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Find the charged enclosed by a sphere of charge density ρ and radius a.
Answer: b Explanation: The charge enclosed by the sphere is Q = ∫∫∫ ρ dv. Where, dv = r2 sin θ dr dθ dφ and on integrating with r = 0->a, φ = 0->2π and θ = 0->π, we get Q = ρ(4πa3/3).
Answer: b
See lessExplanation: The charge enclosed by the sphere is Q = ∫∫∫ ρ dv.
Where, dv = r2 sin θ dr dθ dφ and on integrating with r = 0->a, φ = 0->2π and θ = 0->π, we get Q = ρ(4πa3/3).
The volume integral is three dimensional. State True/False
Answer: a Explanation: Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral.
Answer: a
See lessExplanation: Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral.
The triple integral is used to compute volume. State True/False
Answer: a Explanation: The triple integral, as the name suggests integrates the function/quantity three times. This gives volume which is the product of three independent quantities.
Answer: a
See lessExplanation: The triple integral, as the name suggests integrates the function/quantity three times. This gives volume which is the product of three independent quantities.
The divergence theorem converts
Answer: b Explanation: The divergence theorem is given by, ∫∫ D.ds = ∫∫∫ Div (D) dv. It is clear that it converts surface (double) integral to volume(triple) integral.
Answer: b
See lessExplanation: The divergence theorem is given by, ∫∫ D.ds = ∫∫∫ Div (D) dv. It is clear that it converts surface (double) integral to volume(triple) integral.
If D = 2xy i + 3yz j + 4xz k, how much flux passes through x = 3 plane for which -1<y<2 and 0<z<4?
Answer: c Explanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.
Answer: c
See lessExplanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.
Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.
Answer: b Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Answer: b
See lessExplanation: While evaluating surface integral, there has to be two variables and one
constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
The ultimate result of the divergence theorem evaluates which one of the following?
Answer: d Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus, it is given by, ψ = ∫∫ D.ds= Q, where the divergence theorem computes the charge and flux, which are both the same.
Answer: d
See lessExplanation: Gauss law states that the electric flux passing through any closed surface
is equal to the total charge enclosed by the surface. Thus, it is given by, ψ = ∫∫ D.ds= Q, where the divergence theorem computes the charge and flux, which are both the same.
Find the value of divergence theorem for A = xy2 i + y3 j + y2z k for a cuboid given by 0<x<1, 0<y<1 and 0<z<1.
Answer: c Explanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz+ ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get (1/3)+ 1 + (1/3) = 5/3.
Answer: c
See lessExplanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz+ ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get (1/3)+ 1 + (1/3) = 5/3.
Compute divergence theorem for D= 5r2 /4 i in spherical coordinates between r=1 and r=2.
Answer: c Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Since it is double integral, we need to keep only two variables and one constant compulsorily. Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second integral, with φ = 0Read more
Answer: c
See lessExplanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Since it is double integral, we need to keep only two variables and one constant compulsorily. Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second integral, with φ = 0→2π and θ = 0→ π. The first integral value is 80π, whereas second integral gives -5π. On summing both integrals, we get 75π.
Compute the Gauss law for D= 10ρ3 /4 i, in cylindrical coordinates with ρ= 4m, z=0 and z=5.
Answer: d Explanation: ∫∫ D.ds = ∫∫ (10ρ3/4).(ρ dφ dz), which is the integral to be evaluated. Put ρ =4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π
Answer: d
See lessExplanation: ∫∫ D.ds = ∫∫ (10ρ3/4).(ρ dφ dz), which is the integral to be evaluated. Put ρ =4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π