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
66 views
in Physics by (90.2k points)
closed by
A small sphere falls from rest in a viscous liquid. Due to frication, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.

1 Answer

0 votes
by (90.9k points)
selected by
 
Best answer
Correct Answer - `(dQ)/(dt)propr^(5)`
Terminal velocity `v_(T) = (2r^(2)g)/(9eta)(rho_(s) - rho_(L))`
and viscous force `F = 6pirv_(T)`
Viscous force is the dissipative force. Hence.
`(dQ)/(dt) = Fv_(T) = (6pietarv_(T))(v_(T)) = 6pietarv_(T)^(2)`
`= 6pietar{(2)/(9)(r^(2)g)/(eta)(rho_(s) - rho_(L))}^(2) = (8pig^(2))/(27eta)(rho_(s) - rho_(L))^(2)r^(5) = (dQ)/(dt)propr^(5)`

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

...