An object cools from $100°\mathrm{C}$ to $40°\mathrm{C}$ in $10$ minutes, when the surrounding temperature is $10°\mathrm{C}$. Then the time taken by the object to cool from $70°\mathrm{C}$ to $20°\mathrm{C}$ is
[Take $\ln 2=0.7,\ln 3=1.1,\ln 6=1.8$]

According to Newton's cooling law, rate of cooling is

$\frac{d\theta }{dt}\propto \left(\theta -{\theta }_0\right)$

Here, $\theta =$the temperature of the body

${\theta }_0=$Room temperature(constant)

Let $k$ be the proportionality constant and $\theta$ be the temperature.
Therefore, 

$$\begin{gathered} \frac{d\theta }{dt}=-k\left(\theta -{\theta }_0\right) \\ \Rightarrow {\int }_{{\theta }_1}^{{\theta }_2}\frac{d\theta }{\left(\theta -{\theta }_0\right)}=-k{\int }_0^{t}dt \\ \Rightarrow \ln \left(\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}\right)=kt \end{gathered}$$

Taking ratios,

$$\begin{gathered} \frac{\ln \left({\displaystyle \frac{40-10}{100-10}}\right)}{\ln \left({\displaystyle \frac{20-10}{70-10}}\right)}=\frac{10}{t_2} \\ \Rightarrow t_2=10\times \frac{\ln {\displaystyle 6}}{\ln {\displaystyle 3}}=16.3 \min \end{gathered}$$