A car moving at 72 km/h is brought to rest in 10 seconds. The deceleration is
A. 2 m/s²
B. 20 m/s²
C. 7.2 m/s²
D. 2.5 m/s²
Answer: Option A
Solution (By JKSSB Mock Tests)
72 km/h = 72 × (5/18) = 20 m/s. Initial velocity u = 20 m/s, final v = 0, t = 10 s. Using v = u + at => 0 = 20 + a×10 => a = -2 m/s², so deceleration magnitude is 2 m/s². Conversion factor 5/18 is essential: multiply km/h by 5/18 to get m/s.
Explanation:
Formula: n = sin[(A+δ_m)/2]/sin(A/2). δ_m depends on prism angle A and refractive index n. Memory tip: 'δ_m increases with A and n; used to measure n'. Optics formula frequently tested in competitive exams.
Explanation:
BJT common-emitter current gain β typically 20-200. Small base current controls larger collector current. Memory aid: 'β = collector/base current ratio; value 20-200 for standard transistors'. Electronics parameter frequently tested in competitive exams.
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