Question

The material which is used as a filament for a bulb has high ________.
a) Boiling point
b) Melting point
c) Freezing point
d) pressure

Answer 1

Hint: The key idea behind a bulb to glow is the principle of heating of the filament when a current flows through it.

Complete step by step answer:
Lighting of a bulb has a simple mechanism. It has a filament, which is a substance (conductor) with high resistance with a property of glowing when heated up. We know that when a current flows through a conductor, heat is produced. This heat produced is directly proportional to the square of the current flowing through the conductor.
As it is said that the filament of the bulb glows when it heats up. Therefore, when a current flows through the filament, heat is produced and the bulb glows. However, it should be ensured that the filament does not melt due to excess heating. If the filament melts the circuit will break and there will be no current flowing in the circuit. Therefore, the filament used in the bulb must have a high melting point so that it does not melt easily. Melting point is the temperature at which the substance starts to melt.
Hence, the correct option is (b) melting point.

Additional Information:
The filament used in the bulb is tungsten. It is a rare metal extracted from the ores of wolframite and scheelite. It is a lustrous and silvery white metal. Tungsten has the highest melting point. It is also used in electric contacts, arc-welding electrodes and making bulletproof armours.

Note: Suppose someone thinks of producing a very bright light by taking a material of very large resistance and producing a large amount of heat. To draw a maximum amount of heat from the circuit of constant voltage, taking a large amount of resistance will not be a good idea. Since $V=iR$ (‘V’ is the voltage of the battery or the potential difference, ’i’ is the current and R is the resistance), for constant voltage, if the value of R is large the current will be very small. Power of heat produced (P) is given by $P=Vi$. Therefore, the power of heat produced will be less and there will not be a brighter light.