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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) : A cyclist always bends inwards while negotiating a horizontal curve.(B) : By bending, cyclist lowers his centre of gravity. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 2. |
Is the angular momentum in respect of a planet revolving in elliptical orbit around the Sun conserved? |
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Answer» Solution :Yes. The angular momentum of the PLANET is conserved. A planet rotates around the SUN DUE to gravitational force between the sun and the planet. This is an internal force of that system passing through the sun. As there is no EXTERNAL torque, angular momentum REMAINS constant. |
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| 3. |
A man is at the top of a ladder and another instance he stands on one of the lower steps. In which case does the foot of the ladder have a higher chance of slipping ? |
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Answer» |
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| 4. |
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 | |
| 5. |
A thin uniform rod of mass .M. and length .L. rotates with constant angular velocity .omega. about perpendicular axis passing through its centre. Two bodies each of mass (M)/(3) are attached to its two ends. What is its angular velocity? |
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Answer» `(2)/(3)omega`  Initially angular momentum `L_(1)=Iomega` but `I=(ML^(2))/(12)` `therefore L_(1)=(ML^(2)omega)/(12)….(1)` Angular momentum in final (other) POSITION  `L_(2)=Iomega.+M^(1)R^(2)omega.+M^(1)R^(2)omega.` `=((ML^(2))/(12)+(M)/(3)xx(L^(2))/(4)+(M)/(3)xx(L^(2))/(4))omega.` `=((ML^(2))/(12)+(ML^(2))/(12)+(ML^(2))/(12))omega.` `L_(2)=3(ML^(2))/(12)omega.....(2)` Angular momentum is conserved. `therefore L_(1)=L_(2)` `therefore (ML^(2)omega)/(12)=(3ML^(2)omega.)/(12)` `therefore omega=3omega.` `therefore omega.=(omega)/(3)` |
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| 6. |
To keep a particle moving with constant velocity on a frictionless horizontal surface an external force |
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Answer» Should act CONTINUOUSLY |
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| 7. |
From two diametrically opposite points A and B on earth, moon is observed. The angle thetasubtended at the moon by the two directions of observation is 1^@ 54^1. If radius of earth is 0.638 xx 10^7 m, find the distance of the moon from the earth. |
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Answer» Solution :Distance between POINTS of OBSERVATION = DIAMETER of earth i.e., `b=2r rArr b=2 xx 0.638 xx 10^(7) = 1.276 xx 10^(7)` m ANGLE SUBTENDED at moon `theta= 1^(@) 54. = 1.9^(@)` `rArr theta = 1.9 xx 1.745 xx 10^(-2) = 3.32 xx 10^(-2)` rad Distance of moon, `D= (b)/(theta) rArr D = (1.276 xx 10^(7) )/( 3.32 xx 10^(-2) )= 3.84 xx 10^(8)` m |
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| 8. |
The dams of water reservoir are made thick near the bottom. Why? |
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Answer» <P> Solution :Pressure exerted by liquid column = hpg so as ‘h’ INCREASES p increases so to with stand high pressure DAMS are made thick near the BOTTOM. |
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| 9. |
A particle moves on thex-axis according to the equation x=x_(0) sin 2wt. The motion is simple harmonic. |
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Answer» With AMPLITUDE `x_(0)//2` |
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| 10. |
A uniform steel rod of mass 1.8 kg and length 0.8 m is hung from a nail with the help of a steel wire of area of cross-section 0.01 mm^(2) and unstretched length of 1m as shown in the figure. the centre of mass of the rod lies vertically below the nail. The increase in the distance between the centre of mass of the rod and the nail due to stretcheing of the wire as the rod huns is____mm. (young.s modulus of steel =2xx1011Nm^(-2)and acceleration due ot gravity =10ms^(-2) |
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Answer» 50 |
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| 11. |
Average energy in one time period of a single harmonic oscillator whose amplitude is A, angular velocity is omega and mass is m is, |
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Answer» `1/4 MW^(2)A^(2)` |
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| 12. |
(A) : Value of radius of gyration of a body depends on axis of rotation.(R ) : Radius of gyration is root mean square distance of particle of the body from the axis of rotation. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 13. |
While conducting a resonance column experiment in the laboratory you can hear the maximum sound at a certain height Is resonance column apparatus an open pipe or a closed pipe? |
| Answer» SOLUTION :CLOSED PIPE | |
| 14. |
Two soap bubbles A and B are given. Radius of B is 50% more than radius of A. Soap content (volume) of B as compared to A is |
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Answer» `1.5` |
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| 15. |
While conducting a resonance column experiment in the laboratory you can hear the maximum sound at a certain height which phenamenon is responsible for this ? |
| Answer» SOLUTION :REASONANCE | |
| 16. |
A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming gamma(=C_(P)//C_(V)) to be 1.5, find the new pressure of the gas. |
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Answer» Solution :Let the INITIAL volume be V. After ISOTHERMAL EXPANSION, pressure `=(P)/(4)`, volume = 4V Let P. is the pressure after ADIABATIC compression. The volume then is V. Therefore, `P.V^(gamma)=(P)/(4)(4V)^(gamma),P.=(P)/(4)(4)^((3)/(2))` `".=2P` |
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| 17. |
Four particles each of mass 1 kg are at the four corners of a square of side 1m. The M.I of the system about a normal axis through centre of square is |
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Answer» `6KGM^(2)` |
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| 18. |
An object may have a) varying speed without having varying velocity b) varying velocity without having varying speed c) nonzero acceleration without having varying velocity d) nonzero acceleration without having varying speed. |
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Answer» a, B & C are correct |
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| 19. |
Two discs of moments of inertia I_(1) and I_(2) about their respective axis (normal to the disc and passing through the centre) and rotating with angular speeds omega_(1) and omega_(2) are brought into contact face to face with their axes of rotation coincident. Calculate (i) angular speed of the two disc system (ii) Show that K.E. of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss of energy?Take omega_(1)neomega_(2) |
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Answer» Solution :(i)`OMEGA=(I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2))` (ii)This loss must be due to FRICTION in the CONTACT of the two discs. Note that angular momentum is CONSERVED as torque due to friction is only an internal torque |
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| 20. |
What is viscosity ? What is the cause of it ? |
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Answer» Solution :In alaminar flow , any two consecutive layers of fluid have relative velocity between them. As a result resistive force is produced tangentially at the surface of the layers in contact . This internal force of friction is known as viscous force . The property of fluid due to which relative motion between two consecutive layer is known as viscosity of the fluid.  A layer of liquid sandwiched between two parallel glass plates in which the lower plate is fixed and the UPPER one is MOVING to the right with velocity v Suppose a fluid like OIL enclosed between two glass plates as shown in figure . The bottom plate is fixed while the top plate is moved with a constant velocity v relative to the fixed plate. The fluid in contact with a surface has the same velocity as that of the surfaces . Hence , the layer of the liquid in contact with top surface moves with a velocity v and the layer of the liquid in contact with the fixed surface(stationery ) have velocity zero . The velocities of different layer increases from bottom to top uniformly (zero velocity to velocity v). For any layer of liquid , its upper layer pulls it forward while lower layer pulls it backward . This reslults in fore between the layers. This type of flow is know as LAMINAR. When a fluid is flowing in a pipe or a tube , then velocity of the liquid layer along the axis of the tube is maximum and decrease gradually as we move towardsthe wall where it bocomes zero This is shown in figure.  Velocity distribution for viscous flow in a pipe (Note : In a steady flow different layers side over each other without getting mixed up . This kind of flow is CALLED laminar flow) |
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| 21. |
Explain the difference between 'steady state' and 'variable state' of a thermal conductor. |
| Answer» Solution :During variable STATE, heat flow DEPENDS on `K, RHO and c`, the SPECIFIC heat capacity. But during steady state the heat flow depends only on THERMAL conductivity. | |
| 22. |
(A): If you look out of the window of a fast moving train, the nearby trees, houses etc., seem to move rapidly in a direction opposite to the train is motion, but the distant objects (hill tops, the Moon, the stars etc) seem to be stationary. (B) : When a train moves rapidly, line of sight changes its direction rapidly. On the other hand, in the case of far off objects, the line of sight changes its direction |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 23. |
A hose pipe lying on the ground shoots a stream of water upward at an angle 60^@ to the horizontal at a speed of 20 ms^(-1). The water strikes a wall 20m away at a height of |
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Answer» 14.64 m |
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| 24. |
(1) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts, (2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius, (3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G.., (4) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis. which one of the following pairs of statements is correct ? |
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Answer» SOLUTION :(i) Centre of MASS of a sphere is at.its geometrical centre, of a ring at its centre. (ii) No. Example : ring, hollow CYLINDER ETC. |
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| 25. |
A car of mass m is moving with a velocity v over a bridge. What will be the values of the force F at the highest point of a convex bridge and at the lowest point of a concave bridge? |
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Answer» Solution :A centripetal FORCE is required to move the car over a convex BRIDGE. The value of this centripetal force is ` (mv^(2))/(r )`. Here, r is the radius of curvature of the convex bridge At the highest point of the bridge. The resultant of the weight of the car mg and F will SUPPLY the necessary centripetal force. `therefore "" mg - F = (mv^(2))/(r )` or,` "" F = m(g- (v^(2))/(r))`  (II) A centripetal force is required to move the car over concave bridge also. At the lowest point of the concave bridge, the resultant of the weight of the car mg and f will supply the necessary centripetal force. `therefore"" F - mg = (mv^(2))/(r ) "" or, f = m(g + (v^(2))/(r))` 
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| 26. |
The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. What is the angle between A and B ? |
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Answer» Solution :Let, the unit vectors along `vecA, vecB and vecR " be"" " hata , HATB and hatc` respectively. Then `vecA=Ahata,vecB=Bhatb,vecR=Rhatc and Ahata+Bhatb=Rhatc` ` or, A(VEC(OP))/(OP)+B(vec(OQ))/(OS)=R(vec(OS))/(OS)` [Fig .2.62].  If `hatn` is the unit vector along theline CD, `vec(OP)xxhatn=OP SIN alphahatm=ONhatm,` `vec(OQ)xxhatn=ONhatm, vec(OS)xxhatn=ONhatm` where `hatm` is a unit vector perpendicular to the plane of THEVECTORS. Then , `A/(OP)vec(OP)xxhatn+B/(OQ)vec(OQ)xxhatn=R/(OS)vec(OS)xxhatn` `or, A/(OP) ONhatm+B/(OQ)hatm=R/(OS)ONhatm` `or, A/(OP)+B/(OQ)=R/(OS)`. |
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| 27. |
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig.The coefficient of friction between the box and the surface below it is 0.15.On a straight road, the truck starts from rest and accelerates with 2 m s^(-2) .At what distance from the starting point does the box fall off the truck?(Ignore the size of the box). |
| Answer» SOLUTION : Acceleration of the BOX due to friction` = mu g = 0.15 XX 10 = 1.5 m s^(-2)` .But the acceleration of the truck is greater.The acceleration of the box relative to the truck is` 0.5 m s^(-2)`towards the rear end.The time taken for the box to fall off the truck ` = SQRT( (2 xx 5)/(0.5)) = sqrt20 s ` .During this time, the truck covers a distance ` = 1//2 xx 2 xx 20 = 20m ` | |
| 28. |
When one metre, one kg and one minute are taken as fundamental units, the magnitude of a force is 36 units. What is the value of this force on CGS system? |
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Answer» `10^(3)` dyne  The dimensional formula of force is `[MLT^(-2)]` `:. a= 1, b= 1, c=-2` `n_(2)= n_(1)[(M_(1))/(M_(2))]^(a)[(L_(1))/(L_(2))]^(b)[(T_(1))/(T_(2))]^(c)= 36[(1KG)/(1g)]^(1)[(am)/(1CM)^(1)][(1min)/(1s)]^(-2)` `=36[(1000g)/(1g)]^(1)[(100CM)/(1cm)^(1)][(60S)/(1s)]^(-2)= (36 xx10^(3) xx 10^(2))/(3600)= 10^(3)` Hence, in the CGS system of units the value of given force is `10^(3)` dyne |
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| 29. |
A vertical spring with force constant .k. is fixed on a table. A ball of mass .m. at a height .h. above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance .d.. The net work done in the process is |
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Answer» `MG(h+d)+(1)/(2)KD^(2)` |
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| 30. |
A bird moves in such a way that is has a displacement of 12 m towards east, 5 m towards north and 9 m vertically upwards. Find the magnitude of its displacement. |
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Answer» `5sqrt(2) m` |
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| 31. |
Two satellite of masses 50 kg and 100 kg revolve around the earth in circular orbits of radii 9R and 16R respectively, where R is the radius of the earth. The speed of the two satellites will be in the ratio is x/y. Find the value of x + y |
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Answer» |
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| 32. |
In an isohoric process, find which is relevant among the following : |
| Answer» Solution :`W = 0` | |
| 33. |
Discuss the bending of a cyclist in curves. |
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Answer» Solution :(i) Consider a cyclist negotiating a circular level road(not banked ) of radius r with a speed V about the center O as shown in figure. (ii) Let m be the mass of the system, which includes cycle and cyclist and C be the center of gravity of this system. (III) Let us consider horizontal is along x-axis and vertical is along Z-axis.  (IV) In this rotating frame, the centrifugal force `(mv^2)/ (r )`. acts away from center O and PASSING through the center of gravityC as shown below. (v) As the system is in rotational equilibrium, the net torque must be zero. Thus, `vectau_(" net ") =0` `-mgAB+(mv^2)/(r )BC=0`  (vi) Here, the clock wise moment (mg AB) is taken as negative and the anti clodkwise moment `((mv^2)/(r )BC)` is taken as positive. `mgAB=(mv^2)/(r )BC` (VII) But from `triangleABC, AB=AC sin theta"" & ""BC=AC cos theta`. Therefore, the above equation can be written as, `mg AC sin theta = (mv^2)/(r )AC cos theta` `tan theta = (v^2)/(rg)` `theta = tan^(-1)((v^2)/(rg))` (viii) It shows that while negotiating a circular road of radius r at velocity v, a cyclist has to bend an angle `theta` from vertical, to avoid a fall. |
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| 34. |
Compute the bulk modulus of water from the following staa , Initial volume = 100. 0 litre, Pressure increase = 100.0 atm (1 atm =1.013xx10^(5) Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. |
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| 35. |
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance. |
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Answer» Solution :50 VSD=49 MSD `:.IMSD =(50)/(49)=VSD` and `IVSD=(49)/(50)MSD` MINIMUM inaccuracy = IMSD -IVSD `=IMSD-(49)/(50)MSD=(1)/(50)MSD` Now 1IMSD=0.5 mm (given) `:.` Minimum inccuracy. `=(1)/(50)xx0.5 mm=(1)/(100)=0.01 mm` |
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| 36. |
Two perfect gases at absolute temperatures T_1 and T_2 are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m_1 and m_2 and the number of molecules in the gases are n_1 and n_2. |
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Answer» `T= n_(1) T_(1) + n_(2) T_(2)` |
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| 37. |
A particle of mass 1 kg is executing SHM with an amplitude of 1 m and time period pis. Calculate the kinetic energy of the particle at the moment when the displacemen is 0.8 m from mean position? |
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Answer» SOLUTION :`OMEGA=(2pi)/T=(2pi)/(pi)=2s` we have `v=omegasqrt(A^(2)-x^(2))` `=2sqrt((1)^(2)-(0.8)^(2))=2xx0.6=1.2m//s` KINETIC energy `=1/2mv^(2)=1/2xx1xx(1.2)^(2)=0.72J` |
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| 38. |
Assertion : The dimension of balloon filled with hydrogen increases as it raises aabove. Reason : Balloon is expanded easily. |
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Answer» Both ASSERTION and REASON are TRUE and the reason is the CORRECT EXPLANATION of the assertion. |
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| 39. |
The escape velocity of a body from the surface of the earth depends upon (i) the mass of the earth M, (ii) The radius of the eath R, and (iii) the gravitational constnat G. Show that v=ksqrt((GM)/R), using the dimensional analysis. |
| Answer» Solution :`V=kG^(x)M^(y)R^(Z),[LT^(-1)]=[M^(-1)L^(3)T^(-2)]^(x)M^(y)L^(z)`. SIMPLIFYING `x=1//2, M=-1//2, z=-1//2`. | |
| 40. |
A space station is set up in space at a distance equal to earth.s radius from earth.s surface. Suppose a satellite can be launched from spac station. Let V_1 and V_2be the scape velocites f the satllite on earth.s surface and space station respectively. Then |
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Answer» `V_2=V_1` |
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| 41. |
Can the relative velocities of two bodies be greater than the absolute velocity of either body? |
| Answer» Solution :Yes, when two bodies move in opposite directions, the RELATIVE VELOCITY of each is greater than the INDIVIDUAL velocity of either BODY . | |
| 42. |
Particles of masses 1kg and 3kg are at (10i-7j-3k)ms^(-1) and (7i-9j+6k)ms^(-1) then a instantaneous velociti of their centre of masses |
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Answer» `(1)/(4)(31i-34j+15k)MS^(-1)` |
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| 43. |
A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is r and the speed of revolution is n rotations/second, find the difference in height of the liquid at the centre of vessel and its sides. |
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Answer» `h=(2PI^(2)N^(2)R^(2))/(G)` |
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| 44. |
The displacement of a particle in S.H.M is in opposite phase with |
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Answer» VELOCITY |
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| 45. |
Surface tension of a liquid |
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Answer» SURFACE phenomenon |
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| 46. |
If the earth has a mass nine times and radius twice that of the planet Mars. Given escape speed on the surface of Earth is 11.2 km s^(-1). |
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Answer» Solution :`g_(m) = g_(e) ((M_(s))/(M_(e))) ((R_(e))/(R_(m)))^(2) = g_(e) xx (1)/(4) xx 4 = (4)/(9) g_(e)` On EARTH, `upsilon_(e) = SQRT(2G_(e) R_(e))`, On Mars, `upsilon_(m) = sqrt(2g_(m) R_(m))` `:. (upsilon_(m))/(upsilon_(e)) = upsilon_(e) xx (sqrt(2))/(3) = (11.2 xx 1.414)/(3)` `= 5.28 km s^(-1)` |
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| 47. |
A race car negotiates a curve of radius 50.0 m on a flat road. What is its angular displacement when it covers a linear distance of 78.5 m? |
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Answer» SOLUTION :RADIUS =r=50.0 m Linear distance = s = 78.5 m ANGULAR displacement = `theta` = ? `theta = s/r = (78.5)/50 = 1.54` radian `=360/(2pi) XX 1.54 = 90^(@)` |
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| 48. |
A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. Will it reach the bottom with the same speed in each case? Will it take longer to roll down one plane than the other? If so, which one and why? |
| Answer» Solution :V must be the same i.e. it will reach the bottom with the same speed. TIME TAKEN will be LESS for the plane of larger angle as `t=(1)/(sintheta)` | |
| 49. |
A cubical block of wood of side 10 cm floats at the interface between an oil and water with its lower surface 2 cm below the interface. the heights of the oil and water columns are 10 cm each. Density of oil is 0.8g/c.c. The mass of the block is |
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Answer» 720 gram |
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| 50. |
A bar magnet of length 5.0cm and breadth 1.2 cm is rotated about an axis passing through its center and perpendicular to its plane. Find its moment of inertia if the mass of the magnet is 200g. |
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Answer» SOLUTION :Moment of INERTIA `I=(M(l^(2)+b^(2)))/(12)` `=200xx10^(-3)(((5XX10^(-2))^(2)+(1.2xx10^(-2))^(2))/(12))` `=4.407xx10^(-5)kgm^(2)` |
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