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K-2517 3 Paper III Total Number of Pages : 16 4. The Green’s function for the Poisson equation 0 2 (r) (r) ∈ ρ ∇ φ = − -is (A) 2 2 1 1 2 r r 4 G(r, r ) − π = (B) 2 1 1 2 r r 4 G(r, r ) − π = (C) 2 1 1 2 4 r r 1 G(r, r ) π − = (D) 2 1 r r 1 2 1 G(r, r ) e 4 − − = π 5. The Lagrange polynomial passing through the two points (x0, y0) and (x1, y1) is (A) 1 0 0 1 01 10 x x x x y y x x xx − − + − − (B) 1 1 0 0 0 1 0 1 y x x x x y x x x x − − + − − (C) 0 1 0 0 1 0 1 1 y x x x x y x x x x − − + − − (D) 1 1 0 0 0 0 1 1 y x x x x y x x x x − − + − − 6. If i T is a second rank mixed tensor, it j transforms under the co-ordinate change { } { } i j x x → as (A) i j i j T x x x x T ∂ ∂ ∂ ∂ = α β α β (B) i j j i T x x x x T β α α β ∂ ∂ ∂ ∂ = (C) i j i j T x x x x T α β α β ∂ ∂ ⋅ ∂ ∂ = (D) i j j i T x x x x T β α α β ∂ ∂ ⋅ ∂ ∂ = 7. Let (x0, y0), (x1, y1), .... (xn, yn) be n points in the x – y plane then integral ∫ n 0 x x y dx by the Simpson’s rule is given by (A) [(y y ) 2(y y y ... 3 h 0 + n + 1 + 3 + 5 + y ) 4 (y y ...y )] + n−1 + 2 + 4 + n−2 (B) [(y y ) 3(y y y ... 3 h 0 + n + 1 + 3 + 5 + y ) 4 (y ... y )] + n−1 + 2 + + n−2 (C) [(y y ) 4(y y ... y ) 3 h 0 + n + 1 + 3 + + n−1 + + ++ 2 (y y ... y ) 2 4 n2− ] (D) [(y y ) 2(y y ... y ) 3 h 0 + n + 1 + 3 + + n−1 + + ++ 3 (y y ... y ) 2 4 n2− ] Where h is the equal interval between the points x0 , x1 , x2 ... and n is even.

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