The polarimeter readings in an experiment to measure the inversion of cane sugar were as follows
 

Time (minutes)	0	30	$\infty$
Rotation due to solution (degree)	30	20	-15

Given 
$\left[\begin{array}{cc}\log 2=0.3,& \log 3=0.48\\ \log 7=0.84,& \end{array}\right]$
Identify the correct statement. 

$$\begin{gathered} k=\frac{2.303}{t}\log \left[\frac{r_{\infty }-r_0}{r_{\infty }-r_t}\right] k=\frac{2.303}{30}\log \left[\frac{-15-30}{-15-20}\right] k=\frac{2.303}{30}\log \left[\frac{9}{7}\right] \\ =9.21\times {10}^{-3} t_{1/2}=\frac{0.693}{k} t_{1/2}\approx 75\min t. \\ Let solution becomes optically inactive after a time \left(t\right). So [r_t=0] \\ k=\frac{2.303}{t}\log \left[\frac{-15-30}{-15-0}\right] k=\frac{2.303}{t}\log 3 k=\frac{2.303}{9.21\times {10}^{-3}}\times 0.48 \\ \Rightarrow t\approx 120\min t \end{gathered}$$