A body is thrown vertically upward with speed 40 m/s. Distance travelled in last second of ascent (g=10)
A. 15 m
B. 5 m
C. 35 m
D. 20 m
Answer: Option B
Solution (By JKSSB Mock Tests)
Time of ascent = u/g = 4 s. Distance in last second (4th) = ½g(1)²? Actually, during last second of upward journey, velocity becomes zero. Distance = ½g(1)² = 5 m. Formula: s = u - g/2 (2n-1) with n=4? Let's compute: s_nth = u - (g/2)(2n-1). For n=4 (4th second), s = 40 - 5×7 = 40 - 35 = 5 m. Yes.
Explanation:
Interference (constructive/destructive superposition) is a characteristic wave phenomenon, requiring coherent sources and path difference. Young's double-slit experiment demonstrated light interference, confirming its wave nature. Particle nature is shown by photoelectric effect. Memory aid: 'Interference/diffraction ⇒ wave nature; photoelectric effect ⇒ particle nature'. This conceptual question tests wave-particle duality fundamentals, frequently examined in competitive exams. Always link specific phenomena to the aspect of light they demonstrate; competitive exams often combine both aspects in advanced questions.
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