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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. |
Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incoorect one(s). (a) v_(av)=1/2 [v(t_(1))+v(t_(2))] (b) v_(av)=r(t_(2))-r(t_(1))/(t_(2)-t_(1)) (c) r=1/2 (v(t_(2))-v(t_(1)) (t_(2)-t_(1)) (d) a_(av)=v(t_(2))-v(t_(1))/(t_(2)-t_(1)) |
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Answer» a and b |
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| 2. |
A block of volume V and of density o, is placed in a liquid of density sigma_(1)(sigma_(1) gt simga_(s)) Then the block is moved upward upto a height hand it is still in liquid. The increase in gravitational potential energy of the system is : |
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Answer» `sigma_(B)V_(G)H` |
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| 3. |
Find the resultant of the vectors shown in fig by the component method |
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Answer» Solution :`vec(R_(X)) = 1hat(i) - 5 cos 37^(@) HAT(i) - 6 cos 53 hat(i)` `vec(R_(x)) = 1hat(i) - 4hat(i) - 3.6 hat(i) therefore vec(R_(x)) = -6.6hat(i)` `vec(R_(y)) = 3HAT(j) + 5 sin 37 hat(j) - 6 sin 53^(@) hat(k)` `vec(R_(y)) = 3hat(j) + 3hat(j) - 4.8 hat(j)therefore vec(R_(y)) = 1.2 hat(j)` `R = SQRT((R_(x)^(2) + R_(y)^(2))) = sqrt((-6.6)^(2) + (1.2)^(2)) = 6.7` |
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| 4. |
In a mechanical refrigerator, the low temperature coils are at a temperature of -23^@Cand the compressed gas in the condenser has a temperature of 27^@C . The theoretical coefficient of performance is |
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Answer» 5 |
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| 5. |
The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress ? |
| Answer» Solution :NO, STRESS is a SCALAR quantitiy. REALLY it is a scalar QUANTITY. | |
| 6. |
The M.I. of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1kgm^(2). It is rotating with an angular velocity to 100rads^(-1). A second disc of same mass and radius is joined to it coaxially. Now these two discs together continue to rotate about the same axis. Then the lose in kinetic energy in kilo joules is |
| Answer» ANSWER :A | |
| 7. |
Explain the variation of g with depth from the Earth's surface. |
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Answer» Solution :VARIATION of g with depth: Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines - in NEYVELI). Assume the depth of the mine as d. To CALCULATE g' at a depth d, consider the follwoing points. The part of the Earth which is above the radius `(R_(E) - d)` do not contirbute to the ACCELERATION. The result is proved earlier and is given as `g^(') = (GM')/((R_(E) - d)^(2))` Here M' is the mass of the Earth of radius `(R_(E) - d)` Assuming the density of Earth `rho` to be constant, `rho = M/V` where M is the mass of the Earth and V its volume, Thys, `rho = (M')/(V') , (M')/(V') = M/V and M' = M/V V'` `M' = (M/4/3 pi R_(E)^(3)))(4/3 pi (R_(E) - d)^(3))` `M' = M/(R_E^3) (R_E - d)^(3)` `g' = G M/(R_E^3) (R_E - d)^3 cdot 1/((R_E - d)^(2))` `g' = GM(R_E(1-d/(R_E)))/(R_E^3) = GM ((1 - d/(R_E)))/(R_E^2)` THUS `g' = g(1 - d/ (R_E))` Here also g' < g. As depth increases, g' decreases. 
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| 8. |
Two particles of same time period (T) and amplitude undergo SHM along the same line with initial phase of pi//6. If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time: |
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Answer» `T//8` |
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| 9. |
Melting if ice under atmospheric pressure is a) Isothermal process b)Isobaric process c) Adiabatic process |
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Answer» both a and B are CORRECT |
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| 10. |
A uniform narrow tube closed at one end contains some air confined by a mercury column. The length of the column is 10 cm at 20^@C IF the temperature is increased to 70^@C, then what will be the shift of the mercury column? gamma_p=0.00366^@C^-1 |
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Answer» |
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| 11. |
A loaded spring is made to oscillate on the Earth surface and then in moon. Will there be any change in the time period ? |
| Answer» SOLUTION :No, because T is INDEPENDENT of G | |
| 12. |
A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time. |
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Answer» Solution :For the ball dropped from the building , `U _(1) =0` For the ball thrown up,` u _(2) = 40 m//s` Velocity of the dropped ball after time t, `v _(1) = u _(1) + g t` `v _(1) = g t` (downward) `""…(1)` For the ball thrown up, `u _(2) = 40 m//s` Velocity of the ball after time t `v _(2) = u _(2) -0 g t = (40 - g t) ` upward) `""...(2)` RELATIVE velocity of one ball w.r.t. another ball `= v _(1) - v _(2)` `- g t[-[- (40-g t)] (because `From (1) and (2)) = 40 m.s |
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| 13. |
The displacement of an object attached to a spring executing S.H.M is given by x = 2 xx 10^(-2) cos pi m. The time at which the maximum speed first occurs is |
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Answer» 0.5 SEC |
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| 14. |
A body of mass 'm' is taken from m the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be |
| Answer» Answer :D | |
| 15. |
(A) : The value of acceleration due to gravity does not depend upon mass of the body. (R) : Acceleration due to gravity is a constant quantity. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 16. |
The horizontal range of a projectile fired at an angle of 15^(@) is 50 m. If it is fired with the same used at angle of 45^(@), its range will be |
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Answer» 125 m If the equal speed `(R_1g)/(sin 2theta_1) = (R_2g)/(sin 2 theta_2)` `R_(2) = (2xx sin 90^@)/(sin 30^@) = 100 `m |
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| 17. |
A speck of negligible mass begins to fall and gathers water vapour at a constant rate beta . Calculate its acceleraton and its mass when it falls 2.45m. |
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Answer» |
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| 18. |
What is scientific method ? List the various stages involved in it. |
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Answer» Solution :The scientific method is a step-by-step APPROACH in studying natural phenomena and establishing laws which govern these phenomena. The stages involved in it are : (i) Systematic observation (II) Controlled experimentation (III) Qualitative and quantitative reasoning (iv) Mathematical modeling (V) Prediction and verification or falsification of theories. |
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| 19. |
A rigid body is rotating. Is it necessary that a net torque acts on the body is zero ? |
| Answer» SOLUTION : If the ROTATION is UNIFORM, TORQUE will be ZERO. | |
| 20. |
A steel ball weighting 1 kg is fastened to a cord 1 m long and is released when the cord is horizontal . At the bottom of its path the ball strikes a 5-kg steel block initially at rest ona frictionless surface . The collision is elastic . Find the speed of the ball and speed of the block just after collision . (g = 9.8 m s^(-2)) [Hint : Apply principal of conservation of momentum and conservation of kinetic energy as the collision is elastic .] |
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Answer» |
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| 21. |
The velocities of cylindrical layers of liquid flowing through a tube , situated at distances 0.8cm and 0.82cm from the axis of the tube are 3cms^(-1)and2.5cms^(-1)respectively . Find the viscous force acting between these layers , if the length of the tube is 10 cm and the coefficient of viscosity of the liquid is 8 poise . |
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Answer» Solution :`r_(1)=0.8cm,r_(2)=0.82cm` `DELTAV=3-2.5=0.5cms^(-1)` `Deltax=` DISTANCE between the layers `=0.02cm` `L=10cm` A = AREA of CONTACT of two layers `=2pi((r_(1)+r_(2))/(2))L` `eta=8`poise Now `F_(v)=etaA(Deltav)/(Deltax)` `=eta[2pi(r_(1)+r_(2))/(2)L](Deltav)/(Deltax)` `=8[2xx3.14((0.8+0.82)/(2))10](0.5)/(0.02)` `=16xx3.14xx0.81xx10xx25` `F_(v)=10173.6` dyne |
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| 22. |
Which of the following is true for a satellite in a circular orbit a) it is a freely falling body b)its speed is constant c)it suffers no acceleration d) it does not require energy for motion in the orbit |
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Answer» a, B and C |
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| 23. |
In the given Fig. 7.154 , two masses 2 kg each are attached to a string which passes over a smooth massless pulley P . The surface PQ is smooth and anglePQR = 45^(@) . The surface PR is rough , with coefficient of kinetic friction (1)/(sqrt3) and anglePRQ = 30^(@) .The acceleration of the system is : |
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Answer» `(g (sqrt2 - 1))/(2 sqrt2)` along PR ` T + MU Mg cos 45^(@) - Mg sin 45^(@) = Ma "" …. (II)` Solving EQNS. (i) and (ii) , we get a . 
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| 24. |
When a driver of Shatabdi Express, running with velocity 108 km/hr, sights a goods train going ahead of him at a distance 50 m in the same direction on the same track, running with velocity 72 km/hr, he applies brakes. In order to avoid an accident, what should be the magnitude of the deceleration produced by the brakes? |
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Answer» `5 ms ^(-2) ` Initial velocity of Shatabdi Express `v _(0) = v _(12) = v _(1) -v _(2) ` `v _(0) = 30 - 20 = 10 ms ^(-1)` Now `v ^(2) =v _(0) ^(2) + 2AD` Putting `v =0, v _(0) =10 ms ^(-1), d=50 m,` `0= 100 + 2 xx a xx 50` `THEREFORE a xx 100 =- 100` `therefore a =-1` `therefore -a =1 ms ^(-2)` `therefore` DECELERATION `=1 ms ^(-2)` |
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| 25. |
On reducing the volume at constant temperature, the pressure of the gas increases. Explain it on the basis of kinetic theory. |
| Answer» SOLUTION :On reducing the volume at constant TEMPERATURE , the number of molecules PER unit volume increases. Then the number of collisions per SECOND with the walls of the container and hence pressure the gas increases. | |
| 26. |
A proton in motion makes head on collision with an unknown particle at rest. If the collision is perfectly elastic and proton rebounds back with 4/9 of its initial kinetic energy after collision, the mass of unknown particle is |
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Answer» Equal to mass of proton |
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| 27. |
A source is approaching towoed a wall as shown in figure .We have three observes O_(1),O_(2) and O_(3). observer O_(2) is over the source itself . Let f_(1),f_(2)and f_(3) be the beat frequencires heard byO_(1),O_(2) and O_(3) between direct sound from the source and reflected sound from the wall . then |
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Answer» (a)`f_(3)=0 |
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| 28. |
Given are two tuning forks near one another. One of them is of unknown frequency and the other is of frequency 591 Hz.We can hear beat of maximum intensity I_(0)with frequency 5 Hz. At t = 0 we hear a maxima. Then |
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Answer» UNKNOWN frequency can be 596 Hz |
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| 29. |
Two small spheres of mass 5kg and 15 kg are joined by a rod of length 0.5m and of negligible mass. The M.I. of the system about an axis passing through centre of rod and normal to it is |
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Answer» `10 kg m^(-2)` |
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| 30. |
Two concave lenses L_1 and L_3 are kept in contact with each other. IF the space between the two lenses is filled with a material of smaller retractive index, the magnitude of the local length of the combinations |
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Answer» BECOMES undefined |
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| 31. |
The apparent change of frequency of sound due to Doppler effect is called Doppler _____ [ Fill in the blank ] |
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Answer» |
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| 32. |
An engine pumps up 1 quintal of coal from a mine 100 m deep in 0.5 s. If its efficiency is 60%, power of the engine is (Take g = 10 m//s^2) |
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Answer» 330 kW `= 2 xx 10^5 ` watt Power REQUIRED = `("Actual power")/("EFFICIENCY") = (2 xx 10^(5))/((60)/100) = 330 kW`. |
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| 33. |
A car of mass 400 kg and travelling at 72 km h^(-1) crashes into a truck of mass 4000 kg and travelling at 9 km h^(-1) in the same direction. The car bounces back at a speed of 18 km h^(-1). The speed of the truck after the impact is |
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Answer» `9 km h^(-1)` `u_1 = 72 km h^(-1) = 72 xx 5/18 ms^(-1) = 20 ms^(-1)` `u_2 = 9 km h^(-1) = 9 xx 5/18 = 2.5 ms^(-1)` `v_1 = -18 km h^(-1) = -18 xx 5/18 = -5 ms^(-1)` `v_2 = ?` ACCORDING to law of conservation of LINEAR momentum , we get `m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2v_2` `400 xx 20 + 4000 xx 2.5 = 400 xx (-5) + 4000 xx v_2` `20 + 25 = 5 + 10v_2` `45= 5 + 10v_2` `v_2= 5MS^(-1) = 5 xx 18/5 km h^(-1) = 18 km h^(-1)`. |
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| 34. |
A bob of mass m suspended by a light string oflength L is whirled into a vertical circle as shown figure . What will be the trajectory of the particle , if the string is cut at (a) point B ? (b) point C ? (C ) Point X ? |
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Answer» Solution :When bob is WHIRLED into a vertical circle , the REQUIRED centripetal force is equal to the tension in the STRING .When string is cut , tension becomes zero and centripetal force also hence , bob start to move in a straight line path along the direction of its velocity . (a) at point B , the velocity of B is vertically downward , therefore right , therefore,whenstringis cut at C,bob moveshorizontallytowards right . Also , the bob moves under gravity simultaneously with HORIZONTAL uniformspeed . So it traversed on a parabolic path with VERTEX at C. (c) At point X , the velocity of the bob is along the tangent drawn at point X, therefore when stringis cut at pooint C , bob moves along the tangent at that point X . Alsothe bob move under gravity simultaneously with horizontal uniform speed . So , it taversed on a parabolic path with vertex higher than C . 
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| 35. |
The loudness and pitch of a sound note depends on |
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Answer» INTENSITY and FREQUENCY |
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| 36. |
A reversible engine converts one - sixth of the heat suplied into work . When the temperature of the sink is reduced by 62^@C, the efficiency of the engine is doubled . The temperatures of the source and sink are |
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Answer» `80^@C, 37^@C` |
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| 37. |
The equation of a progressive wave is y=0.05sin(200t-(x)/(2))wherex,y are in metres and t in seconds then (a) velocity of wave is 100ms^(-1) (b) max velocity of particle is 10ms^(-1) (c ) wavelength of wave is 4m |
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Answer» only a and c are true `omega=200rad//sec`, `K=(1)/(2)m^(-1),A-0.05m` `(a) V=(omega)/(K) (b) (V_(rho))_(max)=Aomega` `(c ) lambda=(2pi)/(K)` |
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| 38. |
State the number of significant 0.007 m^(2) |
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Answer» SOLUTION :`(a) 1 to `Only 7 (b) `3 to (2,6,4)` (c) `4 to (2,3,7,0)` (d) `4 to (6,3,2,0)` (E)`4 to (6,0,3,2)` (f) `4 to (6,0,3,2)` |
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| 39. |
A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Letr be the distance of the body from the centre of the star and let its linear velocity be v, angular velocity , omega kinetic energy K, gravitational potential energy U, total energy E and angular momentum I. As the radius r of the orbit increases, determine which of the above quantities increase and which ones decrease. As shown in figure, where a body of mass m is revolving around a star of mass M. Linear velocity of the body, |
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Answer» SOLUTION :`implies` As shown in figure, where a body of mass m is revolving around a star of mass M. Linear velocity of the body,  `v = sqrt(GM)/R` ` :. V prop 1/sqrtr` Therefore , when r increases , v DECREASES. Angular velcity of the body `omega=(2i)/T` According to Kepler.s `3^(rd)` law, `T^(2) prop r^3` ` :. T = kr ^(3/2)` ` :. omega = (2pi)/(kr^(3/2)) "" ( :. omega= (2pi)/T)` `:. omega prop 1/(r^(3/2))` Kinetic energy of the body , `K = 1/2 mv^2 =1/2 m xx (GM)/r = (GMM)/(2r)` `""( :. v = sqrt((GM)/r))` `:. K prop 1/r` Gravitational potential energy of the body , `U = - (GMm)/(r) implies U prop -1/r` Total energy of the body `E = K +U = (GMm)/(2r) + (-(GMm)/r)= - (GMm)/(2r)` ` :. E prop -1/r` Angular momentum of the body, `L = mvr =mrsqrt((GM)/r) =msqrt(GMr) ` `:. L prop sqrtr` From the above equations it is clear that v , W and K decreases with INCREASE in r And U,E and L increases with increase in r. |
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| 40. |
(A) : The angular momentum of a particle w.r.t origin moving parallel to x-axis with constant velocity is constant.(R ) : There is no change in the perpendicular distance of the particle from the origin when it travels prallel to x-axis. |
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Answer» Both 'A' and 'R' and TRUE and 'R' is the correct EXPLANTATION of 'A' |
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| 41. |
Let a satellite of mass m is revolving in a circular orbit of radius r round a planet of mass M such that the system is bound. IF E, U, K and B denote total energy, potential energy, kinetic energy and binding energy respectively. Then match the following columns. {:("Column-I",,"Column-II"),("(A) E",,(P) (GMm)/(2r)),("(B) U",, (Q) (GMm)/(r)),("(C) K",,(R) (GMm)/(r)),("(D) B",,(S) (GMm)/(2r)):} |
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Answer» |
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| 42. |
What provides the restoring force in the following cases ? Displacement of pendulum bob from mean position. |
| Answer» SOLUTION :WEIGHT of BOB. | |
| 43. |
Two gases have the same initial pressure, volume and temperature. They expands to the same final volume, one adiabatically and the other isothermally. |
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Answer» The FINAL TEMPERATURE is GREATER for the isothermal process |
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| 44. |
Give any one difference between a refrigerator and a heat engine. |
| Answer» Solution :A refrigerator is a DEVICE that transfers heat from LOWER to higher TEMPERATURE, whereas a heat engine is used to CONVERT work into heat. | |
| 45. |
A pump draws water from a reservoir and sends it through a horizontal pipe with speed v. Find the relation between power of the pump and velocity of liquid. |
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Answer» <P> Solution :From work - ENERGY theorem`P=(KE "imparted to water")/("TIME")` `=(KE)/("volume of water")xx("volume of water")/("time")` `=(1/2rhov^(2))(AV)` or `P PROP v^(3)` |
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| 46. |
A ballon at STP can lift a total mass of 175 kg attached with it.When the barometer reads 50 cmHg and the temperature becomes -10^@C at an upper point to where the balloon rises, find the maximum mass that can be lifted.Consider the volume of the balloon to be a constant. |
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Answer» Solution :The change in lifting CAPACITY is due to the change in the upthrust,as the density of air CHANGES at higher ALTITUDE due to the change in TEMPERATURE and pressure. Let V=volume of the balloon, `p_1`=density at STP `p_2`=density at 50 CMHG and `-10^@C` ,M=mass it can carry at the given altitude. Hence from Archimedes principle, `Vp_1=175 kg`and `Vp_2=M kg` Now, `p_1/(p_1T_1)=p_2/(p_2T_2) or,p_1/(Vp_1T_1)=p_2/(Vp_2T_2)` Subsituting the corresponding values, we get `76/(175 times 273)=50/(M times 263)` or, `M=(175 times 273 times 50)/(263 times 76)=119.5 kg`. |
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| 47. |
A charge q is distributed uniformly on a ring of radius R.A sphere of equal radius R is constructed with its centre at the ring, Find the flux of the electric field through the surface of the sphere |
| Answer» SOLUTION :`(Q)/(3 in_(0))` | |
| 48. |
A triatomic, diatomic and monoatomic gas is supplied same amount of heat at constant pressure, then |
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Answer» Fractional energy used to change internal energy is maximum in MONOATOMIC gas |
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| 49. |
Which of the following functions represent SHM : sin 2 omega t |
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Answer» SOLUTION :A motion will be SHM if ACCELERATION is directely proportional to its displacement `a=- omega^(2)y` As `y= SIN 2 omega` `rArr v=(dy)/(dt)=2 omega COS 2 omega= omega t` Acceleration `=(d^(2)y)/(dt^(2))=- 4 omega^(2) sin 2 omega t` `=- 4omega^(2)y` So `, y= sin 2 omega t` represent SHM. |
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| 50. |
"The mass of a flywheel is concentrated on the rim ". Why ? |
| Answer» Solution :This is to increase the moment of inertia. Hence its OPPOSITION to any change in UNIFORM ROTATORY motion is large. So when-a flywheel of large M.I. is USED, the engine runs smoother and STEADIER. | |