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1351.

In Young's double slit experiment the point source is placed slightly off the central axis as shown in figure. The plane of slits (S_(1), S_(2)) and screen are separated by a distance of 2m. The separation between the slits is 10 mm. The position of the slit above the axis is 1mm. if the wavelength lambda = 4000 Å in the above arrangement If central maximum is to occur at point 'P' then a sheet of thin film of refractive index mu = 1.5 can be introduced in the path of rays from one the slits to the point 'P'. (t-thickness of film)

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<P>`S_(1)` to P such that `[S S_(1) + S_(1)P+(MU - 1)t] ~ [S S_(2) + S_(2)P] = 0`
`S_(2)` to P such that `[S S_(2) + S_(2)P+(mu - 1)t] ~ [S S_(1) + S_(1)P] = 0`
`S_(1)` to P such that`[S S_(2) + S S_(1)] ~ [S_(1)P + (mu - 1)t + S_(2)P] = 0`
`S_(1)` to P such that`[S S_(1) + S S_(2)] ~ [S_(1)P + (mu - 1)t + S_(1)P] = 0`

ANSWER :A
1352.

Three containers C_(1), C_(2) and C_(3) have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in litres) are taken from each containers and mixed (assume no loss of heat during the process) The value of theta (in ""^(@)C to the nearest integer) is.......

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SOLUTION :According to table and applying law of calorimetry
`3T_(1) + 1 T_(2) = (3 + 1) 55^@` ..............(1)
`2T_1 +2T_3 = (2+2) 60^@.............(2)`
`T_1 + 2 T_3 = (1+3) 75^@` ........(3)
Solving`T_1 = 60^@ , T_2 =40^@ , T_3 = 80^@`
Hence ,
`T_1 + T_2 +2 T_3 = (1+1+2) theta, theta = 65^@ C`
1353.

A block of mass 'm' is released on the top of a frictionless incline as shown in the figure. The time period of the oscillation of the block is

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`2pi SQRT((m)/(K)) + sqrt((2L sin theta)/(g))`
`2pi sqrt((m)/(K)) + sqrt((2(L-l))/(g sin theta))`
`2pi sqrt(((L - l))/(g sin theta)) + sqrt((m)/(K))`
none of these

Answer :D
1354.

When you look at a clear blue sky you see tiny specks and hair like structures floating in your view, called "floaters" This is basically.

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INTERFERENCE pattern
Diffraction pattern
Emission spectra
Absorption spectra

Answer :B
1355.

Assertion A beam of charged particles is employed in the treatment of cancer. Reason Charged particles on passing through a material medium loss their energy by causing ionisation of the atoms along their path.

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If both ASSERTION and Reason are TRUE and Reason is the correct explanation of Assetion.
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
If Assetion is true but Reason is false.
If both Assertion and Reason are false.

Solution :A raditation consists of a beam of a CHARGED particles. When radiation is used for cancer treatement , them on falling upon the concerous , it destryos, cancer CELLS.
1356.

(i) At what temperature would the resistance of copper conductor be double its resistance at 0^(@)C. (ii) Does this temperature hold for all copper conductors regardless of shape and size? Given alpha for Cu=3.9xx10^(-3)""^(@)C^(-1).

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Solution :(i) `alpha=(R_(2)-R_(1))/(R_(1)(t_(2)-t_(1)))=(2R_(0)-R_(0))/(R_(0)(t-0))=(1)/(t)`
`t=(1)/(alpha)=(1)/(3.9xx10^(-3))=256^(@)C`
thus the resistance of copper CONDUCTOR becomes double at `256^(@)C`.
(II) Since a does not depend on size and shape of the conductor. So the above RESULT holds for all copper conductors.
1357.

In Young's double slit experiment the point source is placed slightly off the central axis as shown in figure. The plane of slits (S_(1), S_(2)) and screen are separated by a distance of 2m. The separation between the slits is 10 mm. The position of the slit above the axis is 1mm. if the wavelength lambda = 4000 Å in the above arrangement The thickness of the film in the above case is

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0.035 mm
0.45mm
0.060 mm
0.070 mm

Answer :D
1358.

A coil of 800 turns and 50cm^(2) are makes 10 rps about an axis in its own plane in a magnetic field of 10gauss perpendicular to this axis. What is the instantaneous induced emf in the coil?

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Solution :`A= 50cm^(2) = 50 xx 10^(-4) m^(2)`
n= 10 rps, N= 800
B= 100 gauss `= 100 xx 10^(-4) T= 10^(-2)T`
Now, `E= E_(0 ) sin omega t= NBA omega sin omega t`
`=800 xx 10^(-2) xx 50 xx 10^(-4) xx 2pi xx 10sin (20 pi t)`
or `E= 2.5 sin (20 pi t)` VOLT
1359.

A magnet of polestrength m is dividedinto four equal parts so that the length and breadth of each part is half than that of the original magnet then pole strength of each is

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m
`(m)/(4)`
`(m)/(2)`
4M

Solution :Pole strengthbecomes HALF DUE to breadth reduced ot half as change in length does not AFFECT polestrength
1360.

A horizontal platform executes up and down S.H.M. about amean position. Its period is 2pi s. A massm is resting on the platform, what is the greatest value of amplitude so that the mass 'm' may notleave the platform ?

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4.9 m
9.8 m
2.25 m
19.6 m

Solution :For mass not to leave the platform, the maximum acceleration `a_("max")=g`.
`:.""OMEGA^(2)r=g`
`(4PI^(2))/(T^(2)).r=g`
`r=g XX(T^(2))/(4pi^(2))`
Since `T=2pi s`.
`:.""r=gxx(4pi^(2))/(4pi^(2))=9.8" m"`
Hencecorrect CHOICE is (b).
1361.

For a house hold wiring, copper alloy is used which has resistivity rho=1.7xx10^(-8)Omega-m and electron density n=8.49xx10^(28)m^(-3). If atomic spacing of copper is about 0.2 nm and an electron on an average travels about N xx100 atomic spacings between two collisions, then find N. Given average speed of free electrons of copperis about 1.6xx10^(6)m//s.

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ANSWER :2
1362.

Arrange the follwing electromagnetic radiations in ascending order to their frequencies: (i) Microwaves (ii) Radio waves (iii) X-rays (iv) Gamma rays Write two uses of any one of these.

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SOLUTION :Radio waves `LT` MICROWAVES `lt`X-rays `lt` Gamma rays.
USES of Microwaves : (a) Aircraft navigation, (b) OVENS
1363.

(A): The resistivity of metals increases with increase in temperature (R): The average time of collision of electrons with the ion of lattice decreases with increase in temperature.

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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 :A
1364.

In an experiment of finding dip by the cot-method it is found that the apparent dips in two mutually perpendicular planes are 30^(@) and 20^(@) . Calculate the true dip of the place.

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ANSWER :`17^(@)7'`
1365.

Five identical conducting plates 1, 2, 3, 4 and 5 are fixed parallel to the equidistant from each other as shown in figure. Plate 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant emf V_0

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ANSWER :B
1366.

Consider the situation shown in figure. The switch S is open for a long time and then closed. Find the change in energy stored in the capactiors.

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`Cepsilon^(2)//2`
`Cepsilon//2`
`Cepsilon^(2)//4`
`C//epsilon`

ANSWER :C
1367.

Consider the situation shown in figure. The switch S is open for a long time and then closed. Find the heat developed in the system .

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`Cepsilon^2`
`2Cepsilon^2`
`Cepsilon^2//2`
`Cepsilon^2//4`

ANSWER :D
1368.

यदि A और B दो समुच्चय हो ताकि n(AuuB)=38,n(A)=17 और n(B)=23, तो A के उन अवयवों की संख्या क्या होगीजो B के भी अवयव हैं?

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10
2
21
5

Answer :B
1369.

Consider the situation shown in figure. The switch S is open for a long time and then closed. Find the charge flown through the battery when the switch S is closed.

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`Cepsilon//2`
`Cepsilon//5`
`C//epsilon`
`2C//epsilon`

ANSWER :A
1370.

Consider the situation shown in figure. The switch S is open for a long time and then closed. Find the work done by the battery .

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`2Cepsilon//2`
`Cepsilon^(2)//2`
`Cepsilon//2`
`Cepsilon//3`

ANSWER :B
1371.

If E, M, L and G denote energy, mass, angular momentum and universal Gravitational constant respectively, prove that(EL^(2))/(M^(5)G^(2)) is a dimensionless quantity .

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SOLUTION :Taking dimensional formulac energy `( E)= ML^(2)T^(-2)`
Universal GRAVITATIONAL CONSTANT (G) = `M^(-1) L^(3)T^(-2)`
SUBSTITUTING in `(EL^(2))/(M^(5)G^(2))` we get
`((ML^(2)T^(-2))(ML^(2)T^(-1))^(2))/((ML^(0)T^(-0))^(5)(M^(-1)L^(3)T^(-2))^(2))=(M^(1+2)L^(2+4)T^(-2-2))/(M^(502)L^(0+6)T^(0-4))`
= A dimensionless quantity .
1372.

Whenthe electron in hydrogen atom jumps from the second orbit to the first orbit, the wavelength of the radiation emitted is lambda. When the electron jumps from the third to the first orbit . The wavelenth of the radiation emitted is

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`(9)/(4)LAMBDA`
`(4)/(9)lambda`
`(27)/(32)lambda`
`(32)/(27)lambda`

ANSWER :C
1373.

When the torque value is maximum and when it minimum.

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SOLUTION :when `theta=o,MAX, theta=90^@,MIN`
1374.

A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, then the elongation of wire will be (in mm).

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l
2l
zero
`l/2`

Solution :In CONNECTED system, T =`(2W_(1)W_(2))/(W_(1)+W_(2))`
`rArrT=(2xxWxxW)/(2W)`=W
This tension PRODUCES EXTENSION 1 mm in the STRING. So correct CHOICE is (a).
1375.

Two long parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10A and 5A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.

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Solution :`F= (mu_0 I_1 I_2l)/(2 pi r) = (4 pi XX 10^(-7) xx 10xx5 xx .020)/(2 pi xx 0.05) = 4 xx 10^(-5)N`
The direction of the force is perpendicular to the length of wire Y and acts away from X (repulsion).
1376.

A particle is projected at an angle of 60^(@) with the horizontal from the ground with a velocity 10sqrt3ms^(-1). The angle between velocity vector after 2s and initial velocity vector is (g=10ms^(_2))

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`0^(@)`
`30^(@)`
`60^(@)`
`90^(@)`

Solution :Initial VELOCITY, `v_(i)=2costhetahati+4sinthetahatj=5sqrt3hati+15hatj`
Final velocity VECTOR (after 2S),
`v_(f)=ucosthetahati+(usintheta-"GT")hatj=5sqrt3hati+5hatj`
Now, `v_(i)*v_(f)=25xx3-15xx5=0`
`thereforev_(i)botv_(f)`
1377.

Which effects could be explained by wave theory by wave theory of light.

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SOLUTION :\INTERFERENCE, DIFFRACTION and POLARISATION.
1378.

Two light rays having the same wavelength lambda in vacuum are in phase initially. Then the first ray travels a path L_(1) through a medium of refractive index n_(1) while the second ray travels a path of length L_(2) through a medium of refractive index n_(2). The two waves are then combined to observe interference. The phase difference between the two waves is :

Answer»

`(2PI)/(lambda)(L_(2) - L_(1))`
`(2pi)/(lambda)(n_(1)L_(2) - n_(2)L_(1))`
`(2pi)/(lambda)(n_(2)L_(1) - n_(1)L_(2))`
`(2pi)/(lambda)((L_(1))/(n_(1)) - (L_(2))/(n_(2)))`

Solution :For first ray , optical path = `n_(1)L_(1)`
Optical path for 2ND ray = `n_(2)L_(2)`
path differenc = `n_(1)L_(1) - n_(2)L_(2)`
`THEREFORE` phase DIFFERENCE =`(2pi)/(lambda)(n_(1)L_(1) - n_(2)L_(2))`
1379.

A bodythrown at angle of 15^(@) with the hoizontal whichcertain speed has a horizontal rangeof 10m. If thebodyis thrown with the samespeed but at an angleof 45^(@) with horizontal , findits horizontal range ?

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ANSWER :20M
1380.

मानाf:RrarrR,f(x)=x^2द्वारा परिभाषित होता है तो 16 तथा -3 के पूर्व प्रतिबंध क्रमशः है

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`varphi,{2,-2}`
`{3,-3},varphi`
`{4,-4},varphi`
`{4,-4},{2,-2}`

ANSWER :C
1381.

In the circuit shown in fig the base current I_(B)=5.0muA, base resistor R_(B)=1.0MOmega collector resistor R_(c)=1.0kOmega the collector current I_(C)=5.0mA and the d.c. voltage in the collect circuit V_("CC")=6.0V. (i) Can this circuit be used as an ampifier? (ii) What happens if the resistance R_(C) is 400Omega and I_(B), I_(C) and i_(C) and R_(B) remain same as above?

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ANSWER :(i) No (II) Can be USED as AMPIFIER
1382.

Uttarakhand, Uttar Pradesh, Bihar, West Bengal and Sikkim have common frontiers with

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China
Bhutan
Nepal
Myanmar

Answer :C
1383.

An electric change 10^(3) mu C is placed at the origin (0,0) of X-Y coordinate system. Two points A and B are situated at (sqrt(2),sqrt(2)) and (2,0) respectively. The potential difference between the points A and B will be

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9 V
zero
2 V
4.5 V

Answer :B
1384.

A swimmer is capable of swimming 1.65 ms^(-1)in still water. If she swims directly across a 180m wide river whose current is 0.85 m/s, how far downstream (from a point opposite her starting point) will she reach ?

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92.7 m
40 m
48 m
20 m

Answer :A
1385.

A radioactive substance .A. is being generated at a constant rate C(100xx10^(6) atoms /sec) it disintegrates at a rate of lamda(37dps) to form B. Initially there are no A and B atoms. If the number of atoms of B after one mean life of A is 1xx10^(x) atoms, then find the value of x

Answer»


ANSWER :6
1386.

A parallel plate air capacitor of capacitance C is connected to a cell o f emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

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(A) The energy stored in the capacitor decreases K TIMES.
(B) The change in energy stored is
`(1)/(2) CV^(2) ((1)/(K) -1)`
(C) The charge on the capacitor is not conserved.
(D) The potential difference between the plates decreases K times.

Solution :Once the capacitor is being CHARGED, its charge remains CONSTANT. Charge Q = CV When dielectric slab is placed, its equivalent capacitance, C. = KC Initially, the energy stored in capacitor,
`U = (Q^(2))/(2C)`
Energy stored after placing dielectric slab
`U = (U)/(K) ` and ` V =(Q)/(C)`
and p.d. after placing dielectric slab
`V= (V)/(K)`
1387.

Which property of photoelectric particles was measured from Lenard's experiment?

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ANSWER :SPECIFIC CHARGED
1388.

If the loss in graviational potential energy to falling the sphere by h height and heat loss to surrounding at constant rate H are also taken to account the energy equation will modify to (A) m_(1) s_(1) (theta_(1)-theta) + (m_(1)gh)/(J) = m_(2) s_(2) (theta - theta_(2)) + m_(3) s_(3) (theta - theta_(2)) - H t (B) m_(1)s_(1) (theta_(1) -theta) - (m_(1)gh)/(J) = m_(2)s_(2)(theta -theta_(2)) + m_(3) s_(3) (theta -theta_(2)) + Ht (C) m_(1)s_(1) (theta_(1) -theta) + (m_(1)gh)/(J) = m_(2) s_(2) (theta - theta_(2)) + m_(3) s_(3) (theta -theta_(2)) + Ht (D) m_(1) s_(1) (theta_(1)-theta)-(m_(1)gh)/(J)=m_(2)s_(2)(theta-theta_(2)) +m_(3)s_(3)(theta-theta_(2))-Ht .

Answer»

Solution :HEAT generated `= m_(1) s_(1) (theta_(1)-THETA) + (m_(1)GH)/(J)`
.
1389.

A body starts from a point with a velocity and uniform acceleration a. The direction of acceleration is reversed when the velocity of the body becomes 5u. The velocity of body al point will be

Answer»

`-v`
`-5v`
`-7V`
`-9V`

Solution :Here `v^(2)-U^(2)=2aS` =Constant
`:. (5v)^(2)-v^(2)=v^(2)-(5v)^(2)`
or `v^(2)=49v^(2)` or v=-7v
neglecting+ ve sign
1390.

A man is throwing balls in air. He throws next ball when previous one is at the highest point. If he throws each ball after 2s , then height to which ball rises is

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10 m
20 m
30 m
15 m

Solution :The ball reaches the highest points in 2 s . Now
v= u+at `:.0=u-gxx2`
or `u=2xx10=20 ms^(-1)`
ALSO `h=(v^(2)-u^2)/(2a)=(0-400)/((-10)xx2)=20 m`
1391.

""_(19)^(40)K converts to ""_(18)^(40)Ar by positive beta decay as well as electron capture. Let Q values for the beta decay and electron capture be Q_(1) and Q_(2) respectively in the above reaction

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`Q_(1)=Q_(2)`
`Q_(1)ltQ_(2)`
neutrino EMITTED in positive `beta` decay is monoenergetic
neutrino emitted INCREASES by a factor `(27)/(8)`

ANSWER :D
1392.

The bucket with a ball placed inside is rotated in a verticle circle of radius 1m. What is the minimum speed the bucket should have so that the ball does not fall when it is at the highest point of path

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3. 14 cm/s
31.3 cm/s
3.31 m/s
None

Answer :C
1393.

The dimensions of magnetic intensity are

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[LI]
`[L^(2)I]`
`[L^(-1)I]`
`[L^(-2)I]`

ANSWER :C
1394.

A ratio transmitter operates at a frequency 880 kHz and a power of 10 KW. The number of photons emitted per second is :

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`1.72 xx 10^(31)`
`1.327 xx 10^(25)`
`1.327 xx 10^(37)`
`1.327 xx 10^(45)`

ANSWER :A
1395.

By increasing the temperature, the specific resistance of a conductor and a semiconductor.

Answer»

increases for both
decreases for both
increases, decreases
decreases, increases

SOLUTION :increases, decreases
As the temperature of conductor increases, the RANDOM motion of IONS increases so the motion of electrons becomes more CONSTRAINED, so the specific resistance increases. When increasing the temperature of covalent bond BREAKS and form a larger number of holes electron hence resistance decreases.
1396.

Given following data, Distance of mars from earth =8xx10^7 km lambda=550 nm, eye diameter =5 mm, objective to telescope =5.1m. For two luminous objects on mars

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minimum distance as RESOLVED by an observer with NAKED eye is approximately 1100 km.
minimum distance as resolved by an observer with telescope is 110km
ratio of minimum distance as resolved by an observer with naked eye and with telescope is 1000
ratio of minimum distance as resolved by an observer with telescope and with naked eye is 1000

Answer :C
1397.

Length of potentiometer wire is 5 m. It is connected with a battery of fixed e.m.f Null point is obtained for Daniel cell at 100 cm on it. If the length of the wire is made 7 m, then what will be the position of null point?

Answer»

Solution :Let the e.m.f of battery be E volt, the potential gradient is
`K_1 = E/5 V//m`
When the length of poteniometer wire is 7 m potential gradient is `k_2 = E/7 V//m`
Now, if NULL point is obtained at length `l_2` , then `E_2 1= k_2 l_2 = E/7 l_2 V`
Here same CELL is balanced in two ARRANGEMENTS , HENCE
`E/5 l_1 = E/7 l_2""l_1 = 1 m`
`""l_2 = 7//5 = 1.4 m`
1398.

(A): A person touching a high power line sets stuck with the line. (R): The current carrying wire attract the man towards it.

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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 :C
1399.

In defining the standard of length we have to specify the temperature at which measurement should be made. Are we justified in calling length a fundamental quantity if another physical quantity, temperature has to be specified in choosing a standard?

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Solution :Yes, LENGTH is a fundamental quantity, one METER is the DISTANCE that contains 1650763.73 wavelengths of orange-red light of Kr-86. Hence, the standard meter is independent of temperature. But the length of object varies with temperature and is given by the RELATION `L_(t=)L_0(1+alphat)` Therefore, we usually specify the temperature at which measurement is MAX.
1400.

Which of the following mode is used for the line of sight communication as well as satellite communication?

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GROUND WAVE
SKY wave
space wave
All of these

Answer :C