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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 car tyre circumference 1m and annular diameter 10cm. Find out the volume of air to be introduced into the type at atmospheric pressure,so that the pressure inside it becomes 10 standard atmospheres. |
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| 2. |
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are poisitve or negative : (a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) Work done by gravitational force in the above case. (c ) Work done by friction on a body sliding down an iniclined plane. (d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity. (e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest. |
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Answer» Solution :Work done `W=vecF. vecs=Fs cos theta` where `theta` is smaller angle between force `VEC F` and displacement `vec s`. (a) To lift the bucket, force equal to weight of bucket has to be APPLIED VERTICALLY upwards and the bucket moves along the same direction i.e., `theta=0^(@)` therefore `W=F s cos 0^(@)=F s.` It si positive. (B) The bucket moves in a direction opposite to the gravitational force which ACTS vertically downwards. `:. theta=180^(@), W=F s cos 180^(@)=-Fs.` It is negative. (c) Friction always opposes the relative motion `:. theta =180^(@), W=F s cos 180^(@)=-Fs`. It is negative. (d) As the body moves along the direction of applied force `theta=0^(@), :. W=F s cos 0^(@)=Fs.`. It is positive. (e) The direction of resistive force is opposite to the direction of motion of the bob, i.e., `theta=180^(@).`. Hence, work done, in this case, is negative. |
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| 3. |
Graph below exhibits the anamolous expansion of water. Based on the graph, explain how lakes freeze from the top to botttom rather than from bottom to top. |
Answer» SOLUTION :
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| 4. |
A particle of mass M moves in a circular path of radius r with a constant speed equal to V. Then its centripetal acceleration is |
| Answer» Answer :A | |
| 5. |
Two shirts are warmer than a single shirt of double the thickness. Why ? |
| Answer» Solution :Between the two SHIRTS a thin layer of air is ENCLOSED. STILL air is air is a poor conductor of heat and this still kept the body warmer. | |
| 6. |
A spring of spring constant 5xx10^(3)N//m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by 5 cm is |
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Answer» 12.50 Nm |
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| 7. |
If the mass and radius of the Earth are both doubled, then the accelration due to gravity g |
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Answer» Solution :`M_(E). = 2M_(E) and R_(E).= 2R_(E)` Acceleration due to gravity `g.= (GM_(E).)/(R_(E)2)=(2GM_(E))/(4R_(E)^(2)),g.=(1)/(2)((GM_(E))/(R_(E)^(2)))(g)/(2)` |
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| 8. |
A block of mass 2kg is lying on a rough inclined plane. The force needed to move the block up the plane with uniform velocity by applying a force parallel to the plane is 100N. The force needed to move the block up with an accele-ration of 2 ms^(-2)is |
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Answer» 100N |
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| 9. |
A liquid does not wet the solid surface it the angle of contact is |
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Answer» ZERO |
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| 10. |
When a car rounds a curve to which side of the car does a passenger inside it tend to slide ? |
| Answer» Solution :The PASSENGER TENDS to SLIDE AWAY from the CENTRE of the curve. | |
| 11. |
A vertically-off shore structure is built to withstand a maximum stress of 10^9 Pa is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3km. And ignore ocean currents. |
| Answer» Solution :Pressure at the DEPTH in the SEA is about `3 times 10^7 Pa` The STRUCTURE is suitable since it can withstand far GREATER pressure or stress. | |
| 12. |
A body is moving indirectionally under the influence of a source of constant power . Its displacement in time t is proportional to |
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Answer» `t^(1/2)` ` :. 1/2 mv^(2) = Pt [ :. W = K - K_(0) " here "K_(0)=0] ` ` :. v =sqrt((2Pt)/m)` ` :. (dx)/(dt) =sqrt((2Pt)/m) ""( :. v =(dt)/dt)` Power P=Force F `xx` velocity v ` :. " P = ma " xx " at " [ :. " F = ma , v =at "]` ` :. P= ma^(2)t` ` :.a = sqrt(P/(mt))` Now, from equation (3) of motion of CONSTANT acceleration, In `X = 1/2 sqrt(P/((mt))xxt^(2)` ` :. x = 1/2 sqrt(P/m)xxt^(3/2)` ` :. x prop t^(3/2) " " [ :. " P and m are constant"]` |
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| 13. |
How can the depth of well is measured and also estimate the % of real error . |
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Answer» SOLUTION :(i) Consider a well without water, of some depth d . Take a small object ( for example lemon ) and a stopwatch . When you drop the lemon, start the stop watch . As SOON as the lemon touches the bottom of the well , stop the watch . Note the time taken by the lemon to reach the bottom and denote the time as t. (ii) Since the INITIAL velocity of lemon `u = 0` and the ACCELERATION due to gravity g is constant over the well, by using use the equations of motion for constant acceleration. `s=ut+(1)/(2)at^(2)` (iii) `u=0, s=d, a=g` (Since we choose the y axis downwards), Then `d=(1)/(2)g t^(2)` (iv) Substituting`g = 9.8 ms^(-2)`the depth of the well is obtained . (v) To estimate the error in calculation another method is used to measure the depth of the well . (vi) Take a long rope and hang the rope inside the well TILL it touches the bottom . Measure the length of the rope which is the correct depth of the well `(d_("correct") )`. Then `"error"=d_("correct")-d` `"relative error"=(d_("correct")-d)/(d)` percentage of relative error `=(d_("correct")-d)/(d_("correct"))xx100` |
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| 14. |
Four persons A,B,C and D intially at the corners of a square of side of length d. If every person starts moving with same speed v such that each one faces the other always , the person will meet after time |
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Answer» `d/V ` |
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| 15. |
A block of mass 'M' is lowered with the help of a rope of negligible mass through a distance 'd' with an acceleration of (g)/(3). Work done by the rope on the block is |
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Answer» `(2Mgd)/(3)` |
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| 16. |
If C=332 m s^(-1), say the velocity of sound in dry air at STP, find the velocity of sound in loist air at STP. Molecular weight of air =28.8, that of water vapour =18, gamma of dry air =1.4 and gamma of water vapour =1.33. The standard pressure is 750 mm of mercury and standard temperature is 0^(@)C. the vapour pressure at 0^(@)C is 4.8 mm of mercury. |
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| 17. |
During an experiment an ideal gas is found to obey an additional law VP^(2)= constant. The gas is initially at temperature T and Volume V. Find the resulting temperature when it expands to volume 2V. |
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Answer» Solution :`PV=RTrArr P=(RT)/(V)` `VP^(2)="constant"` `therefore""V((RT)/(V))^(2)="constant (or)"(R^(2)T^(2))/(V)="constant"` `(T^(2))/(V)="constant"` `therefore""(T.^(2))/(2V)=(T^(2))/(V) rArr T. = sqrt2T` |
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| 18. |
Give an example where work done by a force is zero. |
| Answer» Solution :when a BODY moves in a circular path no WORK is doen.In this casethe centripetal force acts at RIGHT angles to the direction of motion. | |
| 19. |
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m^(2). (a) What is his new angular speed? (Neglect friction.) (b) Is kinetic energy conserved in the process? If not, from where does the change come about? |
| Answer» SOLUTION :(a) 59 rev/min, (b) No, the K.E. is INCREASED and it comes from work DONE by MAN in the process. | |
| 20. |
Which of the processes described below areirreversible ? a) The increase in temprature of an iron rodby hammering it b) A gas in a small cantainer at a tempratureT_(1)is brought in contact with a big reservoir at a higher temprature T_(2)which increasesthe temprature of the gas c) A quasi-static isothermal expansion of anideal gas in cylinder fitted with a frictionlesspiston d) An ideal gas is enclosed in a piston cylinderarrangement with adiabatic walls. A weight W is added to the piston, resulting incompression of gas |
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Answer» only a, B are correct |
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| 21. |
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is |
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Answer» the MEAN POSITION |
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| 22. |
A train is moving with acceleration along a straight line with respect to ground. A person in the train finds that |
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Answer» Newton's 2nd LAW is false but newtons 3RD law is true |
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| 23. |
The sun radiates maximum energy at wavelength 4753 Å. Estimate the surface temperatureof the sun, if b = 2.888 xx 10^(-3) mK. |
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| 24. |
A wire is stretched by 0.01 m when it is stretched by certain force. Another wire of the same material but double the length and double the diameter is stretched by the same force. What is the elongation in metres ? |
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Answer» SOLUTION :`Delta l = (Fl)/(pi r^2Y)` `((Delta l)_(2))/((Delta l)_(1)) = ((l_2)/(l_1)) ((r_1)/(r_2))^(2) = 2 XX (1/2)^(2) = 1/2` `(Delta l)_(2) = 1/2 xx 0.01` `(DELTAL)_(2) = 0.005 m`. |
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| 25. |
The temperature of air in closed space is observed to be 15^(@)C and the dew point is 8^(@)C. If the temperature falls to 10^(@)C how will the dew point be affected? (Saturated vapour pressure at 7^(@)C=7.49mm of mercury and that at 8^(@)C=8.02mm) |
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| 26. |
In summer, a mild wind is often found on the shore of a calm river. This is accused due to |
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Answer» DIFFERENCE in thermal CONDUCTIVITY of water and SOIL |
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| 27. |
Two balls of masses 5m and m have radii 2R and R. Their centre of masses are separated by 12R. They move towards each other under their gravitational force. The distance moved by the centre of smaller sphere when the spheres touch each other is |
| Answer» ANSWER :C | |
| 28. |
A body moving along a straight line travels one third of the total distance with a speed of 3.0 m s^(-1). The remaining distance is covered with a speed of 4.0 m s^(-1) for half the time and 5.0 m s^(-1) for the other half of the time. The average speed during the motion is |
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Answer» `4.0 m s^(-1)` `t_1 = (S/3) / 3 = S/9` Let `t_2` be the time taken for each of the remaining two journeys. Then `S-S/3 = 4t_2 + 5t_2` or `(2S)/3 = 9t_2`or`t_2 = (2S)/27` Average speed = Total distance travelled / Total time taken `= S/t_1 + 2t_2` `= (S)/(S/9) + 2xx(2S)/27 = 3.8 m s^(-1)` |
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| 29. |
A spherical body of diametere D is falling in viscous medium. Its terminal velocityis proportional to |
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Answer» `V_(i) PROP D^(1/2)` |
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| 30. |
The mass of a uniform ladder of length 6 m is 20 kg. A mass 60 kg is placed on the ladder at a height of 2 m from the bottom. The position of centre of mass from the bottom is |
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Answer» `1.5 m` |
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| 31. |
If the law of gravitational instead of being inversesquare law, becomes an inverse cube law |
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Answer» Planets will not have elliptic orbits and circular orbits of planets is not possible. |
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| 32. |
The angle between force vecF =(3hati+4hatj-5hatk) unit and displacement vecd = (5hati+4hatj+3hatk) unit is |
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Answer» `cos^(-1) (0.16)` `vecd=5hati+4hatj+3hatk` unit `therefore |vecF|=sqrt((3)^2 +(4)^2 + (-5)^(2))=sqrt50` unit `|vecd|=sqrt((5)^2+(4)^2+(3)^2)=sqrt50` unit Let `theta` be angle between `vecF` and `vecd` `therefore cos theta=(vecF.vecd)/(|vecF||vecd|)=((3hati+4hatj-5hatk).(5hati+4hatj+3hatk))/(sqrt50sqrt50)` `=(15+16-15)/50=16/50 =0.32 "or" theta=cos^(-1) (0.32)` |
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| 33. |
A simple harmonic oscillation is represented by the equation y= 0.4 sin((440t)/(7)+0.61x)Where 'Y' and 'X' are in m and time 't' in seconds respectively. The value of time period in seconds |
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Answer» 0.1 |
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| 34. |
A steel rod of cross-sectional area1m^(2) is acted upon by forces shown in the fig. Determinethe totalelongation of the bar. Take Y=2.0xx10^(11)N//m^(2). |
Answer» Solution :The ACTIONOF forces on eachpartof rod is shown in fig.  We know that the extension DUE to external force F is given by `Delta_(AB)=(Fl)/(AY)` `Deltal_(AB)=((60XX10^(3))XX1.5)/(1xx2xx10^(11)) =4.5xx10^(-7)m` `Deltal_(BC)=((70xx10^(3))xx1)/( 1xx2xx10^(11)) =3.5xx10^(-7)m` and `Deltal_(CD)=((50xx10^(3))xx2)/(1xx2xx10^(11))=5.0xx10^(-7)m` The total extension `Delta=Delta_(AB)+Delta_(BC)+Delta_(CD)` `=4.5xx10^(-7)+3.5 xx 10^(-7)+5.0xx10^(-7)=13xx10^(-7)m` |
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| 35. |
A drop of liquid of density rho is floating half-immersed in a liquid of density d. If sigma is the surface tension, then what is the diameter of the drop of the liquid ? |
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| 36. |
Constant temperature of ends are T_(1) and T_(2) (where T_(1)gtT_(2)) for a conducting rod of 'L' length and cross-sectional area A. What would be the rate of heat transfer in thermal steady state ? |
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Answer» `(dQ)/(dt)=(KA(T_(1)-T_(2)))/(L)` |
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| 37. |
A source of sound attached to the bob of a simple pendulum execute SHM . The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean frequency of the source . The velocity of the source at the mean position is ( velocity of sound in the air is 340 m//s) [Assume velocity of sound lt lt velocity of sound in air ] |
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Answer» `1.4 m//s` `f_(1) - f_(2) = f_(0) V_(0) ((1)/( V_(0) - V) - (1)/( V_(0) + V))` ` = f_(0) V_(0) (V_(0) + V - V_(0) + V)/( V_(0)^(2) - V^(2)) = f_(0) V_(0) xx ( 2V)/( V_(0)^(2)) = f_(0) ( 2V)/( V_(0))` given `( 2 Vf_(0))/(V_(0)) = 0.02 xx f_(0) rArr V = 0.01 V_(0) = 3.4 m//s`. |
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| 38. |
The physical quantity having the same dimensional formula as that of entropy is |
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Answer» LATENT HEAT |
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| 39. |
If the critical angle for the material of a prism is C and the angle of the prism is A, then there will be no emergent ray when |
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Answer» after deviation it will emerge from the second refacting surface |
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| 40. |
A person of mass 72 kg sitting on ice pushes a block of mass of 30kg on ice horizontally with a speed of 12 ms^(-1). The coefficient of friction between the man and ice and between block and ice in 0.02. If g=10 ms^(-2), the distances between man and the block, when they come to rest is |
| Answer» ANSWER :D | |
| 41. |
The velocity of air over the upper surface of the wing of an aeroplane is 40m/s and that on the lower surface is 30m/s. If the area of the wing is 5m^(2) and the mass of the wing is 300kg, the net force acting on the wing is (Density of air = 1.3kg//m^(3) and g=10m//s^(2)) |
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Answer» 725N UPWARD |
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| 42. |
A body of mass 0.3kg is taken up an inclined plane of length 10m and height 5m, and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the (a) work done by the gravitational force over the round trip. (b) work done by the applied force over the upward journey (c ) work done by frictional force over the round trip. (d) kinetic energy of the body at the end of the trip. How is the answer (d) related to the first three answers ? m=0.3kg, OA=10m, BA=5m |
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Answer» Solution :(a) Gravitational force isconservating force.So work DONE by it for the round trip is ZERO. (b) From FIG.  `OA=10m`, `BA=5m` `OB=sqrt(OA^(2)-BA^(2))` `=sqrt(100-25)` `=sqrt(75)=5sqrt(3)` `costheta=(OB)/(OA)=(5sqrt(3))/(10)=(sqrt(3))/(2)` `sintheta=(BA)/(OA)=(5)/(10)=(1)/(2)` Frictional force `=muN=mu mg costheta` For the upward journey force to be applied `=F=mgsintheta+mu mg costheta` Displacement `s=10m` Work done `=(mgsintheta+mu mg costheta)s` `=mgs(sintheta+mu costheta)` `=0.3xx9.8xx10(0.5+0.15xxsqrt(3)//2)` `=18.52J` (c ) Distance travelled during the round trip `=10+10=20m` Work done by force of friction `=-mu mg costhetaxx20` `=-0.15xx0.3xx9.8xxsqrt(3)//2xx20` `=-7.638J` -ve sign shows that work is done against friction. (d) For the downward motion from A to O.  `u=0`, `v=?` Net acceleration `=a=gsintheta-mu g costheta` `=g(sintheta-mucostheta)` `=9.8(0.5-0.15xxsqrt(3)//2)` `=3.627ms^(-2)` `s=10m` `v^(2)=u^(2)+2as` `=0+2xx3.627xx10` `v^(2)=72.54` K.E.`=1//2mv^(2)` `=1//2xx0.3xx72.54` `=10.88J` From (b) and (c ) Net work done `=18.52-7.64=10.88J` from (d) K.E. `=10.88J` THUS net work done `=` Change in K.E. |
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| 43. |
A siren emitting a sound of frequency 1000 vibrations per second is moving with a speed of 10 ms^(-1).What will be the frequency of sound, which a listener will hear when (i)the siren is moving towards him ? (ii)the siren is moving away from him ? Speed of sound in air =340 ms^(-1). |
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| 44. |
These questions have statement I and statement II. Of the four choices given below, choose the one that best describes the two statements. Statement I : Moment of inertia of a circular ring about a given axis is more than moment of inertia of a circular disc of same mass and same size, about the same axis. Statement II : The circular ring is hollow, so its moment of inertia is morethan circular disc which is solid. |
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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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| 45. |
Several spherical drops of a liquid of density 'd' and radius 'r' coalesce to form a single drop of radius 'R'. If all the energy released is converted into K.E. Then the velocity acquired by the drop is |
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Answer» `SQRT((6T)/( d) ((1)/( R) - (1)/( R))) ` |
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| 46. |
At one end of a weightless rope of length 10 m, a car of massM is tied. The other end of the rope is in the hands of a man of mass M. The arrangement stands on a horizontal frictionless surface. The man is at x=0 and the car is at x =10 m. As the man pulls the car with the rope locate the point where they meet. |
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| 47. |
Discuss the simple pendulum in detail. |
Answer» Solution :Construction : A pendulum is mechanical system which exhibits periodic motion. It has a bob with mass m suspended by massless and inextensible string. The other end is fixed on a stand as shownin Figure. (a). At equilibrium, the pendulum doesnot oscillate and is suspended vertically downward. Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released. The bob of the pendulm executes to and fro motion.  Caclulation of time period. Let l be the length of thependulum which is taken as the distance between the point of suspension and the CENTRE of gravity of the job. TWO forces act on the bob of the pendulum at any displaced position, as shown in the figure (d), (i) The gravitational force acting on the body. `(VECF= mvecg)` that acts vertically donwards. (ii) The tensionin the string `vecT` that acts along the string to the point of suspension. Resolving the gravitational force into its components : (i) Normal COMPONENT : It is along the string but in opposition to the direction of tension `F_(as) = mg cos theta`. (ii) Tangential component : It is perpendicular to the string i.e., along tangential direction of are of swing `F_(ps)= mg sin theta`. Hence , the normal compoent of the forece, is along the string, `T-W_(as)=m(v^(2))/(l)` Here v is speed of bob `T- mg cos theta=m (v^(2))/(l) ""...(1)` From the figure, it is observed that the tengential component `W_(ps)` of the gravitationalforce always points towards the equilibrium position. This direction always points the bob from the mean position. Hence, in this case, the tangential force is the restoring force. Applying Newton's second law along tangential directiojn, we have `m(d^(2)s)/(dt^(2))+F_(ps)=0 rArr m=(d^(2)s)/(dt^(2))=-F_(ps)` `m(d^(2)s)/(dt^(2))=-mg sin theta ""...(2)` Where s is the position of bob that is measured along the ARC. Expressing arc length in terms angular displacement i.e. `s= l theta ""......(3)` Then its acceleration, `(d^(2)s)/(dt^(2))=l (d^(2) theta)/(dt^(2)) ""....(4)` Substituting equation (4) in equations (2), we get, `l(d^(2)theta)/(dt^(2))=-g sin theta""...(5)` Because of the presence of `sin theta` in the above differential equation, it is a non-linear differential equation. it is assumed that '' the small oscillation approximation,'' `sin theta ~~ theta`, the above differential equation becomes linear differential equation. `(d^(2)theta)/(dt^(2))=-(g)/(l)theta` It is known as oscillatory differential equation. Hence, the angular frequency of this oscillator (natural frequency of this system ) is `omega^(2)=(g)/(l) ""....(6)` `rArr omega= sqrt((g)/(l)" in rad" s^(-1) ""...(7)` The frequecny of oscillations is `f=(1)/(2pi) sqrt((g)/(l)) "in" Hz ""....(8)` and time period of oscillations is `T=2pi sqrt((l)/(g))` in second `"".....(9)` |
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| 48. |
Find the magnifying power of a compound microscope whose objective has a focal power of 100D and eyepiece has a focal power of 16D when the object is placed at a distance of 1.1cm, from the objective. Assume that the final image is formed at the least distance of distinct vision, 25cm. |
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Answer» SOLUTION :The magnifying power of a COMPOUND microscope when the final image forms at the least distance of distinct vision `m=v_0/u [1+D/f_e]` To find `v_0`, , Power of the objective `p_0=100D` Focal length of the objective `f=f_0=1/p_0=1/100 m=100/100 cm=1cm` `u=u_0=-1.1cm , v=v_0=?` FOra LENS `1/f=1/v-1/u` `1/1=1/v_0-1/(1.1),1/v_0=1/1-1/1.1=0.1/1.1, v_0=1.1cm` Power of the eyepiece `p_e=16D` focal length of the eyepiece `f_e=1/p_e=1/16m=100/16 cm=6.25 cm`, Least distance of distinct vision `D=25cm` `therefore m=11/(-1.1) (1+25/6.25)=-10 times 5=-50` |
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| 49. |
If the mass of the earth is M, its radius is R and gravitational constant G, then the speed required to be given to a projectile so that it does not return to earth is |
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Answer» `sqrt((GM)/R)` |
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| 50. |
The efficiency of carnot engine working between temperatures 200^@ C and 0^@ C is eta_1. The efficiency of this engine is eta_2 when it work between temperatures 0^@ C and -200^@ C. Then eta_1/eta_2=_____ |
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Answer» `1.0` `T_2`=0+273=273 K `T._1`=0+273 =273 K `T._2`=-200+273 =73 K `eta_1=200/473` `eta_2=200/273` `therefore eta_1/eta_2=273/473`=0.577 |
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