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प्रश्न
A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
विकल्प
A/2
3A/2
2A
3A
उत्तर
2A
Explanation -
Let the resistivity of the material be ρ
Resistance (R) of the first cylindrical conductor = `(rho"l")/"A"`
where, l = length of conductor
A = area of cross section
Now, another cylindrical conductor has double length but same resistance.
Let its area of cross section be A'
Its resistance (R') will be = `(rho xx 2"l")/"A"^"'"`
Since resistance is same
R = R'
`(rho xx "l")/"A" = (rho xx 2"l")/"A"^"'"`
Solving it, we get A' = 2A
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संबंधित प्रश्न
The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below:
| V (volts) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 4.0 | 5.0 |
| I (amperes) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.8 | 1.0 |
Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.
State whether the resistivity of a wire changes with the change in the thickness of the wire.
What is the unit of electric charge?
Define one coulomb charge.
Name a device which helps to maintain potential difference across a conductor (say, a bulb).
Name the law which relates the current in a conductor to the potential difference across its ends.
In the circuit shown below:
The potential difference across the 3 Ω resistor is:
What would you suggest to a student if while performing an experiment he finds that the pointer/needle of the ammeter and voltmeter do not coincide with the zero marks on the scales when the circuit is open? No extra ammeter/voltmeter is available in the laboratory.
Suppose there are three resistors A, B, and C having resistances r1, r2, and r3 respectively. If R represents their equivalent resistance, establish the following relation `1/"R" = 1/"r"_1 + 1/"r"_2 + 1/"r"_3`, when joined in parallel.
A battery of 10 volt carries 20,000 C of charge through a resistance of 20 Ω. The work done in 10 seconds is:
