A fuse is rated $8A$. Can it be used with an electrical appliance rated $5kW, 200V$? Give a reason. Name two safety devices which are connected to the live wire of a household electric circuit. How many alpha and beta particles are emitted when the uranium nucleus decays to lead?

Hint:As the power and the voltage of the electrical appliance is given, we can obtain the current produced by the appliance. Due to the emission of alpha particles, the atomic number decreases by 2 and mass number by 4, while due to the emission of beta particles, only the atomic number increases by 1.

Complete step by step answer:
Here, the electric appliance has a power of 5000W. The voltage of the electric appliance is 200V. We know the relationship between power, voltage and current as follows:
\[P = V \times I\]
Here, P is the power of the electric appliance, V is the voltage of the electric appliance and I is the current flowing across the electric appliance.
Thus, putting the values of power and voltage, the current obtained will be as follows:
\[I = \dfrac{{5000}}{{200}} = 25A\]
Now, the current flowing across the electrical appliance is 25A while the current flowing across the fuse is 8A. Thus as the fuse current is less, it cannot be used with the electrical appliance.The two safety devices which are connected to the live wire of a household electric circuit are a safety fuse and a switch.

The atomic number and mass number of uranium is 92 and 238 while those for lead are 82 and 214 respectively. When alpha particles emit, atomic number decreases by 2 and mass number by 4, while for beta particles, only atomic number increases by 1.Hence, 6 alpha particles and 2 beta particles are required.

Note: The Uranium atom breaks down to lead atom because as there is more mass, the number of neutrons gets increased as compared to the number of protons, the nucleus due to this becomes unstable and radioactivity is seen. Such a phenomenon can only be seen in nuclei having high atomic numbers.