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.

For a RLC series circuit , phasors of current i and applied voltage V=V_(0)sin omegat are shown in diagram at t=0, which of the following is/are CORRECT?

Answer»

At `=(pi)/(2omega)`, instantaneous power supplied by source is negative.
From `0lttlt(2pi)/(3omega)`, AVERAGE power supplied by source is positive.
At `t=(5PI)/(6omega)`, instantaneous power supplied by source is negative.
If `omega` is increased slightly, angle between the two PHASORS decreases.

Answer :A::C
2.

The ground (Fig.) shows the number of particles Nt emitted per second by a radioactive source as a function of time t The relationship between N_t and t is. .

Answer»

`N_t = 1000 e^-(20t//s)`
`N_t = 20 e^(-(20 t//s)`
`N_t = 3 e^(-(0.05 t//s)`
`N_t = 20 e^(-(0.05 t//s)`

Solution :(d) The graph can be REPRESENTED by the LINEAR equation as follows
`(1n N_t - 3)/(t - 0) = (0-3)/(60 -0)`
`rArr 1n N_t - 1n 20 = -(1)/(20) t [1n 20 = 3.00]`
`rArr 1n (N_t)/(20) = -0.05 t`
Hence, `N_t = 20 e^(-(0.05t//s))`.
3.

The temperature of the surface of the sun is approximately 6000 K. If we take a big lens and focus the sun rays, can we produce atemperatuer of 8000 K?

Answer»

Solution :No, ACCORDING to second law of thermodynamics heat itself cannot be transferred from a COLD body to a HOT body i.e., the CREATION of temperature of 8000 K by transferring heat from a cold body at 6000 K is not possible by means of a lens. This is the violation of the second law of thermodynamics
4.

Name the unit of electric potential in lesser units

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ANSWER :`1eV=1.6xx10^-19`
5.

Antennas are made from :

Answer»

INSULATORS
METALLIC conductors
Semiconductors
SUPER conductors

Answer :B
6.

A long cylindrical conductor of radius a contains a long cylindrical holr of radius l. The axes of the two cylinders are parallel and are at a distance d apart. A current I is uniformly distributed over the cross-section. Find the magnetic field at the centre of the hole. [Hint: Apply the principle of superposition of magnetic field]

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ANSWER :`(mu_(0)id)/(2pi(a^(2)-B^(2))`
7.

P and 2P are two forces acting on a particle. When the first force is increased by 20 kg wt, the second force is doubled, the direction of the resultant force remains unchanged. Then P is

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50 KG wt
20 kg wt
30 kg wt
40 Kg wt

Answer :B
8.

Find equivalent capacitance between A and B in the combination givenbelow. Each capacitor is of 2muF capacitance.

Answer»

Solution :Calculation of equilvalent capacitance.
Capacitors `C_(2),C_(3)` and `C_(4)` are in parallel
`:.C_(235)=C_(2)+C_(3)+C_(4)`
` :.C_(234)=6muF`
Capacitors `C_(1),C_(234)` and `C_(5)` are in series
`1/(C_(EQ))=1/(C_(1))+1/(C_(234))+1/(C_(5))=1/2+1/6+1/2`
`=7//6muF `
`C_("equialent)=6//7muF`
9.

The electric and magnetic fields of an electromagnetic wave are

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in PHASE and PERPENDICULAR to each other
out of phase and not perpendicular to each other
in phase and not perpendicular to each other
out of phase and perpendicular to each other

Answer :A
10.

The intensity of X-rays from a coolidge tube is plotted against wavelength lambda as shown in the figure. The minimum wavelength found is lambda_(c) and the wavelength of the K_(a)" line is "lambda_(k). As the accelerating voltage is increased.

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`lambda_(k)-lambda_(c)` increases
`lambda_(k)-lambda_(c)` DECREASES
`lambda_(k)` increases
`lambda_(k)` decreases

Solution :Angular momentum `l=sqrt(n(n+1)0.(h)/(2pi)`
For `H_(2)` and Li second excited state (n=3), l is same.
`l_(H)=l_(Li)`
Now, `E_(n)=(-13.6)/(n^(2)). Z^(2)eV`
`E_(n) alpha (Z^(2)/n^(2))`
Now `Z_(H)=1, Z_(Li)=3 & n=3`
`E_(H)/E_(Li)=1/9`
`E_(H) lt E_(Li)`
11.

A riding glass though formed by spherical surfaces has zero power. Give reason.

Answer»

Solution :The radii of curvature of CONVEX and CONCAVE surfaces are equal but ONE is positive and the other negative. These cancel with each other. Hence its POWER becomes zero.
12.

If surface tension of liquid is 72 dyne/cm, then the work done in increasing its surface area by 2 cm^2

Answer»

72 ERGS
`1.44 XX 10^-5J`
144 J
1.44 J

Answer :B
13.

Volume of a litre of water is subjected to a pressure of 10^6 N/m^2 . The change in the volume produced is 1 c.c. The bulk modulus of water is

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a)`10^10 N/m^2`
B)`10^9 N/m^2`
C)`10^11 N/m^2`
d)NONE of these

Answer :B
14.

If F is the force between two point charge submerged in a medium of dielectric constant K, then on withdrawing the mediun the force between the charges becomes.......

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`Fsqrt(K)`
FK
`F/sqrt(K)`
`F/K`

Solution :Force on MEDIUM `=("FORCES in air or VACCUM"(F))/K`
`therefore F =(F.)/K`
`therefore F. =FK`
15.

How much electric flux will come out through a surface vecds = 8hatj kept in an electrostatic field vecE=2hati+3hatj+4hatk?

Answer»

Solution :`vecds=8hatj, vecE=2hati+3hatj+4hatk, therefore" Flux "=E. vecds =(2hati+3hatj+4hatk).8hatj=3 XX 8 ="24 UNITS"`
`[hati, hatj=hatj.hatk=hatk.hati=0 and hati. Hati=hatj.hatj=hatk. Hatk.=1]`
16.

If the mass of neutron is 1.7xx10^(-27) kg , then the de-Broglie wavelength of neutron of energy 3 eV Is………(h=6.6xx10^(-34)JS)

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`1.4xx10^(-11)m`
`1.6xx10^(-10)m`
`1.65xx10^(-11)m`
`1.4xx10^(-10)m`

SOLUTION :`lambda=(h)/(SQRT(2mE))`
`=(6.6xx10^(-34))/(sqrt(2xx1.7xx10^(-27)xx3xx1.6xx10^(-19)))`
`=(6.6xx10^(-34))/(4.04xx10^(-23))`
`=1.63xx10^(-11)m`
`~~1.65xx10^(-11)m`(NEAREST value)
17.

Assertion To locate null defiection the battery key (K_(1)) is pressed first and then the galvanom eter key (K_(2)) Reason if first K_(2) is pressed and then as soon as K_(1) is pressed current suddenly try to increase so due to self induction a large stopping emf is generated in galvanometer which may damage the glavanometer .

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if both ASSERATION and REASON are TRUE and the Reason is a correct EXPLATION of Asseration
if both Asseration and Reason are true and the Reason is a correct explation of Asseration
if Asseration is ture but Reason is false
If both Asseration and Reason are false

Solution :N//A
18.

A parallel beam falls on solid glass sphere of radius R and refractive index mu. What is the distanceof final image after refraction from two surfaces of sphere? What is, the condition for the image to be real?

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`(R(mu-2))/(2(mu -1))` from 2ND surface , `mu gt 2`
`(R(2-mu))/(2((mu - 1)) ` from 2nd surface , `mu LT 2`
`(R(mu - 1))/((2 -mu))` from 2nd surface , `mu lt 2`
`R/2 ((mu+2)/(mu+1))` from 2nd surface , `mu lt 2` for any value of `mu`

Answer :B
19.

_____ is a paramagnetic substance .

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SOLUTION :GOLD, ALUMINIUM
20.

The least distance of distance vision of a person is 75 cm. The focal length of the reading spectacles for such a person should be

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a. 37.5 cm
b. 40 cm
c. 25 cm
d. 50 cm

Answer :A
21.

The displacement of a particle in simple harmonic motion in one time period is,

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a
2a
4a
Zero

Answer :D
22.

Magler Train: This train is based on the Lenz law and phenomena of electromagnetic induction. In this there is a coil on a railway track and magnet on the base of train. So as train is deciated then as is move down coil on track repel it and as it move up then coil attract it. Disadvantage of magler train is that as it slow down the force decreases and as it moves forwatd so due to Lenz law coil attract it backward. Due to motion of train current induces in the coil of track which levitate it. What is the disadvantage of the train?

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TRAIN experience upward force due to LENZ's law.
Friction force create a DRAG on the train
Retardation
By Lenz's law train experience a drag

Answer :D
23.

A bullet fired vertically up from the ground reaches a height 40 m in its path from the ground and it takes further time 2 seconds to reach the same point during descent.The toatl time of flight is

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4s
3s
6s
8s

Answer :C
24.

In Young double slit experiment the intensity of light a point on the screen where the path difference is lambda is K . The intensity of light at a pointwhere the path difference is lambda/3 [ lambda " is the wavelength of light used "]

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`K/4`
`K/3`
`K/2`
K

Answer :A
25.

Magler Train: This train is based on the Lenz law and phenomena of electromagnetic induction. In this there is a coil on a railway track and magnet on the base of train. So as train is deciated then as is move down coil on track repel it and as it move up then coil attract it. Disadvantage of magler train is that as it slow down the force decreases and as it moves forwatd so due to Lenz law coil attract it backward. Due to motion of train current induces in the coil of track which levitate it. What is the advantage of the train?

Answer»


ANSWER :D
26.

Some magnetic flux is changed from a coil of resistance 10Omega. As a result an induced current is developed in it, which varies with time as shown in figure. What is the magnitude of change in flux through the coil ?

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SOLUTION :The INDUCED charge is `q = (Delta phi)/(R )`

But, AREA of i-t CURVE GIVES charge
` therefore Delta phi = R xx ` area of i - t curve
` Delta phi = qR = 1/2 xx 4 xx 0.1 xx 10therefore Delta phi = 2wb`
27.

Magler Train: This train is based on the Lenz law and phenomena of electromagnetic induction. In this there is a coil on a railway track and magnet on the base of train. So as train is deciated then as is move down coil on track repel it and as it move up then coil attract it. Disadvantage of magler train is that as it slow down the force decreases and as it moves forwatd so due to Lenz law coil attract it backward. Due to motion of train current induces in the coil of track which levitate it. Which force causes the train to elevate up

Answer»

ELECTROSTATIC force
Time VARYING ELECTRIC field
magnetic force
Induced electric field

Answer :C
28.

If the weight stretching the sonometer wire be immeresd in water, then what will be effect on the frequency of the wire? If the length be increased by 2% then?

Answer»

SOLUTION :Will DECREASE, will INCREASE by 1%
29.

Iffc is the critical frequency of radio waves for reflection by the ionosphere andNmax is the density of free electrons in the ionosphere

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`f_(C)prop N_("max")`
`f_(c)prop(1)/(N_("max"))`
`f_(c)prop(1)/sqrt(N_("max"))`
`f_(c)propsqrt(N_("max"))`

Answer :D
30.

Which of the following is incorrect processes ?

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`Fe+Al_(2)O_(3)rarr2Al+Fe_(2)O_(3)`
`ZnO+CrarrZn+CO`
` Cr_(2)O_(3)2Alrarr2Cr+Al_(2)O_(3)`
` 2 [Ag(CN)_(2)]^(-)+Znrarr 2Ag+[Zn(CN)_(4)]^(2-)`

SOLUTION : IRON can-not EDUCE alumia
31.

Oxides of Plutonium and Uranium are used as fuel for _______ reactors.

Answer»

PRODUCTION
research
light water
prototype FAST breeder

Answer :D
32.

In the question number 4, work done in bringing a charge of 4xx10^(-9) C form infinity to that point is

Answer»

`2.4xx10^(-4)J`
`1.8xx10^(-4)J`
`3.2xx10^(-5)J`
`4.1xx10^(-5)J`

SOLUTION :WORK DONE , `W=q(V_f-V_i)=4xx10^(-9)xx4.5xx10^4`
`=1.8xx10^(-4)` J
33.

Consider a hydrogen-like atom whose energy in n^(th) excited state is given by E_(n) = (13.6Z^(2))/(n^(2)) When this excited atom makes transition from excited state to ground state most energetic photons have energyE_(max)= 52.224eV and least energetic photons have energy E_(min) = 1.223eV. find the atomic number of atom and the state of excitation.

Answer»

Solution :Maximum energy is liberated for transition `E_(n)to1` and minimum energy for `E_(n)toE_(n-1)`
HENCE, `(E_(1))/(n^(2))-E_(1)=52.224eV""…(1)`
and `(E_(1))/(n^(2))-(E_(1))/((n-1)^(2))=1.224eV""…(2)`
Solving above EQUATIONS simultaneously, we GET
`E_(1)=-54.4eV` and `n=5` Now
`E_(1)=-(13.6Z^(2))/(1^(2))=-54.4eV`.
Hence, Z = 2 i.e,
gas is HELIUM originally excited to n = 5 energy state.
34.

A tank has two outlets (i) a rounded orifice A of diameter D and (ii) a pipe B with well - rounded entry and of length L , as shown in Fig. For a height of water H in the tank, determine the (a) discharge from the outlets A and B , (b) velocities in the two outlets at levels 1 and 2 indicated in Fig .

Answer»

Solution :(a) Rounded orifice A : Applying Bernoulli.s equation to a POINT on the water surface 3 and point 1, we get
`0 + 0 + H = (p_(1))/(GAMMA) + (V_(1 A)^(2))/(2g) + 0`
As the orifice discharges to atmosphere ,
`(p_1)/(gamma) = 0` and `V_(1 A) = sqrt(2 g H)`
The discharge is ,
`Q_(A) = (pi)/(4) D^(2) sqrt(2 gH)`
At point 2 , the pressure is ATMOSPHERIC and hence by applying Bernoulli.s equation between points 3 and 2 , we get
`0 + 0 + (H + L) = 0 + (V_(2 A) ^(2))/(2 g) + 0`
or `V_(2 A) = sqrt(2 g (H + L))`
As the discharge is `Q_(A)` , the diameter at 2 will be smaller than D .
(b) Pipe B : By applying Bernoulli.s equation between points 3 and 2 .
`0 + 0 + (H + L) = 0 + (V_(2)^(2) B)/(2g) + 0`
or `V_(2B)= sqrt(2 g (H + L))`
As the pipe size is uniform from points 1 to 2 , by continuity equation
`V_(1 B) + V_(2 B) = sqrt(2 g (H + L))`
thus , the results are :
`{:(, "Orifice" , "Pipe") , ("Velocity at 1" , sqrt(2 gH) , sqrt(2 g ( H+ L))), ("Velocity at 2" , sqrt(2 g(H+ L)) , sqrt(2 g(H + L))), ("Discharge Q" , (pi)/(4) D^(2) sqrt(2gH) , (pi)/(4) D^(2) sqrt(2g (H + L))):}`
35.

An electric charge q is moving with a velocity v in the direction parallel to the magnetic field B. The magnetic force acting on the charge is

Answer»

qvB
zero
q/B
v/qB

ANSWER :B
36.

A body is projected at angle 30^@ to the horizontal with kinetic energy E. The KE at the top most point is

Answer»

Zero
E/4
E/2
3E/4

Answer :D
37.

Three equal resistors, each equals to r are connected as shown in figure. Then, the equivalent resistance between points A and B is

Answer»

r
3R
`(r )/(3)`
`(2R)/(3)`

ANSWER :C
38.

A child, playing with a long rope, ties one end and holds the other. The rope is stretched taut along the horizontal. The child shakes the enrf he is holding, up and down, in a sinusoidal manner with amplitude 10 cm and frequency 3Hz. Speed of the wave is 15 mis and, at t =0, displacement at the child.s end is maximum positive. Assuming that there is no wave reflected from the fixed end, so that the waves in the rope are plane progressive waves, answer the following questions (also assume that the wave propagates along.the positive x-direction) A wave function that describe the wave in the given situation is

Answer»

y = (0.1m) cos[(2 rad / m) x - (12.5 rad / s) t]
y = (0.1m) cos[(1.26 rad / m) x - (18.8 rad / s) t]
y = (0.1m) sin[(1.5 rad / m) x - (10 rad / s) t]
y = (0.1m) sin[(2 rad / m) x - (4 rad / s) t]

Answer :B
39.

The ratio of earth's surface area as covered by two communication satellite moving synchronously in equatorial plane with orbital speed of 1/sqrt2 times and sqrt3/5 times that of required orbital speed near earths surface is

Answer»

`4//5`
`5//2`
`5//4`
`2//5`

ANSWER :C
40.

A currrent of 5A is passing through a metallic wire of cross sectional area 14xx 10^(-6) m^2 . If the density of the charge carries in the wire is 5 xx 10^(26)//m^3, find the drift speed of the electrons (charge carries).

Answer»

Solution :`I= 5 A , A = 14xx10^(-6) m^2`
density of electron `=5 XX 10^(26)//m^3`
CHARGE of electron `e= 1.6xx10^(-19) C`.
DRIFT speed `v_d= i/("neA")`
`=(5)/(5XX10^(26) xx 1.6 xx 10^(-19( xx 14 xx 10^(-6))`
`=(10^(-1))/(1.6xx 14) = 4.46xx10^(-3) ` m/s
41.

A moving positive charge approaches a negative charge. What will happen to the potential energy of the system?

Answer»

will remain constant
will DECREASE
will increase
may increase or decrease

SOLUTION :`U =(K(q_(1))(-q_(2)))/(R)`
`r UARR implies Udarr`
42.

A child, playing with a long rope, ties one end and holds the other. The rope is stretched taut along the horizontal. The child shakes the enrf he is holding, up and down, in a sinusoidal manner with amplitude 10 cm and frequency 3Hz. Speed of the wave is 15 mis and, at t =0, displacement at the child.s end is maximum positive. Assuming that there is no wave reflected from the fixed end, so that the waves in the rope are plane progressive waves, answer the following questions (also assume that the wave propagates along.the positive x-direction) Phase difference between the child.s end and a point 2.5m from the child.s end will be

Answer»

`PI/2`
`(3PI)/2`
`pi/4`
`pi`

ANSWER :D
43.

Ovserver S detects two flashes of light. A big flash occurs at x_(1)=1200m and, 5.00mus later, a small flash occurs at x_(2)=700m. As detected by observer S', the two flashes occur at a single coordinate x'. (a) What is the speed parameter of S', and (b) is S' moving in the positive or negative direction of the x axis? To S', (c ) which flash occurs first and (d) what is the time interval between the flashes?

Answer»


ANSWER :(a) 0.334; (b) negative; (c ) BIG FLASH; (d) 4.71 `MUS`
44.

A child, playing with a long rope, ties one end and holds the other. The rope is stretched taut along the horizontal. The child shakes the enrf he is holding, up and down, in a sinusoidal manner with amplitude 10 cm and frequency 3Hz. Speed of the wave is 15 mis and, at t =0, displacement at the child.s end is maximum positive. Assuming that there is no wave reflected from the fixed end, so that the waves in the rope are plane progressive waves, answer the following questions (also assume that the wave propagates along.the positive x-direction) Equation of displacement of a point 2.5m from the child.s end can be expressed as

Answer»

y = -(0.1 m) COS (18.8 RAD / s) t
y = (0.1 m) cos (12.5 rad / s) t
y = (0.1 m) SIN (4 rad / s) t
y = -(0.1 m) sin (10 rad / s) t

Answer :A
45.

A child, playing with a long rope, ties one end and holds the other. The rope is stretched taut along the horizontal. The child shakes the enrf he is holding, up and down, in a sinusoidal manner with amplitude 10 cm and frequency 3Hz. Speed of the wave is 15 mis and, at t =0, displacement at the child.s end is maximum positive. Assuming that there is no wave reflected from the fixed end, so that the waves in the rope are plane progressive waves, answer the following questions (also assume that the wave propagates along.the positive x-direction) Velocity of the child.s end at t = 0 is

Answer»

3 m/s
4.5 m/s
zero
12.5 m/s

ANSWER :C
46.

Choose the correct alternative from the clues given at the end of the each statement : The size of the atom in Thomson's model is ....... the atomic size in Rutherford's model. (much greater than/no different from/much less than).

Answer»

SOLUTION :no DIFFERENT from
47.

In a metal in the solid state, such as a copper wire, the atoms are strongly bound to one another and occupý fixed positions. Some electrons (called the conductor electrons) are free to move in the body of the metal while the other are strongly bound to their atoms. In good conductors, the number of free electrons is very large of the order of 10^(28) electrons per cubic metre in copper. The free electrons are in random motion and keep colliding with atoms. At room temperature, they move with velocities of the order of 10^5 m/s. These velocities are completely random and there is not net flow of charge in any directions. If a potential difference is maintained between the ends of the metal wire (by connecting it across a battery), an electric field is set up which accelerates the free electrons: These accelerated electrons frequently collide with the atoms of the conductor, as a result, they acquire a constant speed called the drift speed which is given by V_e = 1/enA where I = current in the conductor due to drifting electrons, e = charge of electron, n = number of free electrons per unit volume of the conductor and A = area of cross-section of the conductor. If no potential difference is maintained between the ends of a conductor

Answer»

the free electrons are at rest
 the free electrons move randomly in all DIRECTIONS
 the RATE at which electrons pass through a CERTAIN cross-section from right to left is equal to the rate at which electrons pass through it from left to right
no CURRENT flows through the conductor.

Answer :B::C::D
48.

Four charges of +q, -q, +q and +q are placed at the corners A, B, C and D of a square. The resultant force on the charge at D is

Answer»

`(q^(2))/(4PI epsi_(0) a^(2)) (sqrt2 - (1)/(2))`
`(2Q^(2))/(PI epsi_(0)a^(2))`
`(q^(2))/(8pi epsi_(0)a^(2)) 2 sqrt2`
zero

Answer :A
49.

According to Ohm's law (R = (V)/(I) ) , as current flowing through a conductor increases, resistance of conductor .....

Answer»

Decreases
INCREASES
remain constant
nothing CAB be said

Solution :REMAINS constant
Knowledge based question
ACCORDING to Ohm.s law,
R = `(V)/(I) = ` constant
thus if I increases, then V ALSO increases and hence R remains constant .
50.

A uniform electric field vecE directed vertically upwards exists in a region of space. A point mass m carrying positive charge q isa state of equilibrium inside this electric field

Answer»

The point mass in in STABLE EQUILIBRIUM
The point mass in in unstable equilibrium
The point mass is in a STATE of neutral equilibrium
Equilibrium is not possible

Solution : INSIDE a UNIFORM electric field the net force will remain zero everywhere. Such type of equilibrium is called neutral equilibrium.