JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 13)
If momentum (P), area (A) and time (T) are
taken to be the fundamental quantities then the
dimensional formula for energy is
[P2AT–2]
$$\left[ {{P^{{1 \over 2}}}A{T^{ - 1}}} \right]$$
$$\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]$$
[PA–1T–2]
Explanation
Let
[E] =
K[P]x[A]y [T]z
[ML2T–2] = [MLT–1]x[L2]y[T]z
[ML2T–2] = [Mx][Lx+2y][T–x+z]
Comparing both side we get,
x = 1
x + 2y = 2
$$ \Rightarrow $$ 1 + 2y= 2 or y = $$\frac{1}{2} $$
z – x = –2 $$ \Rightarrow $$ z–1 = –2 or z = –1
$$ \therefore $$ [E] = $$\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]$$
[ML2T–2] = [MLT–1]x[L2]y[T]z
[ML2T–2] = [Mx][Lx+2y][T–x+z]
Comparing both side we get,
x = 1
x + 2y = 2
$$ \Rightarrow $$ 1 + 2y= 2 or y = $$\frac{1}{2} $$
z – x = –2 $$ \Rightarrow $$ z–1 = –2 or z = –1
$$ \therefore $$ [E] = $$\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]$$
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