Step 1: The angular momentum \( L \) of a particle rotating with respect to a central force is given by: \[ L = r \times p, \] where \( r \) is the position vector, and \( p \) is the linear momentum of the particle.
Step 2: The rate of change of angular momentum is related to the torque \( \tau \) acting on the particle: \[ \frac{dL}{dt} = \tau. \]
Step 3: If the torque \( \tau \) is zero, then the angular momentum remains constant. This happens when there is no external torque acting on the particle.
Step 4: Since central forces always act along the line joining the particle and the center of rotation, they produce zero torque. Therefore, the angular momentum of the particle remains constant.
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.