It is required to prepare a steel metre scale such that the millimetre intervals are to be accurate within 0.0005 mm at a certain temperature. Maximum temperature variationallowable during the rulings of millimetre marks is :(αsteel=13.22×10-6/C o)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
37.8°C
b
45.4°C
c
10.2°C
d
62.6°C
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Here,l=1 mm,∆l=0.0005 mm As, ∆l=l α ∆T So, ∆T=∆lαl=0.000513.22×10-6×1=37.8C o.