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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 heavy uniform chain lies on horizontal table top. If the co-efficient of friction between the chain and the table surface is 0.25, the maximum percentage of the length of the chain that can hang over one edge of the table is |
| Answer» ANSWER :A | |
| 2. |
A particle is found to be at rest when seen from a frame S_(1) and moving with a constant velocity when seen from another frame S_2 a) Both the frames are inertial b) Both the frames are non inertial c) S_(1) is inertial and S_2 is non inertial d) S_(1) is non inertial and S_2 is inertial |
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Answer» Both the FRAMES are INERTIAL |
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| 3. |
Prove that speed of sound in air increases by 0.61 ms^(-1) per degree celsius rise in temperature. |
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Answer» Solution :Since `V prop SQRT(T)` the speed of sound changes directly to the square root of temperature in kelvin. Let `v_(0)` be the speed of sound at temperature at `0^(@)C ` or `273 K`. Let v the speed of sound at any arbitrary temperature T (in kelvin,) then `(v)/(v_(0))= sqrt((T)/(273))= sqrt((273+t)/(273))` `v=v_(0) sqrt(1+(t)/(273))=v_(0)(1+(t)/(546))` (Using binomial expansion) Since `v_(0)=331 MS^(-1)` at `0^(@)C`, v at a particular temperature in `t^(@)C` is `v=(331+0.60t) ms^(-1)` Thus the speed of sound in AIR increases by `0.61 ms^(-1)` perdegree celsius in temperature. |
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| 4. |
The transverse displacement of a string clamped at its both ends is given by y(x,t)=2sin((2pi)/3x)cos(100pit) where x and y are in cm and t is in s. Which of the following statements is correct? |
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Answer» All the points on the STRING between TWO CONSECUTIVE NODES vibrate with same frequency, phase and amplitude. `y(x,t)=2sin((2pi)/3x)cos(100pit)` It represents a STATIONARY wave. Therefore, all the points between consecutive nodes vibrate with same frequency and same phase but different amplitude. |
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| 5. |
The equation of state corresponding to 8 g of O_(2) is |
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Answer» `PV=8RT` |
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| 6. |
The critical angular velocity w of a cylinder inside another cylinder containing a liquid at which its turbulence occurs depends on viscosity eta, density d and the distance x between the wall of the cylinders. Then w is proportional to |
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Answer» `(ETA)/(X^(2)d)` |
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| 7. |
When a body is absorbing latent heat, what will happen to its temperature? |
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Answer» |
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| 8. |
The excess of pressure inside the first soap bubble is three times that inside the second bubble. Then ratio of volume of the first to second bubble is |
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Answer» `1:3` |
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| 9. |
If vec(A)is a vector of magnitude 4 units due east. What is the magnitude and direction of a vector- 4 vec(A) ? |
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Answer» 4 UNITS due EAST |
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| 10. |
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of the damping constant b for the spring and shock absorbar system of one wheel, assuming that each wheel supports 750 kg. |
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Answer» Solution :For each spring system supporting 750 kg of WEIGHT. Time period `T= 2PI sqrt((m)/(k))` `= 2XX 3.14 xx sqrt((750)/(5xx 10^4))` `= 6.28 xx sqrt(150 xx 10^(-4))` `= 6.28 xx sqrt(0.015) = 6.28xx 0.122` `= 0.769 approx 0.77s`. Now DISPLACEMENT of damped oscillation at time t, `x= x_(0) e^(-(bt)/(2m))` `therefore (x_0)/(2) = x_(0) e^(-(bt)/(2m))` `therefore 2= e^((bt)/(2m))` `therefore ln2 = (bt)/(2m)""[therefore " Taking log on both side "]` `2.303 log 2 = (bt)/(2m)` `therefore 2.303 xx 0.3010 = (bt)/(2m)` `therefore (0.693 xx 2m)/(t) =b` `therefore b = (0.693xx 2 xx 750)/(0.77)` `therefore b = 1350 kg s^(-1)`. |
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| 11. |
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of the spring constant k. |
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Answer» SOLUTION :Mass M = 3000 kg Displacement `x= 15 cm = 0.15 cm, g= 10 ms^(-2)` There are four spring systems. If k is spring constant of each spring then TOTAL spring constant of all the four springs in PARALLEL is, `k_(p) = 4k` Now, `Mg = k_(p)x` `Mg = 4KX` `therefore k = (Mg)/(4x) = (3000xx10)/(4XX 0.15)` `therefore k = 5 xx 10^(4) N"/"m`. |
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| 12. |
Two blocks of mass 4 kg and 2 kg are connected by a heavy string and placed on rough horizontal plane. The 2 kg block is pulled with a constant force F. The coefficient of friction between the blocks and the ground is 0.5. What is the value of F so that tension in the string is constant throughout duringthe motion of the blocks ? (g=10 m//s^(2)) |
| Answer» Answer :B | |
| 13. |
A jar is filled with two non-mixing liquide 1 and 2 having densities rho_(1)andrho_(2), respectively A solid ball, made of a material of density rho_(3), is dropped in the jar. If comes to equilibrium in the position shown in the figure. Which of the following is true for rho_(1),rho_(2)andrho_(3) |
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Answer» `rho_(1)gtrho_(3)gtrho_(2)` |
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| 14. |
If the angle of incidence is twice the angle of refraction in a medium of refractive indexmu then the angle of incidence is |
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Answer» `cos^-1(mu/2)` |
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| 15. |
(A) : An object falling through a viscous medium eventually attains a terminal velocity. (R ) : All the raindrops hit the surface of the earth with the same constant velocity. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 16. |
The mass of a ball if four times the mass of another ball. When these balls are separated by a distance of 10 cm, the gravitational force between them is 6.67xx10^(-7)N. Find the masses of the two balls. |
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Answer» 0.5,2 |
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| 17. |
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation ? (a) A tragic sea accident several centuries ago drowned a young Samurai. As a tribute to his bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.(b) After the sea tragedy, fishermen in that area, in a gesture of honour to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. Consequently, the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.[Note : This interesting illustration taken from Carl Sagan's ''The Cosmos' highlights the fact that often strange and inexplicable facts which on the first sight appear 'supernatural' actually turn out to have simple scientific explanations. Try to think out other examples of this kind]. |
| Answer» Solution :(a) The explanation given in OPTION (B) STRIKES as a scientific explanation. | |
| 18. |
The system shown in figure is in equilibrium. Surface PQ of wedge A, having mass M, is horizontal. Block B, having mass 2M, rests on wedge A and is supported by a vertical spring. The spring balance S is showing a reading of sqrt2 Mg. There is no friction anywhere and the thread QS is parallel to the incline surface. The thread QS is cut. Find the acceleration of A and the normal contact force between A and B immediately after the thread is cut. |
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Answer» |
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| 19. |
The optical path of a monochromatic light is the same if it goes through 2.00m of glass or x 1.510 and that of ruby is 1.760 find the value of x is……cm |
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Answer» 1.716 |
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| 20. |
Which one of the following is not a unit of time? |
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Answer» LUNAR month |
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| 21. |
The nature of a physical quantity is described by its dimensions.Which one of the following is a dimensional constant? |
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Answer» RELATIVE ERROR of a measurement |
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| 22. |
In rainy season sand is sometimes thrown on railway tracks. Why? |
| Answer» Solution :The force impressed on the wheel is generally greater than the force of rolling friction, this produces rolling of the wheel on the railway TRACK. On the other hand the kineticfriction is fairly higher that the impressed force and there is no sliding of the WHEELS. But in rainy SEASON on the wet railway track, there is a danger that the kinetic friction MAY drop below the impressed force resulting in a sliding of the wheels. To avoid this sand is sometimes thrown on the railway track to increase the kinetic friction. | |
| 23. |
Explain why (a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. Explain. (b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (c ) Surface tension of liquid is independent of the area of the liquid surface. (d) Detergents should have small angles of contact. (e ) A dorp of liquid under no external force is always spherical in shape. |
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Answer» Solution :(a) Angle of contact between a LIQUID and solid is the angle enclosed between the tagents to the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of the liquid with solid. In CASE of a mercuryin glass capillary tube, the force of cohesion between mercury molecules is greater than the force of adhesion between mercury molecules and glass molecules. Due to it, the surface tensionbetween solid (glass)-liquid (mercury) interface (i.e., `S_(sl)`) is greater than the surface tension between liquid (mercury)-air(i.e., `S_(la)`) interface. It means , `S_(sl) gt S_(la)`. As a result of it, the mercury dies not wet the glass tube. the shape of mercury meniscus in the capillary tube is convex upwards. fig. then the angle of contact between mercury and glass is more than `90^(@)`, i.e., obtuse angle. In case of a water in glasscapillary tube, the force of cohesion between wate molecules in less than the force of adhesion between water molecules and glass modecules. Due to it, the surface tension between solid (glass)-liquid (water) interface (i.e., `S_(sl)`) is less than the surface tension between liquid (water) - air interface (i.e., `S_(la)`) . it means `S_(sl) gt S_(la)`. ltbr. As a result of it, the water wets the glass tube. The shapeof the meniscus in the capillary tube is concave upwards , fig. then the angle of contact between water and glass is less than `90^(@)`, ie., acute angle.  (b) For mercury - glass, angle of contact is obtsue. in ORDER to achieve this obtuse value of angle of contact, the water TENDS to spread. (c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. since this force is independent of the area of liquid surface, therefore, surface tension is also independent of the area of the liquid surface. We know that the cloth has narrow spaces in the form of capillaries. the rise of liquid in a capillary tube is directly proportional to `cos theta`. if `theta` will be large. due to which, capillary rise will be more and so the detergent will penetrate more in cloth. (e ) In the absence of external forces, teh surface of the liquid drop tends to acquire the minumum surface area due to surface tension. since for a given volume, the surface are of sphere is least, hence the liquid drop takes the spherical shape. |
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| 24. |
An object covers 4m in the first two seconds and 4.4m in the next 4 seconds. Find its velocity after 7 seconds from the start. |
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Answer» |
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| 25. |
The dimensional formula of magnetic induction is |
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Answer» `MT^(-1) A^(-1)` |
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| 26. |
Two wires A and B have Young's modulii in the ratio 1:2 and ratio of lengths is 1:1. Under the application of same stress the ratio of elongations is |
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Answer» `1:1` |
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| 27. |
Two boys weighing 30kg and 50kg are sitting on either side of the fulcrum of a see saw at distance 2m and 3m respectively. Where should a man of 60kg must sit to balance the see-saw ? |
Answer» Solution : At equilibrium, `30bar(AO)+60bar(CO)=50bar(BO)` `30xx2+60x=50xx3` `60x=150-60=90` `:.""x=(90)/(60)=1.5m` The man has to SIT at 0.5 METRE from A. |
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| 28. |
Which one of the following statement is correct When the speedof amoving body is doubled |
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Answer» its acceleartio is DOUBLED |
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| 29. |
The value of G depends upon |
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Answer» The mass of BODIES |
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| 30. |
What are damed oscillations? Discuss it by illustration of spring. |
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Answer» Solution :The motion of a simple pendulum swinging in air, dies out eventually. This is because the air drag and the friction at the support oppose the motion of the pendulum. The mechanical energy of the oscillating system opposite to this resistive force is dissipated in the form of heat energy and hence mechanical energy decreases. As a result according to `E= (1)/(2) kA^(2)` the AMPLITUDE gradually decreases. Damped oscillations : ..Oscillations in which the amplitude of the oscillator decreases with time then such TYPE of oscillations is known as damped oscillations...  According to as shown in figure let us consider a block of MASS m connected to an elastic spring of spring constant k oscillating vertically. If the block is pushed down a little and released, it oscillates in vertical plane. Its angular frequency of oscillation is `OMEGA = sqrt((k)/(m))`. In practice, the SURROUNDING medium (air) will exert a damping force on the motion of the block and the mechanical energy of the block spring system will decrease. The energy loss will appear as heat of the surrounding medium (and the block also). The damping force depends on the nature of the surrounding medium. If a block is immersed in a liquid, the magnitude of damping will be much greater and the disspation of energy much faster. The damping force is generally proportional to the velocity of the bob. |
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| 31. |
Two waves are represented by the equationsy _(1) = a sin (omega t + kx + 0.57)m and y_(2) =a cos (omega t + kx) m, where x is in meter and t in sec. The phase difference between them is……. |
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Answer» `1.0` RADIAN `y _(2) =a COS (omega t + kx) m` `therefore y _(2) = a sin ((pi)/(2) + omega t + kx) m ""…(2)` `(because cos theta = sin ((pi)/(2) + theta ))))` From (2), phase of SECOND wave, `theta_(2)= (pi)/(2) + omega t + kx` `theta _(1) = omega t + kx + 0.57` `therefore` Phase difference `= theta _(2) - theta_(1)(pi)/(2) - 0.57` `= (3.14 )/(2) -0.57 =1.57 -0.57 =1 rad` |
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| 32. |
A car starting from rest, accelerates at a constant rate x for sometime after which it decelerates at a constant rate y to come to rest. If the total time elapsed is t, the maximum velocity attained by the car is given by |
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Answer» (a) `(XY)/(X+y)t` |
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| 33. |
There are two point masses A and B, situated at origin and point (5m,0m) respectively . At a certain time v_(A) and v_(B) are respectively the velocities of point masses A and B . Match the situations under Table-1 with their correct option under Table-2 |
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Answer» <P> |
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| 34. |
A three-wheeler starts from rest, accelerates uniformly with 1 ms^(-2) on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ? |
| Answer» Solution :A straight line INCLINED with the time-AXIS for uniformly ACCELERATED motion, parallel to the time- axis for UNIFORM motion. | |
| 35. |
The angle between vecA=veci+vecj and vecB=veci-vecj is |
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Answer» 45° |
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| 36. |
The universally accepted primary thermometer is |
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Answer» liquid thermometer |
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| 37. |
The speed with which a bullet can be fired is 150ms^(-1). Calculate the greatest distance to which it can be prjected and also the maximum height to which it would rise. |
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Answer» Solution :`R=(U^(2)sin2alpha)/(g)=((150)^(2)SIN90^(@))/(9.8)=2295.1m` `H_(max)=(u^(2)SIN^(2)alpha)/(2g)=((150)^(2)sin^(2)45^(@))/(2xx9.8) = 573.97m` |
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| 38. |
0.5 litre of an ideal gas at a pressure of 20 atm expands to a final volume of 0.5 m ^(3). The process obeys the equation PV = a constant, (i) Find the value of this constant (ii) What is the final pressure ? And (iii) Find, the work done by the gas, |
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Answer» SOLUTION :`V _(1) = 0.5` liter `= 0.5 xx 10 ^(-3) m ^(3)` ` P _(1) = 20 atm = 20 xx 101325 Pa` `V _(2) =- 0.5 m ^(3)` `P _(2) = ?` `PV = P _(1) V _(1) = P _(2) V _(2)` ` P _(1) V _(1) = 20 xx 101325 xx 0.5 x 10^(-3)` `P _(2) = ( P _(1) V _(1))/( V _(2)) = ( 20 xx 0.5 xx 10 ^(-3))/( 0.5)` `= 0. 020 atm` Work done `W = RT log _(e) "" (T _(2))/( V _(1))` But `RT = P _(1) V _(1) = P _(2) V _(2)` ` W = 2. 303 P _(1) V _(1) log _(10) "" (V _(2))/( V _(1))` ` 2. 303 xx 20 xx 101325 xx 0.5 xx 10^(-3) xx 0.5 xx 10 ^(-3) xx log _(10) " "(0.5)/( 0.5 xx 10 ^(-3))` `= 2. 33 xx 10 ^(6) J` |
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| 39. |
One mole of a monatomic ideal gas is taken through the cycle shown in figure The pressures and temperatures at A,B etc, are denoted by P_(A),T_(A),P_(B),T_(B)etc respectively.Given T_(A)=1000K, P_(B)=(2//3)P_(A) and P_(C)=(1//3)P_(A) Then choose the incorrect option. |
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Answer» The work done by the gas in the process A`rarr`B is 1869.75J Where `gamma=5//3` for a monatiomic gas. `T_(B)=T_(A)(P_(B))/(P_(A))^(1-(1))/(gamma) =1000(2)/(3))^(2//5)=850 K` Work dine in an adiabatic process is given by `W=(NR(T_(A)-T_(B))/(gamma-1)=(1xx8.31(1000-850))/((5//3)-1))=1869.75 J` `therefore` Process B`rarr`C is isochoric Hence W=0 and `(P_(C))/(T_(C))xT_(B)=((1//3)P_(A))/(2//3)P_(A)) T_(B)=(T_(B))/(2)=(850)/(2)=425 K` From first law of thermodynamics , `Q=DeltaU+W=nC_(V)DeltaT+0=n(3)/(2)R(T_(C)-850)` Hence `Q=1xx(3)/(2)xx8.31(125-850)=-5297.625 J` Negative sign IMPLIES that the system has lost heat. process D`rarr`A is isochoric `(P_(D))/(P_(A))=(T_(D))/(T_(A)) or P_(D)=P_(A)(T_(D))/(T_(A))` Process C`rarr`D is adiabatic Therefore, `(T_(D))/(T_(C))^(gamma)=(P_(A)T_(D))/(P_(C)T_(A))^(gamma-1)` `thereforeT^(1//gmma)=(C)[(P_(A))/(P_(C)T_(A))]^(1-1//gamma)` `rArr T^(3//5)=T_(C)[(P_(A))/((1//3)P_(A))xx1000))]^(2//5)=425 ((3)/(1000))^(2//5)` `T_(D)=500 K` |
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| 40. |
Give any two salient features of static Friction and Kinetic Friction. |
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Answer» Solution :Static friction: Static friction is the force which opposes the initiation of motion of an object on the surface. The MAGNITUDE of static force `f_(s)` lies between. `=lef_(s)lemu_(s)N_(s)` `mu_(s)`- coefficient of static friction N- Noemal force Kinetic frition : The frictionl force excerted bysurface when an object slides is called as kinectic friction . Also called as sliding friction or dynamic friction, (ii) To move an object-pushior pull? Which is easier? Explain When a body is pushed at an ARBITRARY angle `theta (0 "to"(pi)/(2))`,the applied force F can be resolved into two componentes as F `sin theta` parallel to the surface and F `cos theta ` perpendiculerto the surface as shown in figure. The total downward force acting on the body is mg+F `cos theta` . It implies that the normal force acting on the body increas. Since there is no acceleration alon the vertical direction the normal force N is equal to. `N_(PUSH)=mg+F costheta` As a result the maximal static friction also increasesand is equal to `f_(x)^(MAX)=mu_(r)N_(PUSH)=mu_(5)(mg+fcostheta)` Equation (2) SHOWS that a GREATER n force needs to be applied to push the object into motion .  When an object is pulled at an angle `theta`, the applied force is resolved into two components as shown in figure. the total downward force acting on the object is `N_(pull)=mg-Fcostheta`  Equation (3) shows that the normal force is less than `N_(push)`. from equations (1) and (3), it is easier to pull an object than to push to MAKE it move. |
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| 41. |
A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10cm//s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t,the snapshot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is (a) sqrt(3pi)/(50)hatjm//s (b)-sqrt(3pi)/(50)hatjm//s (c)sqrt(3pi)/(50)hatim//s (d) -sqrt(3pi)/(50)hatim//s |
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Answer» Solution :PARTICLE velocity `v_p =-v` (slope of y-x graph) Here, `v+=ve`,as the WAVE is travelling in POSITIVE x-direction. Slope at P is negative. `:.` Velocity of particle is in positive `y`(or `hatj`) direction. `:.` Correct option is (a). |
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| 42. |
A mass of 2 kg attached to a spring of force constant 800N/m makes 100 osicllation. The time taken is (in seconds) |
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Answer» 31.4 |
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| 43. |
A particle moves in a circular path with decreasing speed. What happens to itsangular momentum? |
| Answer» Solution :As `vecL= vecr=xx mvecv` i.e., `vecL` magnitude DECREASES but DIRECTION REMAINS CONSTANT. | |
| 44. |
When an isolated gas expands against vacuum, then its a)Internal energy increases b) No work is done c) Internal energy remains same |
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Answer» both a and B are CORRECT |
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| 45. |
Elevenrodseachof lengthL , cross section A andcoeffeicient of thermalconductivityKare arrangedas shown . Findthe rateof heattransferefrom P to Q( T_(1)gt T_(2)) . |
| Answer» SOLUTION :`(KA(T_(1) - T_(2)))/(L) - (1)/(2)` | |
| 46. |
Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure, (a) Find the displacement of the centre of mass at time t. (b) If the extension of the spring is x0 at time t, find the displacement of the two blocks at this instant: |
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Answer» Solution :(a) The acceleration of the CENTRE of mass is `a_(CM) = F/(2m)` The DISPLACEMENT of the centre of mass at time t will be `x=1/2a_(CM).t^(2) = (Ft^(2))/(4m)` (b) Suppose the displacement of the first BLOCK is `x_(1)` and that of the second is `x_(2)`. Then, `x=(mx_(1) + mx_(2))/(2m)` or `(Ft^(2))/(4m) =(x_(1) + x_(2))/2`(or) `x_(1) + x_(2) =(Ft^(2))/(2m)`.........(i) Further, the extension of the spring is `x_(1)-x_(2)` THEREFORE, `x_(1)-x_(2)=x_(0)`............(ii) From Eqs. (i) and (ii), `x_(1) = 1/2((Ft^(2))/(2m) +x_(0))` and `x_(2) = 1/2((Ft^(2))/(2m) -x_(0))` |
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| 47. |
The equation of a wave travelling on a string is y =4 sin {(pi)/(2) (8 t - (x)/(8)) } , where x , y are in cm and t in seconds . The velocity of the waves is ……….. . |
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Answer» 4 CM, 32 cm, 16 cm/s, 0.5 Hz |
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| 48. |
In the fig shown, a block of mass M is attached to the spring and another block of mass 2M has been placed over it. The system is in equilibrium. The block are pushed down so that the spring compresses further by (9Mg)/K. System is released (a) At what height above the position of release, the block of mass 2M will lose contact with the other block?(b) What is maximum height attained by 2M above the point of release? |
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Answer» (B)`(24 Mg)/k` |
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| 49. |
A ball strikes a horizontal floor at an angle theta = 45^(@) with the normal to floor. The coefficient of restitution between the bal and the floor is e = 1/2. The fraction of its kinetic energy lost in collision is |
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Answer» Solution :Let u be the velocity of ballbefore collision. Speed of the BALL after collision will become `v-sqrt(u^(2)+sin^(2)theta+e^(2)u^(2)cos^(2)theta)=sqrt(((u)/(sqrt(2)))^(2)+((u)/(2sqrt(2)))^(2))=sqrt((5)/(8)).`u `therefore` Fraction of KE lost in collision `((1)/(2)mu^(2)-(1)/(2)MV^(2))/((1)/(2)m u^(2))=1-((v)/(u))^(2)=1-(5)/(8)=(3)/(8)` |
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