Vector Calculus NAT Level - 2

# Vector Calculus NAT Level - 2

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Vector Calculus NAT Level - 2 for IIT JAM 2023 is part of Topic wise Tests for IIT JAM Physics preparation. The Vector Calculus NAT Level - 2 questions and answers have been prepared according to the IIT JAM exam syllabus.The Vector Calculus NAT Level - 2 MCQs are made for IIT JAM 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Vector Calculus NAT Level - 2 below.
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*Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 1

### Evaluate where and S is the part of the plane 2x + 3y + 6z = 12 which is located in the first octant.

Detailed Solution for Vector Calculus NAT Level - 2 - Question 1   = 24

*Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 2

### Find the value of constant (a + b + c) so that the directional derivative of the function f = axy2 + byz + cz2x3 at the point (1, 2, –1) has maximum magnitude 64 in the direction  parallel to y axis :

Detailed Solution for Vector Calculus NAT Level - 2 - Question 2  lies along y axis
So, 4a + 3c = 0
2b – 2c = 0  *Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 3

### Evaluate the along the portion from path (1, 0, 1) to (3, 4, 5) of the curve C, which is the intersection of the surface z2 = x2 + y2 and z = y + 1.

Detailed Solution for Vector Calculus NAT Level - 2 - Question 3 can be expressed as gradient of scalar function   written the common term once The line integral along the portion (1, 0, 1) to (3, 4, 5) = (375 + 108 – 20) – (1)
= 463 – 1
= 462

*Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 4

The work done by the force in moving a particle over circular path x2 + y2 = 1, z = 0 from (1, 0, 0) to (0, 1, 0) is :

Detailed Solution for Vector Calculus NAT Level - 2 - Question 4 on the curve  *Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 5

Evaluate where C is the path shown in figure. Detailed Solution for Vector Calculus NAT Level - 2 - Question 5

Path AO
y
= 1
dy = 0 Path OB
x
2 + y2 = 1    = –0.416

*Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 6

Let C be any curve x2 + y2 + z2 = 4, z > 0 and the vector field find out (Ans. upto three decimal places)

Detailed Solution for Vector Calculus NAT Level - 2 - Question 6   Consider a closed surface consisting of S and S' i.e   *Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 7

The value of the and C is the curve y2 = x joining (0, 0) to (1, 1) is (correct upto three decimal places)

Detailed Solution for Vector Calculus NAT Level - 2 - Question 7 *Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 8

Find the value of Detailed Solution for Vector Calculus NAT Level - 2 - Question 8

Let   *Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 9 along the curve x = sin θ cos θ, y sin2 θ, z = cos θ with θ  increasing from 0 to π/2. Find the value of α + β.

Detailed Solution for Vector Calculus NAT Level - 2 - Question 9 xz dx + y dy + xdz
Along the given curve, we have  Putting values  *Answer can only contain numeric values
Vector Calculus NAT Level - 2 - Question 10

If f(x, y, z) = x2y + y2z + z2x for all (x, y, x) ∈ R3 and then the value of at (2, 2, 2) is :

Detailed Solution for Vector Calculus NAT Level - 2 - Question 10 Hence, ## Topic wise Tests for IIT JAM Physics

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