Step 1: Identify what's being asked clearly:
The question asks for the work done by the battery to establish a current \( I \) through the circuit in time \( t \).
Step 2: Recall the basic concept:
The work done (\( W \)) by a battery or cell to establish a current is given by:
\[ W = \text{EMF} \times \text{Charge} \]
Step 3: Determine the total charge moved by the battery in time \( t \):
The electric charge (\( Q \)) moved is given by:
\[ Q = I \times t \]
Step 4: Substitute this into the formula for work done:
\[ W = \varepsilon \times Q = \varepsilon \times I \times t \]
This gives the total work done by the battery as:
\[ W = \varepsilon I t \]
Important: The internal resistance \( r \) of the battery does not affect this direct formula for the work done by the battery. The battery always provides energy equal to its emf times the charge passed through it.
Final Answer: Option (D) \( \varepsilon I t \)
The work done by the battery to establish the current is given by the formula: \[ W = \varepsilon \times I \times t \] Where:
\( W \) is the work done,
\( \varepsilon \) is the emf of the battery,
\( I \) is the current,
\( t \) is the time.
Thus, the work done by the battery is \( \varepsilon It \).
Final Answer: Option (D) \(εIt\)
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)): 
The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is: 