| ⇦ | 
| ⇨ |
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: - A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- The focal lengths of a converging lens are fᵥ and fᵣ for violet and red lights
- A metal rod of length l cuts across a uniform magnetic field B with a velocity v.
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
- A solid sphere is rotating in free space.If the radius of the sphere is increased
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- The focal lengths of a converging lens are fᵥ and fᵣ for violet and red lights
- A metal rod of length l cuts across a uniform magnetic field B with a velocity v.
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
- A solid sphere is rotating in free space.If the radius of the sphere is increased
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
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.