NEET Practice: Physics 1

1. A body starts from rest and moves with uniform acceleration of 4 m/s². The distance covered in 5 seconds is:

(a) 25 m (b) 50 m (c) 100 m (d) 200 m

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Correct Answer: (b) 50 m

Explanation:
u = 0, a = 4 m/s², t = 5 s
s = ut + ½at² = 0 + ½×4×25 = 50 m

2. A projectile is launched with initial velocity 20 m/s at an angle of 30° to the horizontal. The maximum height reached is (g = 10 m/s²):

(a) 5 m (b) 10 m (c) 15 m (d) 20 m

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Correct Answer: (a) 5 m

Explanation:
H = (u² sin²θ) / (2g) = (400 × 0.25) / 20 = 5 m

3. The dimensional formula of angular momentum is:

(a) [ML²T⁻¹] (b) [MLT⁻¹] (c) [M²L²T⁻²] (d) [ML²T⁻²]

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Correct Answer: (a) [ML²T⁻¹]

Explanation:
Angular momentum L = mvr → [M][L][LT⁻¹] = [ML²T⁻¹]

4. A wire of resistance 10 Ω is stretched to double its length. Its new resistance will be:

(a) 10 Ω (b) 20 Ω (c) 40 Ω (d) 5 Ω

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Correct Answer: (c) 40 Ω

Explanation:
Volume constant → R becomes 4 times = 40 Ω

5. The work done in moving a charge of 2 C across a potential difference of 10 V is:

(a) 5 J (b) 10 J (c) 20 J (d) 0.2 J

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Correct Answer: (c) 20 J

Explanation:
W = qV = 2 × 10 = 20 J

6. A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2. Minimum force required to move the block (g=10 m/s²) is:

(a) 5 N (b) 10 N (c) 15 N (d) 20 N

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Correct Answer: (b) 10 N

Explanation:
Limiting friction = μmg = 0.2 × 5 × 10 = 10 N

7. The equivalent capacitance of two 4 μF capacitors connected in series is:

(a) 8 μF (b) 4 μF (c) 2 μF (d) 1 μF

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Correct Answer: (c) 2 μF

Explanation:
1/C_eq = 1/4 + 1/4 → C_eq = 2 μF

8. The frequency of a simple pendulum is independent of:

(a) Length of pendulum (b) Acceleration due to gravity (c) Mass of the bob (d) Amplitude (small angles)

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Correct Answer: (c) Mass of the bob

Explanation:
T = 2π√(L/g). Frequency does not depend on mass.

9. In a transformer, primary turns = 100, secondary turns = 400, input voltage = 220 V. Output voltage is:

(a) 55 V (b) 110 V (c) 440 V (d) 880 V

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Correct Answer: (d) 880 V

Explanation:
V_s = 220 × (400/100) = 880 V

10. The de Broglie wavelength of an electron accelerated through 100 V is approximately:

(a) 1.23 Å (b) 0.123 Å (c) 12.3 Å (d) 0.0123 Å

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Correct Answer: (a) 1.23 Å

Explanation:
λ ≈ 12.27 / √100 ≈ 1.23 Å