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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. |
In the figure shown, m_(a)=m_(b)=1kg block-A is neutral while block-B carriescharge -1c. Sizes of A and B are negligible. Block-B is released from rest at a distance 1.8m from A. Initially spring is neither compressed nor stretched. If collision between A and B is perfectly inelastic, then velocity of the combined mass just after collision is |
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Answer» `6m//s` |
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| 2. |
A particle of mass 10 gm is executing S.H.M with an amplitude of 0.5 m and angular frequency 10 rad/sec. The maximum value of force acting on the particle during the course of oscillation is |
| Answer» ANSWER :D | |
| 3. |
The minimum force required to move a body up an inclined plane is two times the minimum force required to prevent it from sliding down the plane. If co-efficient of friction between the body and inclined plane is (1)/(sqrt(3)), the angle of inclined plane is |
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Answer» `60^(@)` |
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| 4. |
In the figure shown, m_(a)=m_(b)=1kg block-A is neutral while block-B carriescharge -1c. Sizes of A and B are negligible. Block-B is released from rest at a distance 1.8m from A. Initially spring is neither compressed nor stretched. The amplitude of oscillation of the combined mass |
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Answer» `(2)/(3)m` |
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| 5. |
In the figure shown, m_(a)=m_(b)=1kg block-A is neutral while block-B carriescharge -1c. Sizes of A and B are negligible. Block-B is released from rest at a distance 1.8m from A. Initially spring is neither compressed nor stretched. Equilibrium position of the combinedmass is at X= |
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Answer» `(-2)/(9)m` |
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| 6. |
If the maximum vertical height and horizontal ranges of a projectile are same, the angle of projection will be |
| Answer» Answer :B | |
| 7. |
A 400 kg satellite is in a circular orbit of radius 2R_(E) about the Earth. How much energy is required to transfer it to a circular orbit of radius 4R_(E) ? |
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Answer» Solution :Initially `E_(i) = (-GM_(E)m)/(4R_(E))` while FINALLY `E_(F) = -(GM_(E)m)/(8R_(E))` The CHANGE in the total energy is `Delta E = E_(f) - E_(i)` `= (GM_(E)m)/(8R_(R)) = ((GM_(E))/(R_(E)^(2)))(mR_(E))/(8)` `Delta E = (gm R_(E))/(8) = (9.81 XX 400 xx 6.37 xx 10^(6))/(8)` `= 3.13 xx 10^(9)J` `T^(2) = (4pi^(2) R^(3))/(GM_(E)) M_(E) = (4pi^(2)R^(3))/(G T^(2))` `= (4 xx 3.14 xx 3.14 xx (3.84)^(3) xx 10^(24))/(6.67 xx 10^(-11) xx (27.3 xx 24 xx 60 xx 60)^(2))` `= 6.02 xx 10^(24) kg` Both methods yields almost the same answer, the difference between them being less than 1%. |
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| 8. |
The amount of work done to move charge from onepoint to another is called ___________. |
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Answer» ENERGY |
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| 9. |
(A) : Magnitude of average velocity is equal to average speed, if velocity is constant. ( R) : If velocity is constant, then there in no change in the direction of motion. |
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Answer» Both (A) and ( R) are ture and ( R) is the CORRECT EXPLANATION of (A) |
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| 10. |
In elastic collision, 100% energy transfer takes place when …………….. |
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Answer» `m_(1) = m_(2)` |
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| 11. |
Answer the following : Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? |
| Answer» Solution :For a CONSERVATIVE FORCE work done over a PATH is minus of CHANGE in potential energy. Over a complete ORBIT, there is no change in potential energy. | |
| 12. |
A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then what is the kinetic energy of the combined system (bag + bullet) |
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Answer» |
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| 13. |
A ladder .AB. of weight 300 N and length 5m is lying on a horizontal surface. Its centre of gravity is at a distance of .2m. from end A. A weight of 80 N is attached at end B. The work done in raising the ladder to the vertical position with end .A. in contact with the ground is, |
| Answer» ANSWER :B | |
| 14. |
A car of mass 1250 kg experiences a resistance of 750 N when it moves at 30 ms^(-1). If the engine can develop 30 KW at this speed, the maximum acceleration that the engine can produce is |
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Answer» `0.8 MS^(-2)` |
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| 15. |
A child sits stationary at one end of a long trolley moving uniformly with a speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the centre of mass of the (trolley + child) system ? |
| Answer» Solution :SINCE the force DUE to the body.s motion is PURELY internal. | |
| 16. |
Passage - I : Four identical rods of mass m=6kg each are welded at their ends to form a square and then welded to a massive ring having man m = 4kg and radius R=1m. If the system is allowed to roll down the incline of inclination theta=30^(0) determine the minimum value of the coefficient of static friction that will prevent slipping The minimum value of friction coefficient to prevent slipping is |
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Answer» `5/7` |
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| 17. |
Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (omega is any positive constant): (a) sinomegat-cosomegat (b) sin^(3)omegat (c) 3cos(pi//4-2omegat) (d) cosomegat+cos3omegat+cos5omegat (e) exp(-omega^(2)t^(2)) (f) 1+omegat+omega^(2)t^(2) |
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Answer» SOLUTION :(a) Simple harmonic, `T = (2pi//omega)`, (b) periodic, `T =(2pi//omega)` but not simple harmonic, (c) simple harmonic, `T = (pi//omega)`, (d) periodic, `T = (2pi//omega)` but not simple harmonic, (E) non-periodic, (F) non-periodic (PHYSICALLY not acceptable as the FUNCTION `to oo` as `t to oo` |
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| 18. |
A particle is subjected to two S H Ms x_(1) = A_(1) sin wt and x_(2) = A_(2)sin (wt + pi//4). The resultant SH M will have an amplitude of |
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Answer» `(A_(1)+A_(2))/(2)` |
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| 19. |
Passage - I : Four identical rods of mass m=6kg each are welded at their ends to form a square and then welded to a massive ring having man m = 4kg and radius R=1m. If the system is allowed to roll down the incline of inclination theta=30^(0) determine the minimum value of the coefficient of static friction that will prevent slipping The acceleration of the system will be |
| Answer» ANSWER :C | |
| 20. |
Passage - I : Four identical rods of mass m=6kg each are welded at their ends to form a square and then welded to a massive ring having man m = 4kg and radius R=1m. If the system is allowed to roll down the incline of inclination theta=30^(0) determine the minimum value of the coefficient of static friction that will prevent slipping The moment of inertia of the system bout the centre of ring will be |
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Answer» `20" KG m"^(2)` |
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| 21. |
When a few water drops placed between two glass plates, it is difficult to separate the glass plates? Why? |
| Answer» SOLUTION :DUE to SURFACE TENSION. | |
| 22. |
"We supply electric current continuously to an electric heater. But the temperature of the heater remains constant after some time." Why ? |
| Answer» Solution :The heater attains steady state when the RATE of production of HEAT BECOMES EQUAL to the rate of LOSS by conduction, convection and radiation. | |
| 23. |
How does the coefficient of viscosity of a liquid change with temperature? |
| Answer» Solution :With the INCREASE of temperature, VISCOSITY of liquid decreases. In liquids when temperature increases the magnitudes of the cohesive forces decrease as the MOLECULAR separation increass. As viscosity is DUE to consesive forces, with the increase of temperature, COEFFICIENT of viscosity decreases. | |
| 24. |
A student performs an experiment for determinationof g= (4pi^2L)/(T^2) , L = 1m and the commits an error of Delta Tfor the tajes the time of n oscillations with the stop watch least count Delta T. For which of the following data the measurement of g will be more accurate? |
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Answer» `Delta L = 0.5 , Delta T = 0.1 , N = 20` |
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| 25. |
The vander waal's equation for a gas is (p + a/V^(2))(V - b) = nRT where p, V, R, T and n represent the Pressure, Volume, universal gas constant, absolute temperature and number of moles of a gas respectively, where a and b are constants. The ratiob/a will have the following dimensional formula |
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Answer» `M^(-1)L^(-2)T^(2)` |
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| 26. |
A refrigator with its power on , is kept in a closed room with its door open . The temperature of the room will |
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Answer» rise |
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| 27. |
In a simple harmonic oscillation, the acceleration against displacement for one complete oscillation will be. |
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Answer» an ellipse |
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| 28. |
The length of a pencil is (15.1+-0.1) cm. A small piece of length is cut out so that the length of the remaining pencil is (11.9+-0.2) cm what is the length of the small piece? |
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Answer» |
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| 29. |
A physicist performs an experiment and takes 200 readings. He repeats the same experiment and now takes 800 readings. By doing so the probable error |
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Answer» REMAINS the same |
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| 30. |
A pendulum is hanging from the ceiling of a cage. When the cage is moving up with certain acceleration and when it is moving down with the same acceleration, the tensions in the string are T_1 and T_2respectively. When the cage moves horizontally with the same acceleration, the tension in the string is, |
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Answer» `SQRT(2(T_1^2 + T_2^2) )` |
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| 31. |
The momentum of a system of particles is always conserved trueor false |
| Answer» SOLUTION : True. The total MOMENTUM of a system of particles is ALWAYS constant i.e., CONSERVED. When no external FORCE acts on it. | |
| 32. |
Give the advantages of selecting wavelength of light as a standard of length. |
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Answer» SOLUTION :1. Wavelength is not AFFECTED by temperature, pressure etc. 2. It does not depend on other physical quantities. 3. Wavelength can be reproduced easily. 4. Wavelengt does not CHANGE with time. |
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| 33. |
A scientist measures the time period of a simple pendulum as T in a lift at rest. If the lift moves up with accelerationi as one fourth of the acceleratioin of gravity, the new time period is |
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Answer» `T/4` |
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| 34. |
A massive box is dragged along a horizontal floor by a rope. The rope makes an angle of 60^(@) with the horizontal. Find the work done if the tension in the rope is 200 N and the box is moved through a distance of 10 m. |
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Answer» |
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| 35. |
A steel scale is correct at 0^(@)C. The length of a brass tube measured by it at 40^(@)C is 4.5m. The correct length of the tube at 0^(@)C is (Coefficients of linear expansion of steel and brass are 11 xx 10^(-6)//^(@) Cand 19 xx 10^(-6)//^(@)C respectively). |
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Answer» 4.001m |
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| 36. |
Two bodies are projected simultaneously from the ground with the same speed, the first one inclined at 30^@ and the second one at 60^@ with the horizontal. Then |
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Answer» the first ONE falls in FRONT of the SECOND |
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| 37. |
Two small spheres of masses 10kg and 30kg are joined by a rod of length 0.5m and of negligible mass . The M.I of the system about a normal axis through centre of mass of the system is |
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Answer» `1.875kgm^(2)` |
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| 38. |
Estimate the mean free path for a water molecule in water vapour at 373 K. |
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Answer» Solution :The number density (N) is inversely proportional to absolute TEMPERATURE. `n alpha 1/T RARR (n_(373))/(n_(273))-(273)/(373)` `rArr n_(373)=n_(273) xx (273)/(373)` `n_(373)=2.7 xx 10^(25) xx (273)/(373)=2 xx 10^(25)m^(-3)` Given `d=2 xx 10^(-10)m` Hence, mean free path `lambda=(1)/(SQRT2 pi d^2 n)` `=(1)/(sqrt2 xx 3.14 xx (2 xx 10^(-10))^(2) xx 2 xx 10^(25))=4 xx 10^(-7)m` |
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| 39. |
The minimum force required to move a body up an inclined plane is two times the minimum force required to prevent it from sliding down the plane. If co-efficient of fraction between the body and inclined plane is 1/(sqrt3),find the angle of inclined plane. |
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Answer» SOLUTION :`60^(@)` |
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| 40. |
A. Kerosene oil spreads out over water surface R: Surface tension is a surface phenomenon |
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Answer» Both A and R are true and R is the correct EXPLANATION of A |
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| 41. |
A stone of mass 0.5 kg is attached to a string of length 2m and is whirled in a horizontal circle. If the string can with stand a tension of 9N, the maximum velocity with which the stone can be whirledis |
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Answer» `6 MS^(-1)` |
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| 42. |
Describe the mutual relationship between the fig and wasp and comment on the phenomenon that operates in this relationship. |
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Answer» Solution :(i) Portion OA:In this region, stress is very small such that stress is proportional to strain, which means Hooke's law is valid. The point A is called limit of proportionality because above this point Hooke's law is not valid. The slope of the line OA gives the Young's MODULUS of the wire. (ii) Portion AB:This region is reached if the stress is increased by a very small amount. In this region, stress is not proportional to the strain. But once the stretching force is removed, the wire will regain its ORIGINAL length. This behaviour ends at point B and hence, the point B is known as YIELD point. The elastic behaviour of the material in stress-strain CURVE is OAB. (iii) Portion BC:If the wire is stretched beyond the point B, stress is increases and the wire will not regain its original length after the removal of stretching force. (iv) Portion CD:With further increase in stress (beyond the point C), the strain increases rapidly and reaches the point D. Beyond D, the strain increases even when the LOAD is removed and breaks (ruptures) at the point E. Hence, the maximum stress (here D) beyond which the wire breaks is called breaking stress or tensile strength. The corresponding point D is known as fracture point. The region BDDE represents the plastic behaviour of the material of the wire. |
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| 43. |
A block of mass m is arranged on the wedge as shown in figure . The wedge angle istheta . If the masses of pulley and thread and negligible and friction is absent , find the acceleration of the wedge |
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Answer» Solution :It is obvious that when block m moves downward along the incline of wedge , the wedge moves to the right . As the length of thread is constant , the distance traversed by wedge along the incline is equal to distance traversed by wedge to the right. This implies that acceleration of wedge to the right is equal to downward acceleration of wedge . Let a be the acceleration of wedge to the right. Then the force acting on the block m are  i) weight mg acting vertically downward ii) normal reaction `R_1` iii) tenstion T ( up the incline ) iv) Fictitious force ma to the left . The free body DIAGRAM of mass m is shown in FIGURE . For motion of block .m. on inclined plane `mg sin theta + ma cos theta -T = ma `.......... (1) As mass .m. does not breaks off the inclined plane, therefore for forces on ..m. normal to inclined plane `R_1 + ma sin theta = mg cos theta `.......... (2) The forces acting on the wedge are i) weight Mg downward ii) normal reaction of ground on wedge =`R_2` iii) normal reaction of block on wedge `=R_1` iv) tension (T,T) in string . The free body diagram of wedge is shown in figure .  For maotion of wedge in horizontal DIRECTION `R_1 sin theta = T-T cos theta = Ma`......... (3) For vertical equilibrium of wedge on surface `R_2 = R_1 cos theta + T sin theta + Mg `.......... (4) From (1) , `T= mg sin theta + ma cos theta - ma `........ (5) From (2) , `R_1 = mg cos theta - ma sin theta `............. (6) SUBSTITUTING these values in (3) , we get `(mg cos theta - ma sin theta ) sin theta + (mg sin theta + ma cos theta - ma) (1- cos theta ) = Ma` `or { M + msin^2 theta + m(1- cos theta )^(2)}a` `= ma cos thetasin theta + mg sin theta (1-cos theta )` ` a= ( mg sin theta )/( M +m sin^(2) theta + m(1- 2 cos theta + cos^(2) theta)) = ( mg sin theta )/( M + m(sin^2 theta + 1 -2 cos theta + cos^(2) theta ))` ` :. a=(mg sin theta )/( M + 2m(1-cos theta))` |
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| 44. |
A car starts from rest and accelerates at uniform rate of 6 m s^(-2)for some time, then moves with constant speed for some time and retards at the same uniform rate and comes to rest. Total time for the journey is 24 s and average speed for journey is 20 m s^(-1) , How long does the car move with constant speed? |
| Answer» Answer :D | |
| 45. |
A given mass of a gas expands ffrom a state A to the state B by three paths 1,2 and 3 as shown in T-V indicator diagram. If W_(1), W_(2) and W_(3)respectively be the work done by the gas along the three paths, then |
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Answer» `W_1gtW_2gtW_3` `W_(1)gtW_(2)gtW_(3)` |
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| 46. |
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. |
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Answer» `2.74 XX 10^(-5) ` |
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| 47. |
A man moving with 5m//sobserves rain falling vertically at the rate of 10m//s . Find the speed and direction of the rain with respect to ground . |
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Answer» Solution :`V_(RM)=10m//s`, `V_(M)=5m//s` `vecV_(RM)=vecV_(RU)-vecV_(M)` `rArr vecV_(Ru)=V(rm)-vecV_(M)rArr vecV_(R)=5sqrt5` `tantheta=1/2, THETA ="tan"^(-1)1/2` 
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| 48. |
A diatomic gas (gamma = 1.4) does 300J work whenit is expanded isobarically. The heat given to the gas in this process is |
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Answer» 1050J |
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| 49. |
A particle is moving eastwards with a velocity of 5m/s. In 10s the velocity changes to 5m/s northwards. Find the average accelration is this time |
Answer» Solution :`veca_(AV) = (vec(DELTAV))/(DELTAT)=(vecv_(f)-vecv_(i))/(Deltat)`  `veca_(av)=(5hat(J)-5hat(i))/(10)=(5sqrt(2))/(10)=(1)/(sqrt(2))"m/s"^(2)`. North-west direction. |
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| 50. |
When a motorcyle moving with a unifrom speed 11 m/s is at a distance 24 m from a car, the car starts from rest and moves with a uniform acceleration 2 m//s^(2) away from the motorcyle. If the car begins motion at t = 0, time at which the motorcyle will overtake the car is t = ? |
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Answer» Solution :Let motorcycle meets the car after time t. For motorcycle: `x = 11 t + 24` For car: `x = (1)/(2) at^(2)` `24 + 11 t = (1)/(2) xx 2 xx t^(2)` `t^(2) - 11 t - 24 = 0` `t = 3 sec` |
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