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.

Two equal spheres B and C, each of mass m, are in contact on a smooth horizontal table. A third sphere A of same size as that of B or C but mass m//2 impinges symmetrically on them with a velocity u and is itself brought to rest. Find the velocity acquired by each of the spheres B and C after collision.

Answer»

`u/sqrt(3)`
`(2u)/sqrt(3)`
`u/(2sqrt(3))`
`u/2`

Solution :`u=`velocity of sphere `A` before impact. As the spheres are identical, the triangle `ABC` formed by joining their centres is equilateral. The spheres `B` and `C` will move in direction `AB` and `AC` after impact making can angle of `30^(@)` with the origina LINES of MOTION of BALL `A`.
Let `v` be the speed ofthe balls `B` and `C` after impact.
Momentum conservationgives
`(m/2)u=mvcos30^(@) +mvcos30^(@)`
`u=2sqrt(3)vimpliesv=u/(2sqrt(3))`.......i
From Newton's experimental law, for an oblique collision, we have take components along normal, i.e. along `AB` for balls `A` and `B`.
`v_(B)-v_(A)=-e(u_(B)-u_(A))`
`v-0=-e(0-ucos30^(@))`
`v=eucos30^(@)`.........ii
COMBINING eqn i and ii we get `e=1/3`
Loss in `KE =1/2 m/2 u^(2)-2(1/2mv^(2))`
`=1/4mu^(2)-m(u/(2sqrt(3)))^(2)=1/6mu^(2)`
Discussion
2.

Two equal spheres B and C, each of mass m, are in contact on a smooth horizontal table. A third sphere A of same size as that of B or C but mass m//2 impinges symmetrically on them with a velocity u and is itself brought to rest. The coefficient of restitution between the two spheres A and B (or between A and C) is

Answer»

`1//3`
`1//4`
`2//3`
`3//4`

Solution :`u=`velocity of sphere `A` before impact. As the spheres are identical, the triangle `ABC` formed by joining their centres is equilateral. The spheres `B` and `C` will move in DIRECTION `AB` and `AC` after impact making can ANGLE of `30^(@)` with the origina lines of motion of BALL `A`.
Let `v` be the speed ofthe balls `B` and `C` after impact.
Momentum conservationgives
`(m/2)u=mvcos30^(@) +mvcos30^(@)`
`u=2sqrt(3)vimpliesv=u/(2sqrt(3))`.......i
From Newton's experimental law, for an oblique collision, we have take components along normal, i.e. along `AB` for balls `A` and `B`.
`v_(B)-v_(A)=-e(u_(B)-u_(A))`
`v-0=-e(0-ucos30^(@))`
`v=eucos30^(@)`.........ii
Combining EQN i and ii we get `e=1/3`
Loss in `KE =1/2 m/2 u^(2)-2(1/2mv^(2))`
`=1/4mu^(2)-m(u/(2sqrt(3)))^(2)=1/6mu^(2)`
Discussion
3.

A vessel is half filled with a liquid at 0^(@) C. When the vessel is heated to 100^(@) C, the liquid occupies 3/4 volume of the vessel. Coefficient of apparent expansion of the liquid is

Answer»

0.5/°C
0.05/°C
0.005/°C
0.0005/°C

Answer :C
Discussion
4.

A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms^(-1) relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Answer»

Solution :The child gives an impulse to the trolley at the start and then runs with a constant RELATIVE velocity of `4 MS^(-1)` with respect to the trolley.s NEW velocity. Apply momentum conservation for an observer OUTSIDE. `10.36 ms^(-1), 25.9m`.
Discussion
5.

A satellite close to earth is in orbit above equator with a period of motion of 1.5 hours. If its is above point P on equator at some time, it will be above P a gain time…

Answer»

1.5 hours
1.6 hours if it is ROTATING from WEST to EAST
24/17 hours if it is rotating from east to west
hours if it is rotating from west to east

Answer :B::C
Discussion
6.

The increase in length of a wire on stretching is 0.025%. If its poisson ratio is 0.4, then the percentage decrease in the diameter is :

Answer»

0.01
0.02
0.02
0.04

Answer :1
Discussion
7.

The coordinates of a body moving in a plane at any instant of time t are x=alphat^(2) and y = betat^(2). The velocity of the body is

Answer»

Solution :`x=alphat^(2)rArrv_(x)=(dx)/(dt)=2alphat`
`y = betat^(2).rArr v_(y)=(dy)/(dt)=2betat`
`:."VELOCITY" v = SQRT((2alphat)^(2)+(2betat)^(2))`
`=2tsqrt(alpha^(2)+BETA^(2))`
Discussion
8.

A student was asked a question 'why are there summer and winter for us? He replied as 'since Earth is orbiting in an elliptical orbit, when the Earth is very faraway from the Sun (aphelion) there will be winter, when the Earth is nearer to the Sun (perihelion) there will be winter}. Is this answer correct? If not, what is the correct explanation for the occurrence of summer and winter?

Answer»

Solution :Early astronomers PROVED that EARTH is spherical in shape by looking at the shape of the SHADOW cast by Earth in the Moon during lunar eclipse.
Discussion
9.

Unbalanced system of forces can produce acceleration as well as deformation in a body, but balanced system of forces produces deformation only explain.

Answer»

SOLUTION :In case of an unbalanced system of forces acting on a body, a resultant force acts at the centre of mass . The body accelerates, as per the equation, F= ma. But for a balanced system fo forces the resultantforce is zero andhence acceleration ofthe centre of mass is zero However in both the situations, different PARTS of the body MAY have relative velocities with respect to one another. Hence, deformation may take place. For example, if a rubber ropeis pulled on both ends with equal forces , the rope extends in length ,i.e., it is deformed, but its centre of massremains steady atone place . when the forces applied at theends are unequal, the rope not only extends , but ALSO MOVES in the direction of thegreater force.
Discussion
10.

Deduce the formula of Reynolds number in the form of inertial force and viscous force .

Answer»

Solution :REYNOLDS number `R_(e)=(rhov^(2))/(((etav)/d))`
Multiplying and DIVIDING by A to right side equation,
`R_(e)=(rhoAv^(2))/(((etaAv)/(d)))`
`R_(e) = ("INERTIAL force")/("Viscous force")`
Thus `(R_(e)`represets the ratio of inertial force, that is force due to mass of moving fluid or due to inertia of obstable in its path to viscous force)
HENCE`R_(e)` represents the ratio of inertial force of to viscous force.
Discussion
11.

Two capillary tubes of lengths in .the ratio 2: 1 and radii in the ratio 1 :2 are connected in series. Assume the flow of the liquid through the tube is steady. Then, the ratio of pressure difference across the tubes is

Answer»

`1:8`
`1:16`
`32:1`
`1:1`

ANSWER :C
Discussion
12.

(A) : It is hotter at the same distance over the top of a fire than it is on the sides. (R) : Heat is transmitted equally in all directions by radiation but convection takes heat upwards only

Answer»

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
Discussion
13.

Find the ratio of escape velocities of two planets if values of g on the two planets are 9.8 m//s^(2) and 3.3 m//s^(2) and ther radii are 6400 km and 3400 km respectively.

Answer»

Solution :The ESCAPE velocity in TIMES of g and R is
The ratio of escape velocities `(V_(E1))/(V_(E2)) = sqrt((g_(1)R_(1))/(g_(2)R_(2)))`
`(V_(e1))/(V_(e2)) = sqrt(((9.8)/(3.3))((6400)/(3400))) = (2.364)/(1)`
`V_(e1) : V_(e2) = 2.364:1`.
Discussion
14.

Three particles of the same mass are kept at the vertices of an equilateral triangle. The mass of each particle is m and the length of an arm of the triangle isl. Due to the the mutual gravitational force of attraction, the particles revolve along the circumcircle of the triangle. Find the velocity of each particle.

Answer»

Solution :ABC is an equilateral triangle . At its vertices three particles A,B and C each of mass m , are kept [Fig.1.4].

The point of intersection of the medians of ABC is O, which is also the centre of the circumcircle of the triangle. HENCE, radius of the circumcircle,
`r=OA =2/3AD =2/3AB sin 60^@= 2/3*l*sqrt(3)/2=l/sqrt(3)`. Hence the centripetal force needed by each particle to revolve along the circumcircle with velocity v is
`F_1=(mv^2)/r=(mv^2)/(l/sqrt(3))=(sqrt(3)mv^2)/l`
Now the gravitational force acting on the particle at point can be calculated. The force of gravitation on the particle A due to the particle at B along AB,
`F=(G*m*m)/((AB)^2)=(Gm^2)/(l^2)`
COMPONENT of this force along AE =`cos 60^@` and that along `AD=Fsin 60^@`
Again , gravitational attraction onthe particle at A due to the particle kept at C is F along AC.
Components ofthis force along `AH =F cos 60^@` and that along `AD =F sin 60^@`.
Clearly, components along AH an AE cancel each other Hence , the resultant force `F_2` on the particle kept at A =sum of the components along AD.
`therefore F_2=F sin 60^@ +Fsin 60^@ =2Fsin 60^@`
`=2F*(sqrt(3))/(2)=sqrt(3)F=(sqrt(3)Gm^2)/(l^2)`
As PAR the question, this force of attraction due to gravitation supplies the necessary centripetal force. Hence
`F_1=F_2 or, (sqrt(3)mv^2)/l =(sqrt(3)Gm^2)/(l^2) or, v^2 =(Gm)/lor, v=sqrt((Gm)/l)`.
Discussion
15.

Two solid rubber balls A and B having masses 200 and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 m/s. After collision the two balls come to rest, then find the velocity of B.

Answer»

SOLUTION :Initial linear MOMENTUM of system `= m _(A) v _(A) = m _(B) v _(B)`
`= 0.2 xx 0.3 +0.4 xx v_(B)`
FINALLY both BALLS come to rest
`therefore `Final linear momentum = 0
By the law of conservation of linear momentum `0.2 xx 0.3 +0.4 xx v _(B) =0`
`therefore u _(B)= (0.2xx 0.3)/( 0.4) =- 0.15 m//s`
Discussion
16.

Two particles of same mass fall down on to the surface of the earth one from infinity and the other from an alitutude 3R. The ratio of velocities on reaching the earth is

Answer»

`2 : 3`
`3 : 2`
`2 : SQRT(3)`
`sqrt(3) : 2`

ANSWER :C
Discussion
17.

Two satellites of earth. S_(1) and S_(2) are moving in the same orbit. The mass of S_(1) is four times the mass of S_(2). Which one of the following statement is true ?

Answer»

The POTENTIAL ENERGIES of earth satellites in the two cases are equal.
`S_(1)` and `S_(2)` are moving with the same speed
The KINETIC energies of the two satellites are equal
The time period of `S_(1)` is four times that of `S_(2)`.

Answer :B
Discussion
18.

A uniform piece of metal sheet is cut in the form as shown in the fig. Locate the center of mass of the piece.

Answer»

Solution :
Let the mass per unit area be `SIGMA` We divide the STRUCTURE into parts 1, 2 and 3. The mass of PART 1 is 300 `sigma`, mass of part 2 is 200 `sigma` and that of part 3 in 100 `sigma`. The coordinates of centre of mass of part 1 are (5, 15), that of part 2 are (20,5), and that of part 3 are (15, 25)
Using the Eq.
`x_(cm)=((300sigma)5+(200sigma)20+(100sigma)15)/(600sigma)~~11.7`
`y_(cm)=((300sigma)15+(200sigma)5+(100sigma)25)/(600sigma)~~13.3`
In vector notation `barr_(cm)=11.7hati+13.3hatj`
Discussion
19.

A resonance tube is resonated with tuning fork of frequency 512 Hz. Two successive lengths of the resonated air-column are 16.0 cm and 51.0 cm. The experiment is performed at the room temperature of 40^(@)C calculate the speed of sound at 0^(@)Cand the end correction.

Answer»

Solution :Here `V=512Hz, L_(1)=16.0cm=0.16m, L_(2)=51.0cm=0.51m`
SPEED of SOUND at `40^(@)C`,
`v=2V(L_(2)-L_(1))`
`=2xx512xx(0.51-0.16)`
`=358.4ms^(-1)`
Speed of sound at `0^(@)C` is given by
`v_(o)=v-0.61t=358.4-0.61xx40`
`=350.4-24.4-334ms^(-1)`
End correction
`=(L_(2)-3L_(1))/(2)=(51.0-48.0)/(2)`
=1.5cm.
Discussion
20.

A uniform chain of mass 'm' and length 'L' is kept on a horizontal table with half of its length hanging from the edge of the table. Work done in pulling the chain on to the so that only (1)/(5)th of its length now hangs from the edge is,

Answer»

`(MGL)/(8)`
`(mgl)/(50)`
`(mgl)/(18)`
`(mgl)/(200)`

ANSWER :D
Discussion
21.

Under what condion will a car skid on a leveled circular road ?

Answer»

Solution :On a leveled circular road, if the STATIC FRICTION is not able to provide ENOUGH centripetal force to turn, the VEHICLE will start to skid.
`mu_(5)LT(v^(2))/(rg)`
Discussion
22.

A man sitting on a swing which is moving to an angle of 60^(@) from the vertical is blowing a whistle is 2.0 m from the fixed support point of the whistle sound is kept in front of the swing. The maximum frequency the sound detechor detected is :

Answer»

`2.027 KHZ`
`1.974 kHz`
`9.74kHz`
`1.011kHz`

Solution :Frequency `v= 2 kHz`
VELOCITY of whistle (source) `v_(s)=2m`
Velocity of sound `v=332m/s`
MAXIMUM frequency `v_("MAX")=((V+V_(s))/(V))v`
`=((332+2)/(332))xx2`
`=((334)/(332)xx2`
`1.006xx2`
`2.012 kHz`
Discussion
23.

The ratio of the mean absolute error to the mean value is called as...........

Answer»

ABSOLUTE ERROR
RANDOM error
RELATIVE error
PERCENTAGE error

Solution :Relative error
Discussion
24.

Two waves y_(1)=a sin (pi/2 x-omegat) and y_(2)=a sin (pi/2 x+omegat+pi/3) get superimposed in the region x ge 0. Find the number of nodes in the region 0 le x le 6 m.

Answer»


ANSWER :`3`
Discussion
25.

{:(,"SECTION-A",,"SECTION-B"),("a)","Equation of continuity","e)","Less than critical velocity"),("b)","Bernoullie's theorem","f)","Formation of eddies & vortices"),("c)","Turbulent flow","g)","Law of conservation of mass"),("d)","Stream line flow","h)","Law of conservation of energy"):}

Answer»

a-h, b-f, c-e, d-g
a-g, b-e, c-g, d-f
a-f, b-g, c-h, d-e
a-g, b-h, c-f, d-e

ANSWER :D
Discussion
26.

In the above problem ,whichwill reachthe bottomwith greater velocity?

Answer»

A
B
Both with same VELOCITY
NONE

ANSWER :A
Discussion
27.

Can we use the equations of kinematics to find the height attained by a body projected upward with any velocity .

Answer»

Solution :No, because equations of MOTIONS are APPLICABLE as long as the acceleration is UNIFORM .
Discussion
28.

Explain the variation of 'g' with latitude.

Answer»

Solution :Variation of g with latitude:
Wherever we analyze the motion of object in ROTATING frames we must take into account the centrifugal force. Even though we treat the Earth as an inertial FRAME, it is not exactly correct because the Earth spins about its own axis. So when and object is on the surface of the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been 'mg'. HOWEVER, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by `momega^(2) R'`.
where `lambda` is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
`a_(c) = OMEGA^(2) R' cos lambda = omega^(2) R cos^(2) lambda`
since `R' = R cos lambda`
Therefore, `g' = g - omega^(2) R cos^(2) lambda""......(2)`
From HTE expression (2), we can infer that at equator ,`lambda = 0, g ' = g - omega^(2)R`. The acceleratoin due to gravity is minimum. At poles `lambda = 90, g' = g`, it is maximum. At the equator, g' is minimum.
Discussion
29.

Explain why the temperature of a wire under tension will charge if it snaps suddenly?

Answer»

Solution :During elongation of a WIRE that WORK done remains STORED as elastic potential energy in the wire. When the wire snaps suddenly, that stored potential energy is TRANSFORMED into heat energy resulting in increase in temperature.
Discussion
30.

Is it possible for a body to be accelerated without speeding up or slowing down ?

Answer»

Solution :YES, an OBJECT in UNIFORM CIRCULAR motion is accelerating but its SPEED neither decreasesnot increases .
Discussion
31.

In the figure shown a plank of mass m is lying at rest on a smooth horizontal surface. A cylinder of same mass m and radius r is rotated to an angular speed w0 and then gently placed on the plank. It is found that by the time the slipping between the plank and the cylinder cease, 50%of total kinetic energy of the cylinder and plank system is lost. Assume that plank is long enough and um is the coefficient of friction between the cylinder and the plank. (a) Find the final velocity of the plank. (b) Calculate the magnitude of the change in angular momentum of the cylinder about its centre of mass. (c) Distance moved by the plank by the time slipping ceases between cylinder and plank.

Answer»


ANSWER :(a) `(romega_(0))/(4)`
`(b) (1)/(4) MR^(2) omega_(0)`
(c) `(r^(2) omega_(0)^(2))/(32 mu g)`
Discussion
32.

(A) : The angular velocity of a platets orbiting around the sun increases when they are nearest to the sun. (R) : The angular momentum of body is propotional to angular velocity.

Answer»

Both 'A' and 'R' are true 'R' is the correct explanation of 'A'
Both 'A' and 'R' are true 'R' is not the correct explanation of 'A'
A' is true and 'R' is FALSE
A' is false and 'R' are true

Answer :B
Discussion
33.

A system is taken from state A to state N along two different paths 1 and 2. The heat absorbed and work done by system along these two paths are Q_(1) and Q_(2), andW_(1) and W_(2) respectively. Then

Answer»

`Q_(1) = Q_(2)`
`W_(1) = W_(2)`
`Q_(1) - W_(1) = Q_(2) - W_(2)`
`Q_(1) + W_(1) = Q_(2) + W_(2)`

Answer :C
Discussion
34.

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L meter long. The length of the open pipe will be

Answer»

`L/2m`
`4Lm`
`Lm`
2L m

Solution :(i) For open pipe of LENGTH L. second overtone `=(3V)/(2L.)`
(II) For closed pipe of length L first overtone `= (3v)/(4L)`
As per the STATEMENT, `(3v)/(2L.) = (3v)/(4L)`
`therefore L. = 2L`
Discussion
35.

If the coefficient of volume expansion of glass would have been equal to the coefficient of real expansion of mercury, a mercury thermometer would not be active-explain.

Answer»

Solution :A thermometer shows the apparent expansion of the thermometric LIQUID.
But, the COEFFICIENT of apparent expansion of mercury = the coefficient of real expansion of mercury - the coefficient of volume expansion of glass = 0, if `gamma_(glass)=gamma_(mercury)` (given). Therefore, the mercury within the tube WOULD REMAIN at the same level, no matter what the change in temperature is.
Discussion
36.

Two parallel forces P and Q (P gt Q) are acting at two points, A and B, of a body. If the forces interchange their positions, by how much will the point of action of the resultant of the resultant of the two forces shift along the line AB?

Answer»

Solution :In Fig. SUPPOSE C is the point of ACTION of the resultant in the first case.
`:. PxxAC=QxxBC""cdots(1)`
If the NEW POSITION of the point of action of the resultant is C. then
`QxxAC.=PxxBC. ""cdots(2)`
From (1) we get,
`PxxAC = Q(AB-AC) " ""or", (P+Q)xxAC=QxxAB`
`:. AC=(Q)/(P+Q)xxAB`
Similarly from (2) we get,
`AC.=(P)/(P+Q)xxAB ""cdots(4)`
`:.` Subtracting equation (3) from equation (4) we get,
`AC. -AC =(AB)/((P+Q))(P-Q) " ""or", "CC".= (P-Q)/(P+Q) xxAB`.
Discussion
37.

What is the nature of a fluid flow when the speed of the fluid exceeds critical velocity?

Answer»

SOLUTION :TURBULENT
Discussion
38.

Four particles , each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is:

Answer»

`SQRT((GM)/R)`
`sqrt(2sqrt(2)(GM)/R)`
`sqrt((GM)/R(1+2sqrt(2)))`
`1/2sqrt((GM)/R(1+2sqrt2))`

ANSWER :D
Discussion
39.

Which of the function of time represent (a) simple harmonic. (b) periodic but not simple harmonic. and (c) non periodic motion ? Give period for each case of periodic motion (omega is any positive constant), sinomegat-cosomegat

Answer»

Solution :`sinomegat-cosomegat= sqrt(2)[sinomegatcos((pi)/(4))-cosomegatsin((pi)/(4))]= sqrt(2)sin(OMEGAT-(pi)/(4))`
It is SIMPLE HARMONIC with a time PERIOD `T = (2PI)/(omega)`
Discussion
40.

(A) When a body is thrown into air, at its maximum height its KE is zero always. (B) If a body is moved along a rough circular path in the horizontal plane, work done by the force applied is zero

Answer»

A is TRUE, B is FALSE
A is false, B is true
A & B are true
A & B are false

Answer :D
Discussion
41.

Which of the function of time represent (a) simple harmonic. (b) periodic but not simple harmonic. and (c) non periodic motion ? Give period for each case of periodic motion (omega is any positive constant), 1+omegat+omegat^(2)

Answer»

SOLUTION :`1+omegat+omegat^(2)` is not PERIODIC as FUNCTION increases with increase in t with out REPETITION.
Discussion
42.

Which of the function of time represent (a) simple harmonic. (b) periodic but not simple harmonic. and (c) non periodic motion ? Give period for each case of periodic motion (omega is any positive constant), e^(-omega^(2)t^(2))

Answer»

Solution : `e^(-OMEGA^(2)t^(2))` is not periodic as t INCREASES the FUNCTION`e^(-omega^(2)t^(2))` decreases and TENDS to zero as `t to oo`
Discussion
43.

Which of the function of time represent (a) simple harmonic. (b) periodic but not simple harmonic. and (c) non periodic motion ? Give period for each case of periodic motion (omega is any positive constant), cosomegat+cos3omegat+cos5omegat

Answer»

Solution :`cosomegat+ cos3omegat+ cos5omegat` is a periodic function but not SIMPLE HARMONIC. The time periods of each periodic function are `(2pi)/(OMEGA), (2pi)/(3omega)` and `(2pi)/(5 omega)`. SINCE `(2pi)/(omega)` is the multiple of the other two periods , the given sum is periodic with time period `(2pi)/(omega)`
Discussion
44.

Which of the function of time represent (a) simple harmonic. (b) periodic but not simple harmonic. and (c) non periodic motion ? Give period for each case of periodic motion (omega is any positive constant), 3cos(pi//4-2omegat)

Answer»

Solution :`3COS(pi//4-2omegat)= 3cos(2omegat-pi//4)`it is simple HARMONIC with a time PERIOD `T= (2PI)/(2omega)= (pi)/(omega)`
Discussion
45.

Which of the function of time represent (a) simple harmonic. (b) periodic but not simple harmonic. and (c) non periodic motion ? Give period for each case of periodic motion (omega is any positive constant), sin^(3)omegat

Answer»

Solution :`sin^(3)omegat` is a periodic FUNCTION but not simple harmonic because `aalpha-y` CONDITION is not SATISFIED. Its time period is `T= (2pi)/(OMEGA)`
Discussion
46.

The velocity ofa river current is u. A man crosses the river with velocity v making an angle theta with thedirection of the current. If the width of the river is d, find the time taken to cross the river and the displacement of the man in the direction of the current. What would be the minimum time required for crossing the river ?In what time could he reach exactly the opposite point ?

Answer»


ANSWER :`d/(VSIN THETA);DCOT theta;d/(SQRT(u^2+v^2-2uvcos theta));d/(sqrt(v^2-2uvcos theta))`
Discussion
47.

The acceleration due to gravity at the latitude 45° on the earth becomes zero if the angular velocity of rotation of earth is equal to

Answer»

`SQRT(2/(GR))`
`sqrt(2gR)`
`sqrt((2g)/R)`
`sqrt((5R)/2)`

Answer :C
Discussion
48.

Express cross and dot product of two vectors in Cartesian coordinate.

Answer»

Solution :Let `vecA and vecB` be the two vectors.
`vecA=vecA_(x)hati+vecA_(s)HATJ+vecA_(z)hatk, vecB=vecB_(x)hati+vecB_(y)hatj+vecB_(z)hatk`
Cross product of `vecA and vecB`.
`VECAXXVECB=(vecA_(x)hati+vecA_(y)hatj+vecA_(z)hatk)XX(vecB_(x)hati+vecB_(y)hatj+vecB_(z)hatk)`
`=vecA_(x)vecB_(x)hatixxhati+vecA_(x)vecB_(y)hatixxhatj+vecA_(x)vecB_(z)hatixxhatk`
`+vecA_(y)vecB_(y)hatjxxhati+vecA_(y)vecB_(z)hatjxxhatj+vecA_(z)B_(z)atkxxhatk`
`vecAxxvecB=vecA_(x)vecB_(y)(hatk)+A_(x)B_(z)(-hatj)+vecA_(y)vecB_(x)(-hatk)+vecA_(y)vecB_(y)(0)`
`+vecA_(y)vecB_(z)(hati)+vecA_(z)vecB_(x)(hatj)+vecA_(z)vecB_(z)(0)`
`(vecAxxvecB)=(vecA_(y)vecB_(z)-vecA_(z)vecB_(y))hati+(vecA_(z)vecB_(x)-vecA_(x)vecB_(z))hatj+(vecA_(x)vecB_(y)-vecA_(y)vecB_(x))hatk`
`[because hatixx hati=hatjxxhatj=hatkxxhatk=0`
`hatixxhatj=k, hatixxhatk=-hatj, hatjxxhati=-k,hatjxxhatk=hati, hatk xxhati=+hatjxxhatj=-hatk]`
It can be in DETERMINANT form as,
`vecA xx vecB=|(hati,hatj,hatjk),(A_(x),A_(y),A_(z)),(B_(x),B_(y),B_(z))|`
`=hati(A_(y)B_(z)-B_(y)A_(z))-hatj(A_(x)B_(z)-B_(y)A_(z))+hatk(A_(x)B_(y)-B_(x)A_(y))`
DOT product of `vecA and vecB`.
`vecA=vecA_(x)hati+vecA_(y)hatj+vecA_(z)hatk, vecB=vecB_(x)hati+vecB_(x)hatj_vecB_(z)hatk`
`vecA.vecB=(vecA_(x)hati+vecA_(y)hatj+vecA_(z)hatk).(vecB_(x)hati+vecB_(y)hatj+vecB_(z)hatk)`
`(vecA_(x)vecB_(x)(hati.hatj)+vecA_(x)vecB_(y)(hati.hatj)`
`+vecA_(x)vecB_(z)(hati.hatk)+vecA_(y)vecB_(x)(hatj.hatk)+vecA_(y)vecB_(y) (hati.hatj)`
`+vecA_(y)vecB_(z)(hatj.hatk)+vecA_(z)vecB_(x)(hatk.hati)+vecA_(z)vecB_(y)(hatk.hatj)`
`+A_(z)B_(z)(hatk.hatk)`
`vecA.vecB=vec(A_(x))vec(B_(x))(1)+vec(A_(x))vec(B_(y))(0)+vecA_(x)vec(B_(z))(0)+vec(A_(y))vec(B_(z))(0)+vec(A_(y))vec(B_(y))(1)`
`+vec(A_(y))vec(B_(z))(0)+vec(A_(z))vec(B_(x))(0)+vec(A_(z))v_(y)(0)+A_(z)B_(z)(1)`
`vecA.vecB=vecA_(x)vecB_(x)+vec(A_(y))vec(B_(y))+vecA_(z)vecB_(z)`
`[because hati.hati=hatj.hatj=hatk.hatk=1`
`hati.hatj=hati.hatk=0, hatj.hati=hatj.hatk=0, hatk.hati=hatk.hatj=0]`
Discussion
49.

The angular momentum (M) of a particle relative to a certain point (O), varies with time as vec(M)=vec(a)+vec(b)t^(2) where vec(a) and vec(b) are constant vectors and vec(a) is perpendicular to vec(b). The moment of the force (N) relative to the point O, acting on the particle, when N makes 45^(@) with M is xbsqrt(a//b). Find the value of x ?

Answer»


ANSWER :2
Discussion
50.

A wire of length L and cross section area A is kept on a horizontal surface and one of its end is fixed at point 0. A ball of mass m is tied to its other end and the system is rotated with angular velocity omega.Show that increase in its length. Delta l = (m omega ^(2) L ^(2))/(AY). Y is young’s modulus.

Answer»

Solution :
By rotating the ball of MASS m with angular speed W, the pseudo centifugal force acting on it due to centripetal force
`F = ( mv ^(2))/() = ( ML ^(2) omega ^(2))/( L )`
`F = mL iomega^(2)`
`therefore` ongitudinal stress `sigma = (F)/(A) = (mL omega ^(2))/(A) `
Let increase in length of rod `= Dela L`
`therefore` strain `= (Delta L )/(L)`
Young modulus `Y = ("stress")/("strain") = ( m Lomega ^(2))/(A) //(Delta L )/(L)`
`therefore Delta L = (m L ^(2) omega ^(2))/(A.Y)`
Discussion