KE increased by 100%. Momentum increase percentage:
A. 50%
B. 41.4%
C. 200%
D. 100%
Answer: Option B
Solution (By JKSSB Mock Tests)
KE ∝ p². Double KE ⇒ p_new = √2 p_old ⇒ increase = (√2 - 1)×100% ≈ 41.4%. Memory tip: 'KE ∝ p²; %Δp ≈ ½ %ΔKE for small changes, but use exact for large'. Energy-momentum relation frequently tested in competitive exams.
Explanation:
Newton's Law of Cooling is a practical approximation stating that the rate of heat loss of a body is directly proportional to the difference in temperatures between the body and its immediate environment, provided this temperature difference is relatively small. The equation is dQ/dt = -k(T_body - T_surroundings).
Explanation:
Pascal's law: pressure applied to confined fluid transmits equally. Force multiplication: F₂/F₁ = A₂/A₁. Small force on small area creates large force on large area. Memory aid: 'Pascal = pressure transmission; hydraulic advantage = area ratio'. Application question testing fluid mechanics principles, common in competitive exams linking theory to devices.
Explanation:
Candela (cd) is SI base unit for luminous intensity. Lumen: luminous flux; Lux: illuminance; Watt: power. Memory aid: 'Candela = candle power; base unit for perceived light'. Unit knowledge frequently tested in competitive exams.
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