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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. |
Noforce is required for |
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Answer» an object moving in straight line with constant velocity |
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| 2. |
A particle of mass m= 5 unit is moving with a uniform speed v= 3 sqrt2 unit in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the particle about the orgin is |
| Answer» Answer :A | |
| 3. |
Does friction also occur as actio-reaction pair? |
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Answer» Solution :YES, if there were no FRICTION, it WOULD not have been CARRIED by the WATER flow. |
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| 4. |
The work done by the body which is in S.H.M, against the restoring force is stored in the form of |
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Answer» K.E, |
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| 5. |
A person runs along a circular path of radius 5 m. If he completes half of the circle find the magnitude of the displacement vector, How far the person ran ? |
| Answer» Answer :A | |
| 6. |
The acceleration due to gravity is always down word i.e., along the - ve y direction can be choose this direction as the positive direction for the acceleration due to gravity? |
| Answer» SOLUTION :Yes, the direction can be TAKEN + ve for the .g. when the CASE is FREE fall of a body. | |
| 7. |
If surface tension of water is 0.07 N/m the weight of water supported in a capillary tube of radius 0.1 mm is |
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Answer» `4.4xx10^(-6)N` |
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| 8. |
An ice cart of mass 60 kg rests on a horizontal snow patch with coefficient of static friction 1/3. Assuming that there is no vertical acceleration, then the magnitude of the maximum horizontal force required to move the ice cart is (g = 9.8 ms^(-2)) |
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Answer» 100 N |
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| 9. |
A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is |
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Answer» SOLUTION :Inpulse= CHANGEIN LINEARMOMENTUM ` MV -(-MV) = 2MV` |
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| 10. |
can a room be cooled by leaving the door of a refrigator open ? |
| Answer» SOLUTION :A refrigator TAKES HEAT from the things kept inside it and REJECTS it to the ROOM .Hence the room can not be cooled by leaving its door open . | |
| 11. |
The equation motion of a particle in S.H.M. is a + 16 p^(2)x = 0. In the equation .a. is the linear acceleration (in m//sec^(2)) of the particle at a displacement .x. in meter. The time period of S H M in seconds is |
| Answer» ANSWER :B | |
| 12. |
Is rocket in flight an illustration ofprojectile ? |
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Answer» |
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| 13. |
A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 mcdot s^(-1). A plumb bob is suspended from the roof of the car by a light rigid rod of length 1.0 m . The angle made by the rod with the track is (g = 10 mcdot s^(-2) ) |
| Answer» ANSWER :C | |
| 14. |
An earth satellite is moving round the earth in a circular orbit. For such a satellite which of the following statement is wrong |
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Answer» it is a freely falling BODY |
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| 15. |
Two particles of combined mass M, placed in space with certain separation are released Interaction between the particles in only of gravitational nature and there is no extenal force present Acceleration of one particle with respect to the other when separation bewteen them is R has a magnitude |
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Answer» `(GM)/(2R^(2))` |
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| 16. |
An artifical satellite is orbiting in a circular orbit around the earth. One of the rockets of the satellite is momentarilyfired, the direction of firing of rocket is mentiond in column -I andcorresponding changes (s) are given in column-II match the entires of column -I with the entries of column-II. {:("Column-I",,"Column-II"),("(A) Towards the earth.centre",,"(P) orbitchangesand becomes elliptical"),("(B) Away from the earth.s centre",,"(Q) Orbitplane changes"),("(C) At right angle to the plane of orbit",,"(R) Semi-major axis of orbit increases"),("(D) In forward direction",,"(S) Energy of earth-satellite system increases"):} |
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Answer» |
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| 17. |
A quantity X is given by (me^(4))/(8epsilon_(0)^(2)ch^(3))where m is the mass of electron, e is the charge of electron, epsilon_(0) is the permittivity of free space, c is the velocity of light and h is the Planck's constant. The dimensional formula for X is the same as that of |
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Answer» length |
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| 18. |
A rod of length l and mass M is fixed to one end of a string . The string passes over a massless , frictionless pulley . A small ball of mass m (lt M) is attached to the other side of the string . The ball has an opening through which the string passes and the ball can slide along the string with some friction . Initially , the ball is held at level with the lower end of the rod . Whenset free , both the force of friction , if the ball rises to the level of the upper end of the rod in time t ? |
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Answer» Solution :When the system is set in motion , let `a_(M) , a_(m)` be the ACCELERATION of rod and ball respectively and F the force of friction applied by the string on the ball . The equation of MOTIONS are : `Mg - T = Ma_(M) ""……..(i)` ` F - mg = ma_(m) "" .......(ii)` The string is pulled up and the ball slides down due to its weight . Friction between the ball and string tries to hold it but ULTIMATELY it moves with acceleration `a_(m)` . As the string is massless , the net force on any portion of it must be zero . So , F = T Eqn . (i) can be written as : `Mg - F = Ma_(M) "" ....... (iii)` Adding eqns . (ii) and (iii) , `a_(m) + a_(M) = F((1)/(m) - (1)/(M)) "" ........ (iv)` As the rod moves DOWNWARD while ball upward , so , RELATIVE distance of the rod and ball in time t , `l = (1)/(2) (a_(m) + a_(M))t^(2)` Substituting for `(a_(m) + a_(M))` from eqn. (iv), `l = (1)/(2) F ((1)/(m) - (1)/(M))t^(2)` or `""F = (2 lmM)/(( M - m) t^(2))` |
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| 19. |
Water of volume 2L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27^(@)C" to "77^(@)C? (Specific heat of water is 4.2 kJ / kg and that of the container is negligible). |
| Answer» Answer :D | |
| 20. |
The KE required to project a body of mass .m. from the earth.s surface to infinity is |
| Answer» ANSWER :C | |
| 21. |
The momentum of a body in two perpendicular direction at any time 't' are given by P_x =2t^2+6 and P_y = (3t^2)/(2) + 3 . The force acting on the body at t = 2 sec is |
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Answer» 5 UNITS |
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| 22. |
A simple pendulum of length L_(1) has a time period 'T_(1)' and another pendulum of length L_(2) has time period 'T_(2)' Then the time period of pendulum of length (L_(1) +L_(2)) is |
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Answer» `T_(1)+T_(2)` |
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| 23. |
State the values of amplitude and frequency of a particle having the equation of simple harmonic motion as. |
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Answer» SOLUTION :`y=A sin omega t` Amplitude =A ANGULAR FREQUENCY `=omega` `:.` Frequency `=(omega)/(2PI)` |
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| 24. |
Figures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why |
| Answer» Solution :Fig(a) is incorrect [Reason: at a construction (i.e., where the area of cross SECTION of the TUBE is smaller). FLOW SPEED is larger due to mass conservation. Consequently pressure there is smaller according to Bernoulli.s equation. We ASSUME the fluid to be incompressible]. | |
| 25. |
The diffusion rate of hydrogen has an average value of 28.8 "cm"^(3) S^(-1). The diffusion of another gas under same contains is measured to have an average rate of 7.2 "cm"^(3) S^(-1) the gas is |
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Answer» nitrogen |
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| 26. |
A ballet dancer spins about a vertical axis at 60 rpm with arms outstretched. When her arms folded the angular frequency increases to 90 rpm. Find the change in her moment of inertia. |
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Answer» Solution :By the principle of conservation of angualr momentum `Ixx60=I_(2)xx90` FINAL moment of INERTIA, `I_(2)=(2I)/(3)` Change in moment of inertia `=I-(2I)/(3)=(I)/(3)` Assuming the initial I to be 100% change in moment of inertia `=(100)/(3)%=33(1)/(3)%`. |
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| 27. |
The height of a cylindrical tank is 0.4 m. The upper end of it open whose diameter is 0.16 m. It contains water up to a height of 0.16m. How much would be the time required to empty the tank through a hole of radius 10^(-3) m at the bottom of the tank? |
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| 28. |
The speed of a train increases at a consantrate alpha from zero to vand thenremains constant for an interval and finally decreases to zero at a constant rate beta. If l be the total distance described, and the total time then |
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Answer» `t =l/v + v/2 ( 1/alpha + 1/beta)` |
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| 29. |
Does potential energy of a spring decrease/increase when it is compressed or stretched ? |
| Answer» SOLUTION :When a spring is compressed or STRETCHED , potential energy of the spring increases in both the cases. This is because WORK is done in COMPRESSION as WELL as stretching . | |
| 30. |
The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi Meter.The wider part of the meter has a cross-sectional area equal to that of the artery. A=8mm^2. The narrow part has an area a=4 mm^2. The pressure drop in the artery is 24Pa. What is the speed of the blood in the artery? |
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Answer» SOLUTION :We TAKE the density of blood from table 10.1 to be `1.06 times 10^3 kg m^-3`. The ratio of the areas is `(A/a)=2` USING Eq.(10.17) we OBTAIN `v_1=sqrt((2 times 24 Pa)/(1060 kg m^-3 times (2^2 -1)))=0.123 ms^-1` |
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| 31. |
Assertion : A vector vecA can be resolved into component along with given vectors vecaand vecb lying in the same plane. Reason : vecA = lambda veca + mu vecb where lambda and mu are real numbers. |
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Answer» |
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| 32. |
A carpet of mass 'm' made of inextensible material is rolled along its length in the form of a cylinder of radius 'R' and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a very small push is given to it. Calculate the horizontal velocity of the axis of cylindrical part of the carpet when its radius reduces to R/2 . |
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Answer» Solution :Lose in PE DUE to unrolling when radius changes from `R` to `(R )/(2)` is `=MgR-((MgR)/(8))=(7)/(8)(MgR)` The lose in PE is equal to the gain in KE which is `K=K_(T)+K_(R )=(1)/(2)MV^(2)+(1)/(2)Iomega^(2)` If .V. is the velocity when half the carpet has UNROLLED then as `V=(Romega)/(2)`, `M to (M)/(4)` and `I=(1)/(2)((M)/(4))((R )/(2))^(2)` `:.(7)/(8)MgR=(3)/(16)MV^(2)` on solving `V=((14gR)/(3))^((1)/(2))` |
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| 33. |
The displacement of a particle varies with time according to the relation y= a sin omega t+b cos omega t…………… |
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Answer» the motion is oscillatory but not SHM comparing `y= A sin omega t cos phi +A cos omega t sin phi` `therefore a= A sin phi " and "b= A cos phi` `therefore a^(2)+ b^(2)= A^(2) (cos^(2) phi + sin^(2)phi)= A^(2)` `therefore A= sqrt((a^2)+(b^2))` `implies y= a sin omega t+ b cos omega t` `= A sin omega t sin phi + A cos omega t cos phi` `= A sin (omega t +phi)` `(dy)/(dx)= A omega cos (omega t +phi)` and `(d^(2)y)/(dt^(2))= -A omega^(2) sin (omega t+phi)` `= -omega^(2) (A sin omega t+phi)` `therefore a = -omega^(2)y` Hence it is EQUATION of SHM with amplitude `A= sqrt(a^(2)+b^(2))`. |
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| 34. |
A smooth sphere of mass m is moving on a horizontal plane with a velocity 3hati+hatj when it collides with a vertical wall which is parallel to the hatj vector. If the coefficient of restitution between the sphere and the wall is 1//2. find a. the velocity of the sphere after impact, b. the loss in kinetic energy caused by the impact. c. the impulse that acts on the sphere. |
Answer» Solution :Let `v` be the velocity of the sphere after the impact. Now we separate the velocity components PARALLEL and PERPENDICULAR tothe wall. Using the law of restitution, the COMPONENT of velocity paralel to the wall remains unchanged while the component perpendicular to the wall becomes e times in opposite direction.  Thus `vecv=-3/2hati+hatj` a. THEREFORE, the velocity of the sphere after the impact. `=-3/2hati+hatj` b. the loss in `KE =1/2mu^(2)-1/2mv^(2) ` `=1/2m(3^(2)+1^(2))-1/2m({3/2}^(2)+1^(2))=27/8m` C. `vecJ=/_\vecP=vecP_(f)-m(vecv)-m(vecu)` `=m(-3/2hati+hatj)-m(3hati+hatj)=-9/2mhati` |
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| 35. |
The linear speed of a rocket is not constant in its orbit. Comment. |
| Answer» Solution :The areal velocity of a planet around the sun is CONSTANT but its linear SPEED continuously CHANGES. The linear speed of the planet is large when it passes CLOSE to the sun and is small when the planet is away from the sun. | |
| 36. |
A bullet hits and gets embedded in a solid block resting on a frictionless surface. In this process which is correct ? |
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Answer» momentum alone |
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| 37. |
A person moving on a motor cycle in a ground takes a turn through 60^(@) on his left after every 50m. Then find the magnitude of displacement suffered by him after 9th turn |
| Answer» Answer :A | |
| 38. |
The height and width of each step of a staircase are 20cm and 30cm respectively. A ball rolls off the top of a stair with horizontal velocity v and hits the fifith step. The magnitude of v is (g = 10 ms^(-2)) |
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Answer» `1.5 SQRT(5) ms^(-1)` |
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| 39. |
When you exercise in the morning, by considering your body as thermodynamic system, which of the following is true? |
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Answer» `DeltaU GT 0, W GT0` |
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| 41. |
A boy is hanging from a horizontal branch of a tree. The tension in the arms will be maximum when the angle between the arms is |
| Answer» ANSWER :D | |
| 42. |
(A) : During solar eclipse, the solar spectrum is an emission spectrum (R) : During solar eclipse, the radiation from the elements in chromosphere are only received on earth. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct explanation of 'A' |
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| 43. |
A liquid having coefficient of real expansion .a. is in a vessel of coefficient of linear expansion /3. What laappens to the level of the liquid when heated by 10^(@)C? |
| Answer» SOLUTION :We know `gamma _(G) = gamma _(s) + gamma _(solid) = gamma _(app) + 3 alpha ,` From the date ,` gamma _(a) = a and alpha =a//3 THEREFORE a = gamma _(s) + 3 (a //3) implies gamma _(s) =0`It indicates that the apparent increase in VOLUME is ZERO. Hence the level of the liquid remains constant. | |
| 44. |
A ball is thrown vertically upwards with a certain velocity. Then |
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Answer» during its UPWARD motion WORK done by gravity is negative |
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| 45. |
The ratio of maximum acceleration to maximum displacement of a Simple Harmonic Motion is equal to |
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Answer» SQUARE of its ANGULAR VELOCITY |
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| 46. |
The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction when the angle of inclination of the plane is 60^@ is |
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Answer» `1/3` |
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| 47. |
A gun is mounted on a gun carriage movable on a smooth plane and the gun is elevated at an angle alpha to the horizontal. A shot, fired leaves the gun in a direction inclined at an angle theta to the horizontal. If the mass of the gun and its carriage b n times that of the shot, find tan theta : tan alpha |
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Answer» |
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| 49. |
Two rotating bodies A and B of masses m and 2 m with moments of inertia I_(A) and I_(B) (I_(B) gt I_(A)) have equal kinetic energy of rotation . If L_(A) and L_(B) be their angular momenta respectively , then |
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Answer» `L_(A) = (L_(B))/(2)` Both bodies A and B have EQUAL kinetic energy of rotation `k_(A) = k_(B) IMPLIES (1)/(2) I_(A) omega_(A)^(2) = (1)/(2) I_(B) omega_(B)^(2) implies (omega_(A)^(2))/(omega_(B)^(2)) = (I_(B))/(I_(A)) "" …. (i)` Ratio of ANGULAR momenta , `(L_(A))/(L_(B)) = (I_(A) omega_(A))/(I_(B) omega_(B)) = (I_(A))/(I_(B)) xx sqrt((I_(B))/(I_(A))) ""` [Using eqn. (i)] `= sqrt((I_(A))/(I_(B)) lt 1 (because I_(B) gt I_(A)) therefore L_(B) gt L_(A)` |
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| 50. |
A ladder 'AB' 2.5m long and of weight 150N with its centre of mass at a distance 1m from end 'A' is on the ground. A 40N weight is attached to the end B. The work to be done to arrange the ladder in vertical position with end 'A' contact with the ground is |
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Answer» 190J |
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