⇦ | ⇨ |
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 self induced e.m.f. in a 0.1 H coil when the current in it is changing
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- For transistor action (1) Base, emitter and collector regions should have similar size
- A beam of electron passes undeflected through mutually perpendicular
- A point charge q is situated at a distance r on axis from one end of a thin
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The self induced e.m.f. in a 0.1 H coil when the current in it is changing
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- For transistor action (1) Base, emitter and collector regions should have similar size
- A beam of electron passes undeflected through mutually perpendicular
- A point charge q is situated at a distance r on axis from one end of a thin
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.