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ASTRONOMY NSEA 2016-2017 EXAMINATION
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INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD EXAMINATION IN ASTRONOMY 2016-2017
Date of Examination 27th November 2016 Time 120 min.
[Ques. Paper Code : A422]
1. If 31xcos2 then cosec x =
(a) 3 (b) 3
2 (c) 32 (d)
23
Ans. [d]
Sol. 31xcos2
32xsin 2
32xsin
23ecxcos
2. Which of digits 1, 3, 4, 5, 7, 0 cannot appear at the ten's place of powers of 3 ? (a) 1, 3, 5 only (b) 1, 3, 4, 7 only (c) 1, 3, 5, 7 only (d) 0, 1, 3, 5 only Ans. [c] Sol. Clearly 1, 3, 5, 7 cannot appear at the ten's place of powers of 3. 3. A fish is swimming in still water. At a given instant there is a bird flying vertically above the fish. For the
bird the fish appears to be 15 m below the surface of water while for the fish the bird appears to be 20 m above the surface. Refractive index of water is 4/3. Actual distance between the bird and the fish is -
(a) 35 m (b) 40 m (c) 30 m (d) 25 m Ans. [a] Sol.
fish
y
x
bird
m15y
x = 20
y = 15
20x
3415y
3/420x
y = 20m x = 15m
Actual distance = x + y = 15 +20 = 35 m
Marks : 240
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4. The smallest integer n such that 01.0n1n is -
(a) 2499 (b) 2500 (c) 2501 (d) 2502
Ans. [b]
Sol. 100
1n1n
n100
11n
100
n2n10000
11n
n21009999
299.99n
n 2499.5
n = 2500
smallest
5. A uniform wire of resistance per unit length 1/m is bent in the form of an equilateral triangle. If effective
resistance between adjacent vertices is 2.4, length of each side of the triangle is-
(a) 1.8 m (b) 2.4 m (c) 3.6 m (d) 7.2 m
Ans. [c]
Sol.
R =
R =
R =
A B
3R2R AB
3
24.2
3
)1(24.2
= 3.6 m
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6. ABC is equilateral with each side being of unit length and P is an interior point, then the maximum product of the lengths AP, BP and CP is -
(a) 35
1 (b) 34
1 (c) 33
1 (d) 61
Ans. [c]
Sol. 31
)CPBPAP(3
CPBPAP
So max value of AP BP CP is 3
3CPBPAP
It happens when 3
160sin2
1RCPBPAP
7. The resultant of the forces P and Q is R. If Q is doubled then R gets doubled. If Q is reversed, even then R gets doubled. Then -
(a) 2:3:2:R:Q:P (b) 3:2:2:R:Q:P
(c) 2:3:3:R:Q:P (d) 3:3:2:R:Q:P
Ans. [a]
Sol. cosPQ2QPR 22 ..... (i)
cosPQ4Q4PR2 22
cosPQ2QPR2 22 ..... (ii)
For (1) & (2)
5R2 = 2P2 + 2Q2
P : Q : R
2:3:2
8. The unit's digit of 232015 × 72016 × 132017 is - (a) 1 (b) 3 (c) 7 (d) 9 Ans. [a] Sol. unit place of (23)2015 = 7
unit place of (7)2016 = 1
unit place of (13)2017 = 3
So unit place of product = 1
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9. A block of mass 5 kg is to be dragged along a rough horizontal surface having s = 0.5 and k = 0.3. The horizontal force applied for dragging it is 20 N. (g = 10 m/s2). Select correct statement/s.
(a) Frictional force acting on the block is 20 N (b) Block will be displaced (c) Block will move with acceleration 1m/s2 (d) Block will initially move and then stop. Ans. [a] Sol.
5kg
F= 20N
s = 0.5 k = 0.3 Limiting Fmax = s mg
Fmax = 0.5 × 5 × 10
Fmax = 25 N
a = 0 (F < Fmax )
Fr = F
Frictional Force Fr = 20 N
10. The negation of statement: ' f(x) is continuous for all real numbers x. ' is (a) f(x) is not continuous for all real numbers x (b) f(x) is not continuous for any real number x (c) f(x) is not continuous for every real number x (d) f(x) is not continuous for some real number x Ans. [b] Sol. Negation of the statement is
f(x) is not continuous for any real number x
11. A bullet moving with a speed of 72 m/s comes to a half in a fixed wooden block on travelling 9 cm inside it. If the wooden block (of the same type of wood) were to be 8 cm thick, the bullet would come out of the block with a speed.
(a) 9 ms–1 (b) 8 ms–1 (c) 24 ms–1 (d) 64 ms–1 Ans. [c] Sol. Assuming constant resistive force
as2uv 22 0 = (72)2 – 2 (a) (x1) (x1 = 9 cm)
1
2
x2)72(a
v2 = u2 +2as v2 = (72)2 – 2a (x2)
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v2 = (72)2 – )x(x)72(
21
2
v2 = (72)2
1
2
xx1
v2 = (72)2
981
v2 = (72)2
91
243
72v m/s
12. Let l be a vertical line and m a line that makes an angle of 6 with l. Consider the cone generated by rotating
m around the axis l. If plane L makes an angle of 15° with line l then the intersection of the plane and the
cone is –
(a) A parabola (b) A pair of straight lines
(c) An ellipse (d) A hyperbola
Ans. [d]
Sol.
30° m
L
Hyperbola
15°
30° m
L
Pair of straight line
15°
13. A piece of brass (an alloy of copper and zinc) weight 12.9 g in air. When completely immersed in water,
it weighs 11.3 g. What is the mass of copper contained in the alloy ? The density of copper and zinc are
8.9 g/cm3 and 7.1 g/cm3 respectively.
(a) 6.89 g (b) 4.54 g (c) 8.93 g (d) 7.61 g
Ans. [d]
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Sol. Let the mass of copper contained in alloy = x,
c = density of copper, zn = density of zinc, = density of water = 1 gm/cm3
(mg – vg) = m'g
(m– v) = m'
v = m – m'
v = (12.9 – 11.3)
v = 1.6 gm
6.1)x9.12(x
znc
6.1)x9.12(xznc
6.11.7
)x9.12(9.8
x
1.79.126.1
1.71
9.81x
1.7
54.11.79.89.81.7x
8.1
54.19.8x
x = 7.61 gm
14. The coefficient of x in the expansion of (1 + x)5 correspond to the - (a) 5th row of Pascal's triangle (b) 6th row of Pascal's triangle (c) 7th row of Pascal's triangle (d) 4th row of Pascal's triangle Ans. [c] Sol. Coefficient is 5C1 which corresponds to 6th row of Pascal's triangle.
15. Linked question (15-16) The adjacent figure shows a ramp (30°) holding a block of mass 50 kg. The block attached to a movable
pulley A with an inextensible massless string. The movable pulley is in turn held with the help of another fixed pulley B. The block kept on the ramp is to be raised through a vertical height of 10 cm. By what distance the string should be lowered down vertically, below E ?
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30°
Block of mass 50 kg
B
E A
(a) 20 cm (b) 5 cm (c) 40 cm (d) 10(31/2) cm Ans. [c] Sol.
=30°
50 kg
B
F A
F F E
T
m = x
10cm mgsin30° =30°
T
x sin 30 = 10 cm, x = 20 cm
xE = 2x
= 40 cm
16. Refer to figure in question 15, pulleys in the figure are massless and frictionless. Neglecting friction between
block and the ramp, the force that should be applied vertically downwards, at E, to side the block along the ramp without acceleration is (g = 10 ms–1)
(a) 65 N (b) 125 N (c) 175 N (d) 250 N Ans. [b]
Sol. mg sin 30° = T 2
mgT
T = 2F 2TF
4
10504
mgF
F = 125 N
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17. The figure shows a particular position of a Vernier callipers. The value of x in cm is -
3
Main scale (cm) x 4 5
0 10
Vernier Scale (a) 0.03 (b) 0.15 (c) 3.83 (d) 0.02 Ans. [a] Sol. Position of 2nd VSD = Reading + 2 × VSD
= (3.6 + 0.01 × 5) + 2 × 0.09
= 3.83
x = 3.83 – 3.8 cm
x = 0.03 cm
18. If A and B are two sets then the set A × B is given by - (a) {a × b | a A, b B} (b) {(a, b) | a B, b A} (c) {(a, b) | a A, b B} (d) {ab | a A, b B} Ans. [c] Sol. A × B = {(a, b) | a A, bB}
19. A lens is held directly above a pencil lying on a floor and forms an image of it. On moving the lens vertically through a distance equal to its focal length, it again forms image of same size as that of the previous image. If the length of the pencil is 5.0 cm, the length of the image is
(a) 10.0 cm (b) 15.0 cm (c) 20.0 cm (d) 12.5 cm Ans. [a] Sol.
x
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f1
u1
v1
fu
fuv
Ist virtual image
uvm
xf
fafu
fm
afx
IInd image real m = –a
)fx(f
fa
xfa
1a
aa
a11
1a
aff
fa
a – 1 = 1 a = 2
52H|a|HH
io
i Hi = 10 cm
20. If A = {2, 3}, B = {4, 5} then A × B = (a) {8, 15} (b) {8, 10, 12, 15} (c) {(2, 4), (3, 5)} (d) {(2, 4), (2, 5), (3, 4), (3, 5)} Ans. [d] Sol. A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}
21. A balloon less dense than air is tied at the floor of a truck with a massless, inextensible and flexible string. The truck is observed to be taking a left turn. Select correct statement.
(a) The string will incline towards right (outward, w.r.t. person in the truck) (b) The string will incline towards left (inward, w.r.t. person in the truck) (c) The string will still be vertical as the balloon is less dense than air (d) Buoyant force on the balloon is equal to weight of the balloon Ans. [a] Sol.
r
B
mg Centrifugal
wrt truck
Wrt person in truck the string will incline towards right-outward.
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22. A function F from A to B is - (a) Relation F with (a, b), (c, b) F a = c (b) F A × B
(c) Relation F with (a, b), (a, c) F b = c (d) Relation F with (a, b), (b, c) F (a, c) F Ans. [c] Sol. As function is a relation from A to B
such that every element should have a unique image
so if (a, b), (a, c) F b = c
23. An ice cube with a steel ball bearing trapped inside it is floating above water in a glass. What will happen to the water level in the glass after the ice melts completely ?
(a) Rise (b) Fall (c) Will not change (d) Answer depends upon actual position of the steel ball Ans. [b] Sol.
v1
v2
When ice melts
ice v1 = vwater
vwater =
1icev
vsub g = (m1g + m2g)
vsub = )vv( 2ball1ice
2ball
1ice
sub vvv
Net sub0 vvv1
2ball
1ice
0 vvvv1
2water0 vvvv2
= 21ice
0 vvv
12
vv (level fall)
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24. Which of the following is a mathematically acceptable statement ? (a) It is an even number (b) 13th December is Saturday (c) Common donkey belongs to class orthopoda (d) Alexander was a great king Ans. [c] Sol. Common donkey belongs to class orthopoda.
25. Linked questions (25-29) A star can be considered as a spherical ball of hot gas of radius R. Inside the star, the density of the gas is r
at a radius r and mass of the gas within this region is Mr. The correct differential equation for variation of mass with respect to radius is (refer to the adjacent figure) –
Mr
r+dr
r
dMr
(a) 3r
r r34
drdM
(b) 2r
r r4dr
dM (c) 2
rr r
32
drdM
(d) 3r
r r31
drdM
Ans. [b] Sol.
dr
r
Volume of element = drr4 2
Mass of element = drr4dm r2
r2r4
drdm
26. A star in its prime age is said to be under equilibrium due to gravitational pull and outward radiation pressure (P). Consider the shell of thickness dr as in the figure of question (25). If the pressure on this shell is dP then the correct equation is (G is universal gravitational constant)
(a) r2r
rGM
drdP
(b) r2r
rGM
drdP
(c) r2r
rGM
32
drdP
(d) r2r
rGM
32
drdP
Ans. [a] Sol.
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dr
r
P P+dP
Excess force due to pressure = dP × 4r2
force due to gravitation = 4rr2 dr × 2r
rGM
2r
2r2
rdrGMr4r4dP
r2r
rGM
drdP
as r pressure decrease
r2r
rGM
drdP
27. In astronomy, order of magnitude estimation plays an important role. The derivative drdP can be taken as
difference ratio rP . Consider the star has a radius R, pressure at its centre is Pc and pressure at outer most
layer is zero. If the average mass is 2
M0 and average radius 2
R 0 then the expression for Pc is-
(a) 40
20
c RGM
23P
(b) 40
20
c RGM
43P
(c) 40
20
c RGM
32P
(d) 40
20
c RGM
23P
Ans. [a] Sol.
r
dr
mass of element (dm)
dF = (dm)E
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drr4R
34
MR
GMrdF 2
33
rdrR4
GM3r4
dF R
06
2
2
2
RGMR43P
22
6c
4
2
RGM
83P
40
20
c
2R4
GM83P
2R
R,2
MMputting 00
40
20
c RGM
23P
28. The value of mass and radius of sun are given by M0 = 2 × 1030 kg and R0 = 7 × 105 km respectively. The pressure at the centre is about (G = 6.67 × 10–11 m3 kg–1 s–2)
(a) 2 × 1014 N m–2 (b) 2 × 1015 N m–2 (c) 5 × 1014 N m–2 (d) 7 × 1015 N m–2 Ans. [c]
Sol. 40
20
c RGM
23P
Put M0 = 2 × 1030 kg, R0 = 7 × 105 km
Pc = 5 × 1014 N m–2
29. Assuming that the gas inside the sun behaves very much like the perfect gas, the temperature at the centre of
the sun is nearly (the number density of gas particles = HM
2 , Boltzmann constant kB = 1.4 × 10–23 JK–1 and
mass of proton MH = 1.67 × 10–27 Kg) (a) 3 × 107 K (b) 2 × 107 K (c) 4 × 107 K (d) 6 × 107 K Ans. [b]
Sol. KTVNP
30
0
R34M
P
HMP2
VN
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KTMR
34
2MP
H30
0
0
H30
KM2PM
3R4T
0
H30
KMPM
3R2T
T = 3023
142738
102104.11051067.1)107(14.3
32
T = 2×107 kelvin
30. At the earth's equator a satellite is observed passing directly overhead moving west to east in the sky. Exactly 12 hours later, satellite is again observed directly overhead. The altitude of the satellite is (Radius of earth = 6400 km)
(a) 1.82 × 107 m (b) 1.39 × 107 m (c) 3.59 × 107 m (d) 6.4 × 107 m Ans. [b] Sol. (WS – WE)t = 2
212
T1
T12
ES
121
241
Ts1
TS = 8 hr
Time period = TS = GM
r2 23
= 2
23
gR
r2
2
23
SgR
)hR(2T
Putting all value
h = 1.39 × 107 meter
31. Passage 31 to 33 Two stars, with masses M1 and M2 are in circular orbits around their common centre of mass. The star with
mass M1 has an orbit of radius R1 and the star with mass M2 has an orbit of radius R2. The correct relation is -
(a) 1
2
2
1
MM
RR
(b) 2
1
2
1
MM
RR
(c) 1
2
2
1
MM
RR
(d) 2
1
2
1
MM
RR
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Ans. [a] Sol.
M1 M2 CM
R1 R2
Summation of mass moment about centre of mass is equal to zero
2211 RMRM
1
2
2
1
RR
MM
or 1
2
2
1
MM
RR
32. The time period of each of the star is -
(a) 2
22
212
2
GMR)RR(4T
(b) )MM(G)RR(4T
21
321
22
(c) 1
22
212
2
GMR)RR(4T
(d) 1
12
212
2
GMR)RR(4T
Ans. [c] Sol.
M1 M2 R2R1
CM
1
211
221
21
RVM
)RR(MGM
221
211 )RR(
MGRV
1
1
VR2T
221
21
21
2
21
21
22 )RR(
MGRR4
VR4T
221
2
12
2 )RR(GM
R4T
as M1R1 = M2R2
1
2
2
1
MR
MR
1
2212
22
GM)RR(R4T
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33. The two stars in certain binary system move in circular orbits. The first star alpha has an orbital speed of 36.0 km.s–1. The second star, beta has an orbital speed of 12.0 km.s–1. The orbital period of first star is 137 days. The mass of the two stars are about
(a) 2.1 × 1030 and 6.8 × 1030 kg (b) 1.3 × 1030 and 3.9 × 1030 kg (c) 3.5 × 1030 and 9.2 × 1030 kg (d) 0.8 × 1030 and 2.3 × 1030 kg Ans. [d]
Sol. 3600241372
360024137v
r2
1
1
3600241371036r2
31
1
22
212
2
GMr)rr(4
T
2
2
1
1
rv
rv
21 r
12r36
r1 = 3r2
3r
3rr
GM4T 1
21
11
22
1627.GMr4T
1
31
22
1
1
vr2T
21
21
22
vr4T
1627.GM
r4v
r4
1
31
2
21
21
2
16GM27r
v1
1
121
kg108.0G2716vrM 30
211
1
M2 = 0.8 × 1030× 3 2.3 ×1030 kg Passage question (34-36) 34. Consider a spacecraft in an elliptical orbit around the earth. At the low point or perigee of its orbit, it is
400 km above the earth's surface. At the high point or apogee it is 4000 km above the earth's surface. The period of the space craft's orbit is (g = 9.8 ms–2 and R = 6400 km)
(a) 0.29 hr (b) 1.82 hrs (c) 2.21 hrs (d) 3.56 hrs Ans. [c]
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Sol. GM
a4T32
2
2
322
gRa4T
Rg
a2T2/3
4000 6400 6400 400
2a
V2
V1
Perigee
Apogee
Put a =
2440026400
=2
440012800
= 8600 km T = 2.21 hrs 35. The ratio of speed of the spacecraft at perigee to its speed at apogee is almost equal to (a) 10 : 1 (b) 3 : 2 (c) 2 : 3 (d) 1 : 10 Ans. [b] Sol. mv1r1 = mv2r2
6800
400,10rr
vv
1
2
2
1 = 1.5
= 10
5.1
= 23
36. The speed of the satellite at perigee is (a) 8576 m.s–1 (b) 57,307m.s–1 (c) 5876 m.s–1 (d) 7856 m.s–1 Ans. [c]
Sol. TE of satiable = 321 106800
GMmmv21
a2GMm
2
v1086002
GM106800
GM 21
33
= 21
523 v]286
1681[10g)106400(2
21
52 v172
16818.910642
v1 = 5876 ms–1
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37. Astronomers believe that a large percentage of the mass of the universe is dark matter. In one recent study, the transverse velocity of the large Magellanic cloud(LMC) was measured to be 200 km.s–1. The LMC is believed to orbit the centre of our galaxy at about 17 × 104 ly (1.6 × 1021m). Assuming a circular orbit, percentage of dark matter in our galaxy is about (independent estimate of visible matter is 2 × 1041 kg)
(a) 77 % (b) 82 % (c) 70 % (d) 80 % Ans. [b] Sol. 80% 38. The escape speed from jupiter is approximately 59.5 km.s–1 and its radius is about 12 times that of earth.
From this we may estimate the mean density of jupiter to be about (Radius of earth = 6400 km and escape speed from the earth is 11.2 km.s–1)
(a) 5 times that of earth (b) 0.2 times that of the earth (c) 2.5 times that of the earth (d) 0.4 times that of the earth Ans. [b]
Sol. vesc = RGM2 where M = 3R
34
vesc = 3
R4G2 2 vesc 2R
e
J
e
J
eesc
Jesc
RR
vv
122.115.59
e
J
2
2
e
J
)12(1
2.115.59
= 0.2
J = 0.2 time e 39. The orbit of planet mercury has the largest eccentricity of about 0.2 in the solar system. If the maximum
distance of mercury from the centre of the sun is about 69 million km, its minimum distance from sun is about
(a) 13.8 million km (b) 57.5 million km (c) 46 million km (d) 18 million km Ans. [c] Sol.
ae
Max. distance = a + ae
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Min. distance = a – ae a + ae = 69 mk a = 57.5 mk min. distance = a – ae = 46 Million km 40. As observed from a place in Australia the pole star (a) appears in the southern direction (b) appears at about 30º above the horizon (c) much brighter than that seen from India (d) can never be seen Ans. [d] Sol. pole star can be seen only from northern hemisphere as at is present above north pole of earth. Australia is in
southern hemisphere no pole star can be seen from Australia. 41. Evaporation of (sweat ) water is an essential mechanism in human beings for maintaining normal body
temperature. For human beings, heat of vaporization of water at a body temperature of 37ºC is nearly 2.3 MJ/kg and specific heat capacity is 3.5 kJ /(kg.K).On consuming a certain prescribed diet, the body temperature of an athlete of mass 82 kg is expected to increase by 2ºC. In order to prevent this, he drinks N bottles of mineral water (250 ml water in each ) at 37ºC. Assume that the entire amount of this water is given out as sweat, which vaporizes. N is nearly (density of water = 1000 kg.m–3)
(a) 1 (b) 2 (c) 3 (d) 4 Ans. [a] Sol. Q = msT = mL
82 × 3.5 ×103 ×2 = n
41 × 2.3 ×106
n = 2300
428.382
m = 1 42. The number n = 1 + 12 + 60 + 160 +240 + 192 + 64 is (a) A perfect square and a perfect cube (b) Neither a perfect square nor a perfect cube (c) A perfect cube but not a perfect square (d) A perfect square but not a perfect cube Ans. [a] Sol. n = 1 + 12 + 60 + 160 + 240 + 192 +64 = 729 = 272 = 93 43. A football (assumed to be a hollow sphere) of mass 200g and radius 16cm, is given horizontal kick 4 cm
above its centre. This imparts a speed of 8 cm/s to the football. Angular speed acquired by the fooball in radian/s is
(a) 9/16 (b) 15/16 (c) 3/4 (d) 16/15 Ans. [Bonus] Sol.
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4 cm
J
J = mv J × 4 = I mv × 4 = I
= Imv4
22 Rv6
mR)3/2(mv4
s/rad163
10)16(10108642
22
(Bonus)
44. For sets A,B we have (here XC denote complement of set X) (A × B)C = (a) AC × BC (b) BC × AC
(c) AC × B BC × A AC × BC (d) AC × B × BCAC × BC Ans. [b] Sol. (A × B)C = BC ×AC 45. A small marble (assumed to be a uniform solid sphere ) is released on one end of a parabolic mirror from a
vertical height of 70 cm. First half part of this mirror is rough on which the marble is released. Other half of the mirror is smooth. Throughout its motion the marble never slips. To what vertical height will it rise on the smooth surface?
(a) 98 cm (b) 70 cm (c) 63 cm (d) 50 cm Ans. [d]
Sol. mgh = 2mv107
v2 = 70g7
10gh7
10
70cm Smooth
h Rough
mghmv
21 2 " only translation energy will convert"
mgh70g7
10m21 Rotational K.E remains same
h = 50 m
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46. The number 312 + 29 + 3(3 × 64 + 65) + 26 is (a) A perfect square and a perfect cube (b) A perfect cube but not a perfect square (c) A perfect square but not a perfect cube (d) Neither a perfect square nor a perfect cube Ans. [c] Sol. 312 + 29 + 3(3×64 + 65) + 26 = (753)2 47. Radius and moment of inertia of a smooth pulley are 0.1 m and 1 kg.m2 respectively. A tangential force of
f = 40t – 10t2 sets the pulley into rotation . Direction of its rotation reverses after some time. The time duration after which the direction will reverse is
(a) 6s (b) 8s (c) 4s (d) 12 s Ans. [a] Sol. = f.r = (0.1) = 4t –t2
dtdw = 4t – t2
w = c3t–
2t4 32
at t = 0 ; w = 0
w = 2t2 – 3t3
w = 0 at t = 0 ; t = 5s 48. log10 0.01 + log0.1 10 + log10 0.001 + log0.1 0.001 = (a) log10.2 10.012 (b) log10 0.000001+ 3 (c) – 4 + log2 8 (d) None of the above Ans. [b] Sol. 3
103
10102
10 10log10log10log10log 11
= –2 –1 –3 + 3 = –3 49. A car is fitted with a rear view mirror of focal length 20cm. Another car, 2.8 m behind the first car is 15 m.s–1.
faster than the first car and approaching. The relative speed of image of the second car , with respect to first car at this instant, is
(a) 1/15 m.s–1 (b) 1/10 m.s–1 (c) 1/5 m.s–1 (d) 2/15 m.s–1 Ans. [a]
Sol. dtdv
uv
dtdv
2
2
201
2801
f1
u1
v1
v = 56/3
152802803
563
56
dtdv
s/m151
dtdv
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50. 3log9log16log31272
(a) Is defined but cannot be found (b) Is not defined
(c) Is defined and equals –31 (d) Is defined and equals
323
Ans. [d]
Sol. 3231
3283log9log16log
31272
51. An electric buggy is stationed exactly midway between two plane vertical walls parallel to each other. A man
standing adjacent to buggy blows whistle momentarily. Instantly the buggy starts running towards one of the walls with a velocity 35 m/s. The driver of the buggy records first two echoes of the whistle with a delay of exactly one second. Speed of sound in air at that temperature is 350 m/s. Distance between the walls must be
(a) 433.125 m (b) 866.25 m (c) 1732.5 m (d) 3465 m Ans. [c] Sol.
x x
wall -1 wall -2 at t = 0, whistle is blown.
and buggy start moving towards wall-1
Let driver records 1st echo at instant t.
then distance travelled by buggy = 35 t
distance travelled by sound = x + x – 35 t
i.e. (350) t = 2x – 35 t
385 t = 2x ...... (i)
at instant t + 1, driver record 2nd echo.
then during 0 to (t+1), distance travelled by sound which is reflected by wall-2
= x + x + 35 (t +1)
350 (t + 1) = 2x + 35t + 35
350t + 350 = 2x + 35t + 35
315t + 315 = 2x ....... (ii)
from equation (i) & (ii)
2x = 1732.5
hence distance between walls = 1732.5m 52. The number of points at which |x3 –1| is not differentiable is- (a) 3 (b) 2 (c) 1 (d) 0 Ans. [c]
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Sol. y = |x3 –1| = |x–1| |x2 + x + 1| = |x –1| (x2 + x + 1) is not diff. at x = 1 53. The circuit given below has a long straight wire AB placed horizontal along North – South direction . A
small magnetic needle can be held anywhere near this wire. Choose the correct statement.
A B(South)(North)
SN
(a) North pole of the magnetic needle will deflect towards East, if the compass is just above the wire (b) North pole of the magnetic needle will deflect towards West, if the compass is at exactly same level of
the wire (c) North pole of the magnetic needle will deflect towards East, if the compass is just below the wire (d) Magnetic needle will not deflect, if kept just below the wire Ans. [c] Sol. (i)decal North – south magnetic field below the wire into the plane Thus north pole move inwards that is east Below the wire, magnetic field is directed towards East hence North pole deflect towards East when placed
below the wire.
54. If DA,CD,BC,AB are unit vectors such that 2
1BC.AB then
(a) Points A, B, C, D are concyclic
(b) Quadrilateral ABCD has area 22
1
(c) Quadrilateral ABCD has half of the maximal area for quadrilateral with same perimeter
(d) The area determined by the vectors is 2
1
Ans. [a]
Sol. All the sides are of equal in magnitude then it can be either square or rhombus
2
1BC.AB so º
so it is rhombus it cyclic 55. INSAT series of satellites are launched by India for telecommunication. Such satellites appear stationary at a
particular point in the sky when observed from the earth . Consider the following statements : (i). The satellite always experiences gravitation of the earth (ii). The satellite does not need any fuel for its motion (iii). The satellite does not experience net force (iv). Such satellites have to be positioned vertically above the equator
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(a) Only (ii), (iii) & (iv) are correct (b) Only (i), (ii) & (iv) are correct (c) Only (i) & (iii) are correct (d) Only (i), & (ii) are correct Ans. [b] Sol. A geostationary satellite appears stationary with respect to Earth. It revolve in equatorial orbit. Gravitational
force by Earth on satellite provides required centripetal force of circular motion of satellite 56. The number 38(310 + 65) + 23(212 + 67) is (a) A perfect square and a perfect cube (b) Neither a perfect square nor a perfect cube (c) A perfect cube but not a perfect square (d) A perfect square but not a perfect cube Ans. [c] Sol. 38(310 + 65) + 23(212 + 67) 318 +38.25.35.+23.212 + 23.27.37
318 + 313.25 + 215 + 210.37
(36)3 + (25)3 + 3.(36)2. (25) + 3. (36)(25)2
= (36 + 25)3 57. The optical effects (phenomena) involved when we see a rainbow could be associated with (i) internal reflection (ii) dispersion (iii) total internal reflection (iv) deviation select the correct option (a) (ii), (iii) and (iv) (b) (i), (ii) and (iv) (c) (i) and (iv) (d) (iii) and (iv) Ans. [a] Sol. Rainbow formation involve total internal reflection, dispersion & deviation 58. Which of the following statements are true about periodic functions defined on the set of real numbers A: Sum of two functions with finite period is always a periodic function with finite period B. The period of a function that is sum of two periodic functions with finite period is least common multiple
of the period of two functions (a) A and B are correct (b) A is correct but B is incorrect (c) A is false but B is correct (d) A and B are false Ans. [d] Sol. For statement A, if f(x) = sin x, g(x) = {x} then f(x) + g(x) = sin x +{x} which is not periodic so statement 'A' is false for statement 'B' if f(x) = |sin x|, g(x) = |cos x| which are periodic with period ''
but period of f(x) + g(x) = |sin x| + |cos x| is 2
so statement (B) is also false
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59. Unaware about the fact that analog ammeters and voltmeters can also have zero error, a student recorded following readings while determining resistance by using ohm's law.
Obs No. Voltage/V Current/mA Obs. No. Voltage/V Current/mA
1 1.0 40 4 7.0 160 2 3.0 80 5 9.0 200 3 5.0 120
If the ammeter has no zero error, the zero error in the voltmeter is (a) – 1V (b) –1.5 V (c) 0.5 V (d) 1 V Ans. [d] Sol. Let end correction in voltmeter = a v i according to ohm's law
4080
a1a3
3 + a = 2 + 2a a = 1 60. The inverse function of the function sinx + cos x is
(a) sin–1x + cos–1 x (b) xcosxsin
1
(c)
2xsin 1 (d)
42xsin 1
Ans. [d] Sol. y = sinx + cosx
)4/xsin(2
y
x = sin–1
2y –/4
f–1(x) = sin–1
2x – /4
61. Particle A collides elastically (perfect) with another particle B which was at rest. They disperse in opposite
directions with same speeds. Ratio of their masses respectively must be (a) 1 : 2 (b) 1 : 3 (c) 1 : 4 (d) 2 : 3 Ans. [b] Sol. Let masses of A and B are m1 & m2
u
m1 m2
rest
v m1 m2
v
in elastic collision, relative velocity of separation = relative velocity of approach 2v = u
2uv
and Linear Momentum = constant m2v – m1v = m1 u (m2 – m1)v = m1 u
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(m2 – m1) 2u = m1 u
31
mm
2
1
62.
4
x
xcosxsinlim4
x=
(a) ∞ (b) – ∞ (c) 2
1 (d) 2
Ans. [d]
Sol. )
4x(
xcosxsinlim4
x
)
4xsin(
)4
xsin(2lim
4x
= 212
63. A long rowing boat put upside down, shown in adjacent figure, has to be weighed using only a single bathroom scale. The boat will sag if it is supported only in the middle, and so the scales must be put first at position A with a wooden support at B, and then at position B with the wooden support at A. The readings on the scale are 45 kg and 55 kg respectively. The distance of centre of mass (from point A) and mass of the boat are respectively
5 m 8 m 6 m A B
(a) 4.4 m, 120 kg (b) 4.4 m, 100 kg (c) 4.2 m, 100 kg (d) 4.2 m, 120 kg Ans. [b] Sol. Ist case :
8 m
mg
R2R1 = 450 N
A
Force balance Torque balance
R1 + R2 = mg R2 8 = mg x …(2)
450 + R2 = mg
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R2 = mg – 450 …(1)
by (1) (2)
8
mgx = mg – 450 …(3)
IInd case :
8
mg
R R
B
8–x
Force
R+ R mg
R + 550 = mg …(4)
Torque balance
R 8 = mg (8 – x) …(5)
by (4) (5)
8
)x–8(mg = mg – 550
by mg – 8
mgx = mg – 550
8
mgx = 550 …(6)
by (3) (6)
550 = mg – 450
1000 = mg
m = 100 kg
By eq. (6)
8
1000 x = 55
x = 100
855 = 4.4 m
64. The number of real solutions of the equation (x – 1) (3x – 2) = 1 is (a) 0 (b) 1 (c) 2 (d) More than 2 Ans. [c]
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Sol.
x = 1
(0, 1)
y = 2
y
x 0
3x – 2 = 1–x
1 .
Number of Intersection point = 2
65. A train of mass m is moving on a circular track of radius r with constant speed v. Length of the train is exactly equal to half the circumference of the circular track. Magnitude of its linear momentum is
(a) mv/ (b) 0.5 mv (c) 2mv/ (d) mv Ans. [c]
Sol.
Vy
V
Vx
d
x
y
dPy = dmVy
dPy = M d V sin
Py =
MV
2/
2/–dsin =
MV 2/
2/–]cos[– = 0
dPx = dmVx
= M d V cos
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Px =
MV
2/
2/–dcos =
MV 2/
2/–][sin =
MV [1 + 1] =
MV2
66. The number of integers a, b, c for which 2a2 + b2 – 8c = 7 is (a) 2 (b) Infinite (c) 0 (d) 4 Ans. [c] Sol. 2a2 + b2 – 8c = 7 b = odd 42
2 + 42 + 1
2a2 + 42 – 8c = 6 a = odd 4(2) + (2+1)
a2 + 22 – 4c = 3 c = odd
41 + 1 + 22 – 4c = 3 2a2 + b2 – 8c = 7
41 + 21 – 4c = 2
21 + 2 – 2c = 2 c = odd
21 + 23 – 2c = 2 2(21 + 1)2 + (22 – 1)2 – 8 (23 + 1) = 7
1 + 3 – c = 1 2(421 + 41 + 1) + 42
2 + 42 + 1 – 163 = 15
1 + 3 = 1 + c 821 + 42
2 + 81 + 42 – 163 = 12
221 + 2
2 + 21 + 2 – 43 = 3
Since, 1, 3 are even 221 + 21 – 43 + 2 (2 + 1) = 3
So, 'c' is odd
If a = odd the
a2 = 4 × even + 1
= 4 + 1
67. In SI units we use length, mass and time as fundamental quantities. Another intelligent world may not know these. However, three constants G (universal gravitational constant), c (speed of light in vacuum) and h (Planck's constant) are really universal and can be related to almost all the known interactions. In terms of these fundamental constants, the dimensions of time are
(a)
2
125
21
hcG – (b)
2
1
hcG 2–1 (c)
2
121
hcG –2 (d)
2
123
21
hcG –
Ans. [a] Sol. t = [Gx cy hz]
[M0 L0 T] = [M–1 L3 T–2]x [LT–1]y [ML2 T–1]z M0 L0 T = M–x + z L3x + y+2z T–2x–y–z – x + z = 0 …(1) x = z 3x + y + 2z = 0 …(2) 3x + y + 2x = 0 5x = – y – 2x – y – z = 1 …(3) – 2x + 5x – x = 1 2x = 1 x = 1/2 z = 1/2 y = – 5/2 t = [G1/2 c–5/2 h1/2]
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68. If p (x) = x (x + 1) (x + 2) … (x + 2001) – c then the maximum multiplicity of the roots of p (x) can be (a) 1 (b) 2 (c) 3 (d) 2001 Ans. [b] Sol. Multiplicity of roots is how many times a root is repeated. Ex. In (x – 2)3 (x – 3)2 (x – 4)5 x. 2 has multiplicity 3 3 has multiplicity 2 4 has multiplicity 5 0 has multiplicity 1 So, P(x) = x (x + 1) (x + 2) …. (x + 2001) – c
y = cc
Maximum multiplicity is two.
69. A train is running on a circular track of radius R with a constant speed. The driver is blowing siren of a constant frequency (n0) throughout the circular motion of period T. There is a listener on the diameter of the track at a distance R/2 from centre of the circle. At t = 0, the train siren is farthest from the listener. In the following graphs the frequency, as recorded by the listener, is plotted against time. Which of them is closest to the correct pattern?
(a)
t
n
n0
(b)
t
n
n0
T/2
(c)
t
n
n0
T/3 T
(d)
t
n
n0
Ans. [b] Sol.
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cos
A t = 0
P
DO Bt = T/3
Ct = T/2
R/2
at a general point P apparent frequency is
n = cosV–V
V
s n0
From AB, is increases so frequency is continuously from A to B and from B to C, is decreases so apparent frequency is decreases, but for t < T/2 freq. is always large than actual frequency. So graph (b) is correct.
70. If {x} denotes the fractional part of a real number then dx}x{2
0
2 =
(a) 322 (b)
31 (c) 1 –
32 (d) 1 +
32
Ans. [c]
Sol. I = 2
0
2 dx}x{
I = 2
0
22 dx]x[–x
I = 2
0
2dxx – 2
0
2 dx]x[
I = 2
0
3
3x
–
1
0
2 dx]x[ = 2
1
2 dx]x[
I = 3
22 – 0 – 2
0dx
I = 3
22 – 1–2
I = 3
22 – 2 + 1
I = –32 + 1
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71. Adjacent figure shows a block A, held by a spring balance D and submerged into a liquid in a beaker. The beaker is kept on a weighing balance E. Mass of the beaker plus the liquid is 2.5 kg. Balance D reads 2.5 kg and E reads 7.5 kg. Volume of the block is 0.003 m3. Consider the following statements
Figure
(i) The density of the liquid is 5000/3 kg/m3 (ii) The mass of block A is 7.5 kg (iii) The buoyant force is 5 kg. wt. (iv) If half the volume of the block is pulled out of the liquid, E would read 5 kg. Select correct option(s). (a) (i), (iii) and (iv) (b) (i), (ii) and (iv) (c) (i) and (iv) (d) (i), (ii), (iii) and (iv) Ans. [d] Sol. Let m = mass of liquid + beaker = 2.5 kg
T = reading of spring balance = 2.5 kg = 25N
R = reading of balance E = Normal reaction = 7.5 kg = 75 N
m = mass of block
mg
T Fb
mg
R Fb
for balance of block for balance of beaker + liquid
T + Fb = mg R = Fb + mg
25 + Fb = mg …(1) 75 = Fb + 25
Fb = 50 N
So,
buoyant force is = 5 kg.wt
by eq. (1)
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25 + Fb = mg
25 + 50 = mg
m = 7.5 kg
So, mass of block = 7.5 kg
Fb = 50
Vg = 50
(0.003) (10) = 50
= 3
5000 kg/m3
So, density of liquid = 3
5000 kg/m3
when half of block is pulled Fb became = 25 N
and New F.B.D of liquid + beaker
25 N
R Fb = 25
R = 50 N
= 5 Kg
So, all (i), (ii), (iii) & (iv) are correct.
72. AM-HM inequality for positive real numbers a, b, c states that 3
cba cacbab
abc3
. If a, b are positive
irrational numbers then
(a) ba2
ab9
a + b (b) ba2
ab9
1 (c) b2a
ab9
2a + b (d) ba2
ab18
a + 2b
Ans. [c] Sol. 2a + b = a + a + b
3
baa 3 2ba
3
bba 3 2ab
Multiply both,
9
)b2a)(ba2( ab
2a + b b2a
ab9
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73. Two identical stars with mass M orbit around their centre of mass in circular orbit. If radius of the orbit is R and the stars are always diametrically opposite, then consider the following statements
(i) Their binding force is equal to 2
2
R4GM
(ii) If the stars are heavier and closer, their orbital speed is greater.
(iii) The period of the orbit is T = GMR 3
(iv) The minimum energy required to separate the two stars to infinity is equal to R4
GM2
Select correct statement/s (a) Only (i) and (ii) (b) Only (i), (ii) and (iv) (c) Only (i), (iii) and (iv) (d) Only (i) and (iii) Ans. [b]
Sol.
M F com
F
M
Binding Force = G 2)R2(M.M = 2
2
R4M.G
for circular motion of any one star,
2
2
R4M.G =
RMV 2
V = R4
GM
V RM
when M is large and R is small orbital speed is high,
Time period T = V
R2 =
R4GM
R2 = GMR42
3 =
GMR4
3
Minimum Energy required to separate the stars = – (BE)
= G R2MM
= R4
GM2
(ii) and (iv) are correct Ans. (b)
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74. The number of positive integers n 50 such that 3 3 3 ...nnn is a natural number is
(a) Zero (b) 2 (c) 50 (d) 5 Ans. [b] Sol. 3 = n + , N
3 – = n
( – 1) ( + 1) = n
n = ( – 1) () ( + 1)
Possible value of for n 50
= 2, 3
75. A monkey is holding onto one end of a rope which passes over a frictionless pulley and at the other end is a plane mirror which has a mass equal to the mass of the monkey. At equilibrium the monkey is able to see her image in the mirror. How does the monkey see her image in the mirror as she climbs up the rope?
(a) The image of the monkey moves with double speed of that of the monkey. (b) The image of the monkey moves with half the speed of that of the monkey. (c) The image of the monkey moves as fast as the monkey. (d) The monkey will not be able to see her image. Ans. [c]
Sol.
TT
M
a
mg mg
a
M
Both monkey and mirror are have same mass
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ASTRONOMY NSEA 2016-2017 EXAMINATION
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-5151200 Website : www.careerpoint.ac.in, Email: [email protected]
CAREER POINT
NLM eq. T – mg = ma So, both are always move with same acceleration 76. If i = 1– then i2i is a (a) Purely imaginary number (b) A natural number (c) A real number (d) A complex number with non-zero real and imaginary parts Ans. [c]
Sol. i2i = i2
i 2e
= e–
which is a real number
77. A steel ball is dropped from a height of 1 m on to a hard non-conducting surface. Every time it bounces it reaches 80% of its previous height. All the losses in the energy are accounted only for increasing temperature. Nearly how much is the rise in temperature of the ball just after the third bounce? (g = 10 m/s2, specific heat capacity of material of the ball = 500 J/(kgK))
(a) 0.005 ºC (b) 0.01 ºC (c) 0.015 ºC (d) 0.02 ºC Ans. [b] Sol. Let mass of ball = m
Its initial energy = mgh = (m) (10) (1) = 10m
20% loss = (10m) × 0.2 = 2m …(1)
remaining energy = 8 m
now 20% loss in IInd collision = 8 m × 0.2 = 1.6 m …(2)
remaining energy = 6.4 m
now 20% loss in IIIrd collision = (6.4 m) (0.2) = 1.28 m …(3)
total loss = 2m + 1.6 m + 1.28 m
= 4.88 m
and
4.88 m = mst
t = s88.4 =
50088.4 = 0.0097ºC
After rounding off.
t = 0.01ºC
78. In a track event, a circumcircle and an incircle were drawn for a triangle having sides 50 m, 120 m and 130 m respectively. Ram and Sham were asked to walk on the in-circle and the circumcircle respectively. They started walking with same speed in the same direction (sense of rotation) from a point where they were closest. After how many rounds each, will they be closest again?
(a) Ram 4 and Sham 13 (b) Ram 13 and Sham 4 (c) Ram 5, Sham 15 (d) Ram 5, Sham 12 Ans. [a]
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ASTRONOMY NSEA 2016-2017 EXAMINATION
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-5151200 Website : www.careerpoint.ac.in, Email: [email protected]
CAREER POINT
Sol.
50
120
R
130
r
r = 20 ; R = 65
for Ram, 1 = 20V =
260V13 {V is there speed & V = r}
for Sham, 2 = 65V =
260V4
There for same position Ram will take 4 rounds & sham will take 13 rounds.
79. The angle between the two complex number a = ii and b = 1 is
(a) (b) 0 (c) 2 (d) –
2
Ans. [b]
Sol. a = ii = i
i 2e
= 2–e
b = 1
both are positive real numbers. So angle between them is zero.
80. The number of rectangles that can be formed by joining the points of 4 × 4 grid of equi spaced points is (a) 16 (b) 36 (c) 40 (d) 42 Ans. [b] Sol. 4C2 × 4C2 = 36