Unit-4 Set -A 1. The partial differential equation corresponding to the equation z  f ( x 2  y 2 ) is

z z y 0 x y z z b) y x 0 x y z z c) y x 0 x y a)

x

d)

z z  0 x y

2z 2z  2 z z z 3 2    0 is 2. The partial differential equation 2 2  6 xy x y x x a) b) c) d)

Parabolic Hyperbolic Elliptic None of these

3. The initial solution of the partial differential equation

 2 z z z    0 by separation of x 2 x y

variable is a) z  X ( x)  Y ( y) b) c)

z  X ( x)  Y ( y ) z  X ( x).Y ( y)

d) None of these 4. The solution of the partial differential equation a)

Ae kx e kt / 3

b)

Ae kx e 3kt

c)

Ae k / 3 x e kt / 3

d) None of these

z z  3  0 by separation of variables is x y

5. The general solution of the equation

2  2u  u  2  u  c , u (0, t )  0, u (l , t )  0,    0, is 2 2 t x  t  t 0

2 2 nx c3 e c k t l 2 2 2 2 nx b) c 2 sin c3 e  c n  t / l l nx nct c) c 2 sin c3 cos l l

a) c 2 sin

d) None of these 6. The value of bn of the solution of the equation 2  2u  u  2  u c , u (0, t )  0, u (l , t )  0,    0, u ( x,0)  x is 2 2 t x  t  t 0

a)



2 (1) n n

2 (1) n n 1 c)  (1) n n 1 d) (1) n n b)

7. The solution of one dimensional heat equation 2  2u u 2  u  c , u (0, t )  0, u (3, t )  0,  0, u ( x,0)  5 sin x  2 sin 2x 2 2 t t 0 t x a) u( x, t )  5 sin(x). cos  ct  2 sin(2x). cos 2 ct b) u( x, t )  5 sin(x). cos  ct  2 sin(x). cos  ct c) u( x, t )  5 sin(x). cos  ct  2 sin(2x). cos 2 ct

d) None of these

 2u  2u 8. The second order P.D.E. x 2  y 2  0 is x y a) b) c) d)

Elliptic if x<0,y>0 Hyperbolic if x<0,y<0 Elliptic if x>0,y>0 Hyperbolic if x>0,y>0

9. Let u(x, t) satisfy

u  2 u  , u (0, t )  0, u (1, t )  0, , u ( x,0)  sin 2x than t x 2

a) u( x, t )  sin(2x) e 

2

t

 sin(2x) e u( x, t )   sin(2x) e

b) u ( x, t )  c)

 n 2 2t  n  2t / l 2

d) None of these 10. The general solution of the equation 2  2u  u  2  u  c , u (0, t )  0, u (l , t )  0,    0, u ( x,0)  f ( x) is 2 2 t x  t  t 0

nx c 2n 2 2t / l 2 e l nx c 2n 2 2t / l 2 b)  bn sin e l nx n 2 2t / l 2 c)  bn sin e l a)

b

n

sin

d) None of these 11. The P.D.E. corresponding to the equation z  ( x  a) 2  ( y  b) 2 ,where a and b are arbitrary constants is

 z   z  a) 4 z        x   x 

2

 z   z  b) 4 z        x   x 

2

2

2

 z   z  z      x   x 

2

 z   z  d) z        x   x 

2

2

c)

2

12. Which one of the following is correct? a) Wave equation is elliptic and Laplace equation is Hyperbolic b) Wave equation is Hyperbolic and Laplace equation is elliptic c) Wave equation and Laplace equation are elliptic d) Wave equation and Laplace equation are Hyperbolic 13. Which of the following is the solution of a) u( x, y)  e y 2 x b) u( x, y)  (2 x  y) 3 c)

u( x, y)  ( y  2 x)e y 2 x

d) u( x, y)  ( y  2 x)e y 2 x

u u 2 0 x y

14. The solution of the Laplace equation by separation of variables is

(c1 cos kx  c2 sin kx)(c3 cos ckt  c4 sin ckt)

a)

b) (c1 cos kx  c2 sin kx)(c3 e ky  c4 e  ky )

(c1 cos kx  c2 sin kx)c3 e  k

c)

2

t

d) None of these 15. The solution of Laplace equation

 2u  2u   0, u ( x,0)  0, u ( x,  )  0, u (, y)  0, u (0, y)  3 sin 2 y, is x 2 y 2 a) 3 sin 2 x e 2 y b) 3 sin 2 y e 2 x

2 sin 3x e 3 y

c)

d) 2 sin 3 y e 3 x 16. The solution of the P.D.E. U tt  U yy is a) sin( x  t ) b) sin( x  t )

sin( x  t ) d) sin( x  t )     d2A  w 2 t is 17. If A  i cos wt  j sin wt than the value of 2 dt c)

a) 1 b) 0

 A

c)



d) w 2 A



1. If A  (t 3  1) iˆ  t 2 ˆj  (2t  5) kˆ , than the magnitude of velocity at t=1 sec is a) 17

17 b) c) Both (a) and (b) d) None of these 2. The normal vector of a surface is also known as a) Gradient of the surface b) Curl of the surface c) Divergence of the surface d) None of these 3. The gradient of the surface (xy+yz+zx) at the point (1,-1,1) is

a)

2 iˆ  2 ˆj  2 kˆ

b) 2 iˆ  2 ˆj

2 ˆj  2 kˆ d) 2 ˆj c)

4. The unit normal vector of the surface   xy 3 z 2 at the point (1,1,2) is a)



1

(iˆ  3 ˆj  kˆ)

11 1 ˆ b)  (i  3 ˆj  kˆ) 11 1 ˆ c)  (i  3 ˆj  kˆ) 11 1 d)  (iˆ  3 ˆj  kˆ) 11 

5. The directional derivative of a surface  in direction of a vector d is



a) 



b) ( ).dˆ c)

  ( ).d

d) None of these 6. A vector is said to be rotational if its a) Curl is not zero b) Curl is zero c) Divergence is zero d) Divergence is not zero



7. The vector v  3 y 4 z 2 iˆ  4 x 3 z 2 ˆj  3x 2 y 2 kˆ is a) Not solenoidal b) Not rotational c) Solenoidal d) None of these 8. The value of a for which the vector

 v  (2 xy  3 yz )iˆ  ( x 2  axz  4 z 2 ) ˆj  (3xy  byz )kˆ is irrotational is

a) b) c) d)

3 -3 8 -8



9. The vector potential of the vector v  ( x 2  y 2  x)iˆ  (2 xy  y) ˆj is

x2 3 x2 3 x2 3 x3 3

x y2   xy 2  C 2 3 x y2 b)    xy 2  C 3 3 x y3 c)    xy 2  C 3 3 x2 y2 d)    xy 2  C 2 3  10. The limit of the vector A  (t 3  2t ) iˆ  t 2 ˆj  (2t  7) kˆ as t  1 is a) (2iˆ  ˆj  2kˆ) a)



b) (2iˆ  ˆj  7kˆ)

(3iˆ  ˆj  9kˆ)

c)

d) (3iˆ  ˆj  2kˆ)



11. The vector function V (t ) is said to be continuous at a point t=a if

  V ( t )  V (a) limit

a)

t a





limit V (t )  V (a)

b)

t a

  V ( t )   V (a) limit

c)

t a





limit V (t )  V (a)

d)

t a



12. The length of space curve for the vector A  acost ˆi  asint ˆj  kˆ , 0  t  2 is a) b) c) d) a)

2 2a  2  2a 



13. If F is a conservative force field, than the value of Curl F is a) b) c) d)

0 1 -1 Can’t say



14. The value of a for which the vector A  a( x  y)iˆ  4 yˆj  3k is solenoidal is a) 4 b) -4

c) 3 d) -3





15. The value of x(  ) is, where  is any surface (Where x is vector cross product). a) b) c) d)

1 0 -1 None of these



16. The value Div(Curl A) is a) b) c) d)

1 -1 0 None of these



17. The limit of the vector F  sin t iˆ  cos 2t ˆj  kˆ as t   is

(iˆ  ˆj  kˆ)

a)

b) (iˆ  ˆj  5kˆ)

( ˆj  kˆ)

c)

d) ( ˆj  kˆ)

Unit 6 Set-A





1. If a force F  2 x 2 yi  3xyj displace a particle in xy-plane from (0, 0) to (1, 4) along the curve

y  4x 2 . Than the value of work done is 104 6 104 b) 5 114 c) 5 114 d) 6   2. If  ( F .n )ds  0 than F is said to be a)

s

a) b) c) d)

Irrotational vector Solenoidal vector Both (a) and (b) None of these







3. If F  2 xi  3 yj  4 zk than the value of

 F  dv is, (where v is the region bounded by v

x=0,y=0,z=0,x=1,y=1,z=1)

1 3 i 2 2 1 3 f) i 2 2  1 3 g) i 2 2  1 3 h) i 2 2 e)

  j  2k

  j  2k   j  2k   j  2k

4. If  ( x, y ), ( x, y ),

  are continuous function over a region R bounded by simple closed , y x

curve C in x-y plane, then according to Greens theorem

 

 (dx  dy)    x

a)

C

b)

 

 (dx  dy)    x



  dxdy y 



  dxdy y 



  dxdy y 

R

 

 (dx  dy)    x

C

d)

  dxdy y 

R

C

c)



R

 

 (dx dy)    x

C

R









5. The value of line integral {sin y i  x(1  cos y ) j }.dr , where C is the circular path given by C

x y a a) a b) 2a c) a 2 d) 2a 2 2

2

2

6. The value of line integral

 (3x

2

 8 y 2 )dx  (4 y  6 xy )dy) , where C is the boundary of the

C

region bounded by x  0, y  0 and 2 x  3 y  6 {By using of green’s theorem} is a) b) c) d)

20 -20 21 -21

7. Which of the following is the mathematical expression of stock’s theorem

   F . d r  Curl F ds  

a)

S

   b)  F .dr   Curl F .n ds S

   F . n d r  Curl F ds  

c)

S

d) None of these 8. By using stock’s theorem the value of the

 xdx  ydy  zdz is, where c is x

2

 y2  1

c

a) b) c) d)

0 1 -1 None of these

9. By using stock’s theorem the value of the

 (2 x  y)dx  yz

2

dy  y 2 zdz is, where c is

c

x  y 1 a)   2

2

b)  c)  / 2 d) None of these 10. The

   ˆ  zˆj  xk and s is the surface z  1  x 2  y 2 , z  0 , is curl v . n ds , where v  y i  s

a)   b)  c)  / 2 d) None of these 11. The mathematical form of Gauss divergence theorem is a)

 



 F .dr   Curl F ds S

   b)  F .dr   Curl F .n ds S

  c)  F .n ds   div F dv v

d) None of these 12. The value of

     Div F dv , where F  3 x i  4 y j  5 z k over the region bounded by the  v

sphere x  y 2  z 2  1 is 2

a)

3584  3

3584  4 3584 c)   3 b)

d) None of these 13. By using Gauss divergence theorem the value of



 F.n ds,

    where F  4 yi  6 zj  7 xk is

s

a) b) c) d)

0 1 -1 None of these

14. Gauss divergence gives a relationship between a) Surface and volume integral b) Line and volume integral c) Line and surface integral d) None of these 15. Green’s Theorem is a special case of ; a) Stock’s Theorem b) Gauss Theorem c) All of the above d) None of these 16. The value of line integral

 xydx  xydy ) , where C is the region bounded by C

0  x  1,0  y  1{By using of green’s theorem} is a) b) c) d)

1 -1 0 None of these

17. The value of

x

2

zdx  3xdy  y 3 dz, where c is x 2  y 2  1

c

a) b) c) d)

3  3



None of these

18. The value of line integral

 xdx  ydy ) , where C is the region bounded by 0  x  1,0  y  1 C

{By using of green’s theorem} is e) 1 f) -1 g) 0

h) None of these 19.