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
52 views
in Physics by (87.5k points)
closed by
At room temperature `(27.0^@C)` the resistance of a heating element is `100 Omega`. What is the temperature of the element if the resistance is found to be `117 Omega`, given that the temperature coefficient of the material of the resistor is `(1.70 xx 10^(-4))^@C^(-1)`.

1 Answer

0 votes
by (88.8k points)
selected by
 
Best answer
Room temperature, `T = 27^(@)C`
Resistance of the heating element at T, `R = 100 Omega`
Let `T_(1)` is the increased temperature of the filament. Resistanc of the heating element at `T_(1),R_(1)=117 Omega`
Temperature co-efficient of the material of the filament,
`alpha = 1.70 xx 10^(-1) .^(@)C^(-1)`
`alpha` is given by the relation,
`alpha = (R_(1)-R)/(R(T_(1)-T))`
`T_(1)-T = (R_(1)-R)/(R alpha)`
`T_(1)-27 = (117 xx 100)/(100(1.7 xx 10^(-4)))`
`T_(1)-27 = 1000`
`T_(1)=1027^(@)C`
Therefore, at `1027^(@)C`, the resistance of the element is `117 Omega`.

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

...