Mathematical Methods of Physics MCQ Quiz in বাংলা - Objective Question with Answer for Mathematical Methods of Physics - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

The linear algebra is the study of linear equations and their representation in the vector space. 

Explanation:

The divergence of a vector field \(\vec{F} = \langle F_1, F_2, F_3\rangle\) in \(\mathbb{R}^3 \) is given by the formula: \(\text{div}(\vec{F}) = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}.\)

In this case, we have \(F_1(x,y) = x^3 \), \(F_2(x,y) = 3x^2y \), and  \(F_3(x,y) = 3xy^2 \).

Taking partial derivatives with respect to x, y and z we get: \(\frac{\partial F_1}{\partial x} = 3x^2, \quad \frac{\partial F_2}{\partial y} = 3x^2, \quad \frac{\partial F_3}{\partial z} = 0.\)  Now putting these in the expression of  \(\text{div}(\vec{F}) \) given above we have.

the divergence of \(\vec{F} \) at the point \((1,1)\) is given as: \(\text{div}(\vec{F})(1,1) = \frac{\partial F_1}{\partial x}(1,1) + \frac{\partial F_2}{\partial y}(1,1) + \frac{\partial F_3}{\partial z}(1,1) = 3 + 3 = 6.\)

The correct option is option (1) 6.

Explanation:

f, g entire function such that \(\rm\displaystyle\lim_{z \rightarrow \infty} \frac{f(z)}{z^n}=\lim_{z \rightarrow \infty} \frac{g(z)}{z^{n}}\) = 1 (for some n ∈ N)

Now

(1) Let n = 1, f = z, and g(z) = z + 1

\(\rm\displaystyle\lim_{z \rightarrow \infty} \frac{f(z)}{z} = \frac{z}{z} = 1 \) 

 \(\lim_{z \rightarrow \infty} \frac{g(z)}{z} = \frac{z+1}{z} = 1 + \frac{1}{z}\)= 1

But f(z) ≠ g(z)

Option (1) is false.

option (3) and option (4):

Replacing z by 1/z we get

\(\rm\displaystyle\lim_{z \rightarrow \infty} \frac{f(z)}{z^n} = \lim_{z \rightarrow 0} z^n f\left(\frac{1}{z}\right)\) = 1 ≠ 0

⇒ f(1/z) has a pole of order n at 0

⇒ f(z) has pole of order n at z = ∞

As we know that entire fn has a pole if and only if it is a non-constant polynomial and order of pole is degree of polynomial.

⇒ f(z) = a0 + a1z + a2z2 + … + an−1zn−1 + zn

similarly, g(z) = b0 + b1z + b2z2 + … + bn−1zn−1 + zn

⇒ f − g = (a0 − b0) + (a1 − b1)z + … + (an−1 − bn−1)zn−1 + 0

So f − g is a polynomial of deg. n−1

⇒ option (3) and option (4) are false

Hence option (2) is true

Concept:

Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids rather than using rectangles. This integration works by approximating the region under the graph of a function as a trapezoid, and it calculates the area. This rule takes the average of the left and the right sum.

Calculation:

Given:

h = 1/2

The Trapezoidal rule states

I = \({h\over 2}\)[y₀+y_n+2(y₁+y₂+....)]

\({5\over16}\)= \({1 \over 4}\)[1-\({5\over 4}\)+2(0+a)]

\({5\over 4}\) = [\(1- {5 \over 4} +2a\)]

2a = \({10 \over 4} - 1\)

a = 3/4

The correct answer is option (1).

Explanation:

Here, we have:

\( f(z) = \cosh z \)

The derivatives of f(z) are:

• \( f'(z) = \sinh z \)
• \( f''(z) = \cosh z \)
• \( f'''(z) = \sinh z \)

At z = πi  :

• \( f(π i) = \cosh (π i) = \frac{e^{π i} + e^{-π i}}{2} \)
• \( f'(π i) = \sinh (π i) = \frac{e^{π i} - e^{-π i}}{2} \)
• \( f''(π i) = \cosh (π i) \)
• \( f'''(π i) = \sinh (π i) \)

Using Taylor's series expansion around z = πi :

\( f(z) = f(π i) + (z - π i) f'(π i) + \frac{(z - π i)^2}{2!} f''(π i) + \frac{(z - π i)^3}{3!} f'''(π i) + \ldots \)

Substituting the values:

\( f(z) = \cosh (π i) + (z - π i) \sinh (π i) + \frac{(z - π i)^2}{2!} \cosh (π i) + \frac{(z - π i)^3}{3!} \sinh (π i) + \ldots \)

The third term is \(\frac{(z - π i)^2}{2!} \cosh (π i)\)

The correct option is 1).

Concept:

The Trapezoidal Rule is a numerical method used to approximate the definite integral of a function. It works by dividing the area under the curve into trapezoids, calculating the area of each trapezoid, and summing these areas to find the total approximate area.

The formula for the Trapezoidal Rule is:

\(\int_{a}^{b} f(x) dx = \frac{h}{2} \left[ f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right] \ \)

where:

\( h = \frac{b - a}{n} \ \) is the width of each subinterval,

\(( x_0, x_1, \ldots, x_n ) \ \)are the endpoints of the subintervals,

\( f(x_0), f(x_1), \ldots, f(x_n) \ \) are the function values at these points.

Given Problem: Estimate the area under the curve \(f(x) = \sqrt{x} \ \) from x = 1 to x = 9 using the Trapezoidal Rule with n = 4.

Explanation:

Step-by-Step Solution:

1. Define the function \(f(x) = \sqrt{x} \ \).

2. Determine the interval [a, b]:

a = 1

b = 9

3. Number of subintervals n = 4.

4. Calculate the width of each subinterval h:

\( h = \frac{b - a}{n} = \frac{9 - 1}{4} = 2 \ \)

5. Determine the endpoints of the subintervals:

\( x_0 = 1 \\ x_1 = x_0 + h = 1 + 2 = 3 \\ x_2 = x_0 + 2h = 1 + 4 = 5 \\ x_3 = x_0 + 3h = 1 + 6 = 7 \\ x_4 = x_0 + 4h = 1 + 8 = 9 \)

6. Calculate the function values at these points:

\( f(1) = \sqrt{1} = 1 \\ f(3) = \sqrt{3} \approx 1.732 \\ f(5) = \sqrt{5} \approx 2.236 \\ f(7) = \sqrt{7} \approx 2.646 \\ f(9) = \sqrt{9} = 3 \)

7. Apply the Trapezoidal Rule formula:

\( \int_{1}^{9} \sqrt{x} , dx \approx \frac{h}{2} \left[ f(x_0) + 2(f(x_1) + f(x_2) + f(x_3)) + f(x_4) \right] \ \)

Substitute the values:

\(\int_{1}^{9} \sqrt{x} dx = \frac{2}{2} \left[ 1 + 2(1.732 + 2.236 + 2.646) + 3 \right] \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \left[ 1 + 2(6.614) + 3 \right] \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \left[ 1 + 13.228 + 3 \right] \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \left[ 17.228 \right] \)

Therefore, the estimated area under the curve using the Trapezoidal Rule with n = 4 is 17.228.

The correct answer is option 3.

Concept:

• The Legendre polynomial of degree (n), denoted as \((P_n(x)),\) is defined by the formula: \( [P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}[(x^2 - 1)^n] ]\)
• This formula is called Rodrigues's formula for Legendre polynomials.
• The first few Legendre polynomials are given by:

1. \(P_0(x) = 1\)
2. \(P_1(x) = x\)
3. \(P_2(x) = \frac{1}{2}(3x^2 - 1)\)
4. \(P_3(x) = \frac{1}{2}(5x^3 - 3x)\)
5. \(P_4(x) = \frac{1}{8}(35x^4 - 30x^2 + 3)\)

Explanation:

Substituting x = -1 in the polynomial P₃: 

\(P_3(x) = \frac{1}{2}(5(-1)^3 - 3(-1))\)

\(P_3(x) = \frac{1}{2}(-5 + 3)\)

\(P_3(x) = \frac{1}{2}(-2)\)

\(P_3(x) = -1\)

Explanation:

Given:

• n (Number of trials): 5
• k (Number of successes): 3
• p (Probability of success): \(\frac12\)
• The formula for binomial probability is: \( P(X=k) = C(n, k) \times (p^k) \times ((1-p)^{(n-k)})\)
• We can substitute the values into this formula: \(P(X=3) = C(5, 3) \times (\frac12)^3 \times (\frac12)^2\)
• Calculating the binomial coefficient : \(C(5, 3): C(n, k) = \frac{n!}{k! \times (n-k)!} = \frac{5!}{3! \times (5-3)!} = 10\)
• Substitute this into the equation: \(P(X=3) = 10 \times (\frac12)^3 \times (\frac12)^2 = 10 \times (\frac18) \times (\frac14) = 10 \times (\frac1{32}) = \frac{10}{32}=\frac{5}{16}\)
• So, the probability that the captain will win 3 times and lose 2 times in a series of 5 matches is : \(\frac{5}{16} \)

Concept:

(i) A square matrix is said to be a triangular matrix if it is similar to a triangular matrix

(ii): Let A be a square matrix whose characteristic polynomial factors into linear polynomials, then A is similar to a triangular matrix i.e., there exists an invertible matrix P such that P^-1AP is triangular.

Explanation:

 A ∈ M3(ℝ) and X = {C ∈ GL3(ℝ) | CAC-1 is triangular}

So CAC-1 is similar to A

then CAC-1 is triangular if and only if A is triangularizable

Thus if A is not triangularizable then A = ϕ

(1) is false

The characteristic polynomial of A is of degree 3

So it has at least one real root

(4) is false

If X = Ø then the characteristic polynomial of A has 3 distinct roots on ℂ 

So A is diagonalizable over ℂ 

(3) is correct, (2) is false

Explanation:

• For a rotation matrix R in 3D, the trace of the rotation matrix (sum of the diagonal elements) relates to the angle of rotation \(\theta\) by the formula \(\text{Tr}(R) = 1 + 2\cos(\theta)\), yielding \(\cos(\theta) = \frac{(\text{Tr}(R) - 1)}2\).
• Our rotation matrix given is: \(\left(\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & \frac{2 \sqrt{2}}{3} & \frac{1}{3} \end{array}\right)\)
• Calculating the trace gives us: 

​\( \text{Tr}(R) = -1 + (-1/3) + (1/3) = -1 \implies \cos(\theta) = (-1) \implies \theta =( \pi \text{ or } -\pi) \)

•  The rotation axis can be obtained using: 

​\([ n_x = \frac{\sqrt{(R_{22} - R_{11}) + (R_{33} - R_{11}) + 2}}{2} ] [ n_y = \frac{\sqrt{(R_{33} - R_{22}) + (R_{11} - R_{22}) + 2}}{2} ] [ n_z = \frac{\sqrt{(R_{11} - R_{33}) + (R_{22} - R_{33}) + 2}}{2} ]\) which are the square roots of the elements of the rotation matrix.

• The right combination of the signs (±) is obtained by looking at the off-diagonal elements of the rotation matrix. From the given matrix, we have:

\([ n_x = 0, \quad n_y = \pm\sqrt{1/3}, \quad n_z = \pm\sqrt{2/3} ]\)

• The signs of \(n_y\) and \(n_z\) correspond to \(R_{23} - R_{32}\) and \(R_{12} - R_{21}\) of the rotation matrix, respectively.
• This results in \(n_y=-\sqrt{1/3} \quad and \quad n_z=\sqrt{2/3}\).
• But since the axis direction is determined up to the sense of rotation, we switch them (so that \(n_y\) is positive and \(n_z\) negative) and switch the sign of the angle of rotation, giving us:

\([ n = \left(0, \sqrt{1/3}, \sqrt{2/3}\right), \quad \theta = \pi ]\)

Concept:

 We are using Bayes' Theorem which describes the probability of occurrence of an event related to any condition. It is considered as the case of Conditional Probability. 

Here, we have to find the Probability of Red ball in Jar \(J_1\).

Formula used- \(P(J_1) P(\frac {R} {J_1})\over P(J_1) P(\frac {R} {J_1}) + P(J_2) P(\frac {R} {J_2})\)

Explanation:   

Given,

• Probability of Red ball in Jar \(J_1 = \frac {1} {3} \)
• Probability of red ball in Jar \(J_2 = \frac {1} {2}\)
• Probability relation of both jars is given as \(P(J_1) = 2P(J_2)\)

Now, we know that,

• \(P(J_1) +P(J_2) = 1\)
• \(2P(J_2) + P(J_1) = 1 \)
• \(P(J_2) = \frac {1} {3}\) and \(P(J_1) = \frac {2} {3}\)

Using Bayes' formula, we get,

• \(P(J_1) P(\frac {R} {J_1})\over P(J_1) P(\frac {R} {J_1}) + P(J_2) P(\frac {R} {J_2})\)

• => \(\frac {2} {3} \times \frac {1} {3} \over \frac {2} {3} \times \frac {1} {3} + \frac {1} {3}\times \frac {1} {2}\) \(=\)\(\frac {2} {9}\over \frac {2} {9} + \frac {1} {6}\)\(=\) \(\frac {2} {9}\times \frac {18} {7}\)\(=\)\(\frac {4} {7}\)