A metallic cylindrical conductor is used to produce some heat by applying a constant voltage between it's two ends. It's cold, and you need to double the heat released. Which of the following is the most appropriate thing to be done?
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I doubled resistance instead of power. Anyway it was a good question..
If I double only the radius,then I will get greater power
To increase the heat we have to increase current, to increase current have to reduce the resistance so by increasing area we can reduce the resistance so I think the ans is radius is doubled
Exactly. Thanks. This was my first E&M problem.
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I see. So are you in 11th class or 10th class.
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10th. And you in 12th?
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@Satvik Golechha – Yes. Then you would have only basic knowledge of mechanics.
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@Ronak Agarwal – @Ronak Agarwal what resources do you use for EM and Geometry
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@Mardokay Mosazghi – Honestly speaking I am not very good in geometry, I am good only in analytical geometry, while preparing for IIT-JEE we do topics like Geometry of triangles, also we prepare a lot for co-ordinate geometry and trignometry which makes many of the good geometry problems possible to solve for me.
For Electricity and Magnetism I have only the resources given to me by coaching centre in which I am studying, for the preparation of IIT-JEE.
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@Ronak Agarwal – thanks for the reply, you are a great inspiration to me solving every level 5 EM and Mechanics questions.
@Ronak Agarwal – I know a little bit of class 11, like projectile and relative.
Hey I'm also 14 . I would like to be your friend .. I'm too interested in phu and am in class 9 but completed all class 10 maths and physics .. Can you tell me which book is good yo read the advance phy ?
nice first problem
The second equation relating resistance and resistivity is wrong. The area, 'A' should be in the denominator. But you have substituted correctly in the third step, so the final result is correct
This question requires review. Remember the voltage is constant, doubling the radius reduces the resistivity hence reducing the heat production.
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Simple
Power = R V 2
Also R = A ρ l = π r 2 ρ l
Power = P = ρ l π r 2 V 2
To double the heat you have to double the power hence simply double the radius as well as length because :
= ρ ( 2 l ) π ( 2 r ) 2 V 2 = 2 ρ l π r 2 V 2 = 2 P