Test 4: Vector Analysis

Instruction: Select the correct option corresponding to questions given below in the form shared at the bottom

Total Marks: 75

1) Find the directional derivative of ϕ(x,y,z)=x2yz+4xz2 at the point (1,-2,-1) in the direction of the vector 2→i−→j−2→k.

a) 379
b) 373
c) 376
d) 377

2) Find the angle between the surfaces x2+y2+z2=9 and x2+y2−z=3 at the point (2,−1,2).

a) cos−1(83√21)
b) cos−1(83√22)
c) cos−1(83√23)
d) cos−1(83√24)

3) Find the value of a if the vector →F=(2x2y+yz)→i+(xy2−xz2)→j+(axyz−2x2y2)→k is solenoidal.

a) −6
b) 6
c) 3
d) −3

4) The value of ∇⋅(r∇(1r3)) is given by:

a) 1 b) 0 c) 3r4
d) −3r4

5) Evaluate the line integral ∫C(y2dx−x2dy) around the triangle whose vertices are (1,0),(0,1) (-1,0) in the positive sense.

a) =−23
b) =−43
c) =−−23
d) =−−43

6) Evaluate ∫C[(sinx−y)dx−cosxdyl, where C is the triangle with vertices (0,0),(π2,0) and (π2,1).

a) 2π+π4
b) 2π+π2
c) 2π+π3
d) 4π+π4

7) Evaluate ∬S→F⋅→ndS if →F=4y→i+18z→j−x→k and S is the surface of the plane 3x+2y+6z=6 contained in the first octant.

a) 9
b) 10
c) 11
d) 12

8) Evaluate ∬S→F⋅→ndS if →F=yz→i+zx→j+xy→k and S is part of the surface x2+y2+z2=1 which lies in the first octant.

a) 38
b) 34
c) 58
d) 55

9) Using divergence theorem, evaluate ∬→F⋅→ndS, where →F=4xz→i−y2→j+yz→k and S is the surface of the cube bounded by the planes x=0,x=2,y=0,y=2,z=0,z=2.

a) 24
b) 26
c) 28
d) 30

10) Evaluate ∬Sx3dydz+x2ydzdx+x2zdxdy over the surface z=0,z=h,x2+y2=a2.

a) 54πa4h
b) 34πa4h c) 74πa4h
d) 14πa4h

11) Evaluate ∫C(xydx+xy2dy) by Stoke’s theorem, where C is the square in the xy− plane with vertices (1,0), (−1,0), (0,1), (0,−1).

a) 13
b) 23 c) 1
d) 43

12) Using Stoke’s theorem, evaluate ∫C→F⋅d→r, where →F=y2→i+x2→j−(x+z)→k and C is the boundary of the triangle with vertices at (0,0,0), (1,0,0), (1,1,0).

a) 13
b) 23 c) 1
d) 43

13) If →r+x→i+y→j+z→k and r=|→r|, then ∇(logr) is equal to:

a) 1r2 b) →rr3 c) →rr2
d) 1r

14) The value of b if the surfaces ax2−byz=(a+2)x and 4x2y+z3=4 cut orthogonally at the point (1,−1,2) is given by:

a) 2
b) 1
c) 3
d) 4

15) If →r=x→i+y→j+z→k and r=|→r|, then rn→r is solenoidal for:

a) n=3
b) n=−3
c) n=1
d) n=−1