$\begin{array}{1 1}(a)\;function\; of\;a_o\;alone\\(b)\;function \;of\;n\;alone\\(c)\;function\;of\;a_o\;and\;n\\(d)\;none\end{array}$

For $n^{th}$ order reaction

$nA\rightarrow$ products

$-\large\frac{d[A]}{dt}=$$k_n[A]^n$

$\int-\large\frac{d[A]}{[A]^n}=$$k_n\int dt$

$[A_o]=a_o$

$t=\large\frac{1}{k_n(n-1)a^{n-1}}$$+c$

$t=\large\frac{1}{k_n(n-1)}$$\bigg[\large\frac{1}{a^{n-1}}-\frac{1}{a_o^{n-1}}\bigg]$

When $t=t_{1/2},$$a=\large\frac{a_o}{2}$

So $k_nt_{1/2}=\large\frac{1}{n-1}\big(\frac{1}{a_o}\big)^{n-1}$$(2^{n-1}-1)$

When $t=t_{3/4},$$a=\large\frac{3a_o}{4}$

So $k_nt_{3/4}=\large\frac{1}{n-1}\big(\frac{1}{a_o}\big)^{n-1}$$(\big(\large\frac{4}{3}\big)^{n-1}$$-1)$

$\large\frac{t_{1/2}}{t_{3/4}}=\frac{2^{n-1}-1}{(4/3)^{n-1}-1}$

Hence (b) is the correct answer.

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