A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 $$ \times $$ 10^–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :

Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P₁ and P₂ respectively, then :

In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation $${{dR} \over {d\ell }}$$ of its resistance R with length $$\ell $$ is $${{dR} \over {d\ell }}$$ $$ \propto $$ $${1 \over {\sqrt \ell }}$$. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP ?

The galvanometer deflection, when key K₁ is closed but K₂ is open, equals $$\theta $$₀ (see figure). On closing K₂ also and adjusting R₂ to 5$$\Omega $$, the deflection in galvanometer becomes $${{{\theta _0}} \over 5}.$$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :