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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 satellite of mass, .m. revolves around the earth of radius R at a height x from its surface. If .g. is the acceleration due to gravity on the surface of the earth, than the orbital speed of the satellite is |
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Answer» GX |
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| 2. |
The cylindrical tube open at both ends has fundamental frequency f in air.The tube is dipped vertically in water so that half of it is in water.The fundamental frequency of the air coloumn is now |
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Answer» `N/2` |
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| 3. |
An ideal gas is taken from point A to point C on P-V diagram through two process AOC and ABC as shown in the figure. Process AOC is isothermal |
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Answer» Process AOC REQUIRE more heat than process ABC. |
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| 4. |
A 4-kg disk slides over level ice towards the eact at a velocity of 1 meter per second, as shown. The disk strikes a post and rebounds towards the north at the Same speed. The change in the magnitude of the eastward component of the momentum of the disk is: |
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| 5. |
A body of mass 4 m is lying in xy - plane at rest. It suddenly explodes into three pieces. Two pieces each of mass m move perpendicular to each other with equal speed v. The total kinetic energy generated due to explosion is |
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Answer» `MV^(2)` |
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| 6. |
A body of mass M moving with velocity V explodes into two equal parts. If one comes to rest and the other body moves with velocity v, what would be the value of v ? |
| Answer» Answer :D | |
| 7. |
A block of mass 1 kg is fastended to one end of a copper wire of cross- sectional area 1 mm^(2) and is rotated in a horizontal circle of radius 20 cm. If the breaking stress of copper is 5 xx 10^(8)Nm^(-2), find the maximum number of revolutions the block make in the minutewithout the string breaking. |
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Answer» Solution :Maximum tension on the string = Breaking stress x Area of cross-section `= 5xx10^8 xx1xx10^(-6) = 500 N.` When a body REVOLVES in a horizontal circle, Tension on the string = Centripetal force `500 = mr omega^2` Where m= 1 kg, R = 0.20m, `omega`= ? `500= 1 xx 0.2 xx omega^2` `omega^2 = (500)/(0.2) = 2500` Maximum angular speed, `omega = 50 RAD s^(-1)` `omega = (2pi m)/(t) , 50 = (2pim )/(60)` The maximum number of revolutions, `n = (50 xx 60)/(2pi) = (1500)/(pi) = 477.4 r`pm. |
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| 8. |
The capacity to do work |
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Answer» force |
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| 9. |
A skater wearing his skates is oushing a wall. What will be the work done by the contact force on the skater as he moves away from the wall? |
| Answer» Solution :The skater.s hand is at REST with respect to the wall. So, the displacement of the point of contact of the ZERO and HENCE, the contact force does zero work on the skater. | |
| 10. |
Calculate the energy required to move an earth satellite of mass 10^(3) kg from a circular orbit of radius 2R to that of radius 3R. Given mass of the earth,M=5.98xx10^(24) kg and radius of the earth, R=6.37xx10^(6)m. |
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| 11. |
Three particle, located initially on the vertices of an equilateral triangle of side L, start movingwith a constant tangential acceleration towards each other in a cyclic manner, forming spiral loci that coverage at the centroid of the triangle. The length of one such spiral locus is 2L/k. The value of k is |
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Answer» |
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| 12. |
A disc rotating about its axis with angular speed is placed lightly {without any transnational push) on a perfectly friction-less table. The radius of the disc is R.What are the linear velocities of the point A, B and C on the disc shown in the Fig. B 7:2.60 ? Will the disc roll in the direction indicated ? |
| Answer» SOLUTION :Let the origin of the COORDINATE system be taken as the centre of the disc. Linear speed of `A=R omega_(0)`Linear speed of `B=(-R) omega_(0) (-ve)`because B is below O) linear speed of `C=R//2 omega_(0)`. The table is FRICTIONLESS. So there is no FRICTION and hence the disc will not roll. | |
| 13. |
What is free fall ? Write equations of uniformly accelerated motion for object falling freely. |
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Answer» Solution :The acceleration produced in the object due to gravitational FORCE is called gravitational acceleration. This acceleration is in DOWNWARD direction. If the air RESISTANCE is neglected, then this type of falling of object is said to be free fall. Here, acceleration is considered as negative gravitational acceleration. This is also an uniformly accelerated motion. By putting `v_(0) = 0, a = -g, d = -H` in the kinematic equations, we can DERIVE equations for free fall. `v = v_(0) + at` `therefore v =- g t` `d = v _(0) t +1/2 at ^(2)` `therefore -h = -1/2g t^(2)` `therefore h = 1.2 g t ^(2)` `v ^(2) - v _(0) ^(2) = 2ad` `therefore v^(2)=2 (-g) (-h)` `therefore v ^(2) = 2gh` `therefore v = sqrt (2gh)` |
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| 14. |
A body is displaced by 6 m in Y - direction by a force (5,2,3) N , so work done …….. . |
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Answer» 12 N `= (5,2,3).(0,6,0)` `=(0+12+0)` = 45W |
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| 15. |
An organ pipe A closed at one ends is allowed to vibrate in its first harmonic and another pipe B open at both ends is allowed to vibrate in its third harmonic. Both A and B are in resonance with a given tuning fork. The ratio of the length of A and B is |
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Answer» `(8)/(3)` Frequency of closed organ pipe (at one end) `f_(1) = (v)/(4 l_(1))` Frequency of OPEN organ pipe 3RD harmonic `f_(2) = (3v)/(2l_(2))` If , `f_(1) = f_(2) implies (v)/(4l_(1)) = (3v)/(2l_(2))` Ratio of length A and B, `(l_(1))/(l_(2)) = (2)/(12) = (1)/(6)` |
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| 16. |
Three liquids A, B and C of masses 400g 600g and 800g are at 30°C, 40°C and 50°C respectively. When 'A' and 'B' are mixed resultant temperature is 36°C. When B and C are mixed resultant temperature is 44°C. Then ratio of their specific heats are |
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Answer» `2:1:1` |
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| 17. |
One mole of an ideal monoatomic gas undergoes a process described by the equation PV^(3)=constant. The heat capacity of the gas during this process is |
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Answer» `3/2R` |
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| 18. |
Give two similarities and dissimilarities between friction and viscosity. |
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Answer» Solution :Similarities: (i) viscosity exist when there is relative motion between LAYERS of the LIQUID. Similarity friction exists between two solids surfaces in CONTACT when there is a relative motion between them. (ii) Viscosity depends on the nature of the LAYER of the fluid and friction ALSO depends on the nature of the solid surfaces in contact. Dissimilarities : N/A |
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| 19. |
A particle of mass M is placed at the centre of a uniform spherical shell of equal mas and radius a . Find the gravitational potential at a point P at a distance a/2 from the centre . |
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Answer» SOLUTION :The gravitational POTENTIAL at the POINT P due to the PARTICLE at the centre is `V_(1) = (GM)/(a//2)=-(2GM)/a` The potential at P due to the SHELL is `V_(2) = - (Gm)/a` The net potential at P is `V_(1)+V_(2) = - (3Gm)/a` |
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| 20. |
As shown in Fig. 5.29, a body is projected with velocity u_(1) from the point A. At the same time another body is projected vertically upwards with velocity u_(2) from the point B. What should be the value of u_(1)//u_(2) for both the bodies to collide ? |
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| 21. |
The resultant of two forces 2P and sqrt(2)P is sqrt(10)P. The angle between the forces is |
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Answer» `30^(0)` |
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| 22. |
The equation for the displacement of a particle executing SHM is y = {5 sin 2pi t}cm. Find (i) the velocity at 3 cm from the mean position, (ii) acceleration after 0.5 s after leaving the mean position. |
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Answer» `8picms^(-1)`, ZERO |
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| 23. |
The focal length of the lenses of an astronomical telescope are 50 cm and 5 cm. The length of the telescope when the image is formed at the least distance of distinct vision is |
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Answer» 45 |
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| 24. |
Find out the value(s) of theta of the inclined plane such that the mass m remains at rest on the wedge of mass M_(2) as shown in figure-2.215. Friction between the small block and the wedge plane is mu and all other surface are smooth. |
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Answer» where `p=m(G-mua),q=m(a+mug),r=M_(1)(a-g)" and "a=(M_(1)g)/(M_(1)+M_(2)+m)` `sinalpha_("MIN")=sqrt(4l^(2)k^(2)-4(l^(2)+k^(2))(n^(2)-l^(2))-2kn)/(2(l^(2)+k^(2)))`, where `l= m(a-mug),k=m(g+mua),n=M_(1)(a-g)" and "a=(M_(1)g)/(M_(1)+M_(2)+m)]` |
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| 25. |
From the top of a cliff of height 19.6 m a ball is projected horizontally. Findthe time taken by the ball to reach the ground (g=9.8ms^(-2)) |
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Answer» |
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| 26. |
On a particular day when the atmpspheric temperature is 18.5^(@)C the dew point is found oo be 8.6^(@)C. Determine the relative humidity of the air if the maximum vapour pressure of water be 8.04, 8.61, 15.46 and 16.46mm at 8^(@), 9^(@), 18^(@) and 19^(@) respectively. |
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| 27. |
A force vecF=hati+2hatj+3hatkN acts on a particle and displaces it through a distances vecs=4hati+6hatjm calculate the work done. |
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Answer» SOLUTION :`FORCE "vecF=hati+2hatj+3hatkN` `" DISTANCE" vecS=4hati+6hatjm` `"WORK DONE" =vecF.vecS=(hati+2hatj+3hatk).(4hati+6hatj)=4+12+0=16J` |
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| 28. |
Figure shows a weighing-bridge, with a beaker P with water on one pan and a balancing weight R on the other. A solid ball Q is hanging with a thread outside water, It has volume 40 cn^(3) and weighs 80 g. If the solid is lowered to sink fully in water, but not touching the beaker anywhere, the balancing weight R' will be |
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Answer» same as `R` So `R = F_(B) = rho_(1)(VG) = 40G`. `R' - R = 80G - 40g = 40g` Hence, `R'` will be `40g` more than `R` |
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| 29. |
A vector 3hat(i) + 4hat(j) rotates about its tail through an angle 37^(@) in anticlockwise direction then the new vector is |
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Answer» `-3HAT(i) + 4hat(J)` |
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| 30. |
If a satellite orbits as close to the earth.s surface as possible, (a) its speed is maximum (b) time period of its rotation is minimum (c) the total energy of the earth plus satellite system is minimum (d) the total energy of the earth plus satellite system is maximum |
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Answer» only a is TRUE |
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| 31. |
Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is I. What is the moment of inertia about its diameter? |
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Answer» I `I = MR^(2) "" … (i)` Moment of inertia of the ring about its diameter `I_("diameter") = (MR^(2))/(2) = (1)/(2) (MR^(2)) = (I)/(2) ""` (Using (i)) |
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| 32. |
A square plate of 0.1 meter side moves parallel to a second fixed plate with a velocity of 0.1 m/s both plates being immersed in water. If the viscous force is 0.002 newtons and the coefficient of viscosity is 0.01 poise, distance between the plates in metre is |
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Answer» `0.1` |
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| 33. |
An incompressible fluid flows through a tube at a uniform rate. Radius of the tube of at a point A is 2r and that at point b is r. If the velocity at the point A is v, then velocity at the point B will be |
| Answer» ANSWER :D | |
| 34. |
Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the earth. (Radius of the earth = 6400 km) |
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Answer» Solution :g at height, `g_(H) = (g)/((1+(h)/(R))^(2))` In this problem, `g_(h) = (50)/(100)g = 0.5g` `0.5 g = (g)/((1+(h)/(R))^(2)) rArr (1+(h)/(R))^(2) = (g)/(0.5 g) = 2` `1+(h)/(R) = sqrt(2), (h)/(R) = sqrt(2) - 1 rArr` the height, `h = 0.414 R = 0.414 xx 6400 = 2650km`. |
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| 35. |
This question has statement 1 and statement 2. Of the four choice given after the statements, choose the one that best describes the two statements. If two springs S_(1)" and "S_(2) of the force constant k_(1)" and "k_(2) respectively, are stretched by the same force, it is found that more work is done on spring S_(1) than on spring S_(2). Statement -1 : If streched by the same amount, worked done on S_(1), will be more than that of S_(2). Statement-2 : k_(1) lt k_(2). |
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Answer» statement-1 is FALSE, statement-2 is TRUE |
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| 36. |
Assertion : The concept of surface tension applied to liquid only.Reason : The concept of surface tension cannot be applied to gases. |
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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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| 37. |
The radius of gyration of a solid cylinder of mass M and radius R about its own axis is |
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Answer» `(R)/(sqrt2)` `I = (MR^(2))/(2) "" … (i)` As `I = MK^(2)` (where k is the radius or gyration) `""` … (ii) From (i) and (ii) , we get `k^(2) = (R^(2))/(2) or k = (R)/(sqrt2)` |
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| 38. |
A solid cylinder is rolling down the inclined plane without slipping. Which of the following is/are correct ? |
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Answer» The FRICTION force is DISSIPATIVE |
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| 39. |
When a stone is dropped on the surface of still water, the waves produced are …………. |
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Answer» TRANSVERSE |
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| 40. |
When two waves y _(1) =A sin (omega t + (pi)/(6)) and y _(2) = A cos (omega t) superpose, find amplitude of resultant wave. |
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Answer» A `y _(2) =A cos (omega t)` `= A sin ((pi )/(2) + omega t ) ( because cos theta =sin ((pi)/(2) + theta ))` Phase different between above WAVES is, `RHO = ((pi)/(2) + omega t ) - (omega t + (pi )/(6)) = (pi)/(3)` Now, amplitude of resultant wave is, `A . = sqrt (A _(1) ^(2) + A _(2) ^(2) + 2 A _(1) A _(2) cos rho )` `therefore A.= sqrt (A ^(2) + A ^(2) + 2 A A cos ((pi)/(3)))` `therefore A.= sqrt (2A ^(2) + 2 A ^(2) ((1)/(2))) = sqrt ( 3 A ^(2)) = sqrt3 A` |
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| 41. |
Which of the following examples represent periodic motion? A hydrogen molecule rotating about its center of mass. |
| Answer» Solution :The ROTATING motion of a HYDROGEN molecule about its CENTRE of MASS is PERIODIC. Hence, (c) will be the SHM. | |
| 42. |
A simple pendulum bob is given a horizontal velocity sqrt(7/2 gl)at the bottom. The string slackens after swinging through an angle thetawith the vertical. The theta is: |
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Answer» `30^(@)` |
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| 43. |
A body of 200 g begins to fall from a height where its potential energy is 80 J. Its velocity at a point where kinetic and potential energies are equal is |
| Answer» Answer :D | |
| 44. |
A body of mass 3 kgfall freely from the buildingof height60 m . Its kinetic energy after 3s will be……… |
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Answer» 557 J ` :. V = v_(o) +"gt"= 0 +9.8 xx3` ` :. V = 29.4` m/s ` :. ` KINETIC energy after 3s ` K = 1/2 mv^(2)` ` = 1/2 xx 3xx (29.4)^(2)` ` K = 1297 J ` |
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| 45. |
If v = 3i + 4j + 5k is the instantaneous velocity of a body of mass 1.50 kg. Calculate its kinetic energy. |
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Answer» SOLUTION :V = 3i + 4j + 5k , MASS = 1.5 KG Kinetic energy K.E. `= (1)/(2) mv^(2)` `= (1)/(2) 1.5 (3i + 4j + 5k). (3i + 4j + 5k)` `= (1.5)/(2) (9 + 16 + 25) = (1.5)/(2) xx 50 = 37.5` joules |
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| 46. |
Two inclined planes OA and OB having inclination (with horizontal) 30^(@) and 60^(@) respectively , intersect each other at O. A particle is projected from point 'P' with velocity u=10sqrt(3) m/s along a direction perpendicu lar to plane OA. If the particle strikes plane OB perpendicu lar at Q. Calcu late: Time of flight of the particle: |
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Answer» 8s  Now for x-axis `v_(x)=v_(x)+a_(x)T` `0=10sqrt(3)-gsin 60^(@)T` Since, `s_(x)=u_(x0T+(1)/(2)a_(x)T^(2)` `s_(x)=(10sqrt(3).T-(1)/(2)g sin 60^(@)T^(2))` Along y-axis `OP=(1)/(2)g cos 60^(@)T^(2)` `(h)/(OP)=sin 30^(@)` |
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| 47. |
A pendulum bob of mass .m. is held out in the horizontal position and then released from rest. If the string is of length l (1) what is the velocity of the bob and tension in the string when the bob reached the lowest position.(2) If the breaking strength of the string is .nmg.. Find the maximum angle .theta. with respect to the vertical at which the bob is to be released to avoid breaking of the string. |
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Answer» Solution :(1) From law of conservation of mechanical energy. `0+mgl =(1)/(2) mv^(2)+0 "v"=sqrt(2gl)` The tension in the string at the lowest POSITION `=(mv^(2))/(r )+mg =(m(2gl))/(l)+mg (because r = l)=3MG` (2)`mgl (1-cos theta)=(1)/(2) mv^(2)` or `v^(2)=2gl (1-cos theta)` Tension in the string at the lowest position `T = (mv^(2))/(l)+mg therefore nmg.=(m)/(l)[2gl(1-cos theta)]+mg` on simpliication `cos theta = (3-n)/(2)` |
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| 48. |
The mass of a beaker is (10.1 pm 0.1)when empty and (17.3 pm 0.1)gwhen filled with a liquid. The mass of the liquid with possible limits of accuracy is |
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Answer» `(7.2 PM 0.2) G ` |
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| 49. |
A vessel contains 500g of water at 34^(@)C. What amount of ice at -10^(@)C should be added, so that the ice just melts? Ignore the heat capacity of the vessel. Specific heat capacity of ice and water are 0.5 cal//g^(@)C and1 cal//g^(@)C respectively and specific latent heat of fusion of ice is 80 cal/g. |
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