Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Engineering Mathematics  >  Test: Vector Calculus - Civil Engineering (CE) MCQ

Test: Vector Calculus - Civil Engineering (CE) MCQ


Test Description

10 Questions MCQ Test Engineering Mathematics - Test: Vector Calculus

Test: Vector Calculus for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Vector Calculus questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Vector Calculus MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Vector Calculus below.
Solutions of Test: Vector Calculus questions in English are available as part of our Engineering Mathematics for Civil Engineering (CE) & Test: Vector Calculus solutions in Hindi for Engineering Mathematics course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt Test: Vector Calculus | 10 questions in 30 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Engineering Mathematics for Civil Engineering (CE) Exam | Download free PDF with solutions
Test: Vector Calculus - Question 1

Given the vector A = (cos x)(cos y)âx + (cos x)(cos y) ây, âx & ây denote unit vectors along x,y directions respectively. The curl of A is ________.

Detailed Solution for Test: Vector Calculus - Question 1

Concept:

Let 

The curl of A is evaluated as:

Given

Ax = cos x cos y

Ay = cos x cos y

Az = 0

∇ × A = (- sinx cosy + cosx siny) âz

Test: Vector Calculus - Question 2

The following surface integral is to be evaluated over a sphere for the given steady vector field, F = xi + yj + zk defined with respect to a Cartesian coordinate system having i, j, and k as unit base vectors.

, Where S is the sphere, x2 + y2 + z2 = 1 and n is the outward unit normal vector to the sphere. The value of the surface integral is

Detailed Solution for Test: Vector Calculus - Question 2

Gauss divergence theorem:

It states that the surface integral of the normal component of a vector function  taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function  taken over a volume enclosed by the closed surface ‘S’.

Given:

F = xi + yj + zk

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Vector Calculus - Question 3

For the curve xy3 - yx3 = 6, the slope of the tangent line at the point (1, -1) is:

Detailed Solution for Test: Vector Calculus - Question 3

Concept:

Let y = f(x) be the equation of a curve, then the slope of the tangent at any point say (x1, y1) is given by:

Calculation:

Given curve is xy3 - yx3 = 6

Now by partially differentiating the equation of curve with respect to x we get;

The slope(m) i.e. dy/dx of the tangent at (1, -1) is:

m =  -1

Test: Vector Calculus - Question 4

The parabolic arc y = √x, 1 ≤ x ≤ 2 is revolved around the x-axis. The volume of the solid of revolution is

Detailed Solution for Test: Vector Calculus - Question 4

Revolution about x-axis: The volume of the solid generated by the revolution about the x-axis, of the area bounded by the curve
y = f(x), the x-axis and the ordinates
x = a and x = b is

similarly for revolution about y-axis:

V=∫πx2dy

Calculation:

Given:

V = 3π / 2

Hence the required volume will be
3π / 2.

Test: Vector Calculus - Question 5

The value of the line integral

along a path joining the origin  and the point (1,1,1)  is

Detailed Solution for Test: Vector Calculus - Question 5

Concept:

When two points (x1, y1. z1) and (x1, y1. z2) are mentioned find the relation in terms of the third variable in terms of x,y, and z:

Put the value of z,y, and z and use the end-points of one variable.

Given:

I = ∫(2xy2dx + 2x2ydy + dz), A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

∴ x = y = z = t and t : 0 → 1

∴ 

∴ 

*Answer can only contain numeric values
Test: Vector Calculus - Question 6

The volume determined from ∫∫∫v 8 xyz dv for V = [2, 3] × [1, 2] × [ 0,1 ] will be (in integer) ________.


Detailed Solution for Test: Vector Calculus - Question 6

Given

Integral

∫∫∫v 8 xyz dv 

Limits for x, y and z is given as

[2, 3] × [1, 2] × [0, 1]

Volume of the integral

V = ∫∫∫v 8 xyz dv 

i.e. V = ∫ ∫ ∫V 8 xyz dxdydz

V = 5 × 3 × 1

V = 15 

∴ Volume is 15 

Test: Vector Calculus - Question 7

If 2î + 4ĵ - 5k̂ and î + 2ĵ + 3k̂ are two different sides of rhombus. Find the length of diagonals.

Detailed Solution for Test: Vector Calculus - Question 7

If aî + bĵ + ck̂ and pî + qĵ + rk̂ are 2 different sides of the rhombus.

Suppose  = aî+bĵ+ck̂ and B = = pî+qĵ+rk̂

Then anyone diagonal of the rhombus is given by 

The other diagonal is given by 

The magnitude of the vector 

The magnitude of the vector diagonal D1

D1 = 7 

The other diagonal is given by 

D2 = - 1î - 2ĵ + 8k̂

The magnitude of the vector diagonal D2

∴ The length of diagonals of a rhombus is 7 and √69.

Test: Vector Calculus - Question 8

The vector function expressed by

F = ax(5y − k1z) + ay(3z + k2x) + az(k3y−4x)

Represents a conservative field, where ax, ay, az are unit vectors along x, y and z directions, respectively. The values of constant k1, k2, k3 are given by: 

Detailed Solution for Test: Vector Calculus - Question 8

Concept:

For a vector F = F1i + F2j + F3k

For irrotational (or) conservative field  (or) Null Vector.

Calculation:

Given that,

 is a conservative field.

k3 – 3 = 0 ⇒ k3 = 3

-4 + k1 = 0 ⇒ k1 = 4

k2 – 5 = 0 ⇒ k2 = 5

The required values are: k1 = 4, k2 = 5, k3 = 3

Test: Vector Calculus - Question 9

Green's theorem is used to-

Detailed Solution for Test: Vector Calculus - Question 9

Green's theorem

  • It converts the line integral to a double integral. 
  • It transforms the line integral in xy - plane to a surface integral on the same xy - plane.

If M and N are functions of (x, y) defined in an open region then from Green's theorem

Test: Vector Calculus - Question 10

The value of from A(a, 0, 0) to B(a, 0, 2π b) along r̅ = (a cos t) î + (a sin t) ĵ + b t k̂ is:

Detailed Solution for Test: Vector Calculus - Question 10

The value of  From A(a, 0, 0) to B(a, 0, 2π b) along r̅ = (a cos t) î + (a sin t) ĵ + b t k̂ is: πa (bĵ + ak̂)

65 videos|120 docs|94 tests
Information about Test: Vector Calculus Page
In this test you can find the Exam questions for Test: Vector Calculus solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Vector Calculus, EduRev gives you an ample number of Online tests for practice

Top Courses for Civil Engineering (CE)

65 videos|120 docs|94 tests
Download as PDF

Top Courses for Civil Engineering (CE)