A ball is thrown upward with velocity 30 m/s. Time to reach maximum height (g=10) is MCQ with Answer and Explanation

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A ball is thrown upward with velocity 30 m/s. Time to reach maximum height (g=10) is

A. 1.5 s

B. 2 s

C. 6 s

D. 3 s

Answer: Option D

Solution (By JKSSB Mock Tests)

v = u - gt, at top v=0 => t = u/g = 30/10 = 3 s.

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Question #1

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A student measures length as 3.600 cm using vernier calliper of least count 0.001 cm. Number of significant figures is

A. 3

B. 4

C. 2

D. 1

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Correct Answer: Option B

Explanation:
3.600 has 4 significant figures (3,6,0,0). Trailing zeros after decimal are significant due to precision.

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Question #2

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A wire of resistance 10 Ω is stretched to double its length. Its new resistance is

A. 5 Ω

B. 40 Ω

C. 20 Ω

D. 10 Ω

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Correct Answer: Option B

Explanation:
Resistance R = ρL/A. Volume constant: A ∝ 1/L. Stretching length to 2L reduces area to A/2. New R' = ρ(2L)/(A/2) = 4ρL/A = 4R = 40 Ω. Resistance proportional to (length)² when volume constant.

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Question #3

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Critical angle for glass (n=1.5) to air:

A. 90°

B. 60°

C. 42°

D. 30°

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Correct Answer: Option C

Explanation:
sinθ_c = n₂/n₁ = 1/1.5 = 2/3 ⇒ θ_c = sin⁻¹(0.667) ≈ 42°. Total internal reflection condition. Memory tip: 'θ_c = sin⁻¹(1/n) for denser to air'. Optics calculation frequently tested in competitive exams with standard refractive indices.

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