A 60 W bulb is used for 5 hours daily. How much electrical energy is consumed by the bulb in 30 days?
A. 0.9 kWh
B. 9 kWh
C. 90 kWh
D. 900 kWh
Answer: Option B
Solution (By JKSSB Mock Tests)
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.
Explanation:
Beats (periodic amplitude variation from interference of close frequencies) are used to tune instruments: adjust until beats disappear (frequencies match). Also, in Kundt's tube or other methods, beat frequency helps measure sound speed by comparing known and unknown frequencies. Memory tip: 'Beats: tuning (zero beat = matched frequency); speed measurement (beat frequency = |f₁-f₂|)'. This application question tests wave phenomena uses, frequently appearing in competitive exams. Always link physical phenomena to practical applications; competitive exams emphasize real-world relevance of physics concepts.
Explanation:
Pressure has dimensions [ML⁻¹T⁻²]. Energy per unit volume: Energy is [ML²T⁻²], volume is [L³], so [ML²T⁻²]/[L³] = [ML⁻¹T⁻²], matching pressure. Force per unit length is [MLT⁻²]/[L] = [MT⁻²]. Momentum per unit area is [MLT⁻¹]/[L²] = [ML⁻¹T⁻¹]. Power per unit volume is [ML²T⁻³]/[L³] = [ML⁻¹T⁻³]. This dimensional equivalence explains why pressure appears in Bernoulli's equation alongside energy density. Memory aid: Pressure and energy density both represent stored energy per spatial dimension.
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