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 में से कौन-सा कथन सत्य है?

Answer»

`PI`और 22/7 दोनों परिमेय हैं।
`pi` और 22/7 दोनों अपरिमेय हैं।
`pi` अपरिमेय है और 22/7 परिमेय है।
`pi`परिमेय है और 22/7 अपरिमेय है

Answer :C
Discussion
2.

Explain conductor (metal), insulator and semiconductor by drawing diagrams based on band.

Answer»

Solution :Atomic number of Si is 14. The electron configuration of silicon atom is `1s^(2)2S^(2)2p^(6)3s^(2)3p^(2)`hence K and L shells are completely filled and M shell is incomplete and there are `3s^(2)3p^(2)` valence electron in it.
And atomic number of Ge is 32. The electron configuration of silicon atom is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(2)` hence K, L and M shells are completely filled and N shell is incomplete and there are `4s^(2)4p^(2)` valence electron in it.
Hence both Si and Ge semiconductor are tetravalent.
There are total 4 ELECTRONS in outermost orbit of Si or Ge crystal. The maximum possible number of electron in the outer orbit is `8(2s+6p` electrons).
So, for the 4N valence electrons there are 8N energy states.
These 8N discrete energy levels can either form a continuousband or they may be grouped in different bands depending upon the distance between the atoms in the crystal.
At the distance between the atoms in the crystal lattices of Si and Ge, the energy band of these 8N states is split apart into two which are separated by an energy gap `E_(g)`, which is shown in the figure below.

The energy band positions in a semiconductor at 0 K.
The upper band called the conduction band, consists of infinitely large number of closely spaced energy states.
The lower band called the valence band, consists of closely spaced completely filled energy states.
The lower band which is completely occupied by the 4N valence electrons at temperature of absolute zero is the valence band and there is an upper conduction ban with 4N energy levels, which is completely empty at absolutezero temperature.
The lowest energy level in the conduction band is shown as `E_(c )` and highest energy level in the valence band is shown as `E_(v)`. Above `E_(c )` and below `E_(v )` there are a large number of closely spaced energy levels as shown in figure.
The gap between the top of the valence band and bottom of the conduction band is called the energy band gap (Energy gap `E_(g)`). It is known as forbidden gap.
Depending on the type of material, forbidden may be small, large or zero, depending on this gap the following three types are available.
Case I Metal (conductor) :

According to (i) and (ii) in figure (a) one can have a metal either when the conductionband is partially filled and the valence band is partially empty or when the conduction and valence bands overlap.
When there is overlap electrons from valence band can easily move into conduction band.
When the valence band is partially empty, electrons from its lower level can move to HIGHER level making conduction possible. Therefore, the resistance of such materials is low or the conductivity is high.
Case II Insulator:

According to figure (b), the distance between the valence band and the conduction band is very large means energy gap `E_(g)gt 3eV`, there is no electrons in the conduction band as well as raising the temperature, electrons cannot be send from the valence band to the conduction band therefore, such a material does not carry electricity such material is called insulator.
According to figure (c ), such a material has an energy gap `(E_(g) lt 3eV)`.
The region between valence band and conduction band is small so at room temperature some electrons from valence band can acquire enough energy to cross the energy gap and enter the conduction band.
These electrons (through small in numbers) can move in the conduction band, therefore the conduction of charge flow is reduced. Such materials are called semiconductor.
Energy gap in semiconductor `E_(g) lt 3eV`. The value of `E_(g)` for Si is 1. 1 eVand the value of `E_(g)` for Ge is 0.7 eV.
Discussion
3.

The unit of electric field in tensity is .................. .

Answer»


ANSWER :`NC^(-1)`
Discussion
4.

भूगर्भीय प्लेटों में कितने प्रकार की गति होती हैं?

Answer»

एक
दो
तीन
चार

Answer :C
Discussion
5.

The political and constitutional changes brought about by the French Revolution were:

Answer»

it ENDED the ABSOLUTE monarchy.
It transferred POWER to a body of the French citizens.
It proclaimed that HENCEFORTH people would constitute the nation and shape its destiny.
All the above.

Answer :D
Discussion
6.

In the circuit as shown, it is given that R_1//R_2 = 1//2 and potential at O is V.

Answer»

`100//9 V`
`20 V `
`100//7 V`
`100//3 V`

Solution :c. Given, `R_2 = 2R_1`
For AO , `100 -V = i_1R_1`
For `BO, 100- V = i_2(2R_2)`
For `OD, V-0 = i_3 (20)`
For `OC , V - 50 = (i_1 + i_2/2 - i_3)(50)`
`VR_1 = 50R_1 = ((3i_1)/2 xx 50 -i_3 xx 50) R_1`
From EQS. (i) and (iii),
`VR_(1) = 50R_(1) = 75(100-V) -2.5 VR_(1)`
`3.5 VR_1 - 50R_1 = 7500 - 75V` ,brgt If `Vlt75V,` then `RHSgt0`.
So, `LHS gt0`
`3.5 V gt 50 ("since" R_1!=0)`
`Vgt50/3.5`
From Eq.(v), `V=20 V`.
`70 R_1 - 50R_1 = 7500 - 1500 = 6000`
`R_1 = 300` Omega
`R_2 = 2R_1 = 600`Omega
From Eq. (v),
`3.5 VR_1 + 75V = 7500 + 50 R_1`
For any value of `R_1` Vgt0.
Discussion
7.

AB is a potentiometer wire as in figure. If the value of R is increased, in which direction will the balance point J shift ?

Answer»

SOLUTION :By increasing value of R current in main circuit will DECREASE hence p.d. across WIRE AB will decrease. Hence potential gradient (K) across two end of wire AB decreases.
At certain time for GIVEN cell null point chauges. Hence, potential gradient `(phi)`across AB decreases according to equation,
`E = phi I `
`THEREFORE phi = (E)/(I)` ,
Thus, with increase in I, `phi`decreases. Thus, with increase in current (I) null point wi shift toward B.
Discussion
8.

The magnetic flux linked with a coil is phi = (4t^(2)-6t-1) milliweber. Find the emf induced in the coil at t = 2 s

Answer»

Solution :`PHI= 4t^(2)-6t-1`
Induced emf `epsilon= dphi//dt`
`epsilon= 8t-6= 8 XX 2- 6= 10 "mV"`
Discussion
9.

The field at the surface of the earth is approximately equal to .......... .

Answer»

a) 1 tesla
b) 2 gauss
c) 10^4 tesla
d) 1 gauss

Answer :10 T
Discussion
10.

Fission of nuclei is possible because the binding energy per nuclei in them

Answer»

increases with MASS NUMBER at low mass numbers
decreases with mass number at low mass numbers
increases with mass number at high mass numbers
decreases with mass number at high mass numbers

Solution :For nuclei having A gt 56 , binding energy per NUCLEON GRADUALLY decreases.
Discussion
11.

In half-wave rectifier, maximum percentage of A.C. power that can be converted into D.C. power is

Answer»

0.25
`40.6%`
`81.2%`
0.1

Answer :B
Discussion
12.

Suppose that two tanks, 1 and 2, each with a large opening at the top, contain different liquids. A small hole is made in the side of each tank at the same depth h below the liquid surface, but the hole in tank 1 has half the cross-sectional area of the hole in tank 2. (a) What is the ratio rho_(1) // rho_(2) of the densities of the liquids if the mass flow rate is the same for the two holes? (b) What is the ratio R_(v_(1))//R_(v_(2)) of the volume flow rates from the two tanks? (c) At one instant, the liquid in tank 1 is 16.0 cm above the hole. If the tanks are to have equal volume flow rates, what height above the hole must the liquid in tank 2 be just then?

Answer»

SOLUTION :(a) 2 , (B) 1/2 , (C) 4.00 CM
Discussion
13.

Light is incident from glass s(mu=1.50) to water (mu=1.33). Find the range of the angle of deviation of which there are two angles of incidence.

Answer»

0 to `COS^(-1)(8//9)`
0 to `SIN^(-1)(8//9)`
0 to `cos^(-1)(9//8)`
0 to `sin^(-1)(9//8)`

ANSWER :A
Discussion
14.

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10^7V m^(-1). (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation). For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF ?

Answer»

SOLUTION :`:.` Dielectric strength `= 10^7 Vm^(-1)` and field up to 10% of the dielectric strength is to be applied.
Hence, electric field employed E = 10% of `10^7 =10^6 V m^(-1)`
As voltage rating V = 1 kV = 1000V and for a capacitor`E = V/d`
`:. d = V/E = 1000/10^6 =10^(-3)m`
As `C = (K epsi_0A)/d` and here`C = 50 pF = 50 xx 10^(-12) F and K = 3`
`:.` Area of the plates `A = (C.d)/(K epsi_0) = (50 xx 10^(-12) xx 10^(-3))/(3 xx 8.85 xx 10^(-12)) = 1.9 xx 10^(-3) m^2 = 19 cm^2`.
Discussion
15.

The wavelength of a microwave is 3.0 mm and it's electric field has an amplitdude of 4Vm^(1). Find the frequency of the wave

Answer»

SOLUTION :`lambda=3.0mm=3.0xx10^(-3)` m, AMPLITUDE of magnetic field `B_0=(E_0)/C=(4)/(3xx10^8)=1.33xx10^(-7)T`.
Discussion
16.

Mercury of density rho_(m) is poured into cylinderical communicating vessels of cross-sectional area 2A and A respectively. A solid iron cube of volume V_(0) and relative density 2 is dropped into the broad vessel, and as a result the level of the mercury in it rises. Then, water is poured into the broader vessel until the mercury reaches the previous level in it. Find the height of water column if it does not submerge the block.

Answer»

SOLUTION :`(2v_(0))/(A+v_(0).^(2//3))` PROCESS.
Discussion
17.

In the given circuit values are as follows epsilon_(1)=2V, epsilon_(2)=4V, R_(1)=1Omega and R_(2)=R_(3)=1Omega. Calcualte the Currents through R_(1), R_(2) and R_(3).

Answer»

Solution :
LET `i_(1), i_(2)`are CURRENTS across `R_(1) and R_(3)`.
`(i_(1)+i_(2))` is current across `R_(2)`.
Their DIRECTION are taken as shown
From Kirchoff.s second LAW for AGFBA LOOP
`-i_(1)R_(1)-(i_(1)+i_(2))R_(2)+E_(1)=0`
`i_(1)+i_(1)+i_(2)=2 rArr 2i_(1)+i_(2)=2rarr(1)`
From Kirchooff.s second law for BCDEB loop
`-i_(2)R_(3)-(i_(1)+i_(2))R_(2)+E_(2)=0`
`i_(2)+i_(1)+i_(2)=rArr i_(2)+2i_(2)=4rarr(2)`
Solving equation (1) and (2)
we get `i_(1)=0A, i_(2)=2A`
Thus currents across `R_(1)` is 0,
while across `R_(3)` and R_(2)` are 2A each
Discussion
18.

Use Gauss'law to derive the expression for the electric field(oversetto E)due to a straight uniformaly charged infinite line of charge density lambdaC m ^(-1) (b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge. (c) Find the work done is bringing a charge q from perpendicular distancer_1 to r_2 (r_2 lt r_1)

Answer»

Solution :Elementary work done against the electric field for bringing a charge q SITUATED at a normal distance .r. from the line of charge is given by:
`dW =- FDR =- q E dr =- q [(lambda)/( 2 pi in _0 r) ] dr `
` therefore ` Work done in bringing the charge q from perpendicular distance `r_1 "to" r_2 ` is given as :
`W= INT dW =- underset (r_1) oversetto (r_2) int (q lambda)/( 2 pi in _0 r)dr = -(qr)/( 2 pi in _0)[1N r ]_(r_1) ^(r_2)=-( q lambda)/( 2 pi in _0) 1n (r_2)/(r_1)=(q lambda )/( 2 pi in _0)1n ((r_1)/( r_2_)) `
Discussion
19.

i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working. (ii) Why must both the objective and the eyepiece of a compound microscope have short focal lengths ?

Answer»

Solution :(i) A compound microscope consists of an objective lens O of small focal length and small aperture and an eyepiece E of small focal length but slightly greater aperture as shown in Fig. 9.84. OBJECT AB is placed at a distance slightly greater than the focal length of objective (`u_(0) gt f_(0)`) so that` A.B.`reaL inverted and magnified IMAGE A.B. is formed at a distance v0 on other side of lens. The image A.B., being situated between principal focus and optical centre of eyepiece, behaves as a real object for eyepiece. As shown in figure the final image A"B" is formed at least distance of distinct vision and is a virtual, inverted and highly magnified one. The MAGNIFYING power of a compound microscope is GIVEN by:
`m=-L/f_(0) (1+D/f_(e))`

where L = distance between objective and eyepiece lenses, D = least distance of distinct vision and `f_(0)`and `f_(e)`are the focal lengths of objective and eyepiece lenses respectively.
(ii) To have higher magnifying power a compound microscope must have both the objective and the eyepiece having SHORT focal length. This fact is clear from the formula for magnifying power of microscope given above.
Discussion
20.

Suppose an electron is attracted toward the origin by a force k/r where k is a constant and r is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be r_(n) and the kinetic energy of the electron to be T_(n). Then which of the following is true?

Answer»

`T_(N) ALPHA 1/n, r_(n) alpha n^(2)`
`T_(n) alpha 1/n^(2), r_(n) alpha n^(2)`
`T_(n)` independent of `n, r_(n) alpha n`
`T_(n) alpha 1/n, r_(n) alpha n`

SOLUTION :`(mv^(2))/(r_(n))=k/r_(n)`
`RARR mv^(2)=k`
`1//2 mv^(2)=k//2`
K.E. is independent of n it also `mvr_(n)=(nh)/(2pi)`
`r_(n) alpha n`
Discussion
21.

A gun fires two bullets at 60° and 30° with the horizontal the bullets strike at same horizontal distance. The maximum heights for the two bullets are in the ratio :

Answer»

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

Solution :`R_(1)=R_(2)IMPLIES (u_(1)^(2)sin2xx60^@)/g=(u_(2)^(2).sin2xx30^@)/g`
`implies U_(2)^(2)/u_(1)^(2)=(sqrt(3)/2)/(sqrt(3)/2)=1`
`H_(1)/H_(2)=(u_(1)sin^(2)60^@)/(2G)xx(2g)/(u_(2)^(2)sin^(2)30^@)=(sqrt(3)/2)^(2)/(1/2)^(2)`
`implies H_(1)/H_(2)=3/4xx4/1=3/1`
Discussion
22.

Electric field at a point near a uniformaly charged infinite plane sheet having surface density of charge sigmais given as E= ________________

Answer»


ANSWER :`(SIGMA)/(2 in _0) `
Discussion
23.

Thimmakka now makes a living ____________.

Answer»

from PRIZE money and monthly pension
from monthly pension only
by WORKING at the hospital
by SELLING PLANT products

Answer :A
Discussion
24.

In this figure the force should be applied on mass m = 5kg so that it just won't slip is (Given that car is moving with constant acceleration a = 5 m/s and mu=0.4

Answer»

170 N
180 N
250 N
150 N

ANSWER :D
Discussion
25.

In the circuit as shown, it is given that R_1//R_2 = 1//2 and potential at O is V.

Answer»

`150 Omega`
`300 Omega`
`450 Omega`
`600 Omega`

Solution :d. Given, `R_2 = 2R_1`
For `AO , 100 -V = i_1R_1`
For `BO, 100- V = i_2(2R_2)`
For `OD, V-0 = i_3 (20)`
For `OC , V - 50 = (i_1 + i_2/2 - i_3)(50)`
`VR_1 = 50R_1 = ((3i_1)/2 xx 50 -i_3 xx 50) R_1`
From Eqs. (i) and (III),
`VR_(1) = 50R_(1) = 75(100-V) -2.5 VR_(1)`
`3.5 VR_1 - 50R_1 = 7500 - 75V` ,BRGT If `Vlt75V,` then `RHSgt0`.
So, `LHS gt0`
`3.5 V gt 50 ("since" R_1!=0) `
`Vgt50/3.5`
From Eq.(v), `V=20 V`.
`70 R_1 - 50R_1 = 7500 - 1500 = 6000`
`R_1 = 300 Omega`
`R_2 = 2R_1 = 600Omega`
From Eq. (v),
`3.5 VR_1 + 75V = 7500 + 50 R_1`
For any VALUE of `R_1 Vgt0`.
Discussion
26.

In the circuit as shown, it is given that R_1//R_2 = 1//2 and potential at O is V.

Answer»

`80 Omega`
`16 Omega`
`240 Omega`
any value

SOLUTION :d. Given, `R_2 = 2R_1`
For `AO , 100 -V = i_1R_1`
For `BO, 100- V = i_2(2R_2)`
For `OD, V-0 = i_3 (20)`
For `OC , V - 50 = (i_1 + i_2/2 - i_3)(50)`
`VR_1 = 50R_1 = ((3i_1)/2 xx 50 -i_3 xx 50) R_1`
From Eqs. (i) and (iii),
`VR_(1) = 50R_(1) = 75(100-V) -2.5 VR_(1)`
`3.5 VR_1 - 50R_1` = 7500 - 75V` ,brgt If `Vlt75V,` then `RHSgt0`.
So, `LHS gt0`
`3.5 V GT 50` ("since" R_1!=0)`
`Vgt50/3.5`
From Eq.(v), `V=20 V`.
`70 R_1 - 50R_1 = 7500 - 1500 = 6000`
`R_1 = 300 Omega`
`R_2 = 2R_1 = 600Omega`
From Eq. (v),
`3.5 VR_1 + 75V = 7500 + 50 R_1`
For any value of `R_1` Vgt0.
Discussion
27.

What is Magnetic field ?

Answer»

Solution :(i) Magnetic field is the REGION or space around every magnet within which its influence can be FELT by keeping ANOTHER magnet in that region.
(ii) The magnetic field `vecB ` (or) magnetic inductor at a point is defined as a force experienced by the bar magnet of UNIT pole strength.
(iii) It is also defined as the total number of magnetic lines of force crossing per unit area normally through through a materials .
Discussion
28.

One mole ideal monatomic gas is taken at temperature of 300 K. If volume is doubled keeping its pressure constant. The change in the internal energy is

Answer»

450 R
550 R
650 R
750 R

Solution :SINCE pressure is constant, `VpropT`
`therefore (V_(i))/(T_(i))=(V_(F))/(T_(f))`
or `T_(f)=(V_(f))/(V_(i))T_(i)=2T_(i)=2xx300=600K`
`therefore DeltaU=(f)/(2)nRDeltaT` (f = degrees of freedom)
`=(3)/(2)xx1xxRxx(600-300)=450R`
Discussion
29.

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased

Answer»

the size DECREASES.
the INTENSITY increases.
the size increases.
the intensity decreases.

Solution :We know that amount of DIFFRACTION is determined from ratio `((lamda)/(d))`. Here by increasing d, amount of diffraction decreases and so width of diffraction PATTERN decreases. Thus option (A) is correct.
Here power of radiation made incident on the slit is constant. Hence, after passing through slit, if such light is made incident on smaller region then naturally, intensity obtained in that region would be more because `I=(P)/(A)implies` when P = constant, `Iprop(P)/(A)implies` If A decreases then I A must increase `implies`Option (B) is correct.
Discussion
30.

Inside a conductor electrostatic field is zero'. Explain.

Answer»

SOLUTION :In the static situation, when there is no CURRENT inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor.
A conductor has free electrons, the free charge carriers would EXPERIENCE force and drift .
In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside.
Discussion
31.

Figure shows three blocks of mass m each hanging on a string passing over a pulley. Calculate the tension in the string connecting A to B and B to C?

Answer»

SOLUTION :Net FORCE = 2mg -mg = mg
Total mass = m + m + m = 3m
Acceleration , ` a=(mg)/(3m) =(G)/(3)`
CONSIDERING block A, `T_1 - mg = ma or T_1 = mg + ma(or)`
`T_1 = mg + m((g)/(3)) (or)`
`T_1 = (4)/(3) mg ` , Considering block C ,
`mg-T_2 = ma (or) T_2 = mg-ma(or)`
`T_2 = mg - (mg)/(3) = (2)/(3) mg ` .
Discussion
32.

An imaginary line joining the optical centre of the eye lens and the yellow point is called as

Answer»

PRINCIPAL AXIS
Vision axis
Neutral axis
OPTICAL axis

ANSWER :B
Discussion
33.

An astronaut is floating in space near her shuttle when she realizes that the cord that's supposed to attach her to the ship has become disconnected. Her total mass (body+suit+equipment) is 91kg. She reaches into her pocket, finds a 1 kg metal tool, and throws it out into space with a velocity of 6m/s directly away from the ship. if the ship is 10 m away, how long will it take her to reach it?

Answer»

Solution :Here, the ASTRONAUT+tool are the system. Because of conservation of linear momentum ltbgt `m_("astronaut")v_("astronaut")+m_("tool")v_("tool")=0`
`m_("astronaut")v_("astronaut")=-m_("tool")v_("tool")`
`v_("astronaut")=-(m_("tool"))/(m_("astronaut"))v_("tool")`
`=-(1kg)/(90KG)(-6m//s)=(1)/(15)m//s`
Using distance=rate`xx`time, we FIND
`t=(d)/(v)=(10M)/((1)/(15)m//s)=150s=2(1)/(2)`min
Discussion
34.

The angle of diffraction of the second order maximum of wavelength 5xx10^(-5) cm is 30^(@) in the case of a plane transmission grating. How many lines are there in 1 cm of the grating surface.

Answer»


ANSWER :1000
Discussion
35.

Thesetwhichrepresentstheisotope , isobar, isotonerespectivelyis

Answer»

a) `(""_(1)H^(2),""_(1)H^(3) ), (""_(79)Au^(197 ), ""_(80)Hg ^(198) ) and (""_(2) He^(3 ) , ""_(1) H^(2))`
B) `(""_(2) He^(3),""_(1)H^(1)) ,(""_(79)AU^(197) , ""_(80) Hg^(198) ) and(""_(1)H^(1) ,""_(1) H^3)`
c) `(""_(2) He ^(3), ""_(1) H^3) ,(""_(1) H^2,""_(1)H^(3)) , and(""_(79) Au^(197) , ""_(80)Hg ^(198) )`
d) `(""_(1)H^2 ,""_(1)H^3 ) ,( ""_2He^3 ,""_1 H^3 ) , and(""_(79) Au ^(197),""_(80)Hg ^(198))`

Answer :D
Discussion
36.

A uniform tube of length 60 cm stands vertically with its lower end dipping into water. When the length above water successively taken at 14.8 cm and 48.0 cm, the tube responds to a vibrating tuning fork of frequency 512 Hz. Find the lowest fi-equency to which the tube will respond when it is open at both ends.

Answer»


Solution :From the given resonating LENGTHS we can USE
`48 - 14.8 = ((lambda)/(2))`
`lambda` = 66.4 cm.
THUS for given situaton wavelength is `lambda` is `lambda` = 64.4 cm so speed of sound is
`v = nlambda = 512 xx 0.664`
= 339.97 m/s
Fundamental frequency of tube is
`n_(0) = (v)/(2l) = (399.97)/(2 xx 0.6)` = 283.30 Hz.
Discussion
37.

A current carrying power line carries current from west to east. The direction of magnetic field at a short distance, above it, is

Answer»

NORTH to South
South to North
East to West
West to East

Answer :A
Discussion
38.

The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

Answer»

2 OHM
4 ohm
8 ohm
1 ohm

ANSWER :A
Discussion
39.

A thin conductor rod is placed between two unlike point charges +q_(1), and -q_(2). Then

Answer»

Due to charge INDUCED on the rod AB the POINT charge `+q_(1)`, will be acted upon, in addition to the point charge `-q_(2)`, by the induced charges formed at the ends of the rod.
The TOTAL force ACTING on the charge `+q_(1)`, will increases, as compared to the case WITHOUT rod.
On `q_(1)` attractive force exerted by the +ve charge induced at the end B.
None of these

Answer :A::B
Discussion
40.

In winter, the temperature inside the wall of a room as compared to the temperature of air is the room is

Answer»

LOWER
HIGHER
same
may be lower or higher DEPENDING on ATMOSPHERIC pressure

Answer :B
Discussion
41.

Two identical spheres of mass m and charge q are suspended from a common point by two threads of the same length l and it is found that threads make an angle theta with the vertical when the spheres are in equilibrium . Calculate the value of q.

Answer»

Solution :The force acting on the charged particles suspended from the strings is shown in the figure. There will be tension in the wire due to the force of repulsion between the two charges.

At equilibrium , for charge at A, vertical component of tension in the string will be balanced by the weight of the charge .
`T cos theta =MG` …(i)
Where, `cos theta =x/l`
Also, `T SIN theta =F_E =1/(4piepsilon_0)(q.q)/(2X)^2`, where `F_E` is the force of repulsion between charges at A and B .
`rArr T sin theta =F_E = 1/(4piepsilon_0)q^2/(2l cos theta)^2`…(ii) (`because x = l cos theta` )
On dividing equation (i) by (ii) , we get
`(sin theta)/(cos theta)=1/(4piepsilon_0)xxq^2/(4l^2 cos^2 theta)xx1/(mg)`
`rArr q^2=16 pi epsilon_0 mg l^2 sin^2 theta XX tan theta `
`q=4l sin theta sqrt(piepsilon_0 mg tan theta)`
Discussion
42.

Differentiate between polarisedandunpolarised light

Answer»

SOLUTION :
Discussion
43.

Four particles alpha-particle, deuteron and electron and a CI^(-) ion enter in a transverse magnetic field perpendicular to it with same kinetic energy. Their paths are as shown in figure. Now match the following two columns.

Answer»

<P>

ANSWER :`A to Q; B to p; C to s; D to r`
Discussion
44.

Longitudinal waves do not xhibit

Answer»

REFRACTION
REFLECTION
differection
polarisation

Answer :D
Discussion
45.

Assertion: Intensity of magnetic field due to a solenoid near its ends is only half of its magnitude inside the solenoid. Reason: If solenoid is made very long then magnetic field inside the solenoid becomes uniform.

Answer»

If both ASSERTION and REASON are correct and reason is correct EXPLANATION of the assertion
If both assertion and reason are correct but reason is not correct explanation of the assertion
If assertion is correct but reason is incorrect.
If assertion is incorrect but reason is correct.

Answer :B
Discussion
46.

In a refracting type telescope, the distance between objective and eyepiece is 25 cm for normal adjustment. Find the focal lengths of the objective and eyepiece if the magnifying power of the telescope is 10.

Answer»

SOLUTION :22.73 CM, 2.27 cm
Discussion
47.

In figure a point source S is placed at a height h above the plane mirror in a medium of refractive index mu. Find the distance between the images.

Answer»

Solution :Separation between two IMAGES `(2H)/( MU)`
Discussion
48.

A thin foil of certain stable isotope is irradiated by thermal neutrons falling normally on its surface. Due to the capature of neutrons a radioonuclide with decay constant lambda appears. Find the law describing, accumulation of that radionuclide N(t) per unit area of the foil's surface. The neutron flux density is J, the number of nuclei per unit area of the foil's surface is n, and effective cross-section of formation of active nuclei is sigma.

Answer»

SOLUTION :Rate of FORMATION of the radionuclide is `n.J.sigma` PER unit area per.sec. Rate of DECAY is `lambda N`.
Thus`(dN)/(dt)=n.J.sigma-lambda N` per unit area per second
Then `((dN)/(dt)+lambdaN)E^(lambda t)=n.J.sigma^(lambda t)` or `(d)/(dt)(Ne^(lambda t))=n.J.sigma.e^(lambda t)`
Hence `Ne^(lambda t)= Const+(n.J.sigma)/(lambda)e^(lambda t)`
The number of radionuclide at `t=0` whenthe process starts is zero. So constant`=-(n.J.sigma)/(lambda)`
Then `N=(n.J.sigma)/(lambda)(1-e^(-lambda t))`
Discussion
49.

Characteristic X-rays are produced due to ....

Answer»

transfer of momentum in collision of electron with target ATOMS
TRANSITION of electron from higher to lower electrons orbits in an atom
heating of target
transfer of energy in collision of electrons with atoms in the target.

Solution :When the high energy electron hits the target and takes out an electron from the INNER orbit there is a possibility of an outer electron to fall down to fill up that vaccancy and when such a transition of electron takes place from a higher energy level to a lower energy level, the difference of energy `(E_(2)-E_(1) = hv)` between two LEVELS is given in the form of radiation known as characteristic X-rays.
Discussion
50.

A force vector applied on a mass is represented by vec(F)=6hati + 8hatj+10hatk and it accelerates it with 1 ms^(-1) . What will be the mass of the body ?

Answer»

`10sqrt(2)` kg
`2sqrt(10)` kg
10 kg
20 kg

Solution :Here magnitude of the FORCE acting
`|vec(F)\=[6^(2)+8^(2)+10^(2)]^(1//2)=(200)^(1//2)`
`m=|vec(F)|/|a|=(200)^(2)/1=10sqrt(2)kg`
Discussion