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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. |
A block is lying on an inclined plane which makes an angle of 60^(@) with the horizontal. If coefficient of friction between the block and the plane is 0.25 and g=10 ms^(-2), the acceleration of block when it is moves along the plane will be |
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Answer» `2.5 MS^(-2)` |
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| 2. |
A machine gun has a mass of 20kg. It fires 35 g bullets at the rate of 4 bullets per second, with a speed of 400 ms^(-1). What force must be applied to the gun to keep it in position? |
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Answer» Solution :Mass of gun , M=20 kg Velocity of bullet , `v= 400 ms^(-1)` If V be the velocity of recoil , then `V=(mv)/(M)=(0.035 xx 400)/(20) ms^(-1) = 0.7 ms^(-1)` Time taken to fire one bullet, ` t=(1)/(4)` second If F be the AVERAGE force required to hold the gun in position, then `F xx t`= changein momentum of gun or `F xx t = MV`[`:.` Initial momentum of gun is zero ] `:. F=(MV)/(t)=(20 xx 0.07)/(1//4)N = 56 N` Aliter: Number of bullets fired PER second, n = 4 momentum given to the bullets per second = nmv Same momentum per second is gained by the gun. This gives the force acting on the gun and ALSO the force applied to KEEP the gun in position. So, `F= nmv= 4 xx 0.035 xx 400 = 56N` |
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| 3. |
A body is sliding down an inclined plane of coefficient of friction 0.5. If the normal reaction is twice that of resultant downward force along the incline, find the angle between the inclined plane and the horizontal. |
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Answer» SOLUTION :`45^(@)` |
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| 4. |
A metal rod AB of length 10x has its one end A in ice at 0^@Cand the other end B in water at 100^@C. If a point p on the rod is maintained at 400^@C , then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation o water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point p isa t a distance of lamdaxfrom the ice end A. The value of lamdais : [Neglect any heat losss to the surroundings] |
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| 5. |
If a capillary tube is tilted to 45^(@) and 60^(@) from the vertical then the ratio of length l_(1),l_(2) of liquid columns in it will be |
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Answer» `1:SQRT(2)` |
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| 6. |
If F denotes force and t time, then in the equation F=at^(-1) + bt^(2) dimensions of a and b respectively are |
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Answer» `[LT^(-4)]` and `[LT^(-1)]` `:. [a]= [F][t]= [MLT^(-2)][T]= [MLT^(-1)]` `[B]= ([F])/([t^(2)])= ([MLT^(-2)])/([T^(2)])= [MLT^(-4)]` |
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| 7. |
A vector hat(i) + sqrt(3) hat(j) rotates about its tail through an angle 60^(@) in clockwise direction then the new vector is |
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Answer» `hat(i) + SQRT(3) hat(J)` |
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| 8. |
A circular section of radius r si cut out of a uniform disc of radius R, the centre of the hole being R/2 from the centre of the original disc. Locate the centre of gravity of the remainder. |
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| 9. |
A brass disc fits tightly in a hole in a steel plate.Would you heat or cool system to loosen the disc from the hole? |
| Answer» Solution :Coefficient of linear expansion of BRASS is GREATER than that of steel.So the system SHALL be cooled to LOOSEN the DISC from the hole. | |
| 10. |
O' is the centre of a square ABCD. F_(1),F_(2),F_(3) and F_(4) are four forces acting along sides AB, BC, CD and AD as shown in figure. What should be the maginitude of F_(4) so that, the total torque about 'O' is zero ? |
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Answer» `F_(1)+F_(2)+F_(3)` |
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| 11. |
Water rises a height of 10 cm in a capillary tube. When the same capillary tube is dipped in mercury it falls to a height of 3.4cm. Compare the surface tensions of water and mercury. Assume that the specific gravity of mercury is 13.6, angle of contact of water with glass is 0^(@) and angle of contact of water with glass is 135^(@). |
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| 12. |
A drive of a mail train travelling 108 kmh^(-1) sees a passenger train which is 250 m ahead on the same tarck. The passenger train is travelling in the same direation as the mail train. The velocity of the passenger train is 36^(-1) .the driver of the mail train immediatly applies the brakes so as to produces a constant deaceleration of 1ms^(-2) wherease the passenger train continues with the constant speed (i) calculate and find wheater there will be a collision (ii) if the collision occcurs where willit take place ? |
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| 13. |
Statement A : When a person is on the floor of a lift which is at rest, the resultant force on him is equal to his weightStatement B : When the lift is moving with uniform velocity, then the apparent weight of the man is zero |
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Answer» A & B are CORRECT |
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| 14. |
A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval t_1 . An observer at the top of the tower sees the same ball go up above him and then come back at this level in a time interval t_2 . The height of the tower is |
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Answer» `1/2 "gt"_1 t_2` |
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| 15. |
n' drops of a liquid, each with surface energy E, join to form a single drop. Then |
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Answer» some ENERGY will be RELEASED in the process |
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| 16. |
Subtract 0.2 from 7.26 and express the result with correct number of significant figures |
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Answer» `7.1` |
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| 17. |
Five moles of an ideal gas are taken in a carnot engine working between 100^(@)C and 30^(@)C. The useful work done in 1 cycle is 420J. Calculate ratio of volume of the gas at the end and at the beginning of the isothermal expansion. Take R=8.4J mol e^(-1)K^(-1) |
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Answer» `T_(2)= 30^(@)C=(30+273)K=303K`, `W=420J, (V_(2))/(V_(1))=?, R= 8.4J mol e^(-1)K^(-1)` As `(Q_(1))/(Q_(2))=(T_(1))/(T_(2))=(373)/(303) :. Q_(1)(373)/(303)Q_(2)`....(i) Now, `W=Q_(1)-Q_(2)=(373)/(303)Q_(2)-Q_(2)=(70)/(303)Q_(2)` `:. Q_(2)=(373)/(70)W= (303)/(70)xx420=1818J` From (i), `Q_(1)=(373)/(303)xx1818= 2238J` Work done during ISOTHERMAL expansion `W_(1)=Q_(1)= 2.3026 n RT_(1)"LOG"_(10)(V_(2))/(V_(1))` `2238= 2.3026xx5xx8.4xx373 "log"_(10)(V_(2))/(V_(1))` `"log"_(10)(V_(2))/(V_(1))=(2238)/(2.3026xx5xx8.4xx373)=0.0620` `(V_(2))/(V_(1))= antilog0.0620= 1.153` |
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| 18. |
A vector vecA makes on angle 30^(@) with the y-axis in anticlockwise direction. Another vector vecB makes on angle 30^(@) with the x-axis in clockwise direction. Find angle between vectors vecA and vecB. |
Answer» Solution : From the diagram angle between `VECA and VECB` is `30^(@)+90^(@)+30^(@)=150^(@)` |
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| 19. |
A bullet is fired from a cannon with velocity 500m/s. If the angle of projection is 15° and g=10m/s^(2) , then find the range. |
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| 20. |
{:("Column-I","Column-II"),("A) Regnault's apparatus","P) Anomalous expansion of water"),("B) Jolly' s bulb apparatus","Q) Apparent expansion of liquids"),("C) Hopes apparatus","R) Volume coefficient of gas"),("D) Specific gravity","S) Pressure bottle method coefficient of a gas"):} |
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| 21. |
Which of the following symptoms is likely to effect an astronaut in space (a) swollen feel (b) swollen face, (c ) headache, (d) orientiational problem. |
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Answer» Solution :(a) We KNOW that the legs carry the weight of the body in the normal position due to gravity pull. The astrount in sapce is in weightlessness state. Hence, swollen feet may no effect his working. (b) In the conditions of weightlessness, the face of the astrount is expected to get more supply. Due to it, the astrount may develop swollen face. As eyes, EARS, nose, mouth etc. are all emnedded in the face, hence, swollen face may effect to great extent the seeing/hearing/smalling/eating capabilities of the astrount in space. (c ) Headache is due to mental strain. It will perist whether a PERSON is an astrount in space or he is on earth. It means headache will have the same effect on the astrount in space as on person on earth. (d) Space also has ORIENTATION. We also have the frames of reference in psace. Hence, orientational PROBLEM will affect the astronaut in sapce. |
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| 22. |
In Gallilean telescope, the final image formed is |
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Answer» REAL ,ERECT and enlarged |
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| 23. |
A disc of mass m and radius R is given a linear velocity V_(0) on a rough horizontal surface. Initially the motion of the disc is purely translatory. The work done by friction in changing the motion of the disc from pure translatory to pure rolling is 1/xmv_(0)^(2) where 'x' is |
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| 24. |
A solid sphere of mass 2kgis rolling down an inclined plane of angle of inclination 30^(@). What is the agent which supplies the external torque for the rotational motion of the sphere ? Calculate torque if radius is 0.40m (g=10ms^(-2) |
Answer» SOLUTION : The force of friction between the sphere and the inclined plane SUPPLIES the external torque. Given `m=2kg,theta=30^(@),R=0.40m,g=10ms^(-2)` We know that a `=(gsintheta)/(1+(K^2)/(R^2))` Force=`ma=5/7mgsintheta` `F=5/7xx2xx10xxsin30^(@)` i.e `F=(50)/7=7.143N` |
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| 25. |
Explain in detail about IPR. |
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Answer» Solution :In the year 1824 a young FRENCH ENGINEER Sadi CARNOT proved that a certain reversible engine operated in cycle between hot and cold reservoir can have maximum efficiency. This engine is called Carnot engine. A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine has four parts which are given below. (i) Source: It is the source of heat maintained at constant HIGH temperature `T_(H)`. Any amount of heat can be extracted from it, without changing its temperature. (ii) Sink: It is a cold body maintained at a constant low temperature `T_(L)`. It can absorb any amount of heat. (iii) Insulating stand: It is made of perfectly non - conducting material. Heat is not conducted through this stand. (iv) Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting BOTTOM. A non - conducting and frictionless piston is fitted in it. 
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| 26. |
A simple harmonic osicollator has a time period of 2 sec. The change in phase 0.5 sec after leaving the mean position is (in radian) |
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Answer» `pi` |
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| 27. |
Poad skaters walk on the surface of water. Give reason for it. |
| Answer» SOLUTION :When the INSECT rests on a liquid surface, the vertical component of surface TENSION balances the weight of the insect. The surface of liquid acts as a STRETCHED ciastic membrane. | |
| 28. |
Three liquids of densities rho_(1) , rho_(2) and rho_(3) ( with rho_(1) gt rho_(2) gt rho_(3) ), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact theta_(1) , theta_(2) and theta_(3) obey |
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Answer» `(PI)/(2) gt theta_(1)gt theta_(2) gt theta_(3) ge 0` |
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| 29. |
Define the term plasticity. Give example. |
| Answer» Solution :If a body does not regain its original shape and size after removal of the DEFORMING force, it is SAID to be a PLASTIC body and the PROPERTY is called plasticity. EG: Glass. | |
| 30. |
Explain why the concept of centre of mass is muchmore fundamental than centre of gravity. |
| Answer» Solution :As the mass of a body remains constant anywhere in the UNIVERSE, a body always has a definite CENTRE of mass. But at places where there is no GRAVITY (e.g., in OUTER space, inside an artificial satellite rotating around earth or in case of a FREE falling object ), the weight of the body becomes zero i.e., there is no centre of gravity. Hence the concept of centre of mass is much more fundamental than centre of gravity. | |
| 31. |
(A): A balloon ascends from the surface of earth with constant speed. When it was at a height 50m above the ground, a packet is dropped from it. To an abserver on the balloon, the displacement of the packet, from the moment it is dropped to the moment it reaches the surface of earth is 50m. (R): Displacement vector does not depend upon the reference frame used to measure it. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 32. |
For a particle executing simple harmonic motion, the acceleration is proportional to |
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Answer» DISPLACEMENT from the mean position |
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| 33. |
A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and "x_(2) =A_(2) sin [omega t+(pi)/(3)], then find the displacement at time t=0. |
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Answer» Solution :Here phase difference `phi = (PI)/(3)rad` At t=0, time, `x_(1)= 0" and "x_(2) = A_(2) sin"" (pi)/(3)` RESULTANT displacement `X= x_(1)+ x_(2)` `=0+ A_(2) sin ""(pi)/(3) = (sqrt(3)A_(2))/(2)` Resultant amplitude `A= sqrt(A_(1)^(2) +A_(2)^(2)+ 2A_(1) A_(2) cos ""(pi)/(3))` `= sqrt(A_(1)^(2) +A_(2)^(2)+ 2A_(1) A_(2))` |
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| 34. |
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec average velocity is ................ . |
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Answer» 0 Average velocity, `V=("Total displacement")/("Total time")=(80)/(40)` `V=2 m // s` |
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| 35. |
If vectors vec(A) = hat(i) + 2hat(j) + 4hat(k) and vec(B) = 5 hat(i) represent the two sides of a triangle, then the third side of the triangle can have length equal to: |
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Answer» 6 |
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| 36. |
When surfaces in contact are…..friction between them….instead of….. . |
| Answer» SOLUTION :too SMOOTH , INCREASES , DECREASING . | |
| 37. |
What are the applications of dimensional analysis? Verifys= ut + 1/2 at^2 by dimensional analysis |
Answer» Solution : Given EQUATION is dimensionally correct as the dimensions on the both side are same. APPLICATIONS of dimensional analysis (i) CONVERT a physical QUANTITY from one system of units to another. (ii) CHECK the dimensional correctness of a given physical equation. (iii) Establish relations among various physical quantities. |
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| 38. |
A space-ship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000 kg, Mass of the sun = 2 xx 10^30kg. Mass of Mars =6.4xx 10^23 kg, Radius of Mars = 3395 km. Radius of the orbit of Mars = 2.28 xx 10^11m, G=6.67 xx 10^(-11) Nm^2 kg^(-2) |
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Answer» Solution : Let R.be the radius of orbit of Mars and R. be the radius of Mars. M be the MASS of the Sun and M. be the mass of Mars. If m is the mass of the space-ship, then Potential energy of space-ship due to gravitational attraction of the Sun = -GMm/R Potential energy of space-ship due to gravitational attraction of Mars =-GM.m/R. Since the K.E. of space-ship is zero total energy of space-ship ` = - (GMm)/( R) - (GM.m)/(R.) = - Gm (M/R + (M.)/(R.) )` `therefore `Energy required to rocket out the spaceship from the solar system = -(total energy of spaceship) ` = - [-Gm(M/R +(M.)/(R.) ) ] = Gm (M/R + (M.)/(R.) ) = 6.67 XX 10^(-11) xx 1000 xx [ (2 xx 10^30)/(2.28 xx 10^11) + (6.4 xx 10^23)/(3395 xx 10^3)]` `= 6.67 xx 10^(-8) [ (20)/(2.28) + (6.4)/(33.95) ] xx 10^18 J = 4.89 xx 10^11 J` |
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| 39. |
Consider the situation shown in the figure. Uniform rod of length L can rotat freely about the hinge A in vertical plane. Pulleys and stringa are light and frictionless. If the rod remains horizontal at rest when the system is released then the mass of the rod is |
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Answer» `(4)/(3)M` |
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| 40. |
For the one-dimensional motion, described by x = t - sint |
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Answer» `X (t) gt 0` for all `t gt 0` velocity `v = (dx)/(dt) = (d)/(dt) [t - sin t]` `=1 - COS t` Acceleration `a = (DV)/(dt) = (d)/(dt) [1- cos t] sin t ` As acceleration `a gt 0 ` for all `t gt 0 (because sin t gt 0)` Hence,` x (t) gt 0` for all `t gt 0` Velocity `v =1 - cos t` When, `cos t = 1,` velocity `v =0` `V_(max) =1 - (cos t ) _(min) =1 - (-1) =2` `V _(min) =1- (cos t) _(max) 1-1 =0` Hence, v lies between 0 and 2. Acceleration `a - (dv)/(dt) =- sin t` When `t =0, x=0, v =+ 1, a =0` When `t = 90^(@),x = 1, v =0, a =-1` When `t = 180^(@) , x=0, v =-1, a =1` When `t = 360 ^(@),x =0 , v =0 ,a=0` |
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| 41. |
A virtual cannot be caought on a screen, Yet when we see a virtual image, we bring it to the screen i.e., retina of our eye. This happnes because |
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Answer» we have TWO eyes. |
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| 42. |
Find the distance between Venus and Sun. |
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Answer» Solution :The distance between Earth and Sun is considered as one A.U. When venus is at MAXIMUM elongation, venus makes `90^(@)` to Sun. The trigonometric relation is satisfied by this right ANGLED triangle. `SINTHETA=(r)/(R)` Where R = 1 AU `r=Rsintheta=(1AU)(sin46^@)` Here `sin46^(@)=0.77`. Hence Venus is at a distance of 0.77 AU from Sun. |
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| 43. |
An astronomical telescope has a large aperture to |
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Answer» Reduce SPHERICAL aberration |
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| 44. |
When 82 gm of a liquid at 10°C in a specific gravity bottle is heated to 85°C, 2gm of liquid is expelled. Coefficient of apparent expansion of liquid is |
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Answer» 0.0033/°C |
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| 45. |
The ratio of specific heat capacity to molar heat capacity of a body |
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Answer» is a universal constant |
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| 46. |
The position vector of a particle is vecr=4t^(2)hati+2thatj+3hatk. The speed of the particle t=5s is |
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Answer» `42ms^(-1)` SPEED `v=(dvecr)/(dt)=8thati+2hatj` , at t=5s `v=40+2=42` |
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| 47. |
The pressure at the bottom of a well is four times that at a point 2.5 m below its top. If the atmospheric pressure is equal to the pressure of 10 m of a water column, then what is the depth of the well? |
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| 48. |
A ball is projected at an angle 30° with the horizontal with the velocity 49ms^(-1). The horizontal range is |
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Answer» 122.5 m |
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| 49. |
A stick has a length of 12.132 cm and another stick has a length of 12.4 cm. If the two sticks are placed end to end , what is their total length ? |
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Answer» Solution :(a) Let lengths of the STICKS are named as `l_(1)= 12.132cm, l_(2)= 12.4cm` Here `l_(2)` has one decimal place and `l_(1)` has to be rounded off to have only two decimal places `l_(1) + l_(2) = 12.13 + 12.4= 24.53` This is to be rounded off to have one decimal place only `therefore` The TOTAL length is 24.5cm (b) `l_(1) = 12.132, l_(2) = 12.4` `l_(1)-l_(2) = 12.4- 12.132` Here 12.4 has only decimal place and hence 12.132 shoudl have only two decimal places. `therefore l_(1)-l_(2)= 12.4-12.13= 0.27` This should be rouned off to have only one decimal place `therefore l_(1)-l_(2)= 0.3` Hence DIFFERENCE of their lengths is 0.3cm |
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| 50. |
Under the action of force, 2 kg body moves such that its position x as a function of time t is given by x=(t^(3))/(3), x is in metre and t in second. Calculate the work done by the force in the first 2 second. |
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Answer» Solution :From WORK - energy theorem, `W = Delta KE` `x=t^(3)//3 therefore` VELOCITY `v=(dx)/(dt)=t^(2)` At `t=0, v_(i)=0^(2)=0`, At `t=2, v_(f)=2^(2)=4 m//s` work done `W=(1)/(2)m(v_(f)^(2)-v_(i)^(2))=(1)/(2)xx2(4^(2)-0)=16 J` |
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