31 Years NEET Previous Year Questions: Current Electricity - NEET MCQ

# 31 Years NEET Previous Year Questions: Current Electricity - NEET MCQ

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## 30 Questions MCQ Test Topic-wise MCQ Tests for NEET - 31 Years NEET Previous Year Questions: Current Electricity

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31 Years NEET Previous Year Questions: Current Electricity - Question 1

### In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will [2006]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 1

Current will flow from B to A

Potential drop over the resistance CA will be
more due to higher value of resistance. So
potential at A will be less as compared with at
B. Hence, current will flow from B to A.

31 Years NEET Previous Year Questions: Current Electricity - Question 2

### Kirchhoff’s first and second laws for electricalcircuits are consequences of [2006]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 2

Kirchhoff ' s first law deals with conservation
of electrical charge & the second law deals
with conservation of electrical energy.

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31 Years NEET Previous Year Questions: Current Electricity - Question 3

### Two cells, having the same e.m.f., are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is [2006

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 3

Current in the circuit

P.D. across first cell = E – ir1

31 Years NEET Previous Year Questions: Current Electricity - Question 4

The total power dissipated in watts in the circuit shown here is [2007]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 4

The resistance of 6Ω and 3Ω are in parallel in the given circuit, their equivalent resistance is

31 Years NEET Previous Year Questions: Current Electricity - Question 5

Three resistances P, Q, R each of 2 Ω  and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 Ω is connected in parallel to S the bridge gets balanced. What is the value of S?

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 5

For a balanced Wheatstone bridge, we get

Therefore, the equivalent circuit diagram drawn below.

31 Years NEET Previous Year Questions: Current Electricity - Question 6

The resistance of an ammeter is 13 Ω and itsscale is graduated for a current upto 100 amps.After an additional shunt has been connectedto this ammeter it becomes possible to measurecurrents upto 750 amperes by this meter. Thevalue of shunt-resistance is [2007]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 6

We know

31 Years NEET Previous Year Questions: Current Electricity - Question 7

A steady current of 1.5 amp flows through acopper voltameter for 10 minutes. If the electrochemical equivalent of copper is30 × 10–5 g coulomb–1, the mass of copperdeposited on the electrode will be [2007]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 7

We have, m = ZIt
where, Z is the electrochemical equivalent
of copper.
⇒ m = 30 x10-5 x1.5 x10 x 60
= 0.27 gm.

31 Years NEET Previous Year Questions: Current Electricity - Question 8

A current of 3 amp flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5-Ω resistor is: [2008]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 8

Clearly, 2Ω, 4Ω and ( 1 + 5) Ωresistors are
in parallel. Hence, potential difference is
same across each of them.
∴ I1 × 2 = I2 × 4 = I3 × 6

Given I1 = 3A ∴ I1 × 2 = I3 × 6
Given I1 = 3A.
∴ I1 × 2 = I3 × 6 provides

Now, the potential across the 5Ω resistor is
V = I3 × 5 = 1 × 5 = 5V.
∴ the power dissipated in the 5Ωresistor

31 Years NEET Previous Year Questions: Current Electricity - Question 9

A wire of a certain material is stretched slowlyby ten per cent. Its new resistance and specificresistance become respectively: [2008]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 9

In stretching, specific resistance remains unchanged.
After stretching, specific resistance (rho) will remain same.
Original resistance of the wire,

31 Years NEET Previous Year Questions: Current Electricity - Question 10

A cell can be balanced against 110 cm and 100cm of potentiometer wire, respectively with andwithout being short circuited through aresistance of 10Ω. Its internal resistance is [2008]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 10

Here   hence the lengths 110 cm
and 100 cm are interchanged.
Without being short-circuited through R,
only the battery E is balanced.

When R is connected across E, Ri, V/L, l2

Dividing (i) by (ii), we get

or, 100 R + 100 r = 110 R
or, 10 R = 100 r

31 Years NEET Previous Year Questions: Current Electricity - Question 11

A galvanometer of resistance 50 Ω is connectedto battery of 3V along with a resistance of 2950 Ωin series. A full scale deflection of 30 divisions isobtained in the galvanometer. In order to reducethis deflection to 20 divisions, the resistance inseries should be [2008]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 11

Total internal resistance = (50+2950)Ω
= 3000 Ω
Emf of the cell, ε = 3V

Current for full scale deflection of 30
divisions is 1.0 mA.
∴ Current for a deflection of 20 divisions

Let the resistance be x Ω. Then

= 4500 Ω
But the resistance of the galvanometer
is 50Ω
= (4500 –50) Ω= 4450 Ω

31 Years NEET Previous Year Questions: Current Electricity - Question 12

See the electric circuit shown in the figure.

Which of the following equations is a correct equation for it? [2009]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 12

Applying Kirchhoff ’s rule in loop abcfa
ε1 – (i1 + i2) R – i1 r1 = 0.

31 Years NEET Previous Year Questions: Current Electricity - Question 13

A wire of resistance 12 Ωm-1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite point, A and B as shown in the figure, is [2009]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 13

Total resistance of wire = 12Ω×2π×10−1
= 2.4π
Resistance of each half = 2.4π/2​=1.2π
and as about diameter both parts are in parallel
Req. = 1.2π/2​=0.6πΩ
The correct answer is option D.

31 Years NEET Previous Year Questions: Current Electricity - Question 14

A student measures the terminal potentialdifference (V) of a cell (of emf E and internalresistance r) as a function of the current (I)flowing through it. The slope and intercept, ofthe graph between V and I, then, respectively,equal: [2009]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 14

The terminal potential difference of a cell is
given by V + Ir = E

V= VA – VB
or V = E – Ir

Also for, i = 0 then V = E

∴ slope = – r, intercept = E

31 Years NEET Previous Year Questions: Current Electricity - Question 15

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths ℓ1 cm and ℓ2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to             [2010]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 15

(i) When key between the terminals 1
and 2 is plugged in,
P.D. across R = IR = k l1
⇒ R = k l1 as I = 1A
(ii) When key between terminals 1 and 3 is
plugged in,
P.D. across (X + R) = I(X + R) = k l2
⇒ X + R = k l2
∴ X = k (l2 – l1)
∴ R = kl1 and X = k (l2 – l1)

31 Years NEET Previous Year Questions: Current Electricity - Question 16

In producing chlorine by electrolysis 100 kWpower at 125 V is being consumed. How muchchlorine per minute is liberated? (E.C.E. ofchlorine is 0.367×10–6 kg / C) [2010]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 16

Power = 100KW
Voltage = 125V
Power = Vi
100×103 = 125i
i = 100000/125
i = 800A
We know that first law of electrolysis
m = Zit (Z = E.C.E)
= 0.367/1000×10-6×800×60
= 367×48×10-6
= 17616×106 kg
= 17.6×10-3 kg

31 Years NEET Previous Year Questions: Current Electricity - Question 17

Consider the following two statements:
(a) Kirchhoff's junction law follows from the conservation of charge.
(b) Kirchhoff's loop law follows from the conservation of energy.

Which of the following is correct? [2010]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 17

Junction law follows from conservation
of charge and loop law is the conservation
of energy

31 Years NEET Previous Year Questions: Current Electricity - Question 18

The thermo e.m.f E in volts of a certain thermocouple is found to vary with temperature difference θ in °C between the two junctions according to the relation

The neutral temperature for the thermocouplewill be

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 18

For neutral temperature,

Hence, neutral temperature is 225°C.

31 Years NEET Previous Year Questions: Current Electricity - Question 19

A current of 2A flows through a 2Ω resistorwhen connected across a battery. The samebattery supplies a current of 0.5 A whenconnected across a 9Ω resistor. The internalresistance of the battery is [2011]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 19

Let the internal resistance of the battery be
r. Then the current flowing through the
circuit is given by

⇒ 1.5 r  = 0.5 ⇒ r = 1/3Ω

31 Years NEET Previous Year Questions: Current Electricity - Question 20

If power dissipated in the 9-Ω resistor in the circuit shown is 36 watt, the potential difference across the 2-Ω resistor is [2011]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 20

We have,

Current passing through the 9Ω resistor is

The 9Ω and 6Ω resistors are in parallel,
therefore

where i is the current delivered by the
battery.

Thus, potential difference across 2Ω
resistor is
V = iR
= 5 × 2
= 10V

31 Years NEET Previous Year Questions: Current Electricity - Question 21

The rate of increase of thermo–e.m.f. withtemperature at the neutral temperature of athermocouple [2011]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 21

We have,

At neutral temperature,

31 Years NEET Previous Year Questions: Current Electricity - Question 22

A thermocouple of negligible resistanceproduces an e.m.f. of 40 μV/°C in the linear rangeof temperature. A galvanometer of resistance 10ohm whose sensitivity is 1μA/div, is employedwith the termocouple. The smallest value oftemperature difference that can be detected bythe system will be [2011M]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 22

31 Years NEET Previous Year Questions: Current Electricity - Question 23

In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 23

Current from D to C = 1A
∴ VD – VC = 2 × 1 = 2V
VA = 0     ∴ VC = 1V, ∴ VD – VC = 2
⇒VD – 1 = 2     ∴ VD = 3V
∴ VD – VB = 2 ∴ 3 – VB = 2 ∴ VB = 1V

31 Years NEET Previous Year Questions: Current Electricity - Question 24

A milli voltmeter of 25 milli volt range is to beconverted into an ammeter of 25 ampere range.The value (in ohm) of necessary shunt will be :[2012]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 24

Galvanometer is converted into ammeter,
by connected a shunt, in parallel with it.

Here S << G so
S = 0.001 Ω

31 Years NEET Previous Year Questions: Current Electricity - Question 25

In the circuit shown the cells A and B have negligible resistances. For VA = 12V, R1 = 500Ω and R = 100Ω the galvanometer (G) shows no deflection. The value of VB is : [2012]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 25

Since deflection in galvanometer is zero so
current will flow as shown in the above
diagram.

31 Years NEET Previous Year Questions: Current Electricity - Question 26

If voltage across a bulb rated 220 Volt-100 Wattdrops by 2.5% of its rated value, the percentageof the rated value by which the power woulddecrease is : [2012]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 26

Resistance of bulb is constant

31 Years NEET Previous Year Questions: Current Electricity - Question 27

A ring is made of a wire having a resistance R0 = 12 Ω. Find the points A and B as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the sub-circuit between these points is equal to 8/3 Ω

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 27

Let x is the resistance per unit length then

(y2 + 1 + 2y) ×  8/36 y  (where  y l1/l2

31 Years NEET Previous Year Questions: Current Electricity - Question 28

The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is: [2012M]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 28

The power dissipated in the circuit

v = 10 volt

Substituting the values in equation (i)

31 Years NEET Previous Year Questions: Current Electricity - Question 29

Cell having an emf ε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by : [2012M]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 29

The current through the resistance R

The potential difference across R

Thus V increases as R increases upto certain
limit, but it does not increase further.

31 Years NEET Previous Year Questions: Current Electricity - Question 30

A wire of resistance 4 Ω is stretched to twice itsoriginal length. The resistance of stretched wirewould be [NEET 2013]

Detailed Solution for 31 Years NEET Previous Year Questions: Current Electricity - Question 30

Therefore the resistance of new wire
becomes 16 Ω

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