dimensional formula of work
Answers
Explanation:
Correctness of the equation,
T=\frac{2 \pi \sqrt{L}}{g}T=
g
2π
L
Proof:
Let us prove by using dimensional analysis.
\begin{lgathered}\begin{array}{l}{T=M^{0} L^{0} T^{1}} \\ {\frac{\sqrt{L}}{g}=\left[\frac{M^{0} L^{1} T^{0}}{\left(M^{0} L^{1} T^{-2}\right) ]^{\frac{1}{2}}}\right]} \\ {=\left[M^{0} L^{0} T^{2}\right]^{\frac{1}{2}}} \\ {=M^{0} L^{0} T^{1}}\end{array}\end{lgathered}
T=M
0
L
0
T
1
g
L
=[
(M
0
L
1
T
−2
)]
2
1
M
0
L
1
T
0
]
=[M
0
L
0
T
2
]
2
1
=M
0
L
0
T
1
Now we have the dimensional formula for both LHS and RHS
So, now on equating both LHS and RHS of the equation.
We have
\begin{lgathered}\begin{array}{l}{T=\frac{2 \pi \sqrt{L}}{g}} \\ {M^{0} L^{0} T^{1}=M^{0} L^{0} T^{1}}\end{array}\end{lgathered}
T=
g
2π
L
M
0
L
0
T
1
=M
0
L
0
T
1
LHS = RHS
Hence proved.
!! RAM RAM !!
ANSWER ===√
dimensional formula of work
Work = Force × distance =
(mass× acceleration) × distance =
(M× L× T^-2) ×L =M×L^2×T^-2.
hope it helps uh
#Hopēless_ Romantic ❤❤❤