For a wire $$\frac{R}{\ell}=\frac{1}{2}$$ and length of wire is $$\ell=5 \mathrm{~cm}$$. If potential difference of $$\mathrm{l} \mathrm{V}$$ is applied across it, then current through wire will be ($$R=$$ Resistance)

A current of $$10 \mathrm{~A}$$ is passing through a metallic wire of cross-sectional area $$4 \times 10^{-6} \mathrm{~m}^2$$. If the density of the aluminium conductor is $$2.7 \mathrm{~gm} / \mathrm{cc}$$ considering aluminium gives 1 electron per atom for conduction, then find the drift velocity of the electrons if molecular weight of aluminium is $$27 \mathrm{~gm}$$.

A metal wire has a resistance of $$35 \Omega$$. If its length is increased to double by drawing it, then its new resistance will be

In the circuit in the figure, if no current flows through the galvanometer when the key K is closed, the bridge is balanced. The balancing condition for bridge is

In the circuit, shown the galvanometer $$G$$ of resistance $$60 \Omega$$ is shunted by a resistance $$r=0.02 \Omega$$. The current through $$R$$ is nearly $$1 \mathrm{~A}$$. The value of resistance $$R$$ (in ohm) is nearly

In the adjoining circuit diagram, the readings of ammeter and voltmeter are 2 A and 120 V, respectively. If the value of R is 75$$\Omega$$, then the voltmeter resistance will be

Assertion Terminal voltage of a cell is greater than emf of cell during charging of the cell.

Reason The emf of a cell is always greater than its terminal voltage.

Assertion Bulb generally get fused when they are switched on or off.

Reason When we switch on or off, a circuit current changes in it rapidly.

A cylindrical conductor AB of non-uniform area of cross-section carries a current of 5A. The radius of the conductor at one end A is 0.5 cm. The current density at the other end of the conductor is half of the value at A. The radius of the conductor at the end B is nearly

The effective resistance between p and q in given figure is

Two batteries, one of emf $$18 \mathrm{~V}$$ and internal resistance $$2 \Omega$$ and the other of emf $$12 \mathrm{~V}$$ and internal resistance $$1 \Omega$$ are connected as shown. The voltmeter $$V$$ will record a reading of

Assertion : A potentiometer is preferred over that of a voltmeter for measurement of emf of a cell.

Reason : Potentiometer does not draw any current from the cell.