Write the dimensions of a/b in the relation f=ax+bt2
Answers
f = a√x + bt2, where f = (M L T −2(, x = (L) and t = (T).
f = a√x = bt2
f = a√x
(M L T −2) = (a L½)
= (a) = (M L½ T −2)
F = (bt2)
(M L T −2) = (b T2)
= (b) = (M L T−4)
(a)/(b) = (M L½ T −2)/(M L T−4) = (M0 L−½ T2)
So, (a/b) = (M0 L−½ T2)
Hope it helps!!!
Step-by-step explanation:
The dimensions of the ratio of a and b is [L^{-\frac{1}{2}}T^2][L
−
2
1
T
2
] .
Explanation:
Given that,
F =a\sqrt{x}+bt^2F=a
x
+bt
2
Where, F = force
x = distance
According to principle of homogeneity,
The dimension of every term on right hand side should be same as that on Left hand side.
The dimension formula of a
F = a\sqrt{x}F=a
x
a = \dfrac{F}{\sqrt{x}}a=
x
F
Now, write the dimension formula in the equation (I)
a = \dfrac{[MLT^{-2}]}{[L^{\frac{1}{2}}]}a=
[L
2
1
]
[MLT
−2
]
a = [ML^{\frac{1}{2}}T^{-2}]a=[ML
2
1
T
−2
]
the dimension formula of b
F = bt^2F=bt
2
b = \dfrac{F}{t^2}b=
t
2
F
....(II)
Now, write the dimension formula in the equation (II)
b = \dfrac{[MLT^{-2}]}{[T^{2}]}b=
[T
2
]
[MLT
−2
]
b= [MLT^{4}]b=[MLT
4
]
The dimension of a and b is
\dfrac{a}{b}=\dfrac{ [ML^{\frac{1}{2}}T^{-2}]}{[MLT^{4}]}
b
a
=
[MLT
4
]
[ML
2
1
T
−2
]
\dfrac{a}{b}=[L^{-\frac{1}{2}}T^2]
b
a
=[L
−
2
1
T
2
]
Hence, The dimensions of the ratio of a and b is [L^{-\frac{1}{2}}T^2][L
−
2
1
T
2
] .