What percentage $\mathrm{T}_{1}$ is of $\mathrm{T}_{2}$ for a $10\%$ efficiency of a heat engine? $\mathrm{T}_{1}$ is the temperature of sink and $\mathrm{T}_{2}$ is the temperature of heat reservoir.

For a Carnot heat engine, efficiency is $$\eta = 1 - \frac{T_1}{T_2}.$$ Given $$\eta = 10\% = 0.1.$$ So $$0.1 = 1 - \frac{T_1}{T_2}.$$ Hence $$\frac{T_1}{T_2} = 0.9.$$ Therefore, $T_1$ is $90\%$ of $T_2$. This matches option $A$.