IIT JAM Mathematics Practice Test- 9 - Mathematics MCQ

Resultant force F = 5i + 3j + 2k and displacement d = 2i - 2j + 10k, then Work done W = F.cl = 10 - 6 + 20 = 24 unit.

Take

Required work done = (Force vector). (Displacement vector)

Force vector 

∴ Required work done

The parametric equation of the ellipse is 

x = 3cosθ, y = 2sinθ, z = 0

∴ r = (3cosθ, 2sinθ, 0), dr/dθ = (-3sinθ, 2cos0, 0), and

F = (x - 3y, 2x - y, 0) = (3cosθ - 6sinθ, 6cosθ - 2sinθ, 0).

Now,

F .dr = ( xzi + yj + x²k ).( idx + jdy + kdz) = x z dx + y dy + x² dz 

∴ Along the given curve, we have

Putting values,

= 

Here is P and Q as well as the appropriate derivatives.

Let  be a vector field on an open and simply-connected region D

and  

The vector field f is conservative.
Here the two partial derivatives are equal and so this is a conservative vector field.

For an irrotational vector function

As  are independent vectors, so their coefficients must be zero separately, i.e.,

c + 1 = 0, 4 - a = 0, b - 2 = 0

These equations give

a = 4 ,b = 2, c = - 1

Here S is an open surface so we can apply stokes Theorem , 

= 3y dx - 4 x dy    ( ∵ z = 4 ⇒ dz = 0 )

Let x = 2 cos θ , y = 2 sin θ

⇒ dx = -2 sin θ dθ , dy = 2 cosθ dθ

= i [ - 3z² - 2x² y ] - j [0] + k [ 4 xyz - 2xy ]

we have

= i[- 1 ]- j[- 1 ] + k[-1]

= H + j - k )

Take o.p. on xy - plane that be a rectangle bounded by 0 ≤ x ≤ n , 0 ≤ y ≤ 2

Since x = 0 for this surface

given

we know 

Take

=> (A) is correct

Now

again

Hence

Hence the maximum directional derivative at the point (2,1 ,-1 )=  

Again the magnitude of the directional derivative at the point (2,1 ,-1)   

Now, we could use the techniques we discussed when we first looked at line integrals of vector fields. However that would be particularly unpleasant solution.
This vector field is conseivative and a potential function for the vector field is,

Using this, we know that integral must be independent of path and so all we need to do is, to use the theorem from the previous section to do the evaluation.

where

since

Now the unit vector t in the direction of the vector i - j + k is given by t  = 

∴ The required directional derivative is

Now angle between at ( 2 , -1 , 2 ) be 

The directional derivative is

If the vector A is solenoidal, then div A= 0.

Given

if is identically equal to zero , then all three components of  are zero

We have x²+ y² = 9 point on the circ le is represented as ( 3cos θ , 3 sin θ)

Given W = 3/2

Here both the values are possible .

here are two possible locations for Q so that the work done by  is equal to 3/2

since

so