| ⇦ | 
| ⇨ |
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 monoatomic gas at a pressure P, having a volume V expands isothermally
- A polished metal plate has a rough and black-spot. It is heated to 1400 K
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
- If alternating source of primary coil on transformer is replaced by a Laclanche cell
- A particle of mass m, charge Q and kinetic enery T enters a transverse
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A monoatomic gas at a pressure P, having a volume V expands isothermally
- A polished metal plate has a rough and black-spot. It is heated to 1400 K
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
- If alternating source of primary coil on transformer is replaced by a Laclanche cell
- A particle of mass m, charge Q and kinetic enery T enters a transverse
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.