Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
705 views
in JEE by (1.3k points)
edited by

Two identical balls each having a charge of 2×10-7 and a mass of 100g are suspended from a common point by two insulating strings each 50 cms long. The balls are held at a separation 5 cm apart and then released . find the
a)the electric force on one of the charged balls.
b)the components of the resultant force on it along and perpendicular to the string.
c)the tension in the string
d)the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release

1 Answer

0 votes
by (12.6k points)
 
Best answer

b) Lets suppose that the angle made by the string and the vertical line is x.

Then sin x = 5/2x50 = 1/20

so cos x = 0.9987

Since the tension in the string will balance the force along the string, The net force along the string will be zero.

The force perpendicular to the string will be 

F= mgsinx - Fecosx = 0.1x9.8x0.05 - 0.9987x3.6x10-4 = 0.0486 N

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...