The velocity v (in cm/sec) of a particle is g | vedclass

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(C) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.$1$. For the term $c$: Since $c$

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$a = LT^{-2}, b = L, c = T$

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(C) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.$1$. For the term $c$: Since $c$ is added to $t$ (time),the dimension of $c$ must be the same as the dimension of $t$. Therefore,$[c] = [T]$.$2$. For the term $at$: The dimension of $at$ must be equal to the dimension of velocity $v$. Thus,$[a][T] = [LT^{-1}]$,which gives $[a] = [LT^{-2}]$.$3$. For the term $\frac{b}{t+c}$: The dimension of the entire term must be equal to the dimension of velocity $v$. Since $[t+c] = [T]$,we have $\frac{[b]}{[T]} = [LT^{-1}]$. Therefore,$[b] = [L]$.Thus,the dimensions are $a = LT^{-2}, b = L, c = T$.

The velocity $v$ (in $cm/sec$) of a particle is given in terms of time $t$ (in $sec$) by the relation $v = at + \frac{b}{t + c}$. The dimensions of $a, b,$ and $c$ are:

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