⇦ | ⇨ |
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 solenoid has length 0.4 cm, radius 1 cm and 400 turns of wire. If a current
- The thermo e.m.f. E in volts of a certain thermocouple is found to vary
- In which of the processes, does the internal energy of the system remain constant?
- The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain
- In Young’s double slit experiment, the intensity of light coming from the first slit
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A solenoid has length 0.4 cm, radius 1 cm and 400 turns of wire. If a current
- The thermo e.m.f. E in volts of a certain thermocouple is found to vary
- In which of the processes, does the internal energy of the system remain constant?
- The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain
- In Young’s double slit experiment, the intensity of light coming from the first slit
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.