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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 solid material is supplied with heat at a constant rate. The temperature of the material is changing with the heat input as shown in figure. Study the graph carefully and answer the following questions : (i) What do the horizontal regions AB and CD represent? (ii) if CD = 2AB, what do you infer ? (iii) The slope of OA gt the slope of BC. What does this indicate ? (iv) What does the slope DE represent ? |
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Answer» SOLUTION :(i) In the regions AB and CD the TEMPERATURE of the material remains constant. So AB represents fusion and CD represents vapourisation. (ii) CD= 2AB or `Q_(4) = 2Q_(2)`, it shows that latent HEAT of vapourisation is twice that of latent of fusion. (iii) Let `S_(1) and S_(2)` be the specific heats of SOLID and liquid states respectively, then `Q_(1) = mS_(1) Delta T_(1) " or " S_(1) = (1)/(m((Delta T_(1))/(Q_(1)) ) ) = (1)/( m xx "slope of OA")` Similarly `S_(2) = (1)/(m xx "slope of BC")` Since slope of OA `gt` slope of BC, `therefore S_(1) lt S_(2)`. (iv) If `S_(3)` is the specifice heat of vapour state (region DE), then `S_(2) = (1)/(m xx "slope of BC")` or Slope of DE= `(1)/(mS_(3)) = (1)/("heat capacity")` |
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| 2. |
An iron ball of diameter 6cm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 20^(@) C. The temperature at which (both for ball and plate)the ball just pass through the hole is(alpha_("iron")= 12 xx 10^(-6)//1^(@)C, alpha_("brass") = 18 xx 10^(-6) //1^(@) C) |
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Answer» `68^(@) C` |
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| 3. |
The moon has no atmosphere. Why? |
| Answer» SOLUTION :The escape velocity on the MOON is SMALL `(2.38 kms^-1)`. The air molecules have thermal velocities greater than the escape velocity. Therefore, the air molecules escape away and cannot form atmosphere on the moon. | |
| 4. |
A straight rope of length 'L' is kept on a frictionless horizontal surface and a force 'F' is applied to one end of the rope in the direction of its length and away from that end. The tension in the rope at a distance 'l' from that end is |
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Answer» `(F)/(l)` |
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| 5. |
If A = 6.00i - 8.00j units, B= 8.00i + 3.00j units and C = 26.0i + 19.0 j units, determine a and b when a A + b B + C = 0 |
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Answer» a = 0.9, B `= -3.93` |
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| 6. |
Acceleration of a particle changes when |
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Answer» DIRECTION of VELOCITY changes |
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| 7. |
The distance travelled by a body during last second of its total flight is d when the body is projected vertically up with certain velocity. If the velocity of projection is doubled, the distance travelled by the body during last second of its total flight is |
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Answer» 2 d |
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| 8. |
Is it possible for a constant to have dimensions? |
| Answer» SOLUTION :YES , gravitationalconstanthas DIMENSIONS. | |
| 9. |
A diatomic ideal gas is contained in a rigid container of volume V with walls of total inner surface area A, thickness d and thermal conductivity k. The gas initially is at temperature T_0 and pressure P_0. Find the pressure of the gas after a time t seconds if the surrounding temperature is T_s. All temperatures are in Kelvin scale. |
| Answer» SOLUTION :`P_t (P_0)/(T_0) [(T_s -T_0) E^(- ALPHAT)] " where " ALPHA = (2KA)/(2dR)` | |
| 10. |
Doppler's effect is exhibited by |
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Answer» SOUND WAVES only |
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| 11. |
To simulate car accidents, auto manufactures study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and coliding 6.25 xx 10^(3) N m^(-1). What is the maximum compression of the spring ? |
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Answer» Solution :At MAXIMUM COMPRESSION `x_(m)`, the potentia energy V of the spring is equal to the kinetic K of the MOVING car from the principle of conservation of MECHANICAL energy. `K=UrArr(1)/(2)mu^(2)=(1)/(2)kx_(m)^(2)rArrx_(m)=usqrt((m)/(k))rArrx_(m)=usqrt((m)/(k))` We obtain `x_(m)` = 2.00m |
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| 12. |
Two identical vessels A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The masses of gas in A and B are m_(A) and m_(B) respectively. The gases are allowed to expand isothermally to the same final volume 3 V. The change in pressures of the gas in A and B are found to be Delta P and 1.5 Delta P respectively Then |
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Answer» `9m_(A) = 4m_(B)` |
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| 13. |
A bomb travelling in a parabolic path under the effect of gravity explodes in mid air. The centre of mass of fragments will |
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Answer» Move VERTICALLY UPWARDS and then vertically downwards |
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| 14. |
A block of weight W rests on horizontal floor with coefficient of static friction mu. It is desired to make the block move by applying minimum amount of force. The angle thetafrom the horizontal at which the force should be applied and the magnitude F of the force is |
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Answer» `theta = 0,F = MU W` |
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| 15. |
In the Fig 7.89 shown , m_(2) = 2.5 kg, h = 1.5 m , the system is released fromrest at t = 0 and the mass m_(2) strikes ground at t = 0.82s . The system is placed in its initial position and a 1.2 kgmass is placed on top of the block of mass m_(1) . Released from rest , the mass m_(2) now strikes the ground 1.3 s later .Determine the mass m_(1)and the coefficient of kinetic friction between m_(1) and the shelf . |
Answer»  `a_(0) = sqrt((2h)/(t^(2))) =sqrt((2 xx 1.5)/(0.82 xx 0.82))` As , `"" m_(2) G - T = m_(2)a_(0) ""…… (i) ` `T - mu m_(1) g = m_(1) a_(0) "" ...... (ii)` Adding Eqns. (i) and (ii) `g (m_(2) - mum_(1)) = a_(0) (m_(1) + m_(2)) "" ..... (iii)` Now when `1.2` kg mass is placed , then `a= sqrt((2h)/(t'^(2))) = sqrt((2 xx 1.5)/(1.3xx 1.3))` as `m_(2)g - T = m_(2)a "" ...... (iv)` `T - mu(m_(1) + 1.2)g = a [m_(2) + m_(1) + 1.2] "".....(V)` Adding Eqns. (iv) and (v) `m_(2)g - mu_(g) (m_(1) + 1.2) = a[m_(2) + m_(1) + 1.2] "" .... (vi)` SOLVING Eqns. (iii) and (vi) , we get VALUES of `mu` and`m_(1)`. |
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| 16. |
A solid rubber ball of desity d and radius R falls through air. Assume that the air resistance acting on the ball is KRV, where V is velocity of the ball and K is a constant.The terminal velocity attained by the ball is (neglect buoyance) |
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Answer» `4piR^(2)dg//3K` |
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| 17. |
A neutron moving with a certain kinetic energy collides head on with an atom of mass number A. The fractional kinetic energy retained by it is |
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Answer» `(A-1)/(A+1)` |
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| 18. |
A capilllary glass tube of uniform bore contains a thread of mercury 1 m long at 0^(@)C. When temperature is raised to 100^(@)C, the thread is found to be 16.5 mm longer. If the coefficient of real expansion of mercury is =18.2 times 10^(-5@)C^(-1), then what will be the coefficient of linear expansion of glass? |
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Answer» |
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| 19. |
Which of the following phenomenon is used by the musicians to tune their musical instruments? |
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Answer» Interferance |
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| 20. |
Two wheels of M.I. 3kgm^(2) and 5kgm^(2) are rotating at a rate of 600 rpm and 800 rpm respectively in opposite directions. If they are coupled so as to rotated with the same axis of rotation , the resultant speed of rotation (in rpm) |
| Answer» ANSWER :A | |
| 21. |
A particle is given an initial velocity of 20m/s. acceleration of the particle changes with time and its variation is shown in the following figure. |
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Answer» Answer is 6 . |
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| 22. |
Which of the following examples represent periodic motion? An arrow released from a bow. |
| Answer» SOLUTION :It is not a PERIODIC MOTION. | |
| 23. |
If the length of a cylinder is L = (4.00 +-0.01)cm radius r = (0.250 +- 0.001)cm and mass m = (6.25+- 0.01)g . Calculate the percentage error in determination of density. |
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Answer» SOLUTION :Density of cylinder, `rho=(("MASS")/("volume")V)=(m)/(pir^(2)L) ( :. V=pi r^(2)l)` `:.` Relative ERROR in `rho=(Delta rho)/(rho)=(Deltam)/(M)+2(Deltar)/(r)+(Deltal)/(l)` `=(0.01)/(6.25)+2xx(0.001)/(0.250)+(0.01)/(4.00)` `=0.0016+0.008+0.0025` `=:.(Delta rho)/(rho)=0.0121` `:.` The percentage error in `rho=0.0121xx100=1.2%` |
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| 24. |
Three copper blocks of masses M_1, M_2 and M_3 kg respectively are brought into thermal contact till they reach equilibrium. Before contact they were at T_1, T_2, T_3 (T_1 gt T_2 gt T_3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper) |
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Answer» `T=(T_1+T_2+T_3)/3` `M_3 s(T_3-T)=M_1s(T-T_1)+M_2s(T-T_2)` `therefore M_3sT_3-M_3sT=M_1sT-M_1sT_1 + M_2sT-M_2sT_2` `therefore M_3sT_3+M_1sT_1 + M_2sT_2 = M_1sT+ M_2sT+M_3sT` `therefore M_1T_1+M_2T_2+M_3T_3 =[M_1+M_2+M_3]T` `T=(M_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)` |
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| 25. |
Two point size bodies of same mass are knotted to a horizontal string one at the end, and the other at the mid point of it. The string is rotated in horizontal plane with the other end as centre. If T is tension in the string between centre of circles and first body then the tension in the string between the two bodies is |
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Answer» `T//2` |
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| 26. |
What is rolling motion? |
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Answer» Solution :(i) It is the combination of pure translational and pure rotational motion. (I) The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. (II) Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge ofthe disc. (III) While rolling, the point, undergoes translational motion along with its center of mass and rotational motion with respect to its center of mass. (ii) (I) If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by `2 pi R` (its circumference). (II) Not only the center of mass, but all the points on the disc are displaced by the same `2 pi R` after one full rotation. The only difference is that the center of mass takes a straight path, but, all the other points. (III) Undergo a path which has a combination of the translational and rotational motion. Especially the point on the edge undergoes a path of a cycloid. (IV) As the center of mass takes only a straight line path, its velocity `v_(CM)` is only translational velocity `v_("TRANS") (v_(CM) = v_("TRANS"))`. All the other points have two VELOCITIES. One is the translational velocity `v_("TRANS")`, (which is ALSO the velocity of center of mass) and the other is the rotational velocity `v_("ROT") (v_("ROT") = r omega)`. (v) Here, r is the distance of the point from the center of mass and `omega` is the angular velocity. The rotational velocity `v_("ROT")` is perpendicular to the instantaneous position vector from the center of mass. (vi) The resultant of these two velocities is v. This resultant velocity v is perpendicular to the position vector from the point of contact of the rolling ojbect with the surface on which it is rolling as SHOWN in Figure. 
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| 27. |
The velocity of a particle of mass 10^(-2)kg is given as vecv=(2xhatu+3yhatj+4zhatk) m/sec (where x, y and z are in meter). The amount of work done by applied force in displacing a particle from (1m, 2m, 0) to (-1m, 3m, 1m) is ….. |
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Answer» 0.31 joule |
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| 28. |
Sound waves of wavelength greater than that of audible sound are called ………. |
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Answer» infrasonic WAVES |
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| 29. |
A ball is fallen freely from top of tower of height 100 m and another ball is thrown upwards with 50 m/s at the same time when they will cross each other ? (g = 10 m//s^(2)) |
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Answer» Solution :For the ball falling freely `v _(0) = 0, d = 100 m` `therefore` DISTANCE covered in .t. time `d _(1) = v _(0)t + 1/2 g t^(2)` `d _(1) = 0 + 1/2 (10) t ^(2) = 5t ^(2) ""…(1)` Initial velocity of the ball thrown upwards is `v _(0).=50m//s` and it meets the first ball at `(100 -d _(1)`)m distance `therefore 100 -d _(1) = v _(0) .t = 1/2 g t ^(2)` `therefore 100 - d _(1) = 50 t - 1/2 (10) t ^(2) ` `therefore 100 -d _(1) = 50 t -5t ^(2)` From equation (1), `therefore 100-5t ^(2) = 59 t - 5t ^(2)` `therefore 50 t = 100` `therefore t = (100)/(50)` `therefore t =2s` |
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| 30. |
Give the quantities for which the following are the dimensions: M^(1)T^(-2) |
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Answer» |
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| 31. |
A particle performing uniform circular motion has angular momentum Lwrt centre. If its angular frequency is doubled and its kinetic energy is haved, then the new angular momentum is |
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Answer» `L//4` |
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| 32. |
The bobof mass50 gm of a pendulumis released from a horizontal position A as shownin the figure ..If the lengthof the pendulumis1.5 m , what is itskinetic energyat the lowermostpointB ? (g = 10 m/(s^(2))) |
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Answer» `2.5 xx10^(-11)` J ` :. ` K.E of bob at `B(1/2 MV^(2)) = MGH ` ` = 50 xx10^(-3) xx10 xx1/5` `= 7.5 xx10^(-1) J ` |
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| 33. |
The sample of gas expands from v_1to v_2The amount of workdone by the gas is greatest,when the expansion is, |
| Answer» Solution :isothermal | |
| 34. |
A body of mass 1 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to 1/5th of its original velocity. Find the mass of the second body. |
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Answer» `(2)/(3)KG` |
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| 35. |
When a rod of length 2.5 m and radius 10 mm is subjected to a force the elongation is 3mm and decrease of radius is 0.0025 mm. Find poisson.s ratio of the material of the rod. |
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Answer» |
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| 36. |
A satellite moves around the earth in a circular orbit with speed V. If 'm' is the mass of the satellite, its potential energy is |
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Answer» `MV^(2)` |
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| 37. |
If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. |
| Answer» SOLUTION :KINETIC ENERGY, | |
| 38. |
The motor of an engine is rotating about its axis with angular velocity of 120 r.p.m. it comes to rest in 10s, after being switched off. Assuming constant deceleration, calculate the number of revolutions made by it before coming to rest. |
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Answer» SOLUTION :Here, `omega_(0)=120` r.p.m `=2` r.p.s `=4pirads^(-1)` `omega=0` and `t=10s` `omega=omega_(0)+at` `0=4pi+axx10` or `a=-0.4pirads^(-2)` Also, the angle covered by the motor, `theta=omega_(0)t+(1)/(2)at^(2)` `:. theta=4pixx10+(1)/(2)xx(-0.4pi)xx10^(2)=40pi-20PI` `=20pi`rad Hence, the number of revoluations COMPLETED `n=(theta)/(2pi)=(20pi)/(2pi)=10` |
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| 39. |
A particle is moving along a circular path. The angular velocity, linear velocity, angular acceleration and centripetal acceleration of the particle at any instant are vec(omega), vecv, veca, veca_(c) respectively. Which of the following relation is not correct ? |
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Answer» `VEC omega BOT vec v` |
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| 40. |
block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72 If yolative density of silver is 10, then tension in the string will be a |
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Answer» `37.12N` |
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| 41. |
A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is |
| Answer» Answer :D | |
| 42. |
The phase of simple harmonic motion at t = 0 is called |
| Answer» Answer :D | |
| 43. |
A train is moving with a uniform velocity of (2 hat(i)- hat(j)+ hat(k)) ms^(-1). If the force required to overcome friction is (hat(i)- 3hat(j) + 2hat(k))N, the power of the engine is |
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Answer» 7W |
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| 44. |
Out of the following graphs the graph that represents the variaion of the value of gravitational potential V, with distance r from the centre, due to a hollow spherical shell is |
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Answer»
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| 45. |
Two moles of air, when heated through10 K expand by an amount of 1.66 xx 10^(-3) m^(3)under a constant pressure of 10^(5) N//m^(2). If C_(v)=20.81 J/mole K, then C_(p)is, |
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Answer» 29. 11J/mole K |
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| 46. |
State the expressions for the moment of inertia of a circular ring and a circular dise about an axis passing through its centre and perpendicular to its plane. |
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Answer» SOLUTION :MO.I. of circular ring about an AXIS passing through its CENTRE and perpendicular to its plane is `I=MR^(2)` M.O.I. of a circular disc about an axis passing through its centre centre and perpendicular to its PLAN is `I=(MR^(2))/(2)` |
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| 47. |
cylinder of capacity 20 litres is filled with H_2 gas. The total average kinetic energy of translatory motion of its molecules is 1.5 xx 10^(5) J. The pressure of hydrogen in the cylinder is |
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Answer» `2XX10^(6)N//m^(2)` |
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| 48. |
The wettability of a suface by a liquid depends primarily on: |
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Answer» viscosity |
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| 49. |
State Pascal's lae in fluids. |
Answer» Solution : PASCAL's law states that if the effect of gravity can be generated then the pressure in a fluid in equilibrium is the same everywhere. LET us consider any two points A and B inside the fluid imagined. A cylinder is such that points A and B lie at the centre of the circular surface at the TOP and bottom of the cylinder. Let the fluid inside this cylinder be in equilibrium under the action of forces from OUTSIDE the fluid. The forces ACTING on the circular, top and bottom surfaces are perpendicular to the forces acting on the cylindrical surface. Therefore the forces acting on the faces at A and B are equal and opposite and hence add to zero. As the areas of these two faces are equal, pressure at A = pressure at B. This is the proof of Pasccal's law when the effect of gravity is not taken into account. |
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