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.

निम्न मे से कौन सी संख्या एक अपरिमेय संख्या है

Answer»

`SQRT (4/9)`
`4/5`
`sqrt 7`
`sqrt 81`

ANSWER :C
2.

इनमें से किस राज्य में काली मिट्टी पाई जाती है?

Answer»

जम्मू और कश्मीर
गुजरात
उत्तराखंड
झारखंड

Answer :B
3.

The equivalent capacity between points A and B in figure will be, while capacitance of each capacitor is 3 muF.

Answer»

`2 muF`
`4 muF`
`7 muF`
`9 muF`

ANSWER :D
4.

Two coherent light waves of intensity 5xx10^(-2)W m^(-2) each super - impose and produce the interference pattern on a screen. Ata point where the path difference between the waves is lamda/6, lamda being wavelength of the wave, find the (a) phase difference between the waves. (b) resultant intensity at the point. (c) resultant intensity in terms of the intensity at the maximum.

Answer»

Solution :(a) Here intensity of each wave `I= 5xx 10^(-2) WM^(-2)` and path DIFFERENCE `Deltax=lamda/6` .
(a) Phase difference between the waves `phi=(2PI)/lamda xxDeltax=(2pi)/lamdaxxlamda/6=pi/3" or "60^@`
(b) RESULTANT intensity `I_R=4Icos^2(phi/2)=4xx5xx10^(-2)xx(cospi/6)^2=4xx5xx10^(-2)xx(sqrt3/2)^2`
`=0.15 W m^(-2)`
(c) Maximum intensity `I_0=4Iimplies(I_R)/(I_0)=cos^(2)""phi/2=(cos""pi/2)^2=3/4" or "I_R=3/4I_0=0.75I_0`
5.

The workdone to trun a magnet by 60^(@) from its equilibrium position is W in a uniformthe torque required to hold it in that postion will be

Answer»

`(w)/(2)`
`(w)/sqrt(3)`
`sqrt(3)/(2)w`
`wsqrt(3)`

Solution :`W=MB =(mB)/(2)`
`tau-mBsin 60^(@),tau=mBxxsqrt(3)/(2)=Wxxsqrt(3)`
6.

The relation between the refractive index and the critical angle of a medium is

Answer»

SOLUTION :(i) Refractive index` ( mu ) = ( 1)/( SIN C ) `
where, C is the CRITICAL ANGLE.
(ii) Since, refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelengths ( COLOURS ) of light.
7.

How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light ? Give reason.

Answer»

Solution :Angle of minimum deviation of light RAY from a glass prism increases with increase in refractive index. As refractive index of glass for violet light is more than that for red light, (nv> NR) the angle of minimum deviation for red light is LESS than that for violet light i.e. `(delta)_(R) gt delta_(V))`
8.

Consider the given statement with respect to the figure showing a bar of material placed in an external magnetic filed and choose the correet Statemnet(s).

Answer»

The filed lines are repelled or EXPELLED and the FIELD inside the material is reduced.
When placed in a non-uniform magnetic field, the bar MAGNET will tend to move from low to high field.
When placed in a non-uniform magnetic field, the bar will tend to move from high to low field.
Reduction in the field inside the material is slight being one part in `10^(5)`.

Solution :Figure shows a bar of diamagnetic material placed in extemal magnetic field. The field lines are repelled and field inside the material is reduced. This reduction is slight being one part in `10^(5)`. When placed in non-uniform magnetic field, bar will move from high to low field.
9.

A container of depth 2d is half-filled with a liquid of refractive index sqrt2 and another half with a liquid of refractive index mu. The two liquids do not mix with each other. The apparent depth of the inner surface of the bottom of the container will be (neglect the thickness of thebottom of the container)

Answer»

`(mu)/(d(mu + 2))`
`(d(mu + SQRT2))/(mu sqrt2)`
`(sqrt2mu)/(d(mu + sqrt2))`
`(MUD)/(d+3mu)`

ANSWER :B
10.

A block of mass 'm' rests on a rough inclined plane making an angle 30^(@) with the horizontal, The coefficient of static friction between the block and the plane is 0.8. If force of friction on the block is 10N, the mass of the block is (g = 10 m/s^2)

Answer»

0.2 KG
0.4 kg
0.8 kg
2 kg

ANSWER :D
11.

Whichone of the series of hydrogen spectrum is in the visible region?

Answer»

LYMAN series
Balmer series
Paschen series
Pfund series

Answer :B
12.

How the television signals are propagated, derive the formula for the area that a T.V. signal will cover

Answer»

Solution :Television signal waves have the frequency range 80 MHZ these waves neither follow the curvature of the earth nor get reflected by ionosphere. The propagation of such wave is possible only via satellite or by using all RECEIVER antenna which may direct intercept the transmitted signals coming from the transmitter antenna In order to FIND the HEIGHT of a transmitter for a greater coverage let us consider a TV transmitting antenna of high h located at A on the surface of the earth. The transmitted signal can be received within a circle of radius PB (or PC) on the surface of the earth as shown in the figure.Let R be the radius of the earth, then from the geometry of the fig. we havePB=PC=d, AP=h, `therefore` OP=OA+AP=R+h, and in right angle triangle POB, `OP^(2)=PB^(2)+BO^(2) or ` (R+h)^(2)=d^(2)+R^(2)` or `d^(2)=h^(2)+2Rh,If R, then NEGLECTING `h square H^(2)`, we have `d^(2)=2Rh`.THUS a T.V signal will cover the area=`pi d^(2)=2pi Rh`
13.

Two identical taqnks are placed on the two pans of a beam balance. One tank is empty and open to atomosphere. The second tank evacuated and then filled with helium until the two tanks are balanced The pressure of helium will be (Given that rho_(air)=7.5rho_(He)]

Answer»

15 atm
7.5atm
4atm
1atm

Answer :B
14.

At t=0, an inductor of zero resistance is joined to a cell of emf epsilon through a resistance. The current increases with a time constant tau. After what time will the potential difference across the coil be equal to that across the resistance ?

Answer»

Solution :`V_(L)=V_(R)Ee^(-(Rt)/(L))=E(1-e^(-(Rt)/(L))) IMPLIES 2E^(-(Rt)/(L))=1, e^(-(R)/(L)t)=(1)/(2), e^((R)/(L)t)=2`
`(Rt)/(L)=LN 2"(or)" t=(L)/(R)ln2`
15.

Fertilization involving carting of male gametes by pollen tube is-

Answer»

Porogamy
Siphonogamy
Chalazogamy
Syngonogamy

Answer :B
16.

A charge Q moves parallel to a very long straight wire carrying a current l as shown. The force on the charge is

Answer»

OPPOSITE to OX
Along OX
Opposite to OY
Along OY

Answer :A
17.

The numerical value of the charge on either plates of capacitor 'C' as shown in the figure is

Answer»

`CE `
`(CER_1)/(R_1 +R)`
`(CER_2)/(R_2 +r)`
`(CER_1)/(R_2 +r)`

ANSWER :C
18.

In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell ?

Answer»

`~= 1.57 V`
`~= 1.67`
`~=1.47`
`~=1.37 V`

Solution :`E_(1) alpha l_(1)`
`E_(1)alpha l_(2)`
`(E_(1))/(E_(1)) = (l_(1))/(l_(2)) rArr (1.25)/(E_(2)) = (30)/(40) rArr E_(2) = (5)/(3) = 1.67 V`
19.

A message signal of frequency omega_m is superposed on a carrier wave of frequency omega_c to get an amplitude modulated wave (AM). The frequency of AM wave will be

Answer»

`omega_m`
`omega_c`
`(omega_c + omega_m)/2`
`(omega_c - omega_m)/2`

ANSWER :B
20.

A rotating wheel changes angular speed from 1800 rpm to 3000 rpm in 20 s. What is the angular acceleration assuming to be uniform ?

Answer»

`60pi` rad `s^(-2)`
`90pi` rad `s^(-2)`
`2PI` rad `s^(-2)`
`40pi` rad `s^(-2)`

SOLUTION :`v_(0)=(1800)/(60)=30Hz therefore omega_(0)=2piv_(0)=60pi" rad/s"`
`v=(3000)/(60)=50HZ therefore omega=2piv=100pi""rad/s`
Angular acceleration `a=(omega-omega_(0))/(t)=(100pi-60pi)/(20)=2omega" rad/s"`
21.

In 1 gram of solid, there are 5 xx 10^(21) atoms. If one electron is removed from every one of 0.1% of atoms of the solid, the charge gained by the solid is

Answer»

`+0.018C`
`+0.8C`
`+8 MU C`
`-0.08C`

ANSWER :B
22.

Derive an expression for the impedance of a series LCR, circuit, when an AC voltage is applied to it.

Answer»

Solution :
Let I be the instantaneous current in the CIRCUIT. Since the components are connected in series the current through each component is same.
Let `vecI , vecV _(R), vecV _(L) and vecV_(c )` be the phasors REPRESENTING current and voltage across resistor, inductor and capacitor respectively as shown in phasor diagram .
`vecV _(R)` is in phase with `vecI and V _(RM) = i _(m) R`
`vecV_() ` leads `vecI ` by `pi /2 and V _(Lm) = i _(m) X _(L)`
`vecV _(C)` lags behind `vecI` by `pi/2 ` rad and `V _(CM) = i_(m) X _(C)`
Let `X _(C) gt X _(L)` then th4e phasor OD represents the resultant of `vecV _(L) and vecV _(C) i.e., (vecV _(C) - vecV _(L)).` The DIAGONAL of the parallelogram OE represents the resultant of `vecV _(R), vecV _(L) and vecV _(C).`
Applying Pythagorean Theorem, we get
`OE ^(2) = OA ^(2) + AB ^(2)`
`v _(m ) ^(2) = v _(Rm ) ^(2)+ ( v _(Cm ) - v _(Lm)) ^(2)`
`= (i _(m) R ) ^(2) + (i_(m) X _(C) - i _( m ) x _(L)) ^(2)`
`= i _(m) ^(2) [ R ^(2) + (X _(C) - X _(L))^(2) ]`
`i _(m) ^(2) = ( v _(m) ^(2))/( [ R ^(2) + (X _(c) - X _(L ))^(2) ])`
` i _(m )= ( v _(m))/( sqrt ( R ^(2) + (X _(c ) - X _(L)) ^(2)))`
According to Ohm.s LAW, the term `(sqrt ( R ^(2) + ( X _(c) - X _(L ))^(2)))` plays the role of resistance and it is called impedance.It is denoted by Z.
` Z = sqrt( R ^(2) + ( X _(C) - X _(L)) ^(2))`
`Z = ( v _(m))/( i _(m))`
23.

A car completes the first half of its journey with a velocity upsilon_(1) and the rest half with a velocity upsilon_(2). Then the average velocity of the car for the whole journey is :

Answer»

`v=sqrt(v_1 v_2)`
`v=sqrt((v_2)/(v_1))`
`(2)/(v)=(1)/(v_1)+(1)/(v_2)`
`v=(v_1 +v_2)/(2)`

SOLUTION :`V_(av)=("TOTAL distance")/("total time")=(V_(1)xxt/2+V_2xxt/2)/(t)=(V_(1)+V_2)/(2)`
24.

The frequency response curve ( Fig. ) for the filter circuit used for production of AM wave should be

Answer»

(i) followed by (ii)
(ii) followed by (i)
(iii)
(iv)

Solution :BAND WIDTH of `AM` wave shiould be ` 2 OMEGA _(m) = ( omega_(c) + omega_(m)) - ( omega_(c) - omega_(m))`. It is applicable for case (i) , (ii) and (iii) . The case (iv) will reject the REQUIRED signals for `AM` wave .
25.

Mention the SI unit and value of fundamental charge. Write its smaller units.

Answer»

Solution :Electric charge of electron is called fundamental charge. Its symbol is e and charge is negative.
In the international system (SI) of units, a unit of charge is called a COULOMB and is denoted by the symbol C.
One coulomb is the charge FLOWING through a WIRE in 1 s if the current is 1A (ampere).
In this system, the value of the basic unit of charge is, `e = 1.602192 xx 10^(-19)` C. In general it is TAKEN as e = `1.6 xx 10^(-19)` C.
There are about `6.25 xx 10^18` electrons in a charge of - 1 C.
In electrostatics, smaller charges are :
1 mC (milli coulomb) = `10^(-3)` C
`1 muC` (micro coulombs) `=10^(-6)` C
1nC (nano coulombs) `=10^(-9)` C
26.

Derive n= (sin""(A+D)/(2))/( sin ""(A)/(2)) for the prism. Where the symbols have their usual meaning.

Answer»

SOLUTION :Consider a principal section of prism of reflecting angle A and refractive index n placed in a air MEDIUM. A ray of light PQ is incident on the face AB refraction takes place and finally EMERGES out with an angle `i_2`,

fromthe quadrilateralAQXR
` A +ANGLEX = 180 ^@`
`A = 180- angleX `
fromthe triangle`QX R`
` r_1 +r_2= 180- angleX --- (2)`
from(1)& (2)
`A=r_1+r_2 ---(3)`
fromfigure
` delta= i_1 -r_1+i_2-r_2`
` delta =i_1+i_2 -(r_1 +r_2)`
`delta =i_1 +i_2-A ---(4)`
As the angle of incidence increases, since deviation decreases and reaches a minimum then after it increases as angle incidence increases. At minimum deviation position.
if ` delta =D `
` i_1 =i_2 =i`
eqn (3)& (4)becomes
`A =r +r impliesA=2r`
`implies r =(A)/(2)`
` D=i +i -A implies i =(A +D )/(2 )`
from snells law
`n = ( sin i)/( sin r)`
` n=(sin((A+D)/(2)))/(sin (A/2))`
27.

A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from x=A to x =A//2 is :

Answer»

`(T)/(6)`
`(T)/(4)`
`(T)/(3)`
`(T)/(2)`.

ANSWER :A
28.

In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 4Omega resistance and at 3 m when cell is shunted by a 8Omega resistance. The internal resistance of cell is -

Answer»

`12OMEGA`
`8OMEGA`
`16Omega`
`1Omega`

ANSWER :B
29.

समतल दर्पण के सामने रखी एक घड़ी में 6.00 बजे हैं तो घड़ी के प्रतिबिम्ब में कितना समय दिखेगा ?

Answer»

12
6.3
12.3
6

Answer :D
30.

On a particular day, the maximum frequency reflected from the ionosphere is 8 MHz. On another day, it was found to increase to 9 MHz. Calculate the ratio of maximum electron densities of the ionosphere on the two days. Point out a plausible explanations for this.

Answer»

Solution :Here, `v_c = 8 MHz , v'_c = 9 MHz`
Now, `v_c = 9 (N_(max))^(1//2) and v'_c = 9(N'_(max))^(1//2) :. (N'_(max))/(N_(max)) = ((v'_(c))/v_c)^2 = (9/8)^2 = 1.266`
The increase in the maximum number density of electron `(N_(max))`in ionosphere shows more IONISATION DUE
to high energy radiations from sun reaching there, which is possible when sun rays are falling normally.
For a layer of ionosphere, the refractive index of layer is given by
` mu = mu_0 ( 1 - (81.45N)/v^2)^(1//2)`
As we go deep into the ionosphere, N increase, so `mu` decreases. Due to this, the bending of the incident
beam of e.m. waves away from the normal CONTINUES till it reaches the critical angle stage, after which the
total internal reflection takes place and electromagnetic beam is reflected back. If frequency v is too high,
then N may not be so high to produce enough bending for attainment of critical angle or CONDITIONS of total
reflection.
31.

A steady current flows in a long wire. It is bent into a circular lopp of one turn and the magnetic field at the centre of the coil is B. If the same wire is bent into a circular loop of n turns, the magnetic field at the centre of the coil is

Answer»

n B
`n^(2)B`
`2nB`
`2N^(2)B`

ANSWER :B
32.

Derive a relation for capacitance of a parallel plate capacitor with dielectric slab of thickness t filled between the plates. Or. Write expression for capacitance of a parallel plate capacitor and explain the effect of dielectrics on capacitance.

Answer»

Solution :Capacitance of a parallel PLATE capacitor with vacuum between the two plates is
`C_(0)=(epsi_(0)A)/(d)`
Let a plate A be given +Q charge and plate B be given -Q charge. Introduce dielectric slab, say of thickness t(t If `vecE_(0)` is the total FIELD between A and B, the polarisation in the dielectric INTRODUCES a field `vecE_(p)` in the opposite direction.
`therefore`net electric field inside the dielectric is
`vecE=vecE_(0)-vecE_(p)`.
Potential difference between A and B is

But `(E_0)/(E)=K`, the dielectric constant.
`thereforeE=(E_(0))/(K)`
or `V=E_(0)(d-t)+(E_(0))/(K)xxt`
or `V=E_(0)[d-t+(t)/(K)]`
But `E_(0)=(sigma)/(epsi_(0))=(Q)/(Aepsi_(0))`
`therefore V=(Q)/(Aepsi_(0))[d-t+(t)/(K)]`
But capacity, `C=(Q)/(V)`
`therefore C=(Q)/((Q)/(Aepsi_(0))[d-t+(t)/(K)])`
or `Q=(Aepsi_(0))/(d-t+(t)/(K))`
If whole of the space between plates is filled with dielectric, then t=d
`therefore C=(Aepsi_(0))/(d-d+(d)/(K))`
or `C=(KAepsi_(0))/(d)`.
33.

Two polaroids are oriented with their principal planes making an angle of 60^(@). The percentage of incident unpolarised light, which passes through the system is

Answer»

1
0.5
0.25
`12.5%`

Solution :Intensity of LIGHT passing through the polariser `I_(1)=(I_(0))/(2)` and intensity of light passing through the analyser `I_(2)=I_(1)cos^(2)(60^(@))`
`impliesI_(2)=(I_(0))/(2)xx((1)/(2))^(2)=(I_(0))/(8)=12.5%` of `I_(0)`.
34.

A flow of liquid is streamline if the Reynold number is

Answer»

LESS than 1000
greater than 1000
between 2000 to 3000
between 4000 to 5000

Solution :less than 1000
35.

A particle vibrating simple harmonically has an acceleration of 16 cm s^(-2)when it is at a distance of 4 cm from the mean position. Its time period is

Answer»

1s
2.571 s
3.142 s
6.028 s

Solution :Here, `a = 16 CM s^(-2), x= 4 cm, T=?`
We know, `|a|=omega^(2)x` or `omega^(2) =a/x`
`omega =(2PI)/T RARR 2 = (2pi)/T rArr T =(2pi)/2 = pi s = 3.142 s`
36.

A 25 watt and 100 watt are joined in series and connected to the mains. Which bulb will glow brighter ?

Answer»

Solution :The resistance of 25 watt bulb is 4times than the resistance of 100 watt bulb. As they are connected in SERIES the current C in both is the same. We KNOW HEAT produced is given by `i^2R`. Hence, the bulb having higher resistance will glow brighter than thathaving LOW resistance when connected in series.
37.

Brown, Red and Orange coloured bands on carbon resistor are followed by silver band. The value of resistor is ..... .

Answer»

12 K`Omega PM 5 %`
`12 k Omega pm 10 % `
`320 Omega pm 10 %`
`320 Omega pm 5 %`

Solution :`12 k Omega pm 10 % `
Resistance from Brown and Red = 12 `10^(3) ` for ORANGE and for Silver `pm` 10%
`THEREFORE ` Value of resistance = `(12 xx 10^(3) pm 10 % ) Omega`
38.

For the hydrogen atom, the energy of radiation emitted in the transition from 4th excited state to 2nd excited state, according to Bohr's theory is

Answer»

0.567 eV
0.667 eV
0.967 eV
1.267 eVs

Solution :USING `E_(n)=(-13.6)/(n^(2))eV`
ACCORDING to given question, ltbgt `E_(4TH)=-(13.6)/((4)^(2))=-(13.6)/(16)eV=-0.85eV`
And `E_(2nd)=(13.6)/((2)^(2))=(-13.6)/(4)eV=-3.4eV`
`E_(3rd)=(-13.6)/((3)^(2))=-(13.6)/(9)=-1.51eV`
Energy of radiation EMITTED in the transition
`DeltaE=E_(4th)-E_(2nd)`
`=-0.85+3.4=2.55eV`
`DeltaE=E_(3rd)-E_(2nd)`
`=-15.1+3.4=1.89eV`
`DeltaE=E_(4th)-E_(3rd)`
`=-0.85+1.51=0.66eV`
39.

Nichrome and copper wires of same length and area of cross-section are connected in sereis, current is passed through them. Why does the nichrome wire get heated first ?

Answer»

Solution :Rateofproduction OFHEAT,` P =I^2 R`, forgiven` l, P XX R,therefore rho_ ("nichrome") gt rho _(CU)`
`thereforeR_("Nichrome") gt R_(cu) `ofsame lengthandareaof CROSSSECTION.
40.

Which unknown quantity is measured with the help of Wheatstone's Bridge ?

Answer»

ELECTRIC CURRENT
VOLTAGE
RESISTANCE
emf

Answer :C
41.

An electrical meter of internal resistance 200 gives a full scale deflection when one milliampere current flows through it. The maximum current, that can be measured by using three resistors of resistance 120 each, in milliamperes is :

Answer»

10
8
6
4

Answer :C
42.

The explosion of a fire cracker in the air at a height of 40 m produces a 100 dB sound level at ground below. What is the instantaneous total radiated power? Assume that it radiates as a point source.

Answer»

SOLUTION :201 WATT
43.

(A): It is easier to start a car engine on a warm day than a chilly day (R): with increase in temperature, the internal resistance of the car battery decreases

Answer»

Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'
Both 'A' and 'R' are true and 'R' is not the correct explanation of 'A'
'A' is true and 'R' is false 
'A' is false and 'R' is false 

Answer :B
44.

What kind of lens is an air bubble in water?

Answer»

SOLUTION :CONCAVE LENS. This is because both the convex surfaces FACE towardsdenser (water) medium and air INSIDE the bubble is rarer.
45.

Draw a schematic arrangement for winding of primary and secondary coil in a transformer when the two coils are found on top of each other.

Answer»

SOLUTION :
46.

If 'O' is at equilibrium then the values of the tension T_(1) and T_(2) are x, y, if 20 N is vertically down. Then x, y are

Answer»

20N, 30N
`20 sqrt(3), 20 N`
`20 sqrt(3)N, 20 sqrt(3)N`
`10N, 30 N`

Answer :B
47.

The linear charge density of 10muCcharge, uniformly distributed on the ring of 1m radius, will be ........

Answer»

`6.28 Cm^(-1)`
`6.28 xx 10^(5) cm^(-1)`
`1.59 xx 10^(-6) cm^(-1)`
`10^(5) cm^(-1)`

SOLUTION :Here, `Q = 10 muC`
`therefore Q = 10 xx 10^(-6)` C
`LAMBDA =Q/l = Q/(2pir)(therefore l = 2pih)` for ring
`therefore lambda =1.59 xx 10^(-6) cm^(-1)`
48.

Threshold wavelength for metal is 10,000 Å. If light of wavelength 5461 Å is incident on it, then stopping potential is 1.02 V, then value of Planck's constant is :

Answer»

`6.45xx10^(-34)`J-sec
`6.54xx10^(-34)`J-sec
`6.60xx10^(-3)`J-sec
`6.67xx10^(-34)`J-sec

ANSWER :C
49.

Answer the question in one sentence each (Alternatives are to be noted): After what time will the direction of current in an electric supply line of frequency 50 Hz be reversed? Or, Which physical quantity has unit Wbcdot m^(-2)? Is it scalar or vector?

Answer»

SOLUTION :TIME =`(T)/(2) = (1)/(2)((1)/(50)) = 0.01` s
Or,
Magneitc field, it is a VECTOR quantity
50.

Calculate the impact parameter of 5.0 MeV alpha-particle scattered by 10^(@) when it approaches a gold nuclcus (Z = 79).

Answer»

Solution :`Z_(1) = 2 , Z_(2) , = Z = 79`
Impact parameter, ` B = (Ze^(2) cot (THETA//2))/(4 pi epsilon_(0) ((1)/(2) mv^(2)))`
Here` theta = 10^(@) , E = (1)/(2) mv^(2) = 5.0 xx 1.6 xx 10^(-13) `J
`(1)/(4 pi epsilon_(0)) = 9 xx 10^(-9) Nm^(2) C^(-2)`
`therefore b = (79 xx (1.6 xx 10^(-19))^(2) cot 5^(@) xx 9 xx 10^(@))/(5.0 xx 1.6 xx 10^(-13)) = 2.6 xx 10^(-13)` m