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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. |
If a satellite is revolving around a planet of mass M in an elliptic orbit of semi-major axis a, show that the orbit al speed of the satellite when it is at a distance r from the focus will be given byv^(2)=GM(2/r-1/a) |
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Answer» Solution :Total MECHANICAL energy of the system is `E = - (GMm)/(2a) ` which remains conserved. `implies KE + PE = -(GMm)/(2a)` At a position .R. orbital speed of the satellite is V. Then KE =`1/(2)mv^(2) ,PE =-(GMm)/r` So, `1/(2)mv^(2)-(GMm)/r=-(GMm)/(2a) ` (or ) `v^(2) = GM (2/r-1/a)` |
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| 2. |
A carnot has the reversible processes the following order |
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Answer» ISOTHERMAL EXPANSION, ADIABATIC expansion, isothermal COMPRESSION and adiabatic compression |
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| 3. |
In sliding, the resultant velocity of a point of contact acts along |
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Answer» FORWARD direction |
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| 4. |
A uniform rope of length L rests on a smooth plane. One end of the rope is pulled with a force F. what is the tension in the rope at a distance l from that end ? |
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Answer» SOLUTION :Let the mass of the rope be M Hance, mass per unit length of the uniform rope = `(M)/(L)`. Length of the PART BC of the rope = l . length of the part AC = L - l, and mass of the part AC = `(M)/(L)` (L - l) . Acceleration of the rope DUE to the force F is a = `(F)/(M)`. Let the tension at C be T. Hence, part AC of the rope gains accelerationa due to the tension T. `therefore`T = mass of AC `xx` acceleration of the rope `= (M)/(L) (L - l)CDOT (F)/(M) = (F)/(L) (L-l)`. 
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| 5. |
The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20ms^(-1), its time of flight is (g=10ms^(-2)) |
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Answer» 2S |
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| 6. |
A metal plate of area 0.10 m^(2) is connected to a 0.01 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), A liquid with a film thickness of 0.3 mm is placed between the plate and the table. When released the plate moves to the right with a constant speed of 0.085 ms^(-1). Find the coefficient of viscosity of the liquid. |
Answer» SOLUTION : Here `A=0.10m^(2),m=0.01kg` `dx=0.3mm=0.3xx10^(-3)m,dv=0.085ms^(-1)` The METAL plate moves towards right du tothe tension T in the string which is equal to the weight of the suspended mass m. ASSUMING thatthe mass m moves with uniform velocity or zero acceleration, then the FORCE of viscosity will be, `F=T=mg=0.01xx9.8` `=9.8xx10^(-2)N` Taking velocity gradient to be uniform, then `eta=(F)/(A).(dx)/(dv)=(9.8xx10^(-2)xx0.3xx10^(-3))/(0.10xx0.085)` `=3.45xx10^(-3)PaS` |
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| 7. |
A vessel contains some water. It is moved towards right on a horizontal plane with a constant acceleration 'a'. Which of the following figures represents correct water surface ? |
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Answer»
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| 8. |
The displacement time relation for a particle can be expressed as, y = 0.5 [cos^2 (n pi t) - sin^2 (n pi t)] .The relation shows that: |
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Answer» the particle is executing a SHM with amplitude 0.5 m |
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| 9. |
Why do the machine parts get jammed in winter? |
| Answer» Solution :`eta ` DECREASES DUE to decrease in TEMPERATURE. So the parts get JAMMED. | |
| 10. |
A glass bulb of volume 250 c.c is completely filled with mercury at 20°C. The temperature of the system is raised to 100°C. If the coefficient of linear expansion of glass is 9 xx 10^(-6)//""^(@) C and coefficient of absolute expansion of mercury is 1.8 xx 10^(-4)//""^(@) C, the volume of mercury that overflows is nearly |
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Answer» `3.06` c.c |
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| 11. |
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that: |
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Answer» its kinetic energy is CONSTANT |
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| 12. |
A car moving with velocity of 50kmh^(-1) on a straight road is ahead of a jeep moving with Velocity 75kmh^(-1). How would the relative velocity be altered if jeep is ahead of car? |
| Answer» SOLUTION :No CHANGE. | |
| 13. |
Three moles of an ideal gas (C_P =7/2R)at pressure P_0and temperature T_0is isothermally expandedto twice its initial volume. It is then compressed at a constant pressure to its original volume. a) Sketch P-V and P-T diagram for complete process b) Calculate net work done by the gas c) Calculate net heat supplied to the gas during complete process (Write your answer in terms of gas constant = R) |
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Answer» |
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| 14. |
A thin uniform rod of mass 5 kg and length 1 m is held in horizontal position with the help of strings attached to ends of rod, other ends of strings are held by some external agent. Now end A is pulled down with speed V_(A)=3t and end B is pulled down with speed V_(B)=t Where 't' is time in sec. Choose the correct choice(s) at time t=0 [Take g =10m//s^(2)] |
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Answer» ANGULAR acceleration of rod is `"2 rad/s"^(2)` |
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| 15. |
Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s). a) v_(av)=(1)/(2)[v(t_(1))+v(t_(2))] b) v_(av)=(r(t_(2))-r(t_(1)))/(t_(2)-t_(1)) c) r=(1)/(2)(v(t_(2))-v(t_(1)))(t_(2)-t_(1)) d) a_(av)=(v(t_(2))-v(t_(1)))/(t_(2)-t_(1)) |
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Answer» a and B |
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| 16. |
Magnitude of wave vector of an e.m.wave with frequency 150 MHz is ...... rad/s. |
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Answer» `PI` `therefore k = (omega)/(v) = (2pi f )/ (c ) (because` for e.m. wave we have v =c) `therefore k = (2 xx pi xx 150 xx 10^(6))/( 3xx 10 ^(8))` `therefore k = pi rad //m` |
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| 17. |
The moment of inertia of a Thin rod about and axis passing through the centre and perpendicular to the length is ________. |
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Answer» `(1)/(3)Ml^(2)` |
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| 18. |
The magnitude of gravitational potential energy of a body at a distance 'r' from the center of the earth is V. Its weight at a distance '2r' from the centre of the earth is |
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Answer» `(V)/(R)` |
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| 19. |
Can a motion be oscillatory but not simple harmonic?Explain. |
| Answer» SOLUTION :Yes,when a ball is DROPPED from a height on a perfectly elastic surface,the motion is oscillatory but not SIMPLE HARMONIC as restoring force F=mg=constant and not F`alplha`x, which is an essential condition for SHM. | |
| 20. |
The temperature-entropy diagram of a reversible engine cycle is given in figure. What is its efficiency? |
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Answer» Solution :Entropy CHANGE `=int_(A)^(B)(dQ)/T""T(S-S_(0))=Q` `Q_(1)=T_(0)S_(0)+1/2T_(0)S_(0)=3/2T_(0)S_(0)` : `Q_(2)=T_(0)(2S_(0)-S_(0))=T_(0)S_(0),""Q_(3)=0` `n=W/Q_(1)=(Q_(1)-Q_(2))/Q_(1)=1-Q_(2)/Q_(1)=1-(T_(0)S_(0))/(3/2T_(0)S_(0))=1/3` |
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| 21. |
The ratio of times taken by freely fallingbody to cover first metre, second metre,.. Is |
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Answer» `SQRT(1):sqrt(2):sqrt(3)` |
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| 22. |
A body moves a distance of 20 m along a straight line under the action of a force of 10N. If the work done is 100 J, the angle between force and displacement vectors is, |
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Answer» `0^(@)` |
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| 23. |
A thick spherical shell has two equally thick layers with conductivities for inner and outer layers nk and k respectively. If the interface temperature is arithmetic mean of inner and outer surface temperature of composite shell then value of n is (Assume inner radius R and other radius 2R for composite layer) |
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Answer» |
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| 24. |
A hammer of mass 1 kg strikes on the head of a nail with a velocity of 2 ms^(-1)It drives the nail 0.01 m into a wooden block. Find the force applied by the hammer and the time of impact. |
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Answer» ` 200 N , 10^(-2) sec` |
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| 25. |
The angle of contact of mercury with soda lime glass is 140^(@). A capillary tube of radius 1.0 mm is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside?Surface tension of mercury =0.465Nm^(-1) Density of mercury =13.6xx10^(3)kgm^(-3) |
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Answer» Solution :Capillary descent, `"" H = (2 T cos theta)/(R rho g) = (2 XX (0.465 Nm^(-1))(cos 140^@))/((2 xx 10^(-3) m)(13.6 xx 10^3)(9.8 m s^(-2)))` `implies "" h = -6.89 xx 10^(-4) m` Where, negative sign indicates that there is fall of mercury (mercury is DEPRESSED) in GLASS tube. |
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| 26. |
Impulse of the force exertec by A on B during the collision is equal to |
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Answer» `(SQRT(3)m hat(i) + 3M hat(j))kg.m//s` |
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| 27. |
An ideal gas is taken around ABCA as shown in the above diagram. The work during a cycle is |
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Answer» 2PV |
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| 28. |
A particle of mass m is moving yz- plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting Z- axis at z = a. The change in its angular momentum about the origin as it bounces clastically from a wall at y = constant is |
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Answer» `MVA BARI` |
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| 29. |
A gas is at on atmosphere. To what pressure it should be subjected at constant temperature so as to have to its initial volume? |
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Answer» `2 ATMOSPHERE` |
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| 30. |
A hollow sphere is filled with water . It is hung by a long thread . As the water flows out of a hole at the bottom , the period of oscillation will |
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Answer» Firstincrease and thendecrease |
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| 31. |
A mobile phone tower transmits a wave signal of frequency 900 MHz. Calculate the length of the wave transmitted from the mobile phone tower . |
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Answer» 0.33 m SPEED of wave (C) = `3 xx 10^(8) MS^(-1)` `LAMDA = v/f = (3xx10^(8))/(900xx10^(6)) = 1/3 = 0.33` m |
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| 32. |
The speed of sound in hydrogen at STP is V. The speed of sound in a mixture containing 3 parts of hydrogen and 2 parts of oxygen at STP will be |
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Answer» `V//2` `V PROP (1)/(sqrt(rho))implies(Vmix)/(V)=sqrt((rho)/(rho_(mix)))=sqrt((rho_(1))/(7rho_(1)))` |
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| 33. |
Two slabs A & B having lengths l_(1) and l_(2), respectively, and same cross-section have theraml conductivities K_(1) and K_(2) respectively. They are placed in contact and a constant temperature difference is maintained across the combination. The ratio of the quantities of heat flowing through A and B in a given time is |
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Answer» `(K_(1))/(l_(1)):(K_(2))/(l_(2))` |
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| 34. |
When an equation for velocity and acceleration of CM. |
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Answer» SOLUTION :`vecV_(CM)=(SUM m_(i)vecv_(i))/(sum m_(i))` `veca_(CM)=(sum m_(i)veca_(i))/(sum m_(i))` |
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| 35. |
The length, breadth and height of a rectangular block are found to be 15.12 pm 0.02cm, 7.86 pm 0.01cm" and "4.16 pm 0.02cm, respectively. Compute the percentage error in the volume of the block. |
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Answer» Solution :Given, length, `l= 15.12 pm 0.02 cm` Breadth, `b= 7.86 pm 0.01cm` Height, `h= 4.16 pm 0.02 cm` VOLUME of rectangular BLOCK is `V= l xx b xx h` The percentage error in the volume is given by `(triangle V)/(V)xx100= ((trianglel)/(l)+(TRIANGLEB)/(b)+(triangleh)/(h))xx100` `=((0.02)/(15.12)+(0.01)/(7.86)+(0.02)/(4.16))xx100` `=((2)/(1512)+(1)/(786)+(2)/(416))xx100` `=0.1323+0.1272+0.4808= 0.7403%` `=0.74%` (Rounded off to 2 SIGNIFICANT figures). |
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| 36. |
A body cools from 80^(@)C to 60^(@)C in 2 minutes. In how much time it cools from 60^(@) to 40^(@)C ? The temperature of the surroundings is 10^(@)C |
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Answer» SOLUTION :I case : Mean temperature of the body `(80+60)/2=70^(@)C` Mean excess temperature = 70 - 10 = `60^(@)C` `(d THETA)/dt=K(theta-theta_(0))rArr20/2=K(60)to(1)` II case: Mean temperature of the body `(60+40)/2=50^(@)C` Mean excess temperature = (50 - 10) = `40^(@)C` Let .t. minutes be the time to cool down from `60^(@)C` to `40^(@)C` Then `20/t=K(40)to(2)` Dividing equation (1) by (2) `t/2=60/40` i.e., t = 3 minutes |
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| 37. |
State the condition for satellite to be geostationary. |
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Answer» Solution :(i) ROTATION of satellite should be same as that of Earth. (ii) PERIOD of revolution should be the same as that of Earth. (III) It should revolve at a HEIGHT of nearly at 36000 km above the Earth's surface. |
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| 38. |
A ring of inner and outer radii 8 and 9 cm is pulled out of water surface with a force of [ S.T of water (T) = 70 dyne /cm] |
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Answer» `2.6 xx 10^(-2) N` |
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| 39. |
A disc of radius 0.1 mrolls without sliding on a horizontal suirface with a velocity of 6 m//s. It then ascends a smooth continous track as shown in figure. The height upto which it will ascend is (g=10 m//s^(2)) |
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Answer» 2.4 m |
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| 40. |
Consider the motion of the tip of the minute hand of a clock. In one hour a) the displacement is zero b) the distance covered is zero c) the average speed is zero d) the average velocity is zero |
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Answer» a & B are correct |
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| 41. |
The ratio of times takes by freely falling body to cover first metre, second metre, is |
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Answer» `SQRT(1):sqrt(2):sqrt(3)` |
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| 42. |
P-V diagram of an ideal gas for a process ABC is as shown in the figure. Find total heat absorbed or released by the gas during the process ABC. |
| Answer» SOLUTION :`Q_(ABC) =- 2 P_0 V_0` | |
| 43. |
What is called one dimensional wave? What are called progressive or travelling waves ? What are called harmonic waves ? What is called wave equation ? Write wave equation for progressive harmonic transverse wave propagating along + X-axis. Also explain it by drawing some graphs. |
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Answer» Solution :One dimensional wave : It is a wave propagating in one dimensional medium. For example, a transverse wave propagating in a string kept under proper tension is one dimensional wave. Progressive (or travelling) wave : It is a wave which travels ahead and ahead in the medium In short, it is not getting reflected while propagating in any direction. Harmonic wave : It is a wave, during propagation of which, particles of medium are performing S.H.M. about their respective equilibrium positions. Sinusoidal type of disturbance is required for the production of harmonic wave. Wave equation : An equation with the help of which we can find out displacements of different particles TAKING part in the propagation of wave at different TIMES is called wave equation Some wave terminology for better understanding : Amplitude of wave (a): Amplitude of oscillation of particle of medium taking part in the propagation of wave is called amplitude of wave. It is shown by symbol "a" or "A”. It is always positive. It gives magnitude of maximum displacement on either side of an equilibrium POSITION. Its SI unit is m. Frequency of wave (v, n, f) : Frequency of oscillation of particle of medium taking part in the propagation of wave is called frequency of wave, shown by symbol "V" (pronunciation `to` "nu") OR "n" OR "F". Its SI unit is Hz. Important note : For frequency, now onwards we shall use symbol "f" instead of "v" used in the text book, so that we can identify, speak and write it easily. Moreover we can DIFFERENTIATE it from symbol v of wave-velocity. Periodic time of wave (T) : Periodic time of oscillation of particle of medium taking part in the propagation of wave is called periodic time of wave, shown by symbol "T". Its SI unit is s. Relation between f and T: We know that frequency f gives no. of oscillations in unit time, time taken to complete one oscillation will be `T = 1/f implies f = 1/T.` Angular frequency of wave `(omega)` : Angular frequency of oscillation of particle of medium taking part in the propagation of wave is called angular frequency of wave, shown by symbol `"omega".` Its SI unit is `(rad)/(s). `Relation between 0 and f: `2pi f = (2pi )/(T)` Now, suppose a simple harmonic motion type disturbance is produced in first particle A lying at x = 0 with initial phase zero and amplitude equal to a, as shown in the figure.  When particle A completes one oscillation in time T, suppose disturbance reaches at particle I. Here at t = T, phase of particle A becomes `2pi` rad (Because it has completed one oscillation) whereas phase of particle I is zero (because yet it has not started any oscillation.) Thus, at t = T, phase difference between particles A and I is `2pi` rad. Distance between such two particles is defined as a wavelength, shown by symbol `lamda` which is total width of one trough and one crest. Definition of wavelength of a wave `(lamda):` "Distance between two particles having phase difference of `2pi` rad at a given instant of time, along the direction of propagation of wave is called wavelength of wave." It can also be defined as distance between two consecutive crests or two consecutive troughs. Wave number `(vecv):` "no. of waves in a unit distance along the direction of propagation of wave is called wave number. " It is show by symbol `vecv.` We know that no. of waves in distance `lamda` is 1. So no. of waves in unit distance would be `1/lamda,` which gives wave no. as per its definition. Thus, `vecv = (1)/(lamda).` Its SI unit is `m^(-1).` Wave-vector (or angular wave number or propagation constant (k): "Decrease in phase at unit distacne along the direction of propagation of wave is called wave-vector." Its symbol is k. It is always taken along the direction of propagation of wave. We know that at distacne `lamda,` along the direction propagation of wave, decrease in phase in `2pi` rad. Hence decrease in phase at unit distance in this direction would be `(2pi )/(lamda)` which is taken as wave- vector ( or angular wave number of propagation cnstant) as per itsw definition, shown by symbol k. Hence `k =(2pi)/(lamda).` Its SI unit is `(rad)/(m).` If given one dimentional wave propagates along + X-axis then we can write `veck = (2pi )/(lamda) hati` Wave-velocity (v) : "Distacne travelled by distrubance along the direction of propagation of wave in unit time is called wave velocity". It is shown by symbol v. We know that distance travelled by disturbance in time T is equal to `lamda.` Hence distacne travelled by disturbance in unit time would be `lamda /T` which is difined as wave-velocity. Hence, `v = (lamda )/(T) implies v = f lamda (because f = (1)/(T))` Now multiplying an dividing by `2pi` rad on R.H.S., we get : `v = 2(2pi f ) ((lamda )/( 2pi )) = omega xx (1)/(k )({:(because omega = 2pi f ),(k = (2pi )/(lamda )):})` `therefore v = (omega )/(k) ""...(5)` Wave equation for one dimensional progressive harmonic transvers wave, propagating along +X-axis is : `y =a sin (omega t - kx+ phi)""...(6)` Where `phi` is called initial phase or phase contant or epoch. It is the phase of a particle at `x=0` at time `t=0.` Note When simple harmonic type disturbance is produced in particle at `x=0,` if it starts its S.H.M. from (i) equilibrium position towards positive end od path then `phi=0.` (ii) equilibrium position towards negative end of path then `phi =pi` rad (ii) postive end of path then `phi = pi/2` rad ((iv) negative end of path then `phi = (3pi)/(2) `rad Above equation gives displacement of a particle at distance x from the source of a wave, at time t, after the profuction of S.H.M. type of disturbance in one dimensional elastic medium at `t=0` in a particle at `x=0` having initial phase `phi.` In avove wave equation the angle with the sine function is `(omega t - kx + phi)` which gives the total phase of a prticle at distance x from the source of a wave, at time t. If it is shown by symbol `theta` then, a wave, at time t. If it is shown by symbol `theta` then, `theta = omega t - kx +phi""...(7)` Now let us find out rate of change of `theta` with respect to distacne x at a given time t. `(d theta)/(d x) = (d)/(dt) (omegat -kx + phi) (because` Here,` omega, t , phi therefore (dtheta)/(dx) =0- k(1) +0=-k` are constants) `therefore k =- (d theta)/(dx)""...(8)` Thus, propagation constant gives us decreae in phase with respect to distance along the direction of propagation of wave. To obtain wave equation for a progressive harmonic transvcrsc wave propagating along -X-axis, we have to replace x by -x in equation (6). `therefore y =a sin (omega t + kx +phi""...(9)` To undertand motion of disturbance during the propagation of transvers harmonic travelling wave along + X-axis, CONSIDER following five figures at the regular interval of `T/4` time (where T = periodic time of wave = periodic time of oscillation of particle of medium).  We can see from above figures that as the souce of wave (shown by symbol "." in above figures) moves from `Q_(1) to Q_(2) to Q_(3) to Q_(4) to Q _(5)` during its one oscillation crest (shown by symbol `"x"` in above figures) in getting formed at points `R_(1) to R_(2) to R_(3) to R_(4) to R_(5)` As thishappens, we find that wave pattern is moving along + X-axis. Similar movement also takes along -X-axis. (Please note that here there is no physical shifting of any part of medium along the direction of propagation of wave. When we stop producing distrubance, medium assumes its original shape.) Note : Wave equation `y=a sin (omega t - kx+ phi)` can also be obtained by linear combinationof following two waves having initial phase difference `phi.` `y_(1) = A sin (omega t kx)""...(10)` `y_(2) = B cos (omega t - kx)""...(11)` Here `y =a sin {(omega t - kx)+phi}` `therefore y=a [sin (omega t -kx) cos phi + cos (omega t -kx) + a sin phi cos (omega t -kx)""...(12)` Also we can write `y = y _(1) +y _(2)` `therefore y =A sin (omega t - kx) + B cos (omega t - kx)...(13)` Comparing equation (12) and (13), we get `a cos phi =A""...(14)` `a sin phi = B""...(15)` Squaring and adding, we get: `a ^(2) (cos ^(2) phi + sin ^(2) phi) = A ^(2) + B ^(2) ` `therefore a = sqrt (A ^(2) + B ^(2)) ""...(16)` Taking ratio of equation (15) to (14), `tan phi = (B)/(A)` `therefore phi = tan ^(-1) ((B)/(A))""...(17)` |
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| 44. |
Give the expression for nth harmonic or (n-1)th overtone in the case of a closed pipe system. |
| Answer» Solution :The `n^(TH)` harmonicfor a clsed pipe system `= ( 2n + 1 ) ((V)/(4L))`where the I HARMONIC `f_(0) = (v )/(4L)` and `n 0,1,2,"…....."` &v-velocity of sound in AIR. | |
| 45. |
Calculate the total K.E. of earth, assuming it to be a uniform spherical body moving around the sun with a speed of 30 km s^(-1) and spinning about its own axis? Mass of earth = 6 xx 10^(24) kg,Radius of earth = 6400 km ? |
| Answer» Solution :Total K.E. `=(1//2) mv^(2) + (1//2) XX (2//5) MR^(2) xx (2pi//T)^(2), T = 24h, K.E. =2.7 xx 10^(33) J` | |
| 46. |
(A) : The angular momentum of a particle w.r.t origin moving parallel to x-axis with constant velocity is constant (R) : There is no change in the perpendicular distance of the particle from the origin when it travels parallel to x-axis. |
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Answer» Both 'A' and 'R' are TRUE 'R' is the correct explanation of 'A' |
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| 47. |
Minimum force F required to move the block |
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Answer» `MU MG ` |
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| 48. |
A motor vehicle travelled the first third of a distance s at a speed of v_1=10 kmph, the second third at a speed of v_2=20 kmph and the last third at a speed of v_3=60 kmph. Determine the mean speed of the vehicle over the entire distance s. |
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Answer» 15kmph |
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| 49. |
In gravity- free space, a particle is in contact with the inner surface of a hollow vertical cylinder and moves in horizontalcircular path along the surface. There is some friction between the particle and the surface. The retardation of the particle is |
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Answer» ZERO |
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| 50. |
In one dimensional motion, instantaneous speed v satisfies 0 le v lt v _(0). |
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Answer» The displacement in time T must always take non-negative values As maximum velocity in positive direction is `v_(0)` maximum velocity in opposite direction is also `v_(0).` Maximum displacemtn in ONE direction `=v_(0) T` Maximum displacement in opposite directions`=- v _(0) T` THEREFORE `-v_(0) T lt x lt v _(0)T` |
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