A block B of mass 20 Kg is suspended by a metallic rod of length 2 m, mass 10 Kg cross sectional area 4 mm2 and young's modulus 2×1010 N/m2. Then elongation produced in the rod is
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
8.25 mm
b
12.15 mm
c
4.85 mm
d
6.25 mm
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
δ=100×14×10−6×2×1010+200×24×10−6×2×1010m⇒δ=1800+1200m=6.25 mm