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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Gradient

Test: Gradient for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Gradient questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Gradient MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Gradient below.
Solutions of Test: Gradient questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Gradient solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Gradient | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
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### Gradient of a function is a constant. State True/False.

Detailed Solution for Test: Gradient - Question 1

Explanation: Gradient of any scalar function may be defined as a vector. The vector’s magnitude and direction are those of the maximum space rate of change of φ.

### The mathematical perception of the gradient is said to be

Detailed Solution for Test: Gradient - Question 2

Explanation: The gradient is the rate of change of space of flux in electromagnetics. This is analogous to the slope in mathematics.

### Divergence of gradient of a vector function is equivalent to

Detailed Solution for Test: Gradient - Question 3

Explanation: Div (Grad V) = (Del)2V, which is the Laplacian operation. A function is said to be harmonic in nature, when its Laplacian tends to zero.

4. The gradient of xi + yj + zk is

Detailed Solution for Test: Gradient - Question 4

Explanation: Grad (xi + yj + zk) = 1 + 1 + 1 = 3. In other words, the gradient of any position vector is 3.

Find the gradient of t = x2y+ ez at the point p(1,5,-2)

Detailed Solution for Test: Gradient - Question 5

Explanation: Grad(t) = 2xy i + x2 j + ez k. On substituting p(1,5,-2), we get 10i + j + 0.135k

Curl of gradient of a vector is

Detailed Solution for Test: Gradient - Question 6

Explanation: Gradient of any function leads to a vector. Similarly curl of that vector gives another vector, which is always zero for all constants of the vector. A zero value in vector is always termed as null vector(not simply a zero).

Find the gradient of the function given by, x2 + y2 + z2 at (1,1,1)

Detailed Solution for Test: Gradient - Question 7

Explanation: Grad(x2+y2+z2) = 2xi + 2yj + 2zk. Put x=1, y=1, z=1, the gradient will be 2i + 2j + 2k.

The gradient can be replaced by which of the following?

Detailed Solution for Test: Gradient - Question 8

Explanation: Since gradient is the maximum space rate of change of flux, it can be replaced by differential equations.

When gradient of a function is zero, the function lies parallel to the x-axis. State True/False.

Detailed Solution for Test: Gradient - Question 9

Explanation: Gradient of a function is zero implies slope is zero. When slope is zero, the function will be parallel to x-axis or y value is constant.

Find the gradient of the function sin x + cos y.

Detailed Solution for Test: Gradient - Question 10

Grad (sin x + cos y) gives partial differentiation of sin x+ cos y with respect to x and partial differentiation of sin x + cos y with respect to y and similarly with respect to z. This gives cos x i – sin y j + 0 k = cos x i – sin y j.

## Electromagnetic Fields Theory (EMFT)

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