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
87 views
in Physics by (76.5k points)
closed by
At what temperature , will the rms speed of oxygen molecules be sufficient for escaping from the earth ? Take `m = 2.76 xx 10^(-26) kg, k = 1.38 xx 10^(-23) J//K and v_(e) = 11.2 km//s`.
A. `5.016 xx 10^(4) K`
B. `1.254 xx 10^(4) K`
C. `8.360 xx 10^(4) K`
D. `2.508 xx 10^(4) K`

1 Answer

0 votes
by (76.7k points)
selected by
 
Best answer
Correct Answer - C
`v_(rms) = sqrt((3 kT)/(m)) = v_(e)`
`(3 kT)/(m) = v_(e)^(2)`
`(3 xx 1.38 xx 10^(-23) T)/(2.76 xx 10^(-26)) = (11.2 xx 10^(3))^(2)`
`T = 8.360 xx 10^(4) K`

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

...