A galvanometer of resistance $$100 \Omega$$ when connected in series with $$400 \Omega$$ measures a voltage of upto $$10 \mathrm{~V}$$. The value of resistance required to convert the galvanometer into ammeter to read upto $$10 \mathrm{~A}$$ is $$x \times 10^{-2} \Omega$$. The value of $$x$$ is :

In the given figure $$\mathrm{R}_1=10 \Omega, \mathrm{R}_2=8 \Omega, \mathrm{R}_3=4 \Omega$$ and $$\mathrm{R}_4=8 \Omega$$. Battery is ideal with emf $$12 \mathrm{~V}$$. Equivalent resistant of the circuit and current supplied by battery are respectively :

An electric bulb rated $$50 \mathrm{~W}-200 \mathrm{~V}$$ is connected across a $$100 \mathrm{~V}$$ supply. The power dissipation of the bulb is:

To measure the internal resistance of a battery, potentiometer is used. For $$R=10 \Omega$$, the balance point is observed at $$l=500 \mathrm{~cm}$$ and for $$\mathrm{R}=1 \Omega$$ the balance point is observed at $$l=400 \mathrm{~cm}$$. The internal resistance of the battery is approximately :