Q. A capacitor of capacitance C is charged to a voltage V. If the charge is then removed, what is the potential difference across the capacitor? (2023)
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Solution
If the charge is removed, the potential difference across the capacitor becomes 0 volts.
Correct Answer: A — 0
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Q. A capacitor of capacitance C is charged to a voltage V. If the voltage is halved, what is the new energy stored in the capacitor?
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Solution
The energy stored in a capacitor is given by U = 1/2 CV². If the voltage is halved, the new energy becomes U' = 1/2 C(V/2)² = U/4.
Correct Answer: A — U/4
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Q. A capacitor of capacitance C is charged to a voltage V. What is the charge stored in the capacitor? (2023)
A.
C/V
B.
CV
C.
V/C
D.
C^2V
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Solution
The charge Q stored in a capacitor is given by Q = CV.
Correct Answer: B — CV
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Q. A capacitor of capacitance C is connected to a battery of voltage V. If the battery is removed and the capacitor is connected to another capacitor of capacitance 2C, what is the final voltage across the combination?
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Solution
When the charged capacitor C is connected to an uncharged capacitor 2C, the final voltage is V_final = Q_total / C_eq = V/(1 + 1/2) = V/3.
Correct Answer: B — V/2
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Q. A capacitor stores 5 microfarads of charge at a voltage of 10 volts. What is the energy stored in the capacitor?
A.
0.25 mJ
B.
0.5 mJ
C.
0.75 mJ
D.
1 mJ
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Solution
The energy stored in a capacitor is given by the formula E = 0.5 * C * V^2. Here, E = 0.5 * 5 * 10^-6 F * (10 V)^2 = 0.5 * 5 * 10^-6 * 100 = 0.25 mJ.
Correct Answer: B — 0.5 mJ
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Q. A capillary tube is dipped in water. The height of the water column in the tube is determined by which of the following?
A.
Surface tension and density of the liquid
B.
Only surface tension
C.
Only density of the liquid
D.
Viscosity of the liquid
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Solution
The height of the water column in a capillary tube is determined by both surface tension and the density of the liquid.
Correct Answer: A — Surface tension and density of the liquid
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Q. A capillary tube is dipped in water. What is the shape of the water surface inside the tube?
A.
Flat
B.
Concave
C.
Convex
D.
Irregular
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Solution
The water surface inside the capillary tube is concave due to the adhesive forces between water and the tube material being stronger than the cohesive forces among water molecules.
Correct Answer: B — Concave
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Q. A capillary tube is dipped into water. How high will the water rise in the tube if the radius is 1 mm?
A.
2.5 cm
B.
5 cm
C.
10 cm
D.
15 cm
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Solution
Using the capillary rise formula, h = (2γcosθ)/(ρgr), where γ is surface tension, θ is contact angle, ρ is density, g is acceleration due to gravity, and r is radius.
Correct Answer: B — 5 cm
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Q. A capillary tube is dipped into water. The height to which water rises in the tube is determined by:
A.
Surface tension and density of the liquid
B.
Only surface tension
C.
Only density of the liquid
D.
Viscosity of the liquid
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Solution
The height of the liquid column in a capillary tube is determined by both surface tension and the density of the liquid, as described by the capillary rise formula.
Correct Answer: A — Surface tension and density of the liquid
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Q. A capillary tube is dipped into water. The water rises in the tube due to which of the following?
A.
Surface tension and adhesion
B.
Surface tension and cohesion
C.
Only adhesion
D.
Only cohesion
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Solution
The rise of water in a capillary tube is due to both surface tension (which pulls the liquid up) and adhesion (the attraction between water molecules and the tube's surface).
Correct Answer: A — Surface tension and adhesion
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Q. A capillary tube is dipped into water. What will happen to the water level inside the tube?
A.
It will rise
B.
It will fall
C.
It will remain the same
D.
It will oscillate
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Solution
The water will rise in the capillary tube due to capillary action, which is a result of surface tension.
Correct Answer: A — It will rise
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Q. A capillary tube of radius 0.5 mm is dipped in water. What is the height of the water column raised in the tube? (Surface tension = 0.072 N/m, density of water = 1000 kg/m³)
A.
0.5 m
B.
0.1 m
C.
0.2 m
D.
0.3 m
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Solution
Using the formula h = 2γ/(ρgr), h = 2 × 0.072 N/m / (1000 kg/m³ × 9.81 m/s² × 0.0005 m) = 0.2 m.
Correct Answer: C — 0.2 m
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Q. A car accelerates from rest at a constant rate of 2 m/s². How far does it travel in 5 seconds? (2021)
A.
10 m
B.
20 m
C.
25 m
D.
50 m
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Solution
Using the equation of motion: s = ut + (1/2)at². Here, u = 0, a = 2 m/s², t = 5 s. So, s = 0 + (1/2)(2)(5²) = 25 m.
Correct Answer: C — 25 m
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Q. A car accelerates from rest at a rate of 2 m/s². What is its speed after 5 seconds? (2023)
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using the formula v = u + at, where u = 0, a = 2 m/s², and t = 5 s, we get v = 0 + 2 * 5 = 10 m/s.
Correct Answer: B — 10 m/s
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Q. A car accelerates from rest at a rate of 2 m/s². What is the net force acting on the car if its mass is 1000 kg?
A.
200 N
B.
500 N
C.
1000 N
D.
2000 N
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Solution
Using F = ma, F = 1000 kg * 2 m/s² = 2000 N.
Correct Answer: D — 2000 N
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Q. A car accelerates from rest to a speed of 100 km/h in 10 seconds. What is its acceleration? (2023)
A.
5 m/s²
B.
10 m/s²
C.
15 m/s²
D.
20 m/s²
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Solution
Acceleration = (Final Speed - Initial Speed)/Time = (100 km/h × (1000 m/3600 s))/10 s = 5 m/s²
Correct Answer: A — 5 m/s²
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Q. A car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
50 m
B.
100 m
C.
200 m
D.
400 m
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Solution
Using the formula d = ut + 0.5at², where u = 0, a = (20 m/s) / 10 s = 2 m/s², we get d = 0 + 0.5 * 2 * (10)² = 100 m.
Correct Answer: B — 100 m
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Q. A car accelerates from rest to a speed of 20 m/s in 5 seconds. What is its acceleration?
A.
2 m/s²
B.
4 m/s²
C.
5 m/s²
D.
10 m/s²
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Solution
Acceleration is calculated using the formula a = (final velocity - initial velocity) / time. Here, a = (20 m/s - 0 m/s) / 5 s = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the kinetic energy of the car at this speed? (2000)
A.
200 J
B.
1000 J
C.
2000 J
D.
4000 J
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Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 1000 kg × (20 m/s)² = 0.5 × 1000 × 400 = 200000 J.
Correct Answer: C — 2000 J
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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the work done by the engine?
A.
200,000 J
B.
100,000 J
C.
50,000 J
D.
400,000 J
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Solution
Work Done = Change in Kinetic Energy = 1/2 * m * (v^2) = 1/2 * 1000 * (20^2) = 200,000 J
Correct Answer: A — 200,000 J
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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the work done on the car? (2022)
A.
200,000 J
B.
100,000 J
C.
50,000 J
D.
400,000 J
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Solution
Work Done = Change in Kinetic Energy = 1/2 * m * (v^2 - u^2) = 1/2 * 1000 kg * (20 m/s)^2 = 200,000 J
Correct Answer: A — 200,000 J
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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the kinetic energy of the car at that speed?
A.
200,000 J
B.
100,000 J
C.
50,000 J
D.
400,000 J
Show solution
Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 1000 kg × (20 m/s)² = 0.5 × 1000 × 400 = 200,000 J.
Correct Answer: B — 100,000 J
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Q. A car accelerates from rest to a speed of 20 m/s. If the mass of the car is 1000 kg, what is the kinetic energy of the car at that speed? (2000)
A.
200 J
B.
1000 J
C.
2000 J
D.
4000 J
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Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 1000 kg × (20 m/s)² = 0.5 × 1000 × 400 = 200000 J.
Correct Answer: D — 4000 J
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Q. A car accelerates from rest to a speed of 20 m/s. If the mass of the car is 1000 kg, what is the work done by the engine?
A.
200,000 J
B.
100,000 J
C.
50,000 J
D.
400,000 J
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Solution
Work Done = Change in Kinetic Energy = 1/2 * m * (v^2) = 1/2 * 1000 * (20^2) = 200,000 J
Correct Answer: A — 200,000 J
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Q. A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the acceleration of the car? (2022)
A.
3 m/s²
B.
2 m/s²
C.
1 m/s²
D.
4 m/s²
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Solution
Acceleration (a) = (final velocity - initial velocity) / time = (30 m/s - 0) / 10 s = 3 m/s².
Correct Answer: A — 3 m/s²
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Q. A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
150 m
B.
300 m
C.
400 m
D.
600 m
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Solution
Using the formula d = ut + 0.5at², where u = 0, a = (30 m/s) / 10 s = 3 m/s², we get d = 0 + 0.5 * 3 * (10)² = 150 m.
Correct Answer: B — 300 m
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Q. A car accelerates from rest to a speed of 30 m/s. If the mass of the car is 800 kg, what is the work done on the car?
A.
360,000 J
B.
480,000 J
C.
600,000 J
D.
720,000 J
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Solution
Work done = Change in Kinetic Energy = 0.5 × mass × (final speed² - initial speed²) = 0.5 × 800 kg × (30 m/s)² = 360,000 J.
Correct Answer: B — 480,000 J
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Q. A car accelerates uniformly from rest to a speed of 20 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
100 m
B.
200 m
C.
300 m
D.
400 m
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Solution
Using the formula: distance = initial velocity * time + 0.5 * acceleration * time^2. Here, initial velocity = 0, final velocity = 20 m/s, time = 10 s. Acceleration = (final velocity - initial velocity) / time = 2 m/s². Distance = 0 + 0.5 * 2 * 10² = 100 m.
Correct Answer: B — 200 m
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Q. A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
100 m
B.
125 m
C.
150 m
D.
200 m
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Solution
Distance (s) = ut + (1/2)at^2. Here, a = (v-u)/t = (25-0)/10 = 2.5 m/s². So, s = 0 + (1/2)*2.5*10^2 = 125 m.
Correct Answer: B — 125 m
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Q. A car covers a distance of 240 km in 3 hours. What is its speed? (2022)
A.
70 km/h
B.
80 km/h
C.
90 km/h
D.
100 km/h
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Solution
Speed = Distance / Time = 240 km / 3 h = 80 km/h
Correct Answer: B — 80 km/h
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