Q. Five students P, Q, R, S, and T are ranked based on their scores. P is ranked higher than Q but lower than R. S is ranked lower than T but higher than Q. What is the correct order from highest to lowest rank? (2023)
A.
R, P, S, T, Q
B.
T, S, R, P, Q
C.
T, S, R, Q, P
D.
R, T, S, P, Q
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Solution
The correct order is T, S, R, Q, P based on the given conditions.
Correct Answer:
C
— T, S, R, Q, P
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Q. Five students P, Q, R, S, and T scored different marks in an exam. P scored more than Q but less than R. S scored less than T but more than P. Who scored the highest? (2023)
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Solution
The order based on the scores is R > P > Q and T > S > P. Since T is greater than S and S is greater than P, T is the highest scorer.
Correct Answer:
D
— T
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Q. Five students P, Q, R, S, and T scored marks in a test. P scored more than Q but less than R. S scored less than T but more than P. Who scored the highest?
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Solution
The order of scores is R > P > Q and T > S > P. Thus, T is the highest scorer.
Correct Answer:
D
— T
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Q. Five students scored the following marks in a test: P - 85, Q - 90, R - 75, S - 95, T - 80. If they are ranked from highest to lowest, who is in the second position? (2023)
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Solution
The scores in descending order are S (95), Q (90), P (85), T (80), R (75). Thus, S is first, Q is second.
Correct Answer:
C
— S
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Q. Five students scored the following marks: A - 85, B - 90, C - 75, D - 95, E - 80. If they are ranked from highest to lowest, who is in the second position?
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Solution
The scores in descending order are D (95), B (90), A (85), E (80), C (75). Thus, A is in the second position.
Correct Answer:
A
— A
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Q. Five students scored the following marks: A - 85, B - 90, C - 75, D - 95, E - 80. What is the rank of student C?
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Solution
C scored 75 marks, which is the third highest, so C ranks 3rd.
Correct Answer:
C
— 3
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Q. Five students scored the following marks: P - 85, Q - 90, R - 75, S - 95, T - 80. If they are ranked from highest to lowest, who is in the second position? (2023)
Show solution
Solution
The scores in descending order are S (95), Q (90), P (85), T (80), R (75). Thus, S is first, Q is second.
Correct Answer:
C
— S
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Q. For a charged plane sheet, if the surface charge density is doubled, what happens to the electric field?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
Show solution
Solution
The electric field due to a charged plane sheet is directly proportional to the surface charge density. Therefore, if σ is doubled, E also doubles.
Correct Answer:
B
— It doubles
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Q. For a charged sphere, what happens to the electric field inside the sphere as the radius increases?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
Show solution
Solution
The electric field inside a uniformly charged sphere is zero.
Correct Answer:
D
— Becomes zero
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Q. For a charged spherical conductor, what happens to the electric field inside the conductor when it is charged?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
Show solution
Solution
The electric field inside a charged conductor in electrostatic equilibrium is zero.
Correct Answer:
D
— Becomes zero
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Q. For a circular loop of radius R carrying a current I, what is the magnetic field at the center of the loop?
A.
B = μ₀I/(2R)
B.
B = μ₀I/(4R)
C.
B = μ₀I/(πR)
D.
B = μ₀I/(2πR)
Show solution
Solution
The magnetic field at the center of a circular loop of radius R carrying current I is given by B = μ₀I/(2πR).
Correct Answer:
D
— B = μ₀I/(2πR)
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Q. For a closed loop of wire carrying current, what does the line integral of the magnetic field equal?
A.
Zero
B.
The product of current and resistance
C.
μ₀ times the total current enclosed
D.
The electric field times the area
Show solution
Solution
According to Ampere's Law, the line integral of the magnetic field around a closed loop equals μ₀ times the total current enclosed by the loop.
Correct Answer:
C
— μ₀ times the total current enclosed
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Q. For a closed surface enclosing multiple charges, how is the total electric flux calculated?
A.
Sum of individual fluxes
B.
Product of charges
C.
Sum of enclosed charges divided by ε₀
D.
Average of charges
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Solution
The total electric flux through a closed surface is given by Φ = ΣQ_enc/ε₀, where Q_enc is the total charge enclosed.
Correct Answer:
C
— Sum of enclosed charges divided by ε₀
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Q. For a closed surface enclosing multiple charges, how is the total electric flux related to the enclosed charges?
A.
It is proportional to the sum of the charges
B.
It is inversely proportional to the sum of the charges
C.
It is independent of the charges
D.
It is proportional to the square of the charges
Show solution
Solution
According to Gauss's law, the total electric flux through a closed surface is proportional to the total charge enclosed.
Correct Answer:
A
— It is proportional to the sum of the charges
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Q. For a composite body made of a solid cylinder and a solid sphere, how do you calculate the total moment of inertia about the same axis?
A.
Add the individual moments
B.
Multiply the individual moments
C.
Subtract the individual moments
D.
Divide the individual moments
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Solution
The total moment of inertia of a composite body about the same axis is the sum of the individual moments of inertia.
Correct Answer:
A
— Add the individual moments
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Q. For a composite body made of two solid cylinders of mass M1 and M2 and radius R, what is the total moment of inertia about the same axis?
A.
I1 + I2
B.
I1 - I2
C.
I1 * I2
D.
I1 / I2
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Solution
The total moment of inertia of a composite body is the sum of the individual moments of inertia: I_total = I1 + I2.
Correct Answer:
A
— I1 + I2
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Q. For a convex lens, if the object is at infinity, the image will be formed at: (2020)
A.
At the focus
B.
At the center of curvature
C.
At infinity
D.
At the optical center
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Solution
When the object is at infinity, the rays converge at the focus of the convex lens.
Correct Answer:
A
— At the focus
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Q. For a convex lens, if the object is at the focus, what type of image is formed? (2020)
A.
Real and inverted
B.
Virtual and erect
C.
No image
D.
Real and erect
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Solution
When the object is at the focus of a convex lens, the rays of light emerge parallel, resulting in no image being formed.
Correct Answer:
C
— No image
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Q. For a current-carrying loop, what is the magnetic field at the center if the radius is halved?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
Show solution
Solution
The magnetic field at the center of a loop is inversely proportional to the radius. If the radius is halved, the magnetic field quadruples.
Correct Answer:
C
— It quadruples
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Q. For a cylindrical conductor of radius R carrying current I, what is the magnetic field at a point outside the cylinder?
A.
0
B.
μ₀I/2πr
C.
μ₀I/4πr
D.
μ₀I/πr
Show solution
Solution
For points outside the cylinder, B = μ₀I/2πr.
Correct Answer:
B
— μ₀I/2πr
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Q. For a cylindrical conductor of radius R carrying current I, what is the magnetic field at a point outside the conductor?
A.
0
B.
μ₀I/2πR
C.
μ₀I/4πR
D.
μ₀I/πR
Show solution
Solution
Using Ampere's Law, B = μ₀I/2πR for points outside the cylindrical conductor.
Correct Answer:
B
— μ₀I/2πR
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Q. For a damped oscillator, what is the relationship between the natural frequency and the damped frequency?
A.
Damped frequency is greater
B.
Damped frequency is equal
C.
Damped frequency is less
D.
No relationship
Show solution
Solution
The damped frequency is less than the natural frequency due to the effect of damping.
Correct Answer:
C
— Damped frequency is less
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Q. For a diffraction grating with 500 lines per mm, what is the angle of the first order maximum for light of wavelength 600 nm?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
15 degrees
Show solution
Solution
Using the grating equation d sin θ = nλ, where d = 1/500000 m, n = 1, and λ = 600 x 10^-9 m, we find θ ≈ 30 degrees.
Correct Answer:
A
— 30 degrees
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Q. For a diffraction pattern produced by a single slit, how does the width of the central maximum change if the slit width is halved?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Show solution
Solution
If the slit width is halved, the width of the central maximum increases because the angle for the first minimum increases.
Correct Answer:
A
— Increases
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Q. For a diffraction pattern produced by a single slit, how does the width of the central maximum compare to the other maxima?
A.
Wider than all other maxima
B.
Narrower than all other maxima
C.
Equal to all other maxima
D.
None of the above
Show solution
Solution
The central maximum in a single-slit diffraction pattern is wider than all other maxima.
Correct Answer:
A
— Wider than all other maxima
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Q. For a first-order reaction, if the half-life is 10 minutes, what will be the half-life if the initial concentration is doubled?
A.
10 minutes
B.
5 minutes
C.
20 minutes
D.
15 minutes
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Solution
For a first-order reaction, the half-life is independent of the initial concentration. Therefore, it remains 10 minutes.
Correct Answer:
A
— 10 minutes
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Q. For a first-order reaction, the half-life is independent of the initial concentration. What is the expression for half-life?
A.
t1/2 = 0.693/k
B.
t1/2 = k/0.693
C.
t1/2 = 1/k
D.
t1/2 = k/2
Show solution
Solution
For a first-order reaction, the half-life is given by the expression t1/2 = 0.693/k.
Correct Answer:
A
— t1/2 = 0.693/k
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Q. For a first-order reaction, the half-life is independent of which of the following?
A.
Initial concentration
B.
Rate constant
C.
Temperature
D.
All of the above
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Solution
For a first-order reaction, the half-life is independent of the initial concentration.
Correct Answer:
A
— Initial concentration
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Q. For a gas at 300 K, if the RMS speed is 500 m/s, what will be the RMS speed at 600 K?
A.
500 m/s
B.
707 m/s
C.
1000 m/s
D.
250 m/s
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Solution
RMS speed is proportional to the square root of temperature, so v_rms at 600 K = 500 * sqrt(600/300) = 707 m/s.
Correct Answer:
B
— 707 m/s
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Q. For a gas at 300 K, what is the RMS speed if the molar mass is 0.028 kg/mol?
A.
500 m/s
B.
600 m/s
C.
700 m/s
D.
800 m/s
Show solution
Solution
Using v_rms = sqrt(3RT/M), we calculate v_rms = sqrt(3 * 8.314 * 300 / 0.028) which gives approximately 600 m/s.
Correct Answer:
B
— 600 m/s
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