Home
Class 12
PHYSICS
The resistance of a bulb filmanet is 100...

The resistance of a bulb filmanet is `100Omega` at a temperature of `100^@C`. If its temperature coefficient of resistance be 0.005 per `.^@C`, its resistance will become `200 Omega` at a temperature of

A

`300^@C`

B

`400^@C`

C

`500^@C`

D

`200^@C`

Text Solution

Verified by Experts

Let resistance of bulb filament be`R_0` at `0^@C`, then from expression
`R_theta=R_0[1+a Delta theta]`
We have `100=R_0[1+0.005 times 100] and 200=R_0[1+0.005 times x]`
Where, x is temperature in `^@C` at which resistance become `200Omega` Dividing the above two equations, we get :
`200/100=(1+0.005x)/(1+0.005 times 100) implies x=400^@C`
Promotional Banner

Similar Questions

Explore conceptually related problems

The resistance of the bulb filament is 100Omega at a temperature of 100^(@)C . If its temperature co-efficient of resistance be 0.005 per 5 ""^(@)C , its resistance will becomes 200 Omega at a temperature

The resistance of a metallic conductor is 50Omega at a temperature of 100^(@)C .If its temperature coefficient of resistance is 0.005^(@)C^(-1) ,then its resistance will become 125Omega at a temperature of

The resistance of a coi is 4.2Omega at 100^(@)C and the temperature coefficient of resistance of its material is (0.004)/(.^(@)C) . Its resistance at 0^(@)C is

If the resistance of a conductor is 5 Omega at 50^(@)C and 7Omega at 100^(@)C then the mean temperature coefficient of resistance of the material is

A conducting wire has resistance 16 Omega at 15^0 C and 20 Omega at 100^0 C find temperature of coeffecient of resistance

The resistance of a conductor at 15^(@)C is 16 Omega and at 100^(@)C is 20 Omega . What will be the temperature coefficient of resistance of the conductor ?

A wire has a resistance of 3.1 Omega at 30°C and a resistance 4.5 Omega at 100°C. The temperature coefficient of resistance of the wire

The resistance of a wire at 20^(@)C is 20Omega and at 500^(@)C is 60Omega . At which temperature its resistance will be 25Omega ?