Answer:
Option C
Explanation:
Power of battery, when charged is given by $P_{1}=V_{1}I_{1}$
Electrical energy dissipated id given by $E_{1}=P_{1}t_{1}$
$E_{1}=V_{1}I_{1}t_{1}=15 \times 10 \times 8=1200Wh$
Similarly , the electrical energy dissipated during the discharge of battery is given by $E_{2}=V_{2}I_{2}t_{2}$
=$14 \times 5 \times 15=1050 Wh$
Hence, watt-hour efficiency of the battery
$\Rightarrow$ $\eta= \frac{E_{2}}{E_{1}} \times 1000.875 \times 100=87.5$
