Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
If absolute zero is -273.15 text ° C on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is
Q. If absolute zero is
−
273.15
∘
C
on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is
1111
250
VMMC Medical
VMMC Medical 2009
Report Error
A
−
273.15
∘
F
B
−
453.15
∘
F
C
−
459.67
∘
F
D
−
491.67
∘
F
Solution:
Celsius and Kelvin scale are related as
5
C
=
9
F
−
32
or
5
−
273.15
=
9
F
−
32
⇒
F
=
−
459.67
o
F