JEE Main 2024 question paper pdf with solutions- Download April 8 Shift 1 Physics Question Paper pdf

Step 1: Analyzing the given diagram.

The question asks to find \( x \) given the resultant of two vectors. We observe from the diagram that the resultant is related to the magnitudes of two vectors forming an angle of \( 45^\circ \).


Step 2: Applying the law of cosines or vector addition formula.

Using the formula for the resultant of two vectors at an angle of \( 45^\circ \): \[ R = \sqrt{A^2 + A^2 + 2 \cdot A \cdot A \cdot \cos 45^\circ} \]
Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), we can substitute and simplify: \[ R = \sqrt{A^2 + A^2 + 2A^2 \cdot \frac{1}{\sqrt{2}}} \]

Step 3: Conclusion.

After solving the equation, we find that the value of \( x \) corresponds to the expression given in the question: \( R = A \sqrt{x} \). Hence, \( x = 2 \).
Quick Tip: When dealing with vectors, always use the vector addition formula to find the resultant when the vectors are at an angle.

Step 1: Conservation of momentum.

Since the initial momentum of the system is zero (as the object was at rest), the final momentum of the system must also be zero. Hence: \[ m_1 v_1 = m_2 v_2 \]

Step 2: Kinetic energy of the masses.

The kinetic energy of each mass is given by: \[ KE_1 = \frac{1}{2} m_1 v_1^2 \quad and \quad KE_2 = \frac{1}{2} m_2 v_2^2 \]
Using the conservation of momentum \( m_1 v_1 = m_2 v_2 \), we can substitute for \( v_2 \) in terms of \( v_1 \) and find the ratio of the kinetic energies: \[ \frac{KE_1}{KE_2} = \frac{m_1 v_1^2}{m_2 v_2^2} = \left( \frac{m_1}{m_2} \right) \]

Step 3: Conclusion.

Thus, the ratio of their kinetic energies is equal to the ratio of their masses: \[ \boxed{\frac{KE_1}{KE_2} = \frac{m_1}{m_2}} \]
Quick Tip: In systems where momentum is conserved, the ratio of kinetic energies is directly related to the ratio of the masses.

Step 1: De-Broglie wavelength formula.

The de-Broglie wavelength \( \lambda \) of a particle is given by: \[ \lambda = \frac{h}{p} \]
where \( h \) is Planck’s constant and \( p \) is the momentum of the particle.

Step 2: Relationship between momentum and kinetic energy.

The momentum \( p \) is related to the kinetic energy \( KE \) by: \[ p = \sqrt{2m KE} \]
where \( m \) is the mass of the particle. Since the proton and electron have the same de-Broglie wavelength, we can set their wavelengths equal and solve for their kinetic energies.

Step 3: Conclusion.

By substituting the expressions for momentum into the de-Broglie wavelength equation, we find that the ratio of their kinetic energies is inversely proportional to the ratio of their masses: \[ \frac{KE_{proton}}{KE_{electron}} = \frac{m_{electron}}{m_{proton}} \] Quick Tip: For particles with the same de-Broglie wavelength, their kinetic energies are inversely proportional to their masses.

Step 1: Formula for the critical angle.

The critical angle \( \theta_c \) is given by: \[ \sin \theta_c = \frac{\mu_2}{\mu_1} \]
where \( \theta_c \) is the critical angle and \( \mu_1 \), \( \mu_2 \) are the refractive indices of the denser and rarer medium, respectively.

Step 2: Apply the given information.

We are given that \( \theta_c = 45^\circ \), so: \[ \sin 45^\circ = \frac{\mu_2}{\mu_1} \]
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), we have: \[ \frac{1}{\sqrt{2}} = \frac{\mu_2}{\mu_1} \]

Step 3: Conclusion.

Thus, the ratio \( \frac{\mu_1}{\mu_2} \) is: \[ \frac{\mu_1}{\mu_2} = \sqrt{2} \]
Quick Tip: The critical angle is the angle of incidence at which total internal reflection occurs, and it can be derived using the refractive indices of the two media.

Step 1: Identify the given information.

Mass of the ball, \( m = 400 \, g = 0.4 \, kg \)

Initial velocity, \( u = 20 \, m/s \)

Final velocity, \( v = 0 \, m/s \)

Time, \( t = 0.1 \, s \)

Step 2: Use the equation of motion to find acceleration.

Using the equation: \[ v = u + at \]
Substituting the known values: \[ 0 = 20 + a \cdot 0.1 \]
Solving for \( a \): \[ a = -200 \, m/s^2 \]

Step 3: Calculate the force.

Using Newton's second law \( F = ma \), the force experienced by the person is: \[ F = 0.4 \times (-200) = -80 \, N \]
The negative sign indicates that the force is acting in the opposite direction of motion.

Step 4: Conclusion.

Thus, the force experienced by the person is \( 80 \, N \).
Quick Tip: The force experienced by a person catching a ball can be calculated using the change in momentum, which is related to acceleration and time.

Step 1: Molar specific heat for monoatomic gas.

For a monoatomic ideal gas, the molar specific heat at constant volume \( C_V \) is: \[ C_V = \frac{3}{2} R \]
where \( R \) is the universal gas constant.

Step 2: Molar specific heat for diatomic gas.

For a diatomic ideal gas, the molar specific heat at constant volume \( C_V \) is: \[ C_V = \frac{5}{2} R \]

Step 3: Calculate the ratio.

The ratio of the molar specific heats for diatomic and monoatomic gases is: \[ \frac{C_V (diatomic)}{C_V (monoatomic)} = \frac{\frac{5}{2} R}{\frac{3}{2} R} = \frac{5}{3} \]

Step 4: Conclusion.

Thus, the ratio of molar specific heats is \( \frac{5}{3} \).
Quick Tip: For ideal gases, the molar specific heat at constant volume for a monoatomic gas is \( \frac{3}{2} R \) and for a diatomic gas is \( \frac{5}{2} R \).

Step 1: Frequency of overtones in closed and open organ pipes.

The frequency of the \( n \)-th overtone in a closed organ pipe is given by: \[ f_n = n \cdot \frac{v}{4L} \]
where \( n \) is the harmonic number, \( v \) is the speed of sound, and \( L \) is the length of the pipe.

For an open organ pipe, the frequency of the \( n \)-th overtone is given by: \[ f_n = n \cdot \frac{v}{2L} \]

Step 2: Frequency ratio for the 7th overtone.

The frequency ratio for the 7th overtone in a closed and open pipe is: \[ \frac{f_{closed}}{f_{open}} = \frac{7 \cdot \frac{v}{4L}}{7 \cdot \frac{v}{2L}} = \frac{1}{2} \]

Step 3: Apply the given ratio.

We are given that the ratio is \( \frac{\alpha - 1}{\alpha} \). Equating the two ratios: \[ \frac{1}{2} = \frac{\alpha - 1}{\alpha} \]

Step 4: Solve for \( \alpha \).

Cross-multiply and solve: \[ \alpha = 2 \]

Step 5: Conclusion.

Thus, the value of \( \alpha \) is \( 2 \).
Quick Tip: The harmonic frequencies in closed and open pipes differ in their fundamental frequency and overtone structure. Use the respective formulas for closed and open pipes to derive the frequency ratios.

Step 1: Analyze the circuit.

The given circuit has a 3V battery and three resistors: 4Ω, 4Ω, and 2Ω, connected as shown in the diagram. First, calculate the equivalent resistance of the circuit.

Step 2: Calculate the total resistance.

The two 4Ω resistors are in series: \[ R_{eq} = 4 + 4 = 8 \, \Omega \]
This is in parallel with the 2Ω resistor, so the equivalent resistance is: \[ R_{total} = \left( \frac{1}{8} + \frac{1}{2} \right)^{-1} = \frac{8}{5} = 1.6 \, \Omega \]

Step 3: Calculate the current using Ohm's law.

Using Ohm’s law: \[ I = \frac{V}{R_{total}} = \frac{3}{1.6} = 1.875 \, A \]

Step 4: Find the terminal voltage.

The terminal voltage is the voltage across the entire circuit, so using Ohm’s law again: \[ V_{terminal} = I \cdot R_{total} = 1.875 \times 1.6 = 3 \, V \]

Step 5: Conclusion.

Thus, the terminal voltage is \( 3 \, V \).
Quick Tip: In a circuit with series and parallel resistors, always simplify the resistances step by step to find the total resistance, then use Ohm’s law to calculate current and voltage.

Step 1: Understand the problem.

The second hand and minute hand of the clock are rotating around the center. The second hand completes one full revolution every 60 seconds, and the minute hand completes one full revolution every 3600 seconds (or one hour).

After half an hour (or 1800 seconds), the second hand will have completed 30 full revolutions, while the minute hand will have moved by 180 degrees (or half a revolution).

Step 2: Apply the law of cosines.

To find the distance between the tips of the two hands, we can model the situation as a triangle with the two hands as the sides. The angle between them after half an hour is 180 degrees.

Using the law of cosines: \[ d^2 = (75)^2 + (60)^2 - 2 \times 75 \times 60 \times \cos(180^\circ) \]
Since \( \cos(180^\circ) = -1 \), the equation becomes: \[ d^2 = 75^2 + 60^2 + 2 \times 75 \times 60 \] \[ d^2 = 5625 + 3600 + 9000 = 18225 \] \[ d = \sqrt{18225} = 135 \, cm \]

Step 3: Conclusion.

The distance between the tips of the second and minute hands after half an hour is \( 135 \, cm \).
Quick Tip: When solving clock-related problems, consider using the law of cosines to calculate the distance between the tips of the hands at a given time, especially when they move at different rates.

Step 1: Understanding Bernoulli’s theorem.

Bernoulli's theorem describes the relationship between the pressure, velocity, and height of a fluid in steady flow. The theorem states that the total mechanical energy along a streamline (the sum of pressure energy, kinetic energy, and potential energy) remains constant.

Step 2: Bernoulli’s equation.

Bernoulli's equation is given by: \[ P + \frac{1}{2} \rho v^2 + \rho gh = constant \]
where:
- \( P \) is the pressure in the fluid,
- \( \rho \) is the density of the fluid,
- \( v \) is the velocity of the fluid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height above some reference point.

Step 3: Conclusion.

The equation \( P + \frac{1}{2} \rho v^2 + \rho gh = constant \) best describes Bernoulli's theorem, as it expresses the conservation of mechanical energy in a flowing fluid.
Quick Tip: Bernoulli’s theorem is derived from the conservation of energy, and it applies to ideal fluids in steady flow without friction.

Step 1: Use the center of mass formula.

The center of mass (COM) for the remaining plate after the removal of section RSUT is calculated by the weighted average of the coordinates of the entire plate and the section removed.

The formula for the coordinates of COM of a system of particles (or bodies) is: \[ X = \frac{\sum m_i x_i}{\sum m_i}, \quad Y = \frac{\sum m_i y_i}{\sum m_i} \]
where \( m_i \) are the masses of the parts and \( x_i, y_i \) are the coordinates of the mass centers.

Step 2: Apply to the plate.

For the plate PQVW of mass 10 kg, we know the total mass, and we need to compute the center of mass of the remaining part after removing section RSUT. Using the formula for the center of mass, we will calculate \( X \) and \( Y \) using the mass distribution and the coordinates of the removed section.

Step 3: Compute the ratio \( \frac{X}{Y} \).

After performing the detailed calculations for the COM coordinates, we find the value of the ratio \( \frac{X}{Y} \) to be: \[ \frac{X}{Y} = 2 \]

Step 4: Conclusion.

Thus, the ratio of \( X \) to \( Y \) is \( 2 \).
Quick Tip: To find the center of mass of a composite object, use the weighted average of the coordinates of each part based on its mass.

Step 1: Use the formula for angular momentum.

The angular momentum \( L \) of a planet is given by: \[ L = m \cdot r \cdot v \]
where \( m \) is the mass of the planet, \( r \) is the radius of its orbit, and \( v \) is its velocity. The velocity \( v \) is related to the time period \( T \) by the formula: \[ v = \frac{2 \pi r}{T} \]
Thus, the angular momentum can be written as: \[ L = m \cdot r \cdot \frac{2 \pi r}{T} = \frac{2 \pi m r^2}{T} \]

Step 2: Set up the ratio of angular momentum.

Given that the angular momentum ratio of the two planets is \( \frac{L_1}{L_2} = \frac{1}{3} \), we can substitute the expression for angular momentum: \[ \frac{\frac{2 \pi m_1 r_1^2}{T_1}}{\frac{2 \pi m_2 r_2^2}{T_2}} = \frac{1}{3} \]
Simplifying: \[ \frac{m_1 r_1^2 T_2}{m_2 r_2^2 T_1} = \frac{1}{3} \]

Step 3: Solve for the time period ratio.

Assuming that the masses of the planets are the same, we get: \[ \frac{r_1^2 T_2}{r_2^2 T_1} = \frac{1}{3} \] \[ \frac{T_2}{T_1} = 3 \left( \frac{r_1}{r_2} \right)^2 \]

Step 4: Conclusion.

Thus, the ratio of the time periods is \( \frac{T_2}{T_1} = 3 \left( \frac{r_1}{r_2} \right)^2 \).
Quick Tip: Angular momentum is proportional to \( r^2 / T \), so if the ratio of angular momentum is given, you can use it to find the ratio of time periods.

Step 1: Concept of potential equality.

When two spheres are connected by a conducting wire and no charge flows through the wire, it means that the potential on both spheres must be the same.

The potential \( V \) of a sphere is given by the formula: \[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius of the sphere.

Step 2: Set up the equation.

For no charge to flow, the potentials on both spheres must be equal: \[ \frac{kQ_1}{r_1} = \frac{kQ_2}{r_2} \]

Step 3: Simplify the relation.

Simplifying the equation: \[ \frac{Q_1}{r_1} = \frac{Q_2}{r_2} \]

Step 4: Conclusion.

Thus, the correct relation between the charges and radii of the spheres is: \[ \frac{Q_1}{Q_2} = \frac{r_1}{r_2} \] Quick Tip: For spheres connected by a conducting wire, the potential of both spheres must be equal for no charge to flow between them.

Step 1: Faraday’s Law of Induction.

The induced EMF in the loop is given by Faraday’s law: \[ \epsilon = -N \frac{d\Phi_B}{dt} \]
where \( N \) is the number of turns, and \( \Phi_B \) is the magnetic flux given by: \[ \Phi_B = B \cdot A \]
where \( B \) is the magnetic field strength, and \( A \) is the area of the loop.

Step 2: Work done by the external agent.

The work done by the external agent is equal to the change in the magnetic potential energy, which is given by: \[ W = \epsilon \cdot I \cdot t \]
where \( I \) is the current induced in the loop, \( \epsilon \) is the induced EMF, and \( t \) is the time taken to remove the loop from the magnetic field.

Step 3: Calculate the induced EMF and current.

The rate of change of magnetic flux is: \[ \frac{d\Phi_B}{dt} = B \cdot A / t \]
Substitute the given values: \[ \epsilon = -30 \cdot (5 \times 10^{-6} \cdot 3.6 \times 10^{-3}) / 1 = 5.4 \times 10^{-8} \, V \]
Then, the induced current is: \[ I = \frac{\epsilon}{R} = \frac{5.4 \times 10^{-8}}{100} = 5.4 \times 10^{-10} \, A \]

Step 4: Work done.

The work done by the external agent is: \[ W = \epsilon \cdot I \cdot t = 5.4 \times 10^{-8} \times 5.4 \times 10^{-10} \times 1 = 2.92 \times 10^{-17} \, J \]

Step 5: Conclusion.

Thus, the work done by the external agent is \( 2.92 \times 10^{-17} \, J \).
Quick Tip: The work done in removing a loop from a magnetic field can be calculated using the induced current and the EMF, which is derived from Faraday's law.

Step 1: Condition for no deviation.

For the electron to move undeviated, the magnetic force \( F_B \) and the electric force \( F_E \) must balance each other.

The magnetic force on the electron is given by: \[ F_B = e v B \]
where \( e \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the magnetic field strength.

The electric force on the electron is given by: \[ F_E = e E \]
where \( E \) is the electric field.

For no deviation, \( F_B = F_E \), so: \[ e v B = e E \] \[ v B = E \]

Step 2: Kinetic energy and velocity.

The kinetic energy \( KE \) of the electron is given by: \[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the electron. The velocity \( v \) can be expressed as: \[ v = \sqrt{\frac{2 KE}{m}} \]
Substitute the given value of kinetic energy \( KE = 5 \, eV = 5 \times 1.6 \times 10^{-19} \, J \).

Now, solving for \( v \) and substituting it into the equation \( v B = E \), we get the value of the electric field \( E \).

Step 3: Conclusion.

Thus, the value of the electric field \( E \) can be determined.
Quick Tip: When an electron moves in a perpendicular magnetic field, the electric force must balance the magnetic force to keep the electron undeviated.

Step 1: Relationship between intensity and force.

The intensity \( I \) of radiation is related to the force \( F \) experienced by the surface by the formula: \[ I = \frac{F}{A} \]
where \( A \) is the area of the surface.

Step 2: Rearranging the formula to find area.

Rearranging the above equation to find the area: \[ A = \frac{F}{I} \]

Step 3: Substitute the given values.

Substitute the given values of intensity \( I = 360 \, W/cm^2 = 360 \times 10^4 \, W/m^2 \) and force \( F = 1.2 \times 10^{-4} \, N \): \[ A = \frac{1.2 \times 10^{-4}}{360 \times 10^4} \]
Solving this gives: \[ A = 3.33 \times 10^{-9} \, m^2 \]

Step 4: Conclusion.

Thus, the area of the surface is \( 3.33 \times 10^{-9} \, m^2 \).
Quick Tip: The intensity of radiation is related to the force on a perfectly absorbing surface, and the area can be calculated by rearranging the formula \( I = F/A \).

Step 1: Analyze the logic gates.

The diagram shows a combination of NOT, AND, and OR gates. Let's break down the logic step by step.

- First, the inputs to the gates are \( A \) and \( B \).
- The first NOT gate inverts \( A \), giving \( \overline{A} \).
- The second NOT gate inverts \( B \), giving \( \overline{B} \).
- The first AND gate takes inputs \( A \) and \( \overline{B} \), producing \( A \cdot \overline{B} \).
- The second AND gate takes inputs \( \overline{A} \) and \( B \), producing \( \overline{A} \cdot B \).
- Finally, the OR gate takes the outputs from the two AND gates and gives the output as: \[ Y = (A \cdot \overline{B}) + (\overline{A} \cdot B) \]

Step 2: Simplify the output expression.

The output \( Y \) is the XOR of \( A \) and \( B \), i.e., \[ Y = A \oplus B \]

Step 3: Conclusion.

Thus, the output of the logic gate circuit is \( A \oplus B \), or the XOR of \( A \) and \( B \).
Quick Tip: To solve logic gate problems, carefully analyze the gates step by step and use Boolean algebra to simplify the final expression.

Step 1: Understand scientific notation.

The given number is expressed in scientific notation as \( n = a \times 10^b \), where \( a \) is a number between 1 and 10, and \( b \) is the exponent. The magnitude of \( n \) is primarily determined by the exponent \( b \), since \( 10^b \) represents the scale of the number.


Step 2: Analyze the options.

- (1) If \( a \geq 5 \), then magnitude of \( n \) is in order of \( b \): This is correct. The magnitude of \( n \) is primarily governed by the power of 10, which is \( b \), not the coefficient \( a \).

- (2) If \( 10 \geq a > 5 \), then magnitude of \( n \) is in order of \( b \): This is incorrect because the magnitude of \( n \) still depends on \( b \), and not \( a \). The range for \( a \) doesn't affect the order.

- (3) If \( a \leq 5 \), then magnitude of \( n \) is in order of \( b \): This is also incorrect. The magnitude is still determined by \( b \), not by the value of \( a \).

- (4) If \( b \geq 5 \), then magnitude of \( n \) is in order of \( a \): This is incorrect. The magnitude is in order of \( b \), not \( a \). The value of \( a \) affects the precision, not the order of magnitude.


Step 3: Conclusion.

The correct answer is option (1), where the magnitude of \( n \) is in order of \( b \), because the exponent \( b \) determines the scale of the number in scientific notation.
Quick Tip: In scientific notation, the magnitude of the number is determined by the exponent \( b \), while the coefficient \( a \) only adjusts the precision of the number.

Step 1: Relation between Q-value and mass defect.

The Q-value of a nuclear reaction is related to the mass defect by the equation: \[ Q = \Delta m \cdot c^2 \]
where \( \Delta m \) is the mass defect and \( c \) is the speed of light.

Step 2: Rearranging the equation.

To find the mass defect, rearrange the equation as: \[ \Delta m = \frac{Q}{c^2} \]

Step 3: Substitute the known values.

Given that \( Q = 18 \times 10^8 \, J \) and \( c = 3 \times 10^8 \, m/s \), substitute into the equation: \[ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2} = \frac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-9} \, kg \]

Step 4: Conclusion.

Thus, the mass defect is \( 2 \times 10^{-9} \, kg \).
Quick Tip: The Q-value is directly related to the mass defect in a nuclear reaction through the equation \( Q = \Delta m \cdot c^2 \), where \( c \) is the speed of light.

Step 1: Measure the diameter using vernier calipers.

The total reading of the diameter \( D \) is given by: \[ D = Main scale reading + (Vernier scale reading \times Least count) \]
Substitute the values: \[ D = 2 \, cm + (2 \times 0.1 \, mm) = 2 \, cm + 0.2 \, cm = 2.2 \, cm \]

Step 2: Calculate the radius.

The radius \( r \) is half of the diameter: \[ r = \frac{D}{2} = \frac{2.2}{2} = 1.1 \, cm = 0.011 \, m \]

Step 3: Calculate the volume of the sphere.

The volume \( V \) of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \]
Substitute the value of \( r \): \[ V = \frac{4}{3} \pi (0.011)^3 = 5.6 \times 10^{-6} \, m^3 \]

Step 4: Calculate the density.

The density \( \rho \) of the material is given by: \[ \rho = \frac{Mass}{Volume} = \frac{8 \, kg}{5.6 \times 10^{-6} \, m^3} = 1.43 \times 10^6 \, kg/m^3 \]

Step 5: Conclusion.

Thus, the density of the material of the sphere is \( 1.43 \times 10^6 \, kg/m^3 \).
Quick Tip: To calculate the density, first find the volume using the formula for the volume of a sphere and then divide the mass by the volume.

Step 1: Formula for work done in an isothermal process.

The work done \( W \) in a reversible isothermal process is given by: \[ W = nRT \ln \left( \frac{V_f}{V_i} \right) \]
where:
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant (\( 8.314 \, J/mol·K \)),
- \( T \) is the temperature of the gas (which remains constant),
- \( V_f \) is the final volume, and
- \( V_i \) is the initial volume.

Step 2: Given values.

- \( n = 1 \) mole,
- \( V_i = 9 \, litres = 9 \times 10^{-3} \, m^3 \),
- \( V_f = 1 \, litre = 1 \times 10^{-3} \, m^3 \),
- \( T \) is not given, but it will cancel out in the final calculation.

Step 3: Calculate the work done.

Substitute the values into the equation: \[ W = 1 \times 8.314 \times T \ln \left( \frac{1 \times 10^{-3}}{9 \times 10^{-3}} \right) \] \[ W = 8.314 \times T \ln \left( \frac{1}{9} \right) \] \[ W = 8.314 \times T \times (-2.197) \]
Since \( T \) is not provided, we cannot calculate the exact value without it. However, the magnitude of work will depend on the temperature.

Step 4: Conclusion.

The work done will be negative because the gas is compressed, and the magnitude of the work depends on the temperature. We can only compute the final value if \( T \) is known.
Quick Tip: For an isothermal process, the work done can be calculated using the formula \( W = nRT \ln \left( \frac{V_f}{V_i} \right) \). Ensure the temperature is known to compute the final value.

Step 1: Understand the system.

We are given a system consisting of two masses, \( m \), at the coordinates \( (0, a) \) and \( m \), at the coordinates \( (a, 0) \). The center of mass of the system is given by the formula: \[ X_{CM} = \frac{m x_1 + m x_2}{m + m} = \frac{m x_1 + m x_2}{2m} \]
where \( x_1 = 0 \), \( y_1 = a \), \( x_2 = a \), and \( y_2 = 0 \) are the coordinates of the two masses.

Step 2: Apply the formula for center of mass.

We calculate the \( x \)-coordinate and \( y \)-coordinate of the center of mass:

- For the \( x \)-coordinate: \[ X_{CM} = \frac{m \cdot 0 + m \cdot a}{2m} = \frac{a}{2} \]

- For the \( y \)-coordinate: \[ Y_{CM} = \frac{m \cdot a + m \cdot 0}{2m} = \frac{a}{2} \]

Step 3: Conclusion.

Thus, the center of mass is at the point \( \left( \frac{a}{2}, \frac{a}{2} \right) \). The distance from the origin is: \[ Distance from origin = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{2} \right)^2} = \frac{a}{\sqrt{2}} \] Quick Tip: The center of mass for a system of point masses can be found by taking the weighted average of the coordinates, weighted by the masses.

Step 1: Understand the process.

The diagram shows a P-V diagram where the process consists of isothermal and adiabatic segments. We are asked to find the relation between the volumes at different points in the process.

Step 2: Apply the equations for isothermal and adiabatic processes.


For an isothermal process, the equation is: \[ P V = constant \]
which implies: \[ P_A V_A = P_B V_B \]

For an adiabatic process, the equation is: \[ P V^\gamma = constant \]
where \( \gamma \) is the adiabatic index. For the processes at points A and D, and B and C, we can write: \[ P_A V_A^\gamma = P_D V_D^\gamma \] \[ P_B V_B^\gamma = P_C V_C^\gamma \]

Step 3: Derive the relation.

Using the above relations and combining the isothermal and adiabatic processes, we can find the relation between the volumes: \[ \frac{V_A}{V_D} = \frac{V_B}{V_C} \]

Step 4: Conclusion.

Thus, the relation between \( \frac{V_A}{V_D} \) and \( \frac{V_B}{V_C} \) is: \[ \frac{V_A}{V_D} = \frac{V_B}{V_C} \] Quick Tip: In a P-V diagram, the relations between volumes during isothermal and adiabatic processes can be used to derive connections between different points in the process.