33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Current Electricity
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Hints & Explanations
27
1. (a) : Flow of electrons, n t = 107 /second
Therefore, current( I)
= q ne n
t
= t
= t
×e
= 10 7 × (1.6 × 10 –19 ) = 1.6 × 10 –12 A
2. (b) : The resistance of a wire of length l and area A
and resistivity r is given as
l
R = ρ
A
Given, l′ = nl
As the volume of the wire remains constant
Al Al A
\ A′l′ = Al or A′= = or A′=
l ′ nl n
2
2
∴
3.
ρl′
R′=
ρnl
n ρl
or R′= = =nR
A′
A A
n
l
(b) : Resistance of a wire, R = ρ = 4 Ω
A
...(i)
When wire is stretched twice, its new length be l′. Then
l′ = 2l
On stretching volume of the wire remains constant.
\ lA = l′A′ where A′ is the new cross-sectional area
or
l
A′ = A = =
l ′ l A A
2 2
\ Resistance of the stretched wire is
R ′
′ = l
A′ = 2l
A
= l
4
( / 2)
A
= 4(4 W) = 16 W (Using (i))
∆l
4. (d) : = 01 . ∴ l = 11 .
l
But the area also decreases by 0.1.
Mass = rlA = Vr, ln l + ln A = ln mass.
∴ ∆l
+ ∆ A
= ⇒ ∆ l
= −∆
A
0
l A l A
Length increases by 0.1, resistance increases, area
decreases by 0.1, then also resistance will increase. Total
increase in resistance is approximately 1.2 times, due
to increase in length and decrease in area. But specific
resistance does not change.
5. (b) : Resistance of wire =ρ l A
l l
R ∝ =
A π
2
r
When length and radius are both doubled
2l
1
R1 ∝ ⇒R
R
2 1 ∝
π( 2r)
2
The specific resistance of wire is independent of geometry
of the wire, it only depends on the material of the wire.
6. (c) : According to given parameters in question
l
3 ρ 100
R = ρ ⇒100
Ω= ρ ⇒ = ⋅
A A A 3
Thus total resistance of 50 cm wire is
ρ
R1
A l 100 50
= = × 05 . = Ω.
3 3
The total current in the wire is I = 6
100 A.
Therefore potential difference across the two points on
the wire separated by a distance of 50 m is
50 6
V = IR1
= × =
3 100 1 V
7. (a) : Three wires of lengths and cross-sectional areas
= (l, A), (2l, A/2) and (l/2, 2A).
l
Resistance of awire R ∝
A
For I st wire, R 1 ∝ l/A = R
nd
2l
ForII wire,
R2
∝ =4R
A / 2
rd
l / 2 R
For III wire,
R3
∝ =
2A
4
Therefore resistance of the wire will be minimum for III rd
wire.
8. (b) : Length (l) = 50 cm = 0.5 m;
Area (A) = 1 mm 2 = 1 × 10 –6 m 2 ;
Current (I) = 4A and voltage (V) = 2 volts.
Resistance( R) = V
.
I
= 2
4
= 05Ω
Resistivity( ρ= ) × = . × × −
R
A 05 1 10 6
= 1 × 10 –6 W m
l 05 .
9. (d) : m = l × area × density
Area ∝ m l
2
l l
R ∝ ∝
Area m
l1 2 l2 2 l3 2 R1: R2 : R3
= : :
m1
m2
m3
R1 : R2 : 25 9 1
R3
= 125 15 1
1
: 3
: 5 = : :
10. (b) : Here, v d = 7.5 × 10 –4 m/s, E = 3 × 10 –10 V/m
Mobility, m = vd −4
75 . × 10
=
E −10
3×
10
m = 2.5 × 10 6