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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. |
Why a spring balance does not give correct measurements when it has been used for a long time? |
| Answer» SOLUTION :This is DUE to ELASTIC FATIGUE. | |
| 2. |
The spectrum of a black body at two temperature 127°C and 527°C . Let A_1 and A_2 be the areas under the two curves respectively. The value of (A_2)/( A_1) is |
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Answer» `1:16` |
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| 4. |
A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 "J/kg-K" and 2100 "J/kg-K" respectively. Density of K-oil = 800 "kg//m^3. |
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Answer» 5 min |
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| 5. |
Three uniform circular discs, each of mass 0.5kg and radius 10cm, are kept in contact with each other as shown in the figure. Find the moment of inertia of the system when it rotates about the axis AB. |
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Answer» SOLUTION :`I=I_(1)+I_(2)+I_(3)` `I=(5MR^(2))/(4)+(5MR^(2))/(4)+(MR^(2))/(4)impliesI=(11MR^(2))/(4)` `M=0.5kg`, `r=10cm=10xx10^(-2)m` `:.` MOMENT of INERTIA of the system `I=(11MR^(2))/(4)=(11xx0.5xx(0.1)^(2))/(4)` `=1.375xx10^(-2)kgm^(2)` 
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| 7. |
The resultant ot two forces acting at an angle of15^@ ii10 N and is peroendicular to one of the forces. The other force is . |
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Answer» `20 // SQRT 2 N` ` = A^2 +b^2 - 2 AB ( sqrt 3 //2)` …(i) ` tan 90^@ = ( B sin 150^@)/( A+ B cos ` 15^@) =0` or ` A+B cos 1 50^@ =0` or ` A= B xxsqrt 3//3` …(II) ` From (i), ` 100 = 3/ 4 B^2 + B^@ - ( 3 B^2)/2 = B^2/4` ` or ` B = 20 N`. |
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| 8. |
Distinguish between centre of mass and centre of gravity. |
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Answer» Solution :(i) The center of mass of a body is DEFINED as a point where the entire mass of the body APPEARS to be concentrated. (ii) The center of GRAVITY of a body is the point at which the entire WEIGHT of the body acts irrespective of the POSITION and orientation of the body. |
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| 9. |
The volocity v of the particle at time t is given by v = at + b/(t + c), where a, b and c are constants. The dimensions of a,b, and c are respectively |
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Answer» `LT^(-2), L and T` |
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| 10. |
A sheet of rubber and a sheet of cardboard, each 5 mm thick are pressed together and their outer faces are maintained respectively at 0°C and 18°C. If the thermal conductivities of rubber and cardboard are respectively 0.13 and 0.05 "Wm"^(-1) "K"^(-1). find the quantity of heat which flows in one hour across a piece of composite sheer if area is 100 "cm"^(-2). |
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Answer» |
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| 11. |
The surface tension of water is 72dyne cm^(-1). Its value in S.I system is |
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Answer» 72000 `NM^(-1)` |
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| 12. |
A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 min. The angular retardation in rad s^(-2) is |
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Answer» `(pi)/(2)` Now, `0=omega_(0)-alphat` or Angular retardation, `alpha=(omega_(0))/(t)=(30pi)/(60)"rad s"^(-2) =(pi)/(2) "rad s"^(-2)`. |
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| 13. |
The value of 'g' at a particle point on the surface of the earth is 9.8 ms^(-2). Suppose the earth suddenly shrinks uniformly to half its presence size without losing any mass, the value of g at the same point (assuming that the distance of the point from the centre of the earth does not shrink ) will now be |
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Answer» `4.9 MS^(-2)` |
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| 14. |
Two engines pass each other in opposite directions . One of them blows a whistle of frequency 540 Hz . Find the freqauencies heard by a passenger sitting on the other engine before and after passing each other . Velocity of both engines = 72 kmh^(-1), velocity of sound = 340 m.s^(-1) . |
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Answer» Solution :Velocity of the engine , ` u = 72 km h^(-1) = (72 xx100)/(60xx60 ) m. S^(-1) = 20M s^(-1)` Velocity of SOUND , V = 340 m ` S^(-1)` , Actual frequency , n = 540 Hz . At the TIME of approching each other , velocity of the WHISTLE ` u_(s) = + 20 m S^(-1)` Velocity of the passenger on the other engine ` u_(0) = + 20 m S^(-1)` ` therefore ` Apparent frequency , ` n. (V + u_(0))/(V + u_(s)) xx n ` ` = (340 + 20 )/(340 - 20) xx 540 = (360)/(320 ) xx 540 = 607 .5 Hz ` Again at the time of receding from each other , ` u_(s) = - 20 m S^(-1)" and " u_(o) = - 20S^(-1)` ` therefore ` Apperent frequency , ` n . = (V + u_(0))/(V- u_(s)) xx n ` ` = (340-20)/(340 - (-20)) xx 540 = (320)/(360 ) xx 540 = 480 Hz ` . |
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| 15. |
Statement A : Energy is the cause for doing work and work is the effectStatement B : Energy is a scalar but power is a vector quantity. |
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Answer» A & B is correct |
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| 16. |
A cylinder of fixed capacity 44.8litrescontainshelium gas at standard pressureat temperature . What is the amount of heat need to rest that temperature of the gas by 15.00^@C ? [ R = 8.31 " J mol"^(-1)K^(-1)] |
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Answer» Solution :Volume of 1 mole of He at at S.T.P = 22.4 letres Total volume of He at S.T.P = 44.8 litres `:.` No of mole of He `n= (44.8)/(22.4)=2` MOLAR SPECIFIC heat of He (monoatomic gas) at CONSTANT volume , `C_V=3/2R=3/2xx8.31J mol^(-1) K^(-1)` `:. DELTAT = 15^@C = 15K` Here required `Q = nC_VDeltaT` `=2xx3/2xx8.31xx15` `=373.95J` |
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| 17. |
N(lt100) molecules of a gas have velociteis 1,2,3……N km/s respectively.Then |
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Answer» RMS SPEED and average speed of molecules are same |
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| 18. |
Two solid spheres of the same mass are made of metals of different densities. Which of them has larger moment of inertia about the diameter ? |
| Answer» Solution :The SPHERE with SMALL density will have larger RADIUS and hence LARGE MOMENT of inertia. | |
| 19. |
Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth horizontal plane and is released, it moves with an acceleration of 10 m//s^(2). In this box if 1 kg body is supended with a light string, find the tension in the string in equilibrium position (w.r.t experi-menter) (g=10 m//s^(2)) |
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Answer» SOLUTION :Acceleration of BOX `= 10 m//s^(2)` `T = SQRT((mg)^(2)+(ma)^(2))` `T = sqrt(10^(2)+10^(2))T=10 sqrt(2)N` 
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| 20. |
A piece of wood of density 0.8 gmcmfloats in a liquid of density 0.9 gmcm^(-3)at 0^(@)C. At what temperature does the piece of wood sink in the liquid (gamma _(R ) = 8.5 xx 10 ^(-5) //^(@)C and coefficient of expansion of liquid Y_(R ) = 1.2 xx 10 ^(-3) //^(@)C) |
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Answer» Solution :Let the piece of wood sink in the liquid at `t ^(@)C.` From `d _(0) = d _(t) [1 + gamma _(R) t ]` Density of the liquid at `t ^(@) C: d _(0)` (liquid) `= d _(t) ` (liquid) `[1 + gamma _(R) t ] (or d _(t)` (liquid) `= (d _(0) ("liquid"))/(1 + gamma _(R) t ) and d _(t) ` (wood) ` = (d_(0) ("wood"))/( 1 + gamma _(R) t ) ,` For the wood to sink, Density of liquid at `t ^(@)C=` Density of wood `t ^(@)C` `(d _(0) ("liquid"))/( 1 + gamma _(R) t ) = (d _(0) ("wood"))/( 1 + gamma _(OMEGA t )) = (0.9 )/( 1 + 1.2 xx 10 ^(-3) t) = (0.8)/( 1 + 8.5 xx 10 ^(-5) t ) = 0.9 xx 8.5 xx 10 ^(-5) t = 0.8 + 0.8 xx 1.2 xx 10 ^(-3) t` `0.9 + 0.000 765t = 0.8 + 0.00096t , t[ 0.0096-0.000765] =-.9 -0.8 , t = (0.1)/( 0.008835) = 113.19^(@)C` |
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| 21. |
A coin moves in a circular path in a rough rotating horizontal surface which has an angular acceleration alpha |
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Answer» power delivered by the friction on the coin is positive. |
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| 22. |
A ring of radius R is first rotated with an angular velocity omega_(0) and then carefully placed on a rough horizontal surface. The coefficient of friction between the durface and the ring is mu. Time after which its sngular speed is reduced to half is (omega_(0)R)/(xmug) where 'x' is |
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Answer» |
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| 23. |
Two particles A and B initially at rest move towards each other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2v, the speed of the centre of mass of the system is |
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Answer» 0 |
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| 24. |
A car starting from rest, accelerates at the rate of f through a distance S, then continues at constant speed for time t and declerates at the rate f//2 to come to rest. If the total distance travelled is 15 S then S = |
Answer» Solution : `AB=S=(1)/(2)ft_(1)^(2)......(1)v=ft_(1)` `BC=vt=(ft_(1))t......(2)` `CD=(u^(2))/(2a)=((ft_(1))^(2))/(2(F//2))=2S......(3)` `S=ft_(1)t+2S=15S` `ft_(1)t=12S......(4)` Dividing (4) by (1) `t_(1)=(t)/(6)` Now (1) gives `S=(1)/(2)f""(t^(2))/(36) THEREFORE S=(ft^(2))/(72)` |
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| 25. |
A ring of radius R is first rotated with an angular velocity omega_(0) and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is mu. If the time after which its angular speed is reduced to half is (w_(0)R)/(xmug) then find x. |
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Answer» |
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| 26. |
An electron moves round the nucleus of an atom, in a circular path. What is the direction of the angular momentum of the electron ? |
| Answer» SOLUTION :PERPENDICULAR to the PLANE of the ORBIT. | |
| 27. |
An electron of mass 9 xx 10^(-31) kg revolves around a nucleus in a circular orbit of radius 0.53 Å. What is the angular momentum of the electron? (Velocity of electron is, v=2.2 xx 10^(6) ms^(-1)). |
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Answer» Solution :Mass of the electron, `m=9xx10^(-31)kg` Radius of the electron, `R=0.53 Åxx10^(-10)m` VELOCITY of the electron, `v=2.2xx10^(6)ms^(-1)` ngular momentum of ELECTRONIS, `L= I omega` Electron is CONSIDERED as a point mass. Hence, its moment of inertia is, `I -mr^(2)` The relation, `omega=(v)/(r)` could be used. Angular momentum,`L=mr^(2)xx(v)/(r)=mvr` `=9.1xx10^(-31)xx2.2xx10^(6)xx0.53xx10^(-10)` `L=1.06xx10^(-34) kg m^(2)s^(-1)` `=9.1xx10^(31)xx2.2xx10^(6)xx0.53xx10^(-10)` `L=1.06xx10^(-34) kgm^(2)s^(-1)` |
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| 28. |
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surface of the wing are 70 ms^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5 m^2? Take the density of air to be 1.3 kg m^-3. |
| Answer» SOLUTION :`1.5 TIMES 10^3 N` | |
| 29. |
Satement I:A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. Statement II: For every action there is an equal and opposite reaction. |
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Answer» Statement I is true, statement II is true, statement II is a CORRECT EXPLANATION for statement I. |
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| 30. |
A Carnot engine takes up 200 cal of heat per cycle from a source at temperature 500 K and rejects 150 cal of heat to the sink. What is the temperature of the sink and the efficiency of the engine? |
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Answer» SOLUTION :`(Q_1)/(Q_2) = (T_1)/(T_2) or, T_(2) = T_(1) cdot (Q_2)/(Q_1) = 500 xx (150)/(200) = 375 K` EFFICIENCY, `ETA = 1 - (Q_2)/(Q_1)` `=1 - (T_2)/(T_1) = 1 - (375)/(500) = 1 - 3/4 = 1/4` ` = 1/4 xx 100% = 25%`. |
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| 31. |
Velocity of boat in still water is 13m/s. If water flows in a river with a velocity of 5 m/s. What is the difference in times takes by him to cross the river in the shortest path and in the shortest time. If the width of the river is 156 m. |
| Answer» Answer :A | |
| 32. |
Give the vector form of the relation between a and alpha |
| Answer» Solution : Vector RELATION between ACCELERATION a and angular acceleration is `ALPHA` is `bara=barrxxbaralpha` | |
| 33. |
. (A): Mountain roads rarely go straight up the slope. (R): Slope of mountains are large therefore more chances of vehicle to slip from roads. |
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Answer» Both .A. and .R. are true and .R. is the correct EXPLANATION of .A. |
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| 34. |
1m equal to how many astronomical unit? |
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Answer» Solution :`1.668xx10^(-11) AU` `1.496xx10^(11)m=1AU` then 1m=(?) `(1)/(1.496xx10^(11))=0.6684xx10^(-11)` `:.1m=6.68xx10^(-12)AU` SECOND METHOD : `(1m)/(1AU)=(1)/(1.496xx10^(11))` `:.1m=0.6684xx10^(-11)AU` `:.1m=6.684xx10^(-12)AU` |
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| 35. |
The temperature of an ideal gas is increased from 120 K to 480 K.If at 120 K the rms speed is v,at 480 K it becomes |
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Answer» 4v |
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| 36. |
Compare the components for the following vector equations (a) vecF=mveca Here m is a positive number (b) vecp=0 |
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Answer» Solution :`vecF=mveca` `F_(x)hati+F_(y)hatj+F_(z)hatk=ma_(x)hati+ma_(y)hatj+ma_(z)hatk` By comparing the COMPONENTS, we get `F_(x)=ma_(x),F_(y)=ma_(y),F_(z)=ma_(z)` This implies that one vector EQUATION is equivalent to three scalar EQUATIONS. Case (B): `vecP=0` `p_(x)hati+p_(y)hatj+p_(z)hatk=0hati+0hatj+0hatk` By comparing the components, we get `p_(x)=0,p_(y)=0,p_(z)=0` |
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| 37. |
Three rods of same length and area of cross-section but of thermal conductivities K, 2K and 3K are connected in series in the same order. Free end of the first rod is at 0^(0)C and free end of third rod is at 55^(0)C.In steady state, temperature difference across the middle rod is |
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Answer» `15^(0)C` |
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| 38. |
The net force acting on a body when it is at rest is |
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Answer» gravtitational FORCE |
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| 39. |
At what circumstances weightlessness arise? |
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Answer» Solution :(i) In a freely FALLING lift (II) Inside a spacecraft (III) ATTHE CENTRE of the Earth. |
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| 40. |
How can the phase difference of two particles executing simple harmonic motions ? |
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Answer» Solution :Consider two particles EXECUTING SIMPLE harmonic motions. Their equations for DISPLACEMENT are `Y_(1)=A sin (OMEGA t+phi_(1)) and Y_(2) =A sin (omega t+phi_(2))` then the phase difference is given by `DELTA phi` `Delta phi=(omega t+phi_(2))-(omega t+phi_(1))` `=phi_(2)=phi_(1)` |
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| 41. |
A lift is moving down with an acceleration equal to the acceleration due to gravity. A body of mass M kept on the floor of the lift is pulled horizontally. If the coefficient of friction is muthen the frictional resistance offered by the body is: |
| Answer» Answer :C | |
| 42. |
A ball is dropped on a fixed horizontal plane from a certain height. After recoil from the plane, it rises to a lower height. The correct nature of the height-time graph is |
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Answer»
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| 43. |
What is the effect of gravitational force of attraction acting on the person be indise the satellite and stand on moon? |
| Answer» Solution :The gravititanol force of attraction of the Earth on person inside the satellite PROVIDES the centripetal force necceary to move in an ORBIT. A person standing on the moon POSSESSES WEIGHT DUE to the additional gravitational pull of the moon on the person. | |
| 44. |
What is the total work done in a cyclic process ? |
| Answer» SOLUTION :In CYCLIC process, area of the close curve represents the WORK DONE. | |
| 45. |
Assertion: The skill in swinging to greater heights lies in the synchronisation of the rhythm of pushing against the ground with the natural frequency of the swing. Reason: The phenomenon behind this is resonance. |
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Answer» If both ASSERTION and RESON are true and reason is the correct explanation of assertion |
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| 46. |
A clock regulated by u seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.01 m How much will the clock gain or lose in one day ? |
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Answer» Solution :The time PERIOD of a simple pendulum is `T= 2pi sqrt((l)/(g))= (2pi)/(sqrt(g))(l^(1//2))` differentiating T w.r.t. l and dividing T on both slides we GET `(dT)/(T)= (1)/(2)(dl)/(l)` In CASE of a seconds pendulum , T= 2s and `l= (8)/(pi^(2))= 0.9927` Also dl= 1.01-0.9927= 0.0173m. `:. (dT)/(2)= (1)/(2)((0.0173)/(0.9927))` i.e the change in time period or change in time for oscillation , `dT= (0.0173)/(0.9927)s` `dT= (1)/(2)(dl)/(l)T=(1)/(2)(dl)/(l)86,400` per day `= (1)/(2) (0.0173)/(0.9927) xx 86400= 752.9s` When the lenght INCREASE the time period of the pendulum increase. So, the clock loses time or runs slow. (Note: when the lenght decrease the time period of the pendulum decrease and there will be gain of time or the clock FAST.) |
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| 47. |
How many wavelengths ofKr_(86) are there in one metre? |
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Answer» 2348123.73 |
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| 48. |
Assertion (A) : When a vehicle takes a turn on the road, it travels along a curved path. Reason (R): In a curved path, Velocity of vehicle remains same |
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Answer» Both (A) and (B) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 49. |
An Aluminium wire and a steel wire of the same length and cross section are joined end to end. The composite wire is hung from a rigid support and a load issuspended from the free end. The young's modulus of steel 20/7 times that of the Aluminium. The ratio of increase of length of steel to Aluminium is |
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Answer» `20//7` |
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| 50. |
An aluminium wire and a steel wire of the same length and cross-section are joined end- to-end. The composite wire is hung from a rigid support and a load is suspen-ded from the free end. If the increase in lengthe of the composite wire is 1.35 mm, find the ratio the (i) stress in the two wires amd (ii) strain in the two wires. (Y_(Al)=0.7xx10^(!1)Nm^(-2),Y_("steel")=2=10^(11)Nm^(-2)) |
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Answer» `1:1, 20:7` |
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