Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

When Sunita, a class XII student came to know that her parents are planning to rent out the top floor of their house to a mobile company, she protested. She tried hard to convince her parents that this move would be a health hazard. Ultimately, her parents agreed. (i) In what way can the setting up of transmission tower by a mobile company in a residential colonyprove to be injurious to health? (ii) By objecting to this move of her parents, what value did Sunita display? (iii) Estimate the range of em waves which can be transmitted by the antenna of height 20 m (Given radius of earth = 6400 km).

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Solution :(i) Transmission towers in a residential colony are potential health HAZARDS. Some links have been establised
between radiations from cell phone towers and brain cancer. Further, these high structure can attract and
confuse birds. Millions of birds are killed near communication towers every year, raising health concerns.
(ii) By OBJECTING to this move, Sunita displayed the practival use of the knowledge acquiredby her, and also
her concern for the health of COMMUNITY all around.
(iii) Here, `h = 20 m, R = 6400KM = 6.4 xx 10^6m`
RANGE `= sqrt(2hR) = sqrt(2 xx 20 xx 6.4 xx 10^6) = 16 xx 10^3 m = 16 km`.
2.

The average power dissipation in a pure capacitor in ac circuit is

Answer»

`1/2 CV^2`
`CV^2`
`2CV^2`
Zero

Answer :D
3.

We are in danger of away the permanent good for a momentary pleasure.

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Bartering
Bartle
Barter
None of these

Answer :A
4.

A ray of light is incident at an angle of 60^@ on a sqrt3 cm thick plate (mu = sqrt3). The shift in the path of the ray as it emerges out from the plate is (in cm)

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1
1.2
0.5
1.8

Answer :A
5.

An instantaneous current at any instant in an A.C. series circuit is zero. At this instant, the instantaneous voltage is maximum, then ……….is not connected with the source.

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capacitor
INDUCTOR
resistance
both the inductor and capacitor

Answer :C
6.

Asolid sphere of radius R is charged uniformly .The electrostatic potential V is plotted as a function of distance r from the centre of the sphere. Which of the following best represents the resulting curve ?

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ANSWER :C
7.

A current carrying circular coil, suspended freely in a uniform external magnetic field orients to a position of stable equilibrium. In this state :

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The plane of coil is NORMAL to EXTERNAL magnetic field
The plane of coil is PARALLEL to external magnetic field
Flux through coil is minimum
Torque on coil is maximum

Answer :A
8.

Four initially uncharaged thin, large, plane idential metallic plates A,B,C and D are arranged parallel to each other as shown. Now plates A,B,C and D are given charges Q,2Q,3Q and 4Q respectively. Plates A and D are connected by a mtallic wire while plates B and C connected by other metallic wire then after Electrostatic equilibrium is reached. Charge on plate D after earthing it will be

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ZERO
`(5Q)/(2)`
`(-5Q)/(2)`
6Q

Answer :C
9.

Find resistance value of carbon resistors shown below.

Answer»

Solution :`rArr` RESISTANCE VALUE of (a)
`R = (22 x 10^(2) PM 10 % )Omega`
10.

Four initially uncharaged thin, large, plane idential metallic plates A,B,C and D are arranged parallel to each other as shown. Now plates A,B,C and D are given charges Q,2Q,3Q and 4Q respectively. Plates A and D are connected by a mtallic wire while plates B and C connected by other metallic wire then after Electrostatic equilibrium is reached. Charge on left surface of plate B wll be

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ZERO
`(3Q)/(2)`
`(-9Q)/(2)`
`(5Q)/(2)`

ANSWER :D
11.

The little monkey was....

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RESCUED
CAUGHT
RELEASED
NONE OF THE ABOVE

Answer :A
12.

An automatic apparatus or device that performs functions ordinarily ascribed to human or operate with what appears to be almost human intelligence is called…………...

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Robot
Human
Animals
Reptiles

Answer :a
13.

Bulk modulus of water is 2 xx 10^9 N/m^2. The change in pressure required to increase the density of water by 0.1% is

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a)`2 XX 10^9 N /m^2`
B)`2 xx 10^8 N /m^2`
C)`2 xx 10^6 N /m^2`
d)`2 xx 10^4 N /m^2`

Answer :C
14.

A horizontal straight wire 10m long is extending east and west and is falling with a speed of 5.0 ms^(-1) at right angles to the horizontal component of earth's magnetic field 0.30 xx 10^(-4) Wbm^(-2). What is the instantaneous value of the emf induced in the wire?

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ANSWER :1.5mV
15.

Two large thin metal plates are parallel and closeto eachotherontheir innerfaces the plates have surface chargesdensites of opposite region of the first plate (b)in the outer region of the secondplateand (c ) between the plates

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SOLUTION :Consider two extremely large parallel METALLIC plate-1 and plate-2.
For plate-1, its inner face a, B, c, d has charge `q = sigmaA`. Its outer face e, f, G, h has no charge. Hence, on the left of plate-1, E = 0.
Here, electric field produced by positively charged inner face is,
`E_(1) (to) = sigma/(2epsilon_(0))`...........(1)
For plate-2, charge on its inner face `a_(2), b_(2), c_(2), d_(2)`has charge `q =- sigmaA` . Its outer face `e_2, f_2, g_2, h_2` has no charge. Hence, on the right of plate-2, E = 0.
Here electric field produced by negatively charged inner face is,
`E_(2)(to) = sigma/(2epsilon_(0))`...........(2)
Now, resultant electric field between the plates is:
`E = E_(1) + E_(2), (therefore vecE_(1) || vecE_(2))`
`=sigma/(2epsilon_(0)) + sigma/(2epsilon_(0))`
`=sigma/(epsilon_(0))`
`=(17 xx 10^(-22))/(8.85 xx 10^(-12))`
`therefore E = 1.92 xx 10^(-10) N//C`
16.

A small fish 0.4m below the surface of a lake is viewed through a simple converging ends of focal length 3m. The lens is kept at 0.2m above the water surface such that fish lies on the optical axis of the lens. The image of the fish seen by observer will be at (mu_("water") = 4/3)

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A DISTANCE of 0.2m from the WATER SURFACE
A distance of 0.6m from the water surface
A distance of 0.3m from the water surface
The same LOCATION of fish

Answer :D
17.

The magnetism of magnet loses due to .....

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when it is BROKEN into small pieces
on heating a magnet
dropping it into cold water
applying a REVERSE FIELD of APPROPRIATE strength

Answer :D
18.

When 50Omega resistance is connected to one voltmeter, its range is obtained upto V volt. Now, instead of 50Omega, when 500Omega resistance is connected to this voltmeter, its range is obtained upto 2 V volt. Find resistance of this voltmeter.

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`100Omega`
`200Omega`
`300Omega`
`400Omega`

SOLUTION :We have `R=G(n-1)""...(1)`
(i) `R_(1)=G(n_(1)-1)`
`therefore50=G(n_(1)-1)=Gn_(1)-G""...(2)`
(ii) `R_(2)=G(n_(2)-1)`
`therefore500=G(2n_(1)-1)=2Gn_(1)-G""...(3)`
Multiplying equation (2) by 2, we get,
`100=2Gn_(1)-2G""...(4)`
Subtracting equation (4) from equation (3) we get,
400 = G
`thereforeG=400Omega`
19.

The variation of magnetic susceptibility (x) with temperature (T) for a diamagnetic material is best represented by

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Solution :For a diamagnetic MATERIAL the MAGNETIC susceptibility is negative and INDEPENDENT of temperature . So, graph `(B)` is correct.
20.

The following tablehas3 comumn and 4 rows . . Based on table, thereare Threequestions . Eachquestionhas Fouroptios(A),(B) (C ) and (D) . ONLY Oneof thesefouroptionsis correct. A solidsphereof volumeV and densityrho_(S)istiedbylightinextensiblestringABwith thebottomsurfaceof the vesselas shownin thefigure. The vesselis filledwith a liquidof densityrho _(t).(givenrho_(l) gtrho_(s)) Choosethe correct combinationin which tension in the stringis maximum.

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(II) (ii) (S)
(IV) (i) (S)
(III) (i) (S)
(I) (i) (S)

ANSWER :D
21.

A person hasfarsightedness with the minimum distance he couldsee early is 75 cm. Calculate the poer of the spectacts necessary to recify the defect.

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Solution :Theminimum DISTANCE the person could see elarly is, y = 75 cm
The lens should have a focal length of, `f = (y+25cm)/(y-25cm)`
`f=(75cmxx25cm)/(75cm-25cm)=37.5cm`
it is a CONVEX or CONVERGING lens.
The POWER of the lens is, `P = (1)/(0.375m) = 2.67 DIOPTER`
22.

The following tablehas3 comumn and 4 rows . . Based on table, thereare Threequestions . Eachquestionhas Fouroptios(A),(B) (C ) and (D) . ONLY Oneof thesefouroptionsis correct. A solidsphereof volumeV and densityrho_(S)istiedbylightinextensiblestringABwith thebottomsurfaceof the vesselas shownin thefigure. The vesselis filledwith a liquidof densityrho _(t).(givenrho_(l) gtrho_(s)) Choosethe correct combinations .

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(I) (III) (R )
(II) (iii) (P)
(II) (i) (P)
(I) (ii) (S)

ANSWER :C
23.

At a certain place a small bar magnet completes 30 oscillations per minute. At another place, where the magnetic field is double, its time period will be

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4 s
2 s
`1/2 s`
`sqrt2`s

Solution :At 1st PLACE `B_1 = B, T_1 = (60)/(30) s = 2s ` .At 2nd place `2B_1 = 2B`
`because (T_1)/(T_2) = SQRT((B_2)/(B_1)) implies T_2 = T_1 xx sqrt((B_1)/(B_2)) = 2 xx sqrt((1)/(2)) = sqrt2 `s
24.

At the first minimum adjacent to the central maximum of a single slit differation pattern the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the mid point of the slit is......

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`(pi)/(8)` rad
`(pi)/(4)` rad
`(pi)/(2)` rad
`pi` rad

Solution :In right angle `DeltaS_(1)NS, sin theta=(SN)/(SS_(1))`
`sin theta=(dx_(1))/((a)/(2)) [ :. SS_(1)=(a)/(2)]`

`:. dx_(1)=(a)/(2) sin theta ""...(1)`
where `Deltax_(1)` is a path difference.
Now, for FIRST minimum,
`a sin theta= LAMBDA "" [d=a]`
`:. sin theta=(lambda)/(a) ""...(2)`
`:. Deltax_(1)=(a)/(2)xx(lambda)/(a)=(lambda)/(2)` [`:.` From equation (1) and (2)]
Now phase difference
`Delta phi_(1)=KXX` path difference
`=(2pi)/(lambda)xx deltax_(1)`
`=(2pi)/(lambda)xx(lambda)/(2)=pi` rad
25.

In the question number 73, the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60^(@) under the influence of magneticfield. The magnitude of torque on the coil in the final position is

Answer»

25 N m
`25 sqrt3 N m`
40 N m
`40 sqrt3 N m`

Solution :Torque `|VEC(tau)|=|vecm xx vecB|=m B sin theta`
Here, m =`"10 A m"^(2), B = 5 T`
Now initially `theta=0^(@)`
Thus, initial torquie, `tau_(i)=0`
In FINAL position `theta=60^(@)`
`therefore""tau_(F)=m B sin 60^(@)=10xx5xx(sqrt3)/(2)=25sqrt3"N m"`
26.

Find the torque on the shaft of an electric motor of 20 kW power if its rotor turns at 1440 r.p.m.

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ANSWER :`1.3xx10^2` N.m.
27.

The following tablehas3 comumn and 4 rows . . Based on table, thereare Threequestions . Eachquestionhas Fouroptios(A),(B) (C ) and (D) . ONLY Oneof thesefouroptionsis correct. A solidsphereof volumeV and densityrho_(S)istiedbylightinextensiblestringABwith thebottomsurfaceof the vesselas shownin thefigure. The vesselis filledwith a liquidof densityrho _(t).(givenrho_(l) gtrho_(s)) Choose the correct combination in whichsolid sphere movestowardsrightwith respectt vessel

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(I) (IV) (S)
(III) (iv) (Q)
(II) (iv) (S)
(III ) (iv) (R )

ANSWER :B
28.

A uniform semicircular disc of mass 'm' and radius 'R' is shown in the figure. Find out its moment of inertia about. (a) axis 'AB' ( shown in the figure 0 which passes through geometrical centre and lies in the plane of the disc (b) axis 'CD' which passes through its centre of mass and it is perpendicular to the plane of the disc.

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`(MR^2)/2-M((2R)/PI)^2`
`(MR^2)/2-M((4R)/pi)^2`
`(MR^2)/2+M((4R)/(3PI))^2`
`(MR^2)/2+M((2R)/(pi))^2`

Answer :B
29.

What are permanent magnets 2 Give one example.

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Solution :Substances which at room TEMPERATURE retain their ferromagnetic property for a LONG period of time are called "permanent magnets". COMMONLY USED bar MAGNET is permanent magnet.
30.

When an empty lift is moving down with an acceleration of g/4 ms^(-2), the tension in the cable is 9000N. When the lift is moving up with an acceleration of g/3 ms^(-2), the tension in the cable is,

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16,000 N
18,000 N
12,000 N
15,000 N

Answer :A
31.

The horizontal component of the earth's magnetic field at a certain place is 2.5xx10^(-5)T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 0.5 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west, (b) south to north?

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ANSWER :(a) `1.25xx10^(-5)N`, (B) 0 N
32.

An electric current has both D.C. and A.C. components D.C. component of 8A and A.C. component is given as I=6sin omegat A. So l_(rms) value of resultant current is .....

Answer»

8.05 A
9.05 A
11.58 A
13.58 A

Solution :Resultant current at time t
`I=8+6 sin omegat`
`THEREFORE I_(rms)=sqrt(ltI^2gt)=sqrt(lt(8+6 sin omegat)^2gt)`
`=sqrt(lt64gt + lt 96 sin omegatgt+lt 36 sin^2 omegatgt)`
But on ONE PERIOD `lt64gt`=64
`lt96sinomegatgt`=0
`lt36sin^2omegatgt=36xx1/2=18`
`therefore I_(rms)=sqrt(64+0+18)`
`=sqrt82`
= 9.05 A
33.

यदि y=e^x logx तब (dy/dx) है

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`e^X /x`
`e^x (1/x +XLOGX)`
`e^x (1/x +LOGX)`
`e^x /logx`

ANSWER :C
34.

Length of the bar magnet is aligned along the axis of a circular coil which is placed far away from the magnet. When magnet is moved along the axis of coil then looking from the side of the magnet, it is found that clockwise current flows through the coil. Which of the following is/are possible?

Answer»

South pole of the magnet faces the coil and is moved towards the coil
South pole of the magnet faces the coil and is moved away from the coil
North pole of the magnet faces the coil and is moved towards the coil
North pole of the magnet faces the coil and is moved away from the coil

Solution :We must be aware that clockwise CURRENT in the loop is EQUIVALENT to south pole and anticlockwise current is equivalent to north pole, When south pole of the magnet is moved towards the coil then current in the coil should oppose its motion. Hence clockwise current DEVELOPS from the side of the magnet to DEVELOP south pole to repel south pole of the approaching magnet. Hence, option (a) is correct.
When south pole of the mugnet is moved away from the coil then current in the coil should oppose its motion. Hence anticlockwise current develops from the side of the magnet to develop north pole to attract south pole of the receding magnet. Hence, option (b) is wrong.
When north pole of the magnet is moved towards the coil then current in the coil should oppose its motion. Hence anticlockwise current develops from the side of the magnet to develop north pole to repel north pole of the approaching magnet.Hence, option (C) is wrong.
When north pole of the magnet is moved away from the coil then current in the coil should oppose its motion. Hence clockwise current develops from the side of the magnet to develop north pole to attract south pole of the receding magnet. Hence, option (h) is wrong. Options (a) and (d) are correct.
35.

Maximum frequency of the photon produced by the union of a proton and a antiproton is (given : m_(p)=1.67xx10^(-27)kg):

Answer»

`4.56xx10^(21)HZ`
`4.56xx10^(23)Hz`
`5.46xx10^(25)Hz`
`6.45xx10^(25)Hz`

SOLUTION :`mc^(2)=hv`
`:.v=(mc^(2))/(h)=4*56xx10^(23)Hz`
36.

A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

Answer»

Solution :`P=100W"" lambda=589 nm = 589xx10^(-9)m`
a. `E=h upsilon=h (c )/(lambda)=(6.6xx10^(-34)xx3xx10^(8))/(589xx10^(-9))=0.03xx10^(-17)=3xx10^(-19)J`
b. Let .n. be the number of PHOTONS produced/s
then,`nhupsilon=P ""THEREFORE n=(P)/(lambda upsilon)=(100)/(3xx10^(-14))=3.3xx10^(20)`
37.

In the above question what is the value of tension in the string?

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zero
1N
2N
5N

SOLUTION :SEA the solution above
Hence (d) is the CHOICE
38.

Straight in LC oscillation, the sum of energies stored in capacitor & theinductor is constant in time.

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Solution :i. During LC oscilations in LC circuits, the ENERGY of the system oscillates between the electric field of the capacitor and the magnetic field of the INDUCTOR.
ii. Althogh, these two forms of energy vary with time, the total energy REMAINS constant. It means that with the law of constant. It accordance with the law of conservation of energy.
Total energy, `U=U_(E)+U_(B)=(q^(2))/(2C)+(1)/(2)Li^(2)`
iii. consider 3 different stages of LC oscillations and calculate the total energy of the system.
Case (i)When the charge in the capacitor, q `Q_(m)=` and the current throughthe inductor, i = 0, the total energy is GIVEN by
`U=(Q_(m)^(2))/(2C)+0=(Q_(m)^(2))/(2C)`
The total energy is wholly electrical.
Case (ii)When charge = 0 , current `I_(m),`
`U=0+(1)/(2)LI_(m)^(2)=(1)/(2)LI_(m)^(2)`
`=(L)/(2)xx((Q_(m)^(2))/(LC))"since Im"=Q_(m)omega=(Q_(m))/(sqrt(LC))`
`=(Q_(m)^(2))/(2C)`
Case (iii)When charge = q, current = i, the total energy is
`U=(q^(2))/(2C)+(1)/(2)Li^(2)`
IV. Since `q=Q_(m)=cosomegat,i=(dq)/(dt)=Q_(m)=omegasinomegat.` The negative sign in current indicates that the charge in the capacitor decreases with time.
`U=(Q_(m)^(2)cos^(2)omegat)/(2C)+(Lomega^(2)Q_(m)^(2)sin^(2)omegat)/(2)`
`U=(Q_(m)^(2)cos^(2)omegat)/(2C)+(Lomega_(m)^(2)Q_(m)^(2)sin^(2)omegat)/(2)" since "omega^(2)=(1)/(LC)`
`=(Q_(m)^(2))/(2C)(cos^(2)omegat+sin^(2)omegat)`
`U=(Q_(m)^(2))/(2C)`
Frome above three cases, it is clear that the total energy of the system remains constant.
39.

An object is moving towards a convex mirror fromlarge distance. The image will move with ________ velocity than the object ________ the mirror [Fill in the blanks].

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ANSWER :LESS, TOWARDS
40.

In Boolean expression which gate is expressed as y =bar( A+B )

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OR
NAND
AND
NOR

Answer :D
41.

A stone is thrown horizontally with an initial speed of 30 m//s from a bridge. Find the stone 's total speed when it enters the water 4 seconds later. (Ignore air resistance).

Answer»

`30 m//s`
`40 m//s`
`50 m//s`
`60 m//s`

Solution :After 4 seconds, the stone's vertical speed has changed by `Delta v_(y)=a_(y t)=(10 m//s^(2))(4 s)=40 m//s`. Since `v_(0_(y))=0`, the value of `v_(y)` at t=4 is `40 m//s`. The horizontal speed does not change. Therefore, when the rock hits the water, its velocity has a horizontal COMPONENT of `30 m//s` and a vertical component of `40 m//s`.

If you recognize this as a 3 : 4 : 5 triangle that has been scaled up by a factor of 10, you can easily USE the Pythagorean Theorem to find that the magnitude of the total velocity v, is `50 m//s`.
42.

The majority charge carriers in p-type semiconductors are

Answer»

electrons.
protons.
holes.
positrons.

SOLUTION :`because E_(g) = 2.0` eV and `E_(g) = (hv)/(E) eV""RARR "" v = (E_(g.e))/(h) = (2.0 xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) = 5 xx 10^(14)` HZ
43.

What is the magnification if the mirror is concave in the above question ?

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1/3
2/3
1
3/2

Answer :C
44.

Represent the magnetic moment vector in terms of angular momentum vector of electrons revolving around the nucleus.

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Solution :`mu_(L)=2/(2m_(e))l, " vectorally " vec(MU)_(l)=-e/(2m_(e))vec(l)`
where -ve SIGN indicates that the direction of angular momentum is opposite to the magnetic MOMENT.
45.

A resistor R and a 4muF capacitor are connected in series across a 200 V d.c. supply. Across the capacitor is connected a neon lamp that strikes at 120 V in 5 seconds . Find R.

Answer»

SOLUTION :`V= V_0 (1- e^(-t//tau_c))` where `tau_C= RC`
or `120 = 200 (1- e^(-5//tau_C))`
or , `(e^(-5//tau_C))= 1 -(120)/(200) =0.4`
or, `(e^(5//tau_C) =log_e 2.5`
or , `(5//tau_C)= log_(e) 2.5`
`therefore tau_C = 5/(log_e 2.5) = 5.46 ` sec
or, RC= 5.46
or, `R= (5.46)/(C) = (5.46)/(C) = (5.46)/(4XX10^(-6)) = 1.364 xx 10^6 `ohm = `1.364 M Omega`
46.

We can reduce eddy currents in the core of a transformer

Answer»

by USING laminated iron core.
by using a core of SOFT iron
by using THICK copper wire for WINDING.
by reducing the size of transformer.

ANSWER :A
47.

Pure water cooled to -15^(@)C is contained in a thermally insulated flask. Some ice is added into the flask. The fraction of water frozen into ice is :

Answer»

`(3)/(35)`
`(6)/(35)`
`(2)/(35)`
`(6)/(29)`

SOLUTION :Here if m is the MASS of ICE and M is the mass of water, then
`80M =mxx0.5xx15+(M-m)xx1xx[0-(-15)]`
`=m xx 7.5 +15 M -15m`
or `(m)/(M) =(6)/(35)`.
48.

Half-life of radium is 1600 yr. Its average life is :

Answer»

3200 yr
4800 yr
2308 yr
4217 yr

Answer :C
49.

If the whole earth is to be connected by line of sight communication using space waves (no restriction of antenna size or tower height). What is the minimum number of antennas required?

Answer»

2
3
4
6

Solution :Let `h_(t)` be the height of transmitting antenna or receiving antenna in order to cover the ent IRE surface of earth through COMMUNICATION as SHOWN in FIGURE.

`d_(m)^(2)=(R+h_(T))^(2)+(R+h_(T))^(2)=2(R+h_(T))^(2)`
As `d_(m)=sqrt(2h_(T)R)+sqrt(2h_(T)R)=2sqrt(2h_(T)R)`
`therefore 8h_(T)R=2(R+h_(T))^(2)`
or `4h_(T)=R^(2)+2Rh_(T)+h_(T)^(3)`
or `R^(2)-2h_(T)R+h_(T)^(2)`
or `R^(2)-2h_(T)R+h_(T)^(3)`
or `(R-h_(T))^(3)=0` or `R=h_(T)`
50.

Draw the variation of binding energy per nucleon with mass number of atoms and indicate the stable and unstable regions on the diagram.

Answer»

Solution :The almost HORIZONTAL REGION in the MIDDLE portion of the graph indicates the STABLE region.
The region to the extreme LEFT and the sloping region to the extreme right indicate the unstable regions .