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The acceleration of an electron in an electric field of magnitude 50 V/cm, if e/m value of the electron is 1.76×10¹¹ C/kg will be
Options
(a) 8.8×10¹⁴ m/sec²
(b) 3×10¹³ m/sec²
(c) 5.4×10¹² m/sec²
(d) Zero
Correct Answer:
8.8×10¹⁴ m/sec²
Explanation:
Acceleration = a = eE / m
⇒ a = 1.76 × 10¹¹ × 50 × 10 ¹²
⇒ a = 8.8 × 10 ¹⁴ m/ sec²
Related Questions: - The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field
- A particle moves in xy plane according to the equation x = 4t² + 5t + 16 and y = 5t
- An inclined plane of length 5.60m making an angle of 45⁰ with the horizontal is placed
- The electric field at a distance 3R / 2 from the centre of a charged conducting
- When the rate of flow of charge through a metallic conductor
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field
- A particle moves in xy plane according to the equation x = 4t² + 5t + 16 and y = 5t
- An inclined plane of length 5.60m making an angle of 45⁰ with the horizontal is placed
- The electric field at a distance 3R / 2 from the centre of a charged conducting
- When the rate of flow of charge through a metallic conductor
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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E=50v/cm
=5000v/m
e/m=1.76×10power-11
F=ma
F=qĒ
ma=qĒ
a= e/m×Ē
=1.76×10 power11 × 5000
=8.8×10 power 14 m/sec sq.