Engineering Mathematics Multiple Choice Questions on “Errors and Approximations”.
1. Error is the Uncertainty in measurement.
a) True
b) False
Answer: a
Explanation: In the term of mathematics, “Error tells the person how much correct or certain its measurement is.”
2. Relative error in x is?
a) δx
b) δx⁄x
c) δx⁄x * 100
d) 0
Answer: b
Explanation: Option ‘δx’ is called absolute error.
Option ‘δx⁄x’ is called relative error.
Option ‘δx⁄x * 100’ is called Percentage error.
3. Find the percentage change power in the circuit if error in value of resistor is 1% and that of voltage source is .99%.
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
Answer: a
Explanation: Power is given by P = V2⁄R
Taking log on both sides,
log(P) = 2log(V) – log(r)
Differentiating it ,
δp⁄p = 2δV⁄V – δr⁄r
Multiplying by 100 we get,
%P = 2%V – %r = (2*.99) – 1 = 0.98%.
4. Magnitude of error can be negative or positive.
a) True
b) False
Answer: b
Explanation: Magnitude of error can not be negative.Negative or positive sign only shows the increase or decrease in the quatity.
5. Given the kinetic energy of body is T = 1⁄2 mv2. If the mass of body changes from 100 kg to 100 kg and 500 gm and velocity of a body changes from 1600 mt/sec to 1590 mt/sec. Then find the approximate change in T.
a) 960000 J decrease in value
b) 960000 J increase in value
c) 450000 J decrease in value
d) 450000 J increase in value
Answer: a
Explanation: Given T = 1⁄2 mv2
Now taking log and differentiating,
δT = 0.5[v2 δm + 2mvδv]
Now, v = 1600 mt/sec, m = 100kg, δv = -10, δm = 0.5
Then,
δT = -960000 J => decrese in value of T by 960000 J.
6. The speed of a boat is given by, v = k(1⁄t – at), where k is the constant and l us the distance travel by boat in time t and a is the acceleration of water. If there is an change in ‘l’ from 2cm to 1cm in time 2sec to 1sec. If the acceleration of water changes from 0.95 mt/sec2 to 2 mt/sec2 find the motion of boat.
a) -2
b) 2
c) 0.5
d) -0.5
Answer: a
Explanation: Given, v = k(1⁄t – at)
Differentiating it we get
(δv=kleft [frac{(tδl-lδt)}{t^2} – aδt – tδa right ])
Hence,
(frac{δv}{v}=frac{kleft [frac{(tδl-lδt)}{t^2} – aδt – tδa right ]}{v})
(frac{δv}{v}=frac{left [frac{(tδl-lδt)}{t^2} – aδt – tδa right ]}{(frac{l}{t}-at)})
Put, l = 2cm, t = 2sec, a = 0.95 mt/sec2
and δl = 1cm, δt = 1 secand δa = -1.05 mt/sec2
we get,
δv⁄v = 2.
7. The relative error in the volume of figure having hemispherical ends and a body of right circular cylinder is, if error in radius(r) is 0 and in height(h) is 1.
a) 1/(h + 4⁄3 r)
b) 1/(h + 2⁄3 r)
c) h/(h + 4⁄3 r)
d) r/(h + 4⁄3 r)
Answer: a
Explanation: Given V = πr2 h + 4⁄3 πr3
Now since error in radius is zero , it should be treated as constant, Hence,
(δV=frac{πr^2 δh}{πr^2 (h+frac{4}{3} r)}=frac{1}{(h+frac{4}{3} r)})
8. If n resistors of unequal resistances are connected in parallel,and the percenrage error in all resistors are k then,total error in parallel combination is?
a) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
b) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
c) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
d) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
Answer: b
Explanation: Given 1⁄r = 1⁄a + 1⁄b + 1⁄c +⋯.. + 1⁄n
Differentiating all,
– 1⁄r2 dr = – 1⁄a2 da – 1⁄b2 db – ….- 1⁄n2 dn
Now,
1⁄r2 dr = + 1⁄a2 da + 1⁄b2 db + ….+ 1⁄n2 dn
Multiplying by 100 and putting da = db = ……. = dn =k.
9. The approximate value of function f(x + δx, y + δy) is?
a) f + ∂f⁄∂x dx + ∂f⁄∂y dy
b) ∂f⁄∂x dx + ∂f⁄∂y dy
c) f – ∂f⁄∂x dx + ∂f⁄∂y dy
d) ∂f⁄∂x dx – ∂f⁄∂y dy
Answer: a
Explanation: f(x + δx, y + δy) = f(x,y) + df = f + ∂f⁄∂x dx + ∂f⁄∂y dy.
10. At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.
a) 0.342
b) 0.284
c) 0.154
d) 0.986
Answer: b
Explanation: Tan(z) = h⁄x
h = x Tan(z)
Taking log and then differentiate we get,
∂h⁄h = ∂x⁄x + 1⁄Tan(z) Sec2 (z)δz
Now h = 120 tan(60o) = 120√3
Putting, δx = 1⁄12 ft, δz = π⁄(60*180)
Putting the values we get,
δh = 0.284.
11. Find the approximate value of (1.04)3.01.
a) 1.14
b) 1.13
c) 1.11
d) 1.12
Answer: d
Explanation: Let, f(x,y) = xy
Now,
∂f⁄∂x = yx(y-1) and ∂f⁄∂y = xy log(x)
Putting, x = 1, y = 3, δx = 0.04, δy = 0.01
Now, df = δx ∂f⁄∂x + δy∂f⁄∂y = 0.12
Hence, f(x + δx,y + δy) = (1.04)3.01 = 1.12.
12. Find the approximate value of [0.982+2.012+1.942 ](1⁄2).
a) 1.96
b) 2.96
c) 0.04
d) -0.04
Answer: b
Explanation: Let f(x,y,z) = (x2+y2+z2 )(1⁄2) ……………..(1)
Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06
From (1),
∂f⁄∂x = x⁄f
∂f⁄∂y = y⁄f
∂f⁄∂z = z⁄f
And df = ∂f⁄∂x dx + ∂f⁄∂y dy + ∂f⁄∂z dz = ((xdx + ydy + zdz))/f = (-0.02 + 0.02 – 0.12)/3 = -0.04
Hence,
[0.982+2.012+1.942 ](1⁄2) = f(1, 2, 2) + df = 3-0.04 = 2.96.
13. Find the approximate value of log(11.01-log(10.1)), Given log(10) = 2.30 and and log(8.69) = 2.16, all the log are in base ‘e’.
a) 2.1654
b) 2.1632
c) 2.1645
d) 2.1623
Answer: d
Explanation: Let, f(x,y) = log(x-log(y))
Now by differentiating,
(frac{∂f}{∂x}=frac{1}{x-log(y)}) and (frac{∂f}{∂y}=frac{1}{y(x-log(y))})
Now, putting, x = 11, y = 10, δx=.01 and δy=.1
We get,
(frac{∂f}{∂x}∂f/∂x)=1/8.69 and (frac{∂f}{∂y}∂f/∂y)=1/86.9
Hence, df = 0.0023
Hence, f(x + δx, y + δy) = log(11.01 – log(10.1))= 2.16 + df = 2.1623.
14. Find approximate value of e10.19.09, given e90 = 1.22 * 1039.
a) 2.41 * 1039
b) 2.42 * 1039
c) 2.43 * 1039
d) 2.44 * 1039
Answer: b
Explanation: Let, f(x,y) = exy = exy
Now by differentiating,
∂f⁄∂x = yexy and ∂f⁄∂y = xexy
Now, putting, x = 10, y = 9, δx = .01 and δy = .09
We get,
∂f⁄∂x = 1.09* 1040 and ∂f⁄∂y = 1.22* 1040
Hence, df = 1.27* 1039
Hence, f(x + δx, y + δy) = e10.19.09 = 2.42 * 1039.