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MELUHA INTERNATIONAL SCHOOL
INDIA SR MPC 29-03-2020
SECTION – I (SINGLE CORRECT ANSWER TYPE)
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
MATHS
SYLLABUS: Matrices (inverse and linear equations), Differentiation
1. If ,A B C then the value of the determinant
2
2
2
sin cot 1sin cot 1sin cot 1
A AD B B
C C is:
A) 1 B) -1 C) 0 D) None of these
2. A divisor of the determinant
2
2
2
11
1
a ab acD ba b bc
ca cb c
is:
A) 2 2 21 a b c B) 2 2 2a b c C) 2a b c D) ab bc ca
3. If , and are the roots of the equation 3 0,x px q then the value of the determinant
A) p B) q C) 2 2p q D) None of these
4. The determinant cos sin cos 2
sin cos sincos sin cos
is:
A) 0 B) Independent of C) Independent of D) Independent of and both
5. The determinant
cos cos coscos cos cossin sin sin
x y y z z xx y y z z xx y y z z x
is:
A) 2sin sin sinx y y z z x B) 2sin sin sinx y y z z x
C) 2cos cos cosx y y z z x D) 2cos cos cosx y y z z x
6. The determinant b c a a
b c a bc c a b
is:
A) 2abc B) 3abc C) 4abc D) None of these
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7. The set of homogeneous equations 1 1 0tx t y t z , 1 2 0t x ty t z ,
1 2 0t x t y tz has non-trivial solution for: A) Three values of t B) Two values of t C) One value of t D) No value of t
8. If , ,a b c are real then the value of determinant
2
2
2
11 1
1
a ab acab b bcac bc c
if:
A) 0a b c B) 1a b c C) 1a b c D) 0a b c
9. The equation
22 2 2
2
1 2 1 11 1 2
2 1 3 1 5 1 3 2 01 2 2 3 1 2 3 2 2 3
x x xx x x
x x x x x xx x x x x x
A) Has no real solution B) Has 4 real solutions C) Has two real and two non-real solutions D) Has infinite number of solutions, real or non-real 10. The solution of the following equations: 4 3 2 0,x k y z 2 1 1 0,kx k y k z 2 2 3 2 0k x ky k z For 1,k
, ,x y z is given by:
A) ,3 ,t t t B) 2 , 10 , 4t t t C) 9 , 5 , 7t t t D) 4 ,6 ,t t t
11. If 1sin cos2
x y then2
2
d ydx
at ,4 4
is
A) – 4 B) – 2 C) – 6 D) 0
12. If 21t x x and 2 2x t y then at 2, dyxdx
A) 245
B) 101125
C) 488125
D) 358125
13. If 2f x x and g x f f x then 3
0' 2 1
rg r
is equal to
A) 0 B) 1 C) – 1 D) 2
14. If 1yx
then 4 41 1
dy dxy x
A) 0 B) 1 C) x y D) y x
15. If 3
1 txt
and 2
3 42
tyt
then 3'x y
A) 1 'y B) 1 y C) ' 1y D) 'y 16. The value of '' 1y , if 3 2 22 5 5 0x x y x y when 1 1y is equal to
A) 227
B) 21728
C) 8 D) 22827
17. If ' 3 2f then 2 2
2
3 3lim
2x
f h f hh
is
A) 1 B) 2 C) 3 D) 1/2
18. If tan sec 1tan sec 1
x xf xx x
then 'f x is equal to
A) sec tan secx x x B) sec sec tanx x x C) sec sec tanx x x D) None of these
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19. If 1 1 costan1 cos
xyx
, then dy
dx is equal to
A) 22sec 2x B) 21 2 sec 2x C) 1 2 D) 21 2 sec 2x
20. 32
4 1
3 1 2 12
x for xf x x x x for x
then
A) f x is continuous at 1x and 4x B) f x is differentiable at 4x
C) f x is continuous and differentiable at 1x D) f x is only continuous at 1x
SECTION-II (Numerical Value Answer Type)
21. Let a b c
A p q rx y z
and suppose that det 2A then the det B equals, where
4 24 24 2
x a pB y b q
z c r
22. Let
2 2
2 2
2 2
1 sin cos 4sin 2sin 1 cos 4sin 2 ,sin cos 1 4sin 2
x x xf x x x x
x x x
then the maximum value of f x is:
23. Let
2
2
cos cos sin sincos sin sin cos
sin cos 0f
then
6f
is:
24. If 1 1 costan ,sin
xy xx
then dydx
25. Let 3 8 7y x x and x f t . If 2dydt
and 3 0x at t then the value of dxdt
at t = 0 _____
SECTION – I
(SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
PHYSICS
SYLLABUS: Rotational Motion, Gravitation 26. Moment of inertia of solid sphere about its diameter is I. If that sphere is recast into 8
identical small spheres, then the moment of inertia of small sphere about its diameter is:
A) 8I B)
16I C)
24I D)
32I
27. Three particles each of mass ‘m’ are arranged at the corner of an equilateral triangle of side ‘L’. If one of the masses is doubled, the shift in the centre of mass of the system
A) 3
L B) 4 3
L C) 34
L D) 2 3
L
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28. A uniform thin bar of mass 6m and length 12L is bent to make a regular hexagon. Its moment of inertia about an axis passing through centre of mass and perpendicular to the plane of the hexagon is:
A) 220 mL B) 26 mL C) 2125
mL D) 230 mL
29. A thin rod of length L and mass M is held vertically with one end on the floor and is allowed to fall. Find the velocity of the other end when it hits the floor, assuming that the end on the floor does not slip
A) 3gL
B) 3gL C) 3Lg
D) 3gL
30. A cylinder of mass M, radius R is resting on a horizontal platform (which is parallel to XY plane) with its axis fixed along the Y axis and free to rotate about its axis. The platform is given a motion in X - direction given by cosx A t . There is no slipping between cylinder and platform. The maximum torque acting on the cylinder during its motion is:
A) 212
MRA B) 2MRA C) 22MRA D) 22 cosMRA wt
31. Find the gravitational attraction on mass m due to the quarter ring of same mass m and Radius ‘R’ as shown in figure.
A)
2
2
2 2GmR
B)
2
22Gm
R C)
2
2Gm
R D)
2
2
2GmR
32. If diameter of a planet is four times that of earth. What is the time period of a pendulum on that planet, if it is a second pendulum on earth. Take mean density of the planet equal to that of earth.
A) 4 sec B) 1 sec C) 2 sec D) 3 sec 33. Two particles of masses 1m and 2m are initially at rest at infinite separation. When released,
what is their relative velocity of approach when they are at a separation d.
A)
1 2( )2
G m md
B)
1 2( )G m md
C)
1 22 ( )G m md
D)
1 22
2 ( )G m md
34. Figure shows a fixed solid sphere of mass M and radius R with a narrow smooth tunnel along its diameter. Another particle of mass m is placed at point A as shown. If the particle is released from rest, then its speed when it reaches the center of sphere is
A)
GMR
B)
2GMR
C) 2
GMR
D) 2 GM
R
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35. Two satellites A and B of same mass are orbiting earth at altitudes R and 3R respectively, where R is Radius. The earth taking their orbits to be circular, find the ratio of their kinetic energies.
A) 1 : 3 B) 1 : 1 C) 2 : 1 D) 3 : 1 36. A Satellite of mass 32 10 kg is to be shifted from on orbit of radius 2R to another orbit of
radius 3R. Calculate the minimum energy required for this. [R is the Radius of earth] 210g ms A)
101.066 10 J B) 91.066 10 J C)
102.132 10 J D) 92.132 10 J
37. An artificial satellite is moving in a circular orbit around earth with a speed equal to half the escape Velocity from earth surface. Find the height of satellite above the earth surface.
A) 3200 Km B) 6400 Km C) 10800 Km D) 19600 Km 38. The radius of a planet is 1R and a satellite revolves around it in a circular orbit of
radius 2R .The period of revolution is T. What is the acceleration due to gravity on the surface of the planet?
A)
2 31
2 22
4 RR T
B)
2 31
1
RR T
C)
2 32
2 21
4 RR T
D)
2 31
2 22
RR T
39. A satellite revolves around a planet in an elliptical orbit. Its perigee and apogee are 70.5 10 m and 71.5 10 m respectively. If the minimum speed of satellite in orbit is 35 10 m s , then its maximum speed in orbit is
A) 415 10 m s B) 35 10
3m s C) 41.5 10 m s D) 45 10
3m s
40. The change in the value of ‘g’ at a height ‘h’ above the surface of earth is same as at a depth ‘d’ below the surface of earth. When both ‘d’ and ‘h’ are much smaller than the radius of earth, then which one of the following is correct?
A) 2
hd
B)
32hd
C) 2d h D) d h
41. The distance of the centres of moon and earth is d. The mass of earth is 81 times the mass of the moon. At what distance from the centre of the earth, the gravitional field will be zero.
A) 2d
B)
23d
C)
43d
D)
910
d
42. A point P lies on the axis of a ring of mass M and radius R at a distance R from its centre. A small particle starts from p and reaches the centre under gravitational attraction only. It’s speed at center will be
A)
2GMR
B)
2 112
GMR
C) Zero D)
GMR
43. The gravitational field in a region is given by (4 )E i j N Kg . Work done by this field is Zero when a particle is moved along the line is
A) 4 2y x B) 4 6y x C) 5x y D) All of these 44. Two masses 90 Kg and 160 Kg are at a distance 5 m apart. Find the magnitude of intensity of
the gravitational field at a point which is at a distance 3 m from 90 Kg and 4 m from 160 Kg mass.
A) 110 2 G N Kg B) 110 G N Kg C) 17 G N Kg D) 12 G N Kg 45. Find the height from the earth’s surface where acceleration due to gravity is 25% of it’s value
on the surface of earth. [R=6400Km] A) 3200 Km B) 6400 Km C) 10800 Km D) 19600 Km
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SECTION- II (Numerical Value Answer Type)
46. Two kids weighing 10 kg and 15 kg respectively, are trying to balance a see saw of total length
5.00 m, with it’s fulcrum at the centre. If one of the kids is sitting at an end, then what is the distance in meter from the fulcrum to the other kid _____
47. A small hole is made in a disc of mass M and radius R at a distance4R from center. The is
supported on supported on a horizontal peg through this hole. The moment of inertia of the
disc about the peg is 9a
.The value of a is _____
48. A force of 490 N acts tangentially at the highest point of a sphere of mass 5 kg kept on a horizontal plane. If the sphere rolls without slipping, then the acceleration of center of sphere in 2/m s is _____
49. A solid sphere is projected up along an inclined plane of inclination 030 with a speed 12V ms .If it rolls without slipping, then the maximum distance traversed by it in m is 2( 10 )g ms
50. A child is standing with folded hands at the center of a platform rotating about is central axis. The K.E of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The K.E of the system is now BK. The value of B is _____
SECTION – I (SINGLE CORRECT ANSWER TYPE)
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
CHEMISTRY
SYLLABUS: I-A, II-A, III-A &IV-A Groups, Junior organic Chemistry (GOC) and Alkanes 51. Find out ‘Z’ in the following sequence of reaction Maisture in airAir CuNa x y z 1) 2 3Na CO 2) NaOH 3) 2 2Na O 4) 2Na O 52. Aluminum reacts with NaOH and forms compound ‘X’ If the co-ordination number of Al in
‘X’ is 6,the correct formula, of X is 1) 2 4 2
Al H O OH
2) 2 3 3Al H O OH 3) 2 2 4
Al H O OH
4) 2 6 3Al H O OH
53. Amongst 2, ,LiCl RbCl BeCl and 2MgCl the compounds with the greatest and the least ionic character, respectively are :
1) LiCl and RbCl 2) 2RbCl and BeCl 3) 2 2MgCl and BeCl 4) 2RbCl and MgCl 54. Magnesium burns in air giving two products. Their forumulae are 1) ,MgO MgN 2) 2 3 2,MgO Mg N 3) 3 2,MgO Mg N 4) 2 3 2,Mg O Mg N 55. A piece of magnesium is burnt in air to give ash. Then ash is dissolved in water to liberated a
colorless gas, which turns red litmus to blue.The colorless gas is 1) 3NH 2) 2NO 3) 2N 4) NO 56. Sedimentary rocks laid down under water mainly contain 1) CaO 2) 2Ca OH 3) 3CaCO 4) 4CaSO
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57. The property not associated with refractory nature of Mg is 1) it has high m.p 2) it is a good conductor of heat 3) it is chemically inert as well as electrical insulator 4) forms Grignard’s reagent 58. 3 4 3BCl LiAIH A LiCI AICI Re
2 2, .d heatA H O B H B C In this raction sequence A,B and C compounds respectively are
1) 2 6 2 2, ,B H B O B 2) 2 6 3 3 2 3, ,B H H BO B O 3) 2 6 3 3, ,B H H BO B 4) 4, 3 3 2 3,HBF H BO B O 59. Which amongst the following is also called as a sesqui oxide 1) 2 3B O 2) 2 3Al O 3) 2 3Tl O 4) all 60. Which of the following is a lewis acid 1) 3AlCl 2) 2MgCl 3) 2CaCl 4) 2BaCl 61. Silica reacts with hydride of super halogen to form.’X’ on hydrolysis of X another compound
Y on heating to form Z.The ‘Z’ can also be prepared in the following reaction. 1) 2 4 24 2SiO HF SiF H O 2) 2 2Si O SiO 3) 2 6 4 6K SiF K KF Si 4) 2 2 3 22 2Si NaOH H O Na SiO H O 62. Which glass has the highest percentage of lead? 1) Soda glass 2) Flint glass 3) Jena glass 4) Payrex glass 63. Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of HP is 6. If
ionisation constant of HA is 510 ,the ratio of salt to acid concentration in the buffer solution will be
1) 5:4 2) 1:10 3) 4:5 4) 10:1 64. Decreasing order of ‘P’ Character in the following A : 2 2; : ; :SiO B CO C Graphite 1) A B C 2) B A C 3) B C A 4) A C B 65. The difference in properties of 4CH and 4SiH is due to 1) Large difference in the electronegativity of carbon and silicon 2) Less difference in the size of carbon and Silicon atoms 3) The inability of carbon to expand its octet 4) The inability of silicon to form double bonds 66. The IUPAC name of the alkane is
is 1) 2- Isopropyl-2, 6, 6-trimethylheptane 2) 5 tert-butyl-2isopropyl-2-methypentane 3) 2, 3, 3, 7, 7-Pentamethyloctane 4) 2, 2, 6, 6, 7-Pentamethyloctane
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67. The correct Fischer projection formula of (2R,3R)-2,3-dihydroxy butanoic acid is
68. Among the following, the true property
about
3CH
3C CH
3CH
is
1) non-planar 2) C is sp2-hybridized 3) electrophile can attack C 4) does not undergo hydrolysis 69. Select the most stable carbocation from amongst the following
70. Polarization of electrons in acrolein may be written as
1) 2C H CH C H O
2) 2C H CH C H O
3) 2C H CH C H O
4) 2C H CH C H O
NUMERICAL VALUE QUESTIONS
71. How many of the following elements of I A group as lighter than water (i) Cs (ii) Na (iii) K (iv) Li
72. How many of the following will not give any colour in flame (i) Be (ii) Mg (iii) Na (iv) Li 73. Magnesium oxide when mixed with a saturated solution of magnisium chloride sets to a hard
mass like cement known as “sorel cement”. The composition of Sorel cement is . What is the value of n___
74. On heating calcium ammoniate, ammonia and hydrogen are evolved. How many moles of ammonia are evolved when 1.5 moles of calcium ammoniate are heated___
75. The ratio of copper and tin in bell metal alloy is ______.
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MELUHA INTERNATIONAL SCHOOL
INDIA SR MPC 29-03-2020
KEY SHEET
MATHS
1) 3 2) 1 3) 4 4) 2 5) 2 6) 3 7) 3 8) 4 9) 4 10) 3
11) 1 12) 3 13) 1 14) 1 15) 2 16) 4 17) 1 18) 3 19) 2 20) 1
21) -16 22) 6 23) 1 24) 0.5 25) 0.11
PHYSICS
26) 4 27) 2 28) 1 29) 2 30) 1 31) 1 32) 2 33) 3 34) 2 35) 3 36) 1 37) 2 38) 3 39) 3 40) 3 41) 4 42) 2 43) 1 44) 1 45) 2 46) 1.67 47) 16 48) 140 49) 0.56 50) 0.5
CHEMISTRY
51) 1 52) 3 53) 2 54) 3 55) 1 56) 3 57) 4 58) 2 59) 4 60) 1 61) 2 62) 2 63) 4 64) 4 65) 1 66) 4 67) 2 68) 2 69) 2 70) 4 71) 3 72) 2 73) 5 74) 8 75) 4
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MELUHA INTERNATIONAL SCHOOL
INDIA SR MPC 29-03-2020
HINTS & SOLUTIONS MATHS
1. Use: 3 1 2R R R and 2 2 3R R R 2. Multiply 1R 2y ,a 2R 2y b and 3R 2y c and divide the determinant 2y .abc Now take ,a b and c
common from 1 2,c c and 3.c Now use 1 1 2 3C C C C to get 2 2 2 2 2 2
2 2 2
1 1 11 1
1a b c b b b
c c c
Now use: 1 1 2C C C and 2 2 3C C C to get the value as 1. 3. 3 3 3 3D Suppose given 0 As coefficient of 2 0x
Now, 1 1 1
det 0
4. Directly open 2y 1R to get 2 2cos sin cos 2 1 cos 2 . Which is independent of 5. Expand the determinant using first row and use ,x y A y z B and z x C 0A B C 6. Put 1.a b c 7. D simplifies to 4 2 1 0t 8. Multiply 1R 2y a, 2R 2y 2 and 3R 2y c and divide the determinant 2y .abc Now take ,a b and c
common from 1 2,c c and 3.c Now use 1 1 2 3C C C C to get:
2 2 2 2 2 2
2 2 2
1 1 11 1 1
1a b c b b b
c c c
Now use 1 1 2C C C and 2 2 3C C C We get 2 2 21 1a b c 0a b c 9. 1st two columns of 1st determinant are same as 1st two rows of 2nd. Hence transpose the 2nd. Add the
two determinants and use 1 1 3 0C C C D 10. Put 1k and solve
11. 1sin .cos2
x y
cos .cos sin .sin 0dyx y y xdx
cot .cotdy x ydx
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2
2 22 cos .cot cos .cot .d y dyec x y ec y x
dx dx
Now ,
4 4
1dydx
2
2,
4 4
2 1 2 1 1 4d ydx
12. Given
2
22 2
11 1
x dt xtx dx x
32 2
2 24
1 12
1x xdyx t y x
dx x
32 2
32
2 1 1
1
x x xdydx x
13. , 2
2 2 4 , 2 44, 4
x xg x x x x
x x
' 1 ' 1 ' 3 ' 5 1 1 1 1 0g g g g
14. 21dy xdx
4
2 2 24
1 1 1 1 01
ydy dydx dx x x xx
15. dy dtdy
dx dx dt
16. Differentiating the given expression, we get 2 2 23 4 4 . ' 5 ' 0x xy x y y y at 1,x we have 3 4.1.1 4.1.1. ' 5 ' 1 0y y
4 3 ' 1 0 ' 1 4 3y y
Differentiating again, we have 22 2 26 4 8 . ' 8 . ' 4 ' 4 . '' '' 0x y xy y xy y x y x y y y
Putting 1, 1 ' 1 4 3x y and y , we get 4 4 166 4 8 8 4 3 '' 1 03 3 9
y
22'' 1 8.7
y
17. 2 2
20
3 3lim
2y
f h f hh
2
20
3 31 lim2 h
f h fh
1 ' 3 2 2 12
f
18.
2 2
2 2
tan sec 2 tan sec 11 tan sec
x x xf xx x
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2 2tan sec 2 tan sec 1 tan sec
2 tanx x x x x
x
Thus 2' sec sec tan sec sec tanf x x x x x x x
19. 2
1 12
2sin 2tan tan tan 2 22cos 2
xy x xx
Hence 12
dydx
20. Since g x x is continuous function and 1 1
lim 3 limh h
f x f x
so far f is a continuous function.
In a particular f is continuous at 1x and 4.x f is clearly not differentia2le at 4x . Since g x x is not differentia2le at 0x . Now
0
1 1' 1 lim
h
f h ff
h
0
3 3lim 1h
hh
3 2
0
1 2 1 1 3 1 1 2 3' 1 lim
h
h h hf
h
3 2 2
0
1 2 1 1 2 3' 1 lim 5 2
h
h h h h hf
h
Hence f is not differentia2le at 1x
21. 4 2
det 4 2 4 2 14 2
x a p x a pB y b q y b q
z c r z c r
8 8x y z a b ca b c p q rp q r x y z
8 2 16 22. Use: 1 1 2R R R and 2 2 3R R R and expand to get 2 4sin 2 .f x x
23. Put 6 and find determinant.
24. 1 1 costansin
xyx
2
12sin
2tan2sin cos
2 2
x
y x x
1tan tan2xy
2 2 2 2x xy x or
25. 3 8 7y x x
23 8dy xdx
0 3t x
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23 3 8 19dydx
2 0.105319
dydy dt
dxdxdt
PHYSICS26. 2I MR 3M R 1 3R M
1 3
2 2
1 1
R MR M
2
2 2 2
1 1 1
I M RI M R
2 3
2 2 2
1 1 1
I M MI M M
12 32
II
27.
1 1 2 2 3 3
1 2 3cm
m x m x m xxm m m
And also cmy If one of masses dou2led find ' , 'cm cmx y
2 2' 'cm cm cm cmr x x y y
28. 6 onesideI I
222
612
m Lmr
Where 3r l 220I mL
29. 212 2LMg I
2
212 2 3L MLMg
3gL
V r 3V gL 30. cosx A t
sindxV A tdt
2 cosa A t 2
maxa A max maxT I
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maxaIR
2
212
AMRR
2max
12
T MAR
31. Mass of element of ring
2mdm d
Force on ‘m’ due to dm
2
2
2
2
GmdmdFR
Gm ddFR
Net force on m will 2e
4
4
cosF dF
2
2
2 2GmFR
32. 3
2 2
43
G RGMgR R
43
g G R
g R 1 2 2
2 1 1
1
2 1
2
1
2 4
1sec
Tg
T g RT g R
RT RT ond
33. At separation ’d’
2 2 1 21 1 1 1
1 12 2
Gm mm v m vd
From law of conservation of linear momentum 1 1 2 2m v m v From (1) and (2)
22
11 2
2( )
GMVd m m
and
2
12
1 2
2( )
GMVd m m
Relative velocity 1 2relV V V
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34. potential due to sphere at point ,2AGMA V
R
Potential due to sphere at point 3,2c
GMC VR
Work done (1)A CW V V
21 (2)2
W mv
From (1) and (2) 2GMVR
35. Or2ital speed, GMV
R h
Kinetic energy = 212
mv
2
1 1 2
1 2 1
3 4 22 1
KE V R h R R RKE V R h R R R
36. Total energy of a satellite, 2
GMmEr
Work done in shifting, f iW E E
3 3
10
1 12 2 3
1210 6400 10 2 10
121.066 10
GMmWR
gRmW
W
W J
37. Or2it speed 0GMVR h
Given 01 1 22 2e
GMV VR
1 22
1 12
6400
GM GMR h R
R h Rh r Km
38. Time period, 2 3
2 24 RTGM
Acceleration due to gravity
21
GMgR
2ut
2 32
2
2 32
2 21
4
4
RGMT
RgR T
39. From conservation of angular momentum
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a pL L 1 1 2 2mrv mr v
7 3
2 21 7
1
1.5 10 5 100.5 10
r vVr
41.5 10 /m s 40. For smaller heights,
21nhg g
R
Change in g, 1 ng g g
= 2hR
At depth d,
1ddg gR
Change in g, 2g g gd
= dR
Given
1 2
2
2
g gh d
R Rd h
41.
22
22
81
9 1
910
e m
m m
GM GMx d xM Mx d x
x d xdx
42. Potential at a distance R
2 2 2p
GM GMVRR R
Potential at the centre
pGMV
R
Work done (1)p CW V V m
And 21 (2)2
W mv
From (1) and (2), 2 112
GMVR
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43. Work done on a particle is zero, if force and displacement are perpendicular. 44. Due to 90 Kg, at’p’
1 2
(90) 103
GE G 90kg 5m 160 kg
Due to 160 Kg, 3 m 4 m
2 2
(160) 104
GE G
Resultant field,
2 2 01 2 1 2
1
2 cos90
10 2
E E E E E
E GNKg
45. At a height ‘h’
2
1n
gghR
; 4hgg
2
4 1
2 1
6400
hR
hR
h R Km
46. In rotational equili2rium Torque due to One kid a2out fulcrum = Torque due to other kid a2out fulcrum
1 1 2 2
10 2.5 152.5 2 5 1.67
3 3
m gr m grx
x m
47. From parallel axes theorem
20
2 2
2 2
;4
2 169 9
16
RI I Mr r
MR MRI
MR MRIa
48. acceleration, 107
FaF
2
10 4907 5
140ms
49. From conservation of energy
M E L U H A I N T E R N A T I O N A L S C H O O L
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22
2
22
2
1 12
21 4 175
2 20 5
sin1
Kmv mghR
KVR
hg
h
Distance,
0
sin
sin 307 225 114250.56
hl
h
m
50. Angular momentum is constantss
2
1 2 1
1 1 1
12
1
.21.
2
20.5
LK EI
K EI
KE I IKE I I
KEKE
KE