Question

In the relation, F = a + bx, where F is for
and x is distance. Calculate the dimensions
a and b.
[Ans, a = [M' L'T'], b = [M' L'I?​

Answer 1

Answer: see your answer carefully

By dimensional analysis we consider both the LHS and RHS have the same dimensions.

F=a+bx

In the given equation the LHS is F which stands for force

[F] = [MLT^-2 ]

now in RHS we have two constants a & b , by additive property we find that a and F will have the same dimension .

Thus [a] = [ MLT^ -2]

Now for…. bx , we know x is distance so it has dimension x= [ L ]

Thus we get bx = [ MLT ^-2]

=> b[ L ]= [ MLT ^-2]

=>. b= [ MT ^-2]

a= [MLT ^-2 ] & b=[MT ^-2 ]

Explanation:

Answer 2

$\large{\sf{\pink{\underline{\underline{Answer:-}}}}}$

$\rm{Dimension\;of\;a=[MLT^{-2}]}$

$\rm{Dimension\;of\;b=[M^{1}L^{0}T^{-2}]}$

$\large{\sf{\pink{\underline{\underline{Explanation:-}}}}}$

$\rm{Given\;relation,\;F=a+bx}$

$\rm{Since\;a\;is\;connected\;by\;addition\;symbol\;so\;a\;has\;dimension\;same\;as\;F.}$

$\rm{[F]=[a]=[MLT^{-2}]}$

$\rm{Let\;[b]=[M^{x}L^{y}T^{z}]}$

$\rm{Again\;bx\;has\;dimension\;of\;F.}$

$\rm{So,\;[F]=[b][x]}$

$\rm{\implies [MLT^{-2}]=[M^{x}L^{y}T^{z}][L]}$

$\rm{\implies [MLT^{-2}]=[M^{x}L^{y+1}T^{z}]}$

$\rm{Comparing\;the\;powers\;of\;M,L\;and\;T}$

$\rm{x=1}$

$\rm{y+1=1}$

$\rm{\implies y = 0}$

$\rm{z=-2}$

$\rm{So,[b]=[M^{x}L^{y}T^{z}]=[M^{1}L^{0}T{-2}]}$

$\rm{[b]=[M^{1}L^{0}T^{-2}]}$