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write the dimensions of a/b in the relation F=a root x +bt2 where F is force , x is distance and t is time.

According to the principle of homogeneity,dimensions of every term on R.H.S should be same as that on L.H.S. i.e force.

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F=a root x +bt2

So, 

dimension of F = a root x

m / (s2) = a root (metre)

So 

a = m / (s2) divided by root(metre)

Squaring both sides and solving we get--

a2 = m / s4 -- equation 1st

Now

dimension of F = dimension of bt2

Solve it and you will get 

b2 = m2 / s8 -- equation 2

Now divide equation 2 by q

you will get --

a / b = s4 / m

So converting in dimensions 

a / b = L -1 T -4

Thanks hope you understood it

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THNX NITIN..

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THNX A LOT..

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MLT^-2=aL^1/2

MLT^-2/L^1/2-=a

ML^1/2L^1/2T^-2/L1/2=a

ML^1/2T^-2

MLT^-2=bT^2

MLT^-2/T^2=B

MLT^-4=B

a/b=ML1/2T^-2/MLT^-4

=ML^1/2T^-2/ML^1/2L^1/2T^-4

=L^-1/2T^2

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