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Resistivity Formula can be expressed as, ρ = RA/l, where: R = Resistance of the conductor, A = Cross-section area, and l = Length of the conductor. Resistivity the resistance of a conductor, when its length and surface area are in unity. The SI unit of resistivity is expressed as ohm-meter and is represented by the greek symbol ‘ρ’.
Resistivity Diagram
Discover about the Chapter video:
Current Electricity Detailed Video Explanation:
Read More: NCERT Solutions for Class 12 Physics Current Electricity
MCQs on Resistivity Formula
Ques 1. The resistance of a wire is directly proportional to ____.
- Area of Cross-Section
- None of these
- Both a and b
- Length
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Ans. (d) Length
Explanation: The resistance of a wire is directly proportional to its length. In other words, the resistance of a wire increases with the increase in length. The resistance of a wire is also inversely proportional to its cross-sectional area.
Also Check: Kirchoff’s Law MCQs
Ques 2. What happens in a series combination of two or more than two resistances?
- Current via each resistance is the same
- Voltage via each resistance is the same
- Both current and voltage via each resistance are same
- Neither of them through each resistance is the same
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Ans. a) Current via each resistance is the same
Explanation: In a series combination, the current across the circuit components always remains constant. However, in a parallel combination, the voltage found across the circuit components is constant.
Ques 3. Assuming that the resistance of an ammeter’s coil is R, then the shunt needed to increase its range n-fold should have a resistance of?
- R/n
- R/(n-1)
- R/(n+1)
- nR
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Ans. b) R/(n-1)
Explanation: As per what is given,
Rg = R
I = nIg
The shunt resistance which is required, Rs = \(\frac{I_gR_g}{I - I_g}\)
⇒ Rs = \(\frac{I_gR}{nI_g - I_g} = \frac{R}{n-1}\)
Ques 4. The flow of current via a coil with resistance 900 Ω has to be minimized by 90%. Find the value of the shunt that should be connected across the coil.
- 100 Ω
- 90 Ω
- 10 Ω
- 9 Ω
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Ans. a) 100 Ω
Explanation: Since we are aware that,
I = \(\frac{V}{R} \)
We can also say that, \( \frac{I_1}{I_2} = \frac{R_1}{R_2}\)
R1 = 900 Ω
Thus, the flow of current via 900 Ω resistance minimized by 90%
The current ratio = 10% : 90% = 1:9
\(\therefore \ \frac{1}{9} = \frac{R_2}{R_1}\)
= \(\frac{1}{9} = \frac{R_2}{900}\)
= \(R_2 = \frac{900}{9} = 100\)
Hence, Shunt = 100 Ω.
Ques 5. In a meter bridge or wheatstone bridge that is used for the measurement of resistance, the known and the unknown resistance are interchanged. Then the error that is removed is:
- index error
- end correction
- due to temperature effect
- random error
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Ans. b) End Correction
Explanation: It is assumed, in a meter bridge experiment, that resistance of the given L-shaped plate is supposed to be negligible, but actually it isn’t. The error that is is given rise to due to this is known as end error. In order to remove this, the resistance box and the unknown resistance will have to be interchanged, with further mean reading.
Ques 6. The internal resistance of a cell r has been connected to an external resistance R. Find when the current will be maximum in R?
- R = r
- R < r
- R > r
- R = r/2
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Ans. a) R = r
Explanation: With the given data, the maximum current which is drawn from the cell will be the case if only the external resistance is minimum.
The current in the circuit can be expressed by:
I = \(\frac{E}{R+r}\)
Therefore, the smallest value of the given denominator will offer maximum current. For maximum current, the value of R must be equivalent to zero.
Also Read:
Ques 7. One silver resistance of 2.1 Ω at 27.5°C, alongside another resistance of 2.7 Ω at 100°C. Find the temperature coefficient of resistivity of silver.
- 0.0039
- 0.0059
- 0.0159
- 0.0129
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Ans. (a) 0.0039
Explanation: The temperature coefficient of resistivity of silver can be expressed as 0.0039.
Ques 8. The resistivity of a few metals or alloys declines to zero when they are chilled below a certain temperature. This phenomenon is usually called:
- Partial Conductivity
- Conductivity
- Superconductivity
- Non-conductivity
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Ans. c) Superconductivity
Explanation: The resistivity of a few metals or alloys declines to zero when they are chilled below a certain temperature. This phenomenon is usually called Superconductivity.
Ques 9. What is the unit of resistivity as per the CGS system?
- \(\Omega - cm\)
- C – J
- C – cms-1
- \(\Omega - m\)
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Ans. a) \(\Omega - cm\)
Explanation: After rearranging the equation of R = \(\rho \frac{l}{A}\)
We get, \(\rho = R \frac{A}{l}\)
Upon substituting the CGS units, we can obtain:
ρ = \(\Omega \frac{cm^2}{cm}\)
Thus, \(\Omega - cm\)
Read More: Wheatstone Bridge MCQs
Ques 10. What does the Electrical resistivity of a given metallic wire depends upon?
- Nature of the material
- Shape
- Length
- Thickness
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Ans. a) Nature of the Material
Explanation: Electrical resistivity, in general, does not depend on the geometrical property of any material. Some properties that it does not depend upon are its length, thickness, shape and more. Thus, it only depends on what the nature of the material is.
Ques 11. The length (0.2 m) and area of wire (0.5 m2) are mentioned, alongside the resistance as 3 Ω. Calculate the resistivity.
- 0.33 \(\Omega m\)
- 11 \(\Omega m\)
- 7.5 \(\Omega m\)
- 6.25 \(\Omega m\)
Click here for Answer
Ans. c) 7.5 \(\Omega m\)
Explanation: According to the question,
- R = 3 Ω
- l = 0.2 m and
- A = 0.5 m2
Hence, the Resistivity formula: \(\rho = \frac{RA}{l}\)
After value substitution, we get:
\(\rho = \frac{3 \times 0.5}{0.2} = 7.5 \Omega m\)
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