This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Errors and Approximations”.
1. Error is the Uncertainty in measurement
Explanation: In the term of mathematics, “Error tells the person how much correct or certain its measurement is.”
2. Relative error in x is
c) δx⁄x * 100
Option ‘a’ is called absolute error.
Option ‘b’ is called relative error.
Option ‘c’ 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
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
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
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
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.
Given, v = k(1⁄t – at)
Differentiating it we get
Put, l = 2cm , t = 2sec , a = 0.95 mt/sec2
and δl = 1cm ,δt = 1 secand δa = -1.05 mt/sec2
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)
Given V = πr2 h + 4⁄3 πr3
Now since error in radius is zero , it should be treated as constant, Hence,
Explanation: Given 1⁄r = 1⁄a + 1⁄b + 1⁄c +⋯.. + 1⁄n
– 1⁄r2 dr = – 1⁄a2 da – 1⁄b2 db – ….- 1⁄n2 dn
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
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.
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
Explanation: Let, f(x,y) = xy
∂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)
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
∂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’.
Explanation: Let, f(x,y) = log(x-log(y))
Now by differentiating,
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
∂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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