If X = a²b³/c², and percentage errors in a, b, c are 2% each, the maximum percentage error in X is MCQ with Answer and Explanation
Question
Report Error
Discuss this Question (0)
Practice More Physics Questions
Question #1
Report Error
A 60 W bulb is used for 5 hours daily. How much electrical energy is consumed by the bulb in 30 days?
Correct Answer: Option B
Explanation:
Electrical energy (E) = Power (P) × Time (t). Power = 60 W = 0.06 kW. Time per day = 5 hours. Total time in 30 days = 5 × 30 = 150 hours. Energy = 0.06 kW × 150 h = 9 kWh (or 9 units of electricity). One kWh is the commercial unit of energy.
Electrical energy (E) = Power (P) × Time (t). Power = 60 W = 0.06 kW. Time per day = 5 hours. Total time in 30 days = 5 × 30 = 150 hours. Energy = 0.06 kW × 150 h = 9 kWh (or 9 units of electricity). One kWh is the commercial unit of energy.
Question #2
Report Error
A spring of force constant k is stretched by a length x. The work done in stretching it further by the same length x is:
Correct Answer: Option B
Explanation:
Work done to stretch spring from 0 to x: W₁ = ½kx². Work done from 0 to 2x: W₂ = ½k(2x)² = 2kx². Thus work for additional stretch from x to 2x: ΔW = W₂ - W₁ = 2kx² - ½kx² = ³/₂kx². Spring force is variable (F=kx), so work is integral of F·dx, yielding parabolic energy storage. Memory aid: Elastic potential energy U = ½kx²; always calculate difference for incremental work. This problem tests understanding of work done by variable forces, a common theme in energy conservation questions in competitive exams.
Work done to stretch spring from 0 to x: W₁ = ½kx². Work done from 0 to 2x: W₂ = ½k(2x)² = 2kx². Thus work for additional stretch from x to 2x: ΔW = W₂ - W₁ = 2kx² - ½kx² = ³/₂kx². Spring force is variable (F=kx), so work is integral of F·dx, yielding parabolic energy storage. Memory aid: Elastic potential energy U = ½kx²; always calculate difference for incremental work. This problem tests understanding of work done by variable forces, a common theme in energy conservation questions in competitive exams.
Question #3
Report Error
The energy of a photon is directly proportional to its:
Correct Answer: Option C
Explanation:
Photon energy E = hν, where h is Planck's constant, ν is frequency. Thus E ∝ ν. Since c = νλ, E = hc/λ, so E ∝ 1/λ (inversely proportional to wavelength). Amplitude relates to intensity (number of photons), not individual photon energy. Speed is constant (c) in vacuum. Memory aid: 'Higher frequency (shorter wavelength) photons have more energy (e.g., gamma > radio)'. This quantum physics concept is frequently tested in competitive exams. Always use E = hν for photon energy calculations; avoid confusing wave amplitude with photon energy.
Photon energy E = hν, where h is Planck's constant, ν is frequency. Thus E ∝ ν. Since c = νλ, E = hc/λ, so E ∝ 1/λ (inversely proportional to wavelength). Amplitude relates to intensity (number of photons), not individual photon energy. Speed is constant (c) in vacuum. Memory aid: 'Higher frequency (shorter wavelength) photons have more energy (e.g., gamma > radio)'. This quantum physics concept is frequently tested in competitive exams. Always use E = hν for photon energy calculations; avoid confusing wave amplitude with photon energy.
No comments yet. Be the first to start the discussion!