The given circuit shows a uniform straight wire $$A B$$ of $$40 \mathrm{~cm}$$ length fixed at both ends. In order to get zero reading in the galvanometer $$G$$, the free end of $$J$$ is to be placed from $$B$$ at:
According to the law of equipartition of energy, the number of vibrational modes of a polyatomic gas of constant $$\gamma=\frac{C_p}{C_v}$$ is ($$C_P$$ where $$C_V$$ are the specific heat capacities of the gas at constant pressure and constant volume, respectively):
The output Y for the inputs A and B of the given logic circuit is:
The amplitude of the charge oscillating in a circuit decreases exponentially as $$Q=Q_0 e^{-R t/2 L}$$, where $$Q_0$$ is the charge at $$t=0 \mathrm{~s}$$. The time at which charge amplitude decreases to $$0.50 Q_0$$ is nearly:
[Given that $$R=1.5 \Omega, L=12 \mathrm{~mH}, \ln (2)=0.693$$]