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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. |
To a person travelling due east with a velocity V, the wind appears to blow from an angle theta North to east. When he starts travelling due North with a velocity 2V, the wind appears to blow from an angle beta North east find the time direction of wind. |
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Answer» |
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| 2. |
The escape speed from the surface of earth is given by |
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Answer» `SQRT(2gR_s)` |
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| 3. |
A planet of mass 'm' is in a elliptical orbit about the sun (M lt lt M) with an orbital time period 'T'. If 'A' be the area of the orbit then its angular momentum is |
| Answer» Answer :A | |
| 4. |
A carnot engine has an efficiency 0.4. When the temperature of the source is increased by 20^(@)C and the sink is reduced by 20^(@)C, its efficiency is found to increase to 0.5. Calculate the temperature of source and sink. |
| Answer» Solution :`0.6 = T _(2) //T_(1) , 0.5 = (T_(2) - 20) T _(1) + 20,` SOLVING `T _(1) = 300 K, T _(2) = 180 K` | |
| 5. |
In presence of a resonant body, the sound produced by a body is intensified, is the principle of conservation of energy violated here? |
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Answer» Solution :The principle of conservation of energy is not violated here. Suppose, a tuning fork is set into vibration. The loudness of the sound so produced is not very large. If the vibrating fork is held on a hollow wooden box, the sound is Intensified. in this CASE the air inside the box acts as a resonant body. But it is found that the sound of the tuning fork lasts for a long time in the first case. But in the SECOND case, the sound of the tuning fork is SHORT-lived. The total dissipation of energy In both the cases is the same. In the first case, the rate of dissipation of energy is low but continues for a long time. But in the second case, the rate of dissipation of the. same amount of energy is comparatively high and therefore the sound DIES down in a short time. it does not violate the principle of conservation of energy. |
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| 6. |
Define moment of inertia and write its unit. |
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Answer» Solution :Moment of inertia (M.O.L.): The property of a rotating body to oppose any CHANGE in its STATE of ROTATION is called moment of inertia (I). Mathematically moment of inertia `IMR^(2)`,unit ,Kg-`m^(2)`.Dimensional FORMULA ,`ML^(2)` |
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| 7. |
A force time graph for the motion of a body is as shown in figure. Change in linear momentum between 0 and 6s is |
| Answer» Answer :A | |
| 8. |
To produce deviation without dispersion with the help of crown glass prism of refracting angle (A) and flint glass (A') they must be related as |
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Answer» A GT A' |
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| 9. |
If, in , the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :(a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ? |
| Answer» SOLUTION :Atinstantstringbreakstonewillmoveindirectionof TANGENTTO CIRCULARPATH inuniformcircularmotionvelocityat everyinstantwhenstringbreakbyNewtonfirstlawof motionstonewilltravel in DIRECTION of trangentto circularpath | |
| 10. |
Maximum velocity in SHM is V_(m). The average velocity during motion from one extreme point to the other extreme point will be |
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Answer» `(piV_(m))/(2)` |
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| 11. |
At a certain height above the earth's surface the gravitational is 4% of its value at the surface of the earth. Find the height (R is the surface of the earth) |
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Answer» 2R |
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| 12. |
In studying motion of a body, the starting of motion is more difficult than maintaining it because, the coefficient of static friction and kinetic friction satisfy the relation. |
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Answer» `mu_s gt mu_k` |
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| 13. |
Calculate the area of the triangle for which two of its sides are given by the vectors vecA=5hati-3hatj, vecB=4hati+6hatj. |
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Answer» Solution :AREA of the TRIANGLE `=(1)/(2)|barAxxbarB|` `barAxxbarB=|{:(HATI,hatj,hatk),(5,-3,0),(4,6,0):}|=hati(0)+hatj(0)+hatk(30+12)=42hatk` `|barAxxbarB|=sqrt(42^(2))=42` `:.` Area of the triangle `=(1)/(2)|barAxxbarB|=(1)/(2)xx42` Area =21 squared units |
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| 14. |
Find the theoretical maximum efficiency of a steam engine using an intake steam temperature of 100^(@)C, if it is operated in a place where the exhaust temperature is 0^(@)C. If super heated steam at 200^(@)C is used, what is the maximum possible efficiency? |
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Answer» Solution :`T_(1) = 100^(@)C = 373 K, T_(2) = 0^(@)C = 273 K` Efficiency `= eta = (T_(1) - T_(2))/( T_(1)) = (373 - 273)/(373)` ` = 0. 268 = 26.8%` `T_(1) = 200^(@)C = 473 K, T_(2) = 0^(@)C = 273 K` Efficiency `= eta = ( T_(1) - T_(2))/( T_(1)) = (473 - 273)/(373)= (200)/(373) = 0.536= 53.6%` |
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| 15. |
A systme can be taken from the inital state p_(1), V_(1)to the final statep_(2), V_(2)by two different methods. LetDeltaQandDeltaWWrepresent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods? |
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Answer» `DELTAQ` |
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| 16. |
A straight thin uniform rod of length 4L and man 4M is bent into a square. Its M.I about one side is |
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Answer» `(5)/(3)ML^(2)` |
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| 17. |
Which of the following are vectors ? (i) Torque, (ii) Moment of the force (couple) (iii) Angular momentum. |
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Answer» Solution :TORQUE, moment of force (couple) and ANGULAR momentum are vectors. Torque `vectau=barrxxbarF`- VECTOR :Moment of force (couple) C=`barrxxbarF` - vector Angular momentum `barL=m(barrxxbarv)` - vector |
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| 19. |
What happens to the level of the liquid, when the coefficient of real expansion of the liquid is same as the coefficient of volume expansion of the vessel? |
| Answer» SOLUTION :We know `gamma _(g) = gamma _(s) + gamma _("vessel").` · If the coefficient of real expansion of the liquid is EQUAL to the coefficient of volume expansion of the vessel, coefficient of apparent expansion of the liquid BECOMES ZERO. Owing to this the level of the liquid does not change | |
| 20. |
In an isotopic gas, with no preferred direction of motion of molecules, |
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Answer» `upsilon_(x) = upsilon_(y) = upsilon_(Z)` |
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| 21. |
A car is negotiating a curved road of radius R. The road is banked at an angle theta . The coefficient of friction between the tyres of the car and the road is mu_s. The maximum safe velocity on this road is |
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Answer» `SQRT((g)/(R )( (mu_s + tantheta))/(R(1-mu_s tan theta))` |
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| 22. |
Thermal conductivity of air is less than that of felt, but is better insulator in comparison to air why? |
| Answer» Solution :Felt contains fine HOLES in which air is trapped. This air trapped is not free to move in these holes. So convection is not possible. Besides this as felt is a bad conductor, CONDUCTIONIS also not possible. Like this felt is better insulator in comparison to air. | |
| 23. |
A coin of mass m and radius r having moment of inertia I about the axis passing through its centre and perpendicular to its plane. It is beaten uniformly to form a disc of radius 2r. What will be the moment of inertia about the same axis? |
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Answer» I `I = (mr^2)/(2) "" … (i)` Moment of inertia of formed disc about the same axis is `I. =(m(2r)^(2))/(2) = 4 ((mr^(2))/(2)) = 4I""` (USING (i)) |
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| 24. |
A block of mass m hangs on a vertical spring. Initially the spring is unstretched, it is now allowed to fall rest. Find (a) the distance the block falls if the block is releaed slowely, (b) the maximum distance the block falls before it beigns to move up. |
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Answer» Solution :(a) When the BLOCK falls slowely, it comes to rest at a distance `y_(0)`, which is referred to as the equilibrium position. From, condition of equilibrium, `SUM F_(y)=ky_(0)-mg=0 "or " y_(0)=(mg)/(k)`  (B)When the block is released suddenly, it oscillates about the equilibrium position. Initially the SPEED of the block increases then reaches maximum value and then decreases to zero at the lowest position. In this situation the block oscillates about the equilibrium position. The block is released from rest, there fore its total mechanical energy initially. `E_(i)=U_(g)+U_(s)+KE = 0+0+0` Final total mechanical energy. `E_(f)=-mgy_(m)+(1)/(2)ky^(2)+0` From conservation of energy, `E_(i)=E_(f)` `0=-mgy_(m)+(1)/(2)ky^(2)` or `y_(m)=(2mg)/(k)=2y_(0)` |
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| 25. |
If for a material Young's modulus = 6.6 xx 10^10 N m^-2 and Bulk modulus =1 1 xx 10^10 N m^-2, its Poisson's ratio is |
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Answer» 0.4 |
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| 26. |
A soap bubble of radius R is surrounded by another soap bubble of radius 2R, as shown. Take surface tension = S. |
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Answer» Then the pressure inside the SMALLER soap BUBBLE in EXCESS of the atmospheric pressure is ZERO |
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| 27. |
''Air is a bad conductor of heat, but still we do not feel warm without clothes''. Why? |
| Answer» SOLUTION :When we are without clothes AIR is carried AWAY from our BODY by convection and hence we FEEL cold. | |
| 28. |
A thin uniform bar of mass m and length 2L is held at an angle 30^@ with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string breaks. |
Answer» Solution :Just after CUTTING the string the tension of right will disapear and tension in left string will change. Let us draw free body diagram of rod immediately after the string breaks.  `SigmaF_(x)=0` `:. a_(x)=0`…….i `a_(y)=(Sigma F_(y))/m=(mg-T)/m`........ii  APPLYING torque equation about centre of mass `tau=I_(cm)alpha` `T.(Lcos30^@)=(3(2L)^(2))/12.alpha` `implies alpha=tau/Ii=(TLcos30^@)/((m(2L)^(2))/12)=(3sqrt(3)T)/(2mL)` ..........iii  Now just after the string breaks, acceleration of point `A` in the vertical DIRECTION should be zero. `i.e. a_(y)=Lalpha cos30^@ `.........IV Solving equatioin i, ii, iii and iv we get `T=(2mg)/11` and `a=(3sqrt(3))/11 g/L` |
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| 29. |
A particle oscillate according to y= 7 sin 0.5 pi t. Time taken by particle from equilibrium position to maximum displacement is………. |
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Answer» Solution :COMPARING equation `y= 7sin 0.5 pi t` to general differential equation `y= A sin (OMEGA t)` Amplitude A= 7 unit Angular frequency `omega = 0.5 pi` unit `therefore (2PI)/(T) = (pi)/(2)` `therefore T= 4` unit. |
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| 30. |
When a mass is rotating in a plane about a fixed point , its angular momentum is directed along |
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Answer» a LINE perpendicular to the plane of rotation `implies F_(1) x + F_(2) x- F_(3) x = 0` or `F_(3) = F_(1) + F_(2)` |
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| 31. |
Along a streamline |
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Answer» the VELOCITY of a fluid particle remains CONSTANT |
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| 32. |
The rate of emission of radiation of a black body at temperature 27^(@)C is E_(1). If its temperature is increased to 327^(@)C the rate of emission of radiation is E_(2). The relation between E_(1) and E_(2) is |
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Answer» `E_(2)=24E_(1)` |
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| 33. |
A constant pressure gas thermometer has a volume of 200cc, 273.2cc and 525.1cc at 0^(@) C, 100°c and boiling point of sulphur respectively. The resistance possessed by a platinum wire at the same temperatures as above 2.00 Omega , 2.778 Omega and 5.280 Omega respectively. Estimate the boiling point on the basis of two sets of observations. |
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| 34. |
80 g of water at 30^(@) C is mixed with 50 g of water at 60^(@) C, final temperature of mixture will be |
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Answer» Solution :If m g ICE melts, then Heat lost = Heat gain `80 xx 1 xx (30-0) = m xx 80 rArr m = 30KG` |
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| 35. |
Which of the following is the evidence to showthat there must be a force acting on earth and directed towards the sun ? |
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Answer» DEVIATION of the falling bodies towards east. |
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| 36. |
Two rods of different metals, having the same area of cross-section A, are placed end to end between two massive walls as shown in fig. If the temperature of both the rods are now raised by Deltat^(0)C then (a) Find the force with which the rods acts on cach other at higher temperature. (b) Find the lengths of the rods at the higher temperature. Assume that there is no change in the cross-sectional area of the rods and the rods do not bend. There is no deformation of walls. |
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Answer» Solution : (a) Due to heating the increases in length of the COMPOSITE rod will be `(DELTAL)_(1)=l_(1) alpha_(1)Deltat+l_(2) alpha_(2) Deltat=(l_(1)alpha_(1)+l_(2)alpha_(2))Deltat...(1)` due to compressive force .F from the walls, due to elasticity, the decrease in length will be `(Deltal)_(D)=(Fl_(1))/(AY_(1))+(Fl_(2))/(AY_(2))=(F)/(A)((l_(1))/(Y_(1))+(l_(2))/(Y_(2)))....(2)` As the length of the composite rod remains unchanged the increase in length due to heating MUST be equal to decrease in length due to compression. `(l_(1)alpha_(1)+l_(2)alpha_(2))Delta=(F)/(A)((l_(1))/(Y_(1))+(l_(2))/(Y_(2))) rArr (##AKS_ELT_AI_PHY_XI_V01_C_C01_SLV_023_S01.png" width="80%"> As initially the length of one rod in L, and due to heating its length increases by `(Deltal_(1))_(H)=alpha_(1) l_(1)Deltat`. while due to compression its length decreases by `(Deltal_(1_(C))=(Fl_(1))/(AY_(1))` so its final length `l_(1)^(1)=l_(1)+(Deltal_(1))_(H)-(Deltal_(1))_(C)=l_(1)+l_(1)alpha_(1) Deltat-(Fl_(1))/(AY_(1))` for the other ford `l_(1)^(2)=l_(2)+(Deltal_(2))_(H)-(Deltal_(2))_(C), l_(2)^(1)=l_(2)+l_(2)alpha_(2)Deltat-(Fl_(2))/(AY_(2))` |
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| 37. |
The masses and radii of the earth and moon are M_1, R_1 and M2, R_2 respectively. Their centres are at distance d apart. What is the minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity? |
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Answer» Solution :Potential energy of m when it is midway between `M_1` and `M_2` `U=m(V_1+V_2) =m(-(GM_1)/(d//2) + (-GM_2)/(d//2))=(-2GM)/d[M_1+M_2]` And as potential energy at infinity is ZERO, so work required to shift m from the given position to infinity, W=0 - U=2Gm`(M_1+M_2)//d` As this, work is provided by initial kinetic energy. `1/2m upsilon^2 = (2Gm(M_1+M_2))/d` or `upsilon=2sqrt((G(M_1+M_2))/d)` |
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| 38. |
The acceleration 'a' in m//s^(2) of a particle is given by a= 3t^(2)+2t+2 where t is the time. If the particle starts out with a velocity u=2 m//s at t=0 , then the velocity at the end of 2 second is |
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Answer» 12 m/s |
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| 39. |
Magnitude of the impulse on the ball due to the tension, as it loses contact with the surface after collision is |
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Answer» `(2sqrt(3))/(5sqrt(5)) msqrt(GL)` |
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| 40. |
Assertion:The size of a hydrogen balloon increases as it rises in air.Reason:The material of the balloon can be easily stretched.Choose the correct option: |
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Answer» Both assertion and reason are TRUE and reason explains ASSERTIONS correctly. |
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| 41. |
When a stationary wave is formed, then its frequency is |
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Answer» same as that of the INDIVIDUAL waves |
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| 42. |
The pitch of a screw gauge is 0.5 mm and there are 100 divisions on its circular scale. The instrument reads + 2 divisions, when no object is placed between the studs which are firmly in contact. In measuring the diameter of wire, there are 4 divisions on the pitch scale and 83^(rd) division of the of the circular scale coincides with the base line / reference line. Then the diameter of the wire is |
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Answer» 4.05 MM |
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| 43. |
A1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of initial position of the block is x-0. |
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Answer» `=2.33 J ` (b) Maximum kinetic energy of the block is at a point where FORCE changes its direction. or `F=0` at `X=sqrt(2.5)` `=1.58 m` `:. K_(max) =int_(0)^(1.58)(2.5-X^(2))dx` `:. =2.635 J` |
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| 44. |
Column-I shows different charge distributions. Cloumn-II gives corresponding electric dipole moments. |
Answer»  
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| 45. |
Starting from rest a centrifuge accelerates uniformly at the rate of 40 "rad"//s"^2. What is the number of revolutions made by it when its angular velocity becomes 1800 rev/min ? |
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| 46. |
The maximum rarefaction produced by up-to-date laboratory methods is 10^(-11) mm of mercury. What is the density of the rarest air at 17^(@)C? Molecular weight of air =28 and 760mm of mercury =1.013xx10^(5)Pa. |
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| 47. |
A vehicle of mass 20kg is moving with a velocity of 4ms^(-1) Find the magnitude of the force that is to be applied on the vehicle so that the vehicle have a velocity of 1ms^(-1) after travelling a distance of 20m. |
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| 48. |
The cylinderical tube of a spray pump has a cross-section of 6 cm^(2) one of which has 50 holes each of diameter 1 mm. If the liquid flow inside the tube is 1.2 m per minutes, then the speed of ejection of the liquid through the holes is |
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Answer» 1. `2.1 m s^(-1)` `a_(1)=6 cm^(2)=6xx10^(-4) m^(2)` Number of holes =50 Diameter of each hole, D=1mm, `10^(-3)`m Radius of hole, `r=(D)/(2)=(1)/(2)xx10^(-3)=5xx10^(-4)`m Area of cross-section of each hole `=PIR^(2)=pi(5xx10^(-4))^(2) m^(2)` Speed of liquid inside the tube, `v_(1)=1.2 m "MIN"^(-1)=(1.2)/(60) m s^(-1)=0.02 m s^(-1)` Let the velocity of ejection of the liquid through the hole `v_(2)` As `a_(1)v_(1)=a_(2)v_(2)` or `v_(2)=(a_(1)v_(1))/(a_(2))=(6xx10^(-4)xx0.02)/(50xxpi(5xx10^(-4))^(2))=0.31 m s^(-1)` |
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| 49. |
A lighter particle moving with a speed of 10 ms^(-1) collides with an object of double its mass moving in the same direction with half its speed. Assume that the collision is a one dimensional clastic collision. What will be the speed of both particles after the collision? |
Answer» SOLUTION : Let the mass of the first BODY be m which moves with an initial velocity, `U_(1)=10ms^(-1)` Therefore, the mass of second body is 2M and its initial velocity is`u_(2)=1/2u_(1)=1/2(10ms_(-1))` `v_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m_(2))/(m_(1)-m_(2)))u_(2)` `v_(1)=((m-2m)/(m+2m))10+((2xx2m)/(m+2m))5` `v_(1)=-(1/3)10+(4/3)5=(-10+20)/3=10/3` `v_(1)=3.33ms_(-1)` `v_(2)((2m(1))/(m_(1)+m_(2)))u_(1) + ((m_(2)-m_(2))/(m_(1)+m_(2)))u_(2)` `v_(2)=((2m)/(m+2m))10+((2m-m)/(m+2m))5` `v_(2)=(2/3)10+(1/3)5=(20+5)/3=25/3` `v_(2)=8.33ms(-1)` as the two speeds `v_(1) and v_(2)` are posiitive , they movein the same direction with the veloxities `3.33ms^(-1) and 8.33ms^(-1)`, respectively. |
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