The viscosity of a glass is calculated using the following formula:

$={}_0{e}^{\frac{Q}{R{T}_1}}$

where $=$ viscosity of the glass, $Pa*s$

${}_0=$ pre$-$exponential constant, $Pa*s$

$Q=$ molar activation energy for viscous flow

$R=$ universal molar gas constant $)$

$T=$ absolute temperature

Choose the correct expression of viscosity, ${}_1$ and ${}_2$ at $T_1$ and $T_2,$ respectively

a$.\frac{{}_2}{{}_1}=$exp$[\frac{Q}{R}(\frac{1}{T_1}-\frac{1}{T_2})]$

b$.\frac{{}_2}{{}_1}=$exp$[\frac{Q}{R}(\frac{1}{T_2}-\frac{1}{T_1})]$

c$.\frac{{}_2}{{}_1}=$exp$[\frac{Q}{R}(T_1-T_2)]$

d$.\frac{{}_2}{{}_1}=$exp$[\frac{Q}{R}(\frac{T2}{T_1})]$

A soda$-$lime glass has a viscosity of ${10}^{14\mathrm{.}6}Pa$ at $560C.$

What will be its viscosity at $625C$ if the activation energy for viscous flow is $430$

$k\frac{J}{m}ol?$ Assume $R=8\mathrm{.}314J*mo{l}^{-1}*K^{-1}.$

${}_2=[2]{10}^{12}[Pa*s]$

$*$Input up to the $2$ nd decimal place value.