Central Exam — General Science
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Showing 201–207 of 207 questions
Q.201 Medium Physics
A ball is thrown vertically upward with initial velocity 20 m/s. What maximum height does it reach? (g = 10 m/s²)
A 10 m
B 20 m
C 30 m
D 40 m
Correct Answer:  B. 20 m
Explanation:

At maximum height, v = 0. Using v² = u² - 2gh: 0 = 20² - 2(10)h. 20h = 400. h = 20 m.

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Q.202 Medium Physics
Two capacitors of 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
A 5 μF
B 1.2 μF
C 6 μF
D 0.5 μF
Correct Answer:  B. 1.2 μF
Explanation:

For capacitors in series: 1/C = 1/C₁ + 1/C₂ = 1/2 + 1/3 = 5/6. Therefore C = 6/5 = 1.2 μF.

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Q.203 Easy Physics
Which of the following has the highest refractive index?
A Air
B Water
C Diamond
D Glass
Correct Answer:  C. Diamond
Explanation:

Refractive indices: Air ≈ 1, Water ≈ 1.33, Glass ≈ 1.5, Diamond ≈ 2.42. Diamond has the highest refractive index among these options.

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Q.204 Medium Physics
A body moves in a circle of radius 5 m with constant speed 10 m/s. What is the centripetal acceleration?
A 2 m/s²
B 20 m/s²
C 50 m/s²
D 100 m/s²
Correct Answer:  B. 20 m/s²
Explanation:

Centripetal acceleration a = v²/r = (10)²/5 = 100/5 = 20 m/s².

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Q.205 Easy Physics
A current of 5 A flows through a resistor for 10 seconds. How much charge passes through it?
A 0.5 C
B 2 C
C 50 C
D 500 C
Correct Answer:  C. 50 C
Explanation:

Charge Q = I × t, where I = 5 A and t = 10 s. Q = 5 × 10 = 50 Coulombs.

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Q.206 Hard Physics
An electron undergoes a transition from n=3 to n=1 in a hydrogen atom. Which series does this transition belong to?
A Paschen series
B Balmer series
C Lyman series
D Brackett series
Correct Answer:  C. Lyman series
Explanation:

The Lyman series involves transitions to n=1 from higher energy levels. Balmer series ends at n=2, Paschen at n=3, and Brackett at n=4.

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Q.207 Hard Physics
An object of mass 2 kg experiences a net force of 10 N. If it starts from rest, what will be its velocity after traveling 5 m?
A 5 m/s
B √50 m/s
C 10 m/s
D √5 m/s
Correct Answer:  A. 5 m/s
Explanation:

Using F = ma: a = F/m = 10/2 = 5 m/s². Using v² = u² + 2as: v² = 0 + 2(5)(5) = 50. v = √50 ≈ 7.07 m/s. Checking options again using work-energy: Work = KE; 10×5 = (1/2)(2)v²; v = 5√2 m/s. The closest correct approach gives v² = 50, so v = 5√2 m/s or approximately 7.07 m/s. Option B matches √50.

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