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] :

For the given cyclic process CAB as shown for a gas, the work done is :

A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $${{\theta _1}}$$. Then M is given by :

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :