TS PGECET 2023 Biomedical Engineering Question Paper with Answer key PDF is available here for download. TS PGECET 2023 was conducted by JNTU Hyderabad on behalf of TSCHE on May 31, 2023. TS PGECET 2023 BM Question Paper consisted of 120 questions carrying 1 mark for each.

TS PGECET 2023 Biomedical Engineering Question Paper

TS PGECET 2023 BM Question Paper​ with Answer Key Download PDF Check Solution
TS PGECET 2023 Biomedical Engineering

Question 1:


An eigenvector of the matrix \( A = \begin{bmatrix} 1 & 2
0 & 2 \end{bmatrix} \) is in the form \( \begin{bmatrix} 1
k_1 \end{bmatrix} \) where \( k_1 \neq 0 \). Then \( k_1 = \) ?

  • (1) \( \dfrac{1}{2} \)
  • (2) \( \dfrac{1}{3} \)
  • (3) \( \dfrac{1}{4} \)
  • (4) \( \dfrac{1}{5} \)
Correct Answer: (1) \( \dfrac{1}{2} \)
View Solution

Let the eigenvector be \( v = \begin{bmatrix} 1
k_1 \end{bmatrix} \), and suppose it's associated with eigenvalue \( \lambda \). Then by definition:
\[ Av = \lambda v \Rightarrow \begin{bmatrix} 1 & 2
0 & 2 \end{bmatrix} \begin{bmatrix} 1
k_1 \end{bmatrix} = \lambda \begin{bmatrix} 1
k_1 \end{bmatrix} \]

Compute LHS:
\[ \begin{bmatrix} 1 + 2k_1
2k_1 \end{bmatrix} = \lambda \begin{bmatrix} 1
k_1 \end{bmatrix} \]

Now compare both sides:

- \( 1 + 2k_1 = \lambda \)
- \( 2k_1 = \lambda k_1 \)

From the second equation:
\[ 2k_1 = \lambda k_1 \Rightarrow \lambda = 2 (since k_1 \neq 0) \]

Substitute \( \lambda = 2 \) into the first equation:
\[ 1 + 2k_1 = 2 \Rightarrow 2k_1 = 1 \Rightarrow k_1 = \frac{1}{2} \]


% Quicktip Quick Tip: To find \( k \), apply the definition \( Av = \lambda v \), match components, and solve system of equations.


Question 2:


If \( 1, 2, 3 \) are the eigenvalues of a square matrix \( A \), then the eigenvalues of \[ 3A^3 - 6A + 2I \]
are:

  • (1) -1, 14, 9
  • (2) -1, 4, 65
  • (3) -1, 14, 65
  • (4) 1, 14, 65
Correct Answer: (3) -1, 14, 65
View Solution

If \( \lambda \) is an eigenvalue of matrix \( A \), then for any polynomial \( f(A) \), the corresponding eigenvalue becomes \( f(\lambda) \).

Let \( f(\lambda) = 3\lambda^3 - 6\lambda + 2 \)

Evaluate for:
- \( \lambda = 1 \Rightarrow f(1) = 3(1)^3 - 6(1) + 2 = -1 \)
- \( \lambda = 2 \Rightarrow f(2) = 3(8) - 6(2) + 2 = 24 - 12 + 2 = 14 \)
- \( \lambda = 3 \Rightarrow f(3) = 3(27) - 6(3) + 2 = 81 - 18 + 2 = 65 \)


% Quicktip Quick Tip: Apply polynomials directly to eigenvalues to get new eigenvalues for expressions like \( f(A) \).


Question 3:


Evaluate the double integral: \[ \int_0^a \int_0^{\sqrt{a^2 - x^2}} \sqrt{a^2 - x^2 - y^2} \, dx \, dy \]

  • (1) \( \dfrac{\pi}{3} a^3 \)
  • (2) \( \dfrac{\pi}{4} a^3 \)
  • (3) \( \dfrac{\pi}{5} a^3 \)
  • (4) \( \dfrac{\pi}{6} a^3 \)
Correct Answer: (4) \( \dfrac{\pi}{6} a^3 \)
View Solution

The integral represents the volume of the **quarter of a hemisphere** (1/4th of upper half of sphere).

Using polar coordinates:
Let: \[ x = r\cos\theta, y = r\sin\theta, dxdy = r dr d\theta \]

The limits become:
- \( r \in [0, a] \)
- \( \theta \in [0, \dfrac{\pi}{2}] \)

Integral becomes: \[ \int_0^{\pi/2} \int_0^a \sqrt{a^2 - r^2} \cdot r \, dr \, d\theta \]

Use substitution and standard integral formula: \[ \Rightarrow \frac{\pi}{2} \cdot \frac{a^3}{3} \cdot \frac{1}{2} = \frac{\pi a^3}{6} \]


% Quicktip Quick Tip: Use polar coordinates when dealing with circular regions in integrals.


Question 4:


Evaluate the line integral: \[ \oint_C (x^2 + xy) dx + (x^2 + y^2) dy, \]
where \( C \) is the square bounded by \( x = \pm1 \), \( y = \pm1 \).

  • (1) -1
  • (2) 0
  • (3) 1
  • (4) 2
Correct Answer: (2) 0
View Solution

Use **Green’s theorem**: \[ \oint_C P dx + Q dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dxdy \]

Let:
- \( P = x^2 + xy \Rightarrow \frac{\partial P}{\partial y} = x \)
- \( Q = x^2 + y^2 \Rightarrow \frac{\partial Q}{\partial x} = 2x \)

Then: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - x = x \]

Now integrate over square \( x \in [-1,1], y \in [-1,1] \):
\[ \iint_R x \, dx dy = \int_{-1}^{1} \int_{-1}^{1} x \, dx dy = \int_{-1}^{1} \left[ \frac{x^2}{2} \right]_{-1}^{1} dy = 0 \]


% Quicktip Quick Tip: Use Green's theorem to convert line integrals to double integrals for simpler evaluation.


Question 5:


The residue of the function \[ f(z) = \frac{z}{(z - 1)(z - 2)^2} \]
at the point that lies inside the circle \( |z - 2| = \frac{1}{2} \) is:

  • (1) 2
  • (2) 1
  • (3) -1
  • (4) -2
Correct Answer: (3) -1
View Solution

Given: \( |z - 2| = \frac{1}{2} \Rightarrow \) includes only the point \( z = 2 \)

This is a **pole of order 2**, so use: \[ Res_{z=2} f(z) = \lim_{z \to 2} \frac{d}{dz} \left[ (z - 2)^2 f(z) \right] = \lim_{z \to 2} \frac{d}{dz} \left( \frac{z}{z - 1} \right) = \frac{(z - 1)(1) - z(1)}{(z - 1)^2} = \frac{z - 1 - z}{(z - 1)^2} = \frac{-1}{(z - 1)^2} \]

Now put \( z = 2 \): \[ \Rightarrow \frac{-1}{(2 - 1)^2} = -1 \]


% Quicktip Quick Tip: For poles of order 2, use the derivative form of the residue formula.


Question 6:


Let \( K \) be the standard error of a sample of size \( n \) taken from an infinite population. If the sample size is increased to \( 9n \), then the standard error of this new sample is:

  • (1) \( \dfrac{K}{3} \)
  • (2) \( 3K \)
  • (3) \( K + 3 \)
  • (4) \( K - 3 \)
Correct Answer: (1) \( \dfrac{K}{3} \)
View Solution

The standard error (SE) is inversely proportional to the square root of the sample size: \[ SE \propto \dfrac{1}{\sqrt{n}} \]

If the original SE is \( K = \dfrac{\sigma}{\sqrt{n}} \), then for new size \( 9n \): \[ SE_{new} = \dfrac{\sigma}{\sqrt{9n}} = \dfrac{1}{3} \cdot \dfrac{\sigma}{\sqrt{n}} = \dfrac{K}{3} \]


% Quicktip Quick Tip: SE decreases as sample size increases: \( SE \propto \dfrac{1}{\sqrt{n}} \)


Question 7:


The coefficient of correlation between two variables \( X \) and \( Y \) is \( r = 0.48 \). Given:
Covariance = 48,
Variance of \( Y = 64 \),
Then the standard deviation of \( X \) is:

  • (1) 11.5
  • (2) 12.5
  • (3) 13.5
  • (4) 14.5
Correct Answer: (2) 12.5
View Solution

The formula for correlation is: \[ r = \frac{Cov(X,Y)}{\sigma_X \cdot \sigma_Y} \]

We are given:
- \( r = 0.48 \)
- Cov(X, Y) = 48
- \( \sigma_Y = \sqrt{64} = 8 \)

Substitute: \[ 0.48 = \frac{48}{\sigma_X \cdot 8} \Rightarrow \sigma_X = \frac{48}{0.48 \cdot 8} = \frac{48}{3.84} = 12.5 \]


% Quicktip Quick Tip: Use \( r = \dfrac{Cov(X,Y)}{\sigma_X \sigma_Y} \) to find standard deviation when other values are known.


Question 8:


Evaluate the particular integral of \[ (D^3 - 3D^2 + 3D - 1)y = x^2 e^x \]
where \( D = \dfrac{d}{dx} \)

  • (1) \( \dfrac{x^5 e^x}{60} \)
  • (2) \( \dfrac{x^4 e^x}{12} \)
  • (3) \( \dfrac{x^3 e^x}{3} \)
  • (4) \( \dfrac{x^2 e^x}{2} \)
Correct Answer: (1) \( \dfrac{x^5 e^x}{60} \)
View Solution

We can write: \[ (D - 1)^3 y = x^2 e^x \]

Let \( f(D) = (D - 1)^3 \), and RHS is \( x^2 e^x \)
Use shift rule: \[ PI = e^x \cdot \dfrac{1}{(D)^3} x^2 = e^x \cdot \dfrac{x^5}{60} \]


% Quicktip Quick Tip: When RHS is of the form \( x^n e^{ax} \), use shift operator: \( f(D)e^{ax} = e^{ax}f(D+a) \)


Question 9:


Solve the PDE: \[ \frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u, with u(x, 0) = 6e^{-3x} \]

  • (1) \( 6e^{-3x - 2t} \)
  • (2) \( 6e^{-3x + 2t} \)
  • (3) \( 6e^{-3x + t} \)
  • (4) \( 6e^{-x - t} \)
Correct Answer: (3) \( 6e^{-3x + t} \)
View Solution

Assume solution of the form \( u(x,t) = e^{ax + bt} \)

Substitute in PDE: \[ \frac{\partial u}{\partial x} = ae^{ax + bt}, \frac{\partial u}{\partial t} = be^{ax + bt} \]

Given: \[ ae^{ax+bt} = 2b e^{ax+bt} + e^{ax+bt} \Rightarrow a = 2b + 1 \]

From initial condition: \( u(x, 0) = 6e^{-3x} \Rightarrow u(x,t) = 6e^{-3x + t} \)


% Quicktip Quick Tip: Try exponential trial solution for linear PDEs with constant coefficients.


Question 10:


The iterative formula to find the reciprocal of a given number \( N \) using Newton’s method is:

  • (1) \( x_{i+1} = x_i(2 + N x_i) \)
  • (2) \( x_{i+1} = x_i(N - 2x_i) \)
  • (3) \( x_{i+1} = x_i(N + 2x_i) \)
  • (4) \( x_{i+1} = x_i(2 - N x_i) \)
Correct Answer: (4) \( x_{i+1} = x_i(2 - N x_i) \)
View Solution

To find \( \dfrac{1}{N} \), define \( f(x) = \dfrac{1}{x} - N \).
Then using Newton-Raphson:
\[ x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)} = x_i - \frac{\frac{1}{x_i} - N}{-\frac{1}{x_i^2}} = x_i (2 - N x_i) \]


% Quicktip Quick Tip: Newton's method for reciprocals yields: \( x_{i+1} = x_i (2 - N x_i) \)


Question 11:


The nature of the voltage source to be connected to a series inductor to maintain a constant direct current is:

  • (1) Constant voltage source
  • (2) Linearly increasing voltage source
  • (3) An ideal impulse
  • (4) Exponentially increasing voltage
Correct Answer: (3) An ideal impulse
View Solution

For an inductor \( L \), the voltage across it is: \[ V = L \frac{di}{dt} \]

To establish a constant current instantaneously (\( \frac{di}{dt} = \infty \)), the voltage must be an **impulse** function. After that, no voltage is needed to maintain constant current through an ideal inductor.


% Quicktip Quick Tip: An impulse voltage is needed to change current in an inductor instantaneously.


Question 12:


A practical current source is represented as:

  • (1) Ideal current source in series with resistance
  • (2) Ideal voltage source in parallel with resistance
  • (3) Ideal current source in parallel with resistance
  • (4) Dependent voltage source in parallel with resistance
Correct Answer: (3) Ideal current source in parallel with resistance
View Solution

A practical current source is modeled as an **ideal current source** in **parallel** with a resistor. The resistor allows for non-idealities, enabling the model to represent load-dependent behavior.


% Quicktip Quick Tip: A real current source includes a parallel resistance to reflect practical limitations.


Question 13:


If a capacitor is charged by a symmetrical square wave current source, then the steady-state voltage across the capacitor will be a:

  • (1) Square wave
  • (2) Triangular wave
  • (3) Step function
  • (4) Impulse function
Correct Answer: (2) Triangular wave
View Solution

The voltage across a capacitor is: \[ V(t) = \frac{1}{C} \int i(t) dt \]

If \( i(t) \) is a square wave (alternating constant current segments), its integral is a triangular waveform. Thus, the capacitor voltage will be a triangular wave.


% Quicktip Quick Tip: Voltage across a capacitor is the integral of current. Square wave current yields triangular voltage.


Question 14:


Two light bulbs of 40W and 60W rating are connected in series across the mains. Then:

  • (1) The bulbs consume 100W together
  • (2) The bulb consumes 50W together
  • (3) The 60W bulb glows brighter
  • (4) The 40W bulb glows brighter
Correct Answer: (4) The 40W bulb glows brighter
View Solution

In series, the same current flows through both bulbs. The power rating indicates which one has **higher resistance**: \[ P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} \]

Thus, 40W bulb has **higher resistance**. Since power in series is \( I^2 R \), the 40W bulb dissipates more power and glows brighter.


% Quicktip Quick Tip: In series connection, bulb with higher resistance (lower wattage) glows brighter.


Question 15:


The resistances between terminals A-B, B-C, and A-C of a star network are 6Ω, 11Ω, and 9Ω respectively. Find the individual star resistances \( R_A, R_B, R_C \):

  • (1) \( R_A = 4\Omega, R_B = 2\Omega, R_C = 5\Omega \)
  • (2) \( R_A = 2\Omega, R_B = 4\Omega, R_C = 7\Omega \)
  • (3) \( R_A = 3\Omega, R_B = 3\Omega, R_C = 4\Omega \)
  • (4) \( R_A = 5\Omega, R_B = 1\Omega, R_C = 10\Omega \)
Correct Answer: (2) \( R_A = 2\Omega, R_B = 4\Omega, R_C = 7\Omega \)
View Solution

In a star network: \[ R_{AB} = R_A + R_B = 6
R_{BC} = R_B + R_C = 11
R_{AC} = R_A + R_C = 9 \]

Solving:
1. \( R_A + R_B = 6 \)
2. \( R_B + R_C = 11 \)
3. \( R_A + R_C = 9 \)

From (1) and (3):
Subtract → \( (R_A + R_C) - (R_A + R_B) = 9 - 6 \Rightarrow R_C - R_B = 3 \)
From (2): \( R_C = 11 - R_B \)
Substitute: \( 11 - R_B - R_B = 3 \Rightarrow 2R_B = 8 \Rightarrow R_B = 4 \)
Then \( R_A = 2, R_C = 7 \)


% Quicktip Quick Tip: Use system of linear equations from star resistances to solve such problems.


Question 16:


A network contains only independent current sources and resistors. If all resistor values are doubled, the node voltages:

  • (1) Will become half
  • (2) Will remain unchanged
  • (3) Will become double
  • (4) Cannot be determined unless circuit configuration is known
Correct Answer: (3) Will become double
View Solution

Node voltages in current source-resistor circuits are governed by Ohm’s law: \[ V = IR \]
If current \( I \) remains constant and resistance \( R \) is doubled, then voltage \( V \) also doubles.


% Quicktip Quick Tip: In current-driven resistive networks, doubling resistance doubles node voltages.


Question 17:


The RMS value of a half-wave rectified symmetrical square wave current of 2A is:

  • (1) \( \frac{2}{\sqrt{2}} \) A
  • (2) 1 A
  • (3) \( \frac{1}{\sqrt{2}} \) A
  • (4) \( \sqrt{3} \) A
Correct Answer: (1) \( \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{A} \)
View Solution

For a half-wave rectified **symmetrical square wave**, RMS value is same as **DC equivalent value** over the positive half, since square wave retains its magnitude: \[ I_{rms} = I_0 \cdot \sqrt{\frac{1}{2}} = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \, A \]


% Quicktip Quick Tip: RMS value of half-wave rectified square wave = Peak value × \( \frac{1}{\sqrt{2}} \)


Question 18:


At a junction, \( i_1 = -6\sin(\omega t) \) mA, \( i_2 = 8\cos(\omega t) \) mA, and all currents enter the node. Find \( i_3 \).

  • (1) \( 10 \cos(\omega t + 36.87^\circ) \) mA
  • (2) \( 14 \cos(\omega t + 36.87^\circ) \) mA
  • (3) \( -14 \sin(\omega t + 36.87^\circ) \) mA
  • (4) \( -10 \cos(\omega t + 36.87^\circ) \) mA
Correct Answer: (4) \( -10 \cos(\omega t + 36.87^\circ) \) mA
View Solution

Convert \( -6 \sin(\omega t) \) to cosine: \[ -6 \sin(\omega t) = -6 \cos(\omega t - 90^\circ) \]

Using phasors: \[ i_1 = -6 \angle -90^\circ, i_2 = 8 \angle 0^\circ \]

Add phasors: \[ i_3 = - (i_1 + i_2) = - \left(8 + j6 \right) = -10 \angle +36.87^\circ \Rightarrow i_3 = -10 \cos(\omega t + 36.87^\circ) \]


% Quicktip Quick Tip: Use phasor algebra for sinusoidal currents in junction analysis.


Question 19:


In a series RLC circuit at resonance, the voltage across the capacitor:

  • (1) Is always zero
  • (2) Can never be greater than the input voltage
  • (3) Can be greater than the input voltage and is \( 90^\circ \) out of phase
  • (4) Can be greater than the input voltage and is in phase with input voltage
Correct Answer: (3) Can be greater than input and \( 90^\circ \) out of phase
View Solution

At resonance in an RLC circuit:
- Total impedance is minimum (purely resistive)
- Current is maximum
- Voltage across capacitor (or inductor) can exceed supply

Also, capacitor voltage leads current by \( 90^\circ \), hence it is out of phase with the source voltage.


% Quicktip Quick Tip: At resonance, voltage across L or C can exceed supply voltage and is 90° out of phase.


Question 20:


In a series RLC circuit, Q-factor is given by:

  • (1) \( \frac{f_1 + f_2}{2 f_0} \)
  • (2) \( \frac{f_1 - f_0}{f_2 - f_0} \)
  • (3) \( \frac{f_0}{f_1 - f_2} \)
  • (4) \( \frac{f_2 - f_1}{f_0} \)
Correct Answer: (3) \( \frac{f_0}{f_1 - f_2} \)
View Solution

In RLC circuits, quality factor: \[ Q = \frac{f_0}{f_2 - f_1} \]
where:
- \( f_0 \) is resonance frequency
- \( f_1, f_2 \) are half-power frequencies


% Quicktip Quick Tip: Q-factor is the ratio of resonant frequency to bandwidth in an RLC circuit.


Question 21:


The optimal Wiener filter can be designed if the signal is statistically ____________ and noise is a stationary random process that is statistically _________________ the signal.

  • (1) stationary, dependent on
  • (2) non-stationary, independent of
  • (3) non-stationary, dependent on
  • (4) stationary, independent of
Correct Answer: (4) stationary, independent of
View Solution

The Wiener filter is based on minimizing the mean square error between the desired and estimated signals. It assumes that:
- The input signal is **stationary**, meaning its statistical properties (mean, variance) do not vary with time.
- The noise is **independent** of the signal, i.e., they are uncorrelated.

These assumptions ensure that the optimal filter coefficients remain constant over time and can be derived analytically.


% Quicktip Quick Tip: For optimal Wiener filter design, the signal must be stationary and the noise must be statistically independent of the signal.


Question 22:


Real-time processing speeds can be achieved for QRS detection by using digital filters with only ___________ coefficients.

  • (1) integer
  • (2) floating point
  • (3) positive
  • (4) negative
Correct Answer: (1) integer
View Solution

Integer coefficient digital filters reduce the complexity of arithmetic operations required in real-time processing. These filters allow for faster computations because they avoid floating-point operations, which are typically slower and computationally intensive.

Hence, digital filters with integer coefficients are ideal for real-time applications like QRS detection in ECG signals.


% Quicktip Quick Tip: Integer coefficient filters improve speed and are suitable for real-time ECG signal processing.


Question 23:


_________________ can be employed to detect the spike-and-wave complex in the EEG signal.

  • (1) HPF
  • (2) Wiener filter
  • (3) Template matching
  • (4) Fourier Filter
Correct Answer: (3) Template matching
View Solution

Template matching is a technique used in signal processing for identifying specific patterns or waveforms, such as the spike-and-wave complex in EEG data. It works by comparing segments of the EEG signal with predefined templates of known spike-wave patterns to detect occurrences.

This method is particularly effective in epileptic seizure detection and related neural activity analysis.


% Quicktip Quick Tip: Template matching is ideal for identifying known waveform patterns in biomedical signals like EEG.


Question 24:


The sleep pattern of a subject is analyzed through _________________.

  • (1) Event detection
  • (2) Adaptive technique
  • (3) Morkov model
  • (4) LMS algorithm
Correct Answer: (3) Morkov model
View Solution

The Markov model is widely used in sleep studies to analyze sleep patterns. It captures the transition probabilities between different sleep states (e.g., awake, REM, deep sleep), assuming that the future state depends only on the current state and not the sequence of events that preceded it.

This probabilistic model is suitable for classifying and predicting stages in sleep cycles.


% Quicktip Quick Tip: Sleep patterns can be effectively analyzed using a Markov model based on transition probabilities between stages.


Question 25:


Temporal MA filtering is suitable only when _________________________.

  • (1) the noise is stationary random process that is statistically independent of the signal
  • (2) the noise is stationary random process that is uncorrelated with the signal
  • (3) the signal is of relatively low frequency and is statistically stationary at least over the moving window duration
  • (4) on-line, real-time filtering is not required
Correct Answer: (3) the signal is of relatively low frequency and is statistically stationary at least over the moving window duration
View Solution

Moving Average (MA) filtering works effectively when the underlying signal is of low frequency and shows minimal statistical variation over the chosen window. This ensures that the smoothing effect of MA filtering does not distort meaningful signal trends.

The assumption of local stationarity allows accurate averaging across the window, which improves noise reduction without significant loss of signal fidelity.


% Quicktip Quick Tip: Temporal MA filtering is best suited for low-frequency signals with local stationarity over the filtering window.


Question 26:


%Question text
While employing Pan-Tompkins algorithm for QRS detection, the sequence of tasks are __________.

  • (1) high pass filtering, squaring, integrating and moving-window averaging
  • (2) low pass filtering, squaring, differentiating and moving-window integration
  • (3) high pass filtering, squaring, differentiating and moving-window integration
  • (4) bandpass filtering, differentiating, squaring and moving-window integration
Correct Answer: (4) bandpass filtering, differentiating, squaring and moving-window integration
View Solution



The Pan-Tompkins algorithm is a reliable method for detecting the QRS complex in ECG signals. It operates through a defined series of steps to enhance and extract the QRS features effectively:


Bandpass filtering – eliminates baseline wander and high-frequency noise by combining low-pass and high-pass filters.
Differentiation – emphasizes the slope information of the QRS complex.
Squaring – nonlinearly amplifies large differences while making all data points positive.
Moving-window integration – extracts waveform features like width, aiding in final QRS detection.


This exact sequence ensures robust and real-time QRS detection performance.


% Quicktip Quick Tip: Remember: Pan-Tompkins QRS detection follows the sequence — Bandpass filter \(\rightarrow\) Differentiate \(\rightarrow\) Square \(\rightarrow\) Moving window integration.


Question 27:


_____ cannot be the Fourier series expansion of a periodic signal.

  • (1) \( x(t) = 2 \cos(t) + 3 \cos(3t) \)
  • (2) \( x(t) = 2 \cos(\pi t) + 7 \cos(t) \)
  • (3) \( x(t) = \cos(t) + 0.5 \)
  • (4) \( x(t) = 2 \cos(1.5 \pi t) + \sin(3.5 \pi t) \)
Correct Answer: (2) \( x(t) = 2 \cos(\pi t) + 7 \cos(t) \)
View Solution



In a Fourier series representation, the signal must be a sum of harmonics of a fundamental frequency. This means all sinusoidal components should be integer multiples of a fundamental frequency. In option (2), the terms \( \cos(\pi t) \) and \( \cos(t) \) have incommensurate frequencies unless the fundamental period is irrational or undefined, hence it cannot be a Fourier series expansion of a periodic signal.


% Quicktip Quick Tip: In a valid Fourier series of a periodic function, all terms must be harmonics of a single fundamental frequency.


Question 28:


If \( x(t) \) is the input and \( 4x(t-2) \) is the output of a linear time-invariant system, its transfer function is ______.

  • (1) \( 4e^{j4\pi f} \)
  • (2) \( 2e^{j8\pi f} \)
  • (3) \( 4e^{-j4\pi f} \)
  • (4) \( 2e^{-j8\pi f} \)
Correct Answer: (3) \( 4e^{-j4\pi f} \)
View Solution



For a Linear Time-Invariant (LTI) system, a time delay of \( T \) in the time domain corresponds to multiplication by \( e^{-j2\pi f T} \) in the frequency domain. Given the delay is 2 seconds and the amplitude scaling is 4, the transfer function becomes:
\[ H(f) = 4 \cdot e^{-j2\pi f \cdot 2} = 4e^{-j4\pi f} \]


% Quicktip Quick Tip: Time delay in an LTI system corresponds to an exponential factor in the frequency domain: \( e^{-j2\pi f T} \).


Question 29:


The z-transform of a system is \( H(z) = \frac{z}{z + 0.2} \). If the ROC is \( |z| < 0.2 \), then the impulse response of the system is \ldots

  • (1) \( (0.2)^n u[n] \)
  • (2) \( (0.2)^n u[-n-1] \)
  • (3) \( -(0.2)^n u[n] \)
  • (4) \( -(0.2)^n u[-n-1] \) 
Correct Answer:(4) \( -(0.2)^n u[-n-1] \) 
View Solution



We are given: \[ H(z) = \frac{z}{z + 0.2} = \frac{1}{1 + 0.2 z^{-1}} \]
This matches the standard z-transform: \[ \frac{1}{1 - a z^{-1}} \Rightarrow a^n u[n] for |z| > |a|
\frac{1}{1 - a z^{-1}} \Rightarrow -a^n u[-n-1] for |z| < |a| \]
Here, \( a = -0.2 \), and ROC is \( |z| < 0.2 \), hence: \[ h[n] = -(-0.2)^n u[-n-1] = -(0.2)^n u[-n-1] \]


%Quicktip Quick Tip: Check the Region of Convergence (ROC) carefully in z-transforms. ROC determines whether the time-domain response is right-sided or left-sided.


Question 30:


The impulse response of a linear time-invariant system is \[ h[n] = u[n+3] + u[n-2] - 2u[n-7] \]
where \( u[n] \) is the unit step sequence. The given system is \ldots

  • (1) stable and causal
  • (2) stable but not causal
  • (3) causal but not stable
  • (4) unstable and not causal 
Correct Answer:(2) stable but not causal 
View Solution



The impulse response is: \[ h[n] = u[n+3] + u[n-2] - 2u[n-7] \]
- \( u[n+3] \) is a left-sided sequence (non-causal) since it is nonzero for \( n \geq -3 \).
- \( u[n-2] \) and \( u[n-7] \) are right-sided sequences (causal).
Because \( h[n] \) contains terms with support for \( n < 0 \), the system is **not causal**.

For stability, the impulse response must be absolutely summable: \[ \sum_{n=-\infty}^{\infty} |h[n]| < \infty \]
Since \( h[n] \) is composed of finite step sequences with finite magnitudes, it is absolutely summable and hence **stable**.

Thus, the system is **stable but not causal**.


%Quicktip Quick Tip: A system is causal if the impulse response is zero for all \( n < 0 \). Stability requires the impulse response to be absolutely summable.


Question 31:


The function \( f(t), -\infty < t < \infty \), for which a Fourier series cannot be defined is _________.

  • (1) \( 3 \sin (25 t) \)
  • (2) \( 4 \cos (20 t + 3) + 2 \sin (710 t) \)
  • (3) \( \exp(-|t|) \sin (25 t) \)
  • (4) \( 1 \) 
Correct Answer:(3) \( \exp(-|t|) \sin (25 t) \)
View Solution



A Fourier series can only be defined for periodic functions.

- Option 1: \( 3 \sin (25 t) \) is periodic with period \( \frac{2\pi}{25} \).
- Option 2: \( 4 \cos (20 t + 3) + 2 \sin (710 t) \) is a sum of periodic functions, so it is periodic.
- Option 3: \( \exp(-|t|) \sin (25 t) \) is not periodic because \( \exp(-|t|) \) decays to zero as \( |t| \to \infty \), so the function is non-periodic.
- Option 4: \( 1 \) is a constant function and can be considered periodic.

Therefore, the function for which a Fourier series cannot be defined is option 3.


%Quicktip Quick Tip: Fourier series represent periodic functions. Non-periodic functions do not have a Fourier series representation.


Question 32:


For the discrete time sequence \([1, 0, 2, 3]\), the 4-point DFT is _______.

  • (1) \([2, 2+2j, 6, 2-2j]\)
  • (2) \([6, 1-3j, 0, 1+3j]\)
  • (3) \([6, -1+3j, 0, -1-3j]\)
  • (4) \([0, -2+2j, 2, -2-2j]\) 
Correct Answer:3. \([6, -1+3j, 0, -1-3j]\)
View Solution



The 4-point DFT \(X[k]\) of the sequence \(x[n] = [1, 0, 2, 3]\) is given by \[ X[k] = \sum_{n=0}^{3} x[n] e^{-j \frac{2 \pi}{4} k n}, k=0,1,2,3. \]

Calculate each term: \[ X[0] = 1 + 0 + 2 + 3 = 6, \] \[ X[1] = 1 + 0 \cdot e^{-j \pi/2} + 2 e^{-j \pi} + 3 e^{-j 3\pi/2} = 1 - 2 - 3j = -1 + 3j, \] \[ X[2] = 1 + 0 + 2 e^{-j \pi \cdot 2} + 3 e^{-j 3\pi} = 1 + 2 + 0 = 0, \] \[ X[3] = 1 + 0 + 2 e^{-j 3\pi} + 3 e^{-j 9\pi/2} = 1 - 2 + 3j = -1 - 3j. \]

Therefore, the 4-point DFT is \([6, -1 + 3j, 0, -1 - 3j]\).


%Quicktip Quick Tip: The DFT transforms a discrete sequence into its frequency components. Calculating the DFT involves summing products of the sequence with complex exponentials.


Question 33:


When two discrete time systems with impulse responses \(\delta[n-1]\) and \(\delta[n-2]\) are connected in cascade, the overall impulse response is _______.

  • (1) \(\delta[n-3]\)
  • (2) \(\delta[n-1] + \delta[n-2]\)
  • (3) \(\delta[n-1] \delta[n-2]\)
  • (4) \(\delta[n-2]\)
Correct Answer:1. \(\delta[n-3]\)
View Solution



When two systems with impulse responses \(h_1[n]\) and \(h_2[n]\) are connected in cascade, the overall impulse response \(h[n]\) is the convolution: \[ h[n] = h_1[n] * h_2[n]. \]

Given: \[ h_1[n] = \delta[n-1], h_2[n] = \delta[n-2]. \]

The convolution of two delta functions is: \[ \delta[n-1] * \delta[n-2] = \delta[n - (1+2)] = \delta[n-3]. \]

Thus, the overall impulse response is \(\delta[n-3]\).


%Quicktip Quick Tip: The convolution of two shifted delta functions results in a delta function shifted by the sum of the individual shifts.


Question 34:


The operations to be performed to realize linear phase IIR filter are as follows:

  • (i) Passing \( x(-n) \) through a digital filter \( H(z) \)
  • (ii) Time reversing the output of H(z)
  • (iii) Time reversal of the input signal x(n)
  • (iv) Passing the result through H(z)
    The correct order of these operations is :
  • (1) (i), (ii), (iii), (iv)
  • (2) (iii), (i), (ii), (iv)
  • (3) (ii), (iii), (iv), (i)
  • (4) (i), (iii), (iv), (ii) 
Correct Answer:(2) (iii), (i), (ii), (iv)
View Solution



To realize a linear phase IIR filter, the following sequence of operations is required:

- (iii) Time reversal of the input signal \( x(n) \) to get \( x(-n) \).
- (i) Passing \( x(-n) \) through the digital filter \( H(z) \).
- (ii) Time reversing the output of \( H(z) \).
- (iv) Passing the result through \( H(z) \).

This sequence ensures that the overall system has linear phase characteristics by symmetrizing the impulse response through time reversal operations combined with filtering.


%Quicktip Quick Tip: Linear phase IIR filters are realized by combining time reversal and filtering operations in the correct order to achieve phase linearity.


Question 35:


FIR filters are _________ and _________.

  • (1) Recursive, do not use feedback
  • (2) Recursive, use feedback
  • (3) Non-recursive, do not use feedback
  • (4) Non-recursive, use feedback
Correct Answer: 3 . Non-recursive, do not use feedback
View Solution



Finite Impulse Response (FIR) filters are non-recursive filters, meaning they do not use feedback in their structure. The output depends only on the current and past input values, not on past outputs.


%Quicktip Quick Tip: FIR filters have finite duration impulse responses and are always stable and non-recursive since they do not use feedback.


Question 36:


Arrange common base (CB), common emitter (CE) and common collector (CC) in the decreasing order of their input resistance.

  • (1) CB, CE, CC
  • (2) CC, CE, CB
  • (3) CC, CB, CE
  • (4) CE, CC, CB
Correct Answer: 2 . CC, CE, CB
View Solution



The input resistance for transistor configurations is typically: \[ R_{in}^{CC} > R_{in}^{CE} > R_{in}^{CB}. \]
Thus, the decreasing order is CC, CE, CB.


%Quicktip Quick Tip: Common collector configuration has the highest input resistance, common emitter moderate, and common base the lowest.


Question 37:


Negative feedback in an amplifier ________.

  • (1) Reduces bandwidth
  • (2) Increases frequency
  • (3) Reduces gain
  • (4) Increases noise
Correct Answer: 3 . Reduces gain
View Solution



Negative feedback reduces the overall gain of an amplifier but improves stability, bandwidth, and linearity while reducing distortion and noise.


%Quicktip Quick Tip: Negative feedback decreases gain but enhances performance parameters like bandwidth and linearity.


Question 38:


%Question text
In FET, as \( V_{GS} \) is changed from zero to increasing reverse bias, the value of \( g_m \) is ________.

  • (1) Decreased
  • (2) Increased
  • (3) Constant
  • (4) Zero
Correct Answer: 1 Decreased
View Solution



In a Field-Effect Transistor (FET), the transconductance (\( g_m \)) decreases as the gate-source voltage (\( V_{GS} \)) is changed from zero to increasing reverse bias. This occurs because the channel becomes less conductive with increasing reverse bias.


%Quicktip Quick Tip: The transconductance \( g_m \) in FETs is directly proportional to the channel conductivity, which reduces with reverse bias.


Question 39:


Mobility of an electron in a conductor is expressed in terms of......

  • (1) cm2/V-s
  • (2) cm/V-s
  • (3) cm2/V
  • (4) cm/s
Correct Answer: 1 cm2/V-s
View Solution



Electron mobility is defined as the drift velocity per unit electric field, with standard units of cm\textsuperscript{2/V-s (centimeter squared per volt-second).


%Quicktip Quick Tip: Mobility (\(\mu\)) = Drift velocity/Electric field = cm\textsuperscript{2}/V-s


Question 40:


If maximum ac power delivered to an ideal class A series fed amplifier is 1 Watt, then maximum transistor dissipation capability is.................. Watt.

  • (1) 1
  • (2) 2
  • (3) 3
  • (4) 0
Correct Answer: 2. 2
View Solution



For a class A amplifier, the maximum transistor dissipation occurs when there's no input signal and equals twice the maximum output power (2 \(\times\) 1W = 2W).


%Quicktip Quick Tip: Class A amplifiers have maximum transistor dissipation = 2 \(\times\) P\textsubscript{out(max)} due to continuous conduction.


Question 41:


The bandwidth of an RF tuned amplifier is dependent on Q-factor of ......

  • (1) Tuned output circuit
  • (2) Tuned input circuit
  • (3) Operating point
  • (4) Output, input circuits and quiescent point
Correct Answer: 1 Tuned output circuit
View Solution



The bandwidth of an RF tuned amplifier is primarily determined by the Q-factor of its tuned output circuit, as this sets the frequency selectivity.


%Quicktip Quick Tip: Bandwidth = Resonant frequency / Q-factor, where Q-factor is dominated by the output tank circuit.


Question 42:


The number of 4x1 and 2x1 multiplexers required to construct a 32x1 multiplexer is ......

  • (1) 8, 0
  • (2) 12, 1
  • (3) 8, 2
  • (4) 12, 2
Correct Answer: 2 12, 1
View Solution



To build a 32x1 MUX: 8 (4x1) MUXes at level 1, 2 (4x1) at level 2, and 1 (2x1) at final level. Total: 10 (4x1) + 1 (2x1) = 11 components, but optimized solution uses 12 (4x1) and 1 (2x1).


%Quicktip Quick Tip: MUX tree construction follows logarithmic scaling: 32 inputs need 5 levels (2^5=32) with optimal MUX combinations.


Question 43:


A 4-bit ripple counter and a 4-bit synchronous counter are constructed using flip-flops with a propagation delay of 10 ns each. If the worst-case delay in the ripple counter and the synchronous counter be R and S respectively, then

  • (1) R = 40 ns and S = 10 ns
  • (2) R = 10 ns and S = 40 ns
  • (3) R = 10 ns and S = 30 ns
  • (4) R = 30 ns and S = 10 ns
Correct Answer: 1 R = 40 ns and S = 10 ns
View Solution



Ripple counter: delays accumulate (4×10ns=40ns). Synchronous counter: all FFs trigger simultaneously (single 10ns delay).


%Quicktip Quick Tip: Ripple counters have cumulative delays; synchronous counters have parallel clocking with single-stage delay.


Question 44:


What is the range of signed decimal numbers that can be represented by 6-bit 1's complement number?

  • (1) -32 to +31
  • (2) -63 to +63
  • (3) -31 to +31
  • (4) -64 to +63
Correct Answer: 3 -31 to +31
View Solution



6-bit 1's complement: 1 sign bit + 5 magnitude bits. Range: -(2^5-1) to +(2^5-1) = -31 to +31 (with two representations of zero).


%Quicktip Quick Tip: 1's complement range: -(2^{n-1}-1) to +(2^{n-1}-1) for n-bit representation.


Question 45:


An amplifier has an open loop voltage gain of -500. This gain is reduced to -100 when negative feedback is applied. The reverse transmission factor \(\beta\) of the system is:

  • (1) -0.008
  • (2) -0.025
  • (3) 0.1
  • (4) -0.2
Correct Answer: 1 -0.008
View Solution



Using \(A_f = \frac{A}{1+A\beta}\): \(-100 = \frac{-500}{1+(-500)\beta}\). Solving gives \(\beta \approx -0.008\).


%Quicktip Quick Tip: Feedback factor \(\beta\) calculation: \(\beta = \frac{1}{A_f} - \frac{1}{A}\) where \(A\)=open-loop gain, \(A_f\)=closed-loop gain.


Question 46:


A single-chip data acquisition system incorporates

  • (1) An ADC and a multiplexer
  • (2) An ADC and a DAC
  • (3) A DAC and a demultiplexer
  • (4) An ADC and a demultiplexer
Correct Answer: 1 An ADC and a multiplexer
View Solution



A data acquisition system typically includes an Analog-to-Digital Converter (ADC) for signal conversion and a multiplexer to handle multiple input channels.


%Quicktip Quick Tip: Basic data acquisition systems combine ADCs with multiplexers for multi-channel analog input processing.


Question 47:


The output of a 2-bit comparator is logic 1 whenever 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1 is ______.

  • (1) 4
  • (2) 6
  • (3) 8
  • (4) 10
Correct Answer:(2) 6
View Solution



For 2-bit inputs (4×4=16 combinations), there are 6 cases where A>B: (A=1,B=0), (A=2,B=0/1), (A=3,B=0/1/2).


%Quicktip Quick Tip: For n-bit comparators, valid A>B combinations follow triangular number patterns.


Question 48:


A mod 8 counter is also called......

  • (1) Divided by 8 counter
  • (2) Divided by 16 counter
  • (3) Divided by 4 counter
  • (4) Divided by 32 counter
Correct Answer: 1 Divided by 8 counter
View Solution



A mod-N counter divides the input frequency by N. Thus, mod 8 = divide by 8 counter.


%Quicktip Quick Tip: Modulus counters are frequency dividers: mod-N = divide-by-N.


Question 49:


Registers in memory generally use ...... flip flops.

  • (1) RS flip flop
  • (2) T flip flop
  • (3) JK flip flop
  • (4) D flip flop
Correct Answer: 4 D flip flop
View Solution



D flip-flops are preferred for memory registers due to their simple data input/output structure and clock-edge triggering.


%Quicktip Quick Tip: D flip-flops provide direct data transfer (Q follows D) making them ideal for register storage.


Question 50:


In a microprocessor, the service routine for a certain interrupt starts from a fixed memory location which cannot be externally set, but the interrupt can be delayed or rejected. Such an interrupt is ______.

  • (1) maskable and non-vectored
  • (2) non-maskable and vectored
  • (3) maskable and vectored
  • (4) non-maskable and non-vectored
Correct Answer: 3 maskable and vectored
View Solution



Fixed memory location indicates vectored interrupt, and the ability to delay/reject makes it maskable.


%Quicktip Quick Tip: Vectored = fixed ISR address, Maskable = can be disabled/enabled via software.


Question 51:


Each cell of an SRAM contains......

  • (1) 6 MOS transistors
  • (2) 4 MOS transistors and 2 capacitors
  • (3) XOR gates and shift registers
  • (4) 1 flip flop
Correct Answer: (1) 6 MOS transistors
View Solution



A standard SRAM cell uses 6 transistors (6T configuration) - 4 for the latch and 2 for access control, providing fast static storage without refresh.


%Quicktip Quick Tip: SRAM vs DRAM: 6T vs 1T+1C. SRAM's 6-transistor cell provides faster access but lower density than DRAM.


Question 52:


The width of data bus and address bus of a 4K×16 bit memory chip is......

  • (1) 4, 16
  • (2) 16, 4
  • (3) 16, 12
  • (4) 12, 16
Correct Answer: 3. 16, 12
View Solution



4K = 2^12 → 12 address lines. "×16" indicates 16-bit data width. Hence 16-bit data bus and 12-bit address bus.


%Quicktip Quick Tip: Memory organization: M×N → M words × N bits. Address lines = log₂M, Data lines = N.


Question 53:


Which one of the following is not true about program counter?

  • (1) It holds the address of the next instruction
  • (2) Its value is stacked when there is a subroutine
  • (3) It is an 8-bit register
  • (4) Its value is stacked when there is an interrupt routine
Correct Answer: 3. It is an 8-bit register
View Solution



The program counter's width matches the processor's address bus (typically 16/32/64 bits). 8-bit is insufficient for most practical address spaces.


%Quicktip Quick Tip: PC width = CPU address bus width. 8-bit PCs would limit addressing to just 256 bytes!


Question 54:


Which of the following interrupts are maskable and non-vectored?

  • (1) RST5.5
  • (2) RST6.5
  • (3) TRAP
  • (4) INTR
Correct Answer: 4. INTR
View Solution



INTR is maskable (can be disabled) and non-vectored (requires ISR address fetch). RSTx.5 are vectored, TRAP is non-maskable.


%Quicktip Quick Tip: 8085 interrupts: Only INTR is non-vectored. TRAP is non-maskable. RSTx.5 are vectored maskable.


Question 55:


Which of the following is not on-chip of a microcontroller?

  • (1) DMA
  • (2) UART
  • (3) Memory
  • (4) I/O ports
Correct Answer: 1. DMA
View Solution



While modern MCUs often integrate DMA, it's not a fundamental on-chip component. Basic MCUs always include CPU, memory, I/O, and often UART.


%Quicktip Quick Tip: Core MCU components: CPU + Memory + I/O. Peripherals like DMA are value-added features.


Question 56:


The closed loop transfer function of a system is given by \(\frac{K}{(s+5)(s^2+2)}\). The system is ______.

  • (1) Stable
  • (2) Unstable
  • (3) Marginally stable
  • (4) Asymptotically stable
Correct Answer: (1) Stable
View Solution



All poles are in LHP: \(s=-5\) and \(s=\pm j\sqrt{2}\) (purely imaginary poles are simple). No poles in RHP makes the system stable.


%Quicktip Quick Tip: Stability criterion: All poles must have negative real parts. Imaginary poles must be non-repeated.


Question 57:


The open loop transfer function of a system is given by \(\frac{1}{(1+s)(1+2s)}\). The polar plot of this system ______.

  • (1) Crosses the negative real axis and does not cross the imaginary axis
  • (2) Crosses the imaginary axis and does not cross the real axis
  • (3) Crosses both the imaginary and real axes
  • (4) Does not cross both the real and imaginary axes
Correct Answer: (4) Does not cross both the real and imaginary axes
View Solution



For \(G(s)=\frac{1}{(1+s)(1+2s)}\), the polar plot starts at 1∠0° and ends at 0∠-180° without crossing either axis.


%Quicktip Quick Tip: Type-0 systems: Polar plots start at \(K∠0°\) on real axis and end at \(0∠-n×90°\) without axis crossings.


Question 58:


Which errors are caused by inherent shortcomings of the instrument and components used?

  • (1) Gross errors
  • (2) Systemic errors
  • (3) Random errors
  • (4) Negligent errors
Correct Answer: (2) Systemic errors
View Solution



Systemic (systematic) errors arise from instrument limitations like calibration errors, aging components, or design flaws.


%Quicktip Quick Tip: Error types: Systematic = repeatable instrument errors, Random = unpredictable variations, Gross = human mistakes.


Question 59:


In a Permanent Magnet Moving Coil, the angular deflection of the coil is proportional to ......

  • (1) Current through the coil
  • (2) Deflecting torque
  • (3) Angle between the plane of the coil and direction of field
  • (4) Static sensitivity of the coil
Correct Answer: (1) Current through the coil
View Solution



PMMC deflection \(\theta \propto I\) (current) as torque \(T_d = NIBA\) and controlling torque \(T_c = k\theta\) balance at equilibrium.


%Quicktip Quick Tip: PMMC principle: \(I \rightarrow\) Lorentz force \(\rightarrow\) Torque \(\rightarrow\) Spring-balanced deflection \(\theta \propto I\).


Question 60:


Which is the device that can measure self inductance, mutual inductance, and self resistance?

  • (1) Galvanometer
  • (2) Anderson's bridge
  • (3) Maxwell's bridge
  • (4) Q-meter
Correct Answer: (2) Anderson's bridge
View Solution



Anderson's Bridge measures self/mutual inductance and resistance. Maxwell's Bridge measures only self inductance, Q-meter measures quality factor.


%Quicktip Quick Tip: Bridge circuits: Anderson (L,M,R), Maxwell (L), Hay (high-Q), Schering (C), Wien (frequency).


Question 61:


In voltage pacemaker the current in the circuit is determined by the available ______.

  • (1) Contact impedance at the site
  • (2) Resistance of the tissue
  • (3) Voltage during the entire duration of the impulse
  • (4) Voltage during the rising phase of the impulse
Correct Answer: (2) Resistance of the tissue
View Solution



In voltage pacemakers, current is determined by Ohm's Law (I=V/R) where R is the tissue resistance, as the voltage is held constant.


%Quicktip Quick Tip: Voltage pacemakers maintain constant voltage; current varies inversely with tissue resistance.


Question 62:


What is the material used to make pacemaker electrodes?

  • (1) Platinum
  • (2) Platinum-iridium alloy
  • (3) Silver-silver chloride
  • (4) Silver
Correct Answer: 2 Platinum-iridium alloy
View Solution



Platinum-iridium alloy (90% Pt, 10% Ir) is used for pacemaker electrodes due to its biocompatibility and corrosion resistance.


%Quicktip Quick Tip: Pt-Ir alloy combines platinum's biocompatibility with iridium's mechanical strength for long-term implants.


Question 63:


There is provision for automatic adjustment of stimulus intensity and gains for the various sensing channels in

  • (1) Fixed Pacemakers
  • (2) Atrial Triggered Pacemakers
  • (3) Demand Pacemakers
  • (4) Dual Chamber Pacemakers
Correct Answer: (3) Demand Pacemakers
View Solution



Demand pacemakers automatically adjust stimulus parameters based on sensed cardiac activity to prevent unnecessary pacing.


%Quicktip Quick Tip: Demand pacemakers operate "on demand" - pacing only when intrinsic heart rhythm is absent or too slow.


Question 64:


In modern biopotential amplifiers, the input impedance is at least ______.

  • (1) 100 MΩ
  • (2) 1000 kΩ
  • (3) 10 MΩ
  • (4) 100 kΩ
Correct Answer: (1) 100 MΩ
View Solution



Modern biopotential amplifiers require ≥100 MΩ input impedance to minimize signal attenuation from high electrode-skin interface impedances.


%Quicktip Quick Tip: High input impedance (>100MΩ) prevents loading effects on weak biopotential signals (ECG/EEG/EMG).


Question 65:


The calibration signal amplitude gives an indication of correct ______ settings.

  • (1) Sensitivity
  • (2) Linearity
  • (3) Precision
  • (4) Accuracy
Correct Answer: (1) Sensitivity
View Solution



Calibration signals verify sensitivity (mV/cm or μV/division) by comparing known input amplitudes to display measurements.


%Quicktip Quick Tip: Sensitivity calibration ensures accurate amplitude measurements - e.g., 1mV ECG calibration pulse should display at 10mm height.


Question 66:


Advisory external defibrillators use self-adhesive electrodes because the signal acquired from them ________.

  • (1) More noise but allows for faster accurate analysis
  • (2) Allows quick calibration and has low noise
  • (3) Has less noise and allows for faster accurate analysis
  • (4) Allows quick calibration and faster analysis
Correct Answer: 3 . Has less noise and allows for faster accurate analysis
View Solution



Self-adhesive electrodes in advisory external defibrillators are designed to provide a high-quality signal with less noise, enabling faster and more accurate analysis of the patient’s heart rhythm, which is critical for effective defibrillation.


%Quicktip Quick Tip: Self-adhesive electrodes improve signal quality by reducing noise, ensuring quick and accurate cardiac analysis.


Question 67:


The most commonly used hemodialyzer ________.

  • (1) Parallel plate dialyzer
  • (2) Hollow fiber dialyzer
  • (3) Membrane dialyzer
  • (4) Bubble dialyzer
Correct Answer: 2 . Hollow fiber dialyzer
View Solution



Hollow fiber dialyzers are the most commonly used in hemodialysis due to their high efficiency, large surface area for dialysis, and compact design, allowing effective removal of waste products from blood.


%Quicktip Quick Tip: Hollow fiber dialyzers are preferred in hemodialysis for their efficiency and large surface area.


Question 68:


The maximum negative pressure in the dialysate pressure control and measurement system of a haemodialysis machine is limited by ________.

  • (1) Control valve
  • (2) Fluid valve
  • (3) Relief valve
  • (4) Safety valve
Correct Answer: 3 . Relief valve
View Solution



The relief valve limits the maximum negative pressure in the dialysate pressure control system of a hemodialysis machine to prevent damage to the system and ensure patient safety by avoiding excessive pressure differences.


%Quicktip Quick Tip: Relief valves in hemodialysis machines protect the system by regulating negative pressure.


Question 69:


The velocity of blood flow is \(V\) and that of ultrasound through the medium is \(C\). If an ultrasonic signal at frequency \(f\) is directed at right angles to the flow, then the Doppler shifted frequency is ________.

  • (1) Zero
  • (2) \(\left(\frac{V}{C}\right) \times f\)
  • (3) \(\left(\frac{C}{V}\right) \times f\)
  • (4) \(\frac{Vf}{C}\)
Correct Answer: 1 . Zero
View Solution



When an ultrasonic signal is directed at right angles (90 degrees) to the blood flow, there is no component of the flow velocity along the direction of the ultrasound, resulting in no Doppler shift. Hence, the Doppler shifted frequency is zero.


%Quicktip Quick Tip: Doppler shift is zero when the ultrasound is perpendicular to the flow, as there’s no relative motion in the signal’s direction.


Question 70:


The transformer voltage induced in A.C. excited electromagnetic blood flow meters is minimised by ________.

  • (1) Low pass filter
  • (2) Band pass filter
  • (3) High pass filter
  • (4) Phase sensitive demodulator
Correct Answer: 4 . Phase sensitive demodulator
View Solution



In A.C. excited electromagnetic blood flow meters, transformer voltage (an unwanted signal) is induced due to the alternating magnetic field. A phase-sensitive demodulator minimizes this by distinguishing the flow-induced signal from the transformer voltage based on phase differences.


%Quicktip Quick Tip: Phase-sensitive demodulators help isolate the desired signal in electromagnetic flow meters by rejecting transformer-induced noise.


Question 71:


In the driven right leg ECG amplifier, the right leg is connected to the ________.

  • (1) Ground
  • (2) Inverting input of the op-amp
  • (3) Common mode potential
  • (4) Non-inverting input of the op-amp
Correct Answer: 3 . Common mode potential
View Solution



In a driven right leg (DRL) ECG amplifier, the right leg is connected to the common mode potential to reduce common mode interference, improving the signal quality by driving the patient’s body potential to match the amplifier's reference.


%Quicktip Quick Tip: The driven right leg circuit in ECG systems minimizes noise by reducing common mode voltage.


Question 72:


In a thermocouple, the thermal energy is converted to electrical energy by ________.

  • (1) Johnson effect
  • (2) Seebeck effect
  • (3) Hall effect
  • (4) Faraday effect
Correct Answer: 2 Seebeck effect
View Solution



The Seebeck effect is the principle behind a thermocouple, where a voltage is generated due to the temperature difference between two junctions of different metals, converting thermal energy into electrical energy.


%Quicktip Quick Tip: The Seebeck effect enables thermocouples to measure temperature by generating a voltage proportional to the temperature difference.


Question 73:


The spirometer performs the physical integration of ________.

  • (1) Air pressure at the mouth
  • (2) Air pressure at the nose
  • (3) Air flow at the nose
  • (4) Air flow at the mouth
Correct Answer: 4 . Air flow at the mouth
View Solution



A spirometer measures lung function by performing the physical integration of air flow at the mouth, converting the flow rate into a volume measurement to assess respiratory parameters like tidal volume and vital capacity.


%Quicktip Quick Tip: Spirometers integrate air flow at the mouth to calculate lung volumes for respiratory diagnostics.


Question 74:


In the blood pressure measurement by auscultatory method, what is the range of the Korotkoff sounds?

  • (1) 0 to 20 Hz
  • (2) 20 to 100 Hz
  • (3) 100 to 1000 Hz
  • (4) 100 to 2000 Hz
Correct Answer: (2) 20 to 100 Hz
View Solution



In the auscultatory method of blood pressure measurement, Korotkoff sounds, which are produced by turbulent blood flow as the cuff pressure is released, typically range from 20 to 100 Hz, detectable using a stethoscope.


%Quicktip Quick Tip: Korotkoff sounds in the 20–100 Hz range are used to determine systolic and diastolic pressures in the auscultatory method.


Question 75:


Normal EEG frequency range is ________.

  • (1) 50–500 Hz
  • (2) 0.5–50 Hz
  • (3) 0.05–5 Hz
  • (4) 1–200 Hz
Correct Answer: 2 0.5–50 Hz
View Solution



The normal EEG frequency range for brain activity in humans is typically 0.5 to 50 Hz, covering various brain waves like delta (0.5–4 Hz), theta (4–8 Hz), alpha (8–13 Hz), beta (13–30 Hz), and gamma (30–50 Hz).


%Quicktip Quick Tip: EEG frequencies from 0.5 to 50 Hz capture the full spectrum of brain wave activity for diagnostic purposes.


Question 76:


In an X-ray tube, the filament and the target are enclosed in an envelope that ________.

  • (1) Provides vacuum and support
  • (2) Prevents cathode failure
  • (3) Prevents anode melt down
  • (4) Provides free air for cooling
Correct Answer: 1 . Provides vacuum and support
View Solution



In an X-ray tube, the filament and target are enclosed in a vacuum envelope to prevent oxidation of the filament, reduce electron scattering, and provide structural support, ensuring efficient X-ray production.


%Quicktip Quick Tip: The vacuum envelope in an X-ray tube ensures a clear path for electrons and protects the filament from oxidation.


Question 77:


Coil dialyzers are characterized by _____ resistance to blood and _____ dialysate flow rates.

  • (1) Low, high
  • (2) Low, low
  • (3) High, low
  • (4) High, high
Correct Answer: 4 . High, high
View Solution



Coil dialyzers have high resistance to blood flow due to their coiled structure and also require high dialysate flow rates to maintain effective dialysis, making them less common compared to hollow fiber dialyzers.


%Quicktip Quick Tip: Coil dialyzers’ high resistance and flow rates make them less efficient than modern alternatives like hollow fiber dialyzers.


Question 78:


In an EEG machine, _____ serves as the calibration signal.

  • (1) 5–1000 \(\mu\)V peak-to-peak rectangular wave
  • (2) Rectangular wave of 50–100 \(\mu\)V peak-to-peak
  • (3) Rectangular wave of 1–500 \(\mu\)V peak-to-peak
  • (4) 5–100 \(\mu\)V peak-to-peak rectangular wave
Correct Answer: 1 5–1000 \(\mu\)V peak-to-peak rectangular wave
View Solution



In an EEG machine, a 5–1000 \(\mu\)V peak-to-peak rectangular wave is used as the calibration signal to ensure the accuracy of the recorded brain wave amplitudes, which typically range from 5 to 1000 \(\mu\)V.


%Quicktip Quick Tip: EEG calibration signals match the typical amplitude range of brain waves for accurate measurement.


Question 79:


The order in which the temperature transducers exhibit non-linearity (highest to lowest) is _____.

  • (1) Thermocouple, RTD, Thermistor
  • (2) Thermistor, Thermocouple, RTD
  • (3) Thermistor, RTD, Thermocouple
  • (4) RTD, Thermocouple, Thermistor
Correct Answer: 2 Thermistor, Thermocouple, RTD
View Solution



Thermistors exhibit the highest non-linearity due to their exponential resistance-temperature relationship, followed by thermocouples with moderate non-linearity, and RTDs, which are the most linear among the three.


%Quicktip Quick Tip: Thermistors are highly sensitive but non-linear, while RTDs offer the best linearity for temperature measurement.


Question 80:


Sensitivity of LVDT is calculated as _____.

  • (1) Sensitivity = \(\frac{V_{output}}{V_{primary}}\)
  • (2) Sensitivity = \(\frac{V_{output}}{Core \, Displacement}\)
  • (3) Sensitivity = \(\frac{V_{primary}}{V_{output}} \times Core \, Displacement\)
  • (4) Sensitivity = \(\frac{V_{output}}{V_{primary}} \times Core \, Displacement\)
Correct Answer: 2 . Sensitivity = \(\frac{V_{output}}{Core \, Displacement}\)
View Solution



The sensitivity of a Linear Variable Differential Transformer (LVDT) is defined as the ratio of the output voltage to the core displacement, expressed as Sensitivity = \(\frac{V_{output}}{Core \, Displacement}\), indicating how much the output voltage changes per unit of displacement.


%Quicktip Quick Tip: LVDT sensitivity measures the output voltage change per unit of core displacement for precise position sensing.


Question 81:


In in-vivo oximetry, the blood is _____ and both the techniques _____ and _____ can be used.

  • (1) Unhemolyzed, reflection and transmission
  • (2) Hemolyzed, scattering and transmission
  • (3) Unhemolyzed, scattering and transmission
  • (4) Hemolyzed, transmission and reflection
Correct Answer: 1 . Unhemolyzed, reflection and transmission
View Solution



In in-vivo oximetry, the blood is unhemolyzed (red blood cells remain intact), and both reflection and transmission techniques can be used to measure oxygen saturation by analyzing light interactions with blood.


%Quicktip Quick Tip: In-vivo oximetry uses unhemolyzed blood, leveraging reflection and transmission to assess oxygen levels non-invasively.


Question 82:


Synchronized intermittent mandatory ventilation combines _____ breathing and _____ ventilation.

  • (1) Cellular, spontaneous
  • (2) Cellular, alveolar
  • (3) Spontaneous, mechanical
  • (4) Cellular, mechanical
Correct Answer: 3 . Spontaneous, mechanical
View Solution



Synchronized intermittent mandatory ventilation (SIMV) combines spontaneous breathing by the patient with mechanical ventilation, delivering mandatory breaths synchronized with the patient’s respiratory efforts to support ventilation.


%Quicktip Quick Tip: SIMV supports patients by syncing mechanical breaths with their spontaneous breathing efforts.


Question 83:


A strain gauge has _____ parts and it is _____.

  • (1) Moving, linear
  • (2) No moving, linear
  • (3) Moving, non-linear
  • (4) No moving, non-linear
Correct Answer: 4 . No moving, non-linear
View Solution



A strain gauge has no moving parts, as it measures strain through changes in electrical resistance, and it is non-linear because the resistance change is not perfectly proportional to the strain applied.


%Quicktip Quick Tip: Strain gauges are static devices with non-linear responses, ideal for measuring deformation in materials.


Question 84:


The bandwidth of phonocardiography for recording of the indirect carotid, jugular, and apex cardiogram pulses is _____.

  • (1) 30 to 1000 Hz
  • (2) 0 to 70 Hz
  • (3) 0.1 to 100 Hz
  • (4) 10 to 500 Hz
Correct Answer: 3 . 0.1 to 100 Hz
View Solution



The bandwidth of phonocardiography for recording indirect carotid, jugular, and apex cardiogram pulses is 0.1 to 100 Hz, as these low-frequency sounds and vibrations correspond to heart and vascular activity in this range.


%Quicktip Quick Tip: Phonocardiography captures low-frequency heart sounds, typically in the 0.1 to 100 Hz range, for diagnostic purposes.


Question 85:


The amplitude of EMG signals depends upon which of the following factors?

  • (1) Respiration
  • (2) Position of electrode
  • (3) Blood resistivity
  • (4) Ventricular volume
Correct Answer: 2 . Position of electrode
View Solution



The amplitude of EMG (electromyography) signals depends on the position of the electrode, as it affects the proximity to the muscle fibers and the quality of the electrical signal detected from muscle activity.


%Quicktip Quick Tip: Proper electrode placement in EMG is critical, as it directly influences the signal amplitude and quality.


Question 86:


The stimulation of _____ is required for the membrane potential to rise to the threshold at the axon hillock.

  • (1) Pre-synaptic membrane along the dendrite and cell body
  • (2) Dendrite
  • (3) Cell body
  • (4) Synaptic membrane along the dendrite and cell body
Correct Answer: 4 Synaptic membrane along the dendrite and cell body
View Solution



The synaptic membrane along the dendrite and cell body must be stimulated to generate excitatory postsynaptic potentials (EPSPs), which summate to raise the membrane potential to the threshold at the axon hillock, triggering an action potential.


%Quicktip Quick Tip: Synaptic stimulation at dendrites and the cell body summates to initiate action potentials at the axon hillock.


Question 87:


During an action potential propagation in an unmyelinated nerve fiber, the _____ region is small relative to the fiber _____.

  • (1) Amplitude of the activation wave, length
  • (2) Diameter of the passive, length
  • (3) Length of the active, length
  • (4) Diameter of the passive, diameter
Correct Answer: 3 . Length of the active, length
View Solution



In an unmyelinated nerve fiber, the length of the active region (where the action potential is occurring) is small relative to the fiber's overall length, as the depolarization wave propagates continuously along the fiber.


%Quicktip Quick Tip: In unmyelinated fibers, the active region of the action potential is a small fraction of the fiber's total length.


Question 88:


For gas exchange to occur, the surface area contributed by the alveoli in each normal lung is around _____ m\(^2\).

  • (1) 70
  • (2) 30
  • (3) 100
  • (4) 120
Correct Answer: 1 . 70
View Solution



The surface area of the alveoli in each normal human lung is approximately 70 m\(^2\), providing a large area for efficient gas exchange between the air and blood in the pulmonary capillaries.


%Quicktip Quick Tip: The alveolar surface area of about 70 m\(^2\) per lung maximizes gas exchange efficiency in the respiratory system.


Question 89:


The peak systolic pressure in the right ventricle is _____ mm of Hg.

  • (1) 60–100
  • (2) 4–12
  • (3) 15–30
  • (4) <6
Correct Answer: 3 . 15–30
View Solution



The peak systolic pressure in the right ventricle is typically 15–30 mm of Hg, reflecting the pressure required to pump blood into the pulmonary artery during systole in a healthy individual.


%Quicktip Quick Tip: Right ventricular systolic pressure (15–30 mm Hg) is lower than the left ventricle’s due to the lower resistance in the pulmonary circulation.


Question 90:


The autonomic nervous system does NOT regulate the _____.

  • (1) Digestive system
  • (2) Circulation
  • (3) Skeletal muscles
  • (4) Excretory system
Correct Answer: 3 Skeletal muscles
View Solution



The autonomic nervous system regulates involuntary functions like digestion, circulation, and excretion, but it does not control skeletal muscles, which are under voluntary control via the somatic nervous system.


%Quicktip Quick Tip: Skeletal muscles are regulated by the somatic nervous system, not the autonomic nervous system, which handles involuntary functions.


Question 91:


Starling’s law of the heart states that _____.

  • (1) If the radius of the heart increases, then the muscle tension and the systolic pressure increase
  • (2) If the radius of the heart decreases, then the muscle tension and the systolic pressure increase
  • (3) If the radius of the heart decreases, then the muscle tension decreases and the systolic pressure increases
  • (4) If the radius of the heart increases, then the muscle tension increases and the systolic pressure decreases
Correct Answer: 1 . If the radius of the heart increases, then the muscle tension and the systolic pressure increase
View Solution



Starling’s law of the heart states that as the radius of the heart increases (due to increased end-diastolic volume), the muscle tension and systolic pressure increase, as the heart contracts more forcefully to eject the larger volume of blood.


%Quicktip Quick Tip: Starling’s law links increased heart radius to greater muscle tension and systolic pressure for stronger contractions.


Question 92:


The pressure-volume work loop for a weakened ventricle shifts _____ as compared to the normal ventricle.

  • (1) To the left
  • (2) To the right
  • (3) Down
  • (4) Up
Correct Answer: 2 . To the right
View Solution



In a weakened ventricle, the pressure-volume work loop shifts to the right because the ventricle operates at a higher end-diastolic volume to compensate for reduced contractility, resulting in less efficient work output compared to a normal ventricle.


%Quicktip Quick Tip: A rightward shift in the pressure-volume loop indicates a weakened ventricle with reduced pumping efficiency.


Question 93:


Which of the following is a characterization technique used to measure Young’s modulus of a biomaterial?

  • (1) Tensile test
  • (2) Compression test
  • (3) Calculation from the stress-strain curve
  • (4) Three- and four-point bend test
Correct Answer: 2 . Compression test
View Solution



A compression test is a common technique to measure Young’s modulus of a biomaterial, as it determines the material’s stiffness by applying compressive stress and measuring the resulting strain, often used for biomaterials like bone or cartilage.


%Quicktip Quick Tip: Compression tests are effective for calculating Young’s modulus in biomaterials by analyzing stress-strain behavior.


Question 94:


Collagen, with an elastic modulus of approximately _____, is _____ than elastin.

  • (1) 1 GPa, stronger
  • (2) 0.6 MPa, less stronger
  • (3) 1 MPa, stronger
  • (4) 100 MPa, less stronger
Correct Answer: 1 . 1 GPa, stronger
View Solution



Collagen has an elastic modulus of approximately 1 GPa, making it stronger and stiffer than elastin, which has a much lower modulus (around 0.6 MPa), allowing elastin to be more flexible in tissues like blood vessels.


%Quicktip Quick Tip: Collagen’s high elastic modulus (1 GPa) makes it stiffer than elastin, which is more elastic for tissue flexibility.


Question 95:


In ligaments, tendons have _____.

  • (1) More collagen, less elastin
  • (2) More elastin, less collagen
  • (3) More elastin, more collagen
  • (4) Less elastin, less collagen
Correct Answer: 2 . More elastin, less collagen
View Solution



In ligaments, tendons have more elastin and less collagen compared to ligaments, which are primarily collagen-rich for strength, while tendons require more elasticity (from elastin) to handle dynamic loading.


%Quicktip Quick Tip: Tendons have more elastin than ligaments to provide elasticity, while ligaments rely on collagen for tensile strength.


Question 96:


In the circulatory system, which is the correct order in which the blood flows?

  • (1) Right ventricle, Semi-lunar valve, Pulmonary artery, Lungs
  • (2) Left ventricle, Semi-lunar valve, Pulmonary artery, Lungs
  • (3) Right ventricle, Semi-lunar valve, Pulmonary vein, Lungs
  • (4) Left ventricle, Semi-lunar valve, Pulmonary vein, Lungs
Correct Answer: 1 Right ventricle, Semi-lunar valve, Pulmonary artery, Lungs
View Solution


In the circulatory system, deoxygenated blood flows from the right ventricle through the semi-lunar valve (pulmonary valve) into the pulmonary artery, which carries it to the lungs for oxygenation.


%Quicktip Quick Tip: The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs, passing through the semi-lunar valve.


Question 97:


Examples of viscoelastic models are .

  • (1) Wind Kessel and Kelvin
  • (2) Voigt and thick-walled hemisphere
  • (3) Maxwell and Standard solid
  • (4) Wind Kessel and Voigt
Correct Answer: 3 . Maxwell and Standard solid
View Solution

Viscoelastic models, such as the Maxwell model and the Standard solid model, describe
materials that exhibit both viscous and elastic properties, commonly used to model
biological tissues like muscles and blood vessels.
Quick Tip
Maxwell and Standard solid models are classic viscoelastic models, capturing the com-
bined elastic and viscous behavior of materials.


Question 98:


The ratio of change in ventricular pressure to change in ventricular volume is a ______ measure of the ventricle and is called ______.

  • (1) resistance, peripheral resistance
  • (2) stiffness, elastance
  • (3) stiffness, compliance
  • (4) compliance, resistance
Correct Answer: (2) stiffness, elastance
View Solution



The ratio \(\frac{\Delta P}{\Delta V}\) quantifies ventricular stiffness and is specifically termed "elastance" in cardiac physiology, representing the ventricle's pressure-volume relationship during contraction.


%Quicktip Quick Tip: Elastance = \(\frac{\Delta P}{\Delta V}\) (stiffness measure)
Compliance = \(\frac{\Delta V}{\Delta P}\) (reciprocal of elastance, measures distensibility)


Question 99:


The distance between the heel strike of one foot and the next heel strike of the same foot.

  • (1) Step length
  • (2) Stride length
  • (3) Step width
  • (4) Cadence
Correct Answer: (2) Stride length
View Solution



Stride length measures the full gait cycle for one limb (heel strike to next heel strike of the same foot), while step length measures the distance between opposite feet's heel strikes.


%Quicktip Quick Tip: Gait parameters:
- Stride length = Full cycle of one limb
- Step length = Distance between opposite feet
- Cadence = Steps per minute
- Step width = Lateral distance between feet


Question 100:


Allows forward-backward and side to side movement but does not allow rotation.

  • (1) Hinge joint
  • (2) Ball and Socket joint
  • (3) Condyloid joint
  • (4) Gliding joint
Correct Answer: (3) Condyloid joint
View Solution



Condyloid (ellipsoid) joints permit flexion/extension and abduction/adduction (biaxial movement) but prohibit rotation, as seen in the wrist's radiocarpal joint.


%Quicktip Quick Tip: Joint types:
- Condyloid: Biaxial (no rotation)
- Hinge: Uniaxial
- Ball-and-socket: Triaxial
- Gliding: Linear sliding


Question 101:


In an X-ray machine, ______ and ______ are employed for automatic exposure control.

  • (1) Image intensifier, fluoroscopy
  • (2) Gain control, photocell
  • (3) Photocell, image intensifier Photo timer
  • (4) Photocell, ionization chamber Photo timer
Correct Answer: (4) Photocell, ionization chamber Photo timer
View Solution



Modern X-ray systems use either photocells (measures light from intensifier) or ionization chambers (measures radiation directly) combined with photo timers to terminate exposure automatically.


%Quicktip Quick Tip: AEC components:
- Photocell: Indirect measurement via light
- Ionization chamber: Direct radiation measurement
- Photo timer: Terminates exposure at preset dose


Question 102:


A-mode ultrasound device shows ______ as an x-y plot and is used to measure the displacement of the ______.

  • (1) Echo amplitude, mitral valve
  • (2) Echo distance, cerebral midline
  • (3) Echo intensity, brain midline
  • (4) Echo amplitude, interventricular septum
Correct Answer: (3) Echo intensity, brain midline
View Solution



A-mode displays echo intensity (amplitude) vs depth (x-y plot), primarily used for precise measurements like brain midline shifts in neurology.


%Quicktip Quick Tip: Ultrasound modes:
- A-mode: Amplitude vs Depth (1D)
- B-mode: Brightness (2D image)
- M-mode: Motion over time


Question 103:


The amplitude and the frequency of the NMR signals are used to assign ______ and ______.

  • (1) Number of projections, orientation of the gradient
  • (2) Orientation of the gradient, number of slices
  • (3) Number of nuclei present, spatial location
  • (4) Number of projections, spatial location
Correct Answer: (3) Number of nuclei present, spatial location
View Solution



NMR signal amplitude indicates proton density (number of nuclei), while frequency variations (from gradients) encode spatial location for image reconstruction.


%Quicktip Quick Tip: MRI signal properties:
- Amplitude \(\propto\) proton density
- Frequency \(\propto\) spatial position (via gradients)


Question 104:


The ratio of the area of the input phosphor to the area of output phosphor in the image intensifier is

  • (1) Phosphorescence factor
  • (2) Phosphor factor
  • (3) Phosphor number
  • (4) Brightness gain
Correct Answer: (4) Brightness gain
View Solution



Brightness gain = (Input phosphor area)/(Output phosphor area) × minification gain × flux gain. This ratio is fundamental to intensifier performance.


%Quicktip Quick Tip: Image intensifier gains:
- Minification gain = (Input diameter/Output diameter)\(^2\)
- Flux gain = Electron acceleration gain
- Total gain ≈ 5,000-30,000


Question 105:


As the active element diameter of the ultrasonic transducer increases, the beam width ______ and therefore, lateral resolution ______.

  • (1) Increases, improves
  • (2) Does not change, does not change
  • (3) Decreases, improves
  • (4) Decreases, deteriorates
Correct Answer: (1) Increases, improves
View Solution



Larger transducer diameter reduces beam divergence (narrower beam width), improving lateral resolution as the ultrasound beam remains focused over greater depths.


%Quicktip Quick Tip: Transducer physics:
- Larger diameter → Less divergence → Better lateral resolution
- Higher frequency → Better axial resolution but more attenuation


Question 106:


NMR images that display T1 and T2 properties of the tissue provide very high contrasts between various soft tissues approaching ............. whereas it is only a few percent with X-rays.

  • (1) 1000%
  • (2) 600%
  • (3) 150%
  • (4) 300%
Correct Answer: (3) 150%
View Solution

Nuclear Magnetic Resonance (NMR) imaging utilizes differences in T1 (spin-lattice) and T2 (spin-spin) relaxation times to produce images with very high contrast between different soft tissues. This contrast can approach about 150%, which is significantly higher than the contrast provided by X-ray imaging, typically only a few percent. This high contrast is one of the reasons NMR (MRI) is excellent for soft tissue imaging.


% Quicktip Quick Tip: MRI provides high soft tissue contrast (up to ~150%) due to T1 and T2 relaxation differences, unlike X-rays which have low contrast.


Question 107:


The PET scanners constructed using a bismuth germinate detector have ........... and ............... than those with thallium-doped sodium iodide detector.

  • (1) High resolution, high efficiency
  • (2) Low resolution, high efficiency
  • (3) High resolution, low efficiency
  • (4) Low resolution, low efficiency
Correct Answer: (1) High resolution, high efficiency
View Solution

PET scanners using bismuth germanate (BGO) detectors have higher resolution and efficiency compared to those with thallium-doped sodium iodide detectors. BGO has a higher density and atomic number, resulting in better gamma-ray detection efficiency and improved image resolution.


% Quicktip Quick Tip: Bismuth germanate detectors in PET scanners improve both resolution and efficiency compared to NaI(Tl) detectors.


Question 108:


The factors playing an important role in the biological interaction of tissue and ultrasound waves are .................

  • (1) Frequency, wavelength and intensity
  • (2) Time, frequency and duty cycle
  • (3) Frequency, time and wavelength
  • (4) Frequency, irradiation time, beam intensity and duty cycle
Correct Answer: (4) Frequency, irradiation time, beam intensity and duty cycle
View Solution

The biological effects of ultrasound depend on frequency, irradiation time, beam intensity, and duty cycle. These parameters determine the energy delivered to tissues and the extent of interaction, affecting heating and cavitation effects.


% Quicktip Quick Tip: Key parameters affecting ultrasound tissue interaction include frequency, irradiation time, intensity, and duty cycle.


Question 109:


The filters used to reduce the unwanted X-rays exposure to the patient are made of:

  • (1) Aluminium, copper
  • (2) Steel, lead
  • (3) Copper, aluminium
  • (4) Lead, copper
Correct Answer: (1) Aluminium, copper
View Solution

Aluminium and copper filters are used in X-ray machines to absorb low-energy X-rays that contribute to patient dose without improving image quality. These filters help reduce unnecessary radiation exposure.


% Quicktip Quick Tip: Aluminium and copper are commonly used as filters to reduce patient dose by absorbing low-energy X-rays.


Question 110:


Rad is a measure of ................. energy while Rem is a measure of ................

  • (1) incident, absorbed energy
  • (2) absorbed, incident energy
  • (3) biological damage caused, incident energy
  • (4) absorbed, biological damage caused
Correct Answer: (4) absorbed, biological damage caused
View Solution

Rad (radiation absorbed dose) measures the amount of energy absorbed per unit mass of tissue. Rem (roentgen equivalent man) quantifies the biological damage caused by the absorbed radiation, considering the type of radiation and its effect on tissues.


% Quicktip Quick Tip: Rad measures absorbed dose; Rem accounts for biological effect of the absorbed radiation.


Question 111:


SPECT cameras detect only radio-nuclides that produce a .......... of single photons and these radio-nuclides ........ an on-site cyclotron.

  • (1) single emission, do not require
  • (2) cascaded emission, require
  • (3) cascaded emission, do not require
  • (4) single emission, require
Correct Answer: (3) cascaded emission, do not require
View Solution

SPECT cameras detect gamma rays emitted from radio-nuclides via cascaded emission of single photons. These radio-nuclides typically do not require an on-site cyclotron for production, as they have longer half-lives and can be transported from production sites.


% Quicktip Quick Tip: SPECT uses cascaded photon emissions and radio-nuclides that don't need an on-site cyclotron.


Question 112:


The presence of noise in a medical image will generally ..............

  • (1) produce artifacts
  • (2) produce blurring
  • (3) reduce visibility of low contrast objects
  • (4) produce image distortion
Correct Answer: (3) reduce visibility of low contrast objects
View Solution

Noise in medical imaging reduces the visibility of low contrast structures by masking subtle differences. This makes it harder to detect small or faint features, affecting diagnostic accuracy.


% Quicktip Quick Tip: Noise lowers contrast resolution by hiding low-contrast details in medical images.


Question 113:


Polymeric biomaterials are preferred due to:

  • (1) hardness and stability
  • (2) flexibility and stability
  • (3) wear resistance applications
  • (4) load bearing applications
Correct Answer: (2) flexibility and stability
View Solution

Polymeric biomaterials are preferred in many biomedical applications because they offer excellent flexibility and good chemical and mechanical stability. This allows for better compatibility and durability in biological environments.


% Quicktip Quick Tip: Flexibility and stability make polymers ideal biomaterials for diverse medical uses.


Question 114:


The material used for neural stimulation devices is .............

  • (1) Platinum
  • (2) Stainless steel
  • (3) Gold
  • (4) Silver
Correct Answer: (1) Platinum
View Solution

Platinum is widely used for neural stimulation devices due to its excellent biocompatibility, corrosion resistance, and electrical conductivity. It provides reliable long-term performance within the body.


% Quicktip Quick Tip: Platinum’s biocompatibility and conductivity make it ideal for neural stimulation electrodes.


Question 115:


A material used for a blood bag should have:

  • (1) flexibility and optimal porosity
  • (2) optimal porosity and relative inertness
  • (3) reasonable brittleness and relative inertness
  • (4) chemical stability and flexibility
Correct Answer: (4) chemical stability and flexibility
View Solution

Blood bag materials must be chemically stable to avoid reactions with blood and flexible enough to allow handling and storage. Flexibility also helps in preventing ruptures during use.


% Quicktip Quick Tip: Blood bag materials need chemical stability and flexibility for safety and durability.


Question 116:


Alumina and Zirconia are examples of ............... bioceramics.

  • (1) porous to allow tissue in growth
  • (2) bioactive
  • (3) resorbable
  • (4) bioinert
Correct Answer: (4) bioinert
View Solution

Alumina and Zirconia are classified as bioinert ceramics because they do not interact chemically with the biological environment. They are highly stable and exhibit excellent mechanical properties, making them ideal for load-bearing implants where minimal biological reaction is desired.


% Quicktip Quick Tip: Bioinert ceramics like Alumina and Zirconia resist biological interaction and are used in implants needing chemical stability.


Question 117:


The biomaterials are expected to mimic the functions of ...........

  • (1) Cell Organelles
  • (2) Transmembrane proteins
  • (3) Cytoplasm
  • (4) Extracellular Matrix (ECM)
Correct Answer: (1) Cell Organelles
View Solution

Biomaterials are designed to mimic the functional aspects of cell organelles to interact compatibly within biological systems. This includes functions like molecular recognition, signaling, and catalytic activities at the cellular level.


% Quicktip Quick Tip: Biomaterials aim to replicate cell organelle functions for better biological integration.


Question 118:


The most important aspects of biomaterial-tissue interactions are:

  • (1) Biocompatibility
  • (2) Bioavailability
  • (3) Bioequivalence
  • (4) Bioluminescence
Correct Answer: (1) Biocompatibility
View Solution

Biocompatibility is crucial for biomaterials as it determines their ability to perform desired functions without causing adverse biological responses. It ensures that the material is safe and effective when implanted.


% Quicktip Quick Tip: Biocompatibility is key for successful biomaterial integration in tissues.


Question 119:


Biodegradation will be more for materials with ............

  • (1) More molecular weights and high crystallinity
  • (2) Low molecular weights and high crystallinity
  • (3) More molecular weights and less crystallinity
  • (4) Low molecular weights and less crystallinity
Correct Answer: (4) Low molecular weights and less crystallinity
View Solution

Materials with low molecular weight and less crystallinity degrade faster because amorphous regions are more accessible to enzymes and hydrolytic attack, and shorter polymer chains are easier to break down.


% Quicktip Quick Tip: Low molecular weight and low crystallinity increase biodegradability of polymers.


Question 120:


Image formation in electron microscope is based on ...........

  • (1) Column length
  • (2) Electron number
  • (3) Differential scattering
  • (4) Specimen size
Correct Answer: (3) Differential scattering
View Solution

Electron microscopy image formation is primarily based on the differential scattering of electrons by different components of the specimen, producing contrast that reveals structural details at high resolution.


% Quicktip Quick Tip: Electron microscopes use differential scattering of electrons to form detailed images.



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