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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. |
Calculate the radius or the largest drop of water that might evaporate at 0^(@)C without heat being supplied to it. The surface energy of watee at 0^(@)C is 117xx10^(-3) J m^(-2) and the latent heat capacity of water is 606xx10^(3) clorie par kg at 0^(@)C One calorie equal 4.2 joules. [Hint: Equate change in surface energy to the heat required for evaporation of the mass that will disapper.] |
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| 2. |
In what direction does earth rotate ? |
| Answer» SOLUTION :`implies`It ROTATES from west to EAST about its axis. | |
| 3. |
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15^(@). What is the radius of the loop ? |
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Answer» SOLUTION :If N is the normal force on the wings, `N cos theta = mg "" N sin theta = (MV^(2))/(R )` On SIMPLIFICATION, `Tan theta = (v^(2))/(Rg)` `R=(v^(2))/(g tan theta)=(200xx200)/(10xx tan 15^(@))=15 km` |
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| 4. |
Calculate the time period of the oscillation of a particle of mass m moving in the potential defined as U(x)={{:((1)/(2) kx^(2)",", x lt 0),(mgx"," , g gt 0):} |
Answer» Solution : In the problem oscillation of particle into 2 cases via `V lt 0` (consider it as SHM where time period is considered as `t_(1))` and another one as `x GT 0` (consider it as motion under gravity, where time period is `t_(2))` Note : We want to find total time period, which will be `T=t_(1)+t_(2)` According to conservation of energy `(1)/(2)kx^(2)=(1)/(2) mv^(2)` `:.v^(2)=(kx^(2))/(m)` `rArr v= sqrt((kx^(2))/(m)) ""...(1)` We KNOW that, kinetic energy (E) } `(1)/(2) mv^(2)` `v=sqrt((2E)/(m)) ""....(2)` In SHM (case : 1) `T = 2pi sqrt((m)/(k))` Here , `t_(1)=(T)/(2)` `= (2pi sqrt((m)/(k)))/(2)` `= pi sqrt((m)/(k)) ""...(3)` In case (motion under gravity) `t_(2)=(2V)/(g)` substituting equation (2) have we get, `t_(2)=(2)/(g) sqrt((2E)/(m))` `rArr 2sqrt((2E)/(mg^(2))) ""....(4)` Adding equation (3) & (4) Time period of oscillation, `T=t_(1)+t_(2)` `= pi sqrt((m)/(k))+2 sqrt((2E)/(mg^(2)))` |
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| 5. |
Define time of flight. |
| Answer» SOLUTION :The TIME TAKEN for the projectile to complete its trajectory or time taken by the projectile to hit the GROUND is CALLED time of flight . | |
| 6. |
If two circular discs A and B are of same mass but of radii r and 2r respectively, then the moment of inertia of A is |
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Answer» the same as that of B Radius of circular disc B, `r_(2)=2r` For disc A, `I_(A)=mr^(2)`…(i) For disc B, `I_(B)=m(2r)^(2)=4mr^(2)`…(II) Dividing EQ. (i) by Eq. (ii), we get `:. (I_(A))/(I_(B))=(mr^(2))/(4mr^(2))rArr(I_(A))/(I_(B))=(1)/(4)rArr I_(A)=(1)/(4)I_(B)` So, `I_(A)`= one fourth that of B. |
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| 7. |
While during a carpet , we give a sudden jerk or beat it with a stick . Why ? |
| Answer» Solution :Beacause inertia of rest keeps the DUST in its position and the dust gets REMOVED as the carpet moves AWAY. | |
| 8. |
In column various bodies of same mass and radius R are being lowered on a rough horizontal surface. |
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| 9. |
The minimum speed of a bucket full of water whirled in a vertical circle of radius 10m at the highest point so that the water may not fall is (g= 10 ms^(-2)) |
| Answer» Answer :D | |
| 10. |
A force of 40N acts on a body. If the units of mass and length are doubled and the unit of time is tripled, then the force in the new system becomes |
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Answer» 90 N |
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| 11. |
Force and area are measured as 20 N and 5 m^2with errors 0.05 N and 0.0125 m^2 . The maximum error in pressure is (SI units) |
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Answer» `4 PM 0.0625` |
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| 12. |
Find the centre of mass of the system: |
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Answer» Solution :From the figure we can write, `X_(cm) = (m_(0) XX 0 + m_(0) xx L + m_(0) xx L + m_(0) xx 2L)/(m_(0) + m_(0) + m_(0) + m_(0)) rArr X_( cm) = L` `Y_(CM) = (m_(0) xx 0 + m_(0) xx 0 + m_(0) xx L + m_(0) xx 0)/(m_(0) + m_(0) + m_(0) + m_(0)) rArr Y_(cm) = L/4` `therefore` Co-ordinates of CENTRE of mass are `(X_(cm), Y_(cm)) =(L, L/4)` |
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| 13. |
A steel wire of length 1.5 m and diameter 0.25 cm is loaded with force of 98N.The increase in length of the wire 1.5×10^-4m. Calculate the tensile stress and fractional change in length of the wire. |
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| 14. |
What is magnitude of normal force on a body lying stationary on slope with angle theta |
Answer» SOLUTION : N = MG COSTHETA`
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| 15. |
A space craft of mass 2000 kg moving with a velocity of 600 m//s suddenly explodes into two pieces. One piece of mass 500 kg is left stationary. The velocity of the other part must be (in m//s) |
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Answer» 600 |
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| 16. |
The mean kinetic energy of a perfect gas molecule at temperature TK is |
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Answer» `1/2`RT |
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| 17. |
Two particles of masses 10kg and 30kg are lying on a straight line. The 10kg mass is shifted towards the 30kg mass by a distance of 2cm. By what distance should the 30kg mass be shifted so that the position of their centre of mass does not change ? |
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| 18. |
A disc rolls up an inclined plane of angle 45°. The centre of mass of the sphere has a translational speed of 2 m/s at the bottom of the plane. What is the distance through which the disc travel up the plane ? |
| Answer» SOLUTION :`(1//2)mv^(2) + (1//2)IOMEGA^(2)=mg s SIN 45, (3//4)mv^(2) = mgs//sqrt(2), s = 0.433 m` (distance) | |
| 19. |
What is the difference between the temperature of two bodies, if their temperatures are (50+-0.2)^(@)C and (25+-0.6)^(@)C? |
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Answer» `P-Q=(25-16)+-(0.2+0.7)=9+-0.9`. |
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| 20. |
The satellite of mass m revolving in a circular orbit of radius r around the earth has kinetic energy E. Then its angular momentum will be |
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Answer» `sqrt((E)/(mr^(2)))` |
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| 21. |
The path of the centre of mass of earth - moon system around the sun is |
| Answer» Answer :D | |
| 22. |
Explain why a person in an artificial satellite orbiting the earth, feels himself to be weightless. |
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Answer» Solution :Let the artificial satellite be in an orbitof RADIUS R with respect to the centre of the earth. The NECESSARY centripetal force for keeping the man of mass m moving in the ORBIT is supplied by the resultant of the gravitational force of the earth `mg^.`on the person and the upward reaction R of the plane of the satellite on the person. That is `mg^.-R=(mv^2)/r`.......(1) The necessary centripetal force, to keep the satellite of mass M moving ,is supplied by the gravitational attraction of the earthon the satellite. Hence, `Mg^.=(Mv^2)/ror, g^.=(v^2)/r` .........(2) From (1) and (2) we have, `mg^.-R=mg^. or, R=0` As the upward reaction force is zero, the person feels WEIGHTLESS. |
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| 23. |
The velocity of a particle varies with time as per the law vec(V) = vec(a) + vec(b)t where vec(a) and vec(b) are two constant vectors. The time at which velocity of the particle is perpendicular to velocity of the particle at t= 0 is |
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Answer» `-(|VEC(a)|)/(vec(a).vec(B))` |
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| 24. |
In one dimensional motion, instantaneous speed y satisfies the condition 0 |
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Answer» the displacement in time T must ALWAYS TAKE NON- negative values Having one dimensional MOTION , the distancement x in time T satisfies `-V_(0)< x |
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| 25. |
Two sopheres whose radii are in the ratio 1:2 are moving with theier velocities in the ratio 3:4. If their densities are in the ratio 3:2, the ratio of their kinetic energies is |
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Answer» `27:64` |
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| 26. |
Three liquids of masses m_(1),m_(2) and m_(3) are mixed together. Their specific heats are s_(1), s_(2) and s_(3), and they were at temperature t_(1),t_(2) and t_(3) respectively. Calculate the temperature of the mixture. |
| Answer» SOLUTION :`[(m_(1)s_(1)t_(1)+m_(2)s_(2)t_(2)+m_(3)s_(3)t_(3))/(m_(1)s_(1)+m_(2)s_(2)+m_(3)s_(3))]` | |
| 27. |
A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R. For counteracting the centrifugal force on the vehicle the difference in elevation required in between the outer and inner edges of the road is |
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Answer» `(v^(2)B)/(Rg)` |
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| 28. |
Attimet = 0a 2kg paticlehas positionvectorvec (r )= (4 hat(i) - 2 hat(j))relativeto theorigin Its velocity is given byv=2t^2 (m/s) for t≥0. The torque acting on the particle about the origin at at t =2s , is |
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Answer» `32 hat(k) ` N-m |
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| 29. |
A box is dragged across a surface by a rope which makes an angle 45^(@) with thehorizontal. The tension in the rope is 100 N when the box is dragged 10 m . The work done is |
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Answer» (a) 707.1J `"Work DONE" =T cos theta xx x` `=100cos 45^(@)xx10=707.1 J` 
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| 30. |
The heat absorbed by a carnot engine from the source in each cycle is 500 J and the efficiency of the engine is 20%. Calculate the work done in each cycle ? |
| Answer» SOLUTION :`W = ETA Q_(1) = 0.2 XX 500 = 100 J` | |
| 31. |
A steel rail of length 5 m and area of cross section 40 cm^2 is prevented from expanding along its length while the temperature rises by 10^(@) C. If coefficient of linear expansion and Young's modulus of steel are 1.2 xx 10^(-5) K^(-1) and 2 xx 10^(11) "Nm"^(-2) respectively, the developed in the rail is approximately : |
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Answer» `2 xx 10^(9)` N |
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| 32. |
Jogging every day is good for health. Assume that when you jog a work of 500 kJ is done and 230 kJ of heat is given off. What is the change in internal energy of your body? |
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Answer» Solution : Work done by the SYSTEM (body), W = +500 kJ HEAT released from the system (body), Q = -230 kJ The CHANGE in internal energy of a body `= DeltaU=-230 kJ - 500 kJ = - 730 kJ` |
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| 33. |
What does gradual increase of atmospheric pressure indicate? |
| Answer» SOLUTION :It INDICATES DRY WEATHER. | |
| 34. |
Assertion : A tuning fork is made up of material whose elasticity does not change. Reason : A tuning for is made up of an alloy of steel, nickel and chromium. |
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Answer» If both ASSERTION and reason are CORRECT and reason is a correct EXPLANATION of the assertion. |
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| 35. |
Let barA be a unit vector along the axis of rotation of a purely rotating body and B be a unit vector along the velocity of a particle P of the body away from the axis. The value of barA. barB is |
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Answer» 1 |
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| 36. |
Asection of driange culvert1.23 m in length makes a howling noise when the wind blows across its open ends . a. Determine the frequencies of the first harmonics of the culvert , if it is cylindrical in shape and open at both ends . Takev = 343 m//s as the speed of sound in air . b. What are the three lowest natural frequencies of the culvert if it is blocked at one end ? |
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Answer» SOLUTION :The sound of the wind blowing across the end of the pipe contains many frequencies , and the culvert RESPONDS to the sound by vibrating at the natural frequencies of the air COLUMN . This example is a relatively SIMPLE substitution PROBLEM. a. Let us first find the frequency of the first harmonic of the culvert , modelling it as an air column open at both ends . ` f_(1) = (v)/( 2L) = (343 m//s)/(2 (1.23 m)) = 139 Hz` Find the next harmonics by multiplying by integers . ` f_(2) = 2 f_(2) = 278 Hz` `f_(3) = 3 f_(1) = 417 Hz` b. Find the frequency of the first harmonic of the culvert , modelling it as air column closed at one end : `f_(1) =(v)/( 4L) = ( 343 m//s)/( 4( 1.23 m)) = 69.7 Hz` Find the next two harmonics by multiplying by odd integers . `f_(3) = 3 f_(1) = 209 Hz` `f_(5) = 5 f_(1) = 349 Hz` |
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| 37. |
A cyclist starts from the center O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the center along QO asIf the round trip takes 10 min, what is the net displacement, |
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Answer» 0 |
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| 38. |
A disc is recast into a thin walled cylinder of same radius. Which will have large moment of inertia? |
| Answer» SOLUTION :Holow cylinder will have LARGE MOMENT of inertia because most of its MASS is located at comparatively larger distance from the axis of ROTATION. | |
| 39. |
An iron bar (L_(1) = 0.1 m , A_(1) = 0.02 m^(2), K_(1) = 79 W m^(-1) K^(-1))and a brass bar (L_(2) = 0.1 m, A_(2) = 0.02 m^(2) , K_(2) = 109 W m^(-1) K^(-1)) are soldered end to end as shown in Fig. 11.16. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (1) the temperature of the function of the two bars, (11) the equivalent thermal conductivity of the compound bar, and (111) the heat current through the compound bar. |
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Answer» Solution :Given, `L_(1) = L_(2) = L = 0.1 m, A_(1) = A_(2) = A = 0.02 m^(2) K_(1) = 79 W m^(-1) K^(-1) , K_(2) = 109 W m^(-1) K^(-1), T_(1) = 373 K, " and " T_(2) = 273 K`. Under STEADY state condition, the heat current `(H_(1))` through iron bar is equal to the heat current `(H_(2))` through brass bar. So, `H = H_(1) = H_(2)= (K_(1) A_(1) (T_(1) - T_(0))/L_(1)) = (K_(2)A_(2) (T_(0) - T_(2))/L_(2))` For `A_(1) = A_(2) = A " and " L_(1) = L_(2) = L `, this equation LEADS to `K_(1) (T_(1) - T_(0)) = K_(2) (T_(0) - T_(2))` Thus , the junction temperature `T_(0)` of the two BARS is `T_(0) = ((K_(1)T_(1) + K_(2)T_(2))/((K_(1) + K_(2))))` Using this equation, the heat current H through either bar is `H = (K_(1) A(T_(1) - T_(0)))/L = (K_(2)(T_(0) - T_(2)))/L= ((K_(1)K_(2))/(K_(1) + K_(2))) (A(T_(1)- T_(2)))/(L(1/K_(1) + 1/K_(2)))` Using these equations, the heat current H. through the compound bar of length `L_(1) + L_(2) = 2L` and the equivalent thermal conductivity K. , of the compound bar are given by H. = `(K. A (T_(1) - T_(2)))/(2 L) = H` ` K. = (2 K_(1) K_(2))/(K_(1) + K_(2))` (i) ` T_(0) = ((K_(1) T_(1) + K_(2)T_(2)))/((K_(1) + K_(2)))` ` = ((79W m^(-1) K^(-1))(373 K) + (109 W m^(-1) K^(-1))(273 K))/(79 W m^(-1) K^(-1) + 109 W m^(-1) K^(-1))` ` = 315 K` (ii) ` K. = (2K_(1)K_(2))/(K_(1)+K_(2))` ` = (2xx (79 W m^(-1) K^(-1)) xx (109 W m^(-1) K^(-1)))/(79 W m^(-1) K^(-1) + 109 W m^(-1) K^(-1)) ` ` = 91.6 W m^(-1) K^(-1)` `(iii) H. = H = (K. A (T_(1) - T_(2)))/(2L) ` ` = ((91.6 W m^(-1) K^(-1)) xx (0.02 m^(2)) xx (373 K - 273 K))/(2 xx (0.1 m))` ` = 916.1 W` |
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| 40. |
Two heavy spheres each of mass 100 kg andradius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ? |
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Answer» Solution :`implies` Mass of each shpere M = 100 kg O is the midpoind of AB, `:. AO = OB = 1/2 m = 0.5m` Mass of object placed at point O = m `:.`The gravitational FORCE of sphere A and B on a body of mass m. `F_A=(GMm)/((r/2)^2)=(Gxx100 xxm)/((0.5)^2)rarr`in OA direction and `F_B=(GMm)/((r/2)^2)=(Gxx100 xxm)/((0.5)^2)rarr`in OB direction . These both forces are same in value and opposite in direction hence, effective force on the body of mass placed at midpoint is zero. So, the body is in equilibrium. Now, if the body A or B is DISPLACED a little towards either mass body from its equilibrium position it will not return back to its initial position of equilibrium. Hence, body is in unstable equilibrium. Now, gravitational potential `V=-(GM)/2` `V_(A) =-(GM)/(0.5) and V_(B) = - (GM)/(0.5)` `:.` Gravitational potential at point O , V `=V_A +V_B` `:. V_A=-(GM)/(0.5)-(GM)/(0.2)=-(2GM)/(0.5)` `=-(2xx6.67xx10^(-11)xx100)/(0.5) = - 2.7 xx10^(-8) J kg^(-1)` Negative sign INDICATES that equilibrium is unstable. |
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| 41. |
A book with many priting errors contains 100 are given in List-1 and their dimensional formulae are given in List-2, Match correct pair in lists {:(,"List-1",,"List-2"),(a,pa s,e,L^(2)T^(-2)K^(-1)),(b,NmK^(-1),f,MLT^(-3)K^(-1)),(c,Jkg^(-1)K^(-1),g,ML^(-1)T^(-1)),(d,Wm^(-1)K^(-1),h,ML^(2)T^(-2)K^(-1)):} |
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Answer» a-h, b-g, C-e, d-f |
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| 42. |
A charge of +q coulomb is carried bya wooden block , which when attached to a stretched spring as shown in the adjoining figure and released performs S.H.M of frequency n. If niw a uniform electric field vecE is switched on then the block will performS.H.M of |
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Answer» Same frequency , with its mean POSITION shifted. `vec(F) = qvec(E )` in the direction of electric field. The force F exerted by the electric field is CONSTANT and thus, a constant force is added to the potential applied force on the block, SO, only the mean position will CHANGE but the frequency of OSCILLATION will remain same. |
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| 43. |
Threee spherical balls of masses 1kg , 2kg and 3kg are placed at the corner of an equilateral traingle of side 1m. The magnitude of the gravitaional force exerted by 2kg and 3kg masses on 1kg mass. |
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Answer» 19 G |
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| 44. |
On which factor the properties of radiations of perfect black body depend ? |
| Answer» SOLUTION :These PROPERTIES of RADIATIONS only DEPEND on TEMPERATURE. | |
| 45. |
A block of mass M is held against a rough vertical well by pressing it with a finger. If the coefficient of friction between the block and the wall is jx and acceleration due to gravity is g, calculate the m inim um force required to be applied by the finger to hold the block against the wall. |
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Answer» Solution :Heremass of the block= M Coefficientof frictionbetweenthe BLOCKAND THEWALL `=mu` Lef F be forcerequiredto holdthe blockagainstthe WALL  Figureshowthatweightof blockmg isindownwarddirection and frictionforce is in upwarddirectionforequilibrium f = MG F=N Frictionalforce F =`mu N` COMPARING(1) and (2) mF = Mg `F=(MG)/( mu)` |
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| 46. |
Does an oscillating source always produce sound waves ? |
| Answer» Solution :Sound waves are PRODUCED if the source is OSCILLATING with FREQUENCY 20 Hz to 20,000 Hz, such a source PRODUCES sound waves because they can be heared by human ears, such a range is called an audible range. | |
| 47. |
A uniform thin bar of mass 6m and length 12L is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is |
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Answer» `20 mL^2` r = L TAN `60^(@) = L sqrt3` The desired moment of INERTIA of hexagon about O is `I = 6[I_("one side") ] = 6 [ (m (2L)^(2))/(12) + mr^(2)]` `= 6 [(mL^2)/(3) + m(Lsqrt3)^(2)] = 20 mL^(2)` 
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| 48. |
The CGs unit of visosity is poise (P). Find how many poise are there in 1 MKS unit of viscosity called poiseuille (PI)? |
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Answer» 1 decapoise `=1 Ns m^(-2)` `=(10^(5)" DYN")xxsxx(10^(2) cm)^(-2)` `=10` dyn s `cm^(-2)= 10` poise |
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| 49. |
What is the shift in the colour of light when the temperature increase? |
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