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. |
At the top of a mountain a thermometer read 7°C and barometer reads 70cm of Hg. At the bottom of the mountain the barometer reads 76cm of Hg and thermometer reads 27°C. The density of air at the top of the mountain is times the density at the bottom. |
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Answer» `0.99` |
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| 2. |
What is time of flight? |
| Answer» Solution :The TIME taken for the projectile to complete its TRAJECTORY or time taken by the projecrtile to the GROUND is CALLED time of FIGHT. | |
| 3. |
Four thin rods of length L = 1.0 m each are joined to form a square ABCD. The opposite vertices of the square are joined by mass less rods AC and BD. This square frame is mounted on a horizontal axis through its centre so that the frame can rotate freely in the vertical plane. Masses of rods AB and BC are m = 2kg each and the rod AD and DC have mass M = 4kg each. A monkey of mass m_(0) = 12 kg is at rest on the horizontal rod AD and keeps the system in equilibrium. The monkey takes a sudden jump and rises to a height H from its initial position. Calculate minimum value of H so that the square frame is able to complete a rotation about its central axis. Assume no further contact between monkey and the frame. |
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| 4. |
A capillary tube is immersed vertically in a liquid such that the height of liquid column in it is 'x' This arrangement is taken into a mine of depth 'd' and the height of the liquid column is found to be ‘y'. If 'R' is the radius of the earth, then the depth of mine is |
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Answer» `d = R ((y-X)/( x))` |
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| 5. |
The equation of motion of a projectile is y = ax-bx^(2), where a and b are constants of motion. Match the quantities of column I with the relation of II. |
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Answer» A-p, B-q, C-r, D-s |
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| 6. |
An engineering firm is assigned the job to design the cylindrical pressured water tank for a future colony on Mars, where the acceleration due to gravity is 3.71m*s^(-2). The pressure at the surface of the water will be 130 kPa and depth of the water will be 14.2 m. The pressure of the air in the building outside the tank will be 93 kPa. Find the net downward force on the tank's flat bottom of area 2 m^(2) |
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Answer» 179.4 kN |
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| 7. |
An engineering firm is assigned the job to design the cylindrical pressured water tank for a future colony on Mars, where the acceleration due to gravity is 3.71m*s^(-2). The pressure at the surface of the water will be 130 kPa and depth of the water will be 14.2 m. The pressure of the air in the building outside the tank will be 93 kPa. What is the buoyant force on a wooden block of mass 2 kg and relative density 0.8, height 10 cm? |
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Answer» 5.94 N |
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| 8. |
In the Searle.s experiment of determaination of Young.s modulus of a wire, the quantity which should be measured with greater prectision is |
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Answer» DIAMETER of the wire |
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| 9. |
A simple pendulum has a time period 'T' in vaccum. Its time period when it is completely immersed in a liquid of density one - eighth of the density of the material of the bob is |
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Answer» `SQRT(7/8) T` |
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| 10. |
Acceleration displacement graph of a particle executing S.H.M is as shown in given figure. The time period of its oscillation is (in sec) |
| Answer» ANSWER :B | |
| 11. |
No work is done by a force on an object if a) the force is always perpendicular to its velocity b) the force is always perpendicular to its acceleration .c) the object is stationary but the point of application of the force moves on the object d) the object moves in such a way that the point of application of the force remains fixed. |
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Answer» B, C &d are CORRECT |
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| 12. |
Two equal masses separated by a distance (d) attract each other with a force (F). If one unit of mass is transferred from one of them to the other, the force |
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Answer» does not change |
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| 13. |
Draw the resultant direction of the two unit vectors hatiandhatj Use a 2-dimensional Cartesian ystem. Is hati+hatj a unit vector? |
Answer» Solution :By using the triangular law of addition `hati+hatj` as SHOWN in the following FIGURE, The definition of UNIT vector is `HATA.hatA=1` But here, `(hati+hatj).(hati+hatj)=hati.hati+hati.hatj+hatj.hati+hatj.hatj=1+0+0+1=2` SO, `hati+hatj` is not a unit vector. To make any vector to a unit vector, must divide the vector by its magnitude, `hatA=(hatA)/(|vecA|)` The magnitude of the vector `hati+hatj=sqrt2` Hence, the unit vector is `(hati+hatj)/(sqrt2)`. |
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| 14. |
A horizontal force F is applied on various bodies as shown in Column-I Friction is sufficient to prevent slipping in each case. If f represents frictional force between the body and the object and a represents acceleration of the body. Then match the following columns : |
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Answer» |
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| 15. |
A body of mass 5 kg moving along a straight line is accelerated from 4ms^(-1) to 8ms^(-1) with the application of a force of 10 N in the same direction. Then, |
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Answer» WORK DONE by the FORCE is 120 J |
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| 16. |
During the collision a relatively large force acts on each colliding particle for a relatively _________ time |
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Answer» Short |
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| 17. |
A particle A of mass 10/7 kg is moving in the positive directio of x-axis. At initial position x=0, its velocity is 1ms^(-1), then its velocity (in m/s) at x=10 m is (use the graph given) |
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| 18. |
What is the direction of displacement in a simple harmonic motion ? |
| Answer» SOLUTION :AWAY from the MEAN POSITION. | |
| 19. |
Plane travels 400 m towards north then 300 m towards south and 1200 m vertically upwards, then find resultant displacement ………. |
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Answer» 1200 m `VECR = 400 HATJ - 300 hatj + 1200 hatk` `vecr = 100 hatj + 1200 hatk` `|vecr | = sqrt ((100) ^(2) + (1200) ^(2))` `= sqrt ( 10000 + 1440000) ` `r = sqrt (1450000)` `=1204m` `therefore r ~~1200 m` |
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| 20. |
Which of the following is not the unit of time? a) Light yearb) Astronomical unitc) Parsecd) all of the above |
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Answer» Light year |
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| 21. |
From a tower of height H a particle is thrown vertically upwards with a speed U. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path |
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Answer» `2gH = nu^2 (n-2)` |
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| 22. |
What are the limitations of first law of thermodynamics? How are these overcome in second law of thermodynamics? Explain briefly. |
| Answer» SOLUTION :2nd part : The 2nd law of thermodynamics overcomes these PROBLEMS by the introduction of a new variable called ENTROPY(S), which is a property of all thermodynamic systems. There are techniques to measure the change in entropy `(Delta S)` of a system and its surroundings during a process. If the entropy increases `(Delta S > 0)`, the 2nd law asserts that the process occurs in nature. In the reverse process, the entropy necessarily decreases `(Delta S < 0)`, and that process is not ALLOWED. Moreover, the amount of unavailable energy in a process. is directly proportional to `Delta S`, and can be calculated USING a simple formula. | |
| 23. |
A block of mass m slides down the plane inclined at an angle 60^@with an acceleration g/2. Find the co-efficient of kinetic friction. |
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Answer» 0.01 `F = ma = 2xx 0.5 = 1.5 N` Co-efficient of friction `mu` = ? `therefore muN = F` `mu = F/N = F/(MG) = 1.5/(3xx10) = 0.05` |
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| 24. |
The stress required to double the length of a wire of Young.s modulus .E. is --- |
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Answer» E |
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| 25. |
The linear velocity of a particle 2 cm away from the axis of rotation is 10cm^(-1). What will its angular velocity at 4 cm away from axis? |
| Answer» SOLUTION :From `v=romega,omega=5^("rad/s")` and VELOCITY of all PARTICLE of RIGID BODY is same. | |
| 26. |
A car is moving with a velocity V. The car driver sees a truck at rest at a distance s. To avoid collision which is better, to apply brake or to turn sharply ? |
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| 27. |
For the velocity time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the total distance covered by it in all the seven seconds |
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Answer» `1/2` |
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| 28. |
A simple pendulum is hanging from a per inserted in a vertical wall. Its bob is stretched to horizontal position from wall and is left free to move, the bob hits the wall. If coefficient of restitution is (2)/(sqrt(5)). After how many collisions the amplitude of vibration will become less than 60^(@) |
| Answer» ANSWER :B | |
| 29. |
Define monochromatic emissive power of a body. Write the relation between emissive power and radiant emittance. |
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Answer» SOLUTION :The radiant energy emitted PER second by a unit surface area of the body per unit range of WAVELENGHT is defined as monochromatic emissive power. Radiant emittance, `e = int_(lambda =0) e_(lambda) d_(lambda) `where `e_(lambda)`is theemissive power . |
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| 30. |
Find the external work done by the system inkcal, when 20 kcal of heat is supplied to the system the increase in the internal energy is 8400 J. (1=4200J/kcal) |
| Answer» Answer :B | |
| 31. |
Define Average acceleration . |
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Answer» Solution :The ratio of change in VELOCITY over the time INTERVAL`DELTA t = t_(2) - t_(1)` `vec(a)_("AVG") = (vec(V)_(2) - vec(v)_(1))/(t_(2) - t_(1)) = (Deltavec(v))/(Delta t)` |
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| 32. |
A car is moving with a speed of 30 ms^(-1) on a circular path of radius 500 m. If its speed is increasing at the rate of 2ms^(-2), the net acceleration of the car is |
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Answer» `3.6 MS^(-2)` |
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| 33. |
A circular platform is mounted on a vertical frictionless axle. Its radius is r = 2m and its moment of inertia is I=200kg-m^(2). It is initially at rest. A 70 kg man stands on the edge of the platform and begins to walk along the edge at speed V_(0)=1.0m//s relative to the ground. Find the angular velocity of the platform. |
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Answer» Solution :angular MOMENTUM of MAN = angular momentum of platform in opposite DIRECTION `mv_(0)r=Iomega""omega=0.7`rad/s |
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| 34. |
A block of mass 2kg lying on ice when given a velocity of 6ms^(-1) is stopped by friction in 5s. The coefficient of friction between the block and ice is (g=10 ms^(-2)) |
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Answer» `0.5` |
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| 35. |
If the temperature of the source is increased, the efficiency of the carnot engine |
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Answer» INCREASES |
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| 36. |
Three point masses each of mass 'm' are joined together using a string to form an equilateral triangle of side 'a'. The system is placed on a smooth horizontal surface and rotated with a constant angular velocity omega about a vertical axis passing through the centroid. Then the tension in each string is |
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Answer» `maomega^(2)` |
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| 37. |
What is the resulat of ( vec A+vec B) xx (vec A-vec B) gt |
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Answer» |
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| 38. |
Mercury has an angle of contact equal to 140^@ with soda lime glass. A narrow tube of radius 1.00mm made of this glass is dipped in a trough containing mercury.By what amount does the mercury at the temperature of the experiment is 0.465 Nm^-1. Density of mercury =13.6 times 10^3 kg m^-3 |
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Answer» Solution :RADIUS of tube, r = 1.00 mm = `10^(-3)m`, Surface tension of mercury, `alpha=0.465Nm^(-1)` Angle of CONTACT, `theta=140^(@)thereforeh=(2sigmacostheta)/(rrhog)=-5.24xx10^(-3)m=-5.24mm` NEGATIVE sign shows that the mercury LEVEL is depressed in the tube. |
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| 39. |
A uniform disk of mass m and radius R is projected horizontally with velocity v_(0) on a rough horizontal floor so that it starts off with a purely sliding motion at t=0. After t_(0) seconds it acquires a purely rolling motion as shown in figure. (i) Calculate the velocity of the centre of mass of the disk at t_(0). (ii) Assuming the coefficient of friction to be mu calculate t_(0). Also calculate the work done by the frictional force as a function of time and the total work done by it over a time t much longer than t_(0). |
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Answer» (a) `t le t_(0) v = v_(0) - mu g t` …(1) For Rotational motion, torque is being provided by fricitonal FORCE. `mu mgR = I alpha rArr mu mg R = (1)/(2) mR^(2) alpha` `alpha = (2 mu g)/(R)` so `omega = at rArr omega = (2 mu g)/(R) t` at `t = t_(0), v = R omega` form (1) & (2) `v_(0) - mug t_(0) = 2 mu g t_(0) rArr t_(0) = (v_(0))/(3 mu g)` by elimination `t_(0)` from EQUATION (1) `v = (2 v_(0))/(3)` For `t lt t_(0)` Work done by friction `= DELTA K = K_(f) - K_(i)` =` ((1)/(2) mv^(2) + (1)/(2) I omega^(2)) -(1)/(2) mv_(0)^(2)` =`(1)/(2) m(v_(0) - mu g t)^(2) + (1)/(2) ((1)/(2) mR^(2)) ((2mu g)/(R) t)^(2) - (1)/(2) mv_(0)^(2)` =`(1)/(2) [3 mu^(2) mg^(2) t^(2) - 2 mu mgt v_(0)]` For `t gt t_(0)` `W_("friction") = K_(f) -K_(i)` `W_("friction") = (1)/(2) mv^(2) (1+(K^(2))/(R^(2))) -(1)/(2) mv_(0)^(2)` `W_("friction") = (1)/(2) mv^(2) XX(3)/(2) -(1)/(2) mv_(0)^(2)` By putting `v = (2v_(0))/(3)` `W_("friction") = - (mv_(0)^(2))/(6)`. |
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| 40. |
The operating principle of a hydraulic press is …………… . |
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Answer» PASCAL's Law |
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| 41. |
Passage-I : A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N//m^2. During an observation time of 1 s, an atom travelling with the root mean square speed parallel to one of its edges of the cube, was found to make 500 collisions with a particular wall, without any collision with other atoms. Taking R = ( 25//3) J//mol-K and k= 1.38 xx 10^(-23) J//K. What is the total mass of helium gas in the box |
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Answer» 0.1 g |
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| 42. |
Passage-I : A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N//m^2. During an observation time of 1 s, an atom travelling with the root mean square speed parallel to one of its edges of the cube, was found to make 500 collisions with a particular wall, without any collision with other atoms. Taking R = ( 25//3) J//mol-K and k= 1.38 xx 10^(-23) J//K. What is the temperature of the gas? |
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Answer» 80 K |
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| 43. |
Passage-I : A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N//m^2. During an observation time of 1 s, an atom travelling with the root mean square speed parallel to one of its edges of the cube, was found to make 500 collisions with a particular wall, without any collision with other atoms. Taking R = ( 25//3) J//mol-K and k= 1.38 xx 10^(-23) J//K. What is the average kinetic energy per atom |
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Answer» `0.312 xx 10^(-21) J` |
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| 44. |
If vecA+vecB= vecR and 2vecA+vecB is perpendicular to vecB then |
| Answer» ANSWER :A | |
| 45. |
The reaction time for an automobile driver is 0.7 sec. If the automobile can be decelerated at 5m//s^(2) calculate the total distance travelled in coming to stop from an initial velocity of 8.33 m/s after a signal is observed. |
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Answer» `12.77m` |
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| 46. |
For free expansion of a gas which of the following is true? |
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Answer» `Q = W GT 0` and `Delta E_("int") =0` |
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| 47. |
one end of a taut string of length3 m along the x -axis is fixed at x = 0 . The speed of the waves in the string is 100 m*s^(-1) . The other end of the string is vibrating in the y-direction so that stationary waves are setup in string. Obtain the possible waveforms of these stationary wave |
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Answer» `y (t) = A "SIN" (pi x)/(6)"cos"(50pit)/(3)` |
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| 48. |
For two vectors A and B, |bar(A) + bar(B)| = |bar(A) - bar(B)| is always true when (i) |bar(A)| = |bar(B)| ne 0(ii) A bot B (iii) |bar(A)| = |bar(B)| ne 0 and A and B are parallel or anti parallel (iv) When either A or B is zero. |
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Answer» (i) and (II) are true |
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| 49. |
A soundof frequency 512 Hz is emitted from a stationary source . A train running at a speed of 72 km . h^(-1)passes the source . What will be the frequency of the sound heard by a passenger of the train before and after passing the source ? Neglect the effect of wind . Velocity of sound is 336 .S^(-1) . |
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Answer» Solution :Velocity of the train = 72 km . ` h^(-1)` ` = (72 xx1000)/(60xx60) m . S^(-1) = 20 m. S^(-1)` When the train approaches the source , the velocity of sound RELATIVE to the passenger , ` V. = 336 + 20 = 356 m . S^(-1)` SINCE the source is at REST , the wavelength of sound remains the same . ` THEREFORE` Apparent freuency due to Doppler effect `(V.) /(lambda ) = (V.) /((V)/(n)) = (V. n)/(V) = (356 xx 512)/(336) = 542 .5 Hz ` Again when the train recedes from the source , the velocity of sound relative to the passenger ` Again when the train recedes from the source , the velocity of sound relative to the passenger ` V. =336 - 20 = 316 . s^(-1) ` ` thereforeApparent freaquency ` = (V. n)/(V) = (316 xx 512)/(336) = 481 .5 Hz ` |
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| 50. |
A brass boiler has a base area of 0.15 "m"^(2) and thickness is 1.0 "cm". It boils water at the rate of 6.0 "kg/min". When placed on a gas stove. Estimate the temperature of the part of the flame in contact. With the boiler. Thermal conductivity of brass = 109 "Wm"^(-1) "K"^(-1) |
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Answer» |
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