OnlineCBT Logo OnlineCBT

Physics · JEE | Electric Charges and Fields

हिन्दी

An electron falls through a distance of \( 1.5\text{ cm} \) in a uniform electric field of magnitude \( 2.0 \times 10^4\text{ N/C} \). The time of fall starting from rest is approximately:

  1. A
    \( 1.3 \times 10^{-8}\text{ s} \)
  2. B
    \( 2.9 \times 10^{-9}\text{ s} \)
  3. C
    \( 1.6 \times 10^{-10}\text{ s} \)
  4. D
    \( 4.1 \times 10^{-7}\text{ s} \)
View Answer & Solution

Correct answer · Option B

\( 2.9 \times 10^{-9}\text{ s} \)

Explanation

Acceleration \( a = \frac{eE}{m} \). Distance \( s = \frac{1}{2}at^2 \Rightarrow t = \sqrt{\frac{2ms}{eE}} \approx 2.9 \times 10^{-9}\text{ s} \).
Address Contact

+91 9694780911

E-mail

onlinecbtportal@gmail.com

Helpline Number

+91 9694780911


Click To Download
Get it on Google Play

ऐप डाउनलोड करने
के लिए Google Play पर
उपलब्ध है

WhatsApp Chat