Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

There is a uniform electric field of strength 10^3V/m   along y-axis. A body of mass 1g and charge 10–6C is projected into the field from origin along the positive x-axis with a velocity 10m/s. Its speed in m/s after 10s is (Neglect gravitation)

  • Option 1)

    10

  • Option 2)

    5\sqrt2

  • Option 3)

    10\sqrt2

  • Option 4)

    20

 

Answers (1)

best_answer

As we have learned

If ß is the angle made by v -

\tan \beta = \frac{V_{y}}{V_{x}}= \frac{mv_{y}}{QEt}

 

- wherein

 

 

Body moves along the parabolic path.

For vertical motion : By using v = u + at

ÞV_y= 0+\frac{QE}{m}t= \frac{10^{-6}*10^3}{10^{-3}}*10= 10m/sec

For horizontal motion – It’s horizontal velocity remains the same i.e. after 10 sec, horizontal velocity of body vx = 10 m/sec.

Velocity after 10 secv= \sqrt{v^2_x+v^2_y}= 10\sqrt2 m/sec 

 

 


Option 1)

10

Option 2)

5\sqrt2

Option 3)

10\sqrt2

Option 4)

20

Posted by

Avinash

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question