NEET Previous Year Questions (2014-21): Current Electricity Notes | Study Physics Class 12 - NEET

Q.1. Column- I gives certain physical terms associated with flow of current through a metallic conductor.
Column-II gives some mathematical relations involving electrical quantities.
Match Column-I and Column-II with appropriate relations.        (2021)

A: (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
B: (A)-(R), (B)-(Q), (C)-(S), (D)-(P)
C: (A)-(R), (B)-(S), (C)-(P), (D)-(Q)

D: (A)-(R), (B)-(S), (C)-(Q), (D)-(P)
Ans: C



Q.2. In a potentiometer circuit, a cell of EMF 1.5 V gives a balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?        (2021)
A: 64 cm
B: 62 cm
C: 60 cm

D: 21.6 cm
Ans: C


Q.3. The effective resistance of a parallel connection that consists of four wires of equal length,equal area of cross-section and same material is 0.25Ω. What will be the effective resistance if they are connected in series?       (2021)
A: 1Ω
B: 4Ω
C: 0.25Ω

D: 0.5Ω
Ans: B

R = 1Ω
R_series = 4R
= 4(1)
= 4Ω.

Q.4. Three resistors having resistances r₁, r₂ and r₃ are connected as shown in the given circuit. The ratio  i₃/i₁of currents in terms of resistances used in the circuit is:        (2021)

A:
B:
C:

D:
Ans: D
v = ir
i = v/r
i ∝ i/r
[ v is same for r₂ & r₃]



Q.5. Which of the following graph represents the variation of resistivity (r) with temperature (T) For copper?       (2020)
A:

B:

C:

D:

Ans: A
Relation between resistivity and temperature is given by ρ = ρ₀e^-αT

Q.6. A charged particle having drift velocity of 7.5×10^-4 m s^-1 in an electric field of 3×10^-10 Vm^-1 has a mobility in m² V^-1 s^-1 of :        (2020)
A: 2.5×10^-6 
B: 2.25×10^-15 
C: 2.25×10¹⁵ 
D: 2.5×10⁶
Ans: D


μ = 2.5 × 10⁶
(μ = mobility)

Q.7. The color code of a resistance is given below:

The values of resistance and tolerance, respectively, are :        (2020)
A: 4.7 kΩ, 5%
B: 470Ω, 5%
C: 470 kΩ, 5%
D: 47 kΩ, 10%
Ans: B


R = 47 × 10¹Ω ± 5%
= 470 Ω ± 5%
So, 470, 5%

Q.8. The solids which have the negative temperature coefficient of resistance are :        (2020)
A: semiconductors only
B: insulators and semiconductors
C: metals
D: insulators only
Ans: B
Insulators have a negative temperature coefficient because as temperature increases, the resistance of insulators decreases. The resistivity of the semiconductor material to decrease with the rise in temperature, resulting in a negative temperature coefficient of resistance.

Q.9. Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance.        (2019)
The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be :      (2019)

A: 4 : 9
B: 9 : 4
C: 1 : 2
D: 2 : 1
Ans: B
Solution:
(i) All bulbs are glowing


(ii) Two from section A and one from section B are glowing.



Q.10. In the circuits shown below, the readings of the voltmeters and the ammeters will be      (2019)

A: V₂ > V₁ and i₁ = i₂
B: V₁ = V₂ and i₁ > i₂
C: V₁ = V₂ and i₁ = i₂
D: V₂ > V₁ and i₁ > i₂
Ans: C
Solution:
For ideal voltmeter, resistance is infinite and for the ideal ammeter, resistance is zero.


Q.11. A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now,the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is :-    (2018)
A: 10
B: 11
C: 20
D: 9
Ans: A
Solution:


Q.12. A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n ?   (2018)
A:

Ans: A
Solution:

Q.13. Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is     (2018)
A: 40 Ω
B: 25 Ω
C: 250 Ω
D: 500 Ω
Ans: C
Solution:


Q.14. The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be :    (2017)
A: R/n
B: n²R
C: R/n²
D: nR
Ans: B
Solution:

Q.15. A potentiometer  is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves :-     (2017)
A: Potential gradients
B: A condition of no current flow through the galvanometer
C: A combination of cells, galvanometer and resistances
D: Cells
Ans: B
Solution:
In zero deflection condition, potentiometer draws no current.

Q.16. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is :      (2016)
A: 3:2
B: 5:1
C: 5:4
D: 3:4
Ans: A
Solution: According to question, emf of the cell is directly proportional to the balancing length.
i.e., E∝ ℓ    ... (i)
Now, in the first case, cells are connected in series
That is,
Net EMF = E₁ + E₂
From equation (i),
E₁ + E₂ = 50 cm (given)  ... (ii)
Now, the cells are connected in series in the opposite direction,
Net emf = E₁ - E₂
From equation (i)
E₁ - E₂ = 10       ... (iii)
From equation (ii) and (iii),


Q.17. The charge flowing through a resistance R varies with time t as Q = at - bt², where a and b are positive constants. The total heat produced in R is :     (2016)
A: a³R/b
B: a³R/6b
C: a³R/3b
D: a³R/2b
Ans: B
Solution:
Given,
Charge, Q = at - bt²   ... (i)
We know that,
Current, I = dθ/dt
So, equation (i) can be written as,

For maximum value of t, the current is given by,
a-2bt = 0
Therefore, t = a/2b    ..(iii)
Total heat produced (H) can be given as,

On solving the above equation, we get

Given,
Charge, Q = at - bt²   ... (i)
We know that,
Current, I = dθ/dt
So, equation (i) can be written as,

For maximum value of t, the current is given by,
a-2bt = 0

Total heat produced (H) can be given as,

On solving the above equation, we get


Q.18. A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively.     (2015)
Then :