A $\text{J}$-type thermocouple has an output voltage $V_{\theta }=\left ( 13650+50\: \theta _{x} \right ) \mu \;V$, where $\theta _{x}$ is the junction temperature in Celsius ($^{\circ}C$). The thermocouple is used with reference junction compensation, as shown in the figure. The Instrumentation amplifier used has a gain $G=20$. If $\theta _{Ref}$ is $1 ^{\circ}C$, for an input $\theta _{x}$ of $100 ^{\circ}C$, the output $V_{o}$ of the instrumentation amplifier in millivolt is
- $\text{98 mV}$
- $\text{99 mV}$
- $\text{100 mV}$
- $\text{101 mV}$