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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. |
For an ideal gas a) The change in internal energy in a constantpressure process from temperature T_(1) to T_(2)is equal to nC_(v) (T_(2)-T_(1)), where Cv is the molar heat capacity at constant volume andn is the number of moles of the gas b)The change in internal energy of the gasand the work done by the gas are equal inmagnitude in an adiabatic process c) The internal energy does not change in anisothermal process d) No heat is added or removed in anadiabatic process |
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Answer» only a, B are CORRECT |
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| 2. |
In previous problem, the angle of dip is |
| Answer» Answer :A | |
| 3. |
Drops of liquid of density d are floating half immersed in a liquid of density rho. If the surface tension of liquid is T then the radius of the drop will be (d = density of liquid drop) |
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Answer» `SQRT((3T)/( G ( 2D - RHO)))` |
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| 4. |
During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. This implies that |
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Answer» 150 J of HEAT has been removed from the GAS |
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| 5. |
When objects at different distances are seen by the eye which of the following remains constant ? |
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Answer» The focal LENGTH of the eye lens |
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| 6. |
Two person A and B each of mass 60kg are carrying loads 15kg each. If they reach the top of a building of height 10m in 20s and 40s respectively, the ratio of their efficiencies is |
| Answer» ANSWER :C | |
| 7. |
Statement A: Modification of space by a mass particle is called gravitational field Statement B : Law of gravitation is a consequence of "Action at a distance concept". |
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Answer» A is true, B is false |
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| 8. |
Passage - I : Helium is stored in cylinders, each cylinder having a volume 1.5 m^(3) at a gauge pressure 1.5 xx 10^(6) N//m^(2). A balloon is filled with Helium at atmospheric pressure to a volume 600 m^3, using the gas stored in the cylinders. Temperature of Helium inside the balloon and that of the surrounding air are both 27°C. Given that density of air at 27°C and at atmospheric pressure is 1.2 "kg"//m^(3) and atmospheric pressure is 10^(5) N//m^(2), answer the following questions. [use mass of Helium as 4 g/mol, R=8.3 J/(mol-K) and g=10 m//s^(2) ] Total weight, in addition to the weight of the gas, that can be supported by the filled balloon will be nearly |
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Answer» 1470 GK - WT |
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| 9. |
Passage - I : Helium is stored in cylinders, each cylinder having a volume 1.5 m^(3) at a gauge pressure 1.5 xx 10^(6) N//m^(2). A balloon is filled with Helium at atmospheric pressure to a volume 600 m^3, using the gas stored in the cylinders. Temperature of Helium inside the balloon and that of the surrounding air are both 27°C. Given that density of air at 27°C and at atmospheric pressure is 1.2 "kg"//m^(3) and atmospheric pressure is 10^(5) N//m^(2), answer the following questions. [use mass of Helium as 4 g/mol, R=8.3 J/(mol-K) and g=10 m//s^(2) ] If the balloon is filled with Hydrogen (instead of Helium) at the same temperature and pressure. total weight (in addition to the weight of the gas) that can be supported by the balloon will be |
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Answer» More than that in CASE of Helium |
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| 10. |
A brass disc fits snugly in a hole in a steel plate. Should you heat or cool this system to losen the disc from the hole ? given that alpha _b gt alpha _Fe. |
| Answer» Solution :The TEMP, coefficient of linear EXPANSION for brass is greater than that for steel. On cooling the disc shrinks to a greater extent than the HOLE, and hence brass disc GETS lossened. | |
| 11. |
If vecA=3hat(i)-4hat(j) and vecB = -hat(i)-4hat(j), calculate the direction of vecA + vecB |
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Answer» `TAN^(-1)(4) "with" + x-axis` in clock wise |
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| 12. |
A particle is projected vertically upwards from the surface of earth (radius R) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is |
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Answer» R `:. (1)/(2)m((v_(e))/(sqrt(2)))^(2)=(mgh)/(1+(h)/(R))or(v_(e)^(2))/(4)=(gh)/(1+(h)/(R))` or `(2gR)/(4)=(gh)/(1+(h)/(R))or (R)/(2)=(h)/(1+(h)/(R))` Solving this EQUATION, we get `h=R` Note : Kinetic energy is HALF the value required to escape. THEREFORE, speed is `1//sqrt(2)` times the value required to escape. |
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| 13. |
Calculate the work done by a force of 30 N in lifting load of 2g to a height of 10 m (g = 10 ms^(-2)). |
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Answer» Solution :GIVEN : F=30 N, load(m)=2KG, height=10 m,`g=10ms^(-2)` Gravitationalforce Fmg=30 N The distance moved h=10 m work done on the OBJECT `W=Fh=30xx10=300J` |
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| 14. |
A body is projected horizontally from the top of abuilding of height h. Velocity of projection is u. Find |
Answer» SOLUTION :u=initial velocity at the point A [Fig.2.90]. So, u=horizontal component of initial velocity , 0= vertical component of initial velocity. The horizontal component ,u= CONSTANT , since there is no ACCELERATION in that direction . But the vertical motion is under a constant acceleration g, the acceleration DUE to gravity. Let B be the point where the body strikesthe ground . The correspondingvertical motionis through the distance AO =h. |
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| 15. |
A ring of diameter 0.4 m and of mass 10 kg is rotating about its axis at the rate of 1200 rpm. The angular momentum of the ring is |
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Answer» `60.28 "kgm"^(2) s^(2)` `L = I omega=(Mr^(2))(2piv)` `=10xx(0.2)^(2)xx2xx(22)/(7)xx20=50.28"kg-m"^(2)s^(-2)`. |
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| 16. |
A ball impinges directly upon another ball at rest and is itself brought to rest by the impact. If half of initial kinetic energy is destroyed in the collision. The cocfficient of restitution is, |
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Answer» 0.3 |
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| 17. |
The weight of liquid column in a capillary tube is 11xx10^(-1)N. If the radius of tube is 7 mm, calculate the surface tension of liquid. |
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Answer» SOLUTION :Weight of LIQUID COLUMN `W=11xx10^(-4)N` Radius of capillarry tube `r=7mm=7xx10^(-3)m` Weight of liquid column `W=2pirS` Surface tension `S=W/(2pir)=(11xx10^(-4))/(2xx22/7xx7xx10^(-3))impliesS=0.025N//m` |
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| 18. |
A lead bullet just melts when stopped by an obstacle. Assuming that 25% of heat is absrobed by the obstacle find the velocity of the bullet if the initial temperature was 27^(@)C. Melting point of lead = 327^(@)C, specific heat of lead =0.03 cal g ^(-1) ""^(@)C ^(-1).Latent heat of fusion of lead =6 cal g ^(-1). |
| Answer» Solution :`(3//4) xx (1//2)MV ^(2) = mcd T+ mL , c =126 JKG ^(-1) K ^(-1) , L = 2. 52 xx 10 ^(4) J//kg, v = 409.9 MS ^(-1)` | |
| 19. |
Which of the following four statements is false ? |
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Answer» A body can have zero velocity and STILL be accelerated |
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| 20. |
Passage - I : Helium is stored in cylinders, each cylinder having a volume 1.5 m^(3) at a gauge pressure 1.5 xx 10^(6) N//m^(2). A balloon is filled with Helium at atmospheric pressure to a volume 600 m^3, using the gas stored in the cylinders. Temperature of Helium inside the balloon and that of the surrounding air are both 27°C. Given that density of air at 27°C and at atmospheric pressure is 1.2 "kg"//m^(3) and atmospheric pressure is 10^(5) N//m^(2), answer the following questions. [use mass of Helium as 4 g/mol, R=8.3 J/(mol-K) and g=10 m//s^(2) ] Number of cylinders required to fill the balloon are |
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Answer» 16 |
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| 21. |
The ends of copper rod of length 1 m and area of cross section 1cm^(2) are maintained, at 0^(0)C and 100^(0)C, and steady state achieved. At the centre of rod, there is a source of heat of 25W. The thermal conductivity of copper is 400W/mK |
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Answer» The temperature GRADIENT in right HALF is `412^(0)C//m` |
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| 22. |
Abrass scale is graduated at 0^(@)C. What would be the true length of an object, which when measured with this brass scale at 25^(@)C, reads 40 cm ? The coefficient of linear expansion of brass is 18 xx 10^(-6)" "^(0)C^(-1). |
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Answer» |
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| 23. |
(A) : The horizontal displacement of a projectile varies linearly with time. ( R) : Projectile motion is uniform motion along 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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| 24. |
The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle theta without slipping and slipping down the incline without rolling is |
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Answer» `5 : 7` `a_("slipping") = g sin theta "" …. (i)` Acceleration of the solid sphere rollind down the incline without slipping is `a_("rolling")= (g sin theta)/(1 + (k^(2))/(R^(2))) = (a sin theta)/(1 + 2/5) ( because` For solid sphere , `(k^2)/(R^2) = (2)/(5)`) `= (5)/(7) g sin theta "" ... (ii)` Divide eqn. (ii) by eqn.(i) , we get `(a_("rolling"))/(a_("slipping")) = (5)/(7)` |
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| 25. |
Two 20 kg cannon balls are chained together and fired horizontally with a velocity of 200 m//s from the top of a 30 m wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at t=2s, at a distance of 250 m from the foot fo the wall, and 5 m to the right of line of fire determine the position of the other cannon ball at that instant Neglect the resistance of air. |
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Answer» Solution :As no external force acts in `z` direction, HENCE `z` - coordinate of the centre of MASS of he ball should be zero. To make `z`-coordinate zero other ball should fall symmetricaly with respect to `z` axis. Hence `z`-coordinate of other ball `=-5m`. The balls do not have any exeternal force in `X` direction. Hence in `x` direction the centre of mass shouldmove with constant VELOCITY. `x` coordinate of centre of mass at `t=1.5 s=200xx2=400m` Hence `x_(CM)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` `400=(20xx250+20x_(2))/(20+20)` ` x_(2)=800-250=550m` Height fallen by centre of mass at `t=2s`, `h=1/2xx10xx(2)^(2)=20m` Hence `y` coordinate of centre of mass `=30-20=10n` Hence `y_(CM)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))` `10=(20xx0+20xxy_(2))/(20+20)` `implies y_(2)=20m` |
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| 26. |
By which one of the following working temperature the efficiency of a carnot engine obtained maximum ? |
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Answer» 100 K , 80 K `therefore` For getting maximum efficiency , the ratio of `T_2/T_1` MUST be minimum . From option (A) `T_2/T_1=80/100`=0.8 From option (B) `T_2/T_1=60/80=3/4`=0.75 From option (C) `T_2/T_1=20/40=1/2`=0.50 From option (D) `T_2/T_1=40/60 =2/3`=0.67 `therefore` The minimum value of `T_2/T_1` is in (C ) so efficiency obtained maximum |
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| 27. |
Caculate the compressional force required to prevent the metallic rod of length 1 cm and cross-sectional area A cm^(2) when heated through t^(@)C, from expanding along lengthwise. The young's modulus of elasticity of the metal is E and mean coefficient of linear expansion is alpha per degree celsius : |
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Answer» Solution :The change in natural LENGTH = `Delta1_(t)=1alphat` The natural length of rod at TEMPERATURE `t^(@)C` is `1+lalphat` The decrease in natural length due to developed stress = `Deltal` But the length of rod remains constant. `thereforeDelta1_(t)-Delta1=0""thereforeDelta1=Delta1_(t)=1alphat` `thereforeE=("stress")/("strain")=(F/A)/((-DeltaL)/(l+Deltal_(t)))` `thereforeF=(EADeltal)/(l+Deltal_(t))=(EADeltal_(t))/(l+Deltal_(t))=(-EA1alphat)/(l+lalphat)=-(EAalphat)/((1+alphat))` Here, negative sign INDICATES that the FORCES is compressive in nature. |
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| 28. |
A circular disc of radius 'R' and thickness R/6 has mement of inertia 'I' about an axis passing through its centre and perpendicular to its plane. It melted about recasted into a solid sphere. The M.I. of the sphere about its diameter as axis of rotation is |
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Answer» I |
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| 29. |
Techonology is …………….. For…………… |
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Answer» |
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| 30. |
Define Poise . |
| Answer» Solution :If DYNE tangential force is REQUIRED to maintain the velocity gradient of `1cms^(-1)//cm` between two parallel layers of liquid each of area `1CM^(2)` then coefficient of VISCOSITY is 1 poise. | |
| 31. |
Match Column-I with Column-II : {:("Column-I","Column-II"),("(1) Combination of two vector is maximum.",(a)180^@),("(2) Combination of two vector is minimum.",(b) 90^@),(,(c)0^@):} |
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| 32. |
.m. grams of a gas of a molecular weight M is flowing in an isolated tube with velocity 2V. If the gasflow is suddenly stopped the rise in its temperature is, (gamma = ratio of specific heats, R = universal gas constant , J = Mechanical equivalent of heat) |
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Answer» `(2Mv^2(GAMMA -1 ))/(RJ)` |
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| 33. |
The kinetic energy (KE) of a planet revolving in an elliptical orbit, around the sun, is plotted against the distance r from the sun. Out of the following graphs the one that best represents the variation is |
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Answer»
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| 34. |
A 15gm ball is shot from a spring gun whose spring has a force constant 600 N//m. The spring is compressed by 5cm. The greatest possible horizontal range of the ball for this compression is (g = 10 m//"sec") |
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Answer» Solution :`R_("max")= (u^(2))/(g)`……………(1) But K.E acquired by ball = P.E of SPRING gun `IMPLIES (1)/(2)MU^(2)= (1)/(2)kx^(2) implies u^(2)= (kx^(2))/(m)`……(2) From equation (1) and (2) `R_("max")= (kx^(2))/(mg)= (600 XX (5 xx 10^(-2))^(2))/(15 xx 10^(-3) xx 10)= 10cm` |
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| 35. |
(A) : The work done by the spring force in a cyclic process is zero.(R ) : The spring force is a conservative force. |
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Answer» Both (A) and (R ) are true and (R ) is the correct explanation of (A) |
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| 36. |
The three flat blocks in the figure are positionedon the 30 ^(@)incline anda force parallel to the inclined plane is applied tothe middle block . The upper block is prevented from moving by a wire which attaches it to the fixed support . The coefficient of static friction for each of the three pairs ofcontact surfaces is shwon in the figure . Determine the maximum value which P may have before any slipping takes place . |
Answer»  F.B.D of `m_(2)` `P+ M_(2) G "sin" theta= mu_(1) N_(1) + mu_(2) N_(2)` `P M_(2) g "sin" theta = mu_(1) M_(1) g "cos" theta + mu_(2) " g cos" theta (M_(1) + M_(2))` By solving this equation , we getvalue of P . 
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| 37. |
Assertion : Two persons can not conversant on Moon.Reason: Moon does not have an atmosphere. |
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Answer» Assertin and reason both are correct and reason is the correct explantion of assertion. |
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| 39. |
If you stand on the floor it exerts an upward reaction on you. Then why don't you fo up ? |
| Answer» Solution :Here, two forces act on the BODY:(1) downward force of WEIGHT due to earth.s attraction and (2) reaction of the ground. They are EQUAL and opposite. So, resultant force = 0, and there is no motion of the body. | |
| 40. |
A shell is thrown vertically up. The shell at the highest point explodes into two equal fragments. The centre of mass of the two fragments |
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Answer» goes further up ANDTHEN FALLS |
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| 41. |
A projectile is thrown at an angle of 30^(@) with a velocity of 10m/s, the change in velocity during the time interval in which it reaches the heighest point is |
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Answer» 10 m/s |
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| 42. |
Is it possible for a uniformly accelerating object to change its direction of velocity? |
| Answer» Solution :Yes, it is obsrved when a ball is THROWN up, after reading the MAXIMUM HEIGHT itcomes back and changes the direction of its velocity. | |
| 43. |
Thermal radiation falls in |
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Answer» VISIBLE REGION |
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| 44. |
A lift of mass 1000 kg is suspended by steel wire of maximum safe stress 1.4 xx 10^(8) N//m^(2). Find the minimum diameter of the wire, if the maximum acceleration of the lift is 1.2 ms^(-2) |
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Answer» |
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| 45. |
For study of nanotechnology …... Microscope is developed. |
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Answer» |
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| 46. |
A wire has a mass 0.3pm 0.003g radius 0.5pm0.005mm and length 6pm 0.06cm. The maximum percentage error in the measurement of its density is : |
| Answer» SOLUTION :`4%` | |
| 47. |
Assertion : For a free falling object, the next external force is just the weight of the object . Reason : In this case the downward acceleration of the object is equal to the acceleration due to gravity. |
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Answer» If both ASSERTION and reason are true and reason is the correct EXPLANATION of assertion |
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| 48. |
A body in a room cools from 85^(@)C to 80^(@) C in 5 minutes. Calculate the time taken to cool from 80^(@) C to 75^(@) C if the surrounding temperature is 30^(@)C. |
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| 49. |
The length of a rod as measured in an experiment was found to be 3.48, 3.46, 3.49m, 3.50m and 3.48m. Find the average length, the absolute error in each observation and percentage error. |
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Answer» Solution :AVERAGE length `=(3.48+3.46+3.49+3.50+3.48)/(5)` `=(17.41)/(5)` `=3.482m=3.48m` (Round off to 2 places of decimal POINT) The absolute erros in the different measurements are `DeltaL_(1)=3.48-3.48=0.00m` `DeltaL_(2)=3.48-3.46=002 m` `DeltaL_(3)=3.48-3.49=-0.01 m` `DeltaL_(4)=3.48-3.50=-0.02m` `DeltaL_(5)=3.48-3.48=0.00m` The absolute error `=(Sigma|DeltaL_(i)|)/(5)` `=(0.00+0.02+0.01+0.02+0.00)/(5)` `=(0.05)/(5)=0.01m` `therefore " Correct length"=3.48 PM 0.01m` `"PERCENTAGE error "=(0.01)/(3.48)xx100=0.29%` |
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| 50. |
Explain Joule's Experiment of the mechanical equivalent of heat. |
| Answer» Solution :Joule essentially converted mechanical energy to internal energy. In his experiment POTENTIAL energy is converted to rotational KINETIC energy of PADDLE WHEEL and this rotational kinetic energy is converted to internal energy of water. | |
