A particle A has charge +q and particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through same electrical potential difference, the ratio of their speed VA : VB will becomeA. 2:1 B. 1:2 C. 1:4 D. 4:1
Manvendra Singh chahar , 11 Years ago
Grade Upto college level
2 Answers
Archana
Last Activity: 7 Years ago
Ka/Kb=q×v/4q×v =1/2mVa^2/1/2mVb^2 Va^2/Vb^2=1/4 Va/VB=1/2 or Va:Vb=1:2 It is the ratio between the kinectic energies between the two particles.... Hope this is useful for u guys and this is the correct answer for this question....
Rishi Sharma
Last Activity: 4 Years ago
Dear Student, Please find below in the attached image the solution to your problem.
Thanks and Regards
Provide a better Answer & Earn Cool Goodies
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free
Other Related Questions on engineering entrance exams