Engineering & Architecture Admissions
Q. In a heat engine, if the work done is 200 J and the heat absorbed is 500 J, what is the efficiency?
A.
40%
B.
50%
C.
60%
D.
80%
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Solution
Efficiency = (Work done / Heat absorbed) * 100 = (200 J / 500 J) * 100 = 40%.
Correct Answer: B — 50%
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Q. In a heat engine, if the work output is 200 J and the heat input is 600 J, what is the efficiency?
A.
33.33%
B.
50%
C.
66.67%
D.
75%
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Solution
Efficiency = (Work output / Heat input) × 100 = (200 J / 600 J) × 100 = 33.33%.
Correct Answer: C — 66.67%
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Q. In a heat engine, the work done is equal to:
A.
Heat absorbed from the hot reservoir
B.
Heat rejected to the cold reservoir
C.
Heat absorbed minus heat rejected
D.
Heat absorbed plus heat rejected
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Solution
The work done by a heat engine is equal to the heat absorbed from the hot reservoir minus the heat rejected to the cold reservoir.
Correct Answer: C — Heat absorbed minus heat rejected
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Q. In a hydraulic lift, if the input power is 2000 W and the output power is 1800 W, what is the efficiency of the lift?
A.
90%
B.
80%
C.
70%
D.
75%
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Solution
Efficiency is calculated as (Output Power / Input Power) * 100%. Here, efficiency = (1800 W / 2000 W) * 100% = 90%.
Correct Answer: B — 80%
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Q. In a hydraulic lift, if the input power is 500 W and the efficiency is 80%, what is the output power?
A.
400 W
B.
500 W
C.
600 W
D.
700 W
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Solution
Output power can be calculated using the formula: Output Power = Efficiency * Input Power. Here, Output Power = 0.8 * 500 W = 400 W.
Correct Answer: A — 400 W
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Q. In a hydrogen atom, the energy levels are quantized. What is the formula for the energy of the nth level?
A.
E_n = -13.6/n^2 eV
B.
E_n = -13.6n^2 eV
C.
E_n = -13.6/n eV
D.
E_n = -13.6n eV
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Solution
The energy levels of a hydrogen atom are given by E_n = -13.6/n^2 eV, where n is the principal quantum number.
Correct Answer: A — E_n = -13.6/n^2 eV
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Q. In a hydrogen atom, what is the energy of the electron in the ground state?
A.
-13.6 eV
B.
-3.4 eV
C.
-1.51 eV
D.
0 eV
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Solution
The energy of the electron in the ground state of a hydrogen atom is -13.6 eV.
Correct Answer: A — -13.6 eV
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Q. In a hydrogen atom, what is the energy of the electron in the n=2 state?
A.
-3.4 eV
B.
-13.6 eV
C.
-1.51 eV
D.
-0.85 eV
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Solution
The energy of an electron in a hydrogen atom in the n=2 state is given by E_n = -13.6/n^2 = -13.6/4 = -3.4 eV.
Correct Answer: A — -3.4 eV
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Q. In a hydrogen atom, what is the wavelength of the emitted photon when an electron transitions from n=3 to n=2?
A.
656 nm
B.
486 nm
C.
434 nm
D.
410 nm
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Solution
Using the Rydberg formula, the wavelength for the transition from n=3 to n=2 is approximately 486 nm.
Correct Answer: B — 486 nm
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Q. In a hydrogen atom, which transition emits the photon with the highest energy?
A.
n=2 to n=1
B.
n=3 to n=2
C.
n=4 to n=3
D.
n=5 to n=4
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Solution
The energy of the emitted photon is highest for the transition from n=2 to n=1, as it involves the largest energy difference.
Correct Answer: A — n=2 to n=1
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Q. In a hydrogen atom, which transition would emit the highest energy photon?
A.
n=2 to n=1
B.
n=3 to n=2
C.
n=4 to n=3
D.
n=5 to n=4
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Solution
The energy of the photon emitted is highest for the transition from n=2 to n=1, as it involves the largest energy difference.
Correct Answer: A — n=2 to n=1
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Q. In a hydrogen atom, which transition would emit the photon with the highest energy?
A.
n=2 to n=1
B.
n=3 to n=2
C.
n=4 to n=3
D.
n=5 to n=4
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Solution
The energy of the emitted photon is highest for the transition from n=2 to n=1, as it involves the largest energy difference.
Correct Answer: A — n=2 to n=1
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Q. In a lab experiment, the density of a liquid is measured as 1.2 g/cm³ with an uncertainty of ±0.05 g/cm³. What is the relative error?
A.
4.17%
B.
3.33%
C.
5.00%
D.
2.50%
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Solution
Relative error = (Uncertainty / Measured value) * 100 = (0.05 / 1.2) * 100 = 4.17%.
Correct Answer: A — 4.17%
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Q. In a lab experiment, the speed of sound is measured as 340 m/s with an uncertainty of ±2 m/s. What is the total uncertainty if the speed is used in a calculation involving division by 2?
A.
±1 m/s
B.
±2 m/s
C.
±0.5 m/s
D.
±0.25 m/s
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Solution
Total uncertainty = (2 m/s) / 2 = ±1 m/s.
Correct Answer: A — ±1 m/s
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Q. In a lab experiment, the speed of sound is measured as 340 m/s with an uncertainty of ±5 m/s. What is the percentage uncertainty in this measurement?
A.
1.47%
B.
1.5%
C.
2%
D.
0.5%
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Solution
Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (5 / 340) * 100 ≈ 1.47%.
Correct Answer: B — 1.5%
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Q. In a lottery, there are 10 tickets, and 3 of them are winning tickets. If one ticket is drawn at random, what is the probability that it is a winning ticket?
A.
1/10
B.
1/3
C.
3/10
D.
7/10
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Solution
The probability of drawing a winning ticket is 3/10.
Correct Answer: C — 3/10
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Q. In a lottery, there are 10 tickets, and 3 of them are winning tickets. What is the probability of selecting a winning ticket?
A.
1/10
B.
3/10
C.
1/3
D.
1/5
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Solution
The probability of selecting a winning ticket is the number of winning tickets divided by the total number of tickets, which is 3/10.
Correct Answer: B — 3/10
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Q. In a lottery, there are 10 tickets, of which 2 are winning tickets. What is the probability of selecting a winning ticket?
A.
1/5
B.
1/10
C.
1/2
D.
2/10
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Solution
The probability of selecting a winning ticket is the number of winning tickets divided by the total number of tickets, which is 2/10 = 1/5.
Correct Answer: B — 1/10
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Q. In a lottery, there are 10 tickets, of which 3 are winning tickets. What is the probability of selecting a winning ticket?
A.
1/10
B.
1/3
C.
3/10
D.
7/10
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Solution
The probability of selecting a winning ticket is the number of winning tickets divided by the total number of tickets, which is 3/10.
Correct Answer: C — 3/10
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Q. In a lottery, there are 100 tickets, and 10 of them are winning tickets. If one ticket is drawn at random, what is the probability that it is a winning ticket?
A.
1/10
B.
1/5
C.
1/20
D.
1/50
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Solution
The probability of drawing a winning ticket = Number of winning tickets / Total tickets = 10/100 = 1/10.
Correct Answer: A — 1/10
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Q. In a lottery, there are 100 tickets, and 10 of them are winning tickets. What is the probability of selecting a winning ticket?
A.
1/10
B.
1/5
C.
1/20
D.
1/50
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Solution
The probability of selecting a winning ticket is the number of winning tickets divided by the total number of tickets, which is 10/100 = 1/10.
Correct Answer: A — 1/10
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Q. In a magnetic field, the force on a charged particle is maximum when the particle's velocity is:
A.
Parallel to the field
B.
Perpendicular to the field
C.
At an angle of 45 degrees
D.
At an angle of 90 degrees
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Solution
The magnetic force on a charged particle is given by F = qvB sin(θ), which is maximum when θ = 90 degrees (perpendicular).
Correct Answer: B — Perpendicular to the field
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Q. In a magnetic field, the force on a charged particle is zero when it moves:
A.
Perpendicular to the field
B.
Parallel to the field
C.
At an angle of 30 degrees
D.
At an angle of 90 degrees
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Solution
The magnetic force on a charged particle is zero when it moves parallel to the magnetic field, as the angle θ = 0° results in sin(θ) = 0.
Correct Answer: B — Parallel to the field
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Q. In a material with a resistivity of 2 x 10^-8 Ω·m, what is the resistance of a 10 m long wire with a cross-sectional area of 1 mm²?
A.
0.02 Ω
B.
0.2 Ω
C.
2 Ω
D.
20 Ω
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Solution
Resistance R = ρ(L/A) = 2 x 10^-8 * (10 / (1 x 10^-6)) = 0.2 Ω.
Correct Answer: B — 0.2 Ω
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Q. In a material, if the strain energy density is given by U, what is the expression for the total strain energy stored in a volume V of the material?
A.
U * V
B.
U / V
C.
U + V
D.
U - V
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Solution
The total strain energy stored in a volume V is given by the product of strain energy density U and volume V, i.e., Total Energy = U * V.
Correct Answer: A — U * V
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Q. In a material, if the strain is 0.01 and the Young's modulus is 200 GPa, what is the stress in the material?
A.
2 MPa
B.
20 MPa
C.
200 MPa
D.
2000 MPa
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Solution
Stress = Young's modulus * strain = 200 GPa * 0.01 = 2 GPa = 2000 MPa.
Correct Answer: C — 200 MPa
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Q. In a Michelson interferometer, what happens to the interference pattern if one of the mirrors is moved away from the beam splitter?
A.
Fringes move closer
B.
Fringes move apart
C.
Fringes disappear
D.
No change in pattern
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Solution
Moving one mirror changes the path length of one beam, causing the fringes to move apart or closer depending on the direction of movement.
Correct Answer: B — Fringes move apart
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Q. In a Michelson interferometer, what happens to the interference pattern if one of the mirrors is moved?
A.
The pattern disappears
B.
The pattern shifts
C.
The pattern becomes brighter
D.
The pattern becomes dimmer
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Solution
Moving one of the mirrors changes the path length of one beam, causing a shift in the interference pattern.
Correct Answer: B — The pattern shifts
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Q. In a Michelson interferometer, what happens to the interference pattern if one of the mirrors is moved slightly?
A.
The pattern remains unchanged
B.
The pattern shifts
C.
The pattern disappears
D.
The pattern becomes brighter
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Solution
Moving one of the mirrors changes the path length for one of the beams, causing a shift in the interference pattern.
Correct Answer: B — The pattern shifts
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Q. In a Michelson interferometer, what happens when one of the mirrors is moved slightly?
A.
No change in interference pattern
B.
Fringes shift
C.
Fringes disappear
D.
Fringes become brighter
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Solution
Moving one of the mirrors changes the path length, causing a shift in the interference pattern (fringes).
Correct Answer: B — Fringes shift
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