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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. |
Define: (i) stopping distance of a vehicle (ii) reaction time during a free fall. |
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Answer» Solution :(i) Let at a certain moment the brakes are applied on a vehile running on a road. If the vehicle then travels a distance x until it comes to rest then x is called the stopping distance of the vehicle. (II) Many natural events demand immediate ACTION from the observers. An observer an event, registers it in his BRAIN and then takes the required action. The time spent between the first observation and the action taken is called his/her reaction time. For EXAMPLE suppose an observer sees an object falling freely from the ceilling of a room towards a glass plate kept on a table. He/she would try either to intercept the falling object or to remove the plate from the table. In this CONTEXT if the reaction time is greater than the time of fall, then the glass would break. |
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| 2. |
Hailstones fall from a certain height. If only 1% of the hailstones melt on reaching the ground, find the height from which they fall. (g=10ms^(-2).L=80 "calorie"//g and J=4.2//"calorie") |
| Answer» ANSWER :A | |
| 3. |
Two particles 1 and 2 are allowed to descend on the two frictionless chord OA and OB of a vetical circle , at the same instant from point O. The ratio of the velocities of the particles 1 and 2 respectively , when they reach on the circumference will be (OB is the diameter) |
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Answer» Solution :`OA= d cos alpha _(OA)=G cos alpha ` ALONG OA `rArrV_(A)^(2)=2g cos alpha . D cos alpha ` Along OB `V_(B)^(2) =2gdrArr (V_(A))/(V_(B))= cos alpha` 
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| 4. |
Using Newton's law of gravitation, prove Kepler's III^(rd) law of planetary motion for circular orbits. |
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Answer» Solution :PLANET of mass 'm' moves around the Sun of mass 'M' in a circular orbit of radius 'R' with orbital velocity v. `therefore(mv^2)/(r)=(GMm)/(r^2)` `v^(2)=(GM)/(r)""...(1)` But, Orbital speed `(v)=("circumference")/("period of revolution")=(2pir)/(T)` Squaring, `v^(2)=(4pi^(2)r^(2))/(T^2)""...(2)` Comparing of equations (1) and (2), we get `(GM)/(r)=(4pi^(2)r^(2))/(T^2)` In CROSS MULTIPLICATION, `(GMm)/(r^2)` `T^(2)PROPR^(3)` |
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| 5. |
Particles of masses 100 and 300 gram have position vectors (2hat(i)+5hat(j)+13hat(k)) and (-6hat(i)+4hat(j)+2hat(k)). Position vector of their centre of mass is |
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Answer» `-(16)/(4)hati+(17)/(4)HATJ+(7)/(4)HATK` |
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| 6. |
Find the moment of inertia of a thin uniform rod about an axis perpendicular to its length and passing through a point which is at a distance of l/3 from one end. Also find radius of gyration about that axis. |
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Answer» Solution :(i) Let P be the point at a distance `(l)/(3)` from one end. It is at a distance of `((l)/(2)-(l)/(3))=(l)/(6)` from the centre as SHOWN in the figure. By parallel AXES theorem `I=I_(c )+MR^(2)` `=(Ml^(2))/(12)+M((l)/(6))^(2)` (ii) The radius of gyration, `K = sqrt((l)/(M))=sqrt((Ml^(2))/(9M))=(l)/(3)` 
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| 7. |
An object starting from rest slide on slope with 45^(@). If co-efficient of friction between object and surface is 0.3 x where x is distance covered then at what distance speed will be maximum ? |
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Answer» Solution :Accelerationof objectparallelto slope `a= g SIN theta= mu g cos theta ` `V (dv )/(dx )= g sin theta- mu GCOS theta` `:. , (d v)/(dx ) = ( g sin 45^(@)- 0.3 XG cos 45^(@))/( v )` Fordefine valueofx when`(d v)/(dx ) = 0`thenwillbe MAXIMUM `:. , 0= g - 0.3 kg` `:. , 0.3x =1 rArrx = (1)/(0.3)= 3.33 m` |
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| 8. |
If the vibration frequency of atoms in a crystal depends on the atomic mass m, the atomic spacing r and compressibility C, then it is proportional to |
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Answer» `m^((1)/(2))` |
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| 10. |
Givethe expressionsfor linear,areaandvolumethermal expansions. |
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Answer» Solution :Linear Expansion : In solids, for a SMALL change is temperature `DeltaT`, the fractional change in length `((DeltaL)/(L_(0)))` is DIRECTLY proportional to `DeltaT`. `(DeltaL)/(L_(0))=alpha_(L)DeltaT` `"Therefore,"alpha_(L)=(DeltaL)/(L_(0)DeltaT)` AREA Expansion : For a small change in temperature `DetlaT` the fractional change in area `((DeltaA)/(A_(0)))` of a substance is directly proportional to `DeltaT` and it can be WRITTEN as `(DeltaA)/(A_(0))=alpha_(A)DeltaT` `"Therefore,"alpha_(A)=(DeltaA)/(A_(0)DeltaT)` Volume Expansion : For a small change in temperature `DeltaT` the fractional change in volume `((DeltaV)/(V_(0)))` of a substance is directly proportional to `DeltaT`. `((DeltaV)/(V_(0)))=alpha_(0)DeltaT` `"Therefore,"alpha_(V)=(DeltaV)/(V_(0)DeltaT)` |
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| 11. |
Dimension of physical quantity in different unit system are equal. |
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Answer» |
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| 12. |
The work done in carrying a body from A to B is 20J and from B to C is 15J. The work done in moving the body from A to C assuming the force involved in all the cases is conservative force is |
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Answer» 35J |
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| 13. |
A dumb – bell has a rigid mass less stick and two point masses at its ends. Each mass is m and length of the stick is L. The dumb- bell leans against a frictionless wall, standing on a frictionless ground. It is initially held motionless, with its bottom end an infinitesimal distance from the wall. It is released from this position and its bottom end slides away from the wall where as the top end slides down along the wall. (a) Show that centre of mass of the dumb-bell moves along a circle. (b) When the dumb-bell loses contact with the wall what is speed of the centre of mass? |
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| 14. |
Arrange the vectors substractions so that their magnitude are in decreasing order. If the two vectors vecA and vecB are acting at an angle (|vecA|>|vecB|) (a) 60^(@)""(b)90^(@)( c) 180^(@)""(d)120^(@) |
| Answer» Answer :C | |
| 15. |
In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 ml of alcohol. Then I mL of this solution is diluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow through. The solution spreads over the surface of water forming one molecule thick layer. Now lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule. Read the passage carefully and answer the following questions. (a) Why do we dissolve oleic acid in alcohol ?(b) What is the role of lycopodium powder ? (c) What would be the volume of oleic acid in each ml of solution prepared ? (d) How will you calculate the volume of n drops of this solution of oleic acid ? (e) What will be the volume of oleic acid in one drop of this solution ? |
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Answer» Solution :(a) Because it does not dissolve in water. (b) Lycopodium powder spreads over the surface of water when it is sprinkled UNIFORMLY. When a drop of prepared solution is DROPPED on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. Therefore, the area over which oleic acid spreads can be measured. (d) Volume of n drops of this solution of oleic acid can be CALCULATED by means of a burette by measuring cylinder and the NUMBER of drops. (E) If n drops of the solution make 1 ml, the volume of oleic acid in one drop = `(1)/((400)n)mL` |
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| 16. |
State and explain the theorem of parallel axis with an example |
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Answer» Solution :STATEMENT : The moment of inertia of BODY about any axis is equal to the sum of the moment of inertia of the body about an parallel axis passing through the centre of mass and the product of mass of the body and SQUARE of the distance between the two parallel axis. `i=I_0+Md^2` We know that `I_0=(ML^2)/(12)` and `d=L/2` Hence moment of inertia at one end `=(ML^2)/(12)+M(L/2)^2` `I=(ML^2)/(4)[1/3+1]=(ML^2)/(3)`  |
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| 17. |
A fighter plane enters inside enermy territory, at time t=0, with velocity 250m/s and moves horizontally with constant acceleration a=20m//s^(2). An enemy tank at border, spot the plane and fires shots at an angle 60^(@) with horizontal and with velocity U= 600m/s. The altitude H of plane it can be hit by the shot is 2473h m. The value of 'h' is |
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| 18. |
A thin wire ring of 3cm radius rests flat on surface of a liquid. The pull required to raise the ring before the film breaks is 30.14 xx 10^(-3) N more than it is after. The surface tension of liquid (in "Nm"^(-1)) is |
| Answer» Answer :A | |
| 19. |
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x)=kx^(2). Where k is a positive constant. If the amplitude of oscillation is a, then its time period T is…………… |
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Answer» Proportional to `(1)/(sqrt(a))` |
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| 20. |
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is |
| Answer» Answer :D | |
| 21. |
A block slides down a slope of angle theta with constant velocity. It is then projected up with a velocity of 10 ms^(-1), g = 10 ms^(-2) & theta = 30^(@). The maximum distance it can go up plane before coming to stop is |
| Answer» Answer :B | |
| 22. |
On a hypothetical scale X, the ice point is 40^(@) and the steam point is 120^(@). For another scale Y the ice point and steam points are -30^(@)and130^(@) respectively. If X-reads 50^(@) The reading of Y is |
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Answer» `-5^(@)` |
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| 23. |
(I) In slipping, the rotational motion is less than translational motion. (II) In sliding, the rotational motion is more than translational motion Which one is correct ? |
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Answer» I only |
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| 24. |
Which type of frame of reference does a bus starting from rest represent ? |
| Answer» SOLUTION :Non-inertial FRAME of REFERENCE. | |
| 25. |
A pendulum is hanging from the ceiling of a trolley which moves down a smooth inclined plane. Then |
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Answer» `theta = ALPHA` |
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| 26. |
A thin but very large plank of mass 2 m is placed on a horziontal smooth surface. A solid cylinder of mass m and radius r is given only translational velocity v_(0) and gently placed on the plank as shown in the figure. The coefficient of kinetic friciton between the plank and the cylinder is mu. |
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| 27. |
Two capillary tubes of equal length and inner radii 2r and 4r respectively are added in series and a liquid flows through it. If the pressure difference between the ends of the whole system is 8.5 cm of mercury, find the pressure difference between the two ends of the first capillary tube. 38.cm of Hg |
| Answer» Answer :A | |
| 28. |
A thermally insulated vessel contains an ideal gasof molecular mass M and ratio of specific heats gamma. It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by: |
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Answer» `((gamma-1))/( 2 ( gamma+1) R) MV^(2) K` |
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| 29. |
What is meant by positive work, negative work and zero work? Give one example of each. |
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Answer» Solution :Positive work: If a force ACTING on a body has a component in a direction of the DISPLACEMENT. Eg: When a body falls under free fall. `(theta=0^(@))` Negative work: If a force acting on a body has a component in the opposite direction of the displacement. Eg: When brakes are APPLIED to a moving vehicle, the work DONE by the brake force is negative. Zero work: If the body gets displaced ALONG a direction perpendicular to the direction of applied force. Eg: When a coolie walks on a horizontal platform with load. |
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| 30. |
Two wedges, each of mass m, are placed next to each other on a flat horizontal floor. A cube of mass M is balanced on the wedges as shown in figure. Assume no frictionbetween the cube and the wedges, but a coefficent of static friction mult1 between the wedges and the floor. What is the largest M that can be balanced as shown without motion of the wedges ? |
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Answer» `(m)/(sqrt(2))`  From `FBD` of blcok `2N cos 45^(@)=Mg……(1)` For wedge to REMAIN at rest `N sin 45^(@) lt mu N'……(2)` and `N' =mg+N cos 45^(@) ….(3)` From `1,2, ` and 3 we get `Mle(2mum)/(1-mu)` |
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| 31. |
A stream of water flowing horizontally with a speed of 15 m s^(-1)gushes out of a tube of cross-sectional area 10^(-2) m^2 , and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ? |
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Answer» Solution :MASS of water hitting the wall per second ` = 10^3 kg m^(-3) xx 10^(-2) m^2 xx 15 ms^(-1) = 150 kg s^(-1)` FORCE by the wall = momentum loss of water per second ` =150 kg s^(-1) xx 15 ms^(-1) = 2.25 xx 10^3 N` |
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| 32. |
A sphere has a concentric spherical cavity. The specific gravity of the material of the sphere is 8. the sphere is place on a surface of water. If in the equilibrium position the sphere just sinks in water, the ratio of theradius of cavity to the outer radius of the sphere is |
| Answer» SOLUTION :`((7)/(8))^(1//3)` | |
| 33. |
The moment of inertia of a uniform circular disc about a normal axis through centre is I. The M.I. about a tangential axis parallel to any of its diameters is: |
| Answer» ANSWER :D | |
| 34. |
Define most probable speed of the gas. Write its expression. |
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Answer» |
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| 35. |
A body executing forced oscillations under driving force is in sharp resonance. If damping increases then sharpness of resonance |
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Answer» Decrease |
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| 36. |
The radii of two planets are respectively R_(1)andR_(2) and their densities are respectively rho_(1)andrho_(2). The ratio of the accelerations due to gravity (g_(1)//g_(2)) at their surfaces is: |
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Answer» `g_(1):g_(2) = (rho_(1))/(R_(1)^(2)):(rho_(2))/(R_(2)^(2))` |
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| 37. |
Match Column-I with Column-II : {:("Column-I","Column-II"),("(1) Newton's third law of motion",(a)vecF_12=-vecF_21),("(2) Law of conservation of momentum.",(b)Deltavecp=0),(,(c) vecF_12=vecF_21):} |
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Answer» |
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| 38. |
What is the density of water at a depth where pressure is 80.0 atm. Given that its density at the surface is 1.03xx10^(3)kgm^(-3)? Compressibility of water is 45.8xx10^(-11)Pa^(-1) |
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Answer» SOLUTION :Given density at the surface `=1.03xx10^(3)kgm^(-3)` Compressibility of waer `=45.8xx10^(-11)Pa^(-1)` Bulk Modulus `=1/("compressibility")=1/(45.8xx10^(-11)Pa` `K=2.183xx10^(9)Pa` We know that mass is CONSTANT. As VOLUME changes, density of water changes so that, `(DeltaV)/V=-(Delta rho)/(rho)` but `DeltaP=80-1=79atm=79xx1.013xx10^(5)=80.027xx10^(5)Pa` For compression bulk modulus `K=(-DeltaP)/((DeltaV)/V)` i.e. `K=((DeltaP)/(Delta rho))rho` Change in density `Delta rho=((Delta P)rho)/K=((80.027xx10^(5))(1.03xx10^(3)))/(2.183xx10^(9))` `=37.758xx10^(-1)=3.78kgm^(-3)` Hence density at the depth`=rho + Deltarho` `=1.03xx10^(3)+0.00378xx10^(3)` `=1.03378xx10^(3)` `=rho ' =1.034xx10^(3)kgm^(-3)` |
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| 39. |
Sand is pilled up on a horizontal ground in the form of a regular cone of a fixed base radius R. The coefficient of static friction between sand layers is mu. The maximum volume of sand that can be pilled up, without the sand slipping on the surface is |
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Answer» `(mu R ^(3))/(3pi)` |
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| 40. |
A listener moving with constant velocity passes a statlonary source. Draw a graph to show the change of apparent frequency of the source to the listener with time . Actual frequency of source is n. |
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Answer» Solution :While approaching the stationary source apparent frequency will be ` n. = (V + u_(o))/(V) = n` …(1) [velocity of sound in air = V , velocity of listener = ` u_(o)` ] Whlle receding the stationary source apparent frequency will be ` n.. = (V - u_(0))/(V) n` ...(2) from equation (1) and (2) it is clear that n. and n.. remain constant with time also ` n. gt n gt n..` . The change of apparent frequency with time is SHOWN below . Here OS denoes the time taken by the listener to pass by the source . After then the listener continually moves away from the source . 
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| 41. |
One atomic mass unit is amu =166xx10^(-27)kg. The atomic weight of oxcygen is 16. Find the mass of the atom of oxygen. |
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Answer» `26.56xx10^(-27)KG` `m=16amu=16xx166xx10^(-27)kg=26.58xx10^(-27)kg` |
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| 42. |
The displacement equation of the particle in SHM is y = 10 sin(2 pi t + pi//6)m. The initial displacement of the particle is |
| Answer» Answer :A | |
| 43. |
A train starting from rest travels the first part of its journey with constant acceleration a, second part with constant velocity v and third part with constant retardation a, being brought to rest. The average speed for the whole journely is (7v)/(8). For what fraction of the total time, the train travels with constant velocity? |
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Answer» `1//2` |
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| 44. |
A shearing force of 10^(6) N is applied on the top surface of a fixed cube of side 10 cm. If the top surface is displaced by 0.33 mm, find rigidity modulus of the material of the cube. |
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| 45. |
A book of mass 5 kg is kept on a horizontal table. It is connected to a weight of mass 2 kg by a weightlessstring passing over a smooth pulley . The part of the string on the table is horizontal, and the weight is hanging freely from the pulley. If the coefficient of kinetic friction betweenthe table and the block is 0.2, find the acceleration of the block. What will be the tension in the string ? |
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Answer» SOLUTION :Weight of the block, MG = `5xx9.8` N, `:.` Normal force of the table on the block, R = `5xx 9.8 `N Kinetic friction against the MOTION of the block, f.= `mu R=0.2xx5xx9.8` = 9.8N Let the tension in the string be T and acceleration of the block be a fig. Hence for the motion of the block, T-f. = Ma or, T-9.8= 5a `""CDOTS(1)` The weight of mass m (say ) moves downwards with the same acceleration. `:.` mg-T = ma or, `2xx9.8-T = 2a [ becausem=2] ""cdots(2)` Adding (1) and (2) we get, `2xx9.8`-9.8=7a or, a=`(9.8)/(7)=1.4 "m.s"^(-2)` Substituting this value in (1), we get, T-9.8 = `5xx1.4 ""`or, T =16.8 N. 
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| 46. |
A body of mass 6kg is under a force which causes displacement in it which is given by s=(t^(2))/(4) m, where .t. is time. The work done by the force in 2s is |
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Answer» 12J |
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| 47. |
In a reciprocating steam engine, the average pressure of steam is 10 ^(4) N//m ^(2), The length of the stroke is 1 m and the area of the piston is 0.15 m ^(2).Calculate the power of the engine, if it makes 5 strokes per second. |
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Answer» Solution :`P = 10 ^(4) N//m^(2) , L = 1 m, A = 0.15 m ^(2),` `N =5` strokes/second |
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| 48. |
Momentum of two object with different mass is equal. Which object have more speed ? |
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Answer» Solution :LET `m_(1) ` and `m_(2)`be MASSOF twoobjectwith`m_(2) gt m_(1)` Heremomentum of bothobjectis equal`m_(1) v_(1) = m_(2)v_(2)` `:. , (v_(1))/( v_(2))= (m_(2))/(m_(1))` here`m_(2) gt m_(1)` `v_(1) gt v_(2)` Lighterobject`(m_(1))` willhavemorespeed. |
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| 49. |
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 ms^(-2) The total mass of the crew and passengers is 500kg Give the magnitude and direction of the (a) force on the floor of the helicopter by the crew and passengers (b) action of the rotor of the helicopter on the surrounding air (c) force on the helicopter due to the surrounding air (g = 10ms^(-2)) . |
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Answer» Solution :Here, ` m_(1) = 2000 kg , a = 15 ms^(-2) , m_(2) = 500 kg, g = 10 ms^(-2)` (a) Force on the floor of the HELICOPTER by the crew and passenger ` m_(2) (g + a) = 500 (10+5) = 12500 N = 12.5 xx 10^(3)N` (B) Action of rotor of helicopter on the surrounding air ` (m_(1) + m_(2)) (g+a) = (2000 + 5000) (10+ 5) = 2500 xx 25 = 6.25 xx 10^(4) N ` This force is downwards (C) Force on the helicopter due to surrounding air = reaction of force in (b) ` = 6.25 xx 10^(4) N ` upwards . |
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| 50. |
The speed of a motor decreases from 1200 rpm to 600 rpm in 20s. The total number of rotations it makes before coming to rest is |
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Answer» 400 |
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