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}})