# Engineering Mathematics Questions and Answers – Errors and Approximations

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Errors and Approximations”.

1. Error is the Uncertainty in measurement.
a) True
b) False

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) δxx
c) δxx * 100
d) 0

Explanation: Option ‘δx’ is called absolute error.
Option ‘δxx’ is called relative error.
Option ‘δxx * 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

Explanation: Power is given by P = V2R
Taking log on both sides,
log(P) = 2log(V) – log(r)
Differentiating it ,
δpp = 2δVVδrr
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

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 = 12 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

Explanation: Given T = 12 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(1t – 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

Explanation: Given, v = k(1t – at)
Differentiating it we get
$$δv=k\left [\frac{(tδl-lδt)}{t^2} – aδt – tδa \right ]$$
Hence,
$$\frac{δv}{v}=\frac{k\left [\frac{(tδl-lδt)}{t^2} – aδt – tδa \right ]}{v}$$
$$\frac{δv}{v}=\frac{\left [\frac{(tδl-lδt)}{t^2} – aδt – tδa \right ]}{(\frac{l}{t}-at)}$$
Put, l = 2cm, t = 2sec, a = 0.95 mt/sec2
and δl = 1cm, δt = 1 sec⁡and δa = -1.05 mt/sec2
we get,
δvv = 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 + 43 r)
b) 1/(h + 23 r)
c) h/(h + 43 r)
d) r/(h + 43 r)

Explanation: Given V = πr2 h + 43 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) $$x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}$$

Explanation: Given 1r = 1a + 1b + 1c +⋯.. + 1n
Differentiating all,
1r2 dr = – 1a2 da – 1b2 db – ….- 1n2 dn
Now,
1r2 dr = + 1a2 da + 1b2 db + ….+ 1n2 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

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

Explanation: Tan(z) = hx
h = x Tan(z)
Taking log and then differentiate we get,
∂hh = ∂xx + 1Tan(z) Sec2 (z)δz
Now h = 120 tan(60o) = 120√3
Putting, δx = 112 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

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 ](12).
a) 1.96
b) 2.96
c) 0.04
d) -0.04

Explanation: Let f(x,y,z) = (x2+y2+z2 )(12) ……………..(1)
Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06
From (1),
∂f∂x = xf
∂f∂y = yf
∂f∂z = zf
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 ](12) = 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

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

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.

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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