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. |
An aeroplane takes flight from a place towards another place in directiontheta N of E in a strong wind of velocity U to the west. The speed of plane in still air is V. Distance the two places is L At what angle the plane must be driven wrt line of flight |
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Answer» `SIN^(-1)(u/v sin THETA)` |
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| 2. |
An aeroplane takes flight from a place towards another place in direction thetaN of E in a strong wind of velocity U to the west. The speed of plane in still air is V. Distance the two places is L If T_(1), T_(2)are time of travel of on ward and return journey, thenT_(1) T_(2)= .......... |
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Answer» `(L^(2))/(v^(2)+U^(2))` |
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| 3. |
Rain is falling vertically with a speed of 35ms^(-1). Winds starts blowing after sometime with a speed of 12ms^(-1) in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella? |
Answer» Solution :The velocity of the rain and the wind are represented by the vectors `v_(r) and v_(w)` in the figure and are in the direction specified by the problem. Using the rule of vector addition. We see that the RESULTANT of `v_(r) and v_(w)` is R as shown in the figure. The magnitude of R is. `R=sqrt(v_(r)^(2)+v_(w)^(2)) = sqrt(35^(2)+12^(2))MS^(-1) = 37ms^(-1)` The direction `theta` that R makes the vertically is GIVEN by `tantheta=(v_(w))/(v_(r)) =(12)/(35) = 0.343 or, theta = tan^(-1)(0.343)=19^(@)` Therefore, the boy should hold his umbrella in the vertical plane at an angle of about `19^(@)` with the vertically towards the east. |
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| 4. |
A soundof frequency 1000 Hz is emitted from a roadside siren. A car moving along the road with a unifrom velocity of 50 km . h^(-1) crosses the siren . Calculate the frequency of the siren relative to the pasenger of the car before and after crossing the siren . (Velocity of sound = 340 ms^(-1)) |
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Answer» |
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| 5. |
Can we add velocity-vector to the displacement vector? |
| Answer» Solution :No. The necessary condition for adding VECTORS is that they must REPRESENT PHYSICAL QUANTITIES of same NATURE | |
| 6. |
An open organ pipe of length l vibrates with its fundamental note. The presure variation is maximum. |
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Answer» at the middle of the pipe `L=(lambda)/(2)` |
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| 7. |
Statement A: At absolute zero the substancehas internal energy due to the vibrational motion of atoms with in the molecules. This is zero point energy. Statement B: First law of thermodynamics does not satisfy the condition under which a body can use its heat energy to produce the work. |
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Answer» A is TRUE, B is FALSE |
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| 8. |
A point source of light P is placed at a distance L in front of a mirror of widthd hung vertically on a wall . A man walks infront of the mirror a long a line parallel to the mirror at a distance 2L as shown in the figure. The greatest distance over which he can see the imageof the light source, in the mirror is |
Answer» Solution : Locate the position of image and by the help of shortcut method of locating field of view JOIN the mirror through lines, and that region is the field of view then, by similar TRIANGLE concept the TRIANGLES IPQ and IMN are similar, `(PQ)/(MN)=(3L)/L implies PQ=3d` |
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| 9. |
A packet is dropped from a balloon rising up with uniform velocity 9.8ms^(-1). If the balloon is at a height of 39.2 m from the ground at the time of dropping the stone, the stone reaches the ground after |
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Answer» 2 s |
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| 10. |
The temperatures of two bodies measured by a thermometer are t_(1) = (20 pm 0.5)^(@)C, t_(2) = (50 pm 0.5)^(@)C. Calculate the temperature difference and the error therein. |
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Answer» SOLUTION :`t_(1) = (20 pm 0.5)^(@)C t_(2) = (50 pm 0.5)^(@)C` TEMPERATURE DIFFERENCE t = ? t = `t_(2) - t_(1) = (50 pm 0.5) - (20 pm 0.5)^(@)C` `= (50-20)pm(0.5+0.5)` t = `(30pm1)^(@)C` |
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| 11. |
Pressure exerted by a gas is .............. . |
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Answer» independent of the DENSITY of the GAS |
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| 12. |
The time -period of a simple pendulum is 2s and its can go to and fro from equilibrium position at a maximum distance of 5 cm. if at the start of the motion the pendulum is the position of maximum displacements towars the right of the equilibrium position, then write the displacement equation of the pendulum. |
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Answer» Solution :The displacement in SHM is given by `y=A SIN (omegat+phi_(0))` Given `T=2s, A=5 CM` `omega=(2pi)/(T)=(2pi)/(2)= pi "rad" s^(-1)` `:.y=5 sin (pixx0+phi_(0))` (or)`sin phi_(0)=1 rArr :. Phi_(0)=(pi)/(2)` Hence displacement equation for the pendulum is `y= 5 sin (pi +(pi)/(2))= 5 cost. ` |
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| 13. |
A force F varies with time in accordance with the following figure. The mean force for one cycle. |
| Answer» Answer :D | |
| 14. |
The direction cosines of a vector vecA are cos alpha=(4)/(5sqrt(2)), cos beta = (1)/(sqrt(2)) and cos gamma = (3)/(5sqrt(2)), then the vector vecA is |
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Answer» `4hat(i)+hat(J)+3HAT(K)` |
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| 15. |
When liquid medicine of density rho is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening. We first assume that the drop formed at the opening is spherical because that requires in its surface energy . To determine the size, we calculate the net verticle force due to the surface tension T when the radius of the drop is R. when the force becomes smaller than weight of the drop, the drop gets detached from the dropper. i. If the radius of the opening of the dropper is r, the verticle force due to the surface tension of the drop of radius R (assuming r lt lt R ) is |
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Answer» `2pirT` |
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| 16. |
''A hot liquid' remains hot and a cold liquid remains cold in a thermos flask.'' Explain how ? |
| Answer» Solution :Between the TWO walls of a THERMOS flask VACUUM is created, so tehre is no exchange of heat between the liquid and outside air. The SURFACE is made shinning to reduce loss of heat by radiation. Loss of heat by conduction and convection is also minimised. | |
| 17. |
If the angular speed of earth increases two times then the acceleration of gravity (g) at north pole becomes ....... |
| Answer» ANSWER :C | |
| 18. |
A ballis thrown upward with an initial velocity of 100ms^(-1). After how much time will it return? Draw velocity - time graph for the ball and find from the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15 s. Take g=10ms^(-2). |
Answer» Solution : Here, `u=100ms^(-1),g=-10ms^(-2)` At HIGHEST point, `v=0` As `v=u+g t therefore 0=100-10xxt` `therefore` Time taken to REACH highest point, `t=(100)/(10)=10s` The BALL will return to the ground at `t=20s` Corresponding VELOCITY - time graph of the ball is shown in (i) Maximum height ATTAINED the ball = Area of `DeltaAOB` `=(1)/(2)xx10xx100=500m` (ii) Height attained after `15s` = Area of `DeltaAOB+` Area of `DeltaBCD` `=500+(1)/(2)(15-10)xx(-50)=500-125=375m` |
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| 19. |
If the unit of force is 2N, that of length is 4m and that of velocity is 6ms^(-1), the unit of mass is |
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Answer» `2/9 KG ` |
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| 20. |
Give difference between path length and displacement. |
Answer» SOLUTION :
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| 21. |
When 0.15 kg of ice at 0 ^(@) 0is mixed with 0.30 kg of water at50 ^(@) Cin a container, the resulting temperature is6.7 ^(@) C . Calculate the heat of fusion of ice. (S_("water") = 4186 J kg^(-1) K^(-1)) |
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Answer» SOLUTION :Heat lost by water ` = m s_(w) (theta_(F) - theta_(1))_(w)` ` = (0.30 kg)(4186 J kg^(-1) K^(-1)) (50.0^(@) C - 6.7 ^(@) C)` ` = 54376.14 J` Heat required to melt ice ` = m_(2) L_(f) = (0.15 kg) L_(r) ` Heat required to raise temperature of ice water to final temperature ` = m_(1) s_(w) (theta_(f) - theta_(i))_(I)` ` = (0.15 kg) (4186 J kg^(-1) K^(-1)) (6.7 ^(@) C - 0^(@) C)` ` = 4206.93 J` Heat lost = heat gained `54376.14 J = (0.15 kg) L_(f) + 4206.93 J` ` L_(f) = 3.34 XX 10^(5) J kg^(-1)`. |
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| 22. |
What is the gravitational potential at infinity distance from the centre of earth ? |
| Answer» SOLUTION :`IMPLIES`ZERO. | |
| 23. |
Calculate the temperature at which the oxygen molecules will have the same rms velocity as the hydrogen molecules at 150^@C. Molecular weight of oxygen is 32 and that of hydrogen is 2. |
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Answer» SOLUTION :`v_(rms)=SQRT((3RT)/(M))" As "v_(1 rms)=v_(2 rms)`, therefore `T ALPHA M` `T_(1)/T_2=(M_(1))/M_2, T_(2)=T_(1) (M_2/M_1), T_2=423 [(32)/2]` `T_2=6768k, t_2=6768-273=6495^@C` |
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| 24. |
A body projected obliquely with velocity 19.6 ms^(-1)has its kinetic energy at the maximum height equal to 3 times its potential energy there. Its position after 1 second of projection from the ground is (h= maximum height) |
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Answer» `H/2 ` |
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| 25. |
A body starts from rest from a point distance RQ from the surface of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R —> Radius of the earth) (R_(0) gt R) |
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Answer» `SQRT((GMR_(0))/(R+ R_(0)))` |
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| 26. |
A stone is thrown vertically upwards from some high point P. The velocity of the stone at a height h above P is half that at a depth h below P. The maximum height attained by the stone is |
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Answer» `(7)/(3)H` |
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| 27. |
A particle moves in the plane XY with constant acceleration a directed along he negtive Y-axis. The equation of motion of the particle has the form y=Ax-Bx^(2), where A and B are positive constants. Find the velocity of the particle at the origin of the coordinates. |
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Answer» Solution :Taking X-axis HORIZONTAL and Y-axis vertical. We GAVE `Y=Ax-Bx^(2)` COMPARING with equation of trajectory. `y=x tan alpha-(GX^(2))/(2u^(2)cos^(2)alpha)` `=x tan alpha-(gx^(2))/(2u^(2))(1+tan^(2)alpha)` We have `=tan alpha` and `B=g/(2u^(2))(1+A^(2))` (or) `u^(2)=((1+A)^(2)g)/(2B)` or `u=sqrt(((1+A)^(2)g)/(2B))` |
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| 28. |
A particle is going parallel to x-axis with constant speed V at a distance 'a' from the axis. Find its angular velocity about an axis passing the origin O, at the instant when radial vector of the particle makes angle theta with the x-axis |
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Answer» `(V)/(a)sin^(2)theta` |
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| 29. |
A shell is fired from the ground at an angle with horizontal with a velocity 'v'. At its highest point it breaks into two equal fragments. If one fragment comes back through its initial line of motion with same speed, then the speed of the second fragment will be |
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Answer» `3v COS theta` |
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| 30. |
A solid sphere of 1kg reaches the bottom of an inclined plane after rolling without slipping over a horizontal plane with velocity of 20ms^(-1). The inclination of the plane with the horizontal is 30^(@). What is the velocity of the center of mass of the sphere immediately after it starts moving up the plane? Up to what distance along the plane will it rise? Assume sufficient friction to prevent slipping. |
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Answer» |
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| 31. |
A Geo-stationary satellite bits around the earth in a circular of radius 36, 000 km then what will be the time period of a spy satellite orbiting a few hundred km above the earth's surface (R_("earth") = 6400 km) |
| Answer» Solution :`(T_(2))/(T_(1)) =((a_(2))/(a_(1)))^(3/2) rArrT_(2)=24((6400)/(3600))^(3/2) = 2 ` HOURS | |
| 32. |
In the above problem its velocity after 3 seconds inms^(-1) is |
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Answer» 7.05 |
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| 33. |
A person can swim in still water at 5 m/s. He moves in a river of velocity 3 m/s. First down the stream next same distance up the steam the ratio of times taken are |
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Answer» `1:1` |
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| 34. |
A train of 150 metre length is going towards north direction at a speed of 10 m/s. A parrot flies at the speed of 5 m/s towards south direction parallel to the railways track. The time taken by the parrot to cross the train is |
| Answer» Answer :D | |
| 35. |
Three girlsskatting on a circular ice ground ofradius200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig . 4.20 . Whatis the magnitude of the displacement vector for each ? for which girl is this equal to the actual length of path skate ? |
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Answer» |
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| 36. |
A beam of light of red and green rays is incident obliquely at a point on a face of rectangular glass slab. When coming out on the opposite parallel face, the red and green rays emerge from |
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Answer» TWO POINTS PROPAGATING in two different non PARALLEL directions |
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| 37. |
Calculate work done to move a boy of mass 20 kg along an inclined plane (theta = 45^(@)) with constent velocity through a distance of 10 m. |
Answer» Solution :Here the motion is not accelerated the resultant FORCE parallel to the plane must be zero. So, F = Mg SIN `45^(@) = 0`. Force, F = Mg sin `45^(@)`, distance d = 10 m the workdone. W = Fd COS `theta` `=(Mgsin 45^(@))d cos 0^(@)"" [because theta = 0^(@)" So ", cos 0^(@)=1]` `=20xx10xxsin 45^(@)xx10xx cos 0^(@)` `=20xx10xx(1)/(sqrt(2))xx10xx1=1414J` Workdone, W = 1414 J |
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| 38. |
The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2. |
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Answer» 1.2 MM `=0.01 mm` Now, diameter of ball `=(2xx0.5)+(2.5-5)(0.01)=1.2 mm` |
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| 39. |
Spinning of the Earth about its own axis is known as .............. motion. |
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Answer» linear |
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| 40. |
A spring of force constant k is stretched by a small length x. The work done in stretchign it further by a small length y is |
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Answer» `1/2k(X^(2)+y^(2))` |
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| 41. |
A cyclist travels with a speed of 36 km^(-1)if the angleof inclinatio withvertical is 45^(@)whilehe goes around a circle then the radius of the circle is |
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Answer» 7m `:. R = (v^(2))/( g TAN theta )= (10 xx 10)/( 10 xx tan 45^(@))` `= (100)/(10 xx 1) = 10 m` |
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| 42. |
A stone is thrown horizontally with the velocity u=15m/s from a cover of height H=25m. Find (i) the time during which the stone is in motion, (ii) the distance from the tower base to the point where the stone will touch the ground (iii) the velocity v with which it will touch the ground and (iv) the angle theta the trajectory of the stone makes with the horizontal at the point stone touches the ground (Air resistance is to be neglected). (g=10ms^(-2)) |
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Answer» |
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| 43. |
At the moment t=0, a force F=kt (k is a constant) is applied to a small body of mass m resting on a smooth horizontal plane . The permanent direction of force F , makes an angle beta with the gorizontal . Calculate (i) The velocity of the body at the moment of its breaking off the plane . (ii) The distance traversed by the body upto this moment . |
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Answer» Solution :The free BODY diagram of massm is shown in figure . For vertical equlibrium `F sin beta +N =mg "(or)" As F=kt` `:. Kt sin beta +N=mg`…..(1) EQUATION of MOTION of mass m is `F cos beta =ma` or `kt cos beta = m (dv)/(dt) (as a=(dv)/(dt)) , dv =(dt)/cos beta dt` Integrating `int_(0)^(v) dv =k/m cos beta int_(0)^(t) t dt ` `[v]_(0)^(v)=k/m cos beta [t^(2)/2]_(0)^(t) RARR v=(k cos beta)/(2m) t^(2)`....(2) At the moment of BREAKING off the plane N=0.if `tau` is the corresponding time , then ` `:. "From" (1) k tau sin beta =mg "or" tau=(mg)/(k sin beta)`....(3) `:. ` The velocity at the moment of breaking off the plane is given by putting `t=tau ` in equation (2) , so `(v=v_(0))` `v_(0)=(k cos beta)/(2m) tau^(2)=(k cos beta)/(2m)((mg)/(k sin beta))^(2)` `:. v_(0)=(mg^(2) cos beta)/(2k sin^(2) beta)` (b) Equation (2) may be expressed as `(dx)/(dt)=(k cos beta )/(2m) t^(2)` Integrating `[x]_(0)^(x)=(k cos beta)/(2m)[t^(3)/3]_(0)^(t)` `:. x=(k cos beta)/(6m)t^(3)` When the body breaks off the plane , we have `:. x_(0)=(k cos beta)/(6m) ((mg)/(k sin beta))^(3) =(m^(2)g^(3) cos beta )/(6k^(2) sin ^(3) beta)`
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| 44. |
For the travelling harmonic wave y(x,t) =2.0 cos 2pi [8t-0.0060x + 0.27], where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of, (a) 300 cm, (b) 0.75 m, ( c) lambda/4 |
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Answer» SOLUTION :Here, `y=2.0 COS 2pi (8t - 0.0060x + 0.27)` =2.0 cos `[2pi (8t-0.0060x) + 2pi(0.27)]` Standard equation for a TRAVELLING wave is, `y= r cos[(2pi)/lambda(vt -x) + phi]` Here, `phi =(2pi)/lambda x = 2pi XX 0.006 x` `(2pi)/lambda = 0.006` (a) When x=300 cm, `phi = 2pi xx 0.006 xx 75` `=3.6 pi` rad. (b) When `x= 0.75 m = 75 cm`, `phi = 2pi xx 0.006 xx 75` `=0.9 pi` rad (c ) When `x=lambda/4`, `phi =(2pi)/lambda xx lambda/4 = lambda/2` rad |
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| 45. |
A sphere of mass 50 gm and diameter 2 cm rolls without slipping with a velocfty of 5 cm/s. Calculate the total energy. |
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Answer» Solution :`m = 50 gm = 0.05 KG` r = 1 cm `=10^(-2) m, v = 5 cm//s = 0.05 m//s` TOTAL energy = TRANSLATIONAL K.E. + Rotational K.E. `=1/2mv^(2) + 1/2l omega^(2)` But, `I = 2/3 mr^(2)` and `omega = v/r` `=1/2 mv^(2) + 1/2 xx 2/5 mr^(2) omega^(2) = 1/2mv^(2) + 1/5 mr^(2) xx v^(2)/r^(2)` `=7/10 mv^(2) = 7/10 xx 0.05 xx (0.05)^(2) =1/2 mv^(2) +1/5 mr^(2) xx v^(2)/r^(2)` `=7/10 mv^(2) = 7/10 xx 0.05 xx (0.05)^(2) = 8.75 xx 10^(-5) J` |
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| 46. |
What is the radius of gyration? |
| Answer» SOLUTION :The RADIUS of gyration of an object is the perpendicular DISTANCE from the axis of rotation to an EQUIVALENT point mass, which would have the same mass as well as the same moment of INERTIA of the object. | |
| 47. |
Suppose speed of light (c), force (F) and kinetic energy (K) are taken as the fundamental units, then the dimensional formula for mass will be |
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Answer» `[Kc^(-2)]` Where k is a dimensionless constant. `:. [M^(1)L^(0)T^(0)]= [LT^(-1)]^(x)[MLT^(-2)]^(y)[ML^(2)T^(-2)]^(z)= [M^(y+z)L^(x+y+2z)T^(-x-2y-2z)]` Applying principle of homogeneity of dimensions y+ z= 1...(i) x+y+2z= 0...(II) -x-2y-2z= 0,..(iii) Adding (ii) and (iii)we get -y=0 or y=0 From (i) z= 1-y=1 From (ii)`x= -y-2z=0-2 :. m=c^(-2)F^(0)K^(1)= Kc^(-2)` Hence the dimensional FORMULA for mass is `[Kc^(-2)]` |
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| 48. |
Two satellites S_(1) and S_(2) are revolving round a planet in coplanar and concentric circular orbits of radii R_(1) and R_(2) in the same direction respectively. Their respective periods of revolution are 1hr and 8hr. The radius of the orbit of satellite S_(1) is equal to 10^(4) km. Their relative speed when they are closest, in kmph is |
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Answer» `(pi)/(2) XX 10^(4)` |
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| 49. |
Three objects , A : (a solid sphere) , B : (a thin circular disk ) and C : (a circular ring) , each have the same mass M and radius R . They all spin with the same angular speed omega about their own symmetry axes . The amounts of work (W) required to bring them to rest , would satisfy the relation . |
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Answer» `W_(C) gt W_(B) gt W_(A)` `DELTA W = Delta KE` `Delta W = (1)/(2) I omega^(2)` , where I = MOMENT of inertia For same `omega , Delta W prop I` For a solid sphere , `I_(A) = (2)/(5) MR^(2)` For a THIN CIRCULAR disk , `I_(B) = (1)/(2) MR^(2)` For a circular ring , `I_(C) = MR^(2)` `therefore I_(C) gt I_(B) gt I_(A) therefore W_(C) gt W_(B) gt W_(A)` |
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