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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 non-viscous fluid of constant density of 1000kg//m^(3) flows in stream line motion along a tube of variable cross-section The area of cross-section at two P and Q at lengths 5m are 40 cm^(2) and 20 cm^(2) respectively. If velocity of fluid at P is 3 m//s then find velocity of fluid at Q. |
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Answer» `3m//s` |
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| 2. |
A boy wants to throw a letter wrapped over a stone to his friend across the street 40 m wide. The boy's window is 10 m below the friend's window. How should he throw the ball ? |
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| 3. |
An object is being heated by a heater supplying 60 W heat. Temperature of surrounding is 20^(0)C and the temperature of object becomes constant at 50^(0)C. Now the heater is switched off. The rate at which the object will lose heat when its temperature has dropped to 30^(0)C is 4xxz watt. Then z is equal to ...................... |
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| 4. |
Define amplitude of the wave. |
| Answer» SOLUTION :The maximum displacement of the oscillating PARTICLE on either side of its MEAN POSITION is called its amplitude. | |
| 5. |
A bus of mass 1000 kg is standing stationary on bus stationwhat is itslinearmomentum ? |
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Answer» SOLUTION :`m= 100 kg` `VEC( V )= 0 ` `vec(p ) = MVEC( v)` `=100 (0)` `=0` |
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| 6. |
Define internal energy of a gas. Why do we say that internal energy is an exact differential ? |
| Answer» Solution :The change in INTERNAL energy depends only on the INITIAL and final states of the SYSTEM. So it is a in EXACT differential. | |
| 7. |
In gravitational field the work done in movingbody from one point into another depends on |
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Answer» INTIAL and FINAL position |
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| 8. |
A hollow sphere of inner radius r_(0) and outer radius 2r_(0) is made of a uniform material of constant thermal conductivity K. The temperature with in the ball is maintained at 2T_(0) and outside the ball is T_(0). If temperature at distance (3r_(0))/(2) from centre is (pT_(0))/(3), then find the value of p. |
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| 9. |
Statement - A : Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is positive.Statement - B : Work done by gravitational force in lifting a bucket out of a well by means of a rope tied to the bucket is negative.Statement - C : Work done by friction on a body sliding down an inclined plane is negative. Statement - D : Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is positive.Statement - E : Work done by the resistive force of air on a vibrationg pendulum in bringing it to rest is negative. |
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Answer» A, B, E are TRUE |
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| 10. |
A certain quantity of vapour of a liquid mixed with air is containe din a vessel of constant volume. The pressure shown at 20^(@)C and 80cm of mercury and at 40^(@)C it is 100cm. The vapour pressure of the liquid at 20^(@)C is 15cm of mercury. Calculate the same at 40^(@)C. |
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| 11. |
An air bubble comes from the bottom to the surface of a lake of depth 2.5 m. The surface temperature of the lake is 40^(0)C. The diameter of the bubble at the bottom a,id at the surface are 3. 6 mm and 4 ,nni respectively. Find the temperature of the lake at tlie bottom |
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Answer» <P> Solution :let `P_(1), V_(1) , T_(1) and P_(2) , V_(2) , T_(2)` are the parameters of the air BUBBLE at the bottom and SURFACE of the lake respectively,`P_(1) = P_(0) +h rho g = (10^(5) + 2.5 xx 10^(3) (10)) N//m^(2)` `P_(2) = 10^(5) N//m^(2) "" V_(1) = (4)/(3) pi r_(1)^(3), V_(2)= (4)/(3) pi r_(2)^(3),` `T_(2) = 273 + 40 = 314 `K From the equation , `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` `T_(1) = (P_(1))/(P_(2)) xx (V_(1))/(V_(2)) xx T_(2)` `T_(1) = ((10^(5) + 0.25 xx 10^(5)) xx cancel((4)/(3))cancel(pi)r_(1)^(3))/((10^(5)) (cancel((4)/(3))cancel(pi)r_(2)^(3))) (313)` `1.242 xx 0.7288 xx 313 = 283.3 K ` Temperature at the bottom of the lake = 283.3- 273 = `10.3^(0)` C |
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| 12. |
A farsighted woman has a near point of 1.5m. Calculate the focal length of the lenses for her eyeglasses so that she can read a book held at 25cm. Also find power of the lens. |
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Answer» `f=+20cm and P=+4.4D` |
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| 13. |
IF the telescope is reversed i.e., seen from the objective side |
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Answer» OBJECT will appear very small |
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| 14. |
A spring 40 mm long is stretched by the application of a force . If 10 N force required to stretchthe springthrough40 mm is …….. |
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Answer» 23 J stretching in the springx = 1 mm = `0.001 ` m spring constant `k = F/x =10/(1xx10^(-3)) = 10^(4)` Now springis stretching through a distance `x_(1) = 40 mm = 0.04 m ` The force requiredto stretch is through `x_(1)` is `F_(1) =kx_(1)` ` :. ` The work done by the force `W = 1/2 kx_(1)^(2) =1/2 kx_(1) xxx_(1)` ` = 400/2 xx0.004` = 8 J |
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| 15. |
A sphere of mass 0.3kg moving with a velocity of 4m/s collides with another sphere of mass 0.5kg which is at rest. Assuming the collision to be elastic, their velocities after the impact are |
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Answer» 4m/s and `0 m/s^(-1)` |
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| 16. |
In an experiment to determine the value of acceleration due to gravity g using a simple pendulum , the measured value of lenth of the pendulum is 31.4 cm known to 1 mm accuracy and the time period for 100 oscillations of pendulum is 112.0 s known to 0.01 s accuracy . find the accuracy in determining the value of g. |
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Answer» Solution :(Accuracy is to be taken as the error involved) `l= 31.4 CM and Delta l= 1mm = 0.1cm` `T= (112.0)/(100) = 1.12s and Delta T = 0.01s` Formula for g is `g= 4PI^(2) ((l)/(T^(2)))` `(Delta g)/(g) = ((Delta l)/(l)) + 2 ((Delta T)/(T))` `=((0.1)/(31.4)) + 2 ((0.01)/(1.12)) = 0.003 + 0.02 = 0.023` RELATIVE error in determing g is 0.023 and PERCENTAGE error `= 0.023 xx 100= 2.3%` |
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| 17. |
Velocity of particle at time t=0is 2ms^(-1). A constant acceleration of 2ms^(-1)acts on the particle for 1 second at an angle of 60^(0) with its initial velocity . Find the magnitude velocity and displacement of the particle at end of t=1s. |
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| 18. |
The density versus pressure graph of one mole of an ideal monatomic gas undergoing a cyclic process is shown in figure. The molecular mass of the gas is M. (a) Find the work done in each process. (b) Find heat rejected by gas in one complete cycle. (c) Find the effiency of the cycle. |
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Answer» Process AB `rho prop p`, i.e. It is an isothermal process (T=constant), because `rho=(pM)/(RT)`. `:.` `W_(AB)=RT_AIn((p_(A))/(p_(B)))=RT_AIn(1/2)` `=-(p_0M)/(rho_0)In(2)` `DeltaU_(AB)=0` and `Q_(AB)=W_(AB)=-(p_0M)/(rho_0)In(2)` Process BC is an isobaric process (p=constant) `W_(BC)=p_B(V_(C)-V_(B))=2p_(0)((M)/(rho_(C))-(M)/(rho_B))=2p_0(M/rho_0-M/(2rho_0))=(p_0M)/(rho_0)` `DeltaU_(BC)=C_VDeltaT` `=(3/2R)[(2p_0M)/(rho_0R)-(2p_0M)/(2rho_0R)]=(3p_0M)/(2rho_0)` `Q_(BC)=W_(BC)+DeltaU_(BC)=(5p_0M)/(2rho_0)` Process CA As `rho=` constant `:.` `V=` constant So, it is an isochoric process. `W_(CA)=0` `DeltaU_(CA)=C_(V)DeltaT` `=(3/2R)(T_A-T_C)` `=(3/2R)[((p_0M)/(rho_0R)-(2p_0M)/(rho_0R)]` `=-(3p_0M)/(2rho_0)` `Q_(CA)=DeltaU_(CA)=-(3p_0M)/(2rho_0)` (b) Heat rejected by gas `=|Q_(AB)|+|Q_(CA)|` `=(p_0M)/(rho_0)[3/2+In(2)]` (c) EFFICIENCY of the CYCLE (in fraction) `eta`=" Total work done"/"Heat supplied"=W_(T otal)/(Q_(+ve))` `=((p_0M)/(rho_0)[1-In(2)])/(5/2((p_0M)/(rho_0)))` `=2/5[1-In(2)]` |
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| 19. |
What is stopping distance for vehicle ? |
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Answer» Solution :When brakes are applied to a moving vehicle, the DISTANCE it traavels before stopping is called stopping distance. Suppose, the VELOCITY of moving vehicle is `v_(0).` After applying brakes retardation : -a Distance covered : `d _(s)` (stopping distance) Final velocity `:v =0` In equation` v ^(2) -v _(0) ^(2) = 2ax` `therefore 0- v _(0) ^(2) = 2 (-a ) (d_(s)) (because v = 0, a =-a , s = d _(s))` `therefore v _(0) ^(2) = 2ad_(s)` `therefore d _(s) =(v_(0) ^(2))/(2a)` Here, a is VALUE of retardation and it is constant. Thus, stopping distance is proporational to SQUARE of the initial velocity. `therefore d_(s) prop v _(0) ^(2)` |
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| 20. |
The escape velocity of a body of mass 1kg when projected from the surface of the earth vertically is 11.2kms^(-1) .The escape velocity of a body of lgm when projected at an angle 30^@with the horizontal is |
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Answer» `(11.2)/sqrt2kms^(-1)` |
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| 21. |
A coin is placed on a horizontal platform, which undergoes vertical simple harmonic motion of angular frequency omega . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time |
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Answer» At the highest position of the PLATFORM |
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| 22. |
A man run a distance on the level road. The same man ascends up a hill with the same velocity through the same distance. When does he do more work? |
| Answer» SOLUTION :WORK DONE is more, while MOVING up a HILL. | |
| 23. |
A particle is projected vertically upwards from O with velocity v and a second particle is projected at the same instant from P ( at a heighth above O) with velocity v at an angle of projection theta. The time when the distance between them is minimum h/kv. where 'k' is |
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| 24. |
Two balls, each of m ass 80 g, moving towards each other with a velocity 5 msA^(1), collide and rebound with the sam e speed. What will be the im pulse o f force on each ball due to the oth er ? W hat is th e valu e of mom entum of each ball |
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Answer» Solution :Sopposevelocityof oneball in positiveX- directionis `vec( v)` Beforecollisionitsmomentum `vec(p )_(1)= m vec( v)_(1)` Changein MOMENTUM`Delta vec( p)= vec( p)_(2) - vec( p)_(1)` `Delta vec( p) =- 2 mv hat(j)` `m= 80g=80 xx 10^(-3) KG` `|Delta vec( p) | = 0.8 N SV = 5 ms^(-1)` Butimpulseof force`vec(F )Delta t = vec( p)` Impulseof forceis alos0.8kg `ms^(-1)` |
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| 25. |
At what positions of the simple pendulum its (i) P.E. is maximum (ii) K.E. is zero ? |
| Answer» Solution :(i) PE is maximum at the EXTREME positions (II) K.E. is zero at the extreme positions | |
| 26. |
Underline the correct alternative : Work done by a body against friction always results in a loss of its kinetic/potential energy. |
| Answer» SOLUTION :KINETIC ENERGY | |
| 27. |
A particle moves on a circle of radius r with centripetal accerleration as function of time as a_(e) = k^(2)rt^(2) where k is a positive constant. Find the resultant acceleration. |
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Answer» `KT^(2)` |
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| 28. |
Let |epsi_(0)| denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then |
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Answer» `[epsi_(0)]=[M^(-1)L^(-3)T^(2)A]` `:.[epsi_(0)]=([q_(1)][q_(2)])/([F][r^(2)]) "" [ :. 4pi` is dimensionless] `=((AT)(AT))/((M^(1)L^(1)T^(-2))(L^(2)))` `=M^(-1)L^(-3)T^(-4)A^(2)` |
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| 29. |
A system acted on by a force F moves along x-axis from x_(i) = 800 m to x_(f) = - 4.00 m. The force has only x-component whose variation with x is shown in figure , the curves are arcs of circle. Find the work done on the system. |
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Answer» Solution :As we KNOW, area under force vs. displacement curve represents the WORK done. From x = 8.00 m to x = 0.00 m force component is negative as well as displacement is along negative x-axis, therefore work done is POSITIVE. Area of semicircle `= (1)/(2) pi(-10.00) (-4.00) = 62.80 J` From x = 0 m to x = -4.00 m, force component is positive and displacement is along negative x-axis , therefore work done is negative. Area under QUARTER circle `= (1)/(4) pi(10.0) (-4.0) = - 31.40 J` The total work done is the algebraic sum of the AREAS : W = 31.40 J |
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| 30. |
(A) Optical Dangers are made of red colour ( R) Scattering of light is minimum for red colour |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 31. |
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength lambda_(B) corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 mum. If the temperature of A is 5802 K |
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Answer» The TEMPERATURE of B is 1934 K |
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| 32. |
A plot of pressure versus temperature and extrapolation of lines for low density gases |
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| 33. |
A billiards player hits a stationary ball by an identical ball to pocket the target ball in a corner pocket that is at an angle of 35 ^(@) with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important, the angle made by the target ball with repect to the incoming ball is |
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Answer» `35 ^(@)` `O = m_(1)v_(1) sin theta_(1) + m_(2)v_(2) sin theta_(2)` |
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| 34. |
In previous question discuss the case when body move downward, upwards and remainsat same position when we increases temperature. |
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Answer» Solution :LET `f =` fraction of volume of body submerged in liquid. `f =("volume of body submerged in liquid")/("TOTAL volume of body")` `f_(1) = (v_(1))/(v_(0)) at theta_(1)^(@)C` `f_(2) = (v_(2))/(v_(0)(1+3 alpha_(s)Delta theta)) at theta_(2)^(@)C` for equilibrium `mg B = v_(1) d_(1)g = v_(2)d_(2)g`. so `v_(2) = (v_(1)d_(1))/(d_(2)) :. d_(2) = (d_(1))/(1+gamma_(L)Delta theta) = v_(1) (1+ gamma_(L) Delta theta) :. f_(2) = (f_(1)(1+gamma_(L)Delta theta))/(v_(0)(1+3alpha_(s)Deltatheta))` where `Delta theta = theta_(2)- theta_(1)` Case I: Body MOVE downward if `f_(2) gt f_(1)` means `gamma_(L) gt 3 alpha_(s)` Case II: Body move upwards if `f_(2) lt f_(1)` means `gamma_(L) gt 3 alpha_(s)` Case III: Body remains at some position if `f_(2) = f_(1)` means `gamma_(L) = 3 alpha_(s)` |
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| 35. |
If there is a relative motion between a source of light and an observer, a change of colour of the light appears in the eyes of the observer. What is the name of this phenomenon ? |
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| 36. |
Calculate the fraction of a floating object of density (d_0) above the surface of a liquid of density (d) |
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Answer» `(d-d_(0))/(d)` |
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| 37. |
A ball whose density is 0.4 xx 10^(3) "kg/m"^(3) falls into water from a height of 9 cm. To what depth does the ball sink |
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Answer» 9 cm |
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| 38. |
The orbital velocity of a body close to the earth's surface is |
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Answer» `8 "kms"^(-1)` `v_(o)=SQRT(RG)=sqrt(6400xx10^(3)xx10)=sqrt(64xx10^(6))=8xx10^(3) " ms"^(-1)` `v_(o)=8 " kms"^(-1)`. |
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| 39. |
A point sized sphere of mass 'm' is suspended from a point a string a string of length 'l'. It is then pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is |
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Answer» `mlsqrt(GL)` |
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| 40. |
A cylindrical vessel contains water up to a height H. A hole is made on the wall of vessel. If the jet of water coming out from the hole traverses the maximum distance from the base of the vessel, then the hole is situated |
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Answer» at a DEPTH H from the UPPER surface of water |
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| 41. |
A rod of Young.s modulus 20 GPa undergoes a linear strain of 6xx10^(-4). Then increase in its energy density is --- |
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Answer» `3600Jm^(-3)` |
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| 42. |
Two bodies are projected at angles 30^(@) and 60^(@) to the horizontal from the ground such that the maximum heights reached by them are equal. Then (a) Their times of flight are equal (b) Their horizontal ranges are equal ( c) The ratio of their initial speeds of projection is sqrt(3):1 (d) Both take same time to reach the maximum height. Mark the answer as |
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Answer» if a, B, C and d are CORRECT |
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| 43. |
In the previous problem 3 the magnitude of the momentum transferred during the hit is . |
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Answer» <P>zero `|Deltabar(p)|=sqrt((-0.9)^(2)+(-1.2)^(2))kgms^(-1)` `=sqrt(0.81+1.44)kgms^(-1)=1.5 kgms^(-1)` |
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| 44. |
Two chambers A and B of equal volumes contain ideal gas at the same temperature but pressures 2P_(0)" and "P_(0) respectively. If B is heated and maintained at a temperature 4 times the initial and the two chambers are connected by a small pipe of negligible volume, find the equilibrium pressure. |
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| 45. |
If vec(A), vec(B), vec(C) represents the three sides of an equilateral triangle taken in the same order then find the angle between (i) vec(A) and vec(B) (ii) vec(B) and vec(C) (iii) vec(A) and vec(C). |
| Answer» Solution :From the diagram the ANGLE between the vectors `VEC(A)` and `vec(B)` is `120^(@)`, the angle between `vec(B)` and `vec(C)` is `120^(@)`, the angle between `vec(A)` and `vec(C)` is `120^(@)` | |
| 46. |
A block placed on an inclined plane making an angle 17^(@) with the horizontal slides down without any acceleration. What is theacceleration of the block if the inclimation is increased to 30^(@) ? |
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| 47. |
A ball of mass 2kg hanging from a spring oscillates with a time period 2p seconds. Ball is removed when it is in equilibrium position, the spring shortens by |
| Answer» Answer :A | |
| 48. |
An athlete in the olympic games covers a distance of 100 m in 10s. His kinetic energy can be estimated to be in the range |
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Answer» 200J - 500j |
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| 49. |
How much heat should be given to an Aluminium sphere of 200 g to carry it from 26^(@)C to 66^(@)C temperature ? What will be the heat capacity of the Aluminium sphere ? C=0.215" cal g"^(-1)C^(0-1) |
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Answer» Solution :`m=200g` `T=T_(F)-T_(i)=67-27=40^(@)C` `Q=?""H_(C)=?` `Q=mCDeltaT` `=200xx0.215xx40` `Q=1720` cal HEAT capacity of ALUMINIUM sphere `H_(C)=(Q)/(DeltaT)=(1720)/(40)` `="43 cal "^(@)C^(@-1)` |
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| 50. |
Why are circular roads banked? Deduce an expression for maximum speed of a vehicle which can be achieved while taking a turn on the banked curved road neglecting friction. |
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Answer» Solution :The breadths of circular roads are sometimes kept slanted with the horizontal. This is called banking. In a banked road, the normal reaction of a vehicle would have a horizontal component. This contributes to the centripetal FORCE of circular motion, and thus helps a vehicle to have a greater safe speed. In `theta` = banking angle, N = normal reaction, Ncos `theta` = vertical component of N, which balances the weight mg of the vehicle, Nsin`theta` = horizontal component of N, which supplies the centripetal force `(mv^(2))/(R )` for the vehicle moving with velocity v in a path of radius r. `therefore "" " Ncos"theta = " mg" "" and " Nsin " theta = (mv^(2))/(r )` Then, `("Nsin"theta)/("Ncos"theta) = ((mv^(2))//r)/(mg) ` or,tan`theta = (v^(2))/(rg) ""or, v = sqrt("rgtan"theta)` this is the maximum velocity that a vehicle may achieve in the curved path. In this treatment, we neglected the effect of friction [ However, friction plays a very important role in motions ALONG curved paths. for example, if the road is not banked , `theta` = 0 , then our formula gives v = 0 . But in practice, vehicles can turn in curved paths even in the absence of banking . in that case, the entire centripetal force is provided by friction] . 
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