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. |
The ratio of maximum velocity and maximum acceleration of a particle of SHM having amplitude A and angular frequency omega……….. |
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Answer» `omega` |
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| 2. |
Under the action of force, 2kg body moves such that its position x as a function of time t is given by x= (t^(3))/(30x is in metre and t in second. Calculate the work done by the force in the first 2 second |
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Answer» Solution :From work energy theorem `W= Delta KE` `x= t^(3)//3 therefore` velocity `v= (dx)/(dt) = t^(2)` At t=0, `v_(i) = 0^(2)= 0` At t= 2, `v_(F)= 2^(2)= 4m//s` word DONE `W= (1)/(2) m(v_(f)^(2)-v_(i)^(2))` `=(1)/(2) xx 2(4^(2)-0)= 16J` |
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| 3. |
Assertion : Electrons do not experience strong nuclear force. Reason : Strong nuclear force is charge-independent force. |
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Answer» If both assertion and reason are true and reason is the correct EXPLANATION of assertion. |
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| 4. |
A wooden block with a coin placed on its, top floats in water with 1 as the length immersed and h as the height of water column. After some time the coin falls into water. Then |
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Answer» 1 DECREASES and H INCREASES |
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| 5. |
A gaseousmixtureconsistsof16gofheliumand16gofoxygenthe ratioof twospecificheatsof themixtureis |
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Answer» Solution :Number of MOLES of helium `(n)=16/4=4,` Number of moles of OXYGEN `(n^(.))=16/32=1/2` For For the monoatomic helium gas, degrees of freedom, f=3 So `C_(V)=f/2R=3/2R` `C_(V)("MIXTURE")=(nC_(V)+n^(.)C_(V))/(n+n^(.))=(4xx3/2R+1/2xx5/2R)/((4+1/2))=29/18R therefore C_(P)-C_(V)=R` `GAMMA("mixture")=C_(P)/C_(V)=1+R/(C_(V)("mixture"))=1+R/((29/18R))=1+18/29=47/29` `gamma("mixture")=1.62` |
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| 6. |
Weighing the Earth : You are given the following data: g = 9.81 ms^(-2), R_(E) = 6.37 ×x 10^(6) m, the distance to the moon R =3.84 ×x 10^(8) m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth M_(E) in two different ways. |
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Answer» SOLUTION :From Eq. (8.12) we have `M_(E) = (g R_(E)^(2))/(G)` = `(9.81 xx (6.37 xx 10^(6))^(2))/(6.67 xx 10^(11))` = 5.97`xx 10^(24)`kg. The MOON is a satellite of the Earth. From the derivation of Kepler’s third law ` T^(2) = (4 pi^(2) R^(3))/(GM_(E))` `M_(E) = (4 pi^(2) R^(3))/(GT^(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 yled almost the same answer. the difference between them being LESS than 1%. |
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| 7. |
Calculate the external work done by the system in Kcal, when 40kCal of heat is supplied to the system, 8400 J of internal work is done. |
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Answer» Solution :Heat supplied to the system at CONSTANT pressure = change in internal ENERGY + External WORK done The change in internal energy = Internal work done= `8400J=8400/4200` KCAL = 2 KCal. `:. 40 KCal = 2 kCal` + External work done The external work done = `40-2=38 kCal ` |
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| 8. |
For a gas, the difference between the two specific heats is 41503 Kg^(-1) K^(-1)and the ratio of specificheats is 1.4. What is the specific heat of the gas at constant volume in J Kg^(-1) K^(-1)? |
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Answer» 8475 |
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| 9. |
A stone tied to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tensions in the rope is 20 N, mass of the stone in kg is (take, g= 10 ms^(-2)) |
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Answer» `0.75` `T_(max)-T_(min)=2MG` Here, `T_(max)-T_(min)=20` Now, mass of the STORE F = ma `m=(F)/(a)=(20)/(2xx10)=1kg` |
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| 10. |
A physical quantity Q is related to measurable quantities L,M and N as Q = KL^(-2)M^2N^3 , where k is constant. If the % errors in the measurements of L,M and N are respectively 1%, 2% and 1%, then the % error in the measurement of Q is |
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Answer» 0.09 |
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| 11. |
Heat is supplied to a diatomic gas at constant pressure. The ratio of Delta Q : Delta U : Delta W is |
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Answer» <P>`5:3:2` At constant pressure, `DELTA Q = nC_(P)Delta T = n((7)/(2)R)Delta T` `Delta U = nC_(V) Delta T = n((5)/(2)R)Delta T` and `Delta W = Delta Q - Delta U = (7)/(2)nRDelta T - (5)/(2) nRDelta T = nRDelta T` `:. Delta Q : Delta U : Delta W = (7)/(2):(5)/(2):1 = 7:5:2` |
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| 12. |
If the shear modulusand bulk modulusof a materialare 0.42 xx 10^(11) Nm^(-2) and 0.56 xx 10^(11) Nm^(-2)respectivelythen young.s modulusof that material is ----- |
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Answer» `10^(10) NM^(-2)` |
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| 13. |
A black body initially at 27^(@)C is heated to 327^(@)C.How many times is the total radiation emitted at the higher temperature than that emitted at the lower temperature ? What is the wavelength of the maximum energy radiation at the higher temperature ? Wien's constant = 2.898xx10^(-3) mK. |
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Answer» Solution :`T = 27+273 = 300 K, T_(2)=327+273=600` Let `E_(1)` and `E_(2)` be the radiation emitted at temperatures `T_(1)K` and `T_(2)K` `E_(1)/E_(2)=(sigmaT_(1)^(4))/(sigmaT_(2)^(4))=300^(4)/600^(4)= 1/16` `E_(2)=16 E_(1)` The radiation emitted at `327^(@)`C is 16 times that at `27^(@)`C `lamda_(m)=b/T=(2.989xx10^(-3))/600=4.83xx10^(-6)` m |
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| 14. |
If the potential energy of the particle is alpha-(beta)/(2)x^(2), then force experienced by the particle is: |
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Answer» `F=(beta)/(2)X^(2)` `=(d)/(dx)(alpha-(beta)/(2)x^(2))` `=(-beta(2X))/(2)=-betax` |
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| 15. |
A body projected vertically with a velocity 'u' from the ground. Its velocity a) At half of maximum height u//2 b) At 3//4^(th) of maximum height (u)/(sqrt(2)) c) At 1//3^(rd) of maximum height sqrt((2)/(3))u d) At 1//4^(th) of maximum height (sqrt(3))/(2)u |
| Answer» Answer :C | |
| 16. |
A: A man is sitting in a boat which is floating on a pond. If the man drinks some water from the pond, the level of the water in the pond decreases. R: According to Archimede.s principle the weight displaced by body is equal to the weight of the body. |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION of A |
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| 17. |
The density of the core of planet is rho_(1) and that of the outer shell is rho_(2). The radius of the core that of the planet are R and 2R respectively. Gravitastional intensity at the surface of the planet is same as at a depth R. Find the ratio (rho_(1))/(rho_(2)) , assuming them to be uniform independently. |
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Answer» `4rho_(1)=rho_(1)+7rho_(2) rArr (rho_(1))/(rho_(2))=(7)/(3)` |
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| 18. |
What is meant by dimensional formula ? |
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Answer» Solution :Dimensional formula is a formula which shows how and which of the fundamental UNITS are REQUIRED to represent the unit of a physical quantity. In this physical quantity is EXPRESSED in terms of the quantities raised to suitable POWERS. Eg : `[M^(0)LT^(-2)]` is the dimensional formula of acceleration. |
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| 19. |
Explain why a fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel. |
| Answer» Solution :This is DUE to law of conservation of linear MOMENTUM. The vessel gets a moment which is equal and opposite to the forward momentum of water. This RATE of CHANGE of momentum prodaces back ward thrust. | |
| 20. |
Certain amount of heat supplied to an ideal gas under isothermal conditions will result in |
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Answer» a RISE in temperature |
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| 21. |
Which fundamental law forms the basis of equation of continuity ? |
| Answer» SOLUTION :LAW of CONSERVATION of ENERGY . | |
| 22. |
Define the different types of potential energy. |
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Answer» SOLUTION :(i) The energy possessed by the BODY DUE to gravitational force gives RISE to gravitational potential energy. (ii) The energy due to spring force and other similar forces give rise to ELASTIC potential energy. (iii) The energy due to electrostatic force on charges gives rise to electrostatic potential energy. |
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| 23. |
If the momentum of a body is increased by 25%,its kinetic energy is increased by |
| Answer» ANSWER :C | |
| 24. |
A simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to |
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Answer» AIR FRICTION |
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| 25. |
A glass sphere of radius 20cm has a small air bubble 4cm below its centre. The sphere is viewed from outside and along a vertical line through the bubble. The apparent depth of the bubble below the surface of sphere is |
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Answer» 13.33 |
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| 26. |
A lorry and a car moving with same momentum are brought to rest by applying the same retarding force. Then |
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Answer» LORRY will COME to REST in a SHORTER distance |
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| 27. |
If two bodies stick together after collision and move as a single body, the collision is said to be |
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Answer» PERFECTLY inelastic |
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| 28. |
In SHM , kinetic energy is ………. . |
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Answer» `1/2msquare^(2)X^(2)` |
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| 29. |
The radius of a planet is R_1 and a satellite revolve round of in a circle of radius R_2. The time period of revolution is T. The acceleration due to the gravitation of the planet at its surface. |
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Answer» Solution :ORBITAL VELOCITY of the satellite `V_0=SQRT(g^1R_2)` Where `g^1=g(R_1/R_2)^2` angular velocity `omega=V_0/R_2=sqrt(g^1/R_2)=sqrt((g(R_1)^2)/R_2^3)` But TIME period of revolution `T=(2pi)/omega=2pisqrt(R_2^3/(gR_1^2))` (or)`T^2=4pi^2 R_2^3/(gR_1^2)` (or) `g=(4pi^2)/T^2 R_2^3/R_1^2` |
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| 30. |
A simple pendulum has time period T_(1). The point of suspension is now moved upwards according to the relation y=Kt^(2),(K=1m*s^(-2)) where y is the vertical displacement. The time period now becomes T_(2). The ratio of T_(1)^(2)//T_(2)^(2) is (g=10m*s^(-2)) |
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Answer» `6/5` |
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| 31. |
On which two physical quantities escape velocity of body thrown on the earth does not depend? |
| Answer» SOLUTION :`IMPLIES` MASS of BODY and angle of projection. | |
| 32. |
Which one of the following motions on a smooth planesurface does not involve force |
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Answer» accereated motion in a straightline |
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| 33. |
. A bullet hits a block at rest on a frictionless horizontal surface and gets embedded into it. The physical quantity associated with the block that remains unchanged is |
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Answer» LINEAR momentum |
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| 34. |
The acceleration of a moving body can be found from : |
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Answer» area under VELOCITY- TIME GRAPH |
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| 35. |
A block of mass m is placed on a circular track and then it is given a velocity upsilon vertically downwards at position A on track. If block moves on track with constant speed, then |
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Answer» Coefficient of friction between block and circular track as function of angle `theta` is `MU=(SINTHETA)/(cottheta+v^(2)/(Rg))` |
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| 36. |
In a cyclic process shown in the figure on ideal gas is adiabatically taken from B to A, the work done on the gas during the process B rarr A is 30J, when the gas is taken from A rarr Bthe heat absorbed by the gas is 20J. What is the change in internal ebergy of the gas in the process A rarr B. |
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Answer» SOLUTION :`W_(BA)= -30J,Q_(BA) = 0` `:. DELTA U_(BA) = -W_(BA) = 30J` In a cyclic process `DeltaU=0` Now, `Delta U_(AB) = -Delta U_(BA) = -30J`
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| 37. |
Two resistors of 10KOmega and 20KOmegaare conne cted in series. If tolerance of each resistor is 10% then tolerance of the combination will be |
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Answer» `5%` |
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| 38. |
“The most incomprehensible thing about the world, is that it is comprehensible. ” Explain. |
| Answer» Solution :The PHYSICAL world AROUND us is very complex. Even then all the physical phenomena can be understood and EXPLAINED in TERMS of a few basic laws. This is what Einstein meant to say. | |
| 39. |
A point moves rectilinearly with deceleration which depends on the velocity v of the particle as a = k sqrt(v) where k = + ve constant. At the initial moment the velocity of the point is equal to v_(0). What distance will it cover before its stops. |
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Answer» `s= (2v_(0)^(3//2))/(3K)` |
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| 40. |
Moment of inertia of a thin uniform ring about an axis passing through the center of gravity and perpendicular to the plane is |
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Answer» `MR^(2)` |
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| 41. |
What is an impure spectrum ? |
| Answer» Solution :An impure spectrum is that in which different COLOURS overlap ONE ANOTHER. | |
| 42. |
If a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. Consider a drop of mass 1.00 g falling from a height 1.00 km. It hits the ground with a speed of 50.0 ms^(-1). (1) What is the work done by the gravitational force ? (2) What is the work done by the unknown resistive force ? |
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Answer» Solution :(1) The change in kinetic energy of the drop is `Delta K=(1)/(2)m UPSILON^(2)-0=(1)/(2)xx10^(-3)xx50xx50=1.25 J` WORK done by the gravitational force is `W_(g)=mgh=10^(-3)xx10xx10^(3)=10.0J` (2) The work done by the resistive force on the raindrop. Thus `W_(r )=Delta K - W_(g)=1.25-10=-8.75 J` |
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| 43. |
If the gravitational force between two objects were proportional to 1/R (and not as 1//R^2) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to, |
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Answer» `1//R^(2)` |
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| 44. |
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L +l). The elastic potential energy stored in the extended wire is: |
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Answer» `(1)/(2) MGL` `=(1)/(2) xx` (force) `xx` (INCREASE in length) `=(1)/(2) xx mg xx (L + l -L)` `=(1)/(2) mgl` |
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| 45. |
A uniform bar of length 6a & mass 8m lies on a smooth horizontal table. Two point masses 'm & 2m' moving in the same horizontal plane with speeds 2v & 'v' respectively strike the bar as shown & stick to the bar after collision then |
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Answer» Velocity of the centre of MASS is zero |
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| 46. |
Calculate the rms velocity of oxygen molecules at S.T.P The molecular weight of oxygen is 32. |
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Answer» Solution :HereP = 1 atm = `10.13 xx10^5Nm^(-2)` `M = 32 g = 32 xx10^(-3) kg` V = Molar VOLUME of S.T.P = 22.4 LITRE `=22.4 xx10^(-3) m^3` Root MEAN square velocity of OXYGEN molecules at S.T.P `v_(rms)=sqrt((3PV)/M)` `=sqrt((3xx1.013xx10^5xx22.4xx10^(-3))/(32xx10^(-3)))` `=461.23 ms^(-1)` |
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| 47. |
State Hooke's law and verify it with the help of an experiment. |
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Answer» Solution :Hooke's LAW states that for a small deformation, when the STRESS and strain are proportional to each other. It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross sectional area A spring is suspended from a fixed point O. A pan and a pointer are ATTACHED at the free end of the wire as shown in Figure 7.2. Using a vernier scale arrangement the extension produced on the wire is measured. From experiment it is KNOWN that for a given load, the corresponding stretching force is F and the elongation produced on the wire is `DeltaL`. It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and `DeltaL` on the Y-axis. This graph is a straight line passing through the origin as shown in Figure 7.3. `thereforeDeltaL=(slope)F` Multiplying and dividing by volume, `V=AL`, `F(slope)=(AL)/(AL)DeltaL` REARRANGING, we get `(F)/(A)=((L)/(A(slope)))(DeltaL)/(L)` Therefore, `(F)/(A)prop((DeltaL)/(L))` From the equations, we get Stress, `sigma=("Force")/("Area")=(F)/(A)` Strain, `epsilon=("Change in size")/("Original size")=(Deltal)/(l)` `sigmapropepsilon` i.e., the stress is proportional to the strain in the elastic limit. |
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| 48. |
A stone is projected from the top of a tower with velocity 20ms^(-1) making an angle 30^(@) with the horizontal. If the total time of flight is 5s and g=10ms^(-2). |
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Answer» the height of the TOWER is 75 m |
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| 49. |
Lubrication reduces friction because |
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Answer» Lubricant molecules ACT as BALL BEARINGS |
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