Let [Y] = [V] a[ F]b[A]c [ML−1 T−2]=[LT−1]a[MLT−2]b[LT−2]c [ML−1 T−2]=[Mb La+b+cT−a−2 b−2c] Comparing power both side of similar terms we get, b=1,a+b+c=−1,−a−2b−2c=−2 solving above equations we get a=−4,b=1,c=2 so [Y]=[V−4FA2]=[V−4 A2 F]
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :
JEE Main 2019 — Physics Mechanics
2019mcqeasy
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :
Official previous-year question
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
V−2 A2 F−2
- B
V−2 A2 F2
- C
V−4 A−2 F
- D
V−4 A2 F
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