Curl of vector F^{\to} = x^{2}z^{2}i^{\^}−2xy^{2}zj^{\^} + 2y^{2}z^{3}k^{\^} is

Correct Answer is : (4yz3+2xy2)i^+2x2zj^−2y2zk^

(4yz^{3} + 2xy^{2}) i^{\^} + 2x^{2}zj^{\^}−2y^{2}zk^{\^}

(4yz^{3} + 2xy^{2}) i^{\^}−2x^{2}zj^{\^}−2y^{2}zk^{\^}

2xz^{2}i^{\^}−4xyzj^{\^} + 6y^{2}z^{2}k^{\^}

2xz^{2}i^{\^} + 4xyzj^{\^} + 6y^{2}z^{2}k^{\^}

GATE Mechanical Engineering (ME) 2014 Shift 2 Solved Paper

Question : 13 of 65

Marks: +1, -0

Curl of vector

F↖{→} = x^{2}z^{2}i↖{\^}−2xy^{2}zj↖{\^} + 2y^{2}z^{3}k↖{\^}

Go to Question:

Prev Question

Next Question