Q. What is the effect of increasing the molar mass of a gas on its average kinetic energy at a constant temperature?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
The average kinetic energy of gas molecules is independent of molar mass and is solely dependent on temperature. Therefore, it remains the same at constant temperature.
Correct Answer:
C
— Remains the same
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Q. What is the effect of increasing the radius of a rolling object on its moment of inertia?
A.
It decreases the moment of inertia.
B.
It increases the moment of inertia.
C.
It has no effect on the moment of inertia.
D.
It depends on the mass of the object.
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Solution
The moment of inertia increases with the square of the radius, as seen in the formula I = k m r^2, where k is a constant depending on the shape.
Correct Answer:
B
— It increases the moment of inertia.
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Q. What is the effect of increasing the resistance (R) in an RC charging circuit on the time constant (τ)?
A.
τ increases
B.
τ decreases
C.
τ remains the same
D.
τ becomes zero
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Solution
Increasing the resistance in an RC circuit increases the time constant τ, since τ = R * C.
Correct Answer:
A
— τ increases
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Q. What is the effect of increasing the resistance in an RC circuit on the time constant?
A.
It decreases the time constant
B.
It has no effect
C.
It increases the time constant
D.
It doubles the time constant
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Solution
Increasing the resistance in an RC circuit increases the time constant, as τ = R * C.
Correct Answer:
C
— It increases the time constant
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Q. What is the effect of increasing the temperature of a gas on its average molecular speed?
A.
Average speed decreases
B.
Average speed remains the same
C.
Average speed increases
D.
Average speed becomes zero
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Solution
Increasing the temperature of a gas increases the average kinetic energy of its molecules, which in turn increases their average speed.
Correct Answer:
C
— Average speed increases
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Q. What is the effect of increasing the temperature on the speed of sound in air?
A.
It decreases
B.
It increases
C.
It remains the same
D.
It becomes zero
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Solution
The speed of sound in air increases with an increase in temperature.
Correct Answer:
B
— It increases
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Q. What is the effect of increasing the wavelength of light on the angle of refraction when entering a medium?
A.
It increases the angle
B.
It decreases the angle
C.
It has no effect
D.
It causes total internal reflection
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Solution
Increasing the wavelength generally increases the angle of refraction due to the change in refractive index.
Correct Answer:
A
— It increases the angle
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Q. What is the electric field (E) at a distance of 2 meters from a point charge of 5 microcoulombs?
A.
1125 N/C
B.
450 N/C
C.
225 N/C
D.
900 N/C
Show solution
Solution
Using Coulomb's law, E = k * |q| / r^2, where k = 8.99 x 10^9 N m^2/C^2, q = 5 x 10^-6 C, and r = 2 m. E = (8.99 x 10^9) * (5 x 10^-6) / (2^2) = 1125 N/C.
Correct Answer:
A
— 1125 N/C
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Q. What is the electric field (E) at a distance of 2m from a point charge of 10μC?
A.
2250 N/C
B.
500 N/C
C.
4500 N/C
D.
1000 N/C
Show solution
Solution
Using Coulomb's law, E = k * |Q| / r^2 = (8.99 x 10^9 N m²/C²) * (10 x 10^-6 C) / (2 m)^2 = 2247.5 N/C, approximately 2250 N/C.
Correct Answer:
A
— 2250 N/C
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Q. What is the electric field (E) due to a point charge (Q) at a distance (r)?
A.
E = k * Q / r^2
B.
E = k * Q * r^2
C.
E = Q / (4 * π * ε * r)
D.
E = Q / (4 * π * ε * r^2)
Show solution
Solution
The electric field due to a point charge is given by E = k * Q / r^2.
Correct Answer:
A
— E = k * Q / r^2
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Q. What is the electric field (E) due to a uniformly charged infinite plane sheet with surface charge density (σ)?
A.
E = σ / (2 * ε)
B.
E = σ / ε
C.
E = σ / (4 * π * ε)
D.
E = 0
Show solution
Solution
The electric field due to a uniformly charged infinite plane sheet is E = σ / (2 * ε).
Correct Answer:
A
— E = σ / (2 * ε)
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Q. What is the electric field at a distance of 2 m from a point charge of +5 µC?
A.
0.56 N/C
B.
1.12 N/C
C.
2.24 N/C
D.
4.48 N/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (2 m)^2 = 1.12 N/C.
Correct Answer:
C
— 2.24 N/C
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Q. What is the electric field between two parallel plates separated by 0.1 m with a potential difference of 100 V?
A.
1000 N/C
B.
500 N/C
C.
100 N/C
D.
10 N/C
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Solution
E = V/d = 100 V / 0.1 m = 1000 N/C.
Correct Answer:
A
— 1000 N/C
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Q. What is the electric field due to a point charge of -4 µC at a distance of 0.25 m?
A.
-5760 N/C
B.
-1440 N/C
C.
5760 N/C
D.
1440 N/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (4 x 10^-6 C) / (0.25 m)^2 = -5760 N/C.
Correct Answer:
A
— -5760 N/C
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Q. What is the electric field inside a parallel plate capacitor with a potential difference of 100 V and plate separation of 0.01 m?
A.
10000 N/C
B.
1000 N/C
C.
100 N/C
D.
10 N/C
Show solution
Solution
E = V/d = 100 V / 0.01 m = 10000 N/C.
Correct Answer:
A
— 10000 N/C
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Q. What is the electric field strength at a distance of 0.2 m from a point charge of +5 µC?
A.
112.5 N/C
B.
112.5 kN/C
C.
25 N/C
D.
25 kN/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (0.2 m)^2 = 112.5 N/C.
Correct Answer:
A
— 112.5 N/C
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Q. What is the electric field strength at a distance of 2 m from a +5 µC point charge?
A.
1.125 N/C
B.
2.25 N/C
C.
0.5625 N/C
D.
0.75 N/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * |5 x 10^-6 C| / (2 m)^2 = 1.125 N/C.
Correct Answer:
B
— 2.25 N/C
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Q. What is the electric field strength at a distance of 2 m from a charge of +5 µC?
A.
1.125 N/C
B.
2.25 N/C
C.
3.75 N/C
D.
4.5 N/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * |5 x 10^-6 C| / (2 m)^2 = 1.125 N/C.
Correct Answer:
A
— 1.125 N/C
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Q. What is the electric field strength at a distance of 2 m from a point charge of +5 µC?
A.
1.12 N/C
B.
0.56 N/C
C.
2.25 N/C
D.
0.75 N/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (2 m)^2 = 1.12 N/C.
Correct Answer:
A
— 1.12 N/C
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Q. What is the electric field strength between two parallel plates separated by 0.1 m with a potential difference of 100 V?
A.
1000 N/C
B.
500 N/C
C.
100 N/C
D.
2000 N/C
Show solution
Solution
The electric field (E) is given by E = V/d. Here, E = 100V / 0.1m = 1000 N/C.
Correct Answer:
A
— 1000 N/C
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Q. What is the electric potential (V) across a capacitor (C) charged to a charge (Q)?
A.
V = Q/C
B.
V = C/Q
C.
V = Q*C
D.
V = C^2/Q
Show solution
Solution
The electric potential across a capacitor is given by the formula V = Q/C, where Q is the charge and C is the capacitance.
Correct Answer:
A
— V = Q/C
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Q. What is the electric potential (V) at a distance (r) from a point charge (Q)?
A.
V = k * Q / r
B.
V = k * Q * r
C.
V = Q / (4 * π * ε * r^2)
D.
V = Q / (4 * π * ε * r)
Show solution
Solution
The electric potential due to a point charge is given by V = k * Q / r, where k is Coulomb's constant.
Correct Answer:
A
— V = k * Q / r
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Q. What is the electric potential at a point 1 m away from a +1 C charge?
A.
9 N/C
B.
1 V
C.
8.99 V
D.
0 V
Show solution
Solution
Electric potential V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 C) / (1 m) = 8.99 V.
Correct Answer:
C
— 8.99 V
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Q. What is the electric potential at a point 1 m away from a charge of +1 µC?
A.
9 kV
B.
1 kV
C.
0.9 kV
D.
0.1 kV
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 1 m = 8.99 kV.
Correct Answer:
A
— 9 kV
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Q. What is the electric potential at a point 1 m away from a charge of +3 µC?
A.
9 V
B.
27 V
C.
18 V
D.
36 V
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (3 x 10^-6 C) / (1 m) = 26.97 V, approximately 27 V.
Correct Answer:
B
— 27 V
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Q. What is the electric potential at a point 3 m away from a charge of +1 µC?
A.
3000 V
B.
9000 V
C.
300 V
D.
900 V
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 3 m = 3000 V.
Correct Answer:
C
— 300 V
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Q. What is the electric potential at a point 3 m away from a charge of +2 µC?
A.
0.6 V
B.
1.2 V
C.
2.4 V
D.
4.8 V
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (2 x 10^-6 C) / 3 m = 5.99 V.
Correct Answer:
B
— 1.2 V
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Q. What is the electric potential energy of a charge of +2 µC placed in an electric field of 500 N/C at a distance of 0.1 m?
A.
0.1 mJ
B.
0.2 mJ
C.
0.3 mJ
D.
0.4 mJ
Show solution
Solution
Electric potential energy U = q * E * d = (2 x 10^-6 C) * (500 N/C) * (0.1 m) = 0.1 mJ.
Correct Answer:
B
— 0.2 mJ
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Q. What is the electric potential energy of a charge of +3 µC placed in an electric field of 2000 N/C at a distance of 0.5 m?
A.
3 J
B.
1.5 J
C.
0.3 J
D.
0.6 J
Show solution
Solution
Potential energy U = q * E * d = (3 x 10^-6 C) * (2000 N/C) * (0.5 m) = 3 J.
Correct Answer:
B
— 1.5 J
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Q. What is the electric potential energy of a system of two charges of +1 µC and -1 µC separated by 0.1 m?
A.
-0.09 J
B.
0.09 J
C.
0.18 J
D.
0.36 J
Show solution
Solution
Potential energy U = k * q1 * q2 / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) * (-1 x 10^-6 C) / 0.1 m = -0.09 J.
Correct Answer:
B
— 0.09 J
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