Caprolactone on reduction with LAH or  + Pt or Pd gives:

The equivalent weight  of  is Half its molecular weight when it is converted to

A 5V battery with internal resistance 2 and a 2V battery with internal resistance 1 are connected to a 10 resistor as shown in the figure

The current in the 10 resistor is

In the electrical network shown in the figure, the potential difference across 3 resistance will be

The current and through the resistor and in the circuit diagram with are respectively.

 The Products are:

In the set up shown, the two slits S₁ and S₂ are not equidistant from the slit S

The central fringe at O is then

A group of N cells whose emf varies directly with the internal resistance as per the equation are connected as shown in the figure. The current I in the circuit is

In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2 resistance is

Two similar accumulators each of emf E and internal resistance r are connected as shown in the following figure. Then, the potential difference between x and y is

A current of A enters one corner one corner P of an equilateral triangle having 3 wires of resistance 2each and leaves by the corner R. then the current and are

For the circuit shown in the figure the potential difference between A and B will be (in volt)

If A = and (aI₂ + bA)² = A, then-

Hence Choice 2 is correct

If A = then (A– 2I) (A – 3I) equals-

If A = and A² – KA– I₂ = 0, then value of K is-

If A = then Aⁿ equal to-