Title MP DPP Sheet 08.pdf ✅

1 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 IIT JAM PHYSICS 2025 (Online Batch) SECTION: MATHEMATICAL PHYSICS Daily Practice Problem (DPP) Sheet 8: Vector Calculus (Divergence Theorem, Stokes Theorem & Green’s Theorem) ; PART - A: MULTIPLE CHOICE QUESTIONS (MCQ) Q.1. If ˆ ˆ 2 ˆ A xzi y j yzk    4  then the value of the integral .ˆ S A ndS   (where S is the surface of a unit cube with two opposite corners at (0,0,0) and (1,1,1) respectively) is equal to (a) 3/2 (b) 5/2 (c) 6 (d) 9/2 Q.2. Consider the following vector field:   V xz i yz j z x y k 2 2 2 2 ˆ ˆ ˆ      . The flux of the vector field through the surface of the region defined by    a x a ,    b y b and 0  z c (a)   1 2 2 3 abc a b  (b)   2 2 abc a b  (c)   4 2 2 3 abc a b  (d)   2 2 2abc a b  Q.3. The flux of the vector field       ˆ ˆ 2 ˆ F x z i xz y j y z k       2 3 2  through the surface of the sphere having centre at (3,-1,2) and radius 3 units, will be (a) 54 (b) 108 (c) 216 (d) 324 Q.4. Let R be the region in the x-y plane enclosed by a simple closed curve C. The area of the region R is (a)   1 2 C y dx x dy   (b)   1 2 C x dy y dx   (c)   1 2 C x dy y dx   (d)   C y dx x dy   Q.5. The flux of the vector field F Axi Byj Czk ˆ ˆ ˆ     through the surface of a solid hemisphere having center at the origin and radius 3 units, will be (a) 9  A B C    (b) 18  A B C    (c) 36  A B C    (d) 108  A B C   
2 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 Q.6. The value of the integral 3 2 2 S x dydz x y dzdx x z dxdy    where S is the surface of the cylinder defined by the equations 2 2 x y z z     4, 0, 4 , is (a) 16 (b) 32 (c) 40 (d) 80 Q.7. Let C is a simple closed curve enclosing an area and   x y z , , is a scalar point function which is differentiable in nature. The value of the integral  . C   dr     is (a) 0 (b) 1 (c) 3 (d) Insufficient Information Q.8. The flux of the vector field ˆ ˆ ˆ F xi yj zk    4 7  through the surface of a cone of radius 1/ 2 unit and height 2 4 unit, will be (a) 4 (b) 2 (c) 4 (d) 2 Q.9. Let C be the unit circle centered at the origin, travelled anticlockwise. The value of the integral   2 2 3 3 x C e y dx x dy     (a)  / 2 2  (b)  / 2 2  (c) 2 (d) 1.5 Q.10. The flux of vector field 2 2 2 ˆ ˆ V r r r r    cos cos cos sin     ˆ  through the surface of sphere 2 2 2 2 x y z R z x y       , 0, 0, 0 will be (a) 4 4  R (b) 4 4  R  (c) 3 4  R  (d) 3 4  R Q.11. Given a vector   1 3 3 3 ˆ ˆ ˆ 3 F y i x j z k      and nˆ is the unit normal vector to the surface of the hemisphere 2 2 2 x y z z     1; 0 . The value of the integral  . ˆ S  F n dS    evaluated on the curved surface of the hemisphere, will be (a)  (b)  / 2 (c)  / 4 (d)  / 3 Q.12. The value of integral cos . sin sin cos  C   x y xy dx x y dy      where C is an unit circle centered at the origin, is (a) 0 (b)  / 2 (c)  (d) 2
3 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 Q.13. The value of integral 4 3 3 2    C   y x dx x y dy       where C is the curve passing through the point 0, 2  and having slope 2 / x y , is (a)  / 2 (b) 2 (c)  (d) 2 PART - B: Numerical Answer Type (NAT) Questions Q.14. The value of the integral     2 2 2 2 2 C   x y dx x y dy       where ‘C’ is the boundary of the surface enclosed by x-axis and the semicircle   1/2 2 y x  1 , is _______________________________ [Your answer should be upto TWO DECIMAL PLACES] Q.15. The value of the following integral:   3 3 3 S x dydz y dzdx z dxdy    where‘S’ is the surface of the sphere 2 2 2 2 x y z a    ) will be _________________ [in the units of 5 a ] [Your answer should be upto ONE DECIMAL PLACES] Q.16. The flux of vector field   ˆ ˆ 2 ˆ F xi yj z k    1  through the surface of cylinder defined by the equations 2 2 x y z z     4, 0, 1, is _______________ [in the units of  ] [Your answer should be AN INTEGER] Q.17. The flux of the vector field F y z i z x j x y k 2 2 2 2 2 2 ˆ ˆ ˆ     through the surface of region bounded by the part of the sphere 2 2 2 x y z   1 above x-y plane and z = 0 plane, will be ______________________ [Your answer should be upto TWO DECIMAL PLACES] Q.18. The value of the integral   2 2 2 C x y dx xydy    where C is rectangle bounded by the lines x x   0, 2 , y y   0, 3 is ________________________ [Your answer should be AN INTEGER] Q.19. The value of the integral   2 C xy dy y dx   where C is a square, cut from the first quadrant by the lines x y   1, 1 is ________________________ [Your answer should be upto ONE DECIMAL PLACES]
4 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 Q.20. `The flux of the vector field   3 2 3 2 2 3 ˆ ˆ ˆ 4 2 3 y F xz i xz j x z y x k             where 1 2 S S S   such that 2 2 2 1 S x y z z : 1, 0     and 2 2 2 S x y z : 1, 0    , will be __________ (in the unit of  ) [Your answer should be upto ONE DECIMAL PLACES] Answer Key PART - A: MULTIPLE CHOICE QUESTIONS (MCQ) 1. (a) 2. (c) 3. (b) 4. (b) 5. (b) 6. (d) 7. (a) 8. (a) 9. (d) 10. (a) 11. (b) 12. (a) 13. (b) PART - B: NUMERICAL ANSWER TYPE (NAT) QUESTIONS 14. (1.30 to 1.35) 15. (2.4) 16. (4) 17. (0.25 to 0.27) 18. (-36) 19. (1.5) 20. (0.4)