To find the electric energy consumed by a 60 W bulb operating for 10 hours a day, we need to use the formula for electrical energy consumption:
\[ \text{Energy (kWh)} = \frac{\text{Power (W)} \times \text{Time (hours)}}{1000} \]
Given that the power of the bulb is 60 W and the time it operates per day is 10 hours, we substitute these values into the formula:
\[ \text{Energy} = \frac{60 \times 10}{1000} \]
\[ \text{Energy} = \frac{600}{1000} \]
\[ \text{Energy} = 0.6 \text{ kWh} \]
Thus, the electric energy consumed in operating the bulb for 10 hours a day is 0.6 kWh.
We are given:
Now, electric energy consumed is given by:
\[ \text{Energy (in kWh)} = \text{Power (in kW)} \times \text{Time (in hours)} \]
\[ = 0.06 \times 10 = 0.6 \, \text{kWh} \]
Correct Answer: 0.6
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in V/Ω, is __________ (round off to two decimal places).

An ideal low pass filter has frequency response given by \[ H(j\omega) = \begin{cases} 1, & |\omega| \leq 200\pi \\ 0, & \text{otherwise} \end{cases} \] Let \( h(t) \) be its time domain representation. Then h(0) = _________ (round off to the nearest integer).