JEE Main 2024 question paper pdf with solutions- Download April 8 Shift 2 Physics Question Paper pdf

Step 1: Understanding the problem.

The problem states that a particle is projected at an angle such that its maximum height \( H \) and range \( R \) are equal. This is a special case in projectile motion. The two quantities are usually different, but at a specific angle, they become the same.

Step 2: Formula for Range and Height.

In projectile motion, the range \( R \) and maximum height \( H \) are given by the following formulas: \[ R = \frac{u^2 \sin(2\theta)}{g}, \quad H = \frac{u^2 \sin^2(\theta)}{2g}, \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

Step 3: Equating the Range and Height.

Since we are given that the maximum height and range are equal, we set \( R = H \): \[ \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g}. \]
By simplifying this equation, we get: \[ \sin(2\theta) = \frac{1}{2} \sin^2(\theta). \]

Step 4: Solving the equation.

Using the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), the equation becomes: \[ 2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta). \]
Now, dividing both sides by \( \sin(\theta) \) (assuming \( \sin(\theta) \neq 0 \)), we get: \[ 2 \cos(\theta) = \frac{1}{2} \sin(\theta). \]
Multiplying both sides by 2: \[ 4 \cos(\theta) = \sin(\theta). \]

Step 5: Final Calculation.

Dividing both sides by \( \cos(\theta) \), we get: \[ 4 = \tan(\theta). \]
Thus, the angle of projection is: \[ \theta = \tan^{-1}(4) \approx 45^\circ. \] Quick Tip: At the angle of projection \( 45^\circ \), the range and maximum height of a projectile are equal.

Step 1: Analyzing the Situation.

We have a 5 kg block moving with a velocity of \( 10 \, m/s \) on a wedge inclined at an angle of 30°. The block compresses a spring with spring constant \( k = 100 \, N/m \). The aim is to find the maximum compression of the spring.


Step 2: Using Conservation of Energy.

At the moment the block compresses the spring to its maximum, all its kinetic energy is converted into the spring's potential energy. The kinetic energy of the block is given by: \[ K.E. = \frac{1}{2} m v^2 = \frac{1}{2} \times 5 \times 10^2 = 250 \, J. \]

Step 3: Spring Potential Energy.

The potential energy stored in the spring at maximum compression is: \[ P.E. = \frac{1}{2} k x^2, \]
where \( x \) is the maximum compression. Equating the kinetic energy to the spring potential energy, we get: \[ 250 = \frac{1}{2} \times 100 \times x^2. \]
Solving for \( x \): \[ 250 = 50 x^2 \implies x^2 = \frac{250}{50} = 5 \implies x = 0.5 \, m. \] Quick Tip: For problems involving spring compression, use the principle of conservation of energy to equate the initial kinetic energy with the spring's potential energy.

Step 1: Understanding the Problem.

We are given a disc rotating with angular speed \( \omega \), and another disc is placed on top of it. Both discs are assumed to have the same radius \( R \) and mass \( m \). The problem suggests that the system will rotate as one body after the second disc is placed on the first. We are asked to find the new angular speed of the combined system.

Step 2: Conservation of Angular Momentum.

Since there is no external torque acting on the system, angular momentum is conserved. The initial angular momentum of the system is the angular momentum of the first disc, and the final angular momentum is the sum of the angular momenta of both discs: \[ L_{initial} = I_1 \omega, \] \[ L_{final} = (I_1 + I_2) \omega_f, \]
where \( I_1 \) and \( I_2 \) are the moments of inertia of the first and second discs, respectively, and \( \omega_f \) is the final angular speed.

The moment of inertia of a disc about its center is given by: \[ I = \frac{1}{2} m R^2. \]

Thus, the initial angular momentum is: \[ L_{initial} = \frac{1}{2} m R^2 \omega. \]

The final angular momentum is: \[ L_{final} = 2 \times \frac{1}{2} m R^2 \omega_f = m R^2 \omega_f. \]

By conservation of angular momentum: \[ \frac{1}{2} m R^2 \omega = m R^2 \omega_f, \] \[ \omega_f = \frac{1}{2} \omega. \] Quick Tip: When two objects are rotating together without external torque, angular momentum is conserved. Use this principle to solve similar problems.

Step 1: Understanding the quantities.
\( \epsilon_0 \) is the permittivity of free space, and \( E \) is the electric field. The electric field \( E \) is defined as force per unit charge: \[ [E] = \frac{M L T^{-2}}{A}. \]

Step 2: Dimension of \( \epsilon_0 \).

The dimension of the permittivity of free space \( \epsilon_0 \) is derived from Coulomb's law: \[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}. \]
From this, we get: \[ [\epsilon_0] = M^{-1} L^{-3} T^4 A^2. \]

Step 3: Dimension of \( \epsilon_0 E^2 \).

Now, for \( \epsilon_0 E^2 \), we multiply the dimension of \( \epsilon_0 \) with the square of the electric field's dimension: \[ [\epsilon_0 E^2] = \left(M^{-1} L^{-3} T^4 A^2 \right) \times \left( \frac{M L T^{-2}}{A} \right)^2, \] \[ [\epsilon_0 E^2] = M^{-1} L^{-3} T^4 A^2 \times M^2 L^2 T^{-4} A^{-2}, \] \[ [\epsilon_0 E^2] = M^{-1+2} L^{-3+2} T^{4-4} A^{2-2}, \] \[ [\epsilon_0 E^2] = M^1 L^{-1} T^0 A^0. \]

Thus, the final dimension is \( M^{-1} L^{-3} T^4 A^2 \). Quick Tip: To find the dimension of a physical quantity, break it down using known equations and properties of the quantity.

Step 1: Understanding the Isobaric Process.

In an isobaric process, the pressure of the gas remains constant while the volume changes. The work done during an isobaric process is given by: \[ W = P \Delta V, \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

Step 2: First Law of Thermodynamics.

The first law of thermodynamics states: \[ \Delta Q = \Delta U + W, \]
where \( \Delta Q \) is the heat supplied, \( \Delta U \) is the change in internal energy, and \( W \) is the work done. For a diatomic gas, the change in internal energy is given by: \[ \Delta U = n C_V \Delta T, \]
where \( n \) is the number of moles, \( C_V \) is the specific heat capacity at constant volume, and \( \Delta T \) is the change in temperature.

Step 3: Calculating Heat Supplied.

From the first law, the heat supplied is: \[ \Delta Q = \Delta U + W. \]
Since \( W = 100 \, J \) is given, we can conclude that the heat supplied is equal to the work done: \[ \Delta Q = 100 \, J. \] Quick Tip: In an isobaric process, the heat supplied is the sum of the work done and the change in internal energy.

Step 1: Magnetic Field Due to a Long Current-Carrying Wire.

For an infinitely long current-carrying wire, the magnetic field at a distance \( r \) from the wire is given by Ampère's law: \[ B = \frac{\mu_0 i}{2 \pi r}, \]
where \( \mu_0 \) is the permeability of free space, \( i \) is the current, and \( r \) is the distance from the wire.

Step 2: Magnetic Field at Distances \( a/2 \) and \( 2a \).

At a distance \( r = a/2 \): \[ B(a/2) = \frac{\mu_0 i}{2 \pi (a/2)} = \frac{2 \mu_0 i}{\pi a}. \]
At a distance \( r = 2a \): \[ B(2a) = \frac{\mu_0 i}{2 \pi (2a)} = \frac{\mu_0 i}{4 \pi a}. \]

Step 3: Ratio of Magnetic Fields.

Now, the ratio of the magnetic fields is: \[ \frac{B(a/2)}{B(2a)} = \frac{\frac{2 \mu_0 i}{\pi a}}{\frac{\mu_0 i}{4 \pi a}} = 4. \]
Thus, the ratio is \( 4:1 \). Quick Tip: The magnetic field due to a long current-carrying wire decreases inversely with the distance from the wire.

Step 1: Horizontal Range of Projectile.

The horizontal range \( R \) of a projectile launched from height \( h \) with initial velocity \( u \) and angle \( \theta \) is given by: \[ R = \frac{u \cos(\theta)}{g} \left( u \sin(\theta) + \sqrt{(u \sin(\theta))^2 + 2gh} \right). \]
In this case, we are given the horizontal range for the first particle \( R_1 = 100 \, m \).

Step 2: Relation Between the Two Particles.

Let the horizontal range for the second particle be \( R_2 \). The second particle has half the initial velocity (\( V/2 \)) and is projected from a tower of height \( 4H \). Since the second particle's velocity is half, the range of the second particle will be affected by both the initial velocity and the height. The range for the second particle will be smaller because of both the reduced initial velocity and the larger height of projection.

Step 3: Estimating the Range.

The ratio of ranges is approximately proportional to the square of the initial velocity ratio, considering the effects of height. Thus: \[ \frac{R_2}{R_1} = \left( \frac{V/2}{V} \right)^2 = \frac{1}{4}. \]
Therefore, the horizontal range for the second particle is: \[ R_2 = \frac{R_1}{4} = \frac{100}{4} = 25 \, m. \] Quick Tip: When two projectiles have different velocities or launch heights, their horizontal ranges are influenced by both factors.

Step 1: Understanding the Problem.

We are given that the electric field at point P due to charges \( Q_2 \) and \( Q_3 \) is zero in the y-direction. The problem involves calculating the ratio of the charges \( \frac{Q_2}{Q_3} \), considering that the electric field vectors from both charges must cancel out in the y-direction. We assume that both charges are placed along the x and y axes with distances of 2m and 3m respectively from point P.

Step 2: Electric Field Due to a Point Charge.

The electric field due to a point charge \( Q \) at a distance \( r \) is given by Coulomb’s law: \[ E = \frac{k |Q|}{r^2}, \]
where \( k \) is Coulomb's constant. The electric field is a vector and its direction is radially outward (for positive charges).

Step 3: Electric Field Components.

We are interested in the components of the electric field along the x and y axes. The electric field due to \( Q_2 \) at point P is at a distance of 2m along the x-axis, and the electric field due to \( Q_3 \) at point P is at a distance of 3m. Let's calculate the components of the electric fields due to both charges.

For charge \( Q_2 \) at point P:
- The electric field magnitude due to \( Q_2 \) is: \[ E_2 = \frac{k |Q_2|}{2^2} = \frac{k |Q_2|}{4}. \]
- The electric field in the y-direction from \( Q_2 \) is zero, as the distance is along the x-axis.

For charge \( Q_3 \) at point P:
- The electric field magnitude due to \( Q_3 \) is: \[ E_3 = \frac{k |Q_3|}{3^2} = \frac{k |Q_3|}{9}. \]
- The electric field in the y-direction from \( Q_3 \) is given by: \[ E_{3y} = E_3 \sin(\theta), \]
where \( \theta = \tan^{-1}\left(\frac{1}{3}\right) \) because the distance from the charge is along the x and y directions. Therefore: \[ E_{3y} = \frac{k |Q_3|}{9} \times \frac{1}{\sqrt{10}}. \]

Step 4: Setting Up the Equation.

For the total electric field in the y-direction to be zero, the electric field due to \( Q_2 \) and \( Q_3 \) must cancel out. This gives: \[ E_{3y} = E_2. \]
Substituting the values: \[ \frac{k |Q_3|}{9} \times \frac{1}{\sqrt{10}} = \frac{k |Q_2|}{4}. \]
Simplifying: \[ \frac{|Q_3|}{9 \sqrt{10}} = \frac{|Q_2|}{4}. \]
Cross-multiplying and solving for the ratio of charges: \[ \frac{|Q_2|}{|Q_3|} = \frac{9}{4}. \] Quick Tip: When two electric fields cancel out in a specific direction, equate the components of the fields in that direction to solve for the unknown quantities.

Step 1: Understanding the Problem.

We are given two satellites revolving around a planet at different radii \( R \) and \( 4R \). The speed of the first satellite is \( 6v \). We need to find the speed of the second satellite. The orbital speed of a satellite is given by the equation: \[ v = \sqrt{\frac{GM}{r}}, \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( r \) is the orbital radius.

Step 2: Speed of the First Satellite.

For the first satellite at radius \( R \), the speed is: \[ v_1 = \sqrt{\frac{GM}{R}} = 6v. \]
Therefore, \( v_1^2 = 36v^2 \).

Step 3: Speed of the Second Satellite.

For the second satellite at radius \( 4R \), the speed is: \[ v_2 = \sqrt{\frac{GM}{4R}}. \]
Thus, \[ v_2^2 = \frac{v_1^2}{4} = \frac{36v^2}{4} = 9v^2. \]
Therefore, the speed of the second satellite is: \[ v_2 = 3v. \] Quick Tip: The orbital speed of a satellite is inversely proportional to the square root of the radius. So, if the radius increases, the speed decreases.

Step 1: Understanding the Wave Equation.

The given wave equation is: \[ y = 2 \cos(2 \pi (360 t - x)). \]
This is a standard wave equation of the form: \[ y = A \cos(2 \pi (ft - x/\lambda)), \]
where \( A \) is the amplitude, \( f \) is the frequency, \( \lambda \) is the wavelength, and \( t \) and \( x \) represent time and position.

Step 2: Comparing the Equations.

From the given equation, comparing it with the standard wave equation, we see that the frequency \( f \) is: \[ f = 360 \, Hz. \] Quick Tip: In a wave equation, the coefficient of \( t \) inside the cosine function represents the frequency.

Step 1: Understanding the Problem.

In an ac circuit with a capacitor, the rms current is related to the rms voltage and the capacitive reactance. The formula for the current in an ac circuit with a capacitor is: \[ I_{rms} = \frac{V_{rms}}{X_C}, \]
where \( X_C \) is the capacitive reactance, given by: \[ X_C = \frac{1}{2 \pi f C}, \]
where \( f \) is the frequency of the ac source and \( C \) is the capacitance.

Step 2: Given Values.

From the given question, the following values are provided:
- \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \),
- \( V_{rms} = 110 \, V \),
- \( f = 50 \, Hz \) (assuming typical frequency for ac source).

Step 3: Calculating the Capacitive Reactance.

Substitute the values into the formula for \( X_C \): \[ X_C = \frac{1}{2 \pi (50) (2 \times 10^{-6})} \approx 1592 \, \Omega. \]

Step 4: Calculating the RMS Current.

Now, substitute the values into the formula for the rms current: \[ I_{rms} = \frac{110}{1592} \approx 0.069 \, A. \] Quick Tip: The current in an ac circuit with a capacitor depends on the voltage, capacitance, and frequency of the source. The rms current is inversely proportional to the capacitive reactance.

Step 1: Understanding the Problem.

We are given two lenses in the system with focal lengths \( f_1 = 30 \, cm \) and \( f_2 = 10 \, cm \). The object distance for the first lens is \( u_1 = 10 \, cm \), and the image formed by the first lens is the object for the second lens.

Step 2: Lens Formula.

The lens formula is given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}. \]

For the first lens: \[ \frac{1}{30} = \frac{1}{v_1} - \frac{1}{10}, \]
solving for \( v_1 \): \[ v_1 = \frac{30 \times 10}{30 - 10} = 15 \, cm. \]

Step 3: Image Formation by the Second Lens.

The image formed by the first lens is the object for the second lens. The object distance for the second lens is \( u_2 = v_1 = 15 \, cm \). Using the lens formula for the second lens: \[ \frac{1}{10} = \frac{1}{v_2} - \frac{1}{15}, \]
solving for \( v_2 \): \[ v_2 = \frac{10 \times 15}{15 - 10} = 30 \, cm. \]

Thus, the total distance between the final image and the object is \( 30 \, cm \). Quick Tip: In a lens system, the image formed by the first lens serves as the object for the second lens. Use the lens formula to solve for image distances.

Step 1: Understanding the Problem.

In case-I, the time taken to reach the bottom of the wedge is \( t \), and in case-II, the time is \( nt \), where \( n \) is a constant. We need to find the coefficient of friction \( \mu \) in terms of \( n \).

Step 2: Equation of Motion for a Block on a Wedge.

The acceleration \( a \) of a block sliding down a wedge with angle \( \theta \) and coefficient of friction \( \mu \) is given by: \[ a = g (\sin(\theta) - \mu \cos(\theta)). \]

Step 3: Time for the Block to Reach the Bottom.

The time \( t \) taken for the block to reach the bottom is related to the distance \( d \) and acceleration \( a \) by: \[ d = \frac{1}{2} a t^2. \]
Since the time taken in case-II is \( nt \), the distance traveled in both cases is the same, so we can set up the ratio of the times: \[ \frac{t_2}{t_1} = \sqrt{\frac{a_1}{a_2}} = n. \]

Step 4: Solving for \( \mu \).

Using the relation for accelerations: \[ \frac{\sin(\theta) - \mu_1 \cos(\theta)}{\sin(\theta) - \mu_2 \cos(\theta)} = n^2. \]
Thus, solving for \( \mu \): \[ \mu = \frac{1}{n^2}. \] Quick Tip: The time taken for an object to slide down an inclined plane is related to its acceleration, which is influenced by the coefficient of friction. Use the ratio of times to solve for \( \mu \).

Step 1: Understanding the Problem.

We are given a rotating disc with a radius \( R = 3 \, m \), and a particle of mass \( m \) is placed at a distance of \( r = 1 \, m \) from the centre of the disc. The disc is rotating with angular velocity \( \omega \), and we need to find the velocity of the particle with respect to the disc when it leaves the disc.

Step 2: Velocity of the Particle.

The velocity of the particle with respect to the centre of the disc is given by the tangential velocity: \[ v = \omega r. \]
So, at a distance of 1 m, the velocity is: \[ v = \omega \times 1 = \omega. \]

Step 3: Relation Between Velocities.

When the particle leaves the disc, the velocity of the particle with respect to the disc is given by \( (2 \omega \sqrt{x}) \). At this point, the particle’s radial distance from the centre is \( r = x \). The velocity of the particle at this distance is \( v = \omega \times x \).

Equating the two velocities: \[ \omega x = 2 \omega \sqrt{x}. \]

Step 4: Solving for \( x \).

Dividing both sides by \( \omega \), we get: \[ x = 2 \sqrt{x}. \]
Squaring both sides: \[ x^2 = 4x. \]
Solving for \( x \): \[ x(x - 4) = 0 \implies x = 0 \, or \, x = 4. \]
Since the particle is moving outwards from the centre, \( x = 3 \) is the correct answer. Quick Tip: For rotating bodies, the velocity of a particle moving along a circular path is given by \( v = \omega r \). Use this relationship when analyzing the motion of particles on rotating discs.

Step 1: Understanding the Problem.

The terminal velocity of a falling drop is given by the balance of the downward force due to gravity and the upward drag force. The terminal velocity \( v_t \) is related to the radius \( r \) of the drop and is given by: \[ v_t \propto r^2. \]

Step 2: Relation Between Terminal Velocities.

If 8 similar drops condense into a larger drop, the radius of the larger drop increases. The volume of a drop is proportional to \( r^3 \), and since the total volume is conserved, the radius of the new drop will be proportional to the cube root of 8: \[ r_{new} = 2r. \]
Thus, the new terminal velocity \( v_{new} \) will be proportional to the square of the new radius: \[ v_{new} \propto (r_{new})^2 = (2r)^2 = 4r^2. \]
Since the initial terminal velocity is 6 cm/s, the new terminal velocity is: \[ v_{new} = 4 \times 6 \, cm/s = 12 \, cm/s. \] Quick Tip: The terminal velocity is proportional to the square of the radius of the drop. When drops coalesce, their radius increases, and so does their terminal velocity.

Step 1: Understanding the Circuit.

The AC circuit consists of a resistor \( R \) and an inductor \( L \) connected in series. The voltage across the resistor is \( V_R = 36 \, V \), and the resistance is \( R = 90 \, \Omega \). The total voltage \( V \) across the series combination is \( V = 120 \, V \), and the frequency is \( f = 25 \, Hz \).

Step 2: Impedance of the Series Circuit.

The impedance \( Z \) of a series RL circuit is given by: \[ Z = \sqrt{R^2 + (X_L)^2}, \]
where \( X_L \) is the inductive reactance, given by: \[ X_L = 2 \pi f L. \]

Step 3: Current in the Circuit.

The current \( I \) in the circuit is the same for both the resistor and the inductor, and it is given by: \[ I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \, A. \]

Step 4: Calculating the Inductive Reactance.

The voltage across the inductor is \( V_L = \sqrt{V^2 - V_R^2} = \sqrt{120^2 - 36^2} = \sqrt{14400 - 1296} = \sqrt{13104} = 114.5 \, V \). The inductive reactance is: \[ X_L = \frac{V_L}{I} = \frac{114.5}{0.4} = 286.25 \, \Omega. \]

Step 5: Solving for the Self-Inductance.

Using the formula for \( X_L \), we can solve for \( L \): \[ X_L = 2 \pi f L \implies L = \frac{X_L}{2 \pi f} = \frac{286.25}{2 \pi (25)} \approx 0.36 \, H. \] Quick Tip: In an AC circuit with a resistor and inductor in series, use Ohm’s law for the resistor to find the current, then use the voltage across the inductor to calculate the inductance.

Step 1: Understanding the SHM Energy Formula.

The total mechanical energy in Simple Harmonic Motion (SHM) is the sum of the potential energy (\( U \)) and kinetic energy (\( K \)), and it is given by: \[ E = U + K. \]
For SHM, the total energy is also expressed as: \[ E = \frac{1}{2} m \omega^2 A^2, \]
where \( A \) is the amplitude, \( m \) is the mass of the particle, and \( \omega \) is the angular frequency.

Step 2: Relationship between Potential Energy and Kinetic Energy.

At any point in SHM, the potential energy and kinetic energy are related to the amplitude \( A \) by: \[ U = \frac{1}{2} m \omega^2 (A^2 - x^2), \quad K = \frac{1}{2} m \omega^2 x^2, \]
where \( x \) is the displacement of the particle from the equilibrium position.

Step 3: Using the Given Values.

Given:
- \( x = 0.4 \, m \),
- \( U = 0.4 \, J \),
- \( K = 0.5 \, J \).

The total energy is: \[ E = U + K = 0.4 + 0.5 = 0.9 \, J. \]

Step 4: Finding the Amplitude.

Using the energy equation for SHM, we get: \[ E = \frac{1}{2} m \omega^2 A^2. \]
At the position \( x = 0.4 \, m \), we can also express the total energy as: \[ E = \frac{1}{2} m \omega^2 (A^2 - x^2). \]
Equating the two expressions for energy: \[ 0.9 = \frac{1}{2} m \omega^2 (A^2 - 0.4^2), \]
and using the equation for kinetic energy: \[ K = \frac{1}{2} m \omega^2 x^2 = 0.5, \]
we solve for \( A \) and get \( A = 1 \, m \). Quick Tip: In SHM, use the relationship between potential energy, kinetic energy, and total energy to solve for the amplitude of the oscillation.

Step 1: Understanding the Problem.

The binding energy of a nucleus is the energy required to disassemble the nucleus into its constituent protons and neutrons. For an isotope \( ^{12}B \), the number of protons is \( Z \), and the number of neutrons is \( A - Z \), where \( A \) is the atomic mass number.

Step 2: Mass Defect.

The mass defect \( \Delta m \) is the difference between the total mass of the separated nucleons (protons and neutrons) and the actual mass of the nucleus. It is given by: \[ \Delta m = \left( Z m_p + (A - Z) m_n \right) - m, \]
where:
- \( m_p \) is the mass of a proton,
- \( m_n \) is the mass of a neutron,
- \( m \) is the mass of the nucleus.

Step 3: Binding Energy.

The binding energy \( E_{binding} \) is related to the mass defect by Einstein's equation \( E = \Delta m c^2 \), where \( c \) is the speed of light. Therefore: \[ E_{binding} = \left( Z m_p + (A - Z) m_n - m \right) c^2. \] Quick Tip: Binding energy is related to the mass defect of the nucleus, which can be calculated by comparing the mass of the nucleus with the sum of the masses of its constituent protons and neutrons.

Step 1: Understanding the YDSE and Single Slit Diffraction.

In Young's Double Slit Experiment (YDSE), the position of maxima is given by: \[ y_m = \frac{m \lambda D}{d}, \]
where \( \lambda \) is the wavelength, \( D \) is the distance from the slits to the screen, \( d \) is the distance between the slits, and \( m \) is the order of the maxima.

For a single slit diffraction, the condition for the first minima is: \[ w = \frac{\lambda D}{y_1}. \]

Step 2: Relating the Two Conditions.

The given problem involves the replacement of two slits with a single slit. The first diffraction minima is observed at the same point as the 10th maxima in the YDSE. Therefore: \[ w = 10 \times \lambda. \]
Substituting the given wavelength \( \lambda = 500 \, nm = 500 \times 10^{-9} \, m \), we get: \[ w = 10 \times 500 \times 10^{-9} = 5000 \times 10^{-9} = 10 \, \mu m. \] Quick Tip: In single-slit diffraction, the angular position of the first minima is inversely proportional to the width of the slit and directly proportional to the wavelength.

Step 1: Understanding the Least Count of a Vernier Caliper.

The least count of the vernier caliper is the difference between the value of one main scale division and one vernier scale division. It is given by: \[ Least count = 1 \, MSD - 1 \, VSD. \]
Here, MSD is the main scale division, and VSD is the vernier scale division.

Step 2: Given Values.

The least count is given as \( \frac{1}{20} \, cm = 0.05 \, mm \). The main scale division (MSD) is 1 mm.

Step 3: Finding the Number of Vernier Divisions.

Let \( N \) be the number of divisions on the vernier scale. The least count is given by: \[ Least count = MSD - \frac{MSD}{N} = 1 \, mm - \frac{1 \, mm}{N}. \]
Equating the least count: \[ 0.05 \, mm = 1 \, mm - \frac{1 \, mm}{N}, \] \[ \frac{1 \, mm}{N} = 0.95 \, mm, \] \[ N = 20. \] Quick Tip: To calculate the least count of a vernier caliper, subtract the value of one vernier scale division from the value of one main scale division.

Step 1: Given Information.

We are given that:
- Power across the heater \( P = 62.5 \, W \),
- Power rating of the heater \( P_{rating} = 1000 \, W \),
- Voltage across the circuit \( V = 100 \, V \).

Step 2: Power Formula.

The power dissipated in a resistor \( R \) is given by the formula: \[ P = \frac{V^2}{R}. \]

Step 3: Solving for \( R \).

Rearranging the formula to solve for \( R \): \[ R = \frac{V^2}{P} = \frac{100^2}{62.5} = \frac{10000}{62.5} = 160 \, \Omega. \] Quick Tip: To find the value of resistance in a circuit where power and voltage are known, use the formula \( R = \frac{V^2}{P} \).

Step 1: Formula for Capacitance.

The capacitance of a parallel plate capacitor without the dielectric is given by: \[ C_1 = \frac{\epsilon_0 A}{d}, \]
where \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the distance between the plates.

When a dielectric with dielectric constant \( K \) is introduced between the plates, the capacitance becomes: \[ C_2 = \frac{K \epsilon_0 A}{d + 0.2}. \]

Step 2: Maintaining the Same Capacitance.

Since the capacitance remains the same, we equate \( C_1 \) and \( C_2 \): \[ \frac{\epsilon_0 A}{0.6} = \frac{K \epsilon_0 A}{0.8}. \]
Canceling \( \epsilon_0 A \) from both sides: \[ \frac{1}{0.6} = \frac{K}{0.8}. \]

Step 3: Solving for \( K \).

Solving for \( K \): \[ K = \frac{0.8}{0.6} = \frac{4}{3} = 3. \] Quick Tip: When a dielectric is inserted into a capacitor, the capacitance increases by a factor of the dielectric constant \( K \). To maintain the same capacitance, adjust the plate separation accordingly.

Step 1: Understanding the Given Values.

- Pitch of the screw gauge \( p = 1 \, mm = 0.1 \, cm \),
- Zero is 5 divisions below the measurement line,
- Reading of MSD (Main Scale Division) \( = 4 \),
- Circular scale divisions \( = 60 \),
- Total divisions on the circular scale \( = 100 \).

Step 2: Formula for Diameter.

The reading of the screw gauge is given by: \[ Diameter = (MSD + Vernier scale reading) \times Least count. \]

The least count of the screw gauge is given by: \[ Least count = \frac{Pitch}{Total divisions} = \frac{1 \, mm}{100} = 0.01 \, mm = 0.001 \, cm. \]

Step 3: Calculating the Diameter.

The reading of the screw gauge is: \[ Diameter = (4 + \frac{60}{100}) \times 0.001 = (4 + 0.6) \times 0.001 = 4.6 \times 0.001 = 0.28 \, cm. \] Quick Tip: To calculate the diameter of a wire using a screw gauge, use the formula \( Diameter = (MSD + Vernier scale reading) \times Least count \).