression for Total Magnification:
The total magnification \( m \) of a compound microscope is the product of the linear magnification of the objective \( m_o \) and the angular magnification of the eyepiece \( m_e \):
\[ m = m_o \times m_e \]
Since the objective forms a real image, \( m_o = -\frac{v_o}{u_o} \).
Since the eyepiece acts as a simple magnifier forming the image at the near point, \( m_e = \left(1 + \frac{D}{f_e}\right) \).
Thus:
\[ m = -\frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right) \]

(a) Numerical Estimation:
Given:
Object distance for objective, \( u_o = -1.5\text{ cm} \)
Focal length of objective, \( f_o = +1.25\text{ cm} \)
Focal length of eyepiece, \( f_e = +5\text{ cm} \)
Least distance of distinct vision, \( D = 25\text{ cm} \)

First, find the image distance \( v_o \) for the objective lens using the lens formula:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
\[ \Rightarrow \frac{1}{1.25} = \frac{1}{v_o} - \frac{1}{-1.5} \]
\[ \Rightarrow \frac{1}{v_o} = \frac{1}{1.25} - \frac{1}{1.5} = \frac{100}{125} - \frac{10}{15} = \frac{4}{5} - \frac{2}{3} = \frac{2}{15} \]
\[ \Rightarrow v_o = +7.5\text{ cm} \]

Now, calculate the magnifying power \( m \):
\[ m = -\frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right) = -\left(\frac{7.5}{-1.5}\right)\left(1 + \frac{25}{5}\right) \]
\[ \Rightarrow m = 5 \times (1 + 5) = 5 \times 6 = 30 \]
The magnifying power of the microscope is \( 30 \) (or \( -30 \) to indicate an inverted final image).
In simple words: The compound microscope uses two lenses to multiply the magnification. By using the lens equation, we find the first lens magnifies the object 5 times, and the second lens magnifies that image 6 times, giving a total magnification of 30.

Exam Tip: Keep track of the negative sign in the final magnification value, which mathematically indicates that the final image is inverted with respect to the original object.

Question 49. (a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
(b) The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer: **(a) de Broglie's Explanation of Bohr's Quantization Postulate:**
According to de Broglie, a revolving electron in a circular orbit behaves as a standing wave. For the wave to be stable and not destroy itself by destructive interference, the circumference of the circular orbit must be equal to an integral number of wavelengths:
\[ 2\pi r = n\lambda \] --- (1)
According to de Broglie’s hypothesis, the wavelength of a moving electron is:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \] --- (2)
Substituting equation (2) into (1):
\[ 2\pi r = n \left( \frac{h}{mv} \right) \]
Rearranging this equation to solve for angular momentum (\( mvr \)):
\[ m v r = \frac{n h}{2\pi} \]
This is exactly Bohr's second postulate of quantization of angular momentum. Thus, de Broglie's wave hypothesis provides a clear physical explanation for Bohr's postulate.

**(b) Kinetic and Potential Energies in Ground State:**
The total energy of the electron in the ground state is \( E = -13.6\text{ eV} \).
We know the relationships between total energy \( E \), kinetic energy \( K \), and potential energy \( U \):
1. **Kinetic Energy (\( K \)):**
\[ K = -E = -(-13.6\text{ eV}) = 13.6\text{ eV} \]
2. **Potential Energy (\( U \)):**
\[ U = 2E = 2(-13.6\text{ eV}) = -27.2\text{ eV} \].
In simple words: (a) An electron can only orbit where its wave fits perfectly around the circle without overlapping and destroying itself. (b) In the lowest energy state, the electron has 13.6 eV of moving energy (kinetic) and -27.2 eV of stored electric energy (potential).

Exam Tip: Remember that kinetic energy can never be negative. If your calculation yields a negative kinetic energy, re-check your signs immediately.

Question 50. (a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.
(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.
(c) How is the magnetic field inside a given solenoid made strong?
Answer: (a) Derivation using Ampere's Circuital Law:
Consider a long straight solenoid of length \( L \) having \( N \) total turns, so the number of turns per unit length is \( n = \frac{N}{L} \). Let a steady current \( I \) flow through the solenoid.
To find the magnetic field \( B \) inside the solenoid, we construct a rectangular Amperian loop \( PQRS \) of length \( l \), where:
- Side \( PQ \) lies parallel to the axis inside the solenoid.
- Sides \( QR \) and \( SP \) are perpendicular to the solenoid axis.
- Side \( RS \) lies entirely outside the solenoid.

According to Ampere's Circuital Law:
\( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}} \)

We can split the line integral of \( \vec{B} \) over the closed loop \( PQRS \) into four parts:
\( \oint_{PQRS} \vec{B} \cdot d\vec{l} = \int_{P}^{Q} \vec{B} \cdot d\vec{l} + \int_{Q}^{R} \vec{B} \cdot d\vec{l} + \int_{R}^{S} \vec{B} \cdot d\vec{l} + \int_{S}^{P} \vec{B} \cdot d\vec{l} \)

Evaluating each term:
1. For side \( PQ \), the magnetic field is uniform and parallel to the path, so:
\( \int_{P}^{Q} \vec{B} \cdot d\vec{l} = \int_{P}^{Q} B dl \cos 0^\circ = B \int dl = B l \)
2. For sides \( QR \) and \( SP \), the magnetic field is perpendicular to the displacement vector (\( \theta = 90^\circ \)), so:
\( \int_{Q}^{R} \vec{B} \cdot d\vec{l} = 0 \quad \text{and} \quad \int_{S}^{P} \vec{B} \cdot d\vec{l} = 0 \)
3. For side \( RS \) outside the solenoid, the magnetic field is negligible (\( B \approx 0 \)), so:
\( \int_{R}^{S} \vec{B} \cdot d\vec{l} = 0 \)

Adding these together gives the total path integral:
\( \oint_{PQRS} \vec{B} \cdot d\vec{l} = B l \)

The total current enclosed by the Amperian loop is the current in each turn multiplied by the number of turns inside the loop length \( l \):
\( I_{\text{enclosed}} = (n l) I \)

Applying Ampere's Law:
\( B l = \mu_0 (n l I) \)
\( B = \mu_0 n I \)

(b) Comparison Between a Toroid and a Solenoid:
- Solenoid: A solenoid is a long coil of wire wrapped in a cylindrical shape with two open ends. Its magnetic field is nearly uniform inside and exits at the ends, forming external loops.
- Toroid: A toroid is essentially a solenoid bent into a continuous closed ring (doughnut shape). It has no open ends. Its magnetic field is entirely confined inside the core of the ring, with zero magnetic field in the inner empty region and outside the toroid.

Magnetic Field Lines Comparison:

(c) Making the Magnetic Field Inside a Solenoid Stronger:
The magnetic field inside can be intensified by:
1. Placing a high magnetic permeability core, like soft iron, inside the solenoid coil (which multiplies the field strength by the relative permeability \( \mu_r \)).
2. Increasing the number of turns per unit length (\( n \)) by winding the wire tighter.
3. Increasing the magnitude of the electrical current (\( I \)) passing through the coils.
In simple words: Ampere's Law helps calculate the magnetic field by tracing a loop around the current. For a solenoid, the field behaves like a straight bar magnet. To make this magnet stronger, you can increase the current, pack more turns of wire together, or put a soft iron rod down the center.

Exam Tip: Be sure to write the formula \( B = \mu_0 n I \) and explicitly list the three methods to make the field stronger, as each method is worth distinct marks on school examinations.