$(a)\;0.0018/^{0}C\qquad(b)\;0.00018/^{0}C\qquad(c)\;0.018/^{0}C\qquad(d)\;0.0016/^{0}C$

Thermodynamics

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Answer : $\;0.00018/^{0}C$

Explanation :

When 2 liquid columns balance each other , the processes exerted by them are equal to one another .

$h_{1} \rho_{1} g = h_{2} \rho_{2} g$

$\large\frac{\rho_{1}}{\rho_{2}}=\large\frac{h_{2}}{h_{1}}$

$\large\frac{\rho_{1}}{\rho_{2}}=\large\frac{0.811}{0.8} $

$1+ \bigtriangleup t=1.018$

$r=0.00018/^{0}C$

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