Q. What is the equivalent resistance of two 4Ω resistors in parallel? (2019)
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Solution
Using the formula 1/R_eq = 1/R1 + 1/R2, we get 1/R_eq = 1/4 + 1/4 = 1/2, thus R_eq = 2Ω.
Correct Answer: A — 2Ω
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Q. What is the equivalent resistance of two resistors of 4Ω and 12Ω in parallel? (2022)
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Solution
Using the formula 1/R_eq = 1/R1 + 1/R2, we get 1/R_eq = 1/4 + 1/12 = 1/3, so R_eq = 3Ω.
Correct Answer: A — 3Ω
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Q. What is the escape velocity from the surface of a planet with a radius of 4,000 km and a mass of 6 × 10^24 kg? (G = 6.67 × 10^-11 N m²/kg²)
A.
10,000 m/s
B.
11,200 m/s
C.
12,000 m/s
D.
13,000 m/s
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Solution
Escape velocity = √(2GM/R) = √(2 * 6.67 × 10^-11 * 6 × 10^24 / 4 × 10^6) = 11,200 m/s
Correct Answer: B — 11,200 m/s
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Q. What is the escape velocity from the surface of the Earth? (g = 9.8 m/s², R = 6.4 × 10^6 m)
A.
11.2 km/s
B.
9.8 km/s
C.
7.9 km/s
D.
15 km/s
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Solution
Escape velocity (v) = √(2gR) = √(2 * 9.8 * 6.4 × 10^6) ≈ 11.2 km/s
Correct Answer: A — 11.2 km/s
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Q. What is the escape velocity from the surface of the Earth? (R = 6.4 × 10^6 m, g = 9.8 m/s²)
A.
11.2 km/s
B.
9.8 km/s
C.
7.9 km/s
D.
12.0 km/s
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Solution
Escape velocity v = √(2gR) = √(2 * 9.8 * 6.4 × 10^6) ≈ 11.2 km/s.
Correct Answer: A — 11.2 km/s
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Q. What is the final temperature when 200 g of ice at 0°C is added to 100 g of water at 80°C? (Assume no heat loss to the surroundings) (2023)
A.
0°C
B.
20°C
C.
40°C
D.
80°C
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Solution
Using heat balance: m_ice * L_f + m_water * c * (T_final - 80) = 0. Solving gives T_final = 20°C.
Correct Answer: B — 20°C
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Q. What is the final temperature when 200 g of ice at 0°C is added to 100 g of water at 80°C? (Specific heat of water = 4.2 J/g°C) (2020)
A.
0°C
B.
20°C
C.
40°C
D.
80°C
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Solution
Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. Solving gives a final temperature of 20°C.
Correct Answer: B — 20°C
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Q. What is the force between two charges of +2μC and -3μC separated by 0.5m? (2022)
A.
0.24 N
B.
0.12 N
C.
0.48 N
D.
0.36 N
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Solution
Using Coulomb's law, F = k * |q1 * q2| / r^2 = (9 × 10^9) * (2 × 10^-6) * (3 × 10^-6) / (0.5)^2 = 0.24 N.
Correct Answer: A — 0.24 N
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Q. What is the force between two point charges of +2 µC and -3 µC separated by a distance of 0.5 m in vacuum?
A.
-24 N
B.
-12 N
C.
12 N
D.
24 N
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Solution
Using Coulomb's law, F = k * |q1 * q2| / r^2 = (9 × 10^9) * |2 × 10^-6 * -3 × 10^-6| / (0.5)^2 = -24 N.
Correct Answer: A — -24 N
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Q. What is the force experienced by a current-carrying conductor of length L in a magnetic field B at an angle θ? (2023)
A.
F = BIL
B.
F = BIL sin(θ)
C.
F = BIL cos(θ)
D.
F = BIL²
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Solution
The force on a current-carrying conductor in a magnetic field is given by F = BIL sin(θ), where θ is the angle between the conductor and the magnetic field.
Correct Answer: B — F = BIL sin(θ)
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Q. What is the force on a charge moving in a magnetic field given by F = qvB sin(θ)? (2022)
A.
It is always zero
B.
It depends on the angle θ
C.
It is constant
D.
It is maximum when θ = 90°
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Solution
The force on a charge moving in a magnetic field is maximum when the angle θ between the velocity vector and the magnetic field is 90°, as sin(90°) = 1.
Correct Answer: D — It is maximum when θ = 90°
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Q. What is the force on a charge moving in a magnetic field given by the equation F = qvB sin(θ)? (2020)
A.
Charge times velocity
B.
Charge times magnetic field
C.
Charge times velocity times magnetic field times sine of angle
D.
Charge times sine of angle
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Solution
The force on a charge moving in a magnetic field is given by F = qvB sin(θ), where q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
Correct Answer: C — Charge times velocity times magnetic field times sine of angle
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Q. What is the formula for calculating the force of gravity between two masses? (2021)
A.
F = ma
B.
F = G(m1m2)/r²
C.
F = mv
D.
F = mgh
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Solution
The formula for gravitational force is F = G(m1m2)/r², where G is the gravitational constant.
Correct Answer: B — F = G(m1m2)/r²
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Q. What is the formula for calculating the total current in a parallel RLC circuit? (2023)
A.
I = V/R
B.
I = V/(1/R + 1/X)
C.
I = V/(R + X)
D.
I = V/R + V/X
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Solution
In a parallel RLC circuit, the total current is calculated using I = V/(1/R + 1/X), where X is the total reactance.
Correct Answer: B — I = V/(1/R + 1/X)
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Q. What is the formula for the average power consumed in an AC circuit with a phase angle φ? (2020)
A.
P = VI
B.
P = VI cos(φ)
C.
P = VI sin(φ)
D.
P = V²/R
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Solution
The average power P in an AC circuit is given by P = VI cos(φ), where φ is the phase angle.
Correct Answer: B — P = VI cos(φ)
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Q. What is the formula for the average power consumed in an AC circuit with a resistor? (2019)
A.
P = VI
B.
P = V²/R
C.
P = I²R
D.
All of the above
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Solution
The average power consumed in an AC circuit can be expressed as P = VI, P = V²/R, and P = I²R.
Correct Answer: D — All of the above
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Q. What is the formula for the resonant frequency of an RLC circuit? (2021)
A.
f = 1/(2π√(LC))
B.
f = R/(2πL)
C.
f = 1/(2πRC)
D.
f = 2π√(LC)
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Solution
The resonant frequency f of an RLC circuit is given by f = 1/(2π√(LC)).
Correct Answer: A — f = 1/(2π√(LC))
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Q. What is the formula for the total impedance in a parallel RLC circuit? (2023)
A.
1/Z = 1/Z1 + 1/Z2 + 1/Z3
B.
Z = R + jX
C.
Z = R
D.
.
X
.
Z = R + jωL - j/(ωC)
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Solution
The total impedance in a parallel circuit is given by 1/Z = 1/Z1 + 1/Z2 + 1/Z3.
Correct Answer: A — 1/Z = 1/Z1 + 1/Z2 + 1/Z3
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Q. What is the frequency of a pendulum that completes 30 oscillations in 60 seconds? (2017)
A.
0.5 Hz
B.
1 Hz
C.
2 Hz
D.
3 Hz
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Solution
Frequency f = number of oscillations / time = 30 / 60 = 0.5 Hz.
Correct Answer: B — 1 Hz
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Q. What is the frequency of a sound wave with a period of 0.01 seconds? (2014)
A.
100 Hz
B.
200 Hz
C.
50 Hz
D.
10 Hz
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Solution
Frequency = 1/Period = 1/0.01 s = 100 Hz.
Correct Answer: A — 100 Hz
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Q. What is the frequency of a tuning fork that vibrates 256 times in 2 seconds? (2019)
A.
128 Hz
B.
256 Hz
C.
512 Hz
D.
64 Hz
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Solution
Frequency (f) = Number of vibrations / Time = 256 / 2 = 128 Hz.
Correct Answer: A — 128 Hz
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Q. What is the frequency of a wave with a period of 0.01 s? (2020)
A.
100 Hz
B.
50 Hz
C.
200 Hz
D.
10 Hz
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Solution
Frequency f = 1/T = 1/0.01 s = 100 Hz.
Correct Answer: A — 100 Hz
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Q. What is the frequency of a wave with a period of 0.1 s? (2020)
A.
10 Hz
B.
5 Hz
C.
20 Hz
D.
15 Hz
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Solution
Frequency f = 1/T = 1/0.1 s = 10 Hz.
Correct Answer: A — 10 Hz
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Q. What is the frequency of a wave with a period of 0.2 seconds?
A.
5 Hz
B.
10 Hz
C.
20 Hz
D.
50 Hz
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Solution
Frequency (f) is the reciprocal of the period (T). Therefore, f = 1/T = 1/0.2 s = 5 Hz.
Correct Answer: B — 10 Hz
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Q. What is the frequency of a wave with a period of 0.5 seconds? (2019)
A.
2 Hz
B.
1 Hz
C.
0.5 Hz
D.
4 Hz
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Solution
Frequency (f) = 1/Period (T) = 1/0.5 s = 2 Hz.
Correct Answer: A — 2 Hz
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Q. What is the frequency of a wave with a speed of 300 m/s and a wavelength of 3 m? (2023)
A.
100 Hz
B.
150 Hz
C.
200 Hz
D.
250 Hz
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Solution
Frequency (f) = speed (v) / wavelength (λ) = 300 m/s / 3 m = 100 Hz.
Correct Answer: B — 150 Hz
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Q. What is the frequency of a wave with a wavelength of 2 m and a speed of 340 m/s?
A.
170 Hz
B.
340 Hz
C.
85 Hz
D.
420 Hz
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Solution
Frequency (f) is given by the formula f = speed / wavelength. Here, f = 340 m/s / 2 m = 170 Hz.
Correct Answer: A — 170 Hz
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Q. What is the gravitational field strength at a distance of 10 m from a mass of 100 kg? (G = 6.67 × 10^-11 N m²/kg²)
A.
6.67 × 10^-10 N/kg
B.
6.67 × 10^-9 N/kg
C.
6.67 × 10^-8 N/kg
D.
6.67 × 10^-7 N/kg
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Solution
g = G * M / r² = (6.67 × 10^-11) * (100) / (10²) = 6.67 × 10^-10 N/kg
Correct Answer: A — 6.67 × 10^-10 N/kg
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Q. What is the gravitational force acting on a 50 kg object at the surface of the Earth?
A.
490 N
B.
500 N
C.
510 N
D.
520 N
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Solution
Weight = mass * g = 50 kg * 9.8 m/s² = 490 N
Correct Answer: A — 490 N
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Q. What is the gravitational force acting on a 50 kg object on the surface of the Earth?
A.
490 N
B.
50 N
C.
5 N
D.
500 N
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Solution
Weight W = m * g = 50 kg * 9.8 m/s² = 490 N.
Correct Answer: A — 490 N
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