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.

What is the critical velocity?

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SOLUTION :CRITICAL velocity is a maximum velocity of a fluid above which a stream LINE flow changes to a turbulent flow, i.e., it is the maximum velocity of a liquid below which its flow REMAINS STREAMLINE.
2.

Consider a point object of mass 'm' moving in a circle of radiusa=1m. For anyinstantaneouspositionof theobject, theta is the angle that theradiallinejoiningtheobjectand thecentremakeswiththeposition X-axisof a cartesiancoordinatesystemwith thecentreof circle Oas theorigin. hati and hatj are unit vectors along X- axisand Y-axis, respectively . Suppose that the motion is a 'Unifrom Circular Motion ' with a constantangularspeed(pi)/(36) rad//sec and that thesense of rotationis counterclockwisewiththeta = 0 at t = 0 . For an object whichmovesin a circle , it isusuallyconvenientto introduce twomutuallyperpendicularunitvectors hatr_(r) and hatr_(t), as shown in thefig 3.57. Here hatr_(r)is theradialunit vectorand hatr_(t) , thetangential unitvector.Answerthe followingquestions : For anyinstantantaneousposition of theobjectedP, theradialunit vector hatr_(r)can be expressedas :

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Solution :
`vecr_(r) = HATI cos THETA + hatj SIN theta `
3.

Assertion : Progressive and transverse waves can not be produced in liquid or gas. Reason: Light wave is progressive (travelling).

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Assertin and reason both are CORRECT and reason is the correct explantion of assertion.
Assertion and reason both area correct but reason is not the correct EXPLANATION of assertion.
Asserton is true but reaosn is false.
Both, assertion and reaosn are false.

Solution :Because liquids and GASES can not withstand (SUSTAIN) shearing STRESS.
4.

If a liquid has dissolved organic matter in it, then how does the surface tension of the liquid change?

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SOLUTION :DECREASES
5.

The ratio of the dimension of planck constant and that of moment of inertia is the dimension of ……….

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time
frequency
velocity
angular momentum

Solution :`("Panck's CONSTANT")/("Moment of inertia")=(h)/(M.I)=((E )/(v))/(mr^(2))`
`"Dimension of "=(h)/(M.I)=((ML^(2)T^(-2))/(T^(-1)))/(ML^(2))=T^(-1)`
6.

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic). and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v_("rms") the largest ?

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A
B
C
All are same

Answer :A
7.

Calculate the electrical energy consumed per second in an electric bulb with tungsten filament of area 0.35 cm^2when its temperature is raised to 2000 K, by passing an electric current through it. The emissivity of filament is 0.45. Stefan's constant, sigma = 5.67 xx 10^5 erg s^(-1) cm^(-2) K^(-4)

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SOLUTION :`4.082 XX 10^5 W//m^2`
8.

What is the principle involved in the construction of pressure cooker ?

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ANSWER :Elevation of boiling POINT with RISE in PRESSURE.
9.

(A) : A person standing on a rotating platform suddenly stretcehed his arms, the platform slows down.(R ) : In case of person on a rotating platform angular momentum is constant.

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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 but (R ) is FALSE
Both (A) and (R ) are false

ANSWER :A
10.

A refrigerator works between 4^@C and 30^@C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is _____ (1 cal = 4.2 Joules)

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23.65 W
236.5 W
2365 W
2.365 W

Solution :`T_1`=4+273=277 K
`T_2`=30+273=303 K
`Q_1`=-600 CAL
`rArr` For REFRIGERATOR `Q_1/Q_2=T_1/T_2`
`therefore (Q_2+W)/Q_2=T_1/T_2 [ because Q_1=Q_2+W]`
`therefore 1+W/Q_2=277/303`
`therefore W=(277/303-1)Q_2`
`=-26/303xx(-600xx4.2)`
`=(26xx600xx4.2)/303`
=216.2 W
`APPROX` 236.5 W, closest value.
11.

A block of mass m_1is pulled along a horizontal frictionless surface by a rope of mass m_2 . If a force F is given at its free end. The net force acting on the block is

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`(m_1 F)/(m_1 - m_2)`
`F`
`(m_2F)/(m_1 + m_2)`
`(m_1F)/(m_1 + m_2)`

SOLUTION :F
12.

Sand drops from a stationary hopper at the rate of 5 kg s^(-1) on to a conveyor belt moving with a constant speed of 2 m s ^(-1) . What is the force required to keep the belt moving and what is the power delivered by the motor moving the belt ? [ Hint: Tangetial force exerted on the belt = u_("rel") (dM)/(dt) where u_("rel") is the tangential relative velocity of the sand .]

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ANSWER :10 N , 20 W
13.

How many seconds are there in a light fermi?

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a) `10^(-15)`
B) `3.0×10^8`
c) `3.33×10^(-24)`
d) `3.3×10^(-7)`

ANSWER :C
14.

Define kinetic energy .Give its unit and dimensional formula and mention works by using it .

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Solution :The energy of a bodyis the ability for capacity of the body to do work .
Half of the product of the mass of a body and the square of its VELOITY is defined as kinetic energy (K) of the body .
` :. K =1/2 MV^(2)` where m = mass of body
v = velocity or speed of the body .
Unit of kinetic energyis same as the unit of work .So ,its SI unit is Joule (J) and CGS unit is erg .
Dimensionalformula of kinetic energy is `[M^(1)L^(2)T^(-2)]`
Hence ,a body of definite mass moving with more speed would have more kinetic energy compared to the kinetic energy of the same body moving with comparatively lesser speed .
Kinetic energy is a SCALAR quantity . The kinetic energy of a body indicate the magnitude of work by the body during its motion .
The kinetic energy of a fast flowing STREAM has been used during its motion .
The kinetc energy of a fast flowing stream has been used to grind grains .
Sailing SHIPS employ the kinetic energy of the wind mill .
15.

A uniform metre rod is bent at the middle in the form of L shape with the bent arms at 90^(@) to each other. The distance of the centre of mass from the bent point is

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`(1)/(4sqrt(2))m`
`(1)/(2sqrt(2))m`
`(1)/(SQRT(2))m`
`(1)/(8sqrt(2))m`

ANSWER :A
16.

When an empty lift is moving down with an acceleration of (g)/(4)ms^(-2), the tension in the cable is 9000 N. When the lift is moving up with an acceleration of (g)/(3)ms^(-2), the tension in the cable is,

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16,000 N
18,000 N
12,000 N
15,000 N

ANSWER :A
17.

A girl is standing on the top edge of an 18 m high building . She tosses a coin upwards with a speed of 7.0ms^(-1). How long does it take for the join to hit the ground ? How fast is the coin going just before it strikes the ground ?(g=10ms^(2))

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ANSWER :`2.70 s,20ms^(-10)`
18.

When a celsius thermometer reads 90°C a faulty Fahrenheit thermometer reads 190°F. The correction to be made in the latter scale is

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`+2^(@) F`
`- 2^(@) F`
`-4^(@) F`
`+ 4^(@) F`

ANSWER :D
19.

A simple pendulum has s time period T_(1) when on the earth's surface and T_(2) when it taken a height 'R' above the earth's surface, where 'R' is the radius of the earth. The value of (T_2)/(T_1) is

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`1`
`SQRT(2)`
`4`
`2`

ANSWER :D
20.

An earth satellite is moved from one stable circular orbit to a farther stable circular orbit. Which one of the following quantities increases

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GRAVITATIONAL force
gravitational PE
linear ORBITAL speed
centripetal acceleration

Answer :B
21.

A space vehicle si in circular orbit about the earth. The mass vehicle is 300 kg and the radious of the orbit is 2R_(e ). It is desired to transfer the vehicle to a circular orbit of radius 4R_(e ). (a) What is the minimum energy required for the transfer (b) If the transfer is accomplished through an elliptical orbit as shown in the figure. What initial and final velocity changes are required take g = 10 m//s^(2) earth's surface and R_(e ) = 6400km (radius of earth)

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Solution :(a) `[(-GMm)/(2(4Re))]-[(-GMm)/(2(2Re))]`
(b) `v_(1) = sqrt((GM)/(2Re)), (1)/(2) mv_(2)^(2) = [(-GMm)/(2((2Re+4Re)/(2)))]-[(-GMm)/(2Re)]`
`DELTA v_(12)=v_(2)-v_(1)`
`(1)/(2)mv_(3)^(2)=[(-GMm)/(2((2Re+4Re)/(2)))]-[(-GMm)/(4Re)]`
`v_(4) = sqrt((GM)/(4Re))`
`Delta v_(34)=v_(3)-v_(4)`
22.

One annular metal ring of inner radius 7 cm and outer radius 14 cm and negligible weight is floating on the surface of a liquid. If surface tension of liquid is 0.08Nm^(-1), calculate the force required to detach it from liquid surface.

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Solution :The CONTACT length of annular ring with liquid surface
`L=2pir_(1)+2pir_(2)""L=2xx(22)/(7)xx(14+7)cm`
`L=132cm=1.32m`
The force required, `F=S(L)=0.08xx1.32`
`F=0.1056N`.
23.

Rotational kinetic energy can be calculated by using

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`(1)/(2)I OMEGA^(2)`
`(L^(2))/(2I)`
`(1)/(2)L omega`
all the above

Answer :A
24.

Volume of some air saturated with water vapour is 80 cm^3 at the pressure of 74 cmHg .Keeping the temperature fixed, if pressure is taken to 146 cmHg, then the volume becomes halved. What will be the pressure ( in cmHg) of water vapour then?

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ANSWER :2
25.

An iron rod of length 50 cm is jointed at an end to an aluminium rod of length 100 cm. All measurements refer to 20^(@) C. The coefficients of linear expansion of iron and aluminium are 12xx 10^(-6) //^(@) C and 24 xx 10^(-6)//^(@) C. Respectively. The average coefficient of composite system is

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`36 XX 10^(-6)//^(@) C`
`12 xx 10^(-6)//^(@) C`
`20 xx 10^(-6)//^(@) C`
`48 xx 10^(-6)//^(@) C`

Answer :C
26.

What is meant by interference of waves?

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SOLUTION :INTERFERENCE is a phenomenon in which two waves superimpose to FORM a resultant wave of GREATER, lower or the same amplitude.
27.

The masses of two spherical planets A and B are M_(A) and M_(B), surface areas are S and 9 S respectively. The mass of a spherical planet C is (M_(A) + M_(B)). If the densities of all the planets A, B, C are same, then the escape velocities V_(A), V_(B), V_(C) from the surface of these planets follow:

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`V_(C)` GT `V_(B)` gt `V_(A)`
`V_(B)` = 3`V_(A)`
`V_(C)` lt `V_(B)` lt `V_(A)`
`V_(C)` gt `V_(B)` lt `V_(A)`

Solution :As area of B is 9 times area of A
DENSITIES of all PLANETS is same, then
`r_(B) = 3r_(A), M_(B) = 27 M_(A)`
`r_(C ) = 3sqrt(28)r_(A) = 3.03 r_(A)`
As `V_(A) = sqrt((2GM_(A))/(r_(A))) , V_(B) = sqrt((2GM_(B))/(r_(B))) , V_(A) = sqrt((2GM_(C ))/(r_(C )))`
`rArr V_(A) = sqrt((2GM_(A))/(r_(A))), V_(B) = sqrt((2G xx 27 M_(A))/(3r_(A))) = sqrt((18 GM_(A))/(r_(A)))`
`V_(C ) = sqrt((2GM_(C ))/(r_(C )))`
`= sqrt((2G xx 28 xx M_(A))/(3.03 r_(A)))`
`= sqrt((18.48 GM_(A))/(r_(A)))`
`V_(C ) gt V_(B) gt V_(A)`
28.

Two earth satellites of same mass are revolving round the earth in circular orbits of radii 6400 km and 19, 200 km from the centre of the earth. Ratio of thir orbital velocities will be

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`1:SQRT(3)`
`sqrt(3):1`
`1:3`
`3:1`

ANSWER :B
29.

Which of the following are not a unit of time ?

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Second
Parsec
YEAR
Light year

Solution :We know that 1 light year`= 9.46 xx 10^(11) m`
= distance that light travels in 1 year with speed
`3xx10^8 m//s`
1 parsec = `3.08 xx 10^(6)` m
= Distance which AVERAGE radius of earth.s orbit subtends an angle of 1 parasecond Here, Second and year REPRESENT time.
30.

(A) : Two particles (starting from rest) move towards each other under a mutual force of attraction. The velocity of their c.m is zero.(R ) : Internal forces do not alter the state of motion of c.m.

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Both 'A' and 'R' and TRUE and 'R' is the CORRECT explantation of 'A'
Both 'A' and 'R' and true and 'R' is notthe correct explantation of 'A'
A' is true and 'R' is FALSE
A' is false and 'R' is true

Answer :A
31.

When a person breaths, his lungs can hold up to 5.5 Litre of air at body temperature 37^(@)C and atmospheric pressure ("1 atm = 101 kPa"). This Air contains 21% oxygen. Calculate the number of oxygen molecules in the lungs.

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Solution :We can treat the air inside the lungs as an ideal gas. To find the NUMBER of molecules, we can use the ideal gas LAW.
PV = NkT
Here volume is given in the Litre. 1 Litre is volume occupied by a CUBE of side 10 CM
1 Litre = 10 cm `xx` 10 cm `xx` 10 cm = `10^(-3)m^(3)`
N =`(PV)/(kT) = (1.01 xx 10^(5) xx 5.5 xx 10^(-3))/(1.38 xx 10^(-23) xx 310)`
= 1.29 `xx 10^(23)` Molecules
Only `21%` of N are oxygen. The total number of oxygen molecules
`= 1.29 xx 10^(23) xx (21)/(100)`
Number of oxygen molecules = `2.7 xx 10^(22)` molecules
32.

A body is projected with a velocity u so that the horizontal range is twice the maximum height. Then the maximum height is

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`(U^(2))/(2G)`
`(u^(2))/(G)`
`(5u^(2))/(4G)`
`(2u^(2))/(5g)`

Answer :D
33.

The total workdone on a particle is equal to change in it.s kinetic energy

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Only if the forces acting on it are CONSERVATIVE
Only if gravitational force ALONE ACTS on it
Only if clastic force alone acts on it
none For any KIND of force acting on it

Answer :D
34.

Find the dimensions of a and b in the formula [P + (a)/(V^(2))[V-b]] = RT where P is pressure and V is the volume of the gas.

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Solution :By the principle of homogeneity, `a//V^(2)` is of the dimensions of pressure a and b is of the dimensions of volume.
`[a] = ["pressure"] [V^(2)] = [ML^(-1)T^(-2)][L^(6)]=[ML^(5)T^(-2)]`
`[b]=[V]=[L^(3)]`
35.

Which of the following is the evidence to show that there must be a force acting on earth and directed towards the sun ?

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Deviation of FALLING bodies TOWARDS east
Revolution of the earth around the sun
Apparent MOTION of the sun ROUND the earth
Phenomenon of day and night

Answer :B
36.

Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is 0.15.

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Solution :DUE tostaticfrictionblockwill getacceleration
`:. Ma= f_(s)`
and `N = mg`
Fromequation( 1)
`ma LE mu_(s) mg`
`:. , a lemu_(s) g`
`a_(max)= mu_(s ) g`
`= 0.15 xx 10`
`:. , a_(max)= 1.5 MS^(-2)`
37.

Three equal masses m are placed at the three vertices of an equilateral triangle of side a.The gravitational force exerted by this system on another particle of mass m placed at the midpoint of a side

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`(Gm^2)/(3a^2)`
`(4Gm^2)/(3a^2)`
`(Gm^2)/(a^2)`
`(Gm^2)/(3a^2)`

ANSWER :B
38.

Two cars A and B are at positions 50 m and 100 m from the origin at t = 0. They start simultaneously with constant velocities 10 m/s and 5 m/s respectively in the same direction. Find the time at which they will overtake one another.

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10 s
20 s
5 s
15 s

Solution :Position of both cars `x _(A0) =50 m and x_(B0) = 100 m` respectively at `t=0` from the origin. Here, `V _(A) = 10 ms ^(-1) and V_(B) = 5 ms ^(-1)`
Now, `x_(B) - x_(A) = (x_(BO) -x _(AO)) + (V_(B) - V_(A))t `
Suppose, at `t=t,` both cars OVERTAKE each other.
`THEREFORE x_(B) =x_(A) and x_(B) -x_(A) =0`
`therefore 0= (100 -50) + (5-10) t`
`therefore t = (50)/(5) = 10 s `
39.

The entrance to a harbour consists of 50 m gap between two points A and B such that B is due east of A. Outside the harbour there is a 8 km//hr current flowing due east. A motor boat is located 300 m due south of A. Neglect size of the boat for answering following questions- (a) Calculate the least speed (V_(min)) that the motor boat must maintain to enter the harbour. (b) Show that the course it must steer when moving at Vmin does not depend on the speed of the current.

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ANSWER :(a) `SQRT((48)/(37)) km//hr`
40.

A hollow conical vessel is filled with water al a constant rate. The rate at which the centre of mass of water in the cone rises is plotted against the height of water in the cone on a log - log scale, by taking them on Y and X - axis respectively. What is the slope of the line ?

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ANSWER :`-2`
41.

P is the centre of mass of a system of four point masses A, B, C and D, which are coplanar but not collinear (a) P may or may not coincide with one of the point masses (b) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C and D three at a time(c ) P must lie on a line joining two of the points A, B, C, D(d) P lies out side the quadrangle ABCD

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a & B are CORRECT
C & d are correct
a & c are correct
b & d are correct

ANSWER :A
42.

The work done by a gas is maximum when it expands

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isothermally
adiabatically
isochorically
isobarically

Solution :The WORK DONE by a GAS is MAXIMUM when it EXPANDS isobarically.
43.

Two balls are dropped simultaneously from two points separated by a vertical height of 6 m. The distance of separation between them after next 2s is

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`9.8m`
`6M`
`12M`
zero

Answer :B
44.

(A) : Damped oscillation indicates loss of energy. (R): The loss in damped oscillation may be due to friction, air resistance etc.

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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 true

Answer :A
45.

To measure the height of the mercury column in a barometer, abrass-scale is attached with it. Coefficient of real expansion of mercury is gamma and coefficient of linearexpansion of brass is alpha. Due torise in temperature, the effective coefficient at which the barometer reading increases is

Answer»

`GAMMA`
`gamma+alpha`
`gamma-alpha`
`gamma-2alpha`

ANSWER :C
46.

Give difinition, equation and unit for waveength and wave number for wave.

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Solution :Wavelength: The linear DISTANCE between any two points or particles having phase difference of `2pi` rad is called the wavelength (1) of the wave. Its SI unit is meter (m).
OR
The distance between consecutive crests/though is called a wavelength oof a wave.
`y (x,0) =a sin KX ""[because t =0 implies PHI =0]`
From this equation, sin function repeats its value after interval of `2pi` phae difference solution of function,
`therefore sin kx = sin (kx + 2n pi)=sin [ K(x + (2n pi)/(k)) ]`
It means displacement at point x and x` + (2n pi)/(k)` are equa where, `n = 1,2,3,..`
For minimum displacement between two points of same displacement, n =1 have to be taken,
`therefore x + (2pi)/(k) -x=` wavelength
`therefore (2pi)/(k) =` wavelength` lamda`
`or k = (2pi)/(lamda)`
where k is called anguar wave no. or wave vector.
Unit of k is raisn `"meter"^(-1)` but as rad is unitless, its SI unit is `m^(-1)`
Its dimensional formuls is `[ M^(0) L^(-1) T ^(0)].`
47.

A body starts from rest and travels a distance S with uniform acceleration, then moves uniformly a distance 2S and finally comes to rest after moving further 5S under uniform retardation. Find the ratio of average velocity to maximum velocity :

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Solution :Graphically : Area of `(V-t)` curve represent displacement

`S=(1)/(2)V_(max)t_(1)ort_(1)=(2S)/(V_(max))`,
`2S=V_(max)t_(2)ort_(2)=(2S)/(V_(max))`
`5S=(1)/(2)V_(max)t_(3)ort_(3)=(10S)/(V_(max))`
`V_(av)=("Total displacement")/("Total time"),V_(av)=(S+2S+5S)/((2S)/(V_(max))+(2S)/(V_(max))+(10S)/(V_(max)))`
`(V_(av))/(V_(max))=(8S)/(14S)=(4)/(7)(OR)`
`(V_(av))/(V_(max))=overset("Total displacement")underset(2(("total displacement"),("during ACCELERATION"),("and retardation"))+(("Displacement"),("During uniform"),("velocity")))rarr`
`therefore (V_(av))/(V_(max))=(8S)/(2(S+5S)+2S)=(8)/(14)=(4)/(7)`
48.

A body of mass 5kg rests on a rough horizontal surface of coefficient of friction 0.2. The body is pulled through a distance of 10m by a horizontal force of 25N. Find the kinetic energy acquired by the body. (g=10ms^(-2))

Answer»

SOLUTION :Mass=5kg, s=10m, u=0, F=25N
RESULTANT force : `F^(.)=F-f_(lui)=25-0.2xx5xxg=15N`
KINETIC energy=workdone by `F^(.)=15xx10=150J`
49.

Values for latent heat in Column - I and its values are given in Column - II. Match the followings

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ANSWER :(a-i), (b-iii)
50.

The position of a particle moving in the XY-plane at any time 't is given by x = (3t^2-6t) m, y = (t^2-2t) m. Select the correct statement about the moving particle from the following

Answer»

The ACCELERATION of the particle is ZERO at t=0 second
The VELOCITY of the particle is zero at t=0 second
The velocity of the particle is zero at t=1 second
The velocity and acceleraton of ARTICLE particle are never zero

Answer :C