The correct answer is (B) : 50
Total gravitational PE of water per second
\(=\frac{mgh}{T}\)
\(=\frac{9×10^4×10×40}{3600}=10^4\) J/sec
50% of this energy can be converted into electrical energy so total electrical energy
\(=\frac{10^4}{2}=5000 W\)
So total bulbs lit can be
\(=\frac{5000 W}{100 W}\)
= 50 bulbs
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Read More: Work and Energy