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.

A laser beam has a power of 100 mW. It has an aperture of 5xx10^(-3)m and emits a wavelength 6943 Å. The beam is focused with a lens of focal length 0.1 m. Calculate the areal spread and intensity of the image.

Answer»


ANSWER :`2.87xx10^(-10)m^(2),3.48xx10^(8)WM^(-2)`
Discussion
2.

Which one of the following phenomena is not explained by Huygen's construction of wavefront?

Answer»

refraction
reflection of light
diffraction
ORIGIN of SPECTRA

Solution :Huygen.s construction of WAVEFRONT cannot explain origin of spectra which can be explained on the BASIS of quantum THEORY.
Discussion
3.

Several identical capacitors were connected in parallel and charged to a voltage φ_(0). Subsequently they were reconnected in series with the aid of a switch. What will be the voltage across the terminals? Will the energy of the system change?

Answer»

Solution :The solution is obvious from Figs. 24.11a and 24.11b, which give the diagrams of the parallel and SERIES connections of four identical capacitors. The energy in the case of parallel CONNECTION is
`W_("PAR")=(C_("par")Psi_("par")^(2))/(2)=(nC_(N)varphi_(0)^(2))/(2)`
and of series connections
`W_("ser")=(C_(ser) varphi_("ser")^(2))/(2)=((C_(0)//n)n^(2) varphi_(0)^(2))/(2)=W_("par")`
Discussion
4.

State Lenz's law.

Answer»

SOLUTION :Lenz.s law STATES that the direction of the INDUCED current is such that it always OPPOSES the cause responsible for its production.
Discussion
5.

Given the frequency of microwave used by microwave oven to cook food.

Answer»

0.951 GHz
0.501 GHz
0.651 GHz
0.5051 GHz

Answer :A
Discussion
6.

Given below are some examples of wave motion. Statein each case if the wave motion is transverse, longitudinal or a combinatin of both. (a) Motion of kink in a longitudinal spring producedby displacing one end of the spring sideways. (b) Waves produced in a cylinder containing a liquid by moving its piston back and forth. (c ) Waves produced by a motorboat sailing in water. (d) Ultrasonic waves in air produced by a vibrating quartz crystal.

Answer»

SOLUTION :(a) TRANSVERSE and longitudinal
(B) Longitudinal
(c) TRANS verse and longitudinal
(d) Longitudina
Discussion
7.

A solid metall sphere of radius R has a charge + 2Q. A hollow spherical shell of radius 3R placed concentric with the spherehas charge - Q. Calculate the potential difference between the spheres.

Answer»

`(Q)/(piepsilon_(0)R)`
`(Q)/(3piepsilon_(0)R)`
`(Q)/(2piepsilon_(0)R)`
`(Q)/(7piepsilon_(0)R)`

ANSWER :B
Discussion
8.

The Bohr model of atom ......

Answer»

predicts the same emission spectra for all types of atoms
assumes that the angular momentum of electrons is quantized.
USES Einstein's photo-electric equation
predicts continuous emission spectra for atoms.

SOLUTION :Angular momentu in BOHAR ATOM `LAMBDA =`n(((h)/(2pi))`
Discussion
9.

ABCD is a square loop made of a uniform conducting wire. A current enters the loop at A and leaves at D . The magnetic field is :

Answer»

ZERO only at the centre of the LOOP
maximum at the of the loop
zero at all points outside the loop
zero at all points INSIDE the loop

Answer :A
Discussion
10.

Give the equation which represents law of gauss for magnetism.

Answer»

`oint VEC(E ).d vec(a)=(Q)/(epsilon_(0))`
`oint vec(B).d vec(a)=0`
`oint vec(B).d vec(l)=-(d phi_(B))/(dt)`
`oint vec(E ).d vec(l)=mu_(0)(i_(c )+i_(d))`

ANSWER :B
Discussion
11.

Three particles are projected vertically upwards from a point from the surface of earth with velocities v_(1)=sqrt((2)/(3)gR,)v_(2)sqrt(gR)" and "v_(3)sqrt((4)/(3)gR.) If h_(1)h_(2)h_(3) are the respective maximum heights reached then h_(1):h_(2):h_(3) is given by :

Answer»

``1:2:3`
`2:3:4`
`1:2:4`
`1:3:5`

ANSWER :B
Discussion
12.

What should be the size of the antenna or aerial ? How the power radiated is related to length of the antenna and wavelength ?

Answer»

Solution :Size of antenna (or) aerial : For TRANSMITTING a signal , we need an antenna . The size of the antenna comparable to the wavelength of the signal (at least `lambda//4`) . So that the antenna properly senses the time VARIATION of the signal . For an e.m waves of FREQUENCY 20 kHz, the wavelength `lambda` is 15 km . Obviously such a long antenna is not possible to construct and operate . There is a need of translating the information contained in our original low frequency base band signal into high frequencies before transmission .
Effective POWER RADIATED by an antenna : A linear (length l) show that the power radiated is proportional to `(l)/(lambda^(2))` . For the same antenna length , the power radiated increases with decreasing `lambda` i.e, increasing frequency .Hence the effective power radiated by a long wavelength base band signal would be small .
Discussion
13.

Three long concentric conducting cylindrical shells have radii R, 2R and 2sqrt2 RInner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:

Answer»

`(1/3in_0)/(LN2)`
`(6piin_0)/(ln2)`
`(piin_0)/(ln2)`
NONE

ANSWER :B
Discussion
14.

When electron with mass m is accelerated with volt V associated de-Broglie wavelength is lambda .When proton with mass M is accelerated with same voltage associated de-Broglie wavelength will be………

Answer»

<P>`LAMBDA((m)/(M))`
`lambda((M)/(m))`
`lambdasqrt((m)/(M))`
`lambdasqrt((M)/(m))`

SOLUTION :de-Broglie wavelength `lambda =(h)/(sqrt(2mVq))`
`therefore lambda prop (1)/(sqrt(M))`[`because` Remaining TERMS are same]
`therefore (lambda_(p))/(lambda_(e))=sqrt((m_(e))/(m_(p)))=sqrt((m)/(M))`
`therefore lambda_(p)=lambda sqrt((m)/(M)) [because lambda_(e)=lambda]`
Discussion
15.

The wavelength involved in the spectrum of deuterium ""_(1)^(2)D are slightly different from that of hydrogen spectrum because ………

Answer»

the size of the TWO nuclei are different
the nuclear FORCE are different in the two cases.
the masses of the two nuclei are different.
the attraction between the electron and the nucleus is different in the two cases.

Solution :Because of different in mass of both atoms, `R_("atom")=((R_(oo))/(m))/(1+(m)/(M_("atom")))` where `R_(oo)=1.09737xx10^(7)m^(-1)`
m=mass of electron
changing `m_("atom")` changes `R_("atom")`
Hence the FORMULA for the wavelength of lines of the spectrum `(1)/(lamda)=R_("atom")((1)/(m^(2))-(1)/(n^(2)))` varies SLIGHTLY according to this formula.
Discussion
16.

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40^(@). What is the refractive index of the material of the prism? The refracting angle of the prism is 60^(@). If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light

Answer»

SOLUTION :`n~=1.53 and D_(m)` for PRISM in WATER `~=10^(@)`
Discussion
17.

A motor cyclist moving with 30 km/ h blows a whistle of 476 Hz towards a cliff. If velocity of sound is 1220 Km/h, the apparent frequency of the echo heard by him is :

Answer»

250 Hz
500 Hz
425 Hz
476 Hz.

Solution :When the ,MOTOR cyclist is approaching STATIONARY wall i. e. observer then
`v. = ((1220)/(1220 - 30)) 476 = 488 Hz.`
` therefore` FREQUENCY of reflected wave i.e., ECHO is 488 Hz and observer is moving towards it.
`therefore v.. = ((1220 + 30)/(1220)) 488`= 500 Hz.
hence the correct choice is (b) .
Discussion
18.

Which is the smallestunit of current ?

Answer»

SOLUTION :STATE AMPERE
Discussion
19.

In an A.C. circuit of frequency 400/pi Hz is of capacitance CmuF and reactance of capacitance is 25Omega then the value of C = .....

Answer»

`25muF`
`50muF`
`400muF`
`100muF`

Solution :`f=400/pi` HZ, `X_C=25Omega`
`omega=2pif =2pixx400/pi`=800 rad/s
`X_C=1/(omegaC)`
`THEREFORE C=1/(omegaX_C)=1/(800xx25)`
`therefore C=1/20000=50xx10^(-6) F = 50 muF`
Discussion
20.

A domain in ferromagnetic iron in the form of a cube of side length 1mum. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The atomic mass of iron 55g/mole and its density is 7.9 g/cm^(3). Assume that each iron has a dipole moment of 9.27 xx 10^(-24) A m^(2).

Answer»

SOLUTION :The volume of the cubic domain is `V=(10^(-6) m)^(3)=10^(-18) m^(3)=10^(-12) cm^(3)`
Its mass is volume `xx " density "=7.9 g cm^(-3) xx 10^(-12) cm^(3)=7.9 xx 10^(-12)g`
It is given that Avagadro number `(6.023 xx 10^(23))` of iron atoms have a mass of 55g. Hence the number of atoms in hte domain is `N=(7.9 xx 10^(-12) xx 6.023 xx 10^(23))/(55)`
`=8.65 xx 10^(10)` atoms
The maximum possible dipole moment `m_("max")` is achieved for the (UNREALISTIC) case when all the atomic moments are PERFECTLY ALIGNED.
Thus, `m_("max")=(8.65 xx 10^(10)) xx (9.27 xx 10^(-24))`
`=8.0 xx 10^(-13)Am^(2)`
The CONSEQUENT magnetisation is
`M_("max")=m_("maxx")//"Domain volume"`
`=8.0 xx 10^(-13) Am^(2)//10^(-18) m^(3)`
`=8.0 xx 10^(5) Am^(-1)`
Discussion
21.

Monochromatic radiation of wavelength 640.2 nm (1nm=10^(-9)m) fro a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. the source is replaced by an iron source and its 427.2 nm line irradiates the sae photocell. predict the new stopping voltage.

Answer»

SOLUTION :Here `lamda=640.2nm=6.402xx10^(-7)m,V_(0)=0.54V and lamda.=427.2nm=4.272xx10^(-7)m`
Using the relation
`hv-phi_(0)=hv-phi_(0)=eV_(0) or (HC)/(lamda)-phi_(0)=eV_(0)`, we have
`hc((1)/(lamda.)-(1)/(lamda))=e(V_(0).-V_(0)) or (V_(0).-V_(0))=(hc)/(e)[(1)/(lamda.)-(1)/(lamda)]`
`therefore V_(0).=-0.54=(hc)/(e)((1)/(lamda.)-(1)/(lamda))=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19))[(1)/(4.272xx10^(-7))-(1)/(6.402xx10^(-7))]=0.97`
`implies V_(0).=0.97+0.54=1.51V`.
Discussion
22.

The power used by the human heart while beating 72 times per minute is given by one of the following if the average work done during one singal beat is 0.5 J.

Answer»

0.06 W
0.6 W
6.0 W
0.30 W

Solution :Here `P=W/t=(0.5xx72)/(60)`
=0.6 W
Discussion
23.

In Young's experiment distance between 5^(th) dark fringe and 3rd bright fringe is x_(5 )-x_(3)=......x

Answer»

2
3
`(2)/(3)`
`(3)/(2)`

Solution :For `5^(th)` DARK fringe,
`x_(5)=(9 lambdaD)/(2D) ""...(1)`
For `3^(rd)`bright fringe,
`x_(3)=(3lambdaD)/(d) ""...(2)`
`:.x_(3)-x_(3)=(9 lambdaD)/(2d)-(3lambdaD)/(d)`
`=(3lambdaD)/(2d)=(3)/(2) beta`
Discussion
24.

A soap film of thickness 00011 mm appears dark when seen by the reflected light of wavelength 580 nm.. What is the index of refraction of the soap solution, if it is known to be between 1-2 and 1.5 ?

Answer»


SOLUTION :Given `d=0.0011xx10^-3m`
for minimum reflection of LIGHT `2md=nl`
`rarr mu=(nlamda)/(2D)=(2nlamda)/(4d)`
`=(580x10^-9xx(2n))/(4xx11xx10)`
`=(58(2n))/44=0.132(2n)`
Given that `mu` has a value in between 1.2 and 1.5.
`rarr when n=5, `mu=0.132xx10=1.32`
Discussion
25.

The x-y plane is the boundary between two transparent media. Medium-1 with z gt0 has refractive indexsqrt2 and medium -2 with zlt 0 has a refractive index sqrt3. A ray of light in medium -1 given by the vector A=6sqrt3 i+8sqrt3j-10k is incident on the plane of separation. If the unit vector in the direction of refracted ray in medium -2 is (1)/(5) (ai+bj-(5)/(sqrt2)k) then the value of ab.

Answer»


SOLUTION :N//a
Discussion
26.

In example 9 if B is the mid point of rod then find e_(OB)le_(BA)

Answer»

SOLUTION :`e_(OB)= (1)/(2)B omega((l^(2))/(4))= B omega l^(2)`
`e_(BA)=(1)/(2)B omega(x^(2))_((l)/(2))^(l)= (1)/(2)B omega xx(3L^(2))/(4)= (3)/(8)B omega l^(2)`
`therefore (e_(OB))/(e_(BA))= (1)/(3)`
Discussion
27.

A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. Calculate the magnification

Answer»

Solution :longitudinal magnification (m) = `("lenght of image"(L.))/("lenght of object"(l))`
Give: length of object, l = `(f)/(3)`
LET, l be the length of image, then, `m=(l.)/(l)=(l.)/(f//3)(or)l=(mf)/(3)`
Image of one end coincides with the object. THUS, the coinciding end must be at center of curvature.
Hence, U = R = 2f
`u = u + (f)/(3)`
`u=u.(f)/(3)=2f-(f)/(3)=(5f)/(3)`
`v=u+(f)/(3)+(mf)/(3)=(5f)/(3)+(f)/(3)+(mf)/(3)=(f(6+m))/(3)`
Mirror equation, `(1)/(v) + (1)/(u) = (1)/(f)`
`(1)/(-((f(6+m))/(3)))+(1)/(-((5f)/(3)))=(1)/(-f)`
After equation, `(3)/(f(6+m))+(3)/(5f)=(1)/(f),(3)/((6+m))=(2)/(5)`
`6 + m = (15)/(2) , m = (15)/(2) - 6`
`m = (3)/(2) = 1.5`
Discussion
28.

The four arms of a Wheastone bridge (Figure) have the following resistance : AB=100Omega, BC=10Omega, CD=5Omega and DA=60Omega A galvanometer of 15Omega resistance is connected across BD. Calculate the current through the galvano - meter when a potential difference of 10 V is maintained across AC,

Answer»

Solution :Considering the MESH BADB, we have
`100I_(1)+15I_(g)-60I_(2)=0`
`(or) 20I_(1)+3I_(g)-12I_(2)=0"….(1)"`
Considering the mesh BCDB, we have
`10(I_(1)-I_(g))-15I_(g)-5(I_(2)-I_(g))=0`
`10I_(1)-30I_(g)-5I_(2)=0rArr 2I_(1)=6I_(g)-I_(2)=0".....(2)"`
Considering the mesh ADCEA,
`60I_(2)+5(I_(2)+I_(g))=10`
`65I_(2)+5I_(g)=10 rArr 13I_(2)+I_(g)=2"...(3)"`
MULTIPLYING Eq. (2) by 10
`20I_(1)-60I_(g)-10I_(2)=0"....(4)"`
From Eq. s (4) and (1) we have
`63I_(g)-2I_(2)=0 rArr I_(2)=31.5I_(g)"....(5)"`
Substituting the value of `I_(2)` into Eq (3), we get
`13(31.5I_(g))+I_(g)=2`
`410.5I_(g)=2""rArr I_(g)=4.87mA`.
Discussion
29.

Which one of the following graphs, shows the variation of magnetic induction B with distance from a long thin and straight wire carryning a current

Answer»




ANSWER :A
Discussion
30.

What is robotics ?

Answer»

Solution :ROBOTICS is an integrated study of mechanical ENGINEERING, ELECTRONIC engineering, computer engineering, and SCIENCE.
Discussion
31.

A force of 100 N produces a change of 0.01% in a length of wire of area of cross -section 1.0 mm^2. Young's modulus of the wire is

Answer»

a)`10^11 N/m^2`
B)`10^8 N/m^2`
C)`10^12 N/m^2`
d)`10^9 N/m^2`

ANSWER :C
Discussion
32.

Identify the set of acidic oxides.

Answer»

`Na_2O,CaO,BAO`
`ZnO,PBO,BeO`
`CO,NO,N_2O`
`Mn_2O_7,CrO_3,V_2O_6`

Solution :ACIDIC OXIDE `to Mn_2O_7,CrO_3,V_2O_5`
(DUE to higher oxidation state of central metal)
Basic oxide `to Na_2O, CaO, BaO`
(due to lower oxidation state of central metal)
Alkaline and alkali metal oxide are always basic.
Neutral oxides `to CO,NO,N_2O`
Amphoteric oxides `to ZnO,PbO, BeO` (acidic as well as basic)
Hence option (d) is correct.
Discussion
33.

Define: (a) Magnetic declination (b)Magnetic dip. Mention the S.I. unit of magnetisation.

Answer»

SOLUTION :The S.I. UNIT of MAGNETISATION is `Am^(-1)` .
Discussion
34.

For what distance is ray optics a good approximation when the aperture is 4mm wide and the wavelength is 500 nm ?

Answer»

32m
64m
16m
8m

Solution :Fresnel.s distance `Z_(F)=(a^(2))/(lamda)=((4mm)^(2))/((500nm))=((4XX10^(-3)m)^(2))/((500xx10^(-9)m))=32m`.
Discussion
35.

2A current is flowing in a circular coil of 'n' turn, having 10cm radius. The flux density at its centre is 0.126 xx 10^(-2) wb//m^(2). The value of 'n' is

Answer»

10
100
1000
1

Answer :B
Discussion
36.

{:("(i) Kirchhoff's Ist Law","(a) Junction rule"),("(ii) Joule's Law","(b) H" =I^(2)RT),("(iii) Kirchhoff's Iind Law", "(c ) Voltage rule"),("(iv) Nichrome", "(d) Alloy"):}

Answer»


Answer :`(i) to (B); (II) to (d);(III) to (a); (IV) to (C )`
Discussion
37.

Explain the working of a zener diode as a voltage regulator.

Answer»

Solution :Circuit diagram :
Any INCREASE / decrease in the input VOLTAGE results in increase / decrease of the voltage drop across RU without any change in the voltage across the ZENER diode.
Discussion
38.

What is a diffraction grating?

Answer»

Solution :A dffraction grating is an optical component with a periodic structure that SPLITS and diffracts light TWO several BEAMS travelling in DIFFERENT direction.
Discussion
39.

Resistors ofhighvalue are madeup ofcarbon. Why ?

Answer»

SOLUTION :HIGH resistivityand LOWTEMPERATURE Coefficientofresistance.
Discussion
40.

In a Young's double slit experiment the distance between the slits and the screen is 1.60 m . Using light of wavelength 6 xx 10^(-7) m, the distance between the centre of the interference pattern and fourth bright fringe on either side is 16 mm . calculate the slit separation .

Answer»


ANSWER :`0.24` MM
Discussion
41.

Water stands upto height h behind the dam as shown in the figure. The front view of the dam gate is also shown in the adjoining figure. Density of water is rho and acceleration due to gravity is g. If atmospheric pressure force is also considered, then the point of application of total force acting on the dam due to water above O is

Answer»

`(h)/(rr4r)`
`(h)/(3)`
h
`(h)/(2)`

Solution :
Force on dam = PRESSURE of water `xx` Area = {(density of water) (ACCELERATION due to gravity) (height of water/2} `xx` {(width of dam) (height of water)}
`F = (RHO g (h)/(2)) xx [(b -h).h]`
`= (rho g)/(2) (BH^(2) - h^(3))`
At equilibrium, atmospheric pressure = pressure on dam
So, `(dF)/(dh) = 0`
`rArr (del)/(del h) {(rho g)/(2) (bh^(2)- h^(3))} = 0`
`rArr (rho g)/(2) (2bh - 3h^(2)) = 0`
`rArr 2bh - 3h^(2) = 0`
`rArr 2b = 3h rArr b= 3h//2`
`b-h = (3h)/(2) - h = h//2` is the location where the total weight of water acts at a particular point.
To find the point of action of total force,
`y_(R ) = (I_(xc))/(y_(c )A) + y_(c )`
where, `y_(R )`= location where point of force acts, A = area,
`y_(c )`= location where total weight acts = h/2 and `I_(xc)` = moment of inertia (here rectangular plate).
`=(1)/(12) Ah^(2) = (1)/(12) bh^(3)`
so, `y_(R ) = ((1//12) bh^(3))/((1//2)h.bh) + (1)/(2)h`
`= (1)/(6) h + (1)/(2) h = (2)/(3)h` (from the top)
So, `y_(R ) = (h)/(3)` (above the base) `= (2)/(3)h` (from the top)
Discussion
42.

An electomagnetic wave is entering from onemedium to another. Then the property which remains unaltered is

Answer»

Wavelength
Frequency
Velocity
None of these

Answer :B
Discussion
43.

A sinusoidal transverse wave is traveling along a string in the negative direction of an x axis. Figure 16-S1 shows a plot of the displacement as a function of position at time t = 0, the scale of the y axis is set by y = 4.0 cm. The string tension is 3.6 N, and its linear density is 28 g/m. Find the (a) amplitude, (b) wavelength, (e) wave speed, and (d) period of the wave (e) Find the maximum transverse speed of a particle in the string. If the wave is of the form y(x,t) =y_(m) sin (kx pm omega t+phi), what are (f) k, (g) omega (h) phi, and (i) the correct choice of sign in front of omega?

Answer»


Answer :(a) 5.0 cm; (b) 0.40; (c) 11.3 m/s (d) 0.035 s; (E) 889 cm/s=8.9 m/s (F) `16m^(-1); (g) 1.8 XX RAD//s ` (h) 0.93 rad; (i) plus sign
Discussion
44.

Three plates A, B and C are placed close to each other with +Q charge given to the middle plate. The inner surfaces to A and C canbe connected to earth through plate D and keys K_(1) and K_(2) . The plates D is a dielectric slab with dielectric constant K_(1) then the charge that will flow through plate D and keys K_(1) and K_(2) . The plate D is a dielectric slab with dielectric constant K, then the charge that will flow though plate D when K_(2) is closed and K_(2) is open is

Answer»

`-Q`
`-Q//2`
`Q(1-(1)/(K))`
`(Q)/(2)(1-(1)/(K))`

ANSWER :c
Discussion
45.

We wish to coat flat glass (n=1.50) with a transparent material (n = 1.45) so that reflection of light at wavelength 500 nm is eliminated by interference. What minimum thickness can the coating have to do this?

Answer»

SOLUTION :86.2 nm
Discussion
46.

The equation of motion of a projectile is y=ax-bx^(3) where a and b are constants of motion. Match the quantities of column I with the relations of column II. {:("Column-I", "Column-II"),((A)"The initial velocity of projection",(p) (a)/(b)),((B)"The horizontal range of projectile",(q)asqrt((2)/(bg))),((C)"The maximum vertical height attained by projectile",(r) (a^(2))/(4b)),((D)"The time of fightof projectile",(s) sqrt((g(1+a^(2)))/(2b))):}

Answer»

A-p, B-q, C-r, D-s
A-s,B-p,C-q,D-r
A-s,B-p,C-r,D-q
A-p, B-s,C-r, D-q

Answer :C
Discussion
47.

Average distance of the earth from the sun is L_(1). If one year of the earth=D days, one year of another planet whose average distance from the sun is L_(2) will be

Answer»

`D((L_(2))/(L_(1)))^(1//2)`DAYS
`D((L_(2))/(L_(1)))^(3//2)`days
`D((L_(2))/(L_(1)))^(2//3)`days
`D((L_(2))/(L_(1)))`days

Solution :According to Kepler.s third law,
`T^(2) prop R^(3). or " T prop r^(3//2)`
`:. (D.)/(D)=((L_(2))/(L_(1)))^(3//2)" or D.=D((L_(2))/(L_(1))^(3//2))" days"`
Discussion
48.

A particle of mass m and positive charge q,moving with a uniform velocity v enters a magnetic field B as shown in figure.(a) Find the radius of the circular arc it describes in the magnetic field. (b)Find the angle subtended by the arc at the centre. (c) How long does the particle stay inside the magnetic field? (d)Solve the three parts of the above problem if the charge q on the particle is negative.

Answer»


Solution :(a)`R=(mv)/(QB)`
(b)Angle Subtended by the arc`=pi/2`
(c)`THETA=omegat`
`pi/2=(QBT)/m rArr t=(pim)/(2qB)`
(d)`r=(mv)/(qB) rArr theta=(3pi)/2 rArr t=0/omega =(3pim)/(2qB)`

Discussion
49.

One mole of an ideal gas undergoes cyclic change as shown. The work done in the process is (taking 1 atmosphere =10^(6)" dyne/"cm^(2)):

Answer»

100J
300J
700J
900J

Solution :Work DONE in complete eycle =AREA under the rectangle ABCD

`=DC xxBC =3" litre"xx3` atmosphere
`=3xx10^(-3) m^(3)xx3xx10^(5)" N/m"^(2)`.
`=9XX10^(2)=900 J`.
Thus, correct CHOICE is (d).
Discussion
50.

In an experiment for determination of refractive index of glass of prism by i - delta plot, it was found that a ray incident at angle 35^(@) , suffers a deviation of 40^(@) and that it emerges at angle 79^(@). In that case which of the following is closest to the maximum possible value of the refractive index?

Answer»

1.5
1.6
1.7
1.8

Solution :`"We KNOW", DELTA = i + e - A`
`or, " " 40^(@) = 35^(@) + 79^(@) - A or, A = 74^(@)`
Now, the minimum value of angle of deviation, `delta_(m) le 40^(@)`
`"In that case", mu le (sin""(A + delta_(m))/(2))/(sin""(A)/(2))`
`or, " " mu le(sin""(74^(@) + 40^(@))/(2))/(sin""(74^(@))/(2))`
`or, " " mu le 1.39`
`therefore " " mu_(max) = 1.5`
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