Saved Bookmarks
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. |
Find the which of the following are correct using dimensional analysis. (i) t^(2)=4pi^(2)a^(3) (ii) t^(2)=(4pi^(2)a^(3))/G (iii) t^(2)=(4pi^(2)a^(3))/(GM) where t is the time period, a is the radius of the orbit of a planet and M is the mass of sun. |
|
Answer» This equation is not correct (iii) `a^(3)//GMtoL^(3)//M^(-1)L^(3)T^(-2)M=[T^(2)]`. This equation is correct. |
|
| 2. |
A piece of wood of volume 0.6 m^(3)floats in water. Find the volume exposed What force is required to meet completely under water? Density of wood =800 kg//m^(3). |
|
Answer» |
|
| 3. |
A tank, which is open at the top, contains a liquid up to a height H. A small hole is made in the side of the tank at a density y below the liquid surface. The liquid emerging from the hole lands at a distance x from the tank. a) If y is increased from zero to H, x will first increase and then decrease. b) x is maximum for y=H//2. c) The maximum value of x is H. d) The maximum value of x will depend on the density of the liquid. |
|
Answer» a, B and C are correct |
|
| 4. |
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming the his dimensity remains same, the stress in the leg will change by a factor of |
|
Answer» 9 As density REMAINS same, `prop` volume So, `m_f/m_i=(9)^3` Also,`A_f/A_i=(9)^2` `Stress=(force)/(area)=(mtimesg)/A` `thereforeS_f/S_i=m_f/m_itimesA_i/A_f=(9)^3times1/(9)^2=9` |
|
| 5. |
A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to force Fsinomegat. If the amplitude of the particle is maximum for omega=omega_1 and the energy of the particle is maximum for omega=omega_2, then (where omega_0 natural frequency of oscillation of particle) |
|
Answer» `omega_1=omega_0` and `omega_2!=omega_0` For amplitude RESONANCE (maximum amplitude), frequency of external force `omega=sqrt(omega_0^2-((b)/(2M))^2)impliesomega_1!=omega_0` |
|
| 6. |
The work done by sun on Earth at any finite interval of time is |
|
Answer» POSITIVE, negative or zero Work done = W = F..d `cos theta "Since " theta = 90^(@)` ` = F. delta (0) "Cos " 90^(@) = 0 ` W=0 |
|
| 7. |
A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observe is f_1. If the train'sspeed is reduced to 17 m/s, the frequency registered is f_2 If the speed of sound is 340 m/s, then find the ratio f_1//f_2 . |
|
Answer» SOLUTION :For the stationary observer `f_1 = V/(v-v_s)xxf` `:. F_1 = 340/(340 - 34) xx F and f_2 = 340/(340-17) xx f` Hence `f_2/f_1 = (340-17)/(340-34)= 19/18 rArr f_1 : f_2= 19 : 18` |
|
| 8. |
The thickness of ice in a lake is 5cm and the atmospheric temperature is -10^(@)C. Calculate the time required for the thickness of ice to grow to 7cm. Thermal conductivity of ice = 4xx10^(-3)cal//cm-s-""^(@)C, density of ice = 0.92g//cm^(3) and latent heat of fusion for ice = 80 cal/g. |
|
Answer» Solution :USING `DELTAT=(rhoL)/(2KT)(x_(2)^(2)-x_(1)^(2))` we have, the required time `Deltat` as `Deltat=(92xx10^(-2)g//cm^(3))/(2xx4xx10^(-3)cal//cm-s-""^(@)C)XX(80cal//gm)/(10^(@)C)xx(7^(2)-5^(2))cm^(2)` = `(92xx80xx24)/8s=22080s=6.13hr`. |
|
| 9. |
A vertical off-shore structure is built to withstand a maximum stress of 10^(9) Pa. Is the structure suitable for putting up on top of an oil well in Bombay High? Take the depth of the son to be roughly 3 km, and ignore ocean currents. |
|
Answer» <P> |
|
| 10. |
Both the earth and the moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon' |
|
Answer» will be ELLIPTICAL |
|
| 11. |
In a carnot heat engine 8000J of heat is absorbed from a source at 400 K and 6500 J of heat is rejected to the sink. The temperature of the sink is ………. . |
|
Answer» 325 K |
|
| 12. |
A small block of mas M slides down from top edge A of a smooth curved surface as shown in the figure. The surface become horizontal at edge B. The maximum possible horizontal range for the body is |
|
Answer» H |
|
| 13. |
Give the quantities for which the following are the dimensions: M^(-1)L^(3)T^(-2) |
|
Answer» |
|
| 14. |
A particle of mass 4g is in a gravitational potential field given by V=(800 x^(2) + 150) erg/g . Its frequency of oscillations is |
|
Answer» `100 pi` Hz |
|
| 15. |
In a damped oscillatory motion a block of mass 200 g is suspended to a spring of force constant 90 N/m in a medium and damping constant is 40 g/s. Find (a) time period of oscillation(b) time taken for its amplitude of oscillation to drop to half to its initial value (c) time taken for its mechanical energy to drop to half of its initial value. |
|
Answer» Solution :MASS m = 200 g = 0.2 kg FORCE constant k = 90 N/m damping constant b= 40 g/s = 0.04 kg/s `sqrt(KM) = sqrt(90xx0.2) = sqrt(18) kg//s` Here `b lt lt sqrt(km)` a) time period `T= 2PI sqrt((m)/(k)) = 2pi sqrt((0.2)/(90)) = 0.3s` b) amplitude `=Ae^(-bt//2m)` Let amplitude is dropped to half of its initial value after the time `T_(1//2)` . Amplitude `A.e^(-b.T_(1//2))/(2m) = A/2` `e^(-b(T_(1//2)))/(2m) = "In" (1/2)` `impliesT_(1//2) = ("In" (2))/(b//2 m) = 2.302xx0.3010xx2m//b` `T_(1//2) =0.693xx(2m)/(b) = 0.693 xx (2xx0.2)/(0.04) =6.93 s` c) Let the energy is dropped to half of its initial value after a time `t_(1//2)` Initial energy `E_0 = 1/2 KA^2` At time `t_(1//2) `, energy `=1/2 E_0 = 1/2 kA^2 e^(-b.t_(1//2))/m = 1/2 (1/2 kA^2)` `e^(-b.t_(1//2))/(m) =1/2 impliest_(1//2) = LN (2) xx m/b = 0.693 xx m/b` `t_(1//2) = 0.693 xx(0.2)/(0.04) =3.46 s` |
|
| 16. |
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is……. |
|
Answer» Acceleration `a= -omega^(2)x` but `|v|=|a|` `therefore omega sqrt(A^(2)-x^(2))= omega^(2)x` `sqrt(A^(2)-x^(2))= omega x` `therefore A^(2)- x^(2) = omega^(2) x^(2)` `therefore omega^(2) = (A^(2)-x^(2))/(x^(2))` `= ((3)^(2)-(2)^(2))/((2)^(2)` `= (9-4)/(4)` `=(5)/(4)` `therefore omega = (sqrt(5))/(2)` Now`omega = (2pi)/(T)"" therefore T= (2pi)/(omega)` `therefore T= (2pi)/((sqrt(5))/(2))= (4pi)/(sqrt(5))`. |
|
| 17. |
Angle between vecAand vecB" is "theta_1 and angle between vecAand vecC" be "theta_2."If "vecA*vecB=vecA*vecC," then "vecB=vecC. |
|
Answer» |
|
| 18. |
What do you understand by the term Reverberation ? |
| Answer» SOLUTION :In a closed room the sound is REPEATEDLY reflected from the walls and it is even heard long after the sound soure cease to function. The residual sound remaining in an enclosure and the PHENOMENON of MULTIPLE reflections of sound is CALLED reverberation. | |
| 19. |
Three sphere of masses m_,m andm_ are located at the vertices of an equilateral triangle having side of same length. Find the moment of inertia of the system about one of side of triangle. |
|
Answer» `(3)/(4)ml^(2)`  Suppose equilateral triangle `DELTAABC` has length l and mass m are ARRANGED on each of its vertices. The moment of inertia about an axis passing one of side `bar(BC)` of triangle is I, then `I=mr^(2)`, where r = distance of sphere A from BC From right angle `DeltaAPB` `l^(2)=r^(2)+((l)/(2))^(2)` `therefore r^(2)=l^(2)-(l^(2))/(4)=(3l^(2))/(4)` `therefore I=mxx(3l^(2))/(4)` `=(3)/(4)ml^(2)` |
|
| 20. |
A body of mass 4 kg is attached to another body of mass 2 kg with a massless rod. If 4 kg mass is at (2hati + 5hatj)m, and 2 kg mass at (4hati + 2hatj)m, find the centre of mass of that system. |
|
Answer» |
|
| 21. |
A thermos bottle containing coffee is vigorously shaken. If coffee is considered a system, (b)has heat been added to it? |
| Answer» SOLUTION :No. THERMOS BOTTLE is an INSULATOR | |
| 22. |
Moment of a couple is called |
|
Answer» impulse |
|
| 23. |
Calculate the distance travelled by a body before coming to rest if it is moving on a rough surface with a velocity 18 kmph and the coefficient of friction between the body and the surface is 0.4 (g=10m//s^(2)) |
|
Answer» SOLUTION :`25//8m` |
|
| 24. |
A heavier sphere moving eastward with a certain velocity 'v' collides with a lighter sphere at rest. If it is perfect elastic head on collision, then after collision |
|
Answer» heavier SPHERE MOVES west ward with same speed |
|
| 25. |
A perfect gas at 27^(@)C is heated at constant pressure so as to double its volume. The temperature of the gas becomes. |
|
Answer» `600^(@)C` |
|
| 26. |
Two capillaries P and Q are dipped in water. The height of water level in capillary P is 2/3 of the height in Q capillary. The ratio of diameters is |
|
Answer» `2:3` |
|
| 27. |
A constant power .P. is applied to a particle of mass .m.. The displacement of the particle when its velocity increases from upsilon_(1) to upsilon_(2) is (ignore friction) |
|
Answer» SOLUTION :Power P = F. `UPSILON = (ma)upsilon` `a=(P)/(m upsilon) rArr upsilon(d upsilon)/(DS)=(P)/(m upsilon)` `upsilon^(2).d upsilon=(P)/(m)" ds" (P)/(m) int_(0)^(s) ds = int_(upsilon_(1))^(upsilon_(2))upsilon^(2).d upsilon` `(P)/(m).s=(1)/(3)(upsilon_(2)^(3)-upsilon_(1)^(3)) "" therefore s=(m)/(3P)(upsilon_(2)^(3)-upsilon_(1)^(3))` |
|
| 28. |
In which of the following cases the planet will have same value of .g. as that earth (a)with mass double to that of earth and radius sqrt2 times that of earth (b)with mass fourtimes that of earth and radius sqrt2 times that of earth (c)with mass one fourth that of the earth and radius half that of earth (d)with mass twice that of earth and radius 1/3 that of earth |
|
Answer» a and B only |
|
| 29. |
Some amount of gas at 27^(@)C is suddenlycompressed to 8 times its initial pressure. If gamma = 1.5, find out the rise in temperature. |
|
Answer» <P> Solution :As the gas is suddenly compressed, the PROCESS is adiabatic.So, `T_(1)^(gamma) p_(1)^(1-gamma) = T_(2)^(gamma)p_(2)^(1-y) or, ((T_1)/(T_2))^(gamma) = ((p_2)/(p_1))^(1 - gamma)` or, `T_(2) = T_(1) ((p_1)/(p_2))^((1-gamma)/gamma)` Here, `T_(1) = 27^(@)C = 300 K , (p_2)/(p_1) = 8`, `gamma = 1.5 = 3/2 or, 1 - gamma = 1 - 3/2 = -1/2` or, `(1-gamma)/(gamma) = -1/3` `:. (1-gamma)/(gamma) = -1/3` `:.T_(2) = 300 XX (1/8)^(-1//3)` `= 300 xx (8)^(1//3)` ` = 300 xx 2 = 600 K` `:.` Rise in temperature = 600 - 300 = 300 K = 300^(@)C`. |
|
| 30. |
Which of the following is not a perfectly inelastic collision ? |
|
Answer» Striking of two glass balls |
|
| 31. |
Assertion:The parts of a machine are jammed in winter.Reason:The viscosity of lubricants used in machines increases at low temperature.Select the correct statement of the follwing statements. |
|
Answer» Both assertion and REASON are TRUE. |
|
| 32. |
The acceleration of a particle is found to be non zero while no force acts on the particle. This is possible if the measurement is made from |
|
Answer» inertial frame |
|
| 33. |
A body of mass 4 kg is executing shm of amplitude 0.4 m and period 2 s. Find (i) the maximum restoring force and (ii) the restoring force at a distance of 0.1 m from extreme position. |
|
Answer» |
|
| 34. |
What is coefficient of performance of a refrigerator ? |
| Answer» Solution :It is the RATIO of heat extracted `Q_2` in the refrigerator to the WORK done W on the REFRIGERANT. It is denoted by `ALPHA`. | |
| 35. |
A sphere is rolled on a rough horizontal surface It graduatly shown down and stops. The force of friction tries to |
|
Answer» increase the LINEAR velocity |
|
| 36. |
In which of the following process, convection does not take place primarily ? |
|
Answer» Sea and land BREEZE |
|
| 37. |
A ball of mass m collides with the ground at an angle a with the vertical If the collision lasts for time t, the average force exerted by the ground on the ball is : (e= coefficient of restitution between the ball and the ground) |
Answer» SOLUTION : `therefore` `Ft = m (eu cos ALPHA + u cos alpha) (or) F = ( m u cos alpha (1 + E ))/( t)` |
|
| 38. |
When the temperature of a gas filled in a closed vessel is increased by 1^(@)C, its pressure increases by 0.4 percent. The initial temperature of gas was |
| Answer» Answer :C | |
| 39. |
The liquid drop of density rho, radius r and surface tension sigma oscillates with time period T. Which of the following expression for T^(2) is correct |
| Answer» Answer :A | |
| 40. |
A seconds pendulum is arranged in a lift. If the lift is moving up with an acceleration g//4. Its new time period will be 1 |
| Answer» Answer :C | |
| 41. |
A system goes from A to B via two processes I and II as shown in the figure. IfDeltaU_(1)and DeltaU_(2)are the changes in the internal energies in the processes I and II respectively, then - |
|
Answer» `DeltaU_(1)=DeltaU_(2)`  `:' (DELTAU)_(I)=(DeltaU)_(II)` & [depends only on path not upon initial and final state] |
|
| 42. |
Define displacement and distance. |
|
Answer» Solution :DISTANCE is the actual path length travelled by an object in the given interval of TIME during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial POSITIONS of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity. |
|
| 44. |
Define frequency of simple harmonic motion. |
|
Answer» SOLUTION :The NUMBER of oscillations produced by the particle per second is called frequency. It is DENOTED by f. SI unit for frequency is `s^(-1)` or HERTZ (Hz). Mathematically, frequency is RELATED to time period by `f=(1)/(T)` |
|
| 45. |
A body of mass M with a small block of mass placed on it reats on a smooth horizontal surface with velocity. The block is given a velocity v as shown To what height (relative to the initial velocity level) will the block rise after breaking off the body ? Neglect friction. |
|
Answer» Solution :Let it rise to a height H. Applying the law of CONSERVATION of energy, we get `(1)/(2) mv^(2) = mgh + (1)/(2) (m + M) v_(1)^(2)""...(1)` where `v_(1)` is the velocity of the combined system. Again, applying the law of conservation of momentum, we get `mv = (m + M) v_(1)` `v_(1) = (mv)/(2(m+M))""...(2)` Substituting the VALUE of `v_(1)` from eq (2) in eq (1) and solving, we get `therefore h = (Mv^(2))/(2(m+M)G)` |
|
| 46. |
The pressure and density of a monoatomic gas(gamma = 5//3)change adiabitically from(P_(1), d_(1)) to (P_(2), d_(2)). If (d_(2))/(d_(1))=8then(P_(2))/(P_(1)) should be |
| Answer» ANSWER :B | |
| 47. |
State the law of conservation of linear momentum. |
| Answer» Solution :If no EXTERNAL force ACTS on a SYSTEM, the linear momentum of the system REMAINS constant. | |
| 48. |
A water tank has a circular hole at its base. A solid cone is used to plug the hole. Exactly half the height of the cone protrudes out of the hole. Water is filled in the tank to a height equal to height of the cone.Caluclate the buoyancy force on the conc. Density of water is rho and volume of cone is V. |
|
Answer» |
|
| 49. |
Explain Doppler effect. |
| Answer» Solution :When the source and the observer are in RELATIVE motion with respect to each other and to the medium in which SOUND propagates, the frequency of the sound wave OBSERVED is DIFFERENT from the frequency of the source. This phenomenon is called Doppler Effect. | |
| 50. |
A particle is projected from the ground with an initial speed of v at a angle theta with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is |
|
Answer» Solution :Average VELOCITY `=("displacement")/("time")` `V_(av)=(sqrt(H^(2)+(R^(2))/4))/(T/2)`…..i  Here H=maximum height `=(V^(2)sin^(2)THETA)/(2g)` R=range `=(v^(2)sin 2 theta)/G` and `T=` time of FLIGHT `=(2vsin theta)/g` Substituting i (i) we get `v_(av)=v/2sqrt(1+3 cos^(2)theta)` |
|