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CONCEPT:

Compound microscope: 

• It is used for much larger magnifications.
• Two lenses are used with one lens compounding the effect of the other.

where fo is the focal length of the objective lens,  fe  is the focal length of the eyepiece, h is the object height, h' is the size of the first image

• A compound microscope uses two lenses i.e., objective lens and eyepiece. 
• Objective lens: lens nearest to the object forming a real, inverted, and magnified image of the object.
• Eyepiece: produces a final image that is enlarged and virtual.

​CALCULATION:

Given: magnification of compound microscope(m) = 20, focal length of eye piece (f_e) = 5 cm and image is formed at near point (v= D) = 25 cm.

• Linear magnification due to eyepiece:

\(⇒ m_{e} = 1+\frac{D}{f} = 1+ (\frac {25}{5} )= 6\)

• Total magnification is given by:

⇒ m = m_o × m_e,

where m_o is the magnification due to the objective lens.

• Magnification due to the objective lens is given by:

\(⇒ m_{o} = \frac {m}{m_{e}} = \frac {20}{6} = 3.33\)

• Hence option 3) is correct.

Given:

First term (a) = 7

Last term (l) = 35

Total terms (n) = 6

Formula used:

Number of arithmetic means = Total terms (n) - 2

Calculation:

Number of arithmetic means = 6 - 2

⇒ Number of arithmetic means = 4

∴ The correct answer is option (2).

Concept:

Matrix Multiplication:

• The multiplication of two matrices is possible if the number of columns of the first matrix equals the number of rows of the second matrix.
• If A is an m × n matrix and X is an n × 1 matrix, the product AX is an m × 1 column matrix.
• The elements of the product are calculated by multiplying the elements of the rows of the first matrix by the corresponding elements of the columns of the second matrix.

Equality of Matrices:

• Two matrices are said to be equal if they have the same order and their corresponding elements are identical.
• If \(\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} c \ d \end{bmatrix}\), then a = c and b = d.

 

Calculation:

Given,

⇒ A = \(\begin{bmatrix} 2 & 4 \ 4 & 3 \end{bmatrix}\)

⇒ X = \(\begin{bmatrix} n \ 1 \end{bmatrix}\)

⇒ B = \(\begin{bmatrix} 8 \ 11 \end{bmatrix}\)

⇒ AX = B

 

⇒ \(\begin{bmatrix} 2 & 4 \ 4 & 3 \end{bmatrix} \begin{bmatrix} n \ 1 \end{bmatrix} = \begin{bmatrix} 8 \ 11 \end{bmatrix}\)

⇒ \(\begin{bmatrix} (2 \times n) + (4 \times 1) \ (4 \times n) + (3 \times 1) \end{bmatrix} = \begin{bmatrix} 8 \ 11 \end{bmatrix}\)

⇒ \(\begin{bmatrix} 2n + 4 \ 4n + 3 \end{bmatrix} = \begin{bmatrix} 8 \ 11 \end{bmatrix}\)

⇒ 2n + 4 = 8

⇒ 2n = 8 − 4

⇒ 2n = 4

⇒ n = 4 ÷ 2

⇒ n = 2

⇒ Also, 4n + 3 = 11

⇒ 4(2) + 3 = 11

⇒ 8 + 3 = 11 (Satisfied)

∴ The value of n is 2.

Concept:

Matrix Equation:

• An equation where the unknowns are matrices or scalars related to matrices.
• Physical Quantity: Matrix
• Definition: A rectangular array of numbers or symbols arranged in rows and columns.
• SI Unit: Dimensionless
• Dimensional Formula: [M⁰L⁰T⁰]

Identity Matrix (I):

• A square matrix in which all the elements of the principal diagonal are 1 and all other elements are 0.
• Physical Quantity: Identity Matrix
• Formula for 2 × 2: \( I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \)
• SI Unit: Dimensionless

Matrix Multiplication:

• The product of two matrices is calculated by multiplying the elements of rows of the first matrix by the elements of columns of the second matrix and summing them.

Scalar Multiplication:

• Multiplying a matrix by a scalar k involves multiplying every individual element of the matrix by that scalar.

 

Calculation:

Given,

Matrix \( A = \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} \)

Equation: \( A^2 - kA + 2I = 0 \)

First, calculate \( A^2 \):

⇒ \( A^2 = \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} \)

⇒ \( A^2 = \begin{bmatrix} 9 - 8 & -6 + 4 \ 12 - 8 & -8 + 4 \end{bmatrix} \)

⇒ \( A^2 = \begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix} \)

Now, substitute in the given equation:

⇒ \( \begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix} - k \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = 0 \)

⇒ \( \begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix} - \begin{bmatrix} 3k & -2k \ 4k & -2k \end{bmatrix} + \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \)

Equating the elements from the first row and

first column:

⇒ 1 − 3k + 2 = 0

⇒ 3 − 3k = 0

⇒ 3k = 3

⇒ k = 1

Verify with elements from the second row and

first column:

⇒ 4 − 4k + 0 = 0

⇒ 4k = 4

⇒ k = 1

∴ Hence, the value of k is 1.

Concept:

In a series LCR circuit,

Current:

I = V/Z

where impedance,

Z = √[R² + (X_L − X_C)²]

At resonance:

X_L = X_C

⇒ Z = R (minimum)

Therefore, current becomes maximum.

Power factor:

cosφ = R/Z

At resonance,

Z = R

⇒ cosφ = 1

Solution:

Statement I:

As frequency increases,

Current first increases, becomes maximum at resonance, and then decreases.

This statement is correct.

Statement II:

At resonance in a series LCR circuit,

X_L = X_C

Hence impedance becomes minimum and equal to resistance.

Therefore,

cosφ = R/R = 1

Thus, power factor at resonance is one.

Hence, Statement II is also correct.

Therefore, both Statement I and Statement II are true.

Concept

• Zener Diode: It is a heavily doped P-N junction diode designed to operate specifically in the reverse breakdown region without being damaged.
• Voltage Regulation: A Zener diode acts as a voltage regulator because, in the breakdown region, the voltage across it remains constant even if the current flowing through it changes significantly.
• Biasing: For a Zener diode to function as a regulator, it must be connected in reverse bias across the input voltage.

Explanation

Analysis of Statement A:

• For voltage regulation, the Zener diode must operate in its breakdown region. This region is only reached when the diode is reverse biased (the n-side is connected to the positive terminal and the p-side to the negative terminal).
• If the Zener diode were forward biased, it would behave like an ordinary silicon diode with a constant voltage drop of approximately \(0.7,V\), which is not useful for high-voltage regulation.
• Therefore, Statement A is incorrect.

Analysis of Statement B:

• In a regulator circuit, the Zener diode is connected in parallel with the load resistance (\(R_L\)).
• The total current from the source divides between the Zener diode (\(I_Z\)) and the load (\(I_L\)).
• The magnitude of the load current depends entirely on the value of the load resistance (\(I_L = V_Z / R_L\)). There is no operational requirement or physical law stating that the load current must be much greater than the Zener current. In many practical designs, the Zener current is kept significant to ensure the diode stays in the breakdown region even if the load changes.
• Therefore, Statement B is incorrect.

Conclusion

Since the Zener diode must be reverse biased for regulation and there is no condition requiring the load current to be much greater than the Zener current, both provided statements are false.

Hence, both A and B are not correct.

CONCEPT:

• Power factor: The ratio of the true power to the apparent power of an a.c. the circuit is called the power factor.
• Its value varies from 0 to 1.

The power factor (p.f) of a series LCR-circuit is given by:

Power factor (P) = Cos Φ  

\(\cos {\rm{Φ }} = \frac{R}{Z} = \frac{R}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}\)

Where R = resistance, Z = Impedance, X_L = Inductive reactance and X_C = Capacitive reactance

CALCULATION:

Given that:

R = 100 Ω,  X_L = 200 Ω, X_c = 100 Ω, V = 220 V and ν = 50 Hz

The power factor of a series LCR-circuit is given by

\(\cos {\rm{Φ }} = \frac{R}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}\)

\( \Rightarrow {\rm{;cosΦ }} = \frac{{100}}{{\sqrt {{{\left( {100} \right)}^2} + {{\left( {200 - 100} \right)}^2}} }} = \frac{{;100}}{{100\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\)

\( \Rightarrow {\rm{Φ }} = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = 45^\circ \)

Concept

• The Joule-Thomson effect (or Joule-Kelvin effect) describes the temperature change of a real gas when it is allowed to expand freely through a multi-porous plug or a constricted nozzle from a region of high pressure to a region of low pressure under adiabatic conditions.
• For an ideal gas, there is no temperature change during this process because there are no intermolecular forces to overcome.
• For real gases, whether the gas heats up or cools down depends on its Inversion Temperature (\(T_i\)).

Key Points

The Joule-Thomson coefficient (\(\mu_{JT}\)) is defined as:

\(\mu_{JT} = \left( \frac{\partial T}{\partial P} \right)_H\)

• If \(\mu_{JT} > 0\), the gas cools down upon expansion (cooling effect).
• If \(\mu_{JT} < 0\), the gas heats up upon expansion (heating effect).
• The Inversion Temperature (\(T_i\)) is the temperature at which \(\mu_{JT} = 0\).

Explanation

The behavior of a real gas during the Joule-Thomson expansion is determined by its initial temperature relative to its inversion temperature:

1. Below Inversion Temperature (\(T < T_i\)): Most gases (like \(N_2, O_2, CO_2\)) at room temperature are below their inversion temperature. When they expand, they perform work to overcome attractive intermolecular forces (van der Waals forces), leading to a decrease in internal energy and thus a cooling effect.
2. Above Inversion Temperature (\(T > T_i\)): Gases like Hydrogen and Helium have very low inversion temperatures. If they expand at room temperature (which is above their \(T_i\)), repulsive forces dominate or the \(PV\) work term increases, causing the gas to heat up.
3. At Inversion Temperature (\(T = T_i\)): The gas shows neither heating nor cooling effect.

Since the outcome (heating or cooling) varies based on the gas's starting thermal state, it is strictly dependent on the initial temperature.

Hence, in the Joule-Thomson effect, the gas heats up or cools down depending upon its initial temperature.

Concept

• A Carnot engine is an ideal heat engine that operates on the Carnot cycle.
• The efficiency of a Carnot engine depends only on the absolute temperatures of the heat source and the heat sink.
• Temperatures used in the efficiency formula must always be in the Kelvin scale.

Formula Used

Efficiency of a Carnot engine:

\(\eta = 1 - \frac{T_{sink}}{T_{source}}\)

Temperature conversion (Celsius to Kelvin):

\(T(K) = T(^\circ C) + 273\)

Calculation

Given:

Efficiency, \(\eta = 25% = \frac{25}{100} = 0.25\)

Temperature of the source, \(T_{source} = 327^\circ C\)

Converting source temperature to Kelvin:

\(T_{source} = 327 + 273 = 600,K\)

Using the efficiency formula:

\(\eta = 1 - \frac{T_{sink}}{T_{source}}\)

\(0.25 = 1 - \frac{T_{sink}}{600}\)

Rearranging the equation to find \(T_{sink}\):

\(\frac{T_{sink}}{600} = 1 - 0.25\)

\(\frac{T_{sink}}{600} = 0.75\)

\(T_{sink} = 0.75 \times 600\)

\(T_{sink} = 450,K\)

Now, converting the sink temperature back to Celsius:

\(T_{sink}(^\circ C) = 450 - 273\)

\(T_{sink} = 177^\circ C\)

Hence, the temperature of the sink must be \(177^\circ C\).

Concept:

A Carnot engine operates on a completely reversible cycle known as the Carnot cycle. It represents an ideal heat engine that achieves the maximum possible efficiency for converting heat energy into mechanical work when operating between two given temperature limits.

The efficiency of a Carnot heat engine is defined as the ratio of work obtained to the heat absorbed from the source.

⇒ η = W / Q

Here, η represents the efficiency of the Carnot engine, W denotes the work done by the engine, and Q is the amount of heat supplied to the system.

Calculation:

Given efficiency of the engine,

⇒ η = 1 / 10

The ratio of rejected heat to work done is expressed as

⇒ β = (1 − η) / η

Substituting the given value of efficiency,

⇒ β = (1 − 1/10) / (1/10)

⇒ β = (9/10) ÷ (1/10)

⇒ β = 9

This ratio can also be written as

⇒ β = Q₂ / W

If the work done by the engine is 10 J, then

⇒ Q₂ = 9 × 10 = 90 J