Mathematical Physics MCQs Question and Answer in Engineering Physics

A real traceless 4 × 4 matrix has to eigen values -1 and +1. The other eigen values are

A) zero and +2 B) -1 and +1 C) zero and +1 D) +1 and +1

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Answer & Solution

Answer: Option B

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If \[l = \oint\limits_C {dz\,ln\left( z \right),} \] where C is the unit circle taken anticlockwise and $$l$$n(z) is the principal branch of the Logarithmic function, which one of the following is correct?

A) $$l$$ = 0 by residue theorem B) $$l$$ is not defined since, $$l$$n(z) has a branch cut C) $$l$$ ≠ 0 D) \[\oint\limits_C {dz\,ln\left( {{z^2}} \right) = 2l} \]

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Answer: Option A

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If a vector field \[\overrightarrow {\mathbf{F}} = x{\mathbf{\hat i}} + 2y{\mathbf{\hat j}} + 3z{\mathbf{\hat k}},\] then \[\overrightarrow \nabla \times \left( {\overrightarrow \nabla \times \overrightarrow {\mathbf{F}} } \right)\] is

A) zero B) \[{{\mathbf{\hat i}}}\] C) \[2{\mathbf{\hat j}}\] D) \[3{\mathbf{\hat k}}\]

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Answer: Option A

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Consider the set of vectors; \[\frac{1}{{\sqrt 2 }}\] (1, 1, 0); \[\frac{1}{{\sqrt 2 }}\] (0, 1, 1) and \[\frac{1}{{\sqrt 2 }}\] (1, 0, 1).

A) The three vectors are orthonormal B) The three vectors are linearly independent C) The three vectors cannot form a basis in a three dimensional real vector space D) \[\frac{1}{{\sqrt 2 }}\] (1, 1, 0) can be written as a linear combination of \[\frac{1}{{\sqrt 2 }}\] (0, 1, 1) and \[\frac{1}{{\sqrt 2 }}\] (1, 0, 1)

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Answer: Option B

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Solution of the differential equation \[x\frac{{dy}}{{dx}} + y = {x^4},\] with the boundary condition that y = 1, at x = 1, is

A) \[y = 5{x^4} - 4\] B) \[y = \frac{{{x^4}}}{5} + \frac{{4x}}{5}\] C) \[y = \frac{{4{x^4}}}{5} + \frac{1}{{5x}}\] D) \[y = \frac{{{x^4}}}{5} + \frac{4}{{5x}}\]

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Answer & Solution

Answer: Option D

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