CBSE Class 12 Physics 2016 Delhi Set 1 Paper

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Question : 26 of 26
 
Marks: +1, -0
(i) Define the term drift velocity.
(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?
(iii) Why alloys like constantan and manganin are used for making standard resistors?
OR
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1m and resistance 10. Calculate the potential gradient along the wire and balance length AO(=l).
Solution:
(i) Definition of drift velocity
(ii) Derivation of expression of resistivity
Factors affecting resistivity
(iii) Reason of using constantan and manganin
(i) Average velocity acquired by the electrons in the conductor in the presence of external electric field.
[Alternatively:
vd=eEτm where τ is the relaxation time.]
(ii) vd=eEτm
We have E=V, where V is potential across the length l of the conductor
vd=eVτm
Current flowing I=neAvd
I=neAvdeVrm
=ne2AVτm
IV=ne2Aτm=1R ....(i)
Also, R=ρA .....(ii)
Comparing (i) and (ii),
ρ=mne2τ
Resistivity of the material of a conductor depends on the relaxation time, i.e., temperature and the number density of electrons.
(iii) Because constantan and manganin show very weak dependence of resistivity on temperature.
OR
(i) Working principle of potentiometer
(ii) Calculation of potential gradient and balance length
(i) When constant current flows through a conductor of uniform area of cross section, the potential difference, across a length l of the wire, is directly proportional to that length of the wire.
[Vl (Provided current and area are constant]
(ii) Current flowing in the potentiometer wire
i=ERtotal =2.015+10=225A
Potential difference across the two ends of the wire
VAB=225×10V=2025=0.8 volt
Hence potential gradient
K=VABlAB=0.81.0=0.8Vm
Current flowing in the circuit containing experimental cell,
=1.51.2+0.3=1A
Hence, potential difference across length AO of the wire
=0.3×1V=0.3V
0.3=K×lAO
=0.8×IAO
lAO=0.30.8m=0.375m
=37.5cm
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