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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. |
A springof young.s modulus2xx10^(11) pa is suspended vertically and subjected to a load of 5 kg and elongation is 2 mm. when the loadis doubled - match the following (g=9.8m//s^(2)) {:("Section-A","Section-B"),("a) Elongating force","e) 4 mm"), ("b) Stress","f) Unchanged"),("c) Elongation", "g) 98 N"), ("d) Young.s modulus","h) Doubled"):} |
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Answer» a-h, b-g, c-f, d-e |
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| 2. |
The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance ? |
| Answer» Answer :A | |
| 3. |
A stone A is dropped from rest from a height h above the ground . A second stone B is simultaneously thrown verticallly up from a point on the ground with velocity u. The line of motion of both the stone is same . Find if the value of v which would enable the stone B to meet the stone A midway between their initial positions Time of travel of each stone = |
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Answer» Solution :DISTANCE TRAVELLED by each stone `=h/2` For stone `A, h/2=1/2"GT"^(2)`i.e.,`t=sqrt(h/g)` For stone `B,h/2=ut-1/2"gt"^(2)=usqrt(h/g)-1/2g(h/g)` `rArrh/2=usqrt(h/g)-h/2"(or)"usqrt(h/g)=h:. U=hsqrt(h/g)=sqrt(GH)` 
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| 4. |
Rekesh asked his granpda, (who was once a scientist) that whether the planets orbitting around the sun are regular shaped i.e. properly spherical in shape. His Grandpa answered No, they are not at all. All are irregularly shaped. Then Rakesh asked him again, how to find that they are irregular in shape. What is his grandpa's explanation? (i) How does a RADAR and SONAR Work? |
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Answer» Solution :Rakesh's grandpa explained, the science behind this question is Doppler effect. In Doppler effect, the velocity of light or sound is `c= nlambda` Where n = frequency of sound (light) `LAMBDA` = Wavelength of sound (light) Scientists would TRANSMIT some light pulses on different parts (points) on the surface of a planet. They would measure the returning time, after the reflection of light from the planet if there is an apparent change in frequencies of light from different parts of the planet's surface, it means the planets are irregularly shaped. For example if the light pulses hit the top of the MOUNTAIN on the planet, the time of returning or shift in frequency will be noted. If the light pusle, incident on a valley or a river or a sea, the returning time or frequency will be more. So using this difference in frequencies or time, we can analyse the planets contain peaks and deep parts on them. Thus planets are irregulary shaped. (i) RADAR - it means Radio Detection and Ranging. To find the speed of an AEROPLANE it is used. The radar sends high frequency radio waves towards an aeroplane. The reflected waves are detected by the receiver of the radar station. The difference in frequencies is used to determine it's speed. SONAR - It means Sound Navigation and Ranging. Sound waves generated from a ship fitted with SONAR are transmitted in water towards an APPROACHING submarine. The frequency W of the reflected waves is measured and hence the speed of the submarine is calculated. |
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| 5. |
Four simple harmonic vibrations x_(1) = 8 sin omega t, x_(2) = 6 sin (omega t + pi//2), x_(3) = 4 sin(omega t + pi) and x_(2) = 2 sin (omega t + 3pi //2) are superimposed on each other. The resulting amplitude is |
| Answer» ANSWER :C | |
| 6. |
In the above problem velocity after 3 seconds is |
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Answer» `20hat(i)+20sqrt(3)HAT(J)` |
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| 7. |
Y,B, G represent respectively the young.s modulus, bulk modulus and rigidity modulus of a body. If rigidity modulus is twice the bulk modulus, then |
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Answer» `Y=(5)/(18)B` |
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| 8. |
A string 25 cm long and having a mass of 2.5 g is under tension. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency , 8 beats per second are heard. It is observed that decreasingin the string, decreases the beat frequency . If the velocity of sound in air be 320 m*s^(-1)find the tension in the string . |
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| 9. |
If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0 s to t = T sec, then T may be called the average life of the pendulum. When the spherical bob of the pendulum of mass 1kg suffers a retardation (due to viscous drag) proportional to into velocity with .b. as the constant of proportionality, the average life time of pendulum is (assuming damping is small) in seconds, |
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Answer» `0.693/B` |
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| 10. |
A uniform spring has an unscratched length L and a force constant k. the spring is cut into two parts of unscratched lengths l_1, l_2and such that l_1 = eta l_2 where etais an integer. The corresponding force constant are k_1and k_2, then identify the wrong statements below, |
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Answer» FORCE HASTO be kept same to find `k_1 and k_2` |
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| 11. |
In a container of negligible mass .m. grams of steam at 100^@C is added to 100 gm of water that has temperature 20^@C . If no heat is lost to the surroundings at equilibrium , match the items given in Column - I with that in Column - II. |
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| 12. |
Droplets of a liquid are generally more spherical in shape than large drops of the same liquid is not because of |
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Answer» Force of SURFACE tension is EQUAL and opposite to the force of GRAVITY |
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| 13. |
A Ship .A. steams down north at a speed of 8 kmph. and ship B due west at a speed of 6 kmph. The velocity of A w.r.t B is . |
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Answer» 10 KMPH, `Tan^(-1) ((4)/(3))` N of E |
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| 14. |
The resultant of two forces whose magnitudes are in the ratio 3:5 is 28 N. If the angle of their inclination is 60^(@), then find the magnitude of each force. |
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Answer» Solution :Let `F_(1)` and `F_(2)` be the two forces. Then `F_(1) = 3x, F_(2) = 5x, R = 28 N` and `theta = 60^(@)` `R = sqrt(F_(1)^(2) + F_(2)^(2) + 2F_(1)F_(2) cos theta)` `rArr 28 = sqrt((3x)^(2) + (5x)^(2) + 2(3x)(5x)cos 60^(@))` `rArr 28 = sqrt(9X^(2) + 25x^(2) + 15x^(2)) = 7x` `rArr x = (28)/(7) = 4` `therefore F_(1) = 3 xx = 12N, F_(2) = 5 xx 4 = 20 N`. |
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| 15. |
Two particles A and B moving in x-y plane are at origin at t=0 . The initial velocity vertors of A and B are U_(A)= 8 hat(i) m//s and U_(B)= 8 hat(j) m//s. Their accelerations are a_(A) = - 2hat(i) m//s^(2) and a_(B)= -2 hat(j) m//s {:("Column-I","Column-II"),("A) The time in seconds at which relative velocity of A wrt B is zero","P) "16sqrt(2)),("B) The distance in meters between A and B when theirrelative velocity is zero","Q) " 8 sqrt(2)),("C) The time in seconds at which they have same position", "R) " 8),("D) The relative velocity in m/s when they are at same position","S) " 4),(,):} |
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| 16. |
A man of mass 70 kgstands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s^(-1) , (b) downwards with a uniform acceleration of 5 m s^(-2), (c) upwards with a uniform acceleration of 5 m s^(-2) . What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ? |
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Answer» SOLUTION :The reading on the scale is a measure of the force on the floor by the man.By the THIRD Law, this is equal and opposite to the normal force N on the man by the floor. (a) `N = 70 xx 10 = 700 N : ` Reading is 70 kg (b) `70 xx 10 - N = 70 xx 5` : Reading is 35 kg (c ) `N -70 xx 10 =70 xx 5` : Reading is 105 kg (d) `70 xx 10 -N = 70 xx 10`: Reading would be ZERO, the scale would be read zero . |
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| 17. |
State in which of the following cases, an observer will not see any Doppler effect ? |
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Answer» Both the SOURCE and abserver remain stationary but a wind BLOWS . |
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| 18. |
What isWien'slaw ? |
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Answer» Solution :Wien.s law STATES that, the WAVELENGTH of MAXIMUM intensity of EMISSION of a black body RADIATION is inversely proportional to the absolute temperature of the black body. `lambda_(m)prop (l)/(T) (or) lambda_(m)=(b)/(t)` |
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| 19. |
An object is projected with a velocity of 20 ms^(-1) making an angle of 45^(@) with horizontal. The equation for the trajectory is h = Ax - Bx^(2) where .h. is height, x is horizontal distance, A and B are constants. The ratio A and B is (g = 10 ms^(-2)) |
| Answer» ANSWER :D | |
| 20. |
A motor car is going dur north at a speed of 50km/h. It makes a 90^(@) left turn without changing the speed. The change in the velocity of the car is aboout. |
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Answer» 50km/h towards west |
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| 21. |
A vehicle of mass 10kg is moving with a velocity of 5 ms^(-1) .To stop it in 1//10 sec the required force in opposite direction is |
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Answer» 5000N |
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| 22. |
Two cylinders Aando m sine but different masses of Am f , Initially at restare rolling down an inclined plan. Which cylinder will reach the bottorat. |
| Answer» Solution :The initial mechanical energy of both the linders is in the FORM of potential energy and hence is greater for cylinder B as mas ofB is greater than MASS of A.On REACHING the bottom of the inclined plan, their potential energies will be converted to translational kinetic energy, `K_(1)=(1)/(2)` and rotational kinetic erg, `K_(r)-(1)/(2)Iomega^(2)`both of which are al proprtional to mass . As long mechanical energy is conserved, both of thenboth of the cylinders will arrive the bottom with some velocity, which is independent of man and hence REACH the bottom at same time | |
| 23. |
Four wires made of Al, Cu, Fe and steel and of lengths l_1 = 100m, l_2 = 150m, l_3 = 250m and l_4 = 350m,break due to their own weights. Arrange these wires with their breaking stresses in ascending orderd_(Al) = 7500 kg//m^2 , d_(Cu) 7800 kg//m^3 , d_(Fe) = 9000kg//m^2 , d_("steel") = 11000 kg//m^3 |
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Answer» St, Fe, AL, Ci |
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| 24. |
The ends of an well-lagged copper bar of length L are held at steady temperature theta_(1)andtheta_(2)(theta_(1)gttheta_(2)) Which graph best represents the relationship between dQ/dt, the rate of flow of heat, and x, the distance from the hot end when all temperatures in the bar are steady ? |
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Answer»
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| 25. |
Which one of the following graphs indicate the variation between the frequency n and square root of density (rho) of a wire, keeping its length , radius and tension constant, is : |
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Answer»
For l, T A to CONSTANT `N prop (1)/(sqrt(rho))` |
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| 26. |
Given samples of 1cm^(3) of Hydrogen and 1cm^(3) of Oxygen both at N.T.P., which sample has larger number of molecules |
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Answer» Hydrogen |
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| 27. |
(A): For an oscillating simple pendulum, the tension in the string is maximum at the means position and minimum at the extreme position. (R): The velocity of oscillating bob in simple harmonic motion is maximum at the mean position. |
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Answer» Both 'A' and 'R' are TRUE and R' is the CORRECT EXPLANATION of 'A' |
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| 28. |
Two indentical trains are running in opposite directions over two tracks along the equator with equal speed. Will both the trains exert the same force on the tracks? |
| Answer» Solution :Both trains will not exert the same force on the tracks. The train, Running towards. The east from the west, possesses a greater ANGULAR velocity than that of the earth with respect to the axis of rotation of the earth. This is because, the earth is also rotating from the west to the east with respect to the same axis. As a result, the outward CENTRIFUGAL force, which is DIRECTLY proportional to `omega^(2)` , will increase. Hence, the apparent weight is the difference of the real weight and the centrifugal force on the train. Consequently, the train will exert comparatively less force on the tracks. on the other hand, the train running towards the west from the east, will have an increase in its apparent weight, and hence, will exert a comparatively greater force on the tracks. | |
| 29. |
Value of temperature gradient is 80^(@) C/m on a rod of 0.5 m length. Temperature of hot end is 30^(@)C, then what is the temperature of cold end ? |
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Answer» Solution :`(T_(1)-T_(2))/(L)=80` `30-T_(2)=80xxL` `=80xx0.5` `30-T_(2)=40` `:.T_(2)=30-40` `:.T_(2)=-10^(@)C` |
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| 30. |
An aircraft executes a horizontal loop at a speed of 200m/s with its wings banked at 15°. What is the radius of the loop ? |
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Answer» |
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| 31. |
An automobile of mass m accelerates, starting from rest. The engine supplies constant power P, show that the velocity is given as a function of time by v=((2Pt)/(m))^(1//2) |
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Answer» Solution :Power,P=constant WORK done up to time t is W =Pt From work energy THEOREM, `W=DeltaKE " or "Pt=(1)/(2)MV^(2)` `:. "" v=((2Pt)/(m))^(1//2)` |
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| 32. |
The kinetic energy of translation of 20gm of oxygen at 47^@C is (molecular wt. of oxygen is 32gm/mol and R=8.3 J/mol/K) |
| Answer» Answer :A | |
| 33. |
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules ? |
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Answer» increases |
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| 34. |
Statements :(a) Position of C.m depends on acceleration due to gravity where as position of C.g independent on acceleration due to gravity. (b) Centre of gravity of a body is defined to know the amount of stability of the body and the center of mass of the body is defined to describe the nature of motion of the body. |
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Answer» a is TRUE, B is FALSE |
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| 35. |
Displacement of particle changes with respect to time according to equation x =ae ^(-alpha t) + be ^(betat)where a, b, a and Bare positive constants, then velocity of particle is ...... |
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Answer» indipendent of `alpha and beta` `therefore (DX)/(dt) =a (e ^(- alpha t )) (-alpha ) + B ( e ^(beta r )) (beta)` `therefore v =- aalpha e ^(- alpha t ) + b beta e ^(beta t )` `= b beta e ^(beta t )- a alpha e ^(- alpha t)` `= b beta e ^(beta t) - (a alpha)/(e ^(alpha t))` THUS, `(a alpha )/(e ^(alpha t))` will increase with increase of t and velocity will increase. |
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| 36. |
If .g. is acceleration due to gravity on the earth.s surface, then the gain in PE of an object of mass .m. raised from the surface of the earth to a height equal to the radius R of the earth is |
| Answer» ANSWER :A | |
| 37. |
A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratios are (V_B//V_A) = 2 and (V_D//V_A ) = 4. If the temperature T_A at A is 27^@ C, calculate, a) The temperature of the gas at point B, b) Heat absorbed or released by the gas in each process.c) The total work done by the gas during the complete cycle. Express your answer in terms of the gas constant R. |
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Answer» Solution : Taking `V_A=V_0, V_B =2V_0 , V_D =4V_0` Process `Ato B` : Isobaric process as V/T = constant ( `:.` VT graph is STRAIGHT line) `[(V_0)/((273 + 27))] =((2V_0))/(T_B)` (or)` T_3 = 600 K = 327^@ C` ` Q_1= nC_pDelta T = 2 `(5/2) R (600 - 300) = 1500 R Heat is absorbed in this process as `Q_1` is positive . Process `B to C `(Isothermal ) [ `:.` temperature is constant] ` Q_2 =W_2=nRT_BIn (V_2 //V_1)` `=2 xx R xx600 xx log ((4V_0)/(2V_0))= 1200 R log 2 = 831.78R` Heat is absorbed in this process as Q, is positive. Process `Cto D` (Isochoric) (`:.`Volume is constant) `Q_3 = nC_v Delta T = 2 xs (3//2R)xx (300 - 600) = -900 R` Heat is released in this process as `Q_3` is negative. Process `Dto A` (Isothermal) ` Q_4 =W_4=nRT_A In (V_2//V_1) = 2 xx R xx 300 xx log (V_0 //4V_0) = -831.78 R` Total work DONE :` W=Q_1 +Q_2 +Q_3 + Q_4 = 1500 R+ 831.78 R-900 R - 831.78 R=+600 R.` |
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| 38. |
An iron boiler with walls 1.25cm thick contains water at atmospheric pressure. The heated surface is 2.5m^(2) in ara and the temperature of the underside is 120^(@)C. Thermal conductivity of iron is 20cals^(-1)m^(-1)K^(-1) and the latent heat of evaporatioin of water 536xx10^(3) cal kg^(-1) (J=4.2Jcal^(-)). Find the mass of water evaporated per hour. |
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| 39. |
The physical quantity that determins whether or not a given system A is in thermal equilibrium with another system B is called |
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Answer» Pressure |
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| 40. |
(A): A body may be moving with uniform speed and non uniform acceleration. (R): A body may have uniform velocity and nonzero acceleration. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 41. |
A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If alpha and beta be the base angles and theta be the angle of projection, prove thattam theta=tanalpha+tanbeta. |
Answer» Solution :The situation is SHOWN in figure. From figure, we have  `tanalpha+tanbeta=(y)/(x)+(y)/(R-x)` `tanalpha+tanbeta=(YR)/(x(R-x))""…….(i)` But EQUATION of trajectory is `y = xtantheta[1-(x)/(R)]` `TANTHETA=(yR)/(x(R-x))""...........(ii)` From Eqs. (i) and (ii),`tantheta=tanalpha+tanbeta` |
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| 42. |
A famous relation in physics relates 'moving mass' m to the 'rest mass' m_(0) of a particle in terms of its speed v and the speed of light,c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: m=(m_(0))/((1-v^(2))^(1//2)) Guess where to put the missing c. |
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Answer» Solution :For CORRECT equation, DIMENSIONAL formula of both sides must be same. Here, dimensions of m and `m_(0)` are same. Therefore, `(1-V^(2))^(1/2)` should be dimensionless. But is not. To make it dimensionless, `v^(2)` must be divided by `C^(2)`. Correct equation `m=(m_(0))/((1-(v^(2))/(c^(2)))^(1/2))` |
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| 43. |
When a solid sphere is rolling along level surface the percentage of its total kinetic energy that is translational is |
| Answer» ANSWER :B | |
| 44. |
In uniform circular motion (i) only the value of velocity is constant. (ii) velocity vector is constant. (iii) direction of velocity is constant. Select the correct one. |
| Answer» SOLUTION :Onlythe VALUEOF VELOCITYIS CONSTANT | |
| 45. |
A cubical block rests on an inclined plane of coefficient of friction mu =(1)/( sqrt(3)) . What should be the angle of inclination so that the block just slides down the inclined plane? |
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Answer» SOLUTION :`As TANTHETA = mu` ` theta= (1)/( sqrt(3))`or ` theta= tan^(-1)((1)/( sqrt(3))) = 30^@` |
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| 46. |
A vertical offshore structure is built to withstand a maximum stress of 10^(9) Pa . Isthe structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km , and ignore oceancurrents . |
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Answer» Solution :Here , h=3km=3000m `RHO` (density of WATER)`=10^(3)kg//m^(3)` The pressure exerted by sea water `P=hrhog` `P=(3000)(1000)(9.8)` `P=2.94xx10^(7)Pa` Pressure in the sea at a depth is about `3xx10^(7)` Pa . Hence , STRUCTURE willbe suitable for putting upon top of an oil well provided the pressure exerted by sea water is LESS than the maximum stress of `10^(9)` Pa it canbear . Pressure DUE to sea water is `3xx10^(7)` Pa since the pressure of sea water is less than the maximum. |
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| 47. |
A faulty thermometer has fixed points marked 5 and 95 what is the correct temperature in centigrade when this thermometer reads 59 |
| Answer» Answer :A | |
| 48. |
An elevator can carry a maximum load of 1800 kg (elevator +passengers ) is moving up with a constant speed of 2ms^(-1).The frictional force opposing the motion is 4000 N . Determine the minimumpowerdeliveredby the motor to theelevator in watss as well as in horse power . |
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Answer» Solution :The downward force on the elevatoris , `F = MG +f "where " f = ` friction force ` = 1800 xx10+ 4000` ` = 18000 +4000` ` = 22000 N ` The MOTOR SUPPLY enough power to balance this force . Hence Power `P = Fv` ` = 22000 xx2 ` ` = 44000 W ` ` = (44000)/(746) hp ` `= 58.98 ` hp ` :. P approx = 59 ` hp . |
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| 49. |
A person of mass 50kg carrying a load of 20kg walks up a staircase. If the width and height of each step are 0.25m and 0.2m respectively, the work done by the man in walking up 20 steps is, (g= 10 ms^(-2)) |
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Answer» 1400J |
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| 50. |
Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5 kg and radius 1 m. |
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Answer» SOLUTION :KINETIC ENERGY K.E. `=(1)/(2)I omega^(2)=(1)/(2).(3MR^(2))/(2).(2pi)^(2)` `= 148.1 J`. |
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