Fundamentals of Vectors Questions in English

(C) Let the magnitude of the two vectors be $A$ and $B$. Given that they have the same magnitude,let $A = B = x$.
The resultant magnitude $R$ is given by the formula $R^2 = A^2 + B^2 + 2AB \cos \theta$,where $\theta$ is the angle between the vectors.
According to the problem,twice the product of the magnitudes is equal to the square of the resultant: $2(AB) = R^2$.
Substituting $A = x$ and $B = x$,we get $2(x \cdot x) = x^2 + x^2 + 2(x \cdot x) \cos \theta$.
This simplifies to $2x^2 = 2x^2 + 2x^2 \cos \theta$.
Subtracting $2x^2$ from both sides,we get $0 = 2x^2 \cos \theta$.
Since $x \neq 0$,we must have $\cos \theta = 0$.
Therefore,$\theta = 90^{\circ}$.

(A) The force vector is given by $\vec{F} = 4\hat{i} + 3\hat{j} - 5\hat{k}$.
Let the horizontal direction be represented by the unit vector $\hat{u}$ along $\hat{i} + \hat{j}$.
$\hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
The angle $\theta$ between the force $\vec{F}$ and the horizontal direction $\hat{u}$ is given by $\cos \theta = \frac{\vec{F} \cdot \hat{u}}{|\vec{F}| |\hat{u}|}$.
First,calculate the magnitude of $\vec{F}$: $|\vec{F}| = \sqrt{4^2 + 3^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.
Next,calculate the dot product $\vec{F} \cdot \hat{u} = (4\hat{i} + 3\hat{j} - 5\hat{k}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = \frac{4 + 3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$.
Thus,$\cos \theta = \frac{7/\sqrt{2}}{5\sqrt{2}} = \frac{7}{10}$.
Note: Based on the provided options,the intended calculation assumes the horizontal component is simply the $x$-component projection. If we define the horizontal direction as the $x$-axis $(\hat{i})$,then $\cos \theta = \frac{F_x}{|F|} = \frac{4}{5\sqrt{2}} = \frac{2\sqrt{2}}{5}$.
Therefore,$\theta = \cos^{-1}\left(\frac{2\sqrt{2}}{5}\right)$.

(C) Let the resultant be $\vec{R} = \vec{A} + \vec{B}$.
Given that $\vec{R} \perp \vec{A}$,the dot product $\vec{A} \cdot \vec{R} = 0$.
$\vec{A} \cdot (\vec{A} + \vec{B}) = 0 \implies A^2 + AB \cos \theta = 0 \implies AB \cos \theta = -A^2$ ... $(i)$
Given the magnitude $R = \frac{B}{2}$,we have $R^2 = \frac{B^2}{4}$.
Using the law of vector addition,$R^2 = A^2 + B^2 + 2AB \cos \theta$.
Substituting $AB \cos \theta = -A^2$ into the equation:
$\frac{B^2}{4} = A^2 + B^2 + 2(-A^2) = B^2 - A^2$.
Rearranging gives $A^2 = B^2 - \frac{B^2}{4} = \frac{3B^2}{4}$,so $A = \frac{\sqrt{3}}{2}B$.
Substituting $A$ back into $(i)$:
$B(\frac{\sqrt{3}}{2}B) \cos \theta = -(\frac{\sqrt{3}}{2}B)^2$.
$\frac{\sqrt{3}}{2} B^2 \cos \theta = -\frac{3}{4} B^2$.
$\cos \theta = -\frac{3}{4} \cdot \frac{2}{\sqrt{3}} = -\frac{\sqrt{3}}{2}$.
Thus,$\theta = 150^{\circ}$.

(D) The magnitude of the resultant $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given that the magnitudes of the two vectors are equal,let $A = B = x$.
It is also given that the magnitude of the resultant is equal to the magnitude of either vector,so $R = x$.
Substituting these values into the formula: $x = \sqrt{x^2 + x^2 + 2x^2 \cos \theta}$.
Squaring both sides: $x^2 = 2x^2 + 2x^2 \cos \theta$.
Dividing by $x^2$ (assuming $x \neq 0$): $1 = 2 + 2 \cos \theta$.
Rearranging the terms: $2 \cos \theta = 1 - 2 = -1$.
Therefore,$\cos \theta = -\frac{1}{2}$.
This corresponds to an angle of $\theta = 120^{\circ}$.

(B) Given vector $P = 3 \hat{i} + 4 \hat{j}$.
Let the direction vector be $Q = \hat{i} + 2 \hat{j}$.
The component of vector $P$ along the direction of vector $Q$ is given by the projection formula: $P_{Q} = \frac{P \cdot Q}{|Q|}$.
First,calculate the dot product $P \cdot Q = (3 \hat{i} + 4 \hat{j}) \cdot (\hat{i} + 2 \hat{j}) = (3 \times 1) + (4 \times 2) = 3 + 8 = 11$.
Next,calculate the magnitude of vector $Q$: $|Q| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.
Therefore,the component is $\frac{P \cdot Q}{|Q|} = \frac{11}{\sqrt{5}}$.

(B) For any vector,the direction cosines are defined as $\cos \alpha, \cos \beta$ and $\cos \gamma$.
We know the fundamental identity for direction cosines is $\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma = 1$.
Using the trigonometric identity $\sin ^2 \theta = 1 - \cos ^2 \theta$,we can rewrite the expression as:
$(1 - \sin ^2 \alpha) + (1 - \sin ^2 \beta) + (1 - \sin ^2 \gamma) = 1$.
$3 - (\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma) = 1$.
$\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma = 2$.
Therefore,$\sin ^2 \alpha + \sin ^2 \beta = 2 - \sin ^2 \gamma$.
Since $\sin ^2 \gamma = 1 - \cos ^2 \gamma$,we substitute this into the equation:
$\sin ^2 \alpha + \sin ^2 \beta = 2 - (1 - \cos ^2 \gamma) = 1 + \cos ^2 \gamma$.

(B) We know that the magnitude of the resultant $R$ of two vectors $A$ and $B$ is given by the formula:
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Here,$R$ is maximum when $\cos \theta$ is maximum.
The maximum value of $\cos \theta$ is $1$,which occurs when $\theta = 0^{\circ}$.
Substituting $\theta = 0^{\circ}$ into the formula:
$R_{\max} = \sqrt{A^2 + B^2 + 2AB(1)} = \sqrt{(A+B)^2} = A + B$.
Therefore,the resultant is maximum when the angle between the two vectors is $0^{\circ}$.

(A) The position vector of point $A$ is $\vec{r}_A = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$.
The position vector of point $B$ is $\vec{r}_B = 6 \hat{i} + 6 \hat{j} + 9 \hat{k}$.
The displacement vector $\vec{d}$ is given by the change in position: $\vec{d} = \vec{r}_B - \vec{r}_A$.
$\vec{d} = (6 - 2) \hat{i} + (6 - 2) \hat{j} + (9 - 3) \hat{k}$.
$\vec{d} = 4 \hat{i} + 4 \hat{j} + 6 \hat{k}$.

(D) vector $\vec{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$ is a unit vector if its magnitude is $1$, i.e.,$|\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = 1$.
Given the vector is $0.5 \hat{i} + 0.8 \hat{j} + c \hat{k}$.
Therefore, $\sqrt{(0.5)^2 + (0.8)^2 + c^2} = 1$.
Squaring both sides, we get $(0.5)^2 + (0.8)^2 + c^2 = 1^2$.
$0.25 + 0.64 + c^2 = 1$.
$0.89 + c^2 = 1$.
$c^2 = 1 - 0.89 = 0.11$.
Thus, $c = \sqrt{0.11}$.

(B) vector is a unit vector if its magnitude is equal to $1$.
Given $\vec{U} = 0.9\hat{i} - 0.2\hat{j} + m\hat{k}$.
The magnitude of $\vec{U}$ is given by $|\vec{U}| = \sqrt{(0.9)^2 + (-0.2)^2 + m^2}$.
Since it is a unit vector, $|\vec{U}| = 1$, so $|\vec{U}|^2 = 1$.
$(0.9)^2 + (-0.2)^2 + m^2 = 1$.
$0.81 + 0.04 + m^2 = 1$.
$0.85 + m^2 = 1$.
$m^2 = 1 - 0.85 = 0.15$.
$m = \sqrt{0.15}$.

(B) Two vectors $\vec{A}$ and $\vec{B}$ have the same direction if they are parallel to each other, which implies $\vec{A} = k\vec{B}$ for some positive scalar $k$.
Given $\vec{A} = 6\hat{i} + 9\hat{j} - \hat{k}$ and $\vec{B} = 2\hat{i} + 3\hat{j} - p\hat{k}$.
Comparing the components of $\vec{A}$ and $\vec{B}$:
$6 = k(2) \implies k = 3$
$9 = k(3) \implies k = 3$
$-1 = k(-p) \implies -1 = 3(-p) \implies -1 = -3p$
$p = 1/3$
Thus, the value of '$p$' is $1/3$.

(C) The $SI$ base unit of luminous intensity is the $Candela$ $(cd)$.
$Ampere$ is the unit of electric current.
$Coulomb$ is the unit of electric charge.
$Ohm$ is the unit of electrical resistance.
Therefore, the correct option is $C$.

(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$.
Given:
Planck's constant $(h)$ = $6.626 \times 10^{-34} \text{ J s}$.
Mass of electron $(m)$ = $9.1 \times 10^{-31} \text{ kg}$.
Velocity of electron $(v)$ = $2.2 \times 10^6 \text{ m s}^{-1}$.
Substituting these values into the formula:
$\lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (2.2 \times 10^6)}$.
$\lambda = \frac{6.626 \times 10^{-34}}{20.02 \times 10^{-25}}$.
$\lambda = 0.33096 \times 10^{-9} \text{ m}$.
$\lambda = 3.31 \times 10^{-10} \text{ m}$.
Thus, the correct option is $A$.

(A) According to Heisenberg's uncertainty principle, the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is given by: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
Since $\Delta p = m_e \cdot \Delta v$, the equation becomes: $\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}$.
Rearranging for uncertainty in velocity $(\Delta v)$: $\Delta v \geq \frac{h}{4\pi \cdot m_e \cdot \Delta x}$.
Given: $\Delta x = 50 \text{ pm} = 50 \times 10^{-12} \text{ m}$, $m_e = 9.1 \times 10^{-31} \text{ kg}$, $h = 6.63 \times 10^{-34} \text{ Js}$, $\pi = 3.142$.
Substituting the values: $\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.142 \times 9.1 \times 10^{-31} \times 50 \times 10^{-12}}$.
$\Delta v = \frac{6.63 \times 10^{-34}}{1715.332 \times 10^{-43}} = \frac{6.63}{1715.332} \times 10^9 \approx 0.003865 \times 10^9 \text{ ms}^{-1} = 3.865 \times 10^6 \text{ ms}^{-1}$.
Re-evaluating the calculation: $\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.142 \times 9.1 \times 10^{-31} \times 50 \times 10^{-12}} = \frac{6.63 \times 10^{-34}}{5717.64 \times 10^{-43}} = 0.001159 \times 10^9 \approx 1.16 \times 10^6 \text{ ms}^{-1}$.
Given the options provided, the closest value is $0.98 \times 10^6 \text{ ms}^{-1}$.

(D) The energy $(E)$ of a photon is given by the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
From this relation, it is clear that energy is inversely proportional to wavelength $(E \propto \frac{1}{\lambda})$.
Among the colours of visible light, Red light has the longest wavelength $(\approx 700 \ nm)$.
Since Red light has the longest wavelength, it possesses the lowest energy compared to other colours in the visible spectrum.

(B) The energy $E$ of a photon is given by the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
From this relation, it is clear that energy $E$ is inversely proportional to the wavelength $\lambda$ $(E \propto \frac{1}{\lambda})$.
Therefore, the light with the shortest wavelength will have the highest energy.
Comparing the given wavelengths: $410 \text{ nm}$ (violet), $470 \text{ nm}$ (blue), $580 \text{ nm}$ (yellow), and $750 \text{ nm}$ (red).
The shortest wavelength is $410 \text{ nm}$, which corresponds to violet light.
Thus, violet light has the highest energy.

(D) The Rydberg formula for the wave number of a spectral line is given by: $\bar{\nu} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the Paschen series, the transition occurs to the energy level $n_1 = 3$.
The lowest energy transition corresponds to the transition from the immediate next energy level, which is $n_2 = 4$.
Substituting these values into the formula:
$\bar{\nu} = R_H (\frac{1}{3^2} - \frac{1}{4^2})$
$\bar{\nu} = R_H (\frac{1}{9} - \frac{1}{16})$
$\bar{\nu} = R_H (\frac{16 - 9}{144})$
$\bar{\nu} = R_H (\frac{7}{144}) \text{ cm}^{-1}$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series, the transition occurs to the ground state, so $n_1 = 1$.
For the shortest wavelength, the transition must occur from the highest possible energy level, i.e.,$n_2 = \infty$.
Substituting these values into the formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H (1 - 0) = R_H$.
Therefore, $\lambda = \frac{1}{R_H} = \frac{1}{109677 \text{ cm}^{-1}}$.
Calculating the value: $\lambda \approx 9.117 \times 10^{-6} \text{ cm} = 911.7 \times 10^{-8} \text{ cm}$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \text{ J}$.
For the $H$-atom, the atomic number $Z = 1$.
For the first orbit $(n = 1)$, $E_1 = -2.18 \times 10^{-18} \times \frac{1^2}{1^2} = -2.18 \times 10^{-18} \text{ J}$.
For the second orbit $(n = 2)$, $E_2 = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -2.18 \times 10^{-18} \times \frac{1}{4} \text{ J}$.
$E_2 = -0.545 \times 10^{-18} \text{ J}$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \text{ J s}$
$m = 9.10938 \times 10^{-31} \text{ kg}$
$v = 2.19 \times 10^6 \text{ m s}^{-1}$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{(9.10938 \times 10^{-31}) \times (2.19 \times 10^6)}$
$\lambda = \frac{6.626 \times 10^{-34}}{19.9495 \times 10^{-25}}$
$\lambda \approx 0.33214 \times 10^{-9} \text{ m}$
$\lambda \approx 332.14 \times 10^{-12} \text{ m}$
Since $1 \text{ pm} = 10^{-12} \text{ m}$, the wavelength is approximately $332 \text{ pm}$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{2.18 \times 10^{-18} \text{ J}}{n^2}$.
For the stationary state corresponding to $n = 2$, we substitute the value of $n$ into the formula:
$E_2 = -\frac{2.18 \times 10^{-18} \text{ J}}{(2)^2}$
$E_2 = -\frac{2.18 \times 10^{-18} \text{ J}}{4}$
$E_2 = -0.545 \times 10^{-18} \text{ J}$
$E_2 = -5.45 \times 10^{-19} \text{ J}$.
Therefore, the energy of the electron in the $n = 2$ state is $-5.45 \times 10^{-19} \text{ J}$.

(C) In Bohr's model of the hydrogen atom, the energy of the $n^{th}$ orbit is given by the formula: $E_n = \frac{E_1}{n^2}$, where $E_1$ is the energy of the ground state $(n=1)$.
Given that the ground state energy $E_1 = -E \text{ J}$.
For the third stationary orbit, $n = 3$.
Substituting the values into the formula: $E_3 = \frac{E_1}{3^2} = \frac{-E}{9} \text{ J}$.
Therefore, the energy of the third stationary orbit is $-E/9 \text{ J}$.

(D) Viscosity is the property of fluids (liquids and gases) that offers resistance to flow.
$(A)$ Gradation of lubricant oils depends on their viscosity.
$(B)$ Blood flow velocity and viscosity are used to indicate cardiovascular health.
$(C)$ The cleansing action of soap involves reducing surface tension and altering the viscosity of the medium to remove dirt.
$(D)$ The thickening of glass panes at the bottom in old buildings is a common misconception; glass is an amorphous solid, not a supercooled liquid, and this phenomenon is not related to viscosity.

(D) To obtain an $n$-type semiconductor, a pentavalent impurity (Group $15$ element) is added to a tetravalent semiconductor like $Ge$ (Germanium).
$P$ (Phosphorus), $As$ (Arsenic), and $Sb$ (Antimony) are all Group $15$ elements and can be used as dopants to create $n$-type semiconductors.
$B$ (Boron) is a trivalent element (Group $13$). Adding a trivalent impurity to $Ge$ results in a $p$-type semiconductor, not an $n$-type semiconductor.
Therefore, $B$ is the dopant that is $NOT$ used to obtain an $n$-type semiconductor.

(B, D) Magnetic substances are classified based on their response to an external magnetic field:
$1$. Paramagnetic substances are weakly attracted by an external magnetic field.
$2$. Diamagnetic substances are weakly repelled by an external magnetic field.
$3$. Ferromagnetic substances are strongly attracted by an external magnetic field.
Comparing these facts with the given options:
- Option $(A)$ is true.
- Option $(B)$ is false because diamagnetic substances are repelled, not attracted.
- Option $(C)$ is true.
- Option $(D)$ is false because diamagnetic substances are not strongly attracted; they are weakly repelled.
Therefore, both $(B)$ and $(D)$ are false statements.

(D) To form an $n$-type semiconductor, a pentavalent impurity (an element from Group $15$ of the periodic table) must be added to an intrinsic semiconductor like germanium $(Ge)$.
$B$ (Boron), $In$ (Indium), and $Ga$ (Gallium) are trivalent elements (Group $13$), which are used to create $p$-type semiconductors.
$P$ (Phosphorus) is a pentavalent element (Group $15$), which provides an extra electron when doped into germanium, thus creating an $n$-type semiconductor.
Therefore, the correct dopant is $P$.

(C) In an optical fiber communication system, an Erbium-doped fiber amplifier $(EDFA)$ is widely used to amplify optical signals. Erbium $(Er)$ ions are doped into the silica fiber core. When these ions are pumped with light at specific wavelengths (typically $980 \ nm$ or $1480 \ nm$), they reach an excited state and provide optical gain through stimulated emission at the $1550 \ nm$ wavelength, which is the low-loss window for optical fibers. Therefore, the correct element is Erbium $(Er)$.