Physics Previous year question Of Current Electricity For NDA

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Previous year question Of Current Electricity For NDA

Previous year question Of Current Electricity

Previous year question

Q 2734001852

Which one of the following physical quantities does NOT affect the resistance of a cylindrical resistor?
NDA Paper 2 2017

(A)

The current through it

(B)

Its length

(C)

The resistivity of the material used in the reststor

(D)

The area of cross-section of the cylinder

Solution:

Correct Answer is `=>` (A) The current through it

Q 2158223104

Consider the following circuit..

The equivalent resistant of the circuit will be.
NDA Paper 2 2016

(A)

`12 Omega`

(B)

` 8 11/12 Omega`

(C)

` 9 1/11 Omega`

(D)

` 24/25 Omega`

Solution:

This circuit consists of two sections, `I` and `II` In first

section, three resistors are in parallel connection

therefore equivalent resistance is

` 1/R = 1/2 +1/4 +1/6 = (6+3+2)/12`

` => R_(1) = 12 /11 Omega`

This resistance is in series with the resistance `8 Omega` of

section `II`.

`:.` Equivalent resistance of the circuit

`R_(eq) =R_(1) + 8`

` = 12/11 + 8 = 100/11 = 9 1/11 Omega`

Correct Answer is `=>` (C) ` 9 1/11 Omega`

Q 2158234104

A simple circuit contains a `12 V` battery and bulb having `24` ohm resistance. When turn on the switch, the ammeter connected to the circuit would read.
NDA Paper 2 2016

(A)

`0.5 A`

(B)

`2 A`

(C)

`4 A`

(D)

`5 A`

Solution:

It is given that emf of the battery, `E = 12 V`

Resistance of the bulb, `R = 24 Omega`

where switch is turned on, current through the bulb is

` I = E/R = 12/24 = 0.5A`

Hence, reading of ammeter is `0.5 A.`

Correct Answer is `=>` (A) `0.5 A`

Q 2128334201

Three resistors with magnitudes `2, 4, 8` ohm are connected in parallel equivalent resistance of the system would be.
NDA Paper 2 2016

(A)

less than `2` ohm

(B)

more than `2` ohm but less than `4` ohm

(C)

`4` ohm

(D)

`14` ohm

Solution:

Equivalent resistance of three resistors of resistances

`2, 4` and `8 Omega` is

` 1/R_(eq) = 1/2 +1/4 +1/8 = (4 + 2 + 1 )/8 = 7/8`

` :. R_(eq) = 8/7 Omega = 1.142 Omega`

Therefore, equivalent resistance is less than `2 Omega`.

Correct Answer is `=>` (A) less than `2` ohm

Q 2374556456

A given conductor carrying a current of 1 ampere. It produces an amount of heat equal to 2000 J. The current through the conductor is doubled the amount of heat produced will be.
NDA Paper 2 2016

(A)

2000 J

(B)

4000 J

(C)

8000 J

(D)

1000 J

Solution:

Let resistance of the coil is R and measuring heat
produced for time t seconds.

`:. H = I^2 RT =(1)^2 (R)(t) => 2000 J = Rt`

Similarly, when current is doubled, the amount of heat
produced is

`H' =(I')^2 RT = (2^2) (Rt) = 4 Rt`

`=4 xx 2000 J = 8000 J` `[ because Rt =2000 J]`

Correct Answer is `=>` (C) 8000 J

Q 2762434335

Which one of the following statements is not correct?
NDA Paper 2 2016

(A)

The SI unit of charge is ampere second

(B)

Debye is the unit of dipole moment

(C)

Resistivity of a wire of length `l` and area of cross-section `a` depends upon both `l` and `a`

(D)

The kinetic energy of an electron of mass `m` kg and charge `q` coulomb, when accelerated through a potential difference of `V` volt, is `eV` joule

Solution:

Correct Answer is `=>` (C) Resistivity of a wire of length `l` and area of cross-section `a` depends upon both `l` and `a`

Q 1658091804

The product of conductivity and resistivity of a conductor resistivitv of a conductor
NDA Paper 2 2015

(A)

depends on pressure applied

(B)

depends on current flowing through conductor

(C)

is the same for all conductors

(D)

varies from conductor-to-conductor

Solution:

Resistivity `alpha = 1/text(Conductivity)`

`therefore` Resistivity `xx` Conductivity = 1

Therefore, the product of conductivity and resistivity of a conductor is same for all conductors.

Correct Answer is `=>` (C) is the same for all conductors

Q 1687380287

Three equal resistances when combined in series are

equivalent to 90 ohm. Their equivalent resistance when

combined in parallel will be
NDA Paper 2 2015

(A)

`10 ohm`

(B)

`30 ohm`

(C)

`270 ohm`

(D)

`810 ohm`

Solution:

Let eaeh resistance is `R`.

According to the question,

`R+ R+ R= 90 Omega`

`=> R=30 Omega`

where these resistors are combined in parallel,

equivalent resistance is

`R_(eq) =R/3 = (30)/3 = 10 Omega`

Correct Answer is `=>` (A) `10 ohm`

Q 1627491381

The resistance of a wire of length `l` and area of

cross-section `a` is `x` ohm. If the wire is stretched to

double its length, its resistance would become
NDA Paper 2 2015

(A)

`2x quad ohm`

(B)

`0.5x quad ohm`

(C)

`4x quadohm`

(D)

`6x quad ohm`

Solution:

According to the question, resistance of wire is given by

`x = (pl)/a ` [p =resistivity, l =length, a = area]

Note If length is doubled, its area of cross-section

decreases but volume remains constant.

i.e. `l xx a= 2l xx A'` [A' =New area]

`=> A'= a/2`

Now, new resistance,

`R= (p2l)/(a//2) = (4pl)/a`

as `(pl)/a = x`

So, `R = 4x`

Correct Answer is `=>` (C) `4x quadohm`

Q 1782267137

During short-circuiting, the current flowing in the
electrical circuit
NDA Paper 2 2014

(A)

reduces substantially

(B)

does not change

(C)

increases instantaneously

(D)

varies continuously

Solution:

Correct Answer is `=>` (A) reduces substantially

Q 3181723627

The Current (I)- Voltage (V) plot of a certain
NDA Paper 2 2014

(A)

a semiconductor

(B)

a conductor which obeys Ohm's law

(C)

a superconductor

(D)

an insulator

Solution:

Correct Answer is `=>` (B) a conductor which obeys Ohm's law

Q 1712167039

Two conducting wires `A` and `B` are made of same material.
If the length of `B` is twice that of `A` and the radius of
circular cross-section of `A` is twice that of `B`, then their
resistances `R_ A` and `R_B` are in the ratio
NDA Paper 2 2014

(A)

`2: 1`

(B)

`1 : 2`

(C)

`1 : 8`

(D)

`1 : 4`

Solution:

Given that `I_B =2I_A` and `r_A = 2r_B`

Now, from `R = rho I/A xx (r_(B^2))/ (r_(a^2)) = I_A/(2I_A) xx (r_(B^2))/(r_(B^2))`

` = 1 : 8`

Correct Answer is `=>` (C) `1 : 8`

Q 1618556409

When you walk on a woolen carpet and bring your

finger near the metallic handle of a door an

electric shock is produced. This is because
NDA Paper 2 2014

(A)

charge is transferred from your body to the handle

(B)

a chemical reaction occurs when you touch the handle

(C)

the temperature of the human body is higher than that of the handle

(D)

the human body and the handle arrive at thermal equilibrium by the process

Solution:

When you walk on a wooden cerpet and bring your

finger near the metallic handle of a door, an electric shock

is produced because charge is transferred from your

body to the handle.

Correct Answer is `=>` (A) charge is transferred from your body to the handle

Q 2333278142

Power required by a boy of mass `30` `kg` to run up a staircase of `40` steps in `10 s` is

(Height of each step is `15 cm`)
(Take, `g =10 m//s^2`)
NDA Paper 2 2013

(A)

`1800 W`

(B)

`180 W`

(C)

`18000 W`

(D)

`18 W`

Solution:

`P=W/t=(mgh)/t`

`=(30 xx 10 xx 40 xx 0.15)/10=180 W`

Correct Answer is `=>` (B) `180 W`

Q 2313778640

Ohm's law can also be taken as a statement for
NDA Paper 2 2013

(A)

conservation of energy

(B)

conservation of electric charge

(C)

conservation of angular momentum

(D)

non-conservation of momentum of the flowing charges

Solution:

Ohm's law follow the law of conservation of energy.
According to ohm's law,

`text(Electric current) (I) = (text(Voltage) (V))/(text(Resistance) (R))`

Correct Answer is `=>` (A) conservation of energy

Q 2303180048

What should be the reading of the voltmeter `V` in the circuit given above?

(All the resistances are equal to `1 Omega` and the battery is of `1.5 V`)
NDA Paper 2 2013

(A)

`1.5 V`

(B)

`0.66 V`

(C)

`1 V`

(D)

`2 V`

Solution:

The given circuit is

Equivalent resistance of the circuit

`=1+( 1 xx 1)/(1+1)`

`=1+1/2=3/2 Omega`

Total voltage `(V) = 1.5 V`

Current ` (I)=V/(R_(eq))`

`=(1.5)/(3//2) =1 A`

Reading of voltmeter `(V) = I xx 1`

`=1 xx 1=1 V`

Correct Answer is `=>` (C) `1 V`

Q 2383480347

An electric heater is rated 1500 W, electric power costs Rs `2` per, kilo-watt-hour, then the cost of power for `10 h` running the heater is
NDA Paper 2 2013

(A)

` Rs 30`

(B)

`Rs 15`

(C)

`Rs 150`

(D)

Rs 25

Solution:

Power of electric heater `(p) = 1500 W`

`1` unit `= 1 kWh = 1. 5 xx 1`

`=1.5` unit

For `10` h,

The unit will be `= 1.5 xx 10 =15` unit

Cost of `1` unit `= Rs 2`

Cost of `15` unit `= Rs 15 xx 2 = Rs 30`

Correct Answer is `=>` (A) ` Rs 30`

Q 2374223156

Ohm's law defines
NDA Paper 2 2013

(A)

a resistance

(B)

current only

(C)

voltage only

(D)

both current and voltage

Solution:

Ohm's law states that the current through a conductor
between two points is directly proportional to the potential
difference across the two points. Introducing the constant of
proportionality, the resistance

`I=V/R` or `V/I=R` (constant)

where, `I->` in ampere, `V ->` in volt, `R ->` ohm (`Omega`)

Correct Answer is `=>` (A) a resistance

Q 2334234152

A current `I` flows through a potential difference `V` in an electrical circuit containing a resistance `R`. The product of `V` and `I` i.e., `VI` may be understood as.
NDA Paper 2 2013

(A)

resistance `R`

(B)

heat generated by the circuit

(C)

thermal power radiated by the circuit

(D)

rate of change of resistance

Solution:

`P =VI`, it means that the thermal power is radiated by
the circuit.

Correct Answer is `=>` (C) thermal power radiated by the circuit

Q 2334245152

Which one among the following is the correct order of power consumption for light of equal intensity?
NDA Paper 2 2012

(A)

CFL Tube < Fluorescent Tube < Incandescent Bulb < Light Emitting Diode

(B)

Light Emitting Diode < CFL Tube < Fluorescent Tube < Incandescent Bulb

(C)

CFL Tube < Fluorescent Tube < Light Emitting Diode < Incandescent Bulb

(D)

Incandescent Bulb < Light Emitting Diode < Fluorescent Tube < CFL Tube

Solution:

A Compact Fluorescent Lamp (CFL), also called compact fluorescent light. energy-saving light and compact fluorescent tube, is a fluorescent lamp designee! to replace an incandescent lamp; some types fit into light fixtures formerly used for incandescent lamps. Compared to (general-service incandescent lamps giving the same amount of visible light, CFLs use one-fifth to one-third the electric power. and last eight to fifteen times longer. A Light Emitting Diode (LED) is a semiconductor light source. LEOs are used as indicator lamps in many devices and are increasingly used for other lighting. Light-emitting diodes are used in applications as diverse as aviation lighting, automotive lighting, advertising, general lighting, and traffic signals. There is a difference of 0.4 W power consumption between the CFL and LED.

Correct Answer is `=>` (B) Light Emitting Diode < CFL Tube < Fluorescent Tube < Incandescent Bulb

Q 2364845755

Which one among the following is the true representation of (i) variable DC potential (ii) rheostat and (iii) AC ammeter respectively?
NDA Paper 2 2012

Solution:

Option (c) is the true representation of (i) variable DC potential (ii) rheostat and (iii) AC ammeter.

Correct Answer is `=>` (C)

Q 2314045850

In India, distribution of electricity for domestic purpose is done in the form of
NDA Paper 2 2012

(A)

`220 V; 50 Hz`

(B)

`110 V; 60Hz`

(C)

`220 V; 60 Hz`

(D)

`110 V; 50 Hz`

Solution:

In India, distribution of electricity for domestic purpose is done in the form of `220V; 50Hz.`

Correct Answer is `=>` (A) `220 V; 50 Hz`

Q 2384756657

Consider the following circuit. The current flowing through each of the resistors connected in the above circuit is
NDA Paper 2 2012

(A)

2 A

(B)

1 A

(C)

9 A

(D)

4 A

Solution:

Equivalent resistance of circuit `1/R_(eq)=1/R_1 +1/R_2`

`1/R_(eq) = 1/3 +1/3=2/3`

`R_(eq) = 3/2`

Now `V=IR`

`I=6/(3/2)`

`I=4A`

Since both resistance have same values so current will be equally divided in both resistances

Current in each resistance `I=V/R=6/3=2A`

Correct Answer is `=>` (A) 2 A

Q 2314267150

The power supply in India is at `220 V`, whereas that in the US is at `11 0 V` Which one among the following statements in this regard is correct?
NDA Paper 2 2012

(A)

`110 V` is safer but more expensive to maintain

(B)

`110 V` is safer and cheaper to maintain

(C)

`110 V` leads to lower power loss

(D)

`110 V` works better at higher latitudes

Solution:

Power loss is given by `P=VI` i.e., `PpropV` So `110 V` leads to lower power loss

Correct Answer is `=>` (C) `110 V` leads to lower power loss

Q 2384580457

The resistance of a wire is `10 Omega`. If it is stretched ten times, the resistance will read
NDA Paper 2 2012

(A)

`1 Omega`

(B)

`10 Omega`

(C)

`100 Omega`

(D)

`1000 Omega`

Solution:

Resistance `= 10^3 = 1000 Omega R prop e^2`

Correct Answer is `=>` (D) `1000 Omega`

Q 2384780657

Two metallic wires A and B are of same material and have equal length. If the cross-sectional area of B is double that of A, then which one among the following is the electrical resistance of B'?
NDA Paper 2 2012

(A)

Twice that of A

(B)

`4` times that of A

(C)

`1/4 ` that of A

(D)

`1/2` that of A

Solution:

Electrical resistance of B is became half of the electrical resistance of A.

Correct Answer is `=>` (D) `1/2` that of A

Q 2364391255

Two copper wires A and B of length 1 and 2 respectively, have the same area of cross-section. The ratio of the resistivities of wires A to the B is
NDA Paper 2 2011

(A)

(B)

(C)

(D)

`1/2`

Solution:

Resistivity,` rho=R A/l=> rho prop A/l`

`(rho_A)/(rho_B)=A_A/A_B xx l_B/l_A= 1/1 xx (2l)/l=2`

Correct Answer is `=>` (B) 2

Q 2345712663

Three resistance coils of restiveness `1 Omega, 2 Omega` and `3 Omega` are connected in series. If the combination is connected to a battery of `9 V`, what is the potential drop across the resistance coil of `3 Omega`?
NDA Paper 2 2011

(A)

`2.0 V`

(B)

`3.0 V`

(C)

`4.5 V`

(D)

`6.0 V`

Solution:

Effective resistance, `R = R_1 + R_2 + R_3`

`=1+2+3=6Omega`

According to Ohm's law, `I=V/R=9/6= 1.5A`

`:.` Potential drop across the resistance coil of `3 Omega`

`V =I xx R`

`=1.5 xx 3=4.5V`

Correct Answer is `=>` (C) `4.5 V`

Q 2375812766

An electric lamp of `1 00 W` is used for `10 h` per day. The units of energy consumed in one day by the lamp is
NDA Paper 2 2011

(A)

`1` unit

(B)

`0.1` unit

(C)

`10` units

(D)

`100` units

Solution:

Energy consumed in a day `= 100 W xx 10 h`

`= 1000` Wh

`= 1 kWh = 1` unit

Correct Answer is `=>` (A) `1` unit

Q 2305012868

Kilowatt-hour is the unit of
NDA Paper 2 2011

(A)

potential difference

(B)

electric power

(C)

electric energy

(D)

electric potential

Solution:

Unit of potential difference = Volt

Unit of electric power= Watt

Unit of electric energy= Kilo -watt hour

Unit of electric potential =Volt

Correct Answer is `=>` (C) electric energy

Q 2315845769

If a heater coil is cut into two equal parts and only one part is used in the heater the heat generated will be
NDA Paper 2 2010

(A)

doubled

(B)

four times

(C)

one-fourth

(D)

halved

Solution:

Electrical power, `P=V^2/R`

`=> P prop 1/R`

But `R prop l`

so, `P prop 1/l`

If a heater coil is cut into two equal parts and only one part is used in heater, then heat generated will be double.

Correct Answer is `=>` (A) doubled

Q 2315067869

The effective resistance of three equal resistances, each of resistance `r`, connected in parallel, is
NDA Paper 2 2010

(A)

`3/r`

(B)

`r/3`

(C)

`3r`

(D)

`r^3`

Solution:

In parallel grouping of resistar,ces, the potential difference across each resistance is same but current in the circuit is distributed amongst various resistances in the inverse ratio of their resistance. i.e., `I_1:I_2:I_3=1/R_1 :1/R_2 :1/R_3`

The equivalent resistance for a parallel combination is given as

`1/R=1/R_1+1/R_2-1/R_3`

According to question, `R_1 = R_2 = R_3 = r`

`:. 1/R=1/r+1/r+1/r`

`=> R=r/3`

Thus, in parallel combination, the equivalent resistance is lesser than even the smallest individual resistance.

Correct Answer is `=>` (B) `r/3`

Q 2365678565

The specific resistance of a conducting wire depends upon
NDA Paper 2 2010

(A)

length of the wire, area of cross-section of the wire and material o1 the wire

(B)

length of the wire and area of cross-section of the wire but not on the material of the wire

(C)

material of the wire only but neither on the length of the wire nor on the area of cross-section of the wire

(D)

length of the wire only but neither on tile area of cross-section of the wire nor on the material of the wire

Solution:

Specific resistance or resistivity is a property of material and does not depend upon its shape and size, its length, thickness and area of cross-section.

Correct Answer is `=>` (C) material of the wire only but neither on the length of the wire nor on the area of cross-section of the wire

Q 2355391264

How many sixty watt ( `60 W`) bulbs may be safely used in a `240 V` supply with `4 A` fuse?
NDA Paper 2 2010

(A)

(B)

(C)

(D)

Solution:

Electric power, `P = V xx I`

`=240 xx 4=960 W`

`:.` Number of bulbs of `60 W = 960/60=16` bulbs.

Correct Answer is `=>` (D) 16

Q 2326812771

A house, served by a `220 V` supply line, is protected by a `9 A` fuse. What is the maximum number of `60 W` bulbs that can be turned on in parallel?
NDA Paper 2 2010

(A)

`11`

(B)

`22`

(C)

`33`

(D)

`44`

Solution:

Current required by each bulb

`I=P/V=60/220 A`

If `n` bulbs are joined in parallel, then

`nI=I_(fuse)`

`n xx 60/220 =9`

or, `n=(9 xx 220)/60=33`

Correct Answer is `=>` (C) `33`

Q 2346712673

A current `I` in a lamp varies with voltage `V` as shown in the figure given above. Which one of the following is the variation of power `P` with current `I` ?
NDA Paper 2 2009

Solution:

`P=V^2/R` or `P=(I^2R^2)/R=I^2R`

or `P prop i^2`

Hence, graph between `P` and `i` will be a hyperbola shown in fig (2)

Correct Answer is `=>` (B)

Q 2376712676

Which one of the following is the resistance that must be placed parallel with `12 Omega` resistance to obtain a combined resistance of `4 Omega` ?
NDA Paper 2 2009

(A)

`2 Omega`

(B)

`4 Omega`

(C)

`6 Omega`

(D)

`8 Omega`

Solution:

Let resistance `x` is in parallel with `12 Omega` resistance

Then, `R=(x xx 12)/(x+12)`

`=> 4=(12x)/(x+12)`

`=> 4x+48=12x`

or, `8x=48`

or `x=6 Omega`

Correct Answer is `=>` (C) `6 Omega`

Q 2316712679

An electric iron of resistance `20 Omega` takes a current of `5 A`. The heat developed in joules in `30` s is
NDA Paper 2 2009

(A)

`5 kJ`

(B)

`10 kJ`

(C)

`15 kJ`

(D)

`20 kJ`

Solution:

Heat developed in an electric iron is

`H=i^2Rt`

Here, `i=5A,R=20 Omega,t=30s`

`:. H=(5)^2 xx 20 xx 20 xx 30=25 xx 20 xx 30`

`=15000 J=15 kJ`

Correct Answer is `=>` (C) `15 kJ`

Q 2316756670

In which of the following cases Ohm's law is not valid?
NDA Paper 2 2008

(A)

Wire bound resistor

(B)

Potentiometer

(C)

Junction diode

(D)

Electric bulb

Solution:

Ohm's law is not a universal law. Substances which obey Ohm's law are known as ohmic resistors, such as pure metals, alloys and mercury. Substances which do not obey Ohm's law are called non-ohmic resistors such as gases, electrolytes, transistors, rectifiers etc.

Correct Answer is `=>` (C) Junction diode

Q 2326367271

Consider the following. Heat produced in a conductor carrying current is independent of
1. current passing through it.
2. thermal conductivity.
3. specific resistance.
Which of the statements given above is/are correct?
NDA Paper 2 2008

(A)

1 and 3

(B)

Only 2

(C)

1 and 2

(D)

2 and 3

Solution:

For a given current, the heat produced is directly proportional to the resistance, i.e., `H prop R`.
For a given resistor and time amount of heat produced is directly proportional to the square of current flowing, i.e., `H prop I^2`.

Correct Answer is `=>` (B) Only 2

Q 2346867773

DC current can be controlled by which one of the following components'?
NDA Paper 2 2008

(A)

Impedance

(B)

Resistance

(C)

Capacitance

(D)

Inductance

Solution:

DC current can be controlled by using resistance

Correct Answer is `=>` (B) Resistance

Q 2346478373

If the current is flowing through a `10 Omega` resistor, then in which one of the following cases maximum heat will be generated?
NDA Paper 2 2008

(A)

5 A in 2 min

(B)

4 A in 3 min

(C)

3 A in 6 min

(D)

2 A in 12 min

Solution:

(a) Heat generated `H = i^2Rt = (5)^2 xx 10 xx 2 xx 60`

`= 30000 J`

(b) Heat gelnerated `H = (4)^2 xx 10 xx 3 xx 60`

`= 28800 J`

(c) Heat generated `H = (3)^2 xx 10 xx 6 xx 60`

`= 32400 J`

(d) Heat generated `H = (2)^2 xx 10 xx 12 xx 60`

`= 28800 J`

Correct Answer is `=>` (C) 3 A in 6 min

Q 2346578473

The rating of an electric lamp is `110 V` To use it on `220 V`, one will have to use which one of the following?
NDA Paper 2 2008

(A)

Transistor

(B)

Resistor

(C)

Transformer

(D)

Generator

Solution:

To use the electric lamp on `220 V`, one should use
resistor in series with the lamp.

Correct Answer is `=>` (B) Resistor

Q 2356778674

The figure shows current in a part of electrical network. What is the value of current `I` ?
NDA Paper 2 2008

(A)

`0.2 A`

(B)

`0.1 A`

(C)

`0.3 A`

(D)

`0.5 A`

Solution:

According to Kirchhoff's law, the algebraic sum of current at a junction is zero.

According to question, At junction `A`

`3 + 1- I_1 = 0`

`=> I_1 = 4A`

At junction B

`4-2 -I_2 = 0`

`=>I_2 = 2 A`

At junction C

`2-I-1.5=0`

`=> I= 0.5` A

Correct Answer is `=>` (D) `0.5 A`

Q 2386878777

A wire has a resistance of `32 Omega`. It is melted and drawn into a wire of half of its original length. What is the resistance of the new wire?
NDA Paper 2 2008

(A)

`32 Omega`

(B)

`16 Omega`

(C)

`8 Omega`

(D)

`4 Omega`

Solution:

Resistance of wire, `R = 32 Omega`

Resistance, `R = rho l/A`

i.e., `R prop l/A`

or `R prop l^2/(Al)`

or `R=l^2/V`

`:. (R')/R=((l')/l)^2` (given, `V` is the same for both conditions)

`=> (R')/32=((l//2)/l)^2=1/4`

or `R'=32/4=8 Omega`

Correct Answer is `=>` (C) `8 Omega`

Q 2356178974

The question consists of two statements one labelled as the Assertion (A) and the other as Reason (R). You are t:o examine these two statements carefully and select the answers to these items using the codes given below.
NDA Paper 2 2008

Assertion : If the filament of a light bulb is not uniform, its life is shortened.

Reason : Resistance of glowing light bulb is less than that of bulb at room temperature.

(A)

Both A and R individually true and R is the correct explanation of A

(B)

Both A and R are individually true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Solution:

Filament of a light bulb should be uniform for its long
life.

Resistance of a metal (filament) increases with increase in
temperature.

Correct Answer is `=>` (C)

Q 2356291174

The dimensions of a rectangular block of carbon are `2 cm xx 2 cm xx 10 cm`. If the resistivity of the carbon is `2 xx 10^( -5) Omega - m`, what is the resistance measured between the two square surfaces?
NDA Paper 2 2007

(A)

`5 xx 10^(-3) Omega`

(B)

`2 xx 10^(-3) Omega`

(C)

`5 xx 10^(-2) Omega`

(D)

`2 xx 10^(-2) Omega`

Solution:

Here, resistivity, `rho = 2 xx 10^(-5) Omega m`

Breadth of block `= 2 cm = 2 xx 10^(-2) m`

Height of block`= 10 cm= 10 xx 10^(-2) m`

Resistance between the two square surface

= Resistivity `xx (text(Length of rectangle))/(text(Area of cross section of rectangle))`

`= 2 xx 10^(-5) xx ((10 xx 10^(-2)))/((2 xx 10^(-2) xx 2 xx 10^(-2))` `Omega`

`= 5 xx 10^(-3) Omega`

Correct Answer is `=>` (A) `5 xx 10^(-3) Omega`

Q 2386291177

Which one of the following is correct?
One unit of electric power is consumed when
NDA Paper 2 2007

(A)

1 A of current flows for 1 s at 220 V

(B)

1 A of current flows for 1 sat 1 V

(C)

100 A of current flows for 1 sat 10 V

(D)

10 A of current flows for 1 h at 100 V

Solution:

One unit of electric power `= 1 kWh = 1000` Wh

When `10 A` current flows for 1 h at `100 V`

Then, electricity consumed `= 10 xx 100 xx 1 Wh`

`= 1000 Wh`

`= 1 kWh`

Correct Answer is `=>` (D) 10 A of current flows for 1 h at 100 V

Q 2336791672

Which one of the following is the correct statement?
For an open circuit, the value of load,
NDA Paper 2 2007

(A)

would be infinite

(B)

would be zero

(C)

depends on the voltage of the source

(D)

depends on other components of circuit

Solution:

For an open circuit, the value of load would be infinite.

Correct Answer is `=>` (A) would be infinite

Q 2386791677

Which one of the following is the correct statement'?
If pieces of copper and germanium are cooled from room temperature to liquid nitrogen temperature, then the
NDA Paper 2 2007

(A)

resistance of each increases

(B)

resistance of each decreases

(C)

resistance of copper decreases while that of germanium increases

(D)

resistance of copper increases while that of germanium decreases

Solution:

Copper (metal) has positive temperature of coefficient while germanium (semiconductor) has negative temperature of coefficient, so resistance of copper decreases while that of germanium increases.

Correct Answer is `=>` (C) resistance of copper decreases while that of germanium increases

Q 2316791679

A wire of resistance `16 Omega` is bent in the form of a circle. What is the effective resistance between diametrically opposite points'?
NDA Paper 2 2007

(A)

`1 Omega`

(B)

`2 Omega`

(C)

`4 Omega`

(D)

`8 Omega`

Solution:

When wire is bent in the form of circle, each branch lias
resistance of `8 Omega` So, equivalent resistance between
diametrically opposite points, i.e., between `A` and `B` is

`1/(R_(AB))=1/8+1/8=1/4`

or `R_(AB)=4 Omega`

Correct Answer is `=>` (C) `4 Omega`

Q 2316891779

If an input current 3 A flows through the circuit shown above, what is the value of the current flowing through the `4 Omega` resistor'?
NDA Paper 2 2007

(A)

1.6 A

(B)

0.8 A

(C)

0.75 A

(D)

0.4 A

Solution:

Equivalent resistance of circuit

`1/R=1/2+1/8+1/4+1/1`

or `R=8/15 Omega`

Now, potential difference of each branch

`V=iR=8/15 xx 3=8/5 V`

Hence, current through ` 4 Omega` resistance

`i=V/R=(8//5)/4=8/20=0.4 A`

Correct Answer is `=>` (D) 0.4 A