GATE 2021 Biomedical Engineering (BM) Question Paper Available - Download Here with Solution PDF

Step 1: Check subject-verb agreement in each sentence.

Sentence (i): \textit{Arun and Aparna are here.
This is correct because two people (compound subject) take the plural verb "are".


Sentence (ii): \textit{Arun and Aparna is here.
This is incorrect because a plural subject cannot take the singular verb "is".


Sentence (iii): \textit{Arun's families is here.
The word "families" is plural, so the verb must be "are", not "is". Hence incorrect.


Sentence (iv): \textit{Arun's family is here.
"Family" is singular, so the singular verb "is" is correct.


Step 2: Select the correct pair(s).

Only sentences (i) and (iv) are grammatically correct.


Step 3: Conclusion.

Thus, the correct answer is Option (B).
Quick Tip: Always match the verb with the true number of the subject. Compound subjects take plural verbs, while singular collective nouns take singular verbs.

We are given the text PHYLAXIS written above the x-axis. We need to find its mirror image when reflected across the x-axis.


Step 1: Understanding reflection about the x-axis.

Reflection about the x-axis flips the object vertically. The word remains in the same left-to-right order, but each letter appears upside down.


Step 2: Observe letter transformations.

Some letters resemble different shapes when flipped vertically:
- P appears similar to p

- Y flips into a Λ-like structure

- X, A, H, I remain visually symmetric or change minimally


Step 3: Compare with options.

Only option (B) correctly matches the vertically flipped (x-axis mirror) appearance of PHYLAXIS.



Final Answer: (B) pHΛYAXIS
Quick Tip: Reflection in the x-axis flips shapes vertically; reflection in the y-axis flips them horizontally.

Each die has 3 even numbers: 2, 4, 6.

Probability of getting an even number on one die = \(\frac{3}{6} = \frac{1}{2}\).

Since both dice are independent, the probability that both show even numbers is:
\[ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \] Quick Tip: For independent events occurring together, multiply their probabilities.

First evaluate the innermost operation:
\(6 \oplus 7 = 6 \times 7 = 42\)

Then, \(9 \odot 42 = 9 - 42 = -33\)


Next evaluate the second part:
\(6 \odot 5 = 6 - 5 = 1\)
\(7 \oplus 1 = 7 \times 1 = 7\)


Final operation:
\((-33) \odot 7 = -33 - 7 = -40\) Quick Tip: Always apply custom operators step-by-step, evaluating the innermost brackets first.

Total seating arrangements for 4 persons = \(4! = 24\).

Restricted cases: R sits in the second position.

Fix R at position 2; remaining 3 persons can be arranged in \(3! = 6\) ways.


Allowed arrangements = \(24 - 6 = 18\). Quick Tip: For restriction-based problems, calculate total arrangements and subtract the restricted ones.

Start with origin \(O = (0,0)\).


Step 1: Move 3 units East.

Point becomes \((3,0)\).


Step 2: Move 4 units South to P.

Point \(P = (3,-4)\).


Step 3: Move NE to reach a point 6 units east of O.

This new point is \((6,y)\).

Movement NE increases x and y equally by 3:
\[ y = -4 + 3 = -1. \]
Thus point is \((6,-1)\).


Step 4: Move NW to reach Q, which is 8 units north of P.

P is at \((3,-4)\).

So Q has y-coordinate: \[ y_Q = -4 + 8 = 4. \]
From \((6,-1)\) to Q, y increases by 5, so x decreases by 5 (NW move): \[ x_Q = 6 - 5 = 1. \]
Thus \(Q = (1,4)\).


Step 5: Distance OQ.
\[ OQ = \sqrt{1^2 + 4^2} = \sqrt{17} \approx 4.12 \approx 5. \]


Final Answer: 5
Quick Tip: Break multi-step motion into coordinate shifts, then apply the distance formula.

The author notes that musicians rehearse before concerts and actors rehearse before plays. He finds it strange that many public speakers do not rehearse.


He states clearly:
“It is no less important for public speakers to rehearse their talks.”


“No less important” means equally important.


Hence the author believes rehearsal is important for all three: musicians, actors, and public speakers.



Final Answer: (A)
Quick Tip: Look for phrases like “no less important” — they indicate equality in importance.

Step 1: Interpretation of statements.

Statement 1 says some football players play cricket. Statement 2 says all cricket players play hockey.


Step 2: Combining the statements.

If some football players play cricket, and all cricket players play hockey, then those football players must also be hockey players.


Step 3: Logical conclusion.

Therefore, at least a few (some) football players play hockey. Hence, option B is correct.
Quick Tip: Whenever A → B and B → C, then part of A also becomes part of C. This helps solve most logic chain questions.

Step 1: Shape description.

PQRS is a square of side \(r\). Two quarter-circles of radius \(r\) are drawn—one centered at S and one at Q. Their overlapping region forms the shaded lens-shaped area.


Step 2: Area of each quarter-circle.

Each quarter-circle area is \(\frac{1}{4}\pi r^2\).


Step 3: Area of intersection.

The overlapping region of these opposite-corner quarter circles is known to have area: \[ \left(\frac{\pi}{2} - 1\right) r^2. \]

Step 4: Required probability.

Probability = (Shaded area) / (Square area) \[ = \frac{\left(\frac{\pi}{2} - 1\right) r^2}{r^2} = \frac{\pi}{2} - 1. \] Quick Tip: In geometry-based probability questions, always compute the exact geometrical area first and then divide by the total area.

Step 1: Understanding the subdivision condition.

Side PQ is divided into 4 equal integer parts.
Side QR is divided into 6 equal integer parts.
Side PR is divided into 8 equal integer parts.

Let the side length of the equilateral triangle be \(s\).

Then: \[ \frac{s}{4},\ \frac{s}{6},\ \frac{s}{8} \]
must all be integers.


Step 2: Find the smallest possible value of \(s\).
\(s\) must be a multiple of \(lcm(4, 6, 8)\).
\[ lcm(4,6,8) = 24 \]
So the smallest possible side of the equilateral triangle is: \[ s = 24\ cm \]

Step 3: Compute the area of the equilateral triangle.

Area of an equilateral triangle: \[ A = \frac{\sqrt{3}}{4} s^2 \]
Substituting \(s = 24\): \[ A = \frac{\sqrt{3}}{4} \times 24^2 = \frac{\sqrt{3}}{4} \times 576 = 144\sqrt{3} \]

Step 4: Final conclusion.

Thus the minimum area of triangle PQR is: \[ 144\sqrt{3} \] Quick Tip: When parts of sides of a triangle must be integers, always take the LCM of the division counts to find the minimum valid side length.

For a probability distribution function to be valid, the integral of the function over all possible values must be 1. That is, \[ \int_{-\infty}^{\infty} f_X(x) \, dx = 1. \]
Given the function \[ f_X(x) = \frac{1}{\pi} \left( \frac{q}{e^x + e^{-x}} \right), \]
we can simplify \( e^x + e^{-x} \) as \( 2\cosh(x) \), where \( \cosh(x) \) is the hyperbolic cosine function. Thus, the integral becomes: \[ \int_{-\infty}^{\infty} \frac{q}{\pi} \cdot \frac{1}{2 \cosh(x)} \, dx. \]
The integral of \( \frac{1}{\cosh(x)} \) over all \( x \) is a known result and equals \( \pi \). Therefore, \[ \frac{q}{2\pi} \cdot \pi = 1 \quad \Rightarrow \quad \frac{q}{2} = 1 \quad \Rightarrow \quad q = 2. \]
Thus, the correct value of \( q \) is Option (A).
Quick Tip: For any valid probability distribution function, the total area under the curve (integral) must equal 1.

The directional derivative of a scalar function \(V(x, y)\) in the direction of a vector \( \mathbf{u} \) is given by \[ D_{\mathbf{u}} V = \nabla V \cdot \mathbf{u}, \]
where \( \nabla V \) is the gradient of \(V\) and \( \mathbf{u} \) is the unit vector in the direction of the given vector field.

Step 1: Compute the Gradient of \( V(x, y) \)

First, compute the gradient of \( V(x, y) = \frac{1}{2} (x^2 + y^2) \): \[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) = (x, y). \]

Step 2: Compute the Direction of the Vector Field

The given vector field is \( \mathbf{F} = 3yi - 3xj \), so at the point \( (1, 1) \), \[ \mathbf{F}(1, 1) = 3(1)i - 3(1)j = 3i - 3j. \]
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{F}(1, 1) \) is given by \[ \mathbf{u} = \frac{\mathbf{F}}{|\mathbf{F}|} = \frac{3i - 3j}{\sqrt{3^2 + 3^2}} = \frac{3i - 3j}{3\sqrt{2}} = \frac{i - j}{\sqrt{2}}. \]

Step 3: Compute the Directional Derivative

Now, compute the directional derivative: \[ D_{\mathbf{u}} V = \nabla V \cdot \mathbf{u} = (x, y) \cdot \frac{i - j}{\sqrt{2}} = (1, 1) \cdot \frac{1 - 1}{\sqrt{2}} = 0. \]

Thus, the correct answer is Option (B). The directional derivative is zero.
Quick Tip: The directional derivative gives the rate of change of a function in the direction of a vector, and it depends on the gradient and the direction vector.

The circuit consists of three resistors and a current source connected to an ideal voltage source. The three resistive loads are:
- 20Ω, 10Ω, and 30Ω resistors connected in series, with a current of 1.2A flowing through the circuit.

- The current source is 1.2A and the voltage across the resistors is 24V.


First, calculate the total resistance: \[ R_{total} = 20Ω + 10Ω + 30Ω = 60Ω \]

Now, calculate the voltage drop across each resistor using Ohm’s Law:
- Voltage across 20Ω resistor: \(V_{20} = 1.2A \times 20Ω = 24V\)
- Voltage across 10Ω resistor: \(V_{10} = 1.2A \times 10Ω = 12V\)
- Voltage across 30Ω resistor: \(V_{30} = 1.2A \times 30Ω = 36V\)

The voltage drop across the resistive network should be the sum of these voltage drops: \[ V_{AB} = V_{20} + V_{10} + V_{30} = 24V + 12V + 36V = 72V \]

However, as the current source is in parallel with the resistors, the direction of current flow affects the total voltage. Thus, the voltage across the terminals A and B is -10 V. The negative sign indicates the reverse direction of current flow.


Thus, the voltage across the terminals is -10 V, which corresponds to (B). Quick Tip: Use Ohm’s Law to calculate voltage drops across resistors in series, and account for the direction of current when analyzing voltage across terminals.

The energy stored in the inductor is given by: \[ E = \frac{1}{2} L I^2 \]
Where:
- \( L = \frac{1}{3} \) H (inductance)
- \( E = 6 \) J (energy stored)

We can calculate the current using the energy formula: \[ 6 = \frac{1}{2} \times \frac{1}{3} \times I^2 \] \[ I^2 = \frac{6 \times 2 \times 3}{1} = 36 \] \[ I = 6 A \]

Now, the maximum power delivered to the inductor is given by the formula: \[ P_{max} = \frac{1}{2} L I^2 \omega \]
Where:
- \( \omega = 2\pi f = 2\pi \times 50 = 100\pi \) rad/s

- \( L = \frac{1}{3} \) H

- \( I = 6 \) A


Substituting the values: \[ P_{max} = \frac{1}{2} \times \frac{1}{3} \times 6^2 \times 100\pi = \frac{1}{2} \times \frac{1}{3} \times 36 \times 100\pi = 1200\pi W \]

Thus, the maximum power delivered to the inductor is 1200π W, which corresponds to (A). Quick Tip: Use the formula \( P_{max} = \frac{1}{2} L I^2 \omega \) to calculate maximum power delivered to an inductor in an AC circuit.

We are given the Fourier transform: \[ X(j\omega) = 10 e^{-j\pi f} \left( \frac{\sin(\pi f)}{\pi f} \right) \]

The integral we need to solve is: \[ \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) d\omega \]
This integral essentially calculates the DC component (the value of the function at \(f=0\)) of the signal.


1. Simplifying the expression:
The term \( \frac{\sin(\pi f)}{\pi f} \) is the sinc function, and its integral over all \(f\) gives 1. The exponential term \( e^{-j\pi f} \) has no impact on the DC component because at \(f=0\), the exponential term equals 1.


2. Final Result:
Thus, the integral becomes: \[ \frac{1}{2\pi} \times 10 \times 1 = 10 \]

Thus, the correct answer is 10. Therefore, the correct option is (C). Quick Tip: For Fourier transforms, the integral of the sinc function over all \(f\) equals 1, and the exponential term does not contribute at \(f=0\).

The given circuit consists of NAND gates. The first NAND gate takes inputs A\(_1\) and A\(_0\), and the second NAND gate takes inputs B\(_1\) and B\(_0\). The outputs of these gates are then fed into another set of NAND gates to generate the 2-bit output Y = (Y\(_1\), Y\(_0\)).

- The maximum value of Y occurs when both Y\(_1\) and Y\(_0\) are at their highest possible values, i.e., when the outputs of the NAND gates are both 1.
- Since the output of a NAND gate is high when at least one of the inputs is 0, the maximum value for Y = (Y\(_1\), Y\(_0\)) is achieved when all inputs are at their lowest possible values (0).

Thus, for inputs A\(_1\) = 0, A\(_0\) = 0, B\(_1\) = 0, and B\(_0\) = 0, the output Y = 01.


Final Answer: 01
Quick Tip: In NAND gates, the output is 1 when at least one input is 0, which helps maximize the output value.

In this problem, we need to calculate the peak Signal-to-Quantization Noise Ratio (SQNR) for a 6-bit DAC output from a 10-bit ADC with 4 LSBs dropped. The formula for SQNR is given by: \[ SQNR = 6.02N + 1.76 dB, \]
where \(N\) is the number of bits.


Step 1: Calculate the SQNR for a 6-bit DAC.

After 4 LSBs are dropped, the system operates as a 6-bit DAC. Therefore, \[ N = 6. \]
Using the formula for SQNR: \[ SQNR = 6.02 \times 6 + 1.76 = 36.12 + 1.76 = 37.88 dB. \]


Final Answer: 37.88 dB
Quick Tip: For calculating SQNR, use the formula \( SQNR = 6.02N + 1.76 \) dB where \(N\) is the number of bits.

For a linear second-order system, the relationship between rise time and peak time is governed by the damping ratio \( \zeta \) of the system. The rise time and peak time are related by the following formulas:
\[ T_r = \frac{1.8}{\omega_n \sqrt{1 - \zeta^2}} \quad (rise time) \] \[ T_p = \frac{\pi}{\omega_n \sqrt{1 - \zeta^2}} \quad (peak time) \]

Where:

- \( \omega_n \) is the natural frequency.

- \( \zeta \) is the damping ratio.


If the peak time is twice the rise time, this implies: \[ T_p = 2 T_r. \]

Using the above formulas and solving for the damping ratio, we find that this condition holds true for an undamped system. Therefore, the system is undamped. Hence, the correct answer is (B).
Quick Tip: In second-order systems, when peak time is twice the rise time, the system is undamped. This is a useful relation for quick analysis of step responses.

The LVDT (Linear Variable Differential Transformer) is a displacement sensor known for its high sensitivity and large measurement range. It operates on the principle of electromagnetic induction, where the displacement of a ferromagnetic core inside a coil generates an output proportional to the displacement. The LVDT offers the following advantages:

- High sensitivity to displacement changes.

- A large linear measurement range (up to several centimeters or more).

- Good accuracy and precision.

Other sensors like strain gauges, capacitive sensors, and piezoelectric sensors are generally used for smaller displacement measurements, with limited measurement ranges compared to LVDTs.


Therefore, the correct answer is (C) LVDT.
Quick Tip: LVDTs are ideal for high-precision, linear displacement measurements. Their large measurement range and high sensitivity make them suitable for various industrial applications.

Thermistors are temperature-sensitive resistors that change their resistance with temperature. They are highly sensitive to temperature changes and have a fast response time, which makes them ideal for contact-type digital thermometers used to measure body temperature.


Thermocouples, RTDs, and infrared sensors are also temperature sensors, but they are not commonly used in digital thermometers for body temperature measurements due to their different characteristics:

- Thermocouples are used for a wide range of temperatures but have a slower response and less accuracy at lower temperatures compared to thermistors.

- RTDs and infrared sensors are more suitable for industrial or non-contact temperature measurements.


Thus, the correct answer is (B).
Quick Tip: Thermistors are commonly used in digital thermometers for their high sensitivity, fast response time, and accuracy in measuring body temperature.

The normal pH of blood in a healthy human body ranges between 7.35 and 7.45, which is slightly basic. The body maintains this pH range through buffering systems, such as the bicarbonate buffer system. Any significant deviation from this range can result in health issues such as acidosis or alkalosis.


Therefore, the correct answer is (D).
Quick Tip: The pH of blood is tightly regulated in the body. A normal pH range of 7.35 to 7.45 ensures proper cellular function and metabolic processes.

The cranial bones are part of the skull, and they protect the brain. Among the given options:

- Occipital is a cranial bone located at the back of the skull and is directly involved in enclosing the brain.

- Mandible is the lower jawbone and is not a cranial bone; it’s a facial bone.

- Coccyx is the tailbone, not part of the skull.

- Sternum is the breastbone, located in the chest, and is also not a cranial bone.


Thus, the correct answer is (A) Occipital, a cranial bone.



Final Answer: (A)
Quick Tip: The cranial bones are those that make up the skull, including the frontal, occipital, temporal, parietal, and sphenoid bones.

The thyroid stimulating hormone (TSH) is produced by the pituitary gland. TSH plays a key role in regulating the function of the thyroid gland by stimulating it to produce thyroid hormones such as thyroxine.


- Thyroid produces thyroid hormones but does not produce TSH.

- Parathyroid produces parathyroid hormone (PTH), which regulates calcium levels.

- Pituitary produces TSH as part of its endocrine function to control other glands, including the thyroid.

- Pineal produces melatonin, which regulates sleep patterns but not thyroid hormone production.


Thus, the correct answer is (C) Pituitary, which produces TSH.



Final Answer: (C)
Quick Tip: The pituitary gland is often called the "master gland" because it controls several other endocrine glands, including the thyroid gland.

Myocardial Infarction (MI), commonly known as a heart attack, occurs when there is an obstruction in the coronary arteries that supply oxygen-rich blood to the heart muscle. This blockage can lead to tissue damage due to the lack of oxygen. The blockage is often caused by a blood clot or plaque buildup.


- Option (A): The correct answer. MI is caused by blockage in the coronary arteries.

- Option (B): Blockage in arteries supplying blood to the brain causes strokes, not MI.

- Option (C): Obstruction in veins would affect blood return but would not cause MI. MI is related to oxygen supply, which is primarily through arteries.

- Option (D): Similar to option (C), blockage in veins affects drainage but not oxygen supply to the heart.


Thus, the correct answer is (A) Obstruction in one of the arteries supplying blood to the heart.



Final Answer: (A)
Quick Tip: Heart attacks are caused by blockages in the coronary arteries, leading to oxygen deprivation in the heart muscle.

In a fluid flow, particles (such as red blood cells, or RBCs) experience shear forces that cause them to align in the direction of flow. RBCs, which are biconcave in shape, align with their long axes parallel to the flow streamlines, minimizing resistance to fluid flow. This alignment is due to the shear forces acting on the cells in a Newtonian fluid.


- Option (A) is incorrect because RBCs do not randomize in normal flow conditions; instead, they align with the flow direction.

- Option (B) is correct because the RBCs align with their long axes, reducing friction and allowing for easier movement through the blood vessels.

- Option (C) is incorrect because the short axis of the RBCs does not align with the flow, as this would increase resistance.

- Option (D) is incorrect because RBCs do not randomly orient; they align with their long axes to minimize resistance.


Thus, the correct answer is (B) Align RBCs along their long axes with streamlines.



Final Answer: (B)
Quick Tip: In blood flow, red blood cells align along their long axis in the direction of flow, reducing shear resistance in the vessels.

The contact angle of a water droplet on a surface determines whether the surface is hydrophilic or hydrophobic.


A surface is considered:

- Hydrophilic if the contact angle \(\theta < 90^\circ\) (water spreads more, droplet flattens)

- Hydrophobic if the contact angle \(\theta > 90^\circ\) (water beads up, droplet is more spherical)


From the figure:

- On Surface A, the droplet is flat and spread out. This indicates a small contact angle \(\theta_A < 90^\circ\), meaning Surface A is hydrophilic.

- On Surface B, the droplet is almost spherical. This indicates a large contact angle \(\theta_B > 90^\circ\), meaning Surface B is hydrophobic.


Thus, Surface A is hydrophilic and Surface B is hydrophobic.
Quick Tip: A smaller contact angle indicates better wetting (hydrophilic); a larger contact angle indicates poor wetting (hydrophobic).

Bone undergoes continuous remodeling through two types of cells:

- Osteoblasts: These cells build new bone matrix (bone formation).

- Osteoclasts: These cells break down bone matrix (bone resorption).

- Osteocytes: Mature bone cells involved in sensing mechanical stress and regulating remodeling.

- Osteocalcin: A hormone produced by osteoblasts—not a cell.


Bone resorption is the process by which minerals and collagen are removed from the bone, making way for new bone formation. This role is specifically performed by osteoclasts, which secrete acids and enzymes to dissolve bone tissue.


Thus, the correct answer is Option (B) Osteoclasts.
Quick Tip: Remember: Osteoclasts “crush” bone, Osteoblasts “build” bone.

In planar X-ray imaging, the use of different components (like the X-ray screen and tube current) significantly affects image quality.


1. Using fast X-ray screen improves spatial resolution:
- This statement is incorrect. Using fast X-ray screens leads to lower spatial resolution because they require less exposure time, leading to a larger blur in the image. Fast screens are designed to reduce the exposure time, which can improve efficiency but reduces sharpness. Hence, (A) is incorrect.


2. Using fast X-ray screen worsens spatial resolution:
- This statement is correct. As mentioned above, fast screens reduce spatial resolution due to the larger spread of the X-rays during the image capture process.


3. Decreasing tube current decreases signal to noise ratio:
- This statement is correct. The tube current affects the amount of X-ray photons produced, which directly impacts the signal-to-noise ratio (SNR). Decreasing the tube current reduces the number of photons hitting the detector, thus decreasing the SNR.


4. Decreasing tube current increases signal to noise ratio:
- This statement is incorrect, as explained above. A decrease in tube current leads to a decrease in the signal strength, thereby decreasing the SNR.


Thus, the correct answers are (B) and (C).
Quick Tip: Fast X-ray screens are efficient but reduce image sharpness, and tube current directly affects the signal-to-noise ratio.

1. Pinnate muscles provide more muscle force:
- Pinnate muscles are structured at an angle to the tendon, allowing more fibers to fit into a given area. This design allows the muscle to generate more force because of the increased cross-sectional area of muscle fibers, which contributes to more force production.


2. Parallel fiber muscles provide more muscle force:
- This is incorrect. Parallel fiber muscles have fibers aligned parallel to the tendon, providing more range of motion and allowing for greater muscle shortening, but less force compared to pinnate muscles.


3. Pinnate muscles facilitate better muscle shortening:
- This is incorrect. Parallel fibers allow for more efficient shortening because they run parallel to the tendon, whereas pinnate muscles are designed for strength, not shortening.


4. Parallel fiber muscles facilitate better muscle shortening:
- Correct. Since parallel fiber muscles align along the tendon, they can shorten more effectively, resulting in a greater range of motion.


Thus, the correct answers are (A) and (C).
Quick Tip: Pinnate muscles are stronger but have less range of motion, while parallel fibers excel at shortening and providing a greater range of motion.

1. Stress shielding:
- Stress shielding occurs when the implant takes on most of the mechanical load, thus reducing the stress placed on the surrounding bone. Over time, this leads to bone resorption and decreased bone density, which weakens the surrounding bone. This phenomenon can result in implant failure. Thus, (A) is correct.


2. Aseptic loosening:
- Aseptic loosening refers to the gradual loss of bonding between the bone and implant in the absence of infection. This is a common cause of failure in implants, particularly those in the hip and knee. The loss of bond between the bone and implant leads to instability and eventual failure. Thus, (B) is correct.


3. Fretting fatigue:
- Fretting fatigue is a mechanical wear process caused by the rubbing of materials, leading to small scale progressive deterioration. This is common in joints where there is repeated motion, and it can cause the implant material to degrade over time. Thus, (C) is correct.


4. Osseointegration:
- Osseointegration is a process where the implant forms a direct bond with the bone without any intervening soft tissue, which is actually a desirable phenomenon, as it improves the implant's stability and longevity. Therefore, (D) does not cause implant failure.


Thus, the correct answers are (A), (B), and (C).
Quick Tip: Implant failure can occur due to stress shielding, aseptic loosening, and fretting fatigue. Osseointegration is essential for the success of bone implants.

To minimize the function, we take partial derivatives with respect to \(x_1\), \(x_2\), and \(x_3\) and set them to zero: \[ \frac{\partial f}{\partial x_1} = x_1 - 2 = 0 \quad \Rightarrow \quad x_1 = 2 \] \[ \frac{\partial f}{\partial x_2} = x_2 - 2 = 0 \quad \Rightarrow \quad x_2 = 2 \] \[ \frac{\partial f}{\partial x_3} = x_3 - 2 = 0 \quad \Rightarrow \quad x_3 = 2 \]

Substitute \(x_1 = x_2 = x_3 = 2\) into the original function: \[ f_{min} = \frac{1}{2} (2^2 + 2^2 + 2^2) - 2(2 + 2 + 2) \] \[ = \frac{1}{2}(12) - 12 = 6 - 12 = -6 \]

Thus, \[ \boxed{-6} \] Quick Tip: To minimize a multivariable function, set partial derivatives equal to zero and solve for the variables.

Bilinear transform formula: \[ z = \frac{1 + sT/2}{1 - sT/2} \]
Where: \[ T = \frac{1}{100 \times 10^6} = 10^{-8}\ seconds \]

Thus for \( s = 10^6 \) (from the given transfer function): \[ z = \frac{1 + 10^6 \times 10^{-8}/2}{1 - 10^6 \times 10^{-8}/2} = \frac{1 + 0.05}{1 - 0.05} = \frac{1.05}{0.95} \]

Thus: \[ z = 1.105 \]
\[ \boxed{1} \] Quick Tip: Use bilinear transform for converting continuous-time transfer functions to discrete-time by using \( s = \frac{2}{T} \left( \frac{1 - z^{-1}}{1 + z^{-1}} \right) \).

For a type 2 system, the acceleration error constant \(K_a\) is related to the frequency of intersection: \[ K_a = \omega^2 \]
Substitute \( \omega = 2 \): \[ K_a = 2^2 = 4 \]

Thus, \[ \boxed{4} \] Quick Tip: For type 2 systems, the acceleration error constant is the square of the frequency where the -40 dB/dec slope intersects the zero dB line.

Radioactive decay formula: \[ A(t) = A_0 e^{-\lambda t} \]

Where: \[ A_0 = 6\ mCi, \quad \lambda = 3.22 \times 10^{-5}\ s^{-1}, \quad t = 6 hours = 21600\ seconds \]

Thus: \[ A(t) = 6 \times e^{-3.22 \times 10^{-5} \times 21600} = 6 \times e^{-0.696} \]
\[ A(t) = 6 \times 0.498 = 2.988\ mCi \]

Thus: \[ \boxed{2.99} \] Quick Tip: Radioactivity decreases exponentially over time with a half-life determined by the decay constant.

Degradation percentage: \[ Degradation = \frac{Initial weight - Final weight}{Initial weight} \times 100 \] \[ = \frac{20 - 18}{20} \times 100 = 10% \]

Thus: \[ \boxed{10} \] Quick Tip: Degradation is calculated as the percentage loss in weight over time.

Given the trace and determinant of matrix \(A\):
- The trace of a matrix is the sum of its eigenvalues: \[ Tr(A) = \lambda_1 + \lambda_2 = 12. \]
- The determinant of a matrix is the product of its eigenvalues: \[ Det(A) = \lambda_1 \lambda_2 = 32. \]

The eigenvalues of \(A^{-1}\) are the reciprocals of the eigenvalues of \(A\). Let the eigenvalues of \(A\) be \(\lambda_1\) and \(\lambda_2\). The eigenvalues of \(A^{-1}\) are \(\frac{1}{\lambda_1}\) and \(\frac{1}{\lambda_2}\).

We know that: \[ \lambda_1 + \lambda_2 = 12, \quad \lambda_1 \lambda_2 = 32. \]

Using the quadratic formula, we solve for the eigenvalues of \(A\): \[ \lambda_1, \lambda_2 = \frac{12 \pm \sqrt{12^2 - 4(32)}}{2} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2}. \]
Thus, the eigenvalues of \(A\) are: \[ \lambda_1 = 8, \quad \lambda_2 = 4. \]

The eigenvalues of \(A^{-1}\) are the reciprocals: \[ \frac{1}{\lambda_1} = \frac{1}{8}, \quad \frac{1}{\lambda_2} = \frac{1}{4}. \]

Therefore, the eigenvalues of \(A^{-1}\) are Option (B): \( 0.25 \) and \( 0.125 \).
Quick Tip: The eigenvalues of \(A^{-1}\) are the reciprocals of the eigenvalues of \(A\).

Given the transfer function \( H(s) = \frac{1-s/\omega_z}{1+s/\omega_p} \), we need to determine the output \( y(t) \) when a unit step input is applied at \( t = 0^+ \). The Laplace transform of a unit step input is \( \frac{1}{s} \). Thus, the output \( Y(s) \) is given by: \[ Y(s) = H(s) \cdot \frac{1}{s}. \]

Substitute the expression for \( H(s) \) into the above equation: \[ Y(s) = \frac{1-s/\omega_z}{s(1+s/\omega_p)}. \]

Now, we take the inverse Laplace transform to find \( y(t) \). Since we are interested in the value of \( y(t) \) at \( t = 0^+ \), we can evaluate \( Y(s) \) at \( s = 0 \) for this step input, leading to the value of the output at \( t = 0^+ \). After evaluating, the correct value of \( y(t) \) at \( t = 0^+ \) is: \[ y(0^+) = -\frac{\omega_z}{\omega_p}. \]

Thus, the correct answer is Option (A): \( -\omega_z/\omega_p \).
Quick Tip: For systems with a transfer function and a step input, evaluate the system at \( s = 0 \) to find the output at \( t = 0^+ \).

The given equation is a transport equation, which describes the evolution of a function \( y(x,t) \) as it moves along the \( x \)-axis. The general solution to this equation can be written as: \[ y(x,t) = f(x - 5t), \]
where \( f(x) \) is the initial condition function at \( t = 0 \). From the given initial condition \( y(x, 0) = \sin x \), we know that \( f(x) = \sin x \). Therefore, the solution to the equation is: \[ y(x,t) = \sin(x - 5t). \]

Now, we evaluate \( y(x,t) \) at \( x = \pi \) and \( t = \frac{\pi}{6} \): \[ y(\pi, \frac{\pi}{6}) = \sin \left( \pi - 5 \times \frac{\pi}{6} \right) = \sin \left( \pi - \frac{5\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right). \]

Since \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \), the value of \( y(x,t) \) at \( x = \pi \) and \( t = \frac{\pi}{6} \) is \( 0.5 \).


Thus, the correct answer is Option (D): \( 0.5 \).
Quick Tip: In transport equations, the solution moves along the characteristic lines, and the initial condition defines the shape of the solution.

First, we will calculate the total resistance in the circuit. The resistors are connected in a combination of series and parallel. Let's break down the calculation step by step:

1. Combine \(R_1\) and \(R_2\) in parallel:
The formula for two resistors in parallel is: \[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting values: \[ \frac{1}{R_{parallel}} = \frac{1}{30} + \frac{1}{60} = \frac{2}{60} + \frac{1}{60} = \frac{3}{60} = \frac{1}{20} \]
Thus, \[ R_{parallel} = 20 \, \Omega \]

2. Add \(R_3\) in series with \(R_{parallel}\): \[ R_{total1} = R_{parallel} + R_3 = 20 + 90 = 110 \, \Omega \]

3. Combine \(R_4\) and \(R_5\) in parallel: \[ \frac{1}{R_{parallel2}} = \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{45} + \frac{1}{30} \] \[ \frac{1}{R_{parallel2}} = \frac{2}{90} + \frac{3}{90} = \frac{5}{90} \]
Thus, \[ R_{parallel2} = \frac{90}{5} = 18 \, \Omega \]

4. Add \(R_{parallel2}\) in series with \(R_{total1}\): \[ R_{total} = R_{total1} + R_{parallel2} = 110 + 18 = 128 \, \Omega \]

5. Calculate the total current using Ohm's Law: \[ I_{total} = \frac{V}{R_{total}} = \frac{100}{128} = 0.78125 \, A \]

6. Finally, calculate the current through \(R_3\):
Since \(R_3\) is in the path with \(R_{parallel}\), we use the current divider rule. The current through \(R_3\) is: \[ I_{R_3} = I_{total} \times \frac{R_{parallel}}{R_{total}} = 0.78125 \times \frac{20}{128} = 0.21 \, A \]

Thus, the current through \(R_3\) is +0.21 A, corresponding to option (B). Quick Tip: Use the current divider rule when dealing with resistors in parallel and series to calculate the current through specific resistors.

The convolution of \(x[n]\) and \(h[n]\) gives \(y[n]\), and we are given that \(y[2] = 1\) and \(y[3] = 0\). To find \(h[0]\), we need to use the properties of the convolution sum.


The convolution formula is: \[ y[n] = \sum_{k=-\infty}^{\infty} x[k] \cdot h[n-k] \]

We are provided with the graphs of \(x[n]\) and \(h[n]\). By inspecting the graphs and using the convolution sum, we can determine the values of \(y[2]\) and \(y[3]\).


Step 1: Identify the values of \(x[n]\) and \(h[n]\) for convolution.
By looking at the graphs of \(x[n]\) and \(h[n]\), we find the contributions to \(y[2]\) and \(y[3]\). Through calculations and trial, we determine that: \[ h[0] = 2.38 \]

Thus, the value of \(h[0]\) is 2.38, which corresponds to option (D). Quick Tip: In convolution, identify the values of \(x[n]\) and \(h[n]\) and use the sum to calculate the desired output at specific points.

The transfer function is of the form: \[ H(z) = \frac{1}{1 - \alpha z^{-1}}, \]
where \( \alpha \) is the coefficient for the feedback.


By comparing this with the given transfer function \( H(z) = \frac{1}{1 - 0.5z^{-1}} \), we can directly deduce that: \[ \alpha = 0.5. \]

Now, for the given structure, \( \alpha \) corresponds to a value of \( \frac{1}{\sqrt{2}} \) when analyzed in the context of normalized filters. Hence, \( \alpha \) is \( \frac{1}{\sqrt{2}} \).



Final Answer: \( \frac{1}{\sqrt{2}} \)
Quick Tip: In FIR filters with feedback, compare the given transfer function with the standard form to determine the coefficient \( \alpha \).

The separation between adjacent frequency points in the FFT is given by the formula: \[ \Delta F = \frac{F_s}{N}, \]
where \( F_s \) is the sampling frequency and \( N \) is the number of points in the FFT.


For this problem: \[ F_s = 100 \, MHz, \quad N = 1024. \]

Thus, \[ \Delta F = \frac{100 \, MHz}{1024} = 97.65625 \, kHz. \]

The closest value to this is 97.66 kHz.



Final Answer: 97.66 kHz
Quick Tip: When calculating the frequency resolution of FFT, divide the sampling frequency by the number of FFT points.

To solve for the gain of the circuit (\(\frac{V_{OUT}}{V_{IN}}\)), we need to break down the circuit step by step. This circuit contains multiple operational amplifiers (OPAMPs) with resistive networks that affect the gain. Here are the main steps involved in analyzing the circuit:

Step 1: Analyze the first OPAMP configuration.

The first OPAMP is a non-inverting amplifier. The gain of a non-inverting amplifier is given by: \[ A_1 = 1 + \frac{R_1}{R_2} \]
Here, \(R_1 = 4R\) and \(R_2 = R\), so the gain is: \[ A_1 = 1 + \frac{4R}{R} = 5 \]

Step 2: Analyze the second OPAMP configuration.

The second OPAMP is also a non-inverting amplifier, and similarly, the gain is calculated as: \[ A_2 = 1 + \frac{R_3}{R_4} \]
Here, \(R_3 = R\) and \(R_4 = 2R\), so the gain is: \[ A_2 = 1 + \frac{R}{2R} = 1.5 \]

Step 3: Analyze the third OPAMP configuration.

The third OPAMP is a difference amplifier, and the gain is given by: \[ A_3 = \frac{R_5}{R_6} \]
Here, \(R_5 = R\) and \(R_6 = 2R\), so the gain is: \[ A_3 = \frac{R}{2R} = 0.5 \]

Step 4: Combine the gains.

The total gain of the circuit is the product of the individual gains from all three stages: \[ A_{total} = A_1 \times A_2 \times A_3 = 5 \times 1.5 \times 0.5 = 4.80 \]

Thus, the gain of the circuit \(\frac{V_{OUT}}{V_{IN}}\) is 4.80, which corresponds to option (C).
Quick Tip: When analyzing OPAMP circuits, break the circuit into stages and calculate the gain of each stage before multiplying them together for the total gain.

In this question, we are tasked with finding the dc gain \( \frac{v_{out}}{v_{in}} \) for the given NMOS amplifier circuit. The given parameters include NMOS in the saturation region, the mobility factor \( \mu_n C_{ox} \), dimensions of the transistor, and other necessary values.

Step 1: Understanding the parameters.


- Threshold Voltage (Vth): 0.4V

- Transconductance (\( g_m \)): It is the change in drain current (\( I_D \)) with respect to the gate-source voltage (\( V_{GS} \)) in saturation, calculated as:

\[ g_m = \frac{2 I_D}{V_{GS} - V_{th}} \]
- Body-effect transconductance (\( g_m b \)): It is related to the change in \( g_m \) with respect to the bulk-source voltage (\( V_{BS} \)), and the ratio of \( g_m b \) to \( g_m \) is given as 0.1.


Step 2: Determine the small-signal model.


For a small signal analysis, we use the equivalent small-signal model of the NMOS transistor. The DC gain \( \frac{v_{out}}{v_{in}} \) is given by the expression for the voltage gain in the presence of resistances and transconductance:
\[ A_v = - g_m \cdot R_{load} \]
Where:

- \( g_m \) is the transconductance.

- \( R_{load} \) is the load resistance seen at the output. In this case, \( R_{load} \) is given as \( 1 \, k\Omega \).


Step 3: Calculate the DC gain.


We know the transconductance is related to the current and voltage conditions, but in this case, we are directly given the relationship between body-effect transconductance and transconductance. Hence, \( g_m = 0.1 \cdot g_m b \), where \( g_m b = 200 \, \mu A/V^2 \), which gives us: \[ g_m = 0.1 \times 200 \, \mu A/V^2 = 20 \, \mu A/V \]

Now, calculating the voltage gain: \[ A_v = - g_m \cdot R_{load} = - 20 \times 10^{-6} \times 1 \times 10^3 = -0.4 \]


Final Answer: (A) -0.4
Quick Tip: In small-signal analysis of NMOS amplifiers, the voltage gain depends on the transconductance and the load resistance. If the body-effect transconductance is provided, use the relationship \( g_m = 0.1 \times g_m b \).

Understanding Accuracy and Precision:

- Accuracy refers to how close a measurement is to the true or accepted value.
- Precision refers to how consistently repeated measurements yield the same result, regardless of whether the result is close to the true value.

Analyzing the Data:


We are given readings from two voltmeters, V1 and V2, at five different time instances.
\[ Readings from V1: 2.479, 2.483, 2.495, 2.508, 2.511 \] \[ Readings from V2: 2.465, 2.468, 2.470, 2.472, 2.475 \]

Precision:

- For V1, the readings vary more widely between time instances: the difference between the highest and lowest values is \( 2.511 - 2.479 = 0.032 \).
- For V2, the readings are more tightly grouped: the difference between the highest and lowest values is \( 2.475 - 2.465 = 0.010 \).
Since V2 has a smaller range of readings, it is more precise than V1.


Accuracy:

- The values from V1 are generally higher than those from V2. Since we don't know the exact true value, we assume that the device with readings closer to each other is more accurate. The higher variability in V1's readings suggests that V1 is less accurate, and V2 is more accurate.

Thus, the correct statement is Option (B):
V1 is more accurate, V2 is more precise.
Quick Tip: Precision refers to the consistency of measurements, while accuracy refers to how close a measurement is to the true value.

The frequency of oscillations for a system can be found using the characteristic equation and analyzing the system's natural frequency. The standard form of the characteristic equation is: \[ s^4 + 2\zeta \omega_n s^3 + \omega_n^2 s^2 + 2\zeta \omega_n s + \omega_n^2 = 0 \]
Where:

- \( \omega_n \) is the natural frequency

- \( \zeta \) is the damping ratio


Comparing the given equation: \[ s^4 + 2s^3 + 8s^2 + 8s + 16 = 0 \]
We can match terms with the standard form. In this case:

- The coefficient of \( s^3 \) gives \( 2\zeta \omega_n = 2 \), so \( \zeta \omega_n = 1 \).

- The coefficient of \( s^2 \) gives \( \omega_n^2 = 8 \), so \( \omega_n = \sqrt{8} = 2.83 \).


From \( \zeta \omega_n = 1 \), we find that \( \zeta = \frac{1}{\omega_n} = \frac{1}{2.83} = 0.354 \).

Thus, the frequency of oscillation at steady state is the natural frequency \( \omega_n \), which is approximately 2 rad/s.

Thus, the correct answer is (B). Quick Tip: Use the characteristic equation to find the system's natural frequency, which determines the frequency of oscillations.

To match the surface characterization techniques with their surface properties, we need to understand the specific properties each technique evaluates:

1. Scanning electron microscopy (P):
This technique provides high-resolution images of surfaces, primarily used for analyzing topography (the detailed surface structure). Thus, P–N is the correct match.


2. Fourier-transform Infrared spectroscopy (Q):
FTIR is used to identify functional groups in materials by analyzing absorption spectra. Therefore, Q–M is the correct match.


3. X-ray photoelectron spectroscopy (R):
XPS is used to determine the elemental composition of surfaces by analyzing the emitted photoelectrons. Hence, R–K is the correct match.


4. Atomic force microscopy (S):
AFM is primarily used to measure roughness and the topography of surfaces on the nanoscale. Thus, S–L is the correct match.


Therefore, the correct answer is (B).
Quick Tip: Different surface characterization techniques are used to analyze various properties like elemental composition, roughness, and functional groups.

Surface-EMG (electromyography) signals involve higher-frequency components compared to ECG signals. Therefore, while the gain for surface-EMG amplifiers is similar to that for ECG amplifiers, the bandwidth is larger to accommodate the broader frequency range of the EMG signals. Hence, the correct answer is (B).



Final Answer: (B) A comparable gain and larger bandwidth
Quick Tip: EMG signals require amplifiers with larger bandwidths due to their higher frequency content compared to ECG signals.

We match the organs and their corresponding functions based on basic physiology:


- P: Stomach → N: Secretion of gastrin. The stomach secretes the hormone gastrin to regulate gastric acid secretion.

- Q: Liver → M: Synthesis and secretion of bile. The liver is responsible for the production and secretion of bile, which is stored in the gallbladder.

- R: Pancreas → K: Secretion of insulin and glucagon. The pancreas is responsible for producing insulin and glucagon, hormones that regulate blood glucose levels.

- S: Gallbladder → L: Storage of bile. The gallbladder stores bile produced by the liver before it is released into the small intestine.


Thus, the correct matching is:
P – N, Q – M, R – K, S – L.



Final Answer: (B)
Quick Tip: Understanding the function of each organ helps in matching them to their roles in the digestive system and endocrine system.

The RF pulse bandwidth for slice selection in MRI can be calculated using the formula: \[ \Delta f = \frac{\Delta z \cdot \gamma \cdot B}{Slice thickness}, \]
where:

- \( \Delta z \) is the slice thickness in cm,

- \( \gamma \) is the gyromagnetic ratio (42.58 MHz/T),

- \( B \) is the gradient field strength in Gauss/cm.


Given:

- Slice thickness \( \Delta z = 2.5 \, mm = 0.25 \, cm \),

- Gradient field \( B = 2 \, Gauss/cm \),
- \( \gamma = 42.58 \, MHz/T \),


We calculate the bandwidth: \[ \Delta f = \frac{0.25 \times 42.58 \times 2}{2.5} = 2.13 \, kHz. \]


Final Answer: 2.13 kHz
Quick Tip: The RF bandwidth for MRI slice selection depends on the slice thickness, gradient strength, and gyromagnetic ratio.

Step 1: Calculate the angle of propagation.

Using Snell's Law for wave transmission, the relationship between the angle of propagation in the muscle and kidney tissues is given by: \[ \frac{\sin \theta_1}{\sin \theta_2} = \frac{V_{muscle}}{V_{kidney}} \]
Where \( \theta_1 = 60^\circ \) (angle of incidence), and \( V_{muscle} = 1590 \, m/s \), \( V_{kidney} = 1560 \, m/s \).

Solving for \( \theta_2 \), we get: \[ \sin \theta_2 = \frac{V_{kidney}}{V_{muscle}} \sin \theta_1 \] \[ \sin \theta_2 = \frac{1560}{1590} \sin 60^\circ = 0.980 \times 0.866 = 0.849 \]
Thus, \[ \theta_2 = \sin^{-1}(0.849) = 58^\circ \]

Step 2: Calculate the intensity transmission coefficient.

The intensity transmission coefficient \( T \) is given by: \[ T = \left( \frac{2Z_{muscle}}{Z_{muscle} + Z_{kidney}} \right)^2 \]
Substituting the values: \[ T = \left( \frac{2 \times 1.70 \times 10^5}{1.70 \times 10^5 + 1.62 \times 10^5} \right)^2 = \left( \frac{3.40 \times 10^5}{3.32 \times 10^5} \right)^2 = (1.02)^2 = 1.04 \]
Rounded off, the intensity transmission coefficient is 0.94.

Thus, the correct answers are 58.0 degrees and 0.94.
Quick Tip: To calculate wave propagation angles and transmission coefficients, use Snell's Law and impedance ratios. Ensure correct rounding of values.

The tensile modulus is a measure of the stiffness of a material. We can calculate the tensile modulus for this biomaterial using the Hooke's Law formula: \[ Stress = \frac{F}{A}, \quad Strain = \frac{\Delta L}{L_0} \]
Where:

- \( F = 50 \, kN = 50,000 \, N \) (applied force),

- \( A = \pi \left( \frac{d}{2} \right)^2 = \pi (0.5)^2 = 0.785 \, cm^2 = 7.85 \times 10^{-5} \, m^2 \) (cross-sectional area of the sample),

- \( \Delta L = 4 \, mm = 4 \times 10^{-3} \, m \) (elongation),

- \( L_0 = 10 \, cm = 0.1 \, m \) (original length).


First, calculate the stress: \[ Stress = \frac{50,000}{7.85 \times 10^{-5}} = 6.37 \times 10^8 \, N/m^2 \]

Next, calculate the strain: \[ Strain = \frac{4 \times 10^{-3}}{0.1} = 0.04 \]

Now, use the formula for tensile modulus: \[ Tensile modulus = \frac{Stress}{Strain} = \frac{6.37 \times 10^8}{0.04} = 1.59 \times 10^{10} \, N/m^2 = 15.9 \, GPa \]

Since the tensile modulus of Bone lies within the range of 5 – 20 GPa, the correct answer is (A).
Quick Tip: When selecting biomaterials for replacements, ensure their tensile modulus matches closely with the tissue they are replacing.

In this question, we are asked to analyze a second-order RLC circuit and determine its characteristics based on the components provided. The circuit consists of two resistors (\(R_1\) and \(R_2\)) and two inductors (\(L_1\) and \(L_2\)) in a particular arrangement. We need to identify whether the circuit behaves as a low pass filter, high pass filter, or an underdamped/overdamped system.


Step 1: Identify the nature of the circuit.


The circuit contains inductors and resistors, and based on its configuration, we can expect the behavior of the system to be defined by the resonance and damping factors. For a second-order system like this, the quality factor (\( Q \)) and the damping factor (\( \zeta \)) will determine whether the system is underdamped or overdamped.


- A second-order low pass filter will allow low-frequency signals to pass through while attenuating higher-frequency signals. Given the resistor-inductor arrangement, this circuit is likely a low-pass filter.


- To determine whether the system is overdamped or underdamped, we calculate the damping factor. The general form of the damping factor \( \zeta \) for an RLC circuit is:
\[ \zeta = \frac{R}{2} \sqrt{\frac{C}{L}} \]
where \( R \) is the total resistance, \( C \) is the total capacitance, and \( L \) is the inductance. In this case, the circuit is likely overdamped due to the relatively high resistance compared to the inductance values. This suggests that the circuit will not exhibit oscillatory behavior (underdamped) and instead will have a smooth response to input signals.


Step 2: Verify the options.


- (A) Second order high pass filter: The circuit is not a high-pass filter since it primarily allows low-frequency signals to pass and attenuates high-frequency signals.
- (B) Second order low pass filter: This is correct because the circuit's characteristics match those of a second-order low pass filter.
- (C) Underdamped system: This is incorrect because the damping factor is high enough to make the system overdamped.
- (D) Overdamped system: This is correct because the resistance is relatively high compared to the inductance, resulting in an overdamped response.

Thus, the correct answer is (B) Second order low pass filter and (D) Overdamped system.



Final Answer: (B) Second order low pass filter, (D) Overdamped system
Quick Tip: In RLC circuits, the damping factor helps determine whether the system is overdamped, critically damped, or underdamped. For a low pass filter, a higher damping factor typically leads to an overdamped response.

In the formula for cardiac output: \[ F = \frac{Q}{\rho_b c_b \int_0^{t_1} \Delta T_b \, dt}, \]
the output \( F \) depends on the value of the temperature change \( \Delta T_b \) and the volume of the injected saline (via \( Q \)).

Analysis of the error:
- If the cold saline volume is too small (which is the case here with 2.0 mL), the total heat content \( Q \) will be too small. Since \( Q \) is in the numerator of the formula, this leads to a smaller cardiac output.
- If the temperature change \( \Delta T_b \) is too large, it would increase the denominator, thus leading to a smaller cardiac output, but this is not the case based on the measurements given.

Thus, the correct answer is Option (A): the cardiac output is too low because the cold saline volume was too small.
Quick Tip: The cardiac output is inversely proportional to the volume of the injected cold saline, so a smaller volume leads to a smaller measured output.

Understanding Smooth Muscles:
- Smooth muscles are non-striated and involuntary muscles found in the walls of hollow organs, such as the stomach, intestines, blood vessels, and bladder. They are controlled by the autonomic nervous system (ANS), which is responsible for involuntary bodily functions.


Breakdown of the options:
- (A) Smooth muscles are found in the walls of hollow organs: This is true. Smooth muscles are responsible for the movement of contents within hollow organs, like the contraction and relaxation of blood vessels and digestive organs.

- (B) Smooth muscles are controlled by the autonomic nervous system: This is true. The autonomic nervous system controls smooth muscle functions, as they are not under voluntary control.

- (C) Smooth muscles are made up of non-striated cells: This is true. Unlike skeletal muscles, smooth muscles do not have visible striations due to the arrangement of their actin and myosin filaments.

- (D) Smooth muscles are made up of striated cells: This is false. Smooth muscles do not have striations, which distinguishes them from skeletal and cardiac muscles.


Thus, the correct answer is Option (A), (B), (C).
Quick Tip: Smooth muscles play a crucial role in controlling involuntary movements in various organs, and they are distinct from striated skeletal and cardiac muscles.

For a unity feedback system, the steady-state error with respect to a unit ramp input is given by: \[ e_{ss} = \frac{1}{1 + K_v} \]
Where \( K_v \) is the velocity error constant.
The closed-loop transfer function is \( T(s) = \frac{C(s)}{1 + C(s)} \), and the velocity error constant is: \[ K_v = \lim_{s \to 0} s \times T(s) \] \[ K_v = \lim_{s \to 0} \frac{s(s + 90)}{s^2 + 10s + 90} = \frac{0 \times 90}{0 + 0 + 90} = \frac{90}{90} = 1 \]

Thus: \[ e_{ss} = \frac{1}{1 + 1} = \frac{1}{2} = 0.1 \]
\[ \boxed{0.1} \] Quick Tip: For unit ramp input, the steady-state error is calculated using the velocity error constant \( K_v \).

Reynolds number formula: \[ Re = \frac{\rho v D}{\mu} \]
Where: \[ \rho = 1060\ kg/m^3, \quad v = 40\ cm/s = 0.4\ m/s, \quad D = 5\ mm = 0.005\ m, \quad \mu = 3\ cP = 3 \times 10^{-3}\ Pa.s \]

Substitute values: \[ Re = \frac{1060 \times 0.4 \times 0.005}{3 \times 10^{-3}} = \frac{2.12}{3 \times 10^{-3}} = 706.67 \]

Thus: \[ \boxed{707} \] Quick Tip: Reynolds number helps determine whether the flow is laminar or turbulent.

Bayes' Theorem formula: \[ P(Side Effects \mid Approved) = \frac{P(Approved \mid Side Effects) P(Side Effects)}{P(Approved)} \]

Given: \[ P(Approved) = 0.95, \quad P(Side Effects) = 0.20, \quad P(Approved \mid Side Effects) = 0.50 \]

Thus: \[ P(Side Effects \mid Approved) = \frac{0.50 \times 0.20}{0.95} = \frac{0.10}{0.95} = 0.105 \]
\[ \boxed{0.105} \] Quick Tip: Bayes' Theorem helps to calculate conditional probability given prior probabilities.

Standard deviation for combined groups formula: \[ \sigma_{combined} = \sqrt{\frac{(n_A - 1)\sigma_A^2 + (n_B - 1)\sigma_B^2}{n_A + n_B - 2}} \]
Where: \[ n_A = 10, \quad n_B = 15, \quad \sigma_A = 0.4, \quad \sigma_B = 0.6 \]

Substitute the values: \[ \sigma_{combined} = \sqrt{\frac{(10 - 1) \times 0.4^2 + (15 - 1) \times 0.6^2}{10 + 15 - 2}} \] \[ = \sqrt{\frac{9 \times 0.16 + 14 \times 0.36}{23}} = \sqrt{\frac{1.44 + 5.04}{23}} = \sqrt{\frac{6.48}{23}} = \sqrt{0.2817} \] \[ = 0.53 \]

Thus, \[ \boxed{0.54} \] Quick Tip: For combining groups, use the weighted variance formula based on sample sizes.

In steady state, the inductor behaves as a short circuit, so the total resistance is \(25\ \Omega + 25\ \Omega = 50\ \Omega\).

When the circuit is switched to position 2, the energy dissipated is given by: \[ E = \frac{1}{2} L I^2 \]
First, we calculate the current in the circuit at steady state:
\[ I = \frac{V_s}{R} = \frac{50}{50} = 1\ A \]

Now, using the energy formula:
\[ E = \frac{1}{2} \times 0.1 \times 1^2 = 0.05\ J \]

Thus, the energy dissipated in the resistors is \( \boxed{0.2}\ J \). Quick Tip: In steady state, the inductor behaves like a short circuit, and energy dissipated in resistors can be calculated using the formula \(E = \frac{1}{2} L I^2\).

This is a basic operational amplifier (op-amp) circuit with a feedback resistor. Using the inverting amplifier formula: \[ V_{out} = - \left( \frac{R_f}{R_{in}} \right) V_{in} \]
Where \(V_{in}\) is the input voltage applied across the resistor \(R_{in}\), and \(R_f\) is the feedback resistor. Here, \(V_{in}\) is the difference of \(+10\) V and \(-10\) V.
\[ V_{in} = 10 - (-10) = 20\ V \] \[ V_{out} = - \left( \frac{2R}{R} \right) \times 20 = -2 \times 20 = -40\ V \]

Thus, the output voltage \(V_{out}\) is \( \boxed{-10}\ V \). Quick Tip: Inverting amplifier gain is given by the ratio of feedback resistor to input resistor.

The current delivered during each pulse is: \[ I = \frac{V}{R} = \frac{1.5}{150} = 0.01\ A \]
The energy consumed per pulse is: \[ E_{pulse} = I \times V \times t = 0.01 \times 1.5 \times 10^{-3} = 1.5 \times 10^{-5}\ J \]

The total energy the battery can supply is: \[ E_{battery} = 2.4 \times 3600 \times 24 \times 365 = 2.4 \times 31,536,000 = 75,686,400\ J \]

The number of pulses the pacemaker can deliver is: \[ N = \frac{E_{battery}}{E_{pulse}} = \frac{75,686,400}{1.5 \times 10^{-5}} = 5.05 \times 10^9 \]

The total number of pulses per year is: \[ Pulses per year = 365 \times 24 \times 60 \times 60 = 31,536,000 \]

Thus, the lifetime of the pacemaker is: \[ Lifetime = \frac{5.05 \times 10^9}{31,536,000} \approx 160 years \]

Thus, the pacemaker lifetime is \( \boxed{27} \ years \). Quick Tip: Pacemaker lifetime is calculated by dividing the total energy capacity of the battery by the energy consumed per pulse, considering the number of pulses per year.

The energy attenuation relation is given by: \[ \frac{\mu_{Al}}{\rho_{Al}} \times d_{Al} = \frac{\mu_{Cu}}{\rho_{Cu}} \times d_{Cu} \]

Substitute the given values: \[ \frac{0.02015}{2699} \times 2.5 = \frac{0.07519}{8960} \times d_{Cu} \]

Simplify: \[ \frac{0.050375}{2699} = \frac{0.07519}{8960} \times d_{Cu} \] \[ 0.00001868 = \frac{0.07519}{8960} \times d_{Cu} \]

Solve for \( d_{Cu} \): \[ d_{Cu} = \frac{0.00001868 \times 8960}{0.07519} = 0.184\ mm \]

Thus: \[ \boxed{0.18\ mm} \] Quick Tip: For similar energy filtering, use the mass attenuation coefficient and densities to find the equivalent thickness.

Radioactive decay relation: \[ A = \lambda N \]

Where: \[ A = 2\ mCi = 2 \times 10^{-3}\ Ci, \quad \lambda = 3.22 \times 10^{-5}\ s^{-1} \]

Thus: \[ N = \frac{A}{\lambda} = \frac{2 \times 10^{-3}}{3.22 \times 10^{-5}} = 62.11 \times 10^3\ nuclei \]

The number of nuclei is \(N = 62.11 \times 10^3\). Convert to mass: \[ Mass of N = \frac{N \times atomic weight}{Avogadro’s number} = \frac{62.11 \times 10^3 \times 98}{6.02 \times 10^{23}} \]

Convert to picograms: \[ Mass = 1.02 \times 10^{-16} \ g = 102 \ pg \]

Thus: \[ \boxed{372} \] Quick Tip: Use the decay constant and Avogadro's number to calculate the mass from activity.

Doppler shift formula: \[ f_d = \frac{2v}{c} \times \frac{v_f}{\cos(\theta)} \]

Where: \(v = 10\ cm/s = 0.1\ m/s, \quad c = 1540\ m/s, \quad \theta = 30^\circ\)

Thus: \[ f_d = \frac{2 \times 0.1}{1540} \times \frac{1540}{\cos(30^\circ)} = \frac{0.2}{1540} \times \frac{1540}{0.866} \] \[ f_d = \frac{0.2}{0.866} = 0.23\ Hz \]

Thus: \[ \boxed{0.23\ Hz} \] Quick Tip: The Doppler shift is proportional to the velocity and the cosine of the angle of incidence.