Mathematical Methods of Physics MCQ Quiz - Objective Question with Answer for Mathematical Methods of Physics - Download Free PDF

Calculation:

Aψ = λψ   ⇒   d²(e^−αx2) / dx² − Bx²e^−αx2 = λe^−αx2

⇒ d/dx (−2αx·e^−αx2) − Bx²e^−αx2 = λe^−αx2

−2αe^−αx2 + (−2α)(−2α)x²e^−αx2 − Bx²e^−αx2 = λe^−αx2

(4α² − B)x²e^−αx2 − 2αe^−αx2 = λe^−αx2

(4α² − B) = 0   ⇒   B = 4α²

Explanation:

f(z) = z⁴ × exp(1 / (2z)) = z⁴ × (1 + 1 / (2z) + 1 / (4z²) + 1 / (48z³) + 1 / (384z⁴) + 1 / (3840z⁵) + ... )

Residue is the coefficient of 1/z = 1 / 3840

Integral = (2πi) / 3840 = πi / 1920

Calculation:

Defined for Re(z) > 0, and extended to the whole complex plane (except non-positive integers) via analytic continuation.

Behaviour at z = 0:

Singularities of the Gamma function occur at non-positive integers: z = 0, −1, −2, ...

These are simple poles.

Specifically, at z = 0, Γ(z) has a simple pole.

Supporting fact:

Near z = 0, the Gamma function behaves like:

    Γ(z) ~ 1/z − γ + …

where γ is the Euler-Mascheroni constant.

This clearly shows a simple pole at z = 0.

Calculation:

Analysis of the Relationship between x and y

Check for Linear Relationship

A linear relationship would have constant first differences in y values.

First differences:

2.1 - 0.1 = 2

8.1 - 2.1 = 6

17.9 - 8.1 = 9.8

32.2 - 17.9 = 14.3

49.7 - 32.2 = 17.5

First differences are not constant → Not linear.

Check for Quadratic Relationship

A quadratic relationship has approximately constant second differences.

Second differences:

(6 - 2) = 4

(9.8 - 6) = 3.8

(14.3 - 9.8) = 4.5

(17.5 - 14.3) = 3.2

These second differences are roughly constant (around 4), allowing for small experimental errors.

The relationship between x and y is best represented by a quadratic relationship, which can be written in the form: y = ax² + bx + c

Calculation:

The integral has a second-order pole at z = 2.

The residue theorem gives

∮_c [5 / (z - 2)²] sin(kz) dz = 2πi × [Residue]

The residue at z = 2 will be

R = (1 / 1!) d/dz [5 sin(kz)] = 5k cos(2k)

⇒ (1 / 2πi) ∮_c [5 / (z - 2)²] sin(kz) dz = 5k cos(2k)

Explanation:

• \(I=\int_0^{\infty} e^{-x} x \sin (x) d x\)

This integration from limit 0 to \(\infty\) can be so0lved using Laplace Transform 

• \(L[f(x)]=\int_0^\infty \mathrm{e}^{-sx}\,\mathrm f (x){d}x\) --------1

• \(L[sinax]=\frac {a} {s^2+a^2}\),
• Here  \(L[sinx]=\frac {1} {s^2+1}\)

Now, \(L[x^nf(x)]=(-1)^n\frac{d^n} {ds^n} [Lf(x)]\)

•  \(L[xsinx]\)\(=(-1)^1\frac {d} {ds} [\frac{1} {s^2+1}]\) (As, \(L[sinx]=\frac {1} {s^2+1}\))
• \(L[xsinx]=\)\((-1)\frac {-2s} {(s^2+1)^2}\)\(=\frac {2s} {(s^2+1)^2}\)

Now from equation 1, we know that s = 1,

• So, \(L[f(x)]=\int_0^\infty \mathrm{e}^{-sx}\,\mathrm f (x){d}x\) \(=\frac {2\times1} {(1^2+1)^2}=\frac {2} {4}=\frac {1} {2}\)

So, the correct answer is \(\frac {1} {2} \).

Explanation:

Given the polynomial \( f(x) = x^3 - x^2 - 1\), the initial interval [a, b] = [1, 2] and the function values at the two endpoints are f(1) = -1 and f(2) = 3. An iteration in the bisection method consists of the following steps:

1. Compute the midpoint c = (a + b) / 2.
2. Compute the value of the function f(c).
3. If f(c) is very close to 0, then c is our root. Else, replace the interval [a, b] with [a, c] if f(a) and f(c) have opposite signs; otherwise, replace the interval with [c, b].

The correct bisection method would go as follows:

Iteration 1:

\(c = \frac{(a + b) }{2} = \frac{(1 + 2)}{2}= 1.5\)

\(f(c) = f(1.5) = (1.5)^3 - (1.5)^2 - 1 = 0.125\)

Since f(a) = -1 and f(c) = 0.125 have different signs, we replace b = c. Now the new interval becomes [1, 1.5].

Iteration 2:

\(c =\frac{ (a + b) }{ 2} = \frac{(1 + 1.5)} { 2} = 1.25\)

\(f(c) = f(1.25) = (1.25)^3 - (1.25)^2 - 1 = -0.421875\)

Since f(a) = -1 and f(c) = -0.421875 have the same sign, we replace a = c. Now the new interval becomes [1.25, 1.5].

Iteration 3:

\(c = \frac{(a + b) }{ 2 }= \frac{(1.25 + 1.5)}{2 }= 1.375\)

\(f(c) = f(1.375) ~= (1.375)^3 - (1.375)^2 - 1 = -0.162109375\)

Since f(a) = -0.421875 and f(c) = -0.162109375 have the same sign, we replace a = c. Now the new interval becomes [1.375, 1.5].

After three iterations, the best estimate for the root x₀ is our most recent midpoint, c = 1.375 or 11/8.

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,

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

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

Concept:

Polynomials can be classified by the number of terms with nonzero coefficients, so that a one-term polynomial is called a monomial, a two-term polynomial is called a binomial, and a three-term polynomial is called a trinomial.

Calculation:

x(t) = a₀ + a₁t + a₂t² + ... + a_n-1t^n-1

x(0) = 0

a₀ = 0

x(t) = a1t + a2t2 + ... + an-1tn-1

Also, x(1) = 1

1 = a₁ + a₂ + ... + a_n-1

t,t²,t³ form basis.

c₁t + c₂t² + c₃t³ = 0

If c₁ = c₂ = c₃ ...= 0

Thus, it does not constitute a vector space.

The correct answer is option (4).

Explanation:

• The recurrence relation is: \((n) J_n(x) - xJ_n'(x) = xJ_{n-1}(x)\)
• Rearranging and multiplying the relation by \(x^{-n}\), we get: \(-x^{-n} J_n'(x) = x^{-(n-1)} J_{n-1}(x) - x^{-n} (n) J_n(x)\),
• Which simplifies to: \(-x^{-n} J_n'(x) = x^{-n} (J_{n-1}(x) - nJ_n(x))\)
• But another recurrence relation involves Bessel functions, which is \(J_{n - 1}(x) + J_{n + 1}(x) = 2nJ_n(x) / x\),
• From above we can replace \((J_{n - 1}(x) = x[J_{n + 1}(x) - 2nJ_n(x) / x])\) in our expression for \(-x^{-n} J_n'(x)\)
• This results in: \(-x^{-n} J_n'(x) = x^{-n} (x[J_{n + 1}(x) - 2nJ_n(x) / x] - nJ_n(x))\)
• After simplifying, it becomes: \(-x^{-n} J_n'(x) = -x^{-n} J_{n + 1}(x)\)
• Solving for the derivative of interest, we get: \([\frac{d}{d x}\left[x^{-n} J_n(x)\right] = -x^{-n} J_{n + 1}(x)]\)

Concept:

Contour integration is the process of calculating the values of a contour integral around a given contour in the complex plane.

Calculation:

f(z) = (\({π \over sin 4z}\))²

Thus the poles are

z₀ = 0, π/4   

4z = nπ, z = 0, π/4

The rest are outside the contour.

Residue at z = 0 is 

[\({ π \over 4z- {4^3z^3\over 3!}+... }\)]²

= [\({ 1 \over 4- {4^3z^2\over 3!}+... }\)]²

= [\({4- {4^3z^2\over 3!}+... }\)]^-2

Residue for z = π/4

z  - π/4 = t

sin (4t + π) = - sin 4t

= \([{t+{π \over 4}\over sin 4 (t + {π \over 4})}]^2\)

(\({t+ {π \over 4}\over sin 4 (t + {π \over 4t})}\))² = \({t^2 + {π^2\over 4} + 2t {π \over 4} \over sin^2 4t}\) 

\({π \over 2}{t \over 16 t^2[1 - ...]^2}\) = \({π \over 32t} [1 - ...]^2 \)

b₁ = \({π \over 32}\)

∫_c \({z^2 \over sin^2 4z}\) dz = 2πi [0 + \({\pi\over 32}\)]

= \({i\pi^2\over 16}\) 

The correct answer is option (3).

Solution-Option-1\((1,\frac {1} {2}) and \frac {1} {2}\)

Concept-  Here, we have to find center of the curve and radius of the curve using equation of circle which is given by-\(x^2+y^2+2gx+2fy+c=0\) with center \((x_0,y_0)=(-g,-f) \) and radius is \(R=\sqrt{f^2+g^2-c^2}\).

Formula Used-

• \(Im(z_1-z_2)=Im(z_1)-Im(z_2)\)
• \(R=\sqrt{f^2+g^2-c^2}\)

Calculation-

• The locus of the curve is \(Im(\frac {\pi(z-1)-1} {(z-1)})=1\)
• \(Im(z_1-z_2)=Im(z_1)-Im(z_2)\)

• Now this locus can be written as, \(Im(\frac {\pi(z-1)} {(z-1)}-\frac{1} {z-1})=1\)

• \(Im(\pi)-Im(\frac {-1} {z-1})\)
• \(Im(\pi)\) has zero value, so, we will see the solution of  \(Im(\frac {-1} {z-1})\)

Now, as we know \(z=x+iy\)

• \(\frac{1} {z-1}=\frac {1} {x+iy-1}=\frac {1} {(x-1)+iy}\)

• \(\frac{1} {z-1}=\frac {1} {x+iy-1}=\frac {1} {(x-1)+iy}\)
• By rationalise above value, we get
• \(\frac {1} {(x-1)+iy}\times\frac {(x-1)-iy} {(x-1)-iy}\)\(=\frac {(x-1)-iy} {(x-1)^2-(iy)^2}=\frac {(x-1)-iy} {(x-1)^2+y^2} \)

Imaginary part of above equation is \(\frac {-y} {(x-1)^2+y^2} \)\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

• \(Im(\frac {-1} {z-1})\)\(=\)\(\frac {y} {(x-1)^2+y^2}=1\)
• \(y=(x-1)^2+y^2\)
• \(y=x^2+1-2x+y^2\)
• \(x^2+y^2-2x-y+1=0\)
• This represents the equation of circle and is similar to \(x^2+y^2+2gx+2fy+c=0\)  with center \((-g,-f)\) and radius of circle is given by formula \(R=\sqrt{f^2+g^2-c^2}\)
• Here, \(g=-1, f=\frac {-1} {2}\)
• Coordinates of center of circle are \((x_0,y_0)=(-g,-f)=(1,\frac{1} {2})\)
• Radius of circle \(R=\sqrt{f^2+g^2-c^2}\)\(=\sqrt{1+(\frac {1} {2})^2-1}=\sqrt{1+\frac{1} {4}-1}=\frac {1} {2}\)

So, the correct answer is \((x_0,y_0)=(1,\frac{1} {2})\) and \(R=\frac {1} {2}\)

Concept-  

The Distance Formula can be used to find the distance between two points on the coordinate plane. It is a literal equation with several variables representing the locations of the points.

Formula Used-

Assume, \((x_1,x_2)\) and \((y_1,y_2)\) are coordinates on axes, and d is the distance between them. Then,

• \(d^2=(x_2-x_1)^2 +(y_2 - y_1)^2\)

Explanation:

• AB is a rod with co-ordinates \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\)
• Let the length of the rod = \(L\)
• The radius of the sphere = \(R\)

\((x_2-0)^2+(y_2-0)^2+(z_2-0)^2=R^2\)

\((x_1-0)^2+(y_1-0)^2+(z_1-0)^2=R^2\)

\((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2=L^2\)

So, there are three constraint equations.

Concept:

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

Calculation:

Probability that the two balls are of different colors 5 W, 4 B

= \({5C_1 \times 4C_2\over 9 C_2}\)

= \({{5!\over 4!\times 1!}\times{{4!}\over 3!\times 1!}\over {9!\over 7!\times 2!}}\)

= \({5 \over 9}\)

The correct answer is option (4).