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Calculate the Green’s value for the functions F = y2 and G = x2 for the region x = 1 and y = 2 from origin.
Answer: c Explanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(2x – 2y)dx dy. On integrating for x = 0->1 and y = 0->2, we get Green’s value as -2.
Answer: c
See lessExplanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(2x – 2y)dx dy. On integrating for x = 0->1 and y = 0->2, we get Green’s value as -2.
The path traversal in calculating the Green’s theorem is
Answer: b Explanation: The Green’s theorem calculates the area traversed by the functions in the region in the anticlockwise direction. This converts the line integral to surface integral.
Answer: b
See lessExplanation: The Green’s theorem calculates the area traversed by the functions in the
region in the anticlockwise direction. This converts the line integral to surface integral.
Which of the following is not an application of Green’s theorem?
Answer: c Explanation: In physics, Green’s theorem is used to find the two dimensional flow integrals. In plane geometry, it is used to find the area and centroid of plane figures.
Answer: c
See lessExplanation: In physics, Green’s theorem is used to find the two dimensional flow
integrals. In plane geometry, it is used to find the area and centroid of plane figures.
Find the value of Green’s theorem for F = x2 and G = y2 is
Answer: a Explanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(0 – 0)dx dy = 0. The value of Green’s theorem gives zero for the functions given.
Answer: a
See lessExplanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(0 – 0)dx dy = 0. The value of Green’s theorem gives zero for the functions given.
Mathematically, the functions in Green’s theorem will be
Answer: c Explanation: The Green’s theorem states that if L and M are functions of (x,y) in an open region containing D and having continuous partial derivatives then, ∫ (F dx + G dy) = ∫∫(dG/dx – dF/dy)dx dy, with path taken anticlockwise.
Answer: c
See lessExplanation: The Green’s theorem states that if L and M are functions of (x,y) in an open region containing D and having continuous partial derivatives then,
∫ (F dx + G dy) = ∫∫(dG/dx – dF/dy)dx dy, with path taken anticlockwise.
The resistivity of a material with resistance 200 ohm, length 10m and area twice that of the length is
Answer: c Explanation: Resistance calculated from Ohm’s law and Stoke’s theorem will be R = ρL/A. To get resistivity, ρ = RA/L = 200 X 20/10 = 400.
Answer: c
See lessExplanation: Resistance calculated from Ohm’s law and Stoke’s theorem will be R =
ρL/A. To get resistivity, ρ = RA/L = 200 X 20/10 = 400.
The conductivity of a material with current density 1 unit and electric field 200 μV is
Answer: d Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
Answer: d
See lessExplanation: The current density is given by, J = σE. To find conductivity, σ = J/E =
1/200 X 10-6 = 5000.
Find the power, given energy E = 2J and current density J = x2 varies from x = 0 and x = 1.
Answer: b Explanation: From Stoke’s theorem, we can calculate P = E X I = ∫ E. J ds = 2∫ x2 dx as x = 0->1. We get P = 2/3 units.
Answer: b
See lessExplanation: From Stoke’s theorem, we can calculate P = E X I = ∫ E. J ds
= 2∫ x2 dx as x = 0->1. We get P = 2/3 units.
The voltage of a capacitor 12F with a rating of 2J energy is
Answer: a Explanation: We can compute the energy stored in a capacitor from Stoke’s theorem as 0.5Cv2. Thus given energy is 0.5 X 12 X v2. We get v = 0.57 volts.
Answer: a
See lessExplanation: We can compute the energy stored in a capacitor from Stoke’s theorem as 0.5Cv2. Thus given energy is 0.5 X 12 X v2. We get v = 0.57 volts.
The energy stored in an inductor 2H and current 4A is
Answer: d Explanation: From Stoke’s theorem, we can calculate energy stored in an inductor as 0.5Li2. E = 0.5 X 2 X 42 = 16 units.
Answer: d
See lessExplanation: From Stoke’s theorem, we can calculate energy stored in an inductor as
0.5Li2. E = 0.5 X 2 X 42 = 16 units.