The position (X) of a particle depends on its velocity(v) and time (t) as given by the relation $x = \frac{P v + Q}{P + T}$ .Find the dimentions of PQ.

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Solution

By principle of homogeneity,

Dimension of T = Dimension of Pv

$[T] = [P] \times [L T^{-} 2]$

$[T] / [L T^{-} 2] = P$

$[L^{-} 1 T^{3}] = P$

Dimension of T = Dimension of Q/P

$[T] = [Q] / [L^{-} 1 T^{3}]$

$[T] \times [L^{-} 1 T^{3}] = Q$

$[L^{-} 1 T^{4}] = Q$

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The position (X) of a particle depends on its velocity(v) and time (t) as given by the relation x=Pv+QP+T.Find the dimentions of PQ.

By principle of homogeneity, Dimension of T = Dimension of Pv [T]=[P]×[LT−2] [T]/[LT−2]=P [L−1T3]=P Dimension of T = Dimension of Q/P [T]=[Q]/[L−1T3] [T]×[L−1T3]=Q [L−1T4]=Q

The position (X) of a particle depends on its velocity(v) and time (t) as given by the relation $x = \frac{P v + Q}{P + T}$ .Find the dimentions of PQ.

By principle of homogeneity,

Dimension of T = Dimension of Pv

$[T] = [P] \times [L T^{-} 2]$

$[T] / [L T^{-} 2] = P$

$[L^{-} 1 T^{3}] = P$

Dimension of T = Dimension of Q/P

$[T] = [Q] / [L^{-} 1 T^{3}]$

$[T] \times [L^{-} 1 T^{3}] = Q$

$[L^{-} 1 T^{4}] = Q$