The figure shows a Wheatstone bridge circuit arrangement with a galvanometer G of resistance 5Ω to detect the presence of current. The ratio of heat dissipated across four resistors, $R_1, R_2, R_3$, and $R_4$ respectively, is:

1 : 4 : 1.5 : 6

The correct answer is Option (2) → 1 : 4 : 1.5 : 6

Top branch resistance: $R_{1}+R_{2}=15+60=75\ \Omega$

Bottom branch resistance: $R_{3}+R_{4}=10+40=50\ \Omega$

Apply $V=1\ \text{(choose unit emf)}$ between A and C.

Top branch current: $I_{t}=\frac{1}{75}=\frac{1}{75}=0.013333\ \text{A}$

Bottom branch current: $I_{b}=\frac{1}{50}=0.02\ \text{A}$

Power in $R_{1}$: $P_{1}=I_{t}^{2}R_{1}=(0.013333)^{2}\times15=0.0026667$

Power in $R_{2}$: $P_{2}=I_{t}^{2}R_{2}=(0.013333)^{2}\times60=0.0106667$

Power in $R_{3}$: $P_{3}=I_{b}^{2}R_{3}=(0.02)^{2}\times10=0.004$

Power in $R_{4}$: $P_{4}=I_{b}^{2}R_{4}=(0.02)^{2}\times40=0.016$

Divide by $P_{1}$ to get ratio:

$P_{1}:P_{2}:P_{3}:P_{4}=1:4:1.5:6$