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
0 votes
72 views
in Physics by (87.6k points)
closed by
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?

1 Answer

0 votes
by (88.8k points)
selected by
 
Best answer
Emf of the cell, `E_(1)=1.25 V`
Balance point of the potentiometer, `l_(1)=35` cm
The cell is replaced by another cell of emf `E_(2)`
New balance point of the potentiometer, `I_(2)=63` cm
The balance condition is given by the relation,
`(E_(1))/(E_(2))=(I_(1))/(I_(2))`
`E_(2)=E_(1)xx(l_(2))/(l_(1))`
`= 1.25 xx (63)/(35) = 2.25 V`
Therefore, emf of the second cell is 2.25 V.

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

...