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NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 1
Multiple Choice Questions
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 2
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 3
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 4
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 5
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 6
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 7
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 8
24. The following n data points, (x1, y1), (x2, y2),...,(xn, yn) are given. For conducting quadratic spline
interpolation the x-data needs to be
equally spaced
in ascending or descending order
integers
positive
25. In cubic spline interpolation,
the first derivatives of the splines are continuous at the interior data points
the second derivatives of the splines are continuous at the interior data points
the first and the second derivatives of the splines are continuous at the interior data points
the third derivatives of the splines are continuous at the interior data points
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 9
26. The following incomplete y vs. x data is given
x 1 2 4 6 7
y 5 11 ???? ???? 32
The data is fit by quadratic spline interpolants given by
,
where a, b, c, and d, are constants. The value of c is most nearly
-303.00
-144.50
-0.0000
14.000
27. The following incomplete y vs. x data is given.
x 1 2 4 6 7
y 5 11 ???? ???? 32
The data is fit by quadratic spline interpolants given by
,
where a, b, c, d, e, f, and g are constants. The value of df/dx at x=2.6 is most nearly
-144.50
-4.0000
3.6000
12.200
1 axxf 21 x
42 ,91422
xxxxf
64 ,2
xdcxbxxf
76 ,928303252
xxxxf
1 axxf 21 x
42 ,91422
xxxxf
64 ,2
xdcxbxxf
76 ,2
xgfxexxf
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 10
28. The following incomplete y vs. x data is given
x 1 2 4 6 7
y 5 11 ???? ???? 32
The data is fit by quadratic spline interpolants given by
,
where a, b, c, d are constants. What is the value of ?
23.50
25.67
25.75
28.00
29. A robot needs to follow a path that passes through six points as shown in the figure. To find the shortest
path that is also smooth you would recommend which of the following?
Pass a fifth order polynomial through the data.
Pass linear splines through the data
Pass quadratic splines through the data
Regress the data to a second order polynomial
21 ,1 xaxxf
42 ,91422
xxxxf
64 ,2
xdcxbxxf
76 ,928303252
xxxxf
5.3
5.1
dxxf
Path of a Robot
0
2
4
6
8
0 5 10 15
x
y
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 11
30. The two-segment trapezoidal rule of integration is exact for integrating at most ______ order polynomials.
first
second
third
fourth
31. The value of by using the one-segment trapezoidal rule is most nearly
11.672
11.807
20.099
24.119
32. The value of by using the three-segment trapezoidal rule is most nearly
11.672
11.807
12.811
14.633
33. The velocity of a body is given by
where t is given in seconds, and v is given in m/s. Use the two-segment Trapezoidal Rule to find the distance
covered by the body from t=2 to t=9 seconds.
935.0 m
1039.7 m
1260.9 m
5048.9 m
34. The shaded area shows a plot of land available for sale. The numbers are given in meters measured from
the origin. Your best estimate of the area of the land in square meters is most nearly
2.2
2.0
dxxex
2.2
2.0
dxxex
51,2)( tttv
145,352
tt
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 12
2500
4775
5250
6000
35. The following data of the velocity of a body as a function of time is given as follows.
Time (s) 0 15 18 22 24
Velocity (m/s) 22 24 37 25 123
The distance in meters covered by the body from t=12 s to t=18 s calculated using using Trapezoidal Rule with
unequal segments most nearly is
162.9
166.0
181.7
436.5
36. The highest order of polynomial integrand for which Simpson’s 1/3 rule of integration is exact is
first
second
third
fourth
37. The value of by using two-segment Simpson's 1/3 rule is most nearly
7.8306
7.8423
2.2
2.0
dxex
60
75
45
25
100 60
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 13
8.4433
10.246
38. The value of by using four-segment Simpson's 1/3 rule is most nearly
7.8036
7.8062
7.8423
7.9655
39. The velocity of a body is given by
where t is given in seconds, and v is given in m/s. Using two-segment Simpson's 1/3 rule, the distance covered
in meters by the body from t=2 to t=9 seconds most nearly is
949.33
1039.7
1200.5
1442.0
40. The value of by using two-segment Simpson’s 1/3 rule is estimated as 702.039. The estimate
of the same integral using four-segment Simpson’s 1/3 rule most nearly is
702.39 + 8/3 [2f(7)-f(11)+2f(15)]
702.39/2 + 8/3 [2f(7)-f(11)+2f(15)]
702.39 + 8/3 [2f(7)+2f(15)]
702.39/2 + 8/3 [2f(7)+2f(15)]
41. The following data of the velocity of a body is given as a function of time.
Time (s) 4 7 10 15
Velocity (m/s) 22 24 37 46
2.2
2.0
dxex
51,2)( tttv
145,352
tt
19
3
)( dxxf
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 14
The best estimate of the distance in meters covered by the body from t=4 to t=15 using combined Simpson’s
1/3rd
rule and the trapezoidal rule would be
354.70
362.50
368.00
378.80
42. If is the value of integral using n-segment Trapezoidal rule, a better estimate of the integral can
be found using Richardson’s extrapolation as
43.The estimate of an integral of is given as 1860.9 using 1-segment Trapezoidal rule.
Given f(7)=20.27, f(11)=45.125, and f(14)=82.23, the value of the integral using 2-segment Trapezoidal rule
would most nearly be
787.32
1072.0
1144.9
1291.5
44. The value of an integral given using 1, 2, and 4 segments Trapezoidal rule is given as 5.3460,
2.7708, and 1.7536, respectively. The best estimate of the integral you can find using Romberg integration is
most nearly
1.3355
1.3813
15
2
2
nn
n
III
3
2
2
nn
n
III
nI 2
n
nn
nI
III
2
2
2
nI b
a
dxxf
19
3
dxxf
b
a
dxxf
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 15
1.4145
1.9124
45.Without using the formula for one-segment Trapezoidal rule for estimating the true error, can
be found directly as well as exactly by using the formula
, for
46. For , the true error, in one-segment Trapezoidal rule is given by
, .
The value of for the integral is most nearly
2.7998
4.8500
4.9601
5.0327
47. Given the velocity vs. time data for a body
t(s) 2 4 6 8 10 25
0.166 0.55115 1.8299 6.0755 20.172 8137.5
The best estimate for distance covered between 2s and 10s by using Romberg rule based on Trapezoidal rule
results would be
33.456 m
36.877 m
37.251 m
81.350 m
12
3ab
Et
"f ba
xexf
xxxf 33
352 xxf
xexxf
25
12
3ab
Et
"f ba
dxe
x
2.7
5.2
2.03
sm /
b
a
dxxftE
b
a
dxxftE
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 16
48. is exactly
49. For a definite integral of any third order polynomial, the two-point Gauss quadrature rule will give the same
results as the
1-segment trapezoidal rule
2-segment trapezoidal rule
3-segment trapezoidal rule
Simpson's 1/3 rule
50. The value of by using the two-point Gauss quadrature rule is most nearly
11.672
11.807
12.811
14.633
51. A scientist uses the one-point Gauss quadrature rule based on getting exact results of integration for
functions f(x)=1 and x. The one-point Gauss quadrature rule approximation for is
52. A scientist develops an approximate formula for integration as
1
1
)5.75.2( dxxf
1
1
)5.75.2(5.2 dxxf
1
1
)55(5 dxxf
1
1
)()5.75.2(5 dxxfx
)()(2
bfafab
2)(
bafab
23
1
223
1
22
ababf
ababf
ab
)()( afab
b
a
xfcdxxf ),()( 11
10
5
)( dxxf
2.2
2.0
dxxex
b
a
dxxf )(
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 17
where
The values of c1 and x1 are found by assuming that the formula is exact for the functions of the form a0x + a1x2 polynomial. Then the resulting formula would therefore be exact for integrating
53. You are asked to estimate the water flow rate in a pipe of radius 2m at a remote area location with a harsh
environment. You already know that velocity varies along the radial location, but you do not know how it
varies. The flow rate Q is given by
To save money, you are allowed to put only two velocity probes (these probes send the data to the central office in New York, NY via satellite) in the pipe. Radial location, r is measured from the center of the pipe, that is r=0 is the center of the pipe and r=2m is the pipe radius. The radial locations you would suggest for the two velocity probes for the most accurate calculation of the flow rate are
0,2
1,2
0,1
0.42,1.58 54. To solve the ordinary differential equation , by Euler’s method, you
need to rewrite the equation as
55. Given
and using a step size of h=0.3, the value of y(0.9) using Euler’s method is most nearly
-35.318
-36.458
.1 bxa
2)( xf2
532)( xxxf 2
5)( xxf
xxf 32)(
2
0
2 rVdrQ
50,5sin2
yyxdx
dy
50,5sin3
1 2 yyx
dx
dy
50,3
5cos
3
13
y
yx
dx
dy
50,sin3
1 yx
dx
dy
53.0,sin532
yxydx
dy
50,sin532
yxydx
dy
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 18
-658.91
-669.05
56. Given
and using a step size of h=0.3, the best estimate of dy/dx(0.9) using Euler’s method is most nearly is
-0.37319
-0.36288
-0.35381
-0.34341
57. The velocity (m/s) of a body is given as a function of time (seconds) by v(t)=200 ln(1+t) -t, t≥0
Using Euler’s method with a step size of 5 seconds, the distance in meters traveled by the body from
t=2 to t=12 seconds is most nearly
3133.1
3939.7
5638.0
39397
58. Euler’s method can be derived by using the first two terms of the Taylor series of writing the value
of , that is the value of at , in terms of and all the derivatives of at . If
, the explicit expression for if the first three terms of the Taylor series are chosen
for the ordinary differential equation
would be
59. A homicide victim is found at 6:00PM in an office building that is maintained at 72˚F. When the
victim was found, his body temperature was at 85 ˚F. Three hours later at 9:00PM, his body
temperature was recorded at 78˚F. Assume the temperature of the body at the time of death is the
normal human body temperature of 98.6˚F.
53.0,31.0
yeydx
dy x
1iy y 1ix iy y ix
ii xxh 1 1iy
70,325
yey
dx
dy x
hyeyy ix
iii 3
2
1 51
22
53
2
1 255
1
hehyeyy ii x
i
x
ii
24
9
4
133
2
1 255
1
hyehyeyy i
x
i
x
iiii
2
332
12
5
1
hyhyeyy ii
x
iii
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 19
The governing equation for the temperature θ of the body is
where
= temperature of the body, ˚F
θa = ambient temperature, ˚F
t = time, hours
k = constant based on thermal properties of the body and air.
The estimated time of death most nearly is
2:11 PM
3:13 PM
4:34 PM
5:12 PM
60. To solve the ordinary differential equation
by the Runge-Kutta 2nd order method, you need to rewrite the equation as
61. Given
and using a step size of h=0.3, the value of y(0.9) using the Runge-Kutta 2nd order Heun's method is
most nearly
-4297.4
-4936.7
-0.21336
-0.24489
62. Given
,
)( akdt
d
50,sin2
yxyxdx
dy
50,sin3
1 2 yxyx
dx
dy
50,3
cos3
1 3
y
xyx
dx
dy
50,sin3
1 yx
dx
dy
53.0,sin532
yxydx
dy
1410
1410
53.0,531.0
yeydx
dy x
50,sin32
yxxydx
dy
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 20
and using a step size of h=0.3, the best estimate of dy/dx(0.9) using the Runge-Kutta 2nd order
midpoint-method most nearly is
-2.2473
-2.2543
-2.6188
-3.2045
63. The velocity (m/s) of a body is given as a function of time (seconds) by
Using the Runge-Kutta 2nd order Ralston method with a step size of 5 seconds, the distance in meters
traveled by the body from t=2 to t=12 seconds is estimated most nearly is
3904.9
3939.7
6556.3
39397
64. The Runge-Kutta 2nd order method can be derived by using the first three terms of the Taylor
series of writing the value of yi+1 (that is the value of y at xi+1 ) in terms of yi (that is the value of y at xi)
and all the derivatives of y at xi . If h=xi+1-xi, the explicit expression for yi+1 if the first three terms of
the Taylor series are chosen for solving the ordinary differential equation
would be
65. A spherical ball is taken out of a furnace at 1200K and is allowed to cool in air. Given the following,
radius of ball = 2 cm
specific heat of ball = 420 J/(kg-K)
density of ball = 7800 kg/m^3
convection coefficient = 350 J/s-m^2-K
The ordinary differential equation is given for the temperature, of the ball
0,1ln200 tttt
70,352
yey
dx
dy x
2
5532
3
1
hhyeyy i
x
iii
2
2521532
22
1
hyehyeyy i
x
i
x
iiii
2
6532
22
1
hehyeyy ii x
i
x
ii
2
56532
22
1
hehyeyy ii x
i
x
ii
841310811020673.2
dt
d
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 21
if only radiation is accounted for. The ordinary differential equation if convection is accounted for in
addition to radiation is
66. To solve the ordinary differential equation ,
by Runge-Kutta 4th order method, you need to rewrite the equation as
67. Given and using a step size of , the value of
using Runge-Kutta 4th order method is most nearly
- 0.25011
-4297.4
-1261.5
0.88498
68. Given , and using a step size of , the best estimate of
Runge-Kutta 4th order method is most nearly
-1.6604
-1.1785
300106026.110811020673.228413
dt
d
300103982.410811020673.228413
dt
d
300106026.12
dt
d
300103982.42
dt
d
50,sin2
yxyxdx
dy
50,sin3
1 2 yxyx
dx
dy
50,3
cos3
1 3
y
xyx
dx
dy
50,sin3
1 yx
dx
dy
4010
50,sin32
yxxydx
dy
53.0,sin532
yxydx
dy3.0h
9.0y
53.0,32
yeydx
dy x 3.0h 9.0dx
dy
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 22
-0.45831
2.7270
69. The velocity (m/s) of a parachutist is given as a function of time (seconds) by
Using Runge-Kutta 4th order method with a step size of 5 seconds, the distance traveled by the body
from to seconds is estimated most nearly as
341.43 m
428.97 m
429.05 m
703.50 m
70. Runge-Kutta method can be derived from using first three terms of Taylor series of writing the
value of , that is the value of at , in terms of and all the derivatives of at . If
, the explicit expression for if the first five terms of the Taylor series are chosen for the ordinary differential equation
, would be
0),17.0tanh(8.55 ttt
2t 12t
1iy y 1ix iy y ix
ii xxh 1 1iy
2
553
22
1
hhyeyy i
x
iii
24
3906253009096
625483
2252153
42
32
222
1
hye
hye
hyehyeyy
i
x
i
x
i
x
i
x
ii
ii
ii
24
24
612
2653
42
32
222
1
he
he
hehyeyy
i
iii
x
xx
i
x
ii
24
24
612
25653
42
32
222
1
he
he
hehyeyy
i
iii
x
xx
i
x
ii
70,352
yey
dx
dy x
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 23
71. A hot solid cylinder is immersed in an cool oil bath as part of a quenching process. This process
makes the temperature of the cylinder, , and the bath, , change with time. If the initial temperature of the bar and the oil bath is given as 600° C and 27°C, respectively, and
Length of cylinder = 30 cm
Radius of cylinder = 3 cm Density of cylinder = 2700 kg/m^3 Specific heat of cylinder = 895 J/kg-K Convection heat transfer coefficient = 100 W/(m^2-K) Specific heat of oil = 1910 J/(kg-K) Mass of oil = 2 kg The coupled ordinary differential equations governing the heat transfer are given by
The following equations are used to answer questions#2, 3, and 4
c b
bc
c
dt
d
4.362
cb
b
dt
d
5.675
bc
c
dt
d
4.362
cb
b
dt
d
5.675
bc
c
dt
d
5.675
cb
b
dt
d
4.362
bc
c
dt
d
5.675
cb
b
dt
d
4.362
hkkkkyy ii 43211 226
1
ii yxfk ,1
Oil
Cylinder
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 24
72. The differential equation is
linear
nonlinear
linear with fixed constants
undeterminable to be linear or nonlinear
73. A differential equation is considered to be ordinary if it has
one dependent variable
more than one dependent variable
one independent variable
more than one independent variable
74. Given
, y(2) most nearly is
0.17643
0.29872
0.32046
0.58024
75. The form of the exact solution to is
76. The following nonlinear differential equation can be solved exactly by separation of variables.
The value of θ(100) most nearly is
-99.99
hkyhxfk ii 12
2
1,
2
1
hkyhxfk ii 23
2
1,
2
1
hkyhxfk ii 34 ,
60,2sin32 yxydx
dy
xxBeAe
5.1
xx BxeAe 5.1
xx BeAe 5.1
xxBxeAe
5.1
10000,811026
dt
d
50,32 yey
dx
dy x
NET/SET PREPARATION MCQ ON NUMERICAL ANALYSIS By S. M. CHINCHOLE
L. V. H. ARTS, SCIENCE AND COMMERCE COLLEGE, PANCHAVATI, NSAHIK - 3 Page 25
909.10
1000.32
1111.10
77. A spherical solid ball taken out of a furnace at 1200K is allowed to cool in air. Given the following radius of ball=2 cm density of the ball=7800 kg/m^3 specific heat of the ball=420 J/kg-K emmittance=0.85 Stefan-Boltzman constant=5.67E-8 J/s-m^2-K^4 ambient temperature=300K convection coefficient to air=350 J/s-m^2-K.
The differential equation governing the temperature, of the ball as a function of time, t is given by
841210811020673.2
dt
d
3001060256.12
dt
d
300106026.11081102067.2128412
dt
d
300106026.11081102067.228412
dt
d