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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. |
Heat cannot by itself flow from a body at lowertemperature to a body at higher temperature is a statement or consequence of |
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Answer» IST law of thermodynamics |
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| 2. |
Three blocks connected together by strings are pulled along a horizontal surface by applying force F. If T_(3)=36 N, What is tension T_(2) ? |
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Answer» SOLUTION :Suppose SYSTEM slides with acceleration .a.. Then for MOTION of blocks, `T_(3)=F=(m_(1)+m_(2)+m_(3))a` `T_(3)=36 a, i.e., a=(T_(3))/(36)=1m//s^(2), T_(3)-T_(2)=27a` From above equation, `36-T_(2)=27xx1 therefore T_(2)=9N` |
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| 3. |
If the ice on the polar caps of the earth metls, how will it affect the during of the day ? Explain. |
| Answer» Solution :If the ice on the polar caps of the melts, WATER going away from the AXIS of rotation and the moment of inertia of the earth about its own axis INCREASE. | |
| 4. |
A simple pendulum is attached to the roof of a lift.If time period of oscillation,when the lift is stationary is T,then frequency of oscillation when the lift falls freely,will be |
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Answer» Zero |
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| 6. |
A particle in rotating disc covers angle theta= 2t-t^(3) |
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Answer» AVERAGE angular velocity is `2 rad//s` |
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| 7. |
The adjacent sides of a parallelogram are vecP=2i-3j+k and vecQ=-2i+4j-k . What is the area of the parallelogram ? |
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Answer» `4` units |
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| 8. |
Mountaineers suffer from nose bleeding (nostril ) at higher altitudes. |
| Answer» Solution :As HEIGHT increases , atmospheric pressure decreases .Atmospheric pressure at mountain is low .Whereas the pressure of blood is UNCHANGED Walls of nose arteries are THIN .THEYCANNOT TOLERATE pressure of blood and they breaks and the blood flows. | |
| 9. |
Suppose the gravitational force varies inverselyas the nth power of distance. Then the time period of planet in circular orbit of radius R around the sun will be proportional to .......... |
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Answer» `R^(((n+1)/2))` `:. F = k/r^n` For ORBIT of radius R `F = k/(R^n)` but `F = momega^2R` `:.k/(R^n)=mR((2PI)/T)^2` `:. k/(R^(n+1)) = (m(2pi)^2)/T^2 ` `:. T^(2) prop R^(n+1)` `:. T prop R^((n+1)/2)` |
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| 10. |
The distance between the centres of carbon and oxygen atoms in the gas molecule is 1.13 Å. The centre of mass of the molecule relative to oxygen atom is |
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Answer» `0.602Å` Mass of CARBON atom =14 Mass of oxygen atom = 16 Let C.M. of MOLECULE lies at a distance of x from oxygen atom IE..,e `m_(1)r_(1)=m_(2)r_(2)` `16X=14(1.13-X)` `30X=15.82` `X=0.527 Å` |
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| 11. |
{:("Column-I","Column-II"),("A) Motion of needle of sewing machine ","P) Periodic oscillatory "),("B) Motion of conical pendulum ","Q) Oscillatory non-eriodic "),("C) Motion of simple pendulum ","R) Oscillatory, linear simple harmonic "),("D) Motion of suspended loaded spring ","S) Oscillatory, Non linear SHM "),("pulled and released ",):} |
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Answer» |
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| 12. |
A body is just floating on the surface of liquid. The density of the body is same as that of the liquid. The body is slightly pushed down. Then it will |
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Answer» COME BACK slowly to its earlier position |
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| 13. |
A' thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. "if "4.0 kg of ice is put in the box, estimate the amount of Ice remaining after 6 h. The outside temperature is 45^(@) C and co-efficient of thermal conductivity of thermacole is 0.01J s^(-1) m^(-1) K^(-1). [Heat of fusion of water= 335 xx 10^(3) J kg^(-1) ] |
| Answer» SOLUTION :`3.7 KG` | |
| 14. |
Name the device used for measuring the mass of atoms and molecules. |
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Answer» MASS Spectrograph |
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| 15. |
One angular metal ring of inner radius 7 cm and outer radius 14 cm and of negligible weight is floating on the surface of a liquid. If surface tension of liquid is 0.08Nm^(-1), calculate the force required to detach it from liquid surface. |
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Answer» SOLUTION :The CONTACT length of annular ring with liquid surface L = `2pir_(1)+2pir_(2)` `L=2xx22/7xx(14+7),L=132cm=1.32m` |
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| 16. |
The readings of a length come out to be 2.63 m , 2.56 m , 2.42 m , 2.71 m and 2.80 m . Calculate the absolute errors and relative errors or percentage errors. What do you think of the actual value of the length and its limits ? |
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Answer» Solution :The mean period of oscillation of the pendulum `T = ((2.63+2.56+2.42+2.71+2.80)s)/(5)` `=(13.12)/(5)`s =2.624s =2.62s As the periods are measured to a resolution of 0.01s, all times are to the second decimal, it is proper to put this mean period also to the second decimal. The errors in the measurements are `2.63s=2.62s = 0.01s` 256s - 2.62s = -0.06s 2.42s-2.62s = -0.20s 2.71s-2.62s = 0.09s 2.80s-2.62s = 0.18s Note that the errors have the same units as the quantity to be measured. The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the MAGNITUDES ) is `Delta T_("mean") =[(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s]//5` `=0.54 s//5` `=0.11s` That means , the period of oscillation of the simple pendulum is `(2.62 pm 0.11) ` s i.e. . It lies between (2.62 + 0.11) s and (2.62 - 0.11) s or between 2.73 s and 2.51 s . As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second . Hence there is no POINT in giving the period to a hundredth. A more correct way will be to write . `T = 2.6 pm 0.1s` Note that the last numeral 6 is unreliable, since it may be anything between 5 and 7 . We INDICATE this by SAYING that the measurement has two significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7 . For this EXAMPLE, the relative error or the percentage error is `deltaa =(0.1)(2.6) xx 100 = 4%` |
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| 17. |
A man of mass m moves with a constant speed on a plank of mass 'M' and length 'L' kept initially at rest on a frictionless horizontal surface, from one end to the other in time 't'. The speed of the plank relative to ground while man is moving, is - |
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Answer» `L/t(M/m)` |
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| 18. |
If the two colliding bodies stick togther after collision such collisions are |
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Answer» ELASTIC COLLISION |
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| 19. |
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. The acceleration on the surface of planet ......... |
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Answer» `9.8 m//sec^2` `G = (GM_e)/(R_e^2)` Acceleration of gravity on the surface of PLANET `g_p=(GM_e)/R_e^2` but `M_p = 2M_e ` here `R_p = 2R_e` `:. g_p =(G(2M_e))/((2R_e^2))=(2GM_e)/(4R_1^2)=1/2(GM_e)/(R_e^2)` `:. g_p =g/2 =(9.8)/2 m//s^2` |
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| 20. |
A particle encute SHM between x = -A and x = +A, The time taken for it to go from 0 to A//2 is T_(1) and to go from A//2 to A is T_(2), then |
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Answer» `T_(1)LT T_(2)` |
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| 21. |
A body is allowed to fall freely under gravity from a height of 10 m. If it loses 25% of its energy on impact with the ground, to what height will it rise after one impact? |
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Answer» 2.5 m `1/2 mv_1^2 = 0.75 mgh = 3/4 mgh" or " v_1^2 = 1.5 gh` The height `h_1` to which the bocy will rise is `h_1 = (v_1^2)/(2g) = (1.5gh )/(2g) = 0.75 h = 0.75 xx 10 = 7.5 m ( :. h = 10M). ` |
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| 22. |
A body is thrown horizontally from the top of a tower. It reaches the ground after 4s at an angle 45^(@) to the ground. The velocity of projection is |
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Answer» `9.8 MS^(-1)` |
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| 23. |
The resistance of a platinum resistance thermometer is found to be 11. 0 ohm when dipped in a triple point cell. When it is dipped in a bath, the resistance is found to be 28.887 ohm. Find the temperature of the bath in ""^(0)C on platinum scale. |
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Answer» SOLUTION :In TERM of triple point of water. `T_(K) = [ 273.16 (R)/(R_(TR)) ] K ` so `T_(K) = 273.16 XX (28.887)/(11.0) = 717.32K` Now as `T_(C) = T_(K) - 273.15` `T_(C) = 717.32 - 273.15 = 444.17^(0)` C |
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| 24. |
A satellite is at a height of 25 , 600km from the surface of the earth . If it.s orbital speed is 3.536jkm/s find its time period . (Radius of the earth = 6400 km) |
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Answer» Solution :`6400KM = 6.4 xx10^(6)m, h = 25` 600 km `=25.6m,v_(0) = 3.536 xx10^(3)m//s` The time PERIOD `T=(2pi(R+h))/V_(0)=(2pi(25.6xx6.4)10^(6))/(3.536xx10^(3))` = `(2pixx32)/(3.536) xx10^(3) = 56,870s` |
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| 25. |
A block is kept on a frictionless inclined surface with angle of inclination alpha. The incline is given an acceleration ‘a’ to keep the block stationary. Then 'a' is equal to |
| Answer» Solution :`ma cos alpha=mg sin alpha` | |
| 26. |
A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 4 J then total kinetic energy of the disc is |
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Answer» 3 J = `(1)/(2) mv^(2) + (1)/(2) MR^(2) XX (v^(2))/(r^(2)) = mv^(2) "" (because I_(Y) = mr^(2) and omega = (v)/(r))` But `K_("ring") =4 J` (given) `thereforemv^(2) = 4 J "" ... (i)` Similarly , `K_("disc") = (1)/(2) mv^(2)+ (1)/(2) I_(d) omega^(2)` `= (1)/(2) mv^(2) + (1)/(2) xx (mr^(2))/(2) xx (v^(2))/(r^(2)) ( because I_(d) = (mr^(2))/(2))` `= (3)/(4) mv^(2) = (3)/(4) xx 4 J = 3J "" ` (Using (i)) |
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| 27. |
At room temperature the rms speed of molecule of a certain diatomic gas is found to be 1930 ms^-1.The gas is |
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Answer» `H_2` |
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| 28. |
Describe the method of measuring angle of repose |
Answer» Solution :Angle of Repose Consider an inclined plane on which an object is placed, as SHOWN in figure. Let the angle which this plane makes with the HORIZONTAL be 0. For small angles of `theta `, the object may not slide down. As `theta `is increased for particular value of `theta`, the object begins to slide down. This value is called aeleon pose. In the mile of repose is the angle of inclined plane with the horizontal such that an object placed on it begins to slide  Let us consider the various force in acting here . The gravitational force mg is revolved into components parallel `(mg sin theta)` and perpendicular `(mg cos theta)` to the inclined plane . The COMPONENT of force parallel to the inclined plane `(mg sin theta)`tries to move the object down. The component of force perpendicular to the inclined plane `(mg cos theta)`is balanced by the Normal force (N) ` N = mg cos theta `....(1) When the object just begins to move, the static friction attains its maximum value ` f_s = f_(s)^(max)` This friction also satisfies the relation ` f_(s)^(max) = mu_s mg sin theta `....(2) Equating the RIGHT hand side of equations ( 1) and (2), ` (f_(s)^(max) )//N = sin theta // cos theta ` From the definition of angle of friction, we also know that ` tan theta =mu_s `....(3) in which `theta`is the angle of friction. Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of SURFACE. |
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| 29. |
Define couple. |
| Answer» Solution :A pair of forces which are equal in magnitude but opposite in DIRECTION and SEPARATED by a PERPENDICULAR distance so that their lines of action do not coincide that causes a turning effect is CALLED a couple | |
| 30. |
Least count of a vernier callipers is 0.01 cm. When the two jaws are in touch with each other, the 5^(th) division of vernier scale coinciding with the main scale lies to the left of the zero of the main scale. Further more, while measuring the diameter of the sphere, the zero mark of the vernier lies in between 2.4 cm and 2.5 cm and ther 6^(th) vernier division coincides with a main scale division. |
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Answer» The instrument has a NEGATIVE error |
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| 31. |
The elongation of a steel wire stretched by a force is 'e'. If a wire of the same material of double the length, and half the diameter is subjected to double the force, its elongation will be |
| Answer» ANSWER :A | |
| 32. |
Restoring force on the bob of a simple pendulum of mass 100 gm when its amplitude is 1^(@) is |
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Answer» 0.017 N |
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| 33. |
The wind is blowing from south at 10 ms^(-1), but to a cyclist it appears blowing from the east at 10 ms^(-1). The velocity of cyclist is |
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Answer» `10sqrt(2) ms^(-1)` towards S - W |
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| 34. |
A glass bulb of volume 200cc is sealed at 21^(@)c. If the pressure of the gas inside the bulb is 1.25 xx 10^(-4) cmof Hg, calculate the number of molecules inside the bulb. (Avagadro's number = 6.0225 xx 10^(23), R = 8.314 xx 10^(-7) erg/mole - K ) |
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Answer» SOLUTION :VOLUME V =200 `cm^(3)`, pressure P = `1.25 xx 10^(-4)` cm of Hg ` 1.25 xx 10^(-4) xx 13.6 xx 980 ("dyne")/(cm^(2))` T= 273 + 27 = 300 K , Avogadro.s number , N = 6.0225`xx 10^(23)`, R = `8.314 xx 10^(7)` erg/mole - K By ideal gas equation , PV = nRT n `= (PV)/(RT) = (1.25 xx 10^(-4) xx 13.6 xx 980 xx 200)/(8.314 xx 10^(7) xx 300)` moles Number of molecules inside the BULB = number of moles X Avogadro.s number `(1.25 xx 10^(-4) xx 13.6 xx 980 xx 200)/(8.314 xx 10^(7) xx 300) xx 6.0225 xx 10^(23) = 8.018 xx 10^(15)` |
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| 35. |
A car is being driven on a tilted ground. The ground makes an angle theta with the horizontal. The driven drives on a circle of radius R. The coefficient of friction between the tires and the ground is mu. (a) What is the largest speed for which the car will not slip at point A? Assume that rate of change of speed is zero. (b) What is the largest constant speed with which the car can be driven on the circle without slipping? |
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Answer» `sqrt(gR(mu"cos"theta - "sin"theta))` |
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| 36. |
When water rises in a capillary tube of radius r to a height h, its potential energy is U_1. If capillarytube of radius 2r is dipped in water, then potential energy of risen water is U_2. Then U_1 :U_2 will be |
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Answer» `1:1` |
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| 37. |
Two persons A and B each of mass 60 kg are carrying loads 15 kg each.If they reach the top of a building of height 10 m in 20s and 40s respectively, the ratio of their efficiencies is |
| Answer» ANSWER :C | |
| 38. |
What is similar from the dynamics point of view between a book lying stationary on the horizontal table and a raindrop falling downward with constant speed |
| Answer» SOLUTION : In bothcaseresultantforce is ZERO. | |
| 39. |
Depict Delta kinetic energy = Delta U in potential energy - displacement graph for a spring. |
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Answer» SOLUTION :(i) A compressed or extended spring will transfer its stored potential energy into kinetic energy of the MASS attached to the spring. (ii) The potential energy - displacement graph is SHOWN in Figure.  (iii) In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly such that the total energy of the SYSTEM remains constant. At the means position, `Delta KE = Delta U` |
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| 40. |
Assertion : Beating a person with a scale or stick is an example for how momentum is converted into force. Reason : If 'm' is the mass of the scale 'v' is the velocity of the scale (movement), this 'mv' changes, from the initial point to the final point (i.e. hand or body) with respect to time As we know (mv)/(t)=ma((V)/(t))=ma=F This is how momentum is converted into force. |
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Answer» ASSERTION and REASON are CORRECT and Reason is correct explanation of Assertion |
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| 41. |
The energy spectrum of a black body exhibits a maximum around a wavelength lambda_(0). The temperature of the black body is now changed such that the energy is maximum around a wavelength( 3 lambda_(0))/( 4). The power radiated by the black body will now increase by a factor of |
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Answer» `256//81` |
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| 42. |
An iron tyre is to be fitted onto a wooden wheel 1m in diameter. The diameter of the tyre is 5 mm smaller than wheel. How much should the temperature of the tyre be increased for this purpose? alphafor iron = 12 xx 10^(-6)per celsius degree. |
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Answer» |
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| 43. |
Obtain the relation between the MKS and CGS unit of work. |
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Answer» Solution :The unit of work in MKS system is JOULE and that in CGS system is erg. The relation between Joule and erg can be OBTAINED as follows. Dimensional formula for work, `W=M^(1)L^(2)T^(-2)` in MKS syste = in CGS system `m(kg)=10^(3)m(gm)` `L(m)=10^(2) L (CM)` `T(s)=10^(0)T(s)` `L^(1)L^(2)T^(-2)=(10^(3)M^(1))(10^(2)L)^(2)(10^(0)T)^(-2)` `=10^()xx10^(4)M^(1)L^(-2)T^(-2)` `=10^(7)M^(1)L^(2)T^(-2)` So, MKS unit of work =`10^(7)` CGS unit of work `:. 1J=10^(7)` erg |
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| 44. |
A cricket ball of mass 150 g has an initial velocity mu=(3hati+4hatj) ms^(-1) and a final velocity v = - (3hat(i) + 4hat(j)) ms^(-1), after being hit. The change in momentum (final momentuminitial momentum) is (in kgms^(-1) ) |
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Answer» zero Giveninitialvelocity`mu = (3 hat(j )+ 4 hat(j ))` Finalvelocityv = - `(3hat(j)+ 4hat(j ))` Nowchange inmomentum = Finalmomentum- initialmomentum `= m(v-u)` `= (0.15)[-6HAT(j ) - 8 hat(j ))` `= -[0.9hat(j ) + 1.20 hat(j)]` |
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| 45. |
When a force (8hat(i)+4hat(j)) newton displaces a particle through (3hat(i)-3hat(j)) metre, the power is 0.6 W. The time of action of the force is |
| Answer» Answer :A | |
| 46. |
Give difference between average speed and average velocity. |
Answer» SOLUTION :
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| 47. |
Answer the following questions based on the P-T phase diagram of CO_(2): CO_(2) at 1 am pressure and temperature- 60^(@) Cis compressed isothermally Does it go through a liquid phase ? |
| Answer» SOLUTION :No, VAPOUR CONDENSES to SOLID DIRECTLY. | |
| 48. |
Three particles start from origin at the same time: one with velocity v_(1) along positive x - axis, the second along the positive y-axis with a velocity v_(2) and the third along the line y=x with such a speed that all the three always stay in a straight line, then the velocity of the third particle is (sqrt(k) V_(1)V_(2))/((V_(1)+V_(2))). |
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Answer» |
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| 49. |
A missile is fired radially from the surface of earth (radius 6.4 xx 10^(6) m) at a satellite, orbiting the earth. The satellite appears stationary vertically upwards from the point where the missile is launched. Its distance from the centre of the earth is 25.4 xx 10^(6)m. Will the missile actually hit the satellite? |
| Answer» SOLUTION :At time of launch of missile, it has the following velocities. (i) An orbital angular velocity equal to orbital angular velocity of earth and (ii) vertical velocity to reach geo-stationary satellite. When missile is moving TOWARDS satellite, no torque is acting on it, hence angular momentum of missile-earth system MUST be CONSERVED. With the increase in height `r`, the moment of inertia of missle, `I = mr^(2)` increase as a result of it, its angular velocity decrease because `I omega = a` constant. As the orbital angular velocity of satellite remains unchanged so the missile will not be hitting the satellite. | |
| 50. |
A body starts from rest from a point distance R_0 from the surface of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R to Radius of the earth) (R_(0) gt R) |
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Answer» `sqrt((GMR_(0))/(R+R_(0)))` |
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