An iron rod is subjected to cycles of magnetisation at the rate of 50Hz. Given the density of the rod is 8 × 103kg/m3 and specific heat is 0.11 × 10–3 cal/ kg0C. The rise in temperature per minute, if the area enclosed by the B – H loop corresponds to energy of 10–2 J is (Assume there is no radiation losses)
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
78 oC
b
88 oC
c
8.1 oC
d
none of these
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
n = 50 per secondn per minute = 50 × 60 = 3000n x area = ms∆t3000 x 10-2 = 4.2 x 8 x 103 x 0.11 x 10-3 ∆t ∆t=8.1 oC