Category: Engineering Mathematics Objective Questions

Fourier Analysis Interview Questions and Answers for Experienced people focuses on “Solution of 1D Heat Equation”.

1. The partial differential equation of 1-Dimensional heat equation is ___________
a) u_t = c²u_xx
b) u_t = pu_xx
c) u_tt = c²u_xx
d) u_t = – c²u_xx
View Answer

Answer: a
Explanation: The one-dimensional heat equation is given by u_t = c²u_xx where c is the constant and u_t represents the one time partial differentiation of u and u_xx represents the double time partial differentiation of u.

2. When using the variable separable method to solve a partial differential equation, then the function can be written as the product of functions depending only on one variable. For example, U(x,t) = X(x)T(t).
a) True
b) False
View Answer

Answer: a
Explanation: When solving a partial differential equation using a variable separable method, then the function can be written as the product of functions depending on one variable only.

3. The one dimensional heat equation can be solved using a variable separable method. The constant which appears in the solution should be __________
a) Positive
b) Negative
c) Zero
d) Can be anything
View Answer

Answer: b
Explanation: Since the problems are dealing on heat conduction, the solution must be a transient solution. Therefore the constant should be negative, i.e., k = – p².

4. When solving the 1-Dimensional heat equation for the conduction of heat along the rod without radiation with conditions:
i) u(x,t) is finite for t tends to infinite
ii) u_x(0,t) = 0 and u_x(l,t) = 0
iii) u(x,t) = x(l-x) for t=0 between x=0 and x=l, which condition is the best to use in the first place?
a) u_x(0,t) = u_x(l,t) = 0
b) u(x,t) = x(l-x) for t=0 between x=0 and x=l.
c) u(x,t) = x(l-x) for x=0 between t=0 and t=l.
d) u(0,t) = u(l,t) = 0
View Answer

Answer: a
Explanation: Boundary conditions are always used first to solve the partial differential equations. Using these boundary conditions, we can remove one constant thus making only one constant remaining to remove. The last constant is removed using the initial conditions.

5. Solve the 1-Dimensional heat equation for the conduction of heat along the rod without radiation with conditions:
i) u(x,t) is finite for t tends to infinite
ii) u_x(0,t) = 0 and u_x(l,t) = 0
iii) u(x,t) = x(l-x) for t=0 between x=0 and x=l.
a) U(x,t) =(frac{l^2}{3}/2 + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{-4l^2}{(2m)^2+π^2} )
b) U(x,t) =(frac{l^2}{3} + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{-4l^2}{(2m)^2+π^2} )
c) U(x,t) =(frac{l^2}{3} + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{4l^2}{(2m)^2+π^2} )
d) U(x,t) =(frac{l^2}{3}/2 + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{4l^2}{(2m)^2+π^2} )
View Answer

Answer: a
Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e^-p2 c²t)
u_x = (-cp sinpx + c’p cospx) (c’’ e^-p2 c²t)
Applying the first condition of second condition,
C’ = 0
Now applying the second condition of the second condition,
(p= frac{nπ}{l} )
Now we have only one constant left. This can be solved using the third condition.
U(x,t) =( frac{a_0}{2} + ∑cos⁡(frac{nπx}{l}) a_n e^{frac{-c^2 n^2 π^2 t}{l^2}} )
(a_0 = frac{2}{l} ∫_0^l x(l-x)dx = frac{l^2}{3} )
(a_n = frac{2}{l} ∫_0^l x(l-x) cosnx dx = frac{-4l^2}{(2m)^2+π^2}. )
(U(x,t) = frac{l^2}{3}/2 + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{-4l^2}{(2m)^2+π^2}.)

6. A rod of 30cm length has its ends P and Q kept 20°C and 80°C respectively until steady state condition prevail. The temperature at each point end is suddenly reduced to 0°C and kept so. Find the conditions for solving the equation.
a) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/10 x
b) u_x(0,t) = 0 = u_x(30,t) and u(x,0) = 20 + 60/30 x
c) u_t(0,t) = 0 = u_t(30,t) and u(x,0) = 20 + 60/10 x
d) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/30 x
View Answer

Answer: d
Explanation: 0 and 30 are the end points. So, at these points the function is zero. Hence u(0,t) = 0 = u(30,t). Next due to steady state conditions, at the beginning that is initial conditions, we have u(x,0) = 20 + 60/30 x.

7. Is it possible to have a solution for 1-Dimensional heat equation which does not converge as time approaches infinity?
a) Yes
b) No
View Answer

Answer: b
Explanation: It is not possible to have a solution which does not converge as time approaches infinity because the solution to a heat equation must be transient.

8. Solve the equation u_t = u_xx with the boundary conditions u(x,0) = 3 sin (nπx) and u(0,t)=0=u(1,t) where 0<x<1 and t>0.
a) (3∑_{n=1}^∞ ) e^{-n2 π2 t} cos⁡(nπx)
b) (∑_{n=1}^∞ ) e^{-n2 π2 t} sin⁡(nπx)
c) (3∑_{n=1}^∞ ) e^{-n2 π2 t} sin⁡(nπx)
d) (∑_{n=1}^∞ ) e^{-n2 π2 t} cos(nπx)
View Answer

Answer: c
Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e^-p2t)
When x=0, c=0 and when x=1, p=nπ.
When t=0, 3 sin (nπx) = (∑_{n=1}^∞ ) b_n e^{-n2 π2 t} sin⁡(nπx)
Therefore b_n =3 for all n
Hence the solution is (3∑_{n=1}^∞ ) e^{-n2 π2 t} sin⁡(nπx).

9. If two ends of a bar of length l is insulated then what are the conditions to solve the heat flow equation?
a) u_x(0,t) = 0 = u_x(l,t)
b) u_t(0,t) = 0 = u_t(l,t)
c) u(0,t) = 0 = u(l,t)
d) u_xx(0,t) = 0 = u_xx(l,t)
View Answer

Answer: a
Explanation: Since the ends are insulated no heat can flow through the ends of the bar. Therefore u_x(0,t) = 0 = u_x(l,t).

10. The ends A and B of a rod of 20cm length are kept at 30°C and 80°C until steady state prevails. What is the condition u(x,0)?
a) 20 + ⁵⁄₂ x
b) 30 + ⁵⁄₂ x
c) 30 + 2x
d) 20 + 2x
View Answer

Answer: b
Explanation: u_xx=0.
The solution to this equation is u=a+bx. Since at x=0, u=30 and at x=20, u=80,
a = 30 and b = (frac{(80-30)}{20} = frac{5}{2} )
Therefore, u= 30 + (frac{5}{2} x).

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Engineering Mathematics Interview Questions and Answers for Experienced people focuses on “Lagrange’s Mean Value Theorem – 2”.

1. Mean Value Theorem tells about the
a) Existence of point c in a curve where slope of a tangent to curve is equal to the slope of line joining two points in which curve is continuous and differentiable
b) Existence of point c in a curve where slope of a tangent to curve is equal to zero
c) Existence of point c in a curve where curve meets y axis
d) Existence of point c in a curve where curve meets x axis
Answer: a
Explanation: Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).

2. If f(a) is euquals to f(b) in Mean Value Theorem, then it becomes
a) Lebniz Theorem
b) Rolle’s Theorem
c) Taylor Series of a function
d) Leibnit’x Theorem
Answer: b
Explanation: According to Mean Value Theorem, If a function exist at pt. ‘a’, ‘b’ and continuous in closed interval [a, b] and differentiable in open interval (a, b) then there exists a point ‘c’, such that c∈(a, b), Where,
f’(c)= [f(b)-f(a)]/(b-a).
Hence, By putting f(b) = f(a) in the statement of Mean Value Theorem, we get
f’(c) = [f(b)-f(a)]/(b-a) = 0. Which is a statement of Rolle’s Theorem.

3. Mean Value theorem is applicable to the
a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, b)
b) Functions continuous in closed interval [a, b] only and having same value at point ‘a’ and ‘b’
c) Functions continuous in closed interval [a, b] and differentiable in open interval (a, b)
d) Functions differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’
Answer: c
Explanation: Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).

4. Mean Value theorem is also known as
a) Rolle’s Theorem
b) Lagrange’s Theorem
c) Taylor Expansion
4) Leibnitz’s Theorem
Answer: b
Explanation: Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a, b) and f’(c) = [f(b)-f(a)]/(b-a).
It is also known as Lagrange’s Theorem.

5. Find the point c in the curve f(x) = x³ + x² + x + 1 in the interval [0, 1] where slope of a tangent to a curve is equals to the slope of a line joining (0,1)
a) 0.64
b) 0.54
c) 0.44
d) 0.34
Answer: b
Explanation: f(x) = x³ + x² + x + 1
f(x) is continuous in given interval [0,1].
f’(x) = 3x²+2x+1
Since, value of f’(x) is always finite in interval (0, 1) it is differentiable in interval (0, 1).
f(0) = 1
f(1) = 4
By mean value theorem,
f’(c) = 3c² + 2c + 1 = (4-1)/(1-0) = 3
⇒ c = 0.548,-1.215
Since c belongs to (0, 1) c = 0.54.

6. Find the equation of curve whose roots gives the point which lies in the curve f(x) = xSin(x) in the interval [0, ^π⁄₂] where slope of a tangent to a curve is equals to the slope of a line joining (0, ^π⁄₂)
a) c = -Sec(c) – Tan(c)
b) c = -Sec(c) – Tan(c)
c) c = Sec(c) +Tan(c)
d) c = Sec(c) – Tan(c)
Answer: d
Explanation: f(x) = xSin(x)
Since f₁(x) = x and f₂(x)=Sin(x) both are continuous in interval [0, ^π⁄₂], the curve f(x)=f₁(x)f₂(x) is also continuous.
f’(x) = xCos(x) + Sin(x)
f’(x) always have finite value in interval [0, ^π⁄₂] hence it is differentiable in interval (0, ^π⁄₂).
f(0) = 0
f(^π⁄₂) = ^π⁄₂

By mean value theorem,
f’(c) = cCos(c) + Sin(c) = (^π⁄₂ – 0)/(^π⁄₂ – 0)=1
Hence, c = Sec(c) – Tan(c) is the required curve.