Central Exam — General Science
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Showing 31–37 of 37 questions
Q.31 Hard Physics
A 5 kg mass is attached to a spring with spring constant 200 N/m. What is the period of oscillation?
A π/10 s
B π/5 s
C π s
D 2π s
Correct Answer:  C. π s
Explanation:

Period T = 2π√(m/k) = 2π√(5/200) = 2π√(1/40) = 2π × (1/(2√10)) = π/√10 ≈ π s (approximately).

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Q.32 Hard Physics
Which of the following best explains why objects appear to move slower underwater?
A Water exerts a stopping force on moving objects
B Light refracts when passing from water to air, changing apparent position and velocity
C Water has higher density than air
D The refractive index of water is 4/3
Correct Answer:  B. Light refracts when passing from water to air, changing apparent position and velocity
Explanation:

Due to refraction at the water-air interface, light bends as it exits water. This causes objects to appear at different positions and moving at different speeds than they actually are—an optical illusion, not actual motion change.

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Q.33 Hard Physics
A satellite orbits Earth at radius r with orbital velocity v. If the orbital radius is doubled, what happens to the orbital velocity?
A It doubles
B It becomes half
C It decreases by factor √2
D It increases by factor √2
Correct Answer:  C. It decreases by factor √2
Explanation:

From orbital mechanics: v = √(GM/r). If r becomes 2r, then v_new = √(GM/2r) = v/√2. Velocity decreases by factor √2.

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Q.34 Hard Physics
A photon with frequency 5 × 10¹⁴ Hz is incident on a metal with work function 2 eV. What is the maximum kinetic energy of the photoelectron? (h = 6.63 × 10⁻³⁴ J·s, 1 eV = 1.6 × 10⁻¹⁹ J)
A 0.07 eV
B 1.05 eV
C 2.08 eV
D 3.31 eV
Correct Answer:  A. 0.07 eV
Explanation:

# Photoelectric Effect - Maximum Kinetic Energy

The photoelectric effect describes how photons transfer their energy to electrons in a metal, with excess energy becoming kinetic energy after overcoming the work function barrier.

Step 1: Calculate the Energy of the Incident Photon

The energy of a photon is determined by its frequency using Planck's equation.

\[E_{photon} = h \times f = 6.63 \times 10^{-34} \times 5 \times 10^{14}\]
\[E_{photon} = 33.15 \times 10^{-20} \text{ J} = 3.315 \times 10^{-19} \text{ J}\]

Step 2: Convert Photon Energy to eV

Convert joules to electron volts using the given conversion factor.

\[E_{photon} = \frac{3.315 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.07 \text{ eV}\]

Step 3: Apply Einstein's Photoelectric Equation

The maximum kinetic energy equals the photon energy minus the work function (threshold energy).

\[KE_{max} = E_{photon} - \phi\]
\[KE_{max} = 2.07 - 2.0 = 0.07 \text{ eV}\]

The maximum kinetic energy of the photoelectron is 0.07 eV.

Answer: (A) 0.07 eV

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Q.35 Hard Physics
Which statement about acoustic impedance is correct?
A Acoustic impedance is directly proportional to frequency
B Acoustic impedance = density × velocity of sound
C Acoustic impedance has SI unit of W/m²
D Acoustic impedance increases with temperature in solids
Correct Answer:  B. Acoustic impedance = density × velocity of sound
Explanation:

Acoustic impedance (Z) = ρ × c, where ρ is density and c is sound velocity. SI unit is kg/(m²·s). It's crucial for understanding sound reflection and transmission.

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Q.36 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.37 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 √2 m/s(≈7.07 m/s)
B √50 m/s
C 10 m/s
D √5 m/s
Correct Answer:  A. 5 √2 m/s(≈7.07 m/s)
Explanation:

This problem requires using Newton's Second Law to find acceleration, then applying kinematic equations to determine final velocity.

Step 1: Find the Acceleration Using Newton's Second Law

Newton's Second Law states that the net force equals mass times acceleration. We need to rearrange to solve for acceleration.

\[F = ma \implies a = \frac{F}{m} = \frac{10}{2} = 5 \text{ m/s}^2\]

Step 2: Apply the Kinematic Equation to Find Final Velocity

Since the object starts from rest (initial velocity = 0) and travels a distance of 5 m with constant acceleration, we use the equation that relates velocity, acceleration, and distance.

\[v^2 = u^2 + 2as\]
\[v^2 = 0^2 + 2 \times 5 \times 5 = 50\]
\[v = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \text{ m/s} \approx 7.07 \text{ m/s}\]

The velocity after traveling 5 m is 5√2 m/s (≈ 7.07 m/s). The answer is (A).

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