Question

The dimensions of a and b in the relation F = a + bx, where, F is the force and x is the distance are
a = [M L T^-2] , b = [M L^0 T^-2]
a = [M L T^-2] , b = [M L T^-2]
a = [M^2 L T^-2] , b = [M L^0 T^-2]
a = [M L T^-2] , b = [M^2 L^0 T^-2]

Answer 1

Answer:

hlo mate

Explanation:

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 ]

hope it helps you

Answer 2

The dimensions of F and x are (A) a = [M L T^-2] , b = [M L^0 T^-2]

Explanation:

According to the dimensional analysis formula, LHS=RHS has a current formula.

We know, F= a+bx

here, F= a and F= bx

Also,

$F = [ MLT^{-2}]$

So, $F = a= [ MLT^{-2}]$

now, F= bx

x is given as distance. distance is [L]

so, $F= [L] b$

⇒$[MLT^{-2}] = [L] b$

⇒ $[MT^{-2} ] = b$

so,  $a= [ MLT^{-2}]$ and  $b= [MT^{-2} ]$

Hence, The dimensions of F and x are (A) a = [M L T^-2] , b = [M L^0 T^-2]

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