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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. |
Two identical rods each of moment of inertia .I. about a normal axis through centre are arranged in the from of a cross. The M.I. of the system about an axis through centre and perpendicular to the plane of system is |
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Answer» 2I |
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| 2. |
If the unit of mass is a kg, the unit of length is betametre and the unit of time is gammasecond, The magnitude of calorie in the new system is (1 Cal = 4.2 J) |
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Answer» `4.2 alpha^2 BETA^2 gamma^2` new UNITS |
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| 3. |
A shell is fired at an angle of 30^(@) to the horizontal with a velocity of 392 ms^(-1). Find the time of flight, horizontal range and maximum height attained. |
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Answer» |
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| 4. |
The powerof a caris 20 hp . The caris facingresistanceof 10 kgwt . Due toair and friction . Calculatethe maximum speedattained by thecar against this resistance . |
| Answer» SOLUTION :152.2 m/s | |
| 5. |
If vec(A) = (2hat(i) + 3hat(j)) and vec(B)= (hat(i)- hat(j)) then component of vec(A) perpendicular to vector vec(B) and in the same plane is |
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Answer» `(5)/(2) (HAT(i) + hat(J))` |
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| 6. |
Out of hot moist air and dry cold air in which does sound travel faster ? |
| Answer» Solution :For hot moist AIR, ratio `(P)/(rho)` is more then that for cold dry air. Hence, speed for SOUND `V = sqrt ((gamma P)/(rho))` is more for hot moist air and so sound WIL travel faster in it. | |
| 7. |
What is the distance of centre of mass of a non uniform rod of length L whose mass per unit length lambda =(Kx^(2))/L, whereK is a constant and x is the distance from one end ? |
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| 8. |
If electric force between point charged varies inversely as the cube of the distance, will Gauss.s law be valid ? |
| Answer» Solution :No. Gauss.a LAW is only true if INVERSE square law of ELECTRIC FORCE is true. | |
| 9. |
Which of the following instrument is used for measuring gauge pressure? |
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Answer» Thermometer |
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| 10. |
Find the ratio of the orbital speed of two satellites one of which is rotating round the earth and the other around Mars close to their surface . (mass of the earth = 6.4xx10^(23) kg. Mass of the Mars= 6.4xx10^(23)Radius of the earth = 6400km and Radius of Mars= 3400km) |
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Answer» SOLUTION :The orbital speed of a satellite close to the SURFACE of a planet, `V_(0)=sqrt((GM)/R)` The RATIO orbital speeds , `V_(01)/V_(02) = V_(01)/V_(02)sqrt(((M_1)/M_2)((R_2)/(R_1)))` `V_(01)/V_(02)sqrt(((6xx10^(24))/(6.4xx10^(23)))((3400)/(6400)))=sqrt((60xx34)/(6.4xx64))=(2.232)/1` |
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| 11. |
Starting from rest a wheel rotates with uniform angular accelration 2pi rad s^(-2). After 4s, if the angular acceleration ceases to act, its angular displacement in the next 4s is |
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Answer» `8 PI rad` |
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| 12. |
If the earth stops rotating about its axis, then by what value will the acceleration due to gravity change at the equator ? |
| Answer» SOLUTION :The VALUE of G INCREASES by `omega^2R`. | |
| 13. |
Calculate the rate of flow of heat through a metal sheet 0.02 m thick and area 50 xx 10^(-4) "m"^(2) with its two sides at 273 K and 293 K respectively. Given K=0.2 cal cm""^(-1) "s"^(-1) "c"^(-1). |
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Answer» |
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| 14. |
The minimum speed of a bucket full of water whirled in a vertical circle of radius 10 m at the highest point so that the water may not fall is (g=10ms^(-2)) |
| Answer» Answer :A | |
| 15. |
The limiting stress of a typical human bone is 0.9 xx 10^(8) "Nm"^(-2) while young's modulus is 1.4 xx 10^(10) "Nm"^(-2). How much energy can be absorbed by two leg bones without breaking if each has a typical length of 50cm and an average cross sectional area of 5"cm"^(2) ? |
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Answer» 145 J |
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| 16. |
A block of mass 1 slides down a curved track that is one quadrant of a circle of radius 1m. Its speed at the bottom is 2m/s. The work done by frictional force is: |
| Answer» ANSWER :B | |
| 17. |
When the amplitude of oscillation of a particle in S.H.M is increased to two times, the time period |
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Answer» Is DOUBLED |
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| 18. |
If 5 L of water at 50^(@)C is mixed with 4L of water at 30^(@)C, what will be the final temperature of water? Take the specific heat capacity of water as 4184 J kg^(-1) K^(-1) |
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Answer» SOLUTION :We can use the equation `T_(f)=(m_(1)s_(1)T_(1)+m_(2)s_(2)T_(2))/(m_(1)s_(1)+m_(2)s_(2))` `m_(1)=5L = 5kg and m_(2)=4L = 4KG, s_(1)=s_(2) and T_(1)=50^(@)C=323K and T_(2)=30^(@)C=303K` `"So"T_(f)=(m_(1)T_(1)+m_(2)T_(2))/(m_(1)+m_(2))=(5xx323+4xx303)/(5+4)=314.11K` `T_(f)=314.11K-273K~~41^(@)C` Suppose if we mix equal number of water `(m_(1)=m_(2))` with `50^(@)C` and `30^(@)c`, then the find temperature is average of TWO TEMPERATURES. `T_(f)=(T_(1)+T_(2))/(2)=(323+303)/(2)=313K=40^(@)C` Suppose if both the water are at `30^(@)C` then the final temperature will also `30^(@)C`. It implies that they are at equilibrium and no heat exchange takes place between each other. |
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| 19. |
What is length? What is its SI unit? |
| Answer» Solution :Length is generallya measured DIMENSION of an OBJECT or distancebetween two POINTS. Its SI unit is METRE(m). | |
| 20. |
Two vibrating tuning forks produce progressive waves given be y_(1) = 4 sin 500 pi t and y_(2) = 2 sin 506 pi t where t is in seconds number of beats produced per minute is ……….. |
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Answer» Solution :`omega_1 = 2pi f_1` `omega_2 = 2pi f_2` `500 PI = 2pi f_1 ` ` 506 pi = 2pi f_2` ` f_1 = 250` ` f_2 = 253` ` f_2 = f_1 = 3`beats PER sec and 3 x 60 = 180 beats per minute |
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| 21. |
Production of continuous X-rays is caused by |
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Answer» TRANSITION of ELECTRONS from higher levels to LOWER levels in target atoms |
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| 22. |
Weight of a piece of metal immersed in alcohol at 0^(@)C is W_(1) and at 59^(@)C is W_(2). The coefficient of volume expansion of the metal is less than the coefficient of volume expansion of alcohol. If the density of the metal is greater than that of alcohol then |
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Answer» `W_(1) GT W_(2)` |
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| 23. |
Explain cavity and cavity radiation. |
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Answer» SOLUTION :Perfect black body : ..The body which absorbs all radiant energy incident on it is called perfect black body... Practically it is impossible to have perfectly black body. For a perfect black body absorptivity `a=1` and emissivity `e=1`.  To obtain perfect black body in practice CONSIDER a vessel with a cavity inside it as shown in figure. Inner WALLS of vessel are irregular and coloured black and a very small hole is provided on its wall. The radiation incident on this hole experiences multiple reflections inside and at each reflection it gets partially absorbed by its walls. The possibility of the radiation COMING out of the same hole is negligible. Thus, radiation gets absorbed completely inside the vessel. The part of the wall facing the hole is such that the radiation entering through the hole does not go out of the hle directly after reflection. In this way, this hole can be regarded as a perfectly black body. When this vessel is HEATED uniformly from outside, the radiation emitted through the hole can be considered as the radiation emitted from a perfectly black body. It is called the ..cavity radiation... |
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| 24. |
The time period of a satellite of earth is 5 hr. If the seperation between the earth and satellite is increased to 4 times the previous value the new time period will become .... |
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Answer» 10 hr `T_2^2/T_1^2=r_1^3/r_1^3` `:. T_1^2/(25)=((4R)/r)^3` `T_2^2 = 25 xx 64 ` `:. T_2= 5xx 8 ` `:. T_2 = 40 hr ` |
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| 25. |
What is an ideal liquid ? |
| Answer» Solution :An incompressible, NONVISCOUS LIQUID which cannot withstand any SHEARING stress is called an IDEAL liquid. | |
| 26. |
The SI unit of magnetic permeability is |
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Answer» `A m^(-1)` |
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| 27. |
A block of mass 15 kg is placed on a long trolly . The coefficient of friction between the block and trolly is 0.18 The trolly accelerates from rest at 0.5 m//s^(2) for 20 seconds and then moves with a unifrom velocity Discuss the motion of the block asviewed by (a) a stationary observer on the ground (b) an observer moving with the trolly. |
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Answer» Solution :Here , ` m = 15 kg , mu = 0.18 , a = 0.5 ms^(-2) , t = 20 s ` Force on the BLOCK due to motion of the trooly` F = ma = 15 xx 0.5 = 7.5 N ` Is is obviously in the forward driction (of motion of the trolly) Force of limitting friction on the block` = F = mu R = mu mg = 0.18 xx 15 xx 9.8 = 26 . 46 N ` This opposes the motion of the blockThe block shall not move The force of stice friction F will adjust itsself EQUAL and opposite to F the applied force HENCE a stationary OBSERVER on the ground the block will appear to be at rest relative to the trolly When trolly moves with unifrom velocity forward force is zero force is zero friction alone is acting on the block (b) An observer moving with the trolly has accelerated motion The observer is theregfore non - inertical The law of INERTIA is no longer valid . |
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| 28. |
The wavelenght of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intensity of the star is (Stefan's constant = 5.67 xx 10^(-8) Wm^(-2)K^(-4), constant b= 2898 mumK) |
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Answer» `5.67 XX 10^(8) "Wm"^(-2)` |
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| 29. |
A boy 1.5m tall with his eye level at 1.38m stands before a mirror fixed on a wall. The minimum length of mirror required to view the complete image of boy is |
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Answer» 0.75m |
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| 30. |
The flywheel of an engine, acted upon by a constant torque at the rate of 3000 rev/ min and develops a power of 10 kW. Find the torque acting on the wheel. |
| Answer» Solution :`tau ="Power"//2pin = 10 xx 10^(3) xx 60//2 xx 3.14 xx 6000 = 31.82 NM` | |
| 31. |
Two blocks m_(1) and m_(2) of 1 kg and 2 kg are connected by a spring of spring constant K on a rough horizontal surface of coefficient of friction 0.4. A horizontal force F is applid on m_(1) so that it could just move block mass m_(2). The value of F |
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Answer» |
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| 32. |
A monoatomic ideal gas, initially at temperature T_1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature. T_2 by releasing the piston suddenly. IF L_1 and L_2 are the lengths of the gas column before and after expansion respectively, then T_1 //T_2 is given by |
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Answer» `((L_1)/( L_2))^(2/3)` |
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| 33. |
A uniform wire of length L us bent in the form of a circle. The shift in its centre of mass is |
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Answer» `(L)/(pi)` |
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| 34. |
What are the two types of projectile motion ? |
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Answer» SOLUTION :(i)Projectile GIVEN an INITIAL velocity in the HORIZONTAL direction (horizontal projection). Projectile given an initial velocity at an ANGLE to the horizontal (angular projection). |
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| 35. |
In the above case, if their final temperatures are T_(1),T_(2) and T_(3) . Then . |
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Answer» `T_(2)gtT_(1)gtT_(3)` |
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| 36. |
A heavy uniform metal chain is kept on the surface of a horizontal table. The coefficient of friction between the chain and the table surface is 0.22. Find the maximum fraction of length of the chain that could hung over on one edge of the table ? |
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Answer» Solution :Let L be the total length of the chain, l that of the hanging PART of the chain and m the massper UNIT length of the chain. When the chain is in equilibrium, Weight of the hanging part `=` Frictional FORCE between chain and the table `mlg=mu m(L-l)g` `l=muL-mul` `l(1+mu)=muL` Fraction of length of chain `=(l)/(L)=(mu)/(1+mu)=(0.22)/(1+0.22)=0.18=18%` |
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| 37. |
How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 xx 10^(8) km. |
| Answer» SOLUTION :`2.0 XX 10^(30)` KG | |
| 38. |
(A) : Average speed is equal to total distance travelled divided by lotit time taken. (R ) : The average speed of an object may be to arithmetic, mean of individual speed. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 39. |
The lengths of left and right arms of a balance are 10.5 cm and 10 cm respectively. Without the scale pans, the balance remains horizontal. Even with two scale pans of different masses the balance remains horizontal. Keeping a body on the right pan, equilibrium is achieved by putting a counterpoising 100-g weight on the left pan. What is the actual mass of the body? |
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Answer» |
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| 40. |
A stone is dropped into water from a bridge 44.1 m above the water. Another stne is thrown vertically downward 1s later. Both strike the water simultaneously. What was the initial speed of the second stone? |
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Answer» Solution :`t=SQRT((2 xx 44.1)/(9.8))s =sqrt(9s)=3S` `44.1=v xx 2+1/2 xx 9.8 xx 2 xx 2` `or 2v=44.1 -4.9 xx4=24.5 or` `v=(24.5)/(2) MS^(-1) =12.25ms^(-1)` |
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| 41. |
What do you mean by upthrust or buoyancy ? |
| Answer» Solution :The upward FORCE exerted BI a fluid that oppose the weight of an immersed OBJECT in a fluid is called upthrust or BUOYANT force. | |
| 42. |
(A) : In order to hit a target, a man should point his rifle in the same direction as target. ( R) : The horizontal range of the bullet is depends only on the angle of p[rojectile with horizontal direction. |
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Answer» Both (A) and ( R) are TURE and ( R) is the correct EXPLANATION of (A) |
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| 43. |
One mole of gas occupies a volume of 200 ml. at 100 mm pressure. What is the volume occupied by two moles of this gas at 400 mm pressure and at same temperature? |
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Answer» <P>50 ml `V_(2)=(P_(1)V_(1))/(P_(2))=(100xx200)/(400)=50ML` `rArr"Volume of 2 moles of the gas at 400 mm PRESSURE "=2xx50=100ml` |
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| 45. |
An object falling through a fluid is observed to have acceleration given by a=g-bv where g = gravitational acceleration and b is constant. After a long time of release, it is observed to fall with constant speed. The value of constant speed is |
| Answer» Answer :A | |
| 46. |
One boy takes 2 min to raise a box to a height of 1 m and another boy takes 3 min to do so. The energy of the two boys is |
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Answer» DIFFERENT |
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| 47. |
SI unit of coefficient of thermal conductivity |
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Answer» `WM^(-1)K^(-1)` |
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| 48. |
Does work done by a body against friction always result in a loss of its kinetic or potential energy ? |
| Answer» SOLUTION :The WORK done by a body against friction always result in the loss of only KINETIC ENERGY. | |
| 49. |
A particle m is projected at an angle with the horizontal with kinetic energy E. The potential energy at the top of its trajector is E/2. The range of the particle is (xE)/(mg), then find the value of x is |
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| 50. |
A circular disc is rotating about its own axis at a uniform angular velocity .omega.. The disc is subjected to uniform angular retardation by which its angular velocity is decreased to (omega)/(2) during 120 rotations. The number of rotations further made by it before coming to rest is |
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Answer» 120 |
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