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IIT JAM Mathematics Practice Test- 17 - Mathematics MCQ


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30 Questions MCQ Test IIT JAM Mathematics Mock Test Series - IIT JAM Mathematics Practice Test- 17

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IIT JAM Mathematics Practice Test- 17 - Question 1

Let  are three arbitrary rector’s. Then the rector's  are

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 1

we know three rectors will be coplanar if their scalar triple product is zero

we know two vector’s are collinear iff on vector is the scalar multiple of other. Here among  None o f vector be the Scalar multiple of other.

IIT JAM Mathematics Practice Test- 17 - Question 2

The unit vector normal to the surface x2y + 2xz = 4 at the point (2,-2,3) is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 2

Given f (x,y,z ) = x2y + 2xz - 4 (surface) then for any given surface we know that the unit vector is defined as 

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IIT JAM Mathematics Practice Test- 17 - Question 3

The derivative of f(x, y) at point (1, 2) in the direction of vector i + j is 2 √2 and in the direction of the vector - 2j is - 3 . Then derivative of f(x, y) in the direction - i - 2j is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 3

∴ Derivative in the direction  

IIT JAM Mathematics Practice Test- 17 - Question 4

The point at which the derivative of the function f(x,y) = x2 - xy - y + y2 vanishes along the direction  is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 4

IIT JAM Mathematics Practice Test- 17 - Question 5

Let  where a is a constant. If the line integral  over every closed curve c is zero, then a is equal to

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 5

Since  and we know that if the line integral is zero then given curve c is closed.

hence

IIT JAM Mathematics Practice Test- 17 - Question 6

If  are coplanar vector’s then  is equal to?

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 6

IIT JAM Mathematics Practice Test- 17 - Question 7

Let  denote the force filed on a particle traversing the path L from (0,0,0) to (1,1,1) along the cuive of intersection of the cylinder y= x2 and the plane z = x the work done by  is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 7

work done

IIT JAM Mathematics Practice Test- 17 - Question 8

For a solenoidal vector field  which of the following is not true ?

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 8


by equation (2) and (3)


 

Now use this in (1), we have


= All A,B,C option are correct but (D) is not correct.

IIT JAM Mathematics Practice Test- 17 - Question 9

If S be the surface of sphere x2 + y2 + z2 = 9.

the integral  is equal to 

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 9

The surface element

where

Let V be the volume enclosed by the sphere (surface) then by Gauss’s div. the 

IIT JAM Mathematics Practice Test- 17 - Question 10

The value of surface integral over the surface of the paraboloid z = 1 - x2 - y2, z > 0, where 

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 10

By the stock’s theorem we know that

where C is a closed curve bounded by the surface  

x2 +y2 = 1

IIT JAM Mathematics Practice Test- 17 - Question 11

If the vector's  satisfy the condition  then  is equal to

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 11

given

then by (1)

IIT JAM Mathematics Practice Test- 17 - Question 12

The line integral of the vector field,  along the boundary of the triangle with vertices (1,0,0), (0,2,0), and (0,0,1), oriented anticlock wise, when viewed from the point (2,4,2) is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 12

we will compute

Let S is the surface of Δ ABC, then the eqnof surface is given by 

Now here to evaluate (1), we can use stokes theorem Then we have

where  is outward going UNV to S  and given by

Now

= i(z) - j (-x) + k(y) = zi + xj + yk

Now take orthogonal projection of S on the xy-plane.

then we have

Then by (2)

IIT JAM Mathematics Practice Test- 17 - Question 13

The value of integral

where C is the curve of intersection of the surface x2 + y2 + z2 = a2 bounded by the plane x + y + z = 0, is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 13

Given line integral is

By stoke's th.


Here

then curve

IIT JAM Mathematics Practice Test- 17 - Question 14

The vector  bisects the angle between the vectors  then the unit vector along 

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 14

⇒ Bisector of C and  is

and

Hence

and unit vector along 

IIT JAM Mathematics Practice Test- 17 - Question 15

The vectors  are

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 15

if  are orthogonal then 

it is not possible for real p and q.
Never orthogonal.

IIT JAM Mathematics Practice Test- 17 - Question 16

Let T be the smallest positive real no. such that the tangent to the helix

at t = T is orthogonal to the tangent at t = π/2

Then the line integral of  along the section of the helix from t = 0 to t  = T is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 16

Let

then  will a tangent vector

Given T1 and T2 are orthogonal

IIT JAM Mathematics Practice Test- 17 - Question 17

Let c1 : x2 + y2 = 1 and be the curves from (-1,0,2) to (1,0,2) drawn counter clock wise in the lower half (y ≤ 0) of the plane z = 2 and let   be a vector point function. Let the line integrals be  then

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 17

given

IIT JAM Mathematics Practice Test- 17 - Question 18

Let S be the boundary of the region consisting of the parabolic cylinder z = 4 - x2 and the planes y = 0 , y = 2 and z = 0 and  be a vector point function, then the value of  is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 18

Here S be a closed surface so we will apply gauss divergence theorem,

then we have


IIT JAM Mathematics Practice Test- 17 - Question 19

The work clone by the force  in moving a particle from the origin O ( 0 ,0 ,0 ) to the poin t D (1,1,0) on t h e z = 0 plane along the paths OABD  as shown in the figure (where the coordinates are measured in meters) is,

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 19

since

So  is conservative vector field. 

Hence

work done in moving particle from origin o to D is

IIT JAM Mathematics Practice Test- 17 - Question 20

If  then calculate the flux of  out of the region through the surface at z = c where region is bounded by -a ≤ x ≤ a, - b ≤ y ≤ b

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 20

and ds = dx/dy

flux across z = c,

Flux across the closed surface

IIT JAM Mathematics Practice Test- 17 - Question 21

The circulation of the field  around the curve C where C is the intersection of the sphere x2+ y2 + z2 = 25 and the plane z = 3, is,

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 21

C is the curve of intersection of surfaces 

let x=4cos θ, y=4sinθ Hence,


IIT JAM Mathematics Practice Test- 17 - Question 22

Let  j and let L be the curve is equal to

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 22

Let  be the position vector of any point on L (in Cartesian from) 

then


we curve in parametric form as


 

⇒ L be the arc of given curve from  
 

Now by (1) it is clear that  is conservative so 

IIT JAM Mathematics Practice Test- 17 - Question 23

Let  be a vector point function and S be an open surface x2 + y2 - 4x + 4z = 0 , z ≥ 0, Then the value of  is equal to 

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 23

Here

now S be an open surface so we will use stoke’s theorem here. 


Here curve C is the intersection of x2 + y2 - 4x + 4z = 0 and z = 0 x2 + y2 - 4x = 0 = x2 - 4x + 4 + y2 = 4 = (x -2)2 + y2 = 4 be a circle with centre (2,0) and radius 2.
Now on C:


Now we can use Green’s Theorem here so Here

M = y2-x2 ; N = x2-y2

IIT JAM Mathematics Practice Test- 17 - Question 24

Let  and  be a scalar function which satisfy Laplace eqn. Then the vector field  is _______.

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 24

If we take,

F. dr = 2xyz (cosx2 + 1)dx + z (sinx2 + x2) dy + y (sinx2 + x2)dz = 2xyz cosx2 dx + z sinx2 dy + y sinx2 dz + 2xyz dx + x2 z dy + x2y dz = d(yz sinx2) + d(x2yz)

and φ is the soln of Laplace eqn

 not a solenoidal vector field.

IIT JAM Mathematics Practice Test- 17 - Question 25

Let be a constant vector and  then is equal to

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 25

we know

Now

Put these values in (1), we have

IIT JAM Mathematics Practice Test- 17 - Question 26

If  then the value of   at the point (-1, 1, 1) is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 26

Now

IIT JAM Mathematics Practice Test- 17 - Question 27

If  then f is

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 27

If f is irrotational then

So First we find curl 

IIT JAM Mathematics Practice Test- 17 - Question 28

Let  is a unit vector, then for the maximum value of the scalar triple product  will be equal to ?

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 28

Now

{ where θ is the angle between 

Hence  is maximum if cos θ = 1 i.e. θ = 0°

IIT JAM Mathematics Practice Test- 17 - Question 29

Let there be two points A and B on the curve y = x2 in the plane OXY satisfying   then length of the vector 

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 29

Let

since given that 

again

IIT JAM Mathematics Practice Test- 17 - Question 30

Find equations for the tangent plane to the surface z = x2 + y2 at the point (2, -1, 5).

Detailed Solution for IIT JAM Mathematics Practice Test- 17 - Question 30

Surface z = x2 + y2

⇒ surface S = x2 + y2 - z = 0 
hence

  (1)

the equation of a plane, passing through a point whose position vector is r0 and which is perpendicular to the normal N 

(r - r0).N = 0

then the required equation is

⇒ 4(x - 2) - 2(y + 1) - 1 (z - 5) = 0

⇒ 4 x - 8 - 2 y - 2 - z + 5 = 0

⇒ 4x - 2y - z = 5

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