JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 9)
If speed V, area A and force F are chosen as
fundamental units, then the dimension of
Young’s modulus will be
FA–1V0
FA2V–1
FA2V–2
FA2V–3
Explanation
Y = k [F]x
[A]y
[V]z
[M1L1T –2] = [MLT–2]x [L2]y [LT–1]z
[M1L1T –2] = [M]x [L]x+2y+z[T]–2x–z
Comparing power of M, L and T
x = 1 ……(1)
x + 2y + z = –1 ……(2)
–2x – z = –2 ……(3)
After solving
x = 1
y = –1
z = 0
$$ \therefore $$ Y = FA–1V0
[M1L1T –2] = [MLT–2]x [L2]y [LT–1]z
[M1L1T –2] = [M]x [L]x+2y+z[T]–2x–z
Comparing power of M, L and T
x = 1 ……(1)
x + 2y + z = –1 ……(2)
–2x – z = –2 ……(3)
After solving
x = 1
y = –1
z = 0
$$ \therefore $$ Y = FA–1V0
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