# AIEEE 2004 Solutions

date post

10-Apr-2015Category

## Documents

view

1.189download

2

Embed Size (px)

description

### Transcript of AIEEE 2004 Solutions

Kochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #1# www.mathiit.com1. From Newton's formula) z / v ( AFx Dimensions ofgradient velocity of ensions dim area of ensions dimforce of ensions dim ] T ML [] T ] L [] MLT [ 1 11 22 2. As given in question, retardation (negative acceleration) x a a = kxwhere k is a proportionality constant kxdtdv kxdtdx.dxdv kxdxdvv `

'|dtdxv Q v dv = kx dxIntegrating, we get fivvx0dx kx dv vwhere vi and vf respectively are initial and final velocities of particle.x02vv22xk2v fi

`

'|

`

'| 2xk2v2v 2 2i2f 2 2i2f mkx21mv21mv21 2initial final mkx21. E . K . E . K Hence, loss in kinetic energy x2PHYSICS&CHEMISTRY SOLUTIONSTodays Mathiitians..... Tomorrows IITians.Solutions of AIEEE 2004 PaperKochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #2# www.mathiit.com3. Second law of motion gives2gt21ut s + or 2gt210 h + ) 0 u ( Q

`

'|gh 2TAt sec,3Tt 23Tg210 x `

'|+ 9T. g21s2gh 218gs

`

'|gh 2T Q m9hs Hence, the position of ball from the groundm9h 89hh 4. ) A B ( ) B A ( r r r r (given) 0 ) A B ( ) B A ( r r r r r [ ] ) B A ( ) A B ( 0 ) B A ( ) B A ( r r r rQr r r r r + 0 ) B A ( 2 r r r 2AB sin = 0 sin = 0 [ ] 0 B | B | , 0 A | A | r rQ or 05. We know in advance that range of projectile is same for complementary angles i.e. for and(900 - ).gsin u 2T1gcos u 2g) 90 sin( u 2T02 Kochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #3# www.mathiit.comandg2 sin uR2Therefore, gcos u 2gsin u 2T T 2 122g) cos sin 2 ( u 2 22g) 2 (sin u 2 gR 2 R T T 2 1 6. For a particle moving in a circle with constant angular speed, velocity vector is always tangent tothe circle and the acceleration vector always points towards the centre of the circle or is always alongradius of the circle. Since, tangential vector is perpendicular to radial vector, therefore, velocityvector will be perpendicular to the acceleration vector. But in no case acceleration vector is tangentto the circle.7. Third equation of motion givesv2 = u2 + 2as 2u s (since v = 0)where a = retardation of body in both the cases 222121uuss ................. (i)Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get22 12060s20

`

'|226012020 s `

'| 4 20 = 80 m8. The force exerted by machine gun on man's hand in firing a bullet = change in momentum persecond on a bullet or rate of change of momentum1200100040 `

'| = 48 NThe force exerted by man on machine gun = 144 NHence, number of bullets fired = 348144Kochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #4# www.mathiit.com9. On releasing, the motion of the system will be according to the figure.m1g - T = m1a ...... (i)and T - m2g = m2a ...... (ii)On solving;gm mm ma2 12 1

`

'|+ ........ (iii)Here,m1 = 5 kg, m2 = 4.8 kg, g = 9.8 m/s2 8 . 98 . 4 58 . 4 5a `

'|+8 . 98 . 92 . 0 =0.2 m/s210. Mass per unit lengthLMm / kg 224 The mass of 0.6 m of chain = 0.62 = 1.2 kgThe centre of mass of hanging part m 3 . 020 6 . 0+Hence, work done in pulling the chain on the tableW = mgh= 1.2100.3= 3.6 J11. Let the mass of block be m.Frictional force in rest positionF = mg sin 30010 = m1021 kg 21010 2m 12. Work done in displacing the particler . F W rr) ji2 ( . ) k2 j3 i5 ( + + = 52+3(-1) + 20= 10 - 3= 7 J13. Let the constant acceleration of body of mass m is a.From equation of motionT T a am 1gm 2gFRm g sin300m g cos300m g300Kochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #5# www.mathiit.comv1 = 0 + at111tva ...... (i)At an instant t, the velocity v of the bodyv = 0 + atttvv11...... (ii)Therefore, instantaneous powerP = Fv= mav ) ma F ( Q

`

'|

`

'| ttvtvm1111[from equations (i) and (ii)]2121tt mv14. When a force of constant magnitude acts on velocity of particle perpendicularly, then there is nochange in the kinetic energy of particle. Hence, kinetic energy remains constant.15. In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere.Therefore, taking the same mass of sphere if radius is increased then moment of inertia, rotationalkinetic energy and angular velocity will change but according to law of conservation of momentum,angular momentum will not change.16. Man will catch the ball if the horizontal component of velocity becomes equal to the constant speedof man i.e.2vcos v 00 21cos 060 cos cos 060 17. Let same mass and same outer radii of solid sphere and hollow sphere are M and R respectively.The moment of inertia of solid sphere A about its diameter2A MR52I ..... (i)Similarly the moment of inertia of hollow sphere (spherical shell) B about its diameter2B MR32I ....... (ii)It is clear from eqs. (i) and (ii). IA < IB18. The gravitational force exerted on satellite at a height x is2eG) x R (m GMF+Kochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #6# www.mathiit.comwhere Me = mass of earthSince, gravitational force provides the necessary centripetal force, so,) x R (mv) x R (m GM 202e++ 20emv) x R (m GM+ 202mv) x R (m gR+ `

'|2eRGMg Q ]]]

+) x R (gRv20 2 / 12) x R (gR]]]

+19. Time period of satellitee3GM) h R (2 T + where R + h = orbital radius of satellite, Me = mass of earthThus, time period does not depend on mass of satellite.20. Gravitational potential energy of body on earth's surfaceRm GMU e At a height h from earth's surface, its value is) h R (m GMU eh+ R 2m GMe ) R h ( Qwhere Me = mass of earthm = mass of bodyR = radius of earth Gain in potential energy = Uh - U

`

'| Rm GMR 2m GM e e Rm GMR 2m GM e e+ R 2m GMeKochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #7# www.mathiit.comR 2m gR2 `

'|2eRGMg QmgR2121. The necessary centripetal force required for a planet to move round the sun = gravitational forceexerted on iti.e. ne2Rm GMRmv2 / 11 neRGMv `

'|Now,2 / 1e1 nGMRR 2vR 2T

`

'| 2 / 1e1 n 2GMR R2

`

'|

`

'| +2 / 1e2 / ) 1 n () GM (R2 2 / ) 1 n (R T +22. Work done in stretching the wire = potential energy storedvolume strain stress21 ALLlAF21 Fl2123. Retarding force acting on a ball falling into a viscous fluidRv 6 F where R = radius of ballv = velocity of balland = coefficient of viscosity R F and v F Or in words, retarding force is proportional to both R and v.24. The excess pressure inside the soap bubble is inversely proportional to radius of soap bubble i.e., r / 1 P r being the radius of bubble. It follows that pressure inside a smaller bubble is greater thanthat inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow fromsmaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.25. The time period of simple pendulum in airKochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Golf Links Road, Kowdiar Gardens, Housing Board Colony, Ph: 0471- 2438271, Mobile: 9387814438. #8# www.mathiit.com

`

'| gl2 t T 0 ...... (i)l, being the length of simple pendulum, In water, effective weight of bobw' = weight of bob in air - upthrust Vg ) ' ( Vg ' Vg g ' m mg Vgeff where = density of bob, = density of water g'geff

`

'| g'1

`

'| ]]]]]]

`

'| g'1l2 t...... (ii)Thus,]]]]]]

'11tt0

`

'|

`

'|3 441000 ) 3 / 4 (1000110t 2 t 2 26. Time period of spring

`

'| km2 Tk, being the force constant of spring.For first spring

`

'| 11km2 t...................... (i)For second spring

`

'| 22km2 t...................... (ii)Kochi Branch: Bldg.No.41/352, Mulloth Ambady lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094, 9388465944Trivandrum Branch: TC.5/1703/30, Go