![\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d32e615a6517b76e65b04fd25ae8bf0a_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...
Answer: Formula: Angle in radians = $ Angle\,in\,\deg \times \frac{\pi }{180} $ Therefore, Angle in radians = $ 36\times \frac{\pi }{180}=\frac{\pi }{5} $
Answer : (i) Equation of line parallel to y - axis is given by x = constant, as the x - coordinate of every point on the line parallel to y - axis is 6 i.e. constant. Now the point lies to the right...
, and 
.Solution: The difference in electronegativity between constituent atoms determines the ionic character of a molecule. As a result, the greater the difference, the greater the ionic character of a...
Solution: The bond pair of electrons are not shared equally when two unique atoms with different electronegativities join to form a covalent bond. The bond pair is attracted to the nucleus of an...
Solution: "Electronegativity refers to an atom's ability to attract a bond pair of electrons towards itself in a chemical compound." Sr. No Electronegativity Electron affinity 1 A tendency to...
Solution: There is a difference in electro-negativities of constituents of the atom in a heteronuclear molecule, which causes polarisation. As a result, one end gains a positive charge, while the...
and 
are triatomic molecules, the shape of the molecule is bent while that of is linear. Explain this on the basis of dipole moment.Solution: $CO_2$ has a dipole moment of 0 according to experimental results. And it's only possible if the molecule's shape is linear, because the dipole moments of the C-O bond are equal and...
Solution: Below is a list of Lewis symbols. To form a cation, a metal atom loses one or more electrons, while a nonmetal atom gains one or more electrons. Ionic bonds are formed between cations and...
Solution: Below is a list of Lewis symbols. To form a cation, a metal atom loses one or more electrons, while a nonmetal atom gains one or more electrons. Ionic bonds are formed between cations and...
The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T
, and 
Solution: Resonance is the phenomenon that allows a molecule to be expressed in multiple ways, none of which fully explain the molecule's properties. The molecule's structure is called a resonance...
can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing ? If not, give reasons for the same.Solution: The positions of the atoms remain constant in canonical forms, but the positions of the electrons change. The positions of atoms change in the given canonical forms. As a result, they...
ion.Solution: However, while the carbonate ion cannot be represented by a single structure, the properties of the ion can be described by two or more different resonance structures. The actual structure...
Solution: Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Solution: During the formation of a molecule, the extent of bonding that occurs between two atoms is represented by the bond strength of the molecule. As the bond strength increases, the bond...
and molecules are distorted tetrahedral, bond angle in water is less than that of Ammonia. Discuss.Solution: Ammonia's central atom (N) has one lone pair and three bond pairs. In water, the central atom (O) has two lone pairs and two bond pairs. As a result, the two bond pairs repel the two lone...
Solution: $BeCl_2$ The central atom does not have a lone pair, but it does have two bond pairs. As a result, its shape is AB2, or linear. $BCl_3$ The central atom has three bond pairs but no lone...
Solution: Ionic bonds are formed when one or more electrons are transferred from one atom to another. As a result, the ability of neutral atoms to lose or gain electrons is required for the...
Solution: “Atoms can combine either by transferring valence electrons from one atom to another or by sharing their valence electrons in order to achieve the closest inert gas configuration by having...
Solution: The lewis dot structures are:
and 
and 
Solution: For S and S2- A sulphur atom has only 6 valence electrons, which is a very small number. As a result, the Lewis dot symbol for the letter S is The presence of a...
Solution: Nitrogen atoms have only five valence electrons in total. As a result, the Lewis dot symbol for N is Bromine, because the atom has only seven valence electrons. As a result,...
Solution: Boron atoms have only three valence electrons, which is a very small number. As a result, the Lewis dot symbols for B are as follows: The oxygen atom has only six valence...
Solution: Only two valence electrons exist in the magnesium atom. As a result, the Lewis dot symbols for Mg are as follows: Only one valence electron exists in the sodium atom. As a...
Answer: "A chemical bond is an attractive force that holds a chemical species' constituents together." For chemical bond formation, many theories have been proposed, including valence shell electron...
As per the given question,
In the given word $ASSASSINATION,$ there are $4\;‘S’.$ Since all the $4\;‘S’$ have to be arranged together so let as take them as one unit. The remaining letters are \[=\text{ }3\text{ }A,\text{...
In this question we get $2$ options that is (i) Either all $3$ will go Then remaining students in class are: \[25\text{ }-\text{ }3\text{ }=\text{ }22\] Number of students remained to be chosen for...
Given there are total $9\;people$ Women occupies even places that means they will be sitting on \[{{2}^{nd}},\text{ }{{4}^{th}},\text{ }{{6}^{th}}and\text{ }{{8}^{th}}\] place where as men will be...
We have a deck of cards has $4\;kings.$ The numbers of remaining cards are $52.$ Ways of selecting a king from the deck \[\Rightarrow {{~}^{4}}{{C}_{1}}=\] Ways of selecting the remaining $4\;cards$...
The student can choose $3$ questions from $part\;I$ and $5$ from $part\;II$ Or $4\;questions$ from $part\;I$ and $4$ from $part \;II$ $5$ questions from $part\;I$ and $3$ from $part \;II$
We know that there are $5$ vowels and $21$ consonants in English alphabets. Choosing two vowels out of $5$ would be done in \[^{5}{{C}_{2}}\] ways Choosing $2$ consonants out of $21$ can be done in...
The number is divisible by $10$ if the unit place has $0$ in it. The $6-digit$ number is to be formed out of which unit place is fixed as $0$ The remaining $5\;places$ can be filled by $1, \;3,\;...
In dictionary words are listed alphabetically, so to find the words Listed before $E$ should start with letter either $A,\;B, \;C \;or \;D$ But the word $EXAMINATION$ doesn`t have $B,\;C \;or \;D$...
Given at most $3\;girls$ In this case the numbers of possibilities are $0\;girl\;and\;7\;boys$ $1\;girl\;and\;6\;boys$ $2\;girl\;and\;5\;boys$ $3\;girl\;and\;4\;boys$ Number of ways to choose...
(i) Given exactly $3$ girls Total numbers of girls are $4$ Out of which $3$ are to be chosen ∴ Number of ways in which choice would be made \[\Rightarrow {{~}^{4}}{{C}_{3}}=\] Numbers of boys are...
In the word $EQUATION$ there are $5$ vowels $(A,\;E,\;I,\;O,\;U)$ and $3$ consonants $(Q,\;T,\;N)$ The numbers of ways in which $5$ vowels can be arranged are \[^{5}{{C}_{5}}\] …………… (i) Similarly,...
The word DAUGHTER has $3$ vowels $A,$ $E,$ $U$ and $5$ consonants $D,$ $G,$ $H,$ $T$ and $R.$ The three vowels can be chosen in \[^{3}{{C}_{2}}\] as only two vowels are to be chosen. Similarly, the...
Given $9$ courses are available and $2$ specific courses are compulsory for every student Here $2$ courses are compulsory out of $9$ courses, so a student need to select \[5\text{ }-\text{ }2\text{...
Given a bag contains $5$ black and $6$ red balls Number of ways we can select $2$ black balls from $5$ black balls are \[\Rightarrow {{~}^{5}}{{C}_{2}}=\] Number of ways we can select $3$ red balls...
Given $17\;players$ in which only $5\;players$ can bowl if each cricket team of $11$ must include exactly $4\;bowlers$ There are $5\;players$ how bowl, and we can require $4\;bowlers$ in a team of...
Given a deck of $52$ cards There are $4\;\;Ace$ cards in a deck of $52\;\;cards.$ According to question, we need to select $1\;\;Ace$ card out the $4\;\;Ace\;\;cards$ ∴ Number of ways to select...
Given $6$ red balls, $5$ white balls and $5$ blue balls We can select $3$ red balls from $6$ red balls in \[\Rightarrow {{~}^{6}}{{C}_{3}}=\] ways Similarly, we can select $3$ white balls from $5$...
Given $5$ boys and $4$ girls are in total We can select $3$ boys from $5$ boys in \[\Rightarrow {{~}^{5}}{{C}_{3}}=\] ways Similarly, we can select $3$ boys from $54$ girls in \[\Rightarrow...
Given $21$ points on a circle We know that we require two points on the circle to draw a chord ∴ Number of chords is are \[\Rightarrow {{~}^{21}}{{C}_{2}}=\] ∴ Total number of chords can be drawn...
Simplifying and computing \[\Rightarrow ~4\text{ }\times \text{ }\left( 2n\text{ }-\text{ }1 \right)\text{ }=\text{ }12\text{ }\times \text{ }\left( n\text{ }-\text{ }2 \right)\] \[\Rightarrow...
As per the given question,
Number of places are as \[1~2~3~4~5~6~7~8~9~10~11~12\] There should always be $4$ letters between $P$ and $S.$ Possible places of $P$ and $S$ are $1$ and $6, 2$ and $7, 3$ and $8, 4$ and $9, 5$ and...
(i) Total number of letters in PERMUTATIONS \[=\text{ }12\] Only repeated letter is $T;$ $2times$ First and last letter of the word are fixed as $P$ and $S$ respectively. Number of letters remaining...
Total number of letters in MISSISSIPPI \[=\text{11}\] Letter Number of occurrence ⇒ Number of permutations = We take that $4 I’s$ come together, and they are treated as $1$ letter, ∴ Total number of...
Number of vowels in MONDAY \[=\text{ }2\text{ }\left( O\text{ }and\text{ }A \right)\] ⇒ Number of permutations in vowel = Now, remaining places \[=\text{ }5\] Remaining letters to be used \[=\text{...
(i) Number of letters to be used \[=\text{ }4\] ⇒ Number of permutations = (ii) Number of letters to be used \[=\text{ }6\] ⇒ Number of permutations =
Total number of different letters in EQUATION \[=\text{ }8\] Number of letters to be used to form a word \[=\text{ }8\] ⇒ Number of permutations =
As per the given question,
As per the given question,
Total number of people in committee \[=\text{ }8\] Number of positions to be filled \[=\text{ }2\] ⇒ Number of permutations =
Total number of digits possible for choosing \[=\text{ }5\] Number of places for which a digit has to be taken \[=\text{ }5=4\] As there is no repetition allowed, ⇒ Number of permutations = The...
Even number means that last digit should be even, Number of possible digits at one’s place \[=\text{ }3\text{ }\left( 2,\text{ }4\text{ }and\text{ }6 \right)\] ⇒ Number of permutations= One of digit...
To find four digit number (digits does not repeat) Now we will have $4$ places where $4$ digits are to be put. So, at thousand’s place = There are $9$ ways as $0$ cannot be at thousand’s place = $9...
As per the given question,
, Solution: As per the given question,
Solution: As per the given question,
Solution: As per the given question,
Consider LHS 3! +4! Computing left hand side, we get $ \begin{array}{l} 3 !+4 !=(3 \times 2 \times 1)+(4 \times 3 \times 2 \times 1) \\ =6+24 \\ =30 \end{array} $ Again consider RHS and computing we...
(i) Consider $8 !$ We know that $8 !=8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $=40320$ (ii) Consider 4!-3! $ 4 !-3 !=(4 \times 3 !)-3 ! $ Above equation can be written as $...
Given $5$ flags of different colours We know the signal requires $2$ flags. The number of flags possible for upper flag is $5.$ Now as one of the flag is taken, the number of flags remaining for...
Given $A$ coin is tossed $3$ times and the outcomes are recorded The possible outcomes after a coin toss are head and tail. The number of possible outcomes at each coin toss is $2.$ ∴The total...
Let the five-digit number be $ABCDE.$ Given that first $2$ digits of each number is $ 67.$ Therefore, the number is $67CDE.$ As the repetition is not allowed and $6$ and $7$ are already taken, the...
Suppose the $4$ digit code be $1234.$ Hence, the number of letters possible is $10.$ Let’s suppose any $1$ of the ten occupies place $1.$ So, as the repetition is not allowed, the number of letters...
Let the $3-digit$ number be $ABC,$ where $C$ is at the unit’s place, $B$ at the tens place and $A$ at the hundreds place. As the number has to even, the digits possible at $C$ are $2$ or $4$ or $6.$...
(i) Let the $3-digit$ number be $ABC,$ where $C$ is at the units place, $B$ at the tens place and $A$ at the hundreds place. Now when repetition is allowed, The number of digits possible at $C$ is...
Given \[{{a}_{i\text{ }j}}~=\text{ }{{e}^{2ix}}~sin\text{ }x\text{ }j\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 2}}\] So, the elements in a \[2\text{ }\times \text{ }2\] matrix are...
Answer: Given, 23n – 7n – 1 23n – 7n – 1 = 8n – 7n – 1 Using binomial theorem, 8n = 7n + 1 8n = (1 + 7) n 8n = nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + … + nCn (7) n 8n = 1 + 7n...
Solution: Given: First Plane: $2 x-y+4 z=5$ [On multiply both the sides by $2.5]$ We obtain, $5 x-2.5 y+10 z=12.5 \ldots$ Second Plane: $5 x-2.5 y+10 z=6 \ldots$ Therefore, $\begin{array}{l}...
Solution: It is known to us that the distance between two parallel planes $A x+B y+C z=d_{1}$ and $A x+B y+C z=d_{2}$ is given as...
Solution: It is known to us that the distance of the point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ from the plane $\mathrm{Ax}+\mathrm{By}+\mathrm{Cz}$ $=\mathrm{D}$ is given...
and 
Solution: The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\overrightarrow{\mathrm{b}}$ is...
and parallel to the planes 
and 
Solution: The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$ It is given that the line passes...
from the point of intersection of the line 
and the plane 
Solution: It is given that, The eq. of line is $\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})...
and 
And which is perpendicular to the plane 
Solution: It is known that, The eq. of any plane through the line of intersection of the planes $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{1}}=\mathrm{d}_{1}$ and...
Solution: It is known to us that the eq. of a plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and perpendicular to a line with direction ratios $A, B, C$ is given...
and 
and parallel to 
-axis.Solution: It is known to us that, The eq. of any plane through the line of intersection of the planes $\vec{r} \cdot \overrightarrow{n_{1}}=d_{1}$ and $\vec{r} \cdot \overrightarrow{n_{2}}=d_{2}$ is...
, then find the value of 
.Solution: It is known to us that the distance of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n}=d$ is given as $\left|\frac{\vec{a} \cdot \vec{n}-d}{|\vec{n}|}\right|$...
Solution: It is known to us that the eq. of a plane passing through $\left(x_{1}, y_{1}, z_{1}\right)$ is given by $A\left(x-x_{1}\right)+B\left(y-y_{1}\right)+C\left(z-z_{1}\right)=0$ Where, A, B,...
Solution: It is known to us that the eq. of a line passing through two points $A\left(x_{1}, y_{1}, z_{1}\right)$ and $B\left(x_{2}, y_{2}, z_{2}\right)$ is given as...
Solution: It is known to us that the vector eq. of a line passing through two points with position vectors $\vec{a}$ and $b$ is given as...
Solution: It is known to us that the vector eq. of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given as...
Solution: It is known to us that the shortest distance between lines with vector equations $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and $\vec{r}=\overrightarrow{a_{2}}+\lambda...
Solution: The eq. of a plane passing through $\left(x_{1}, y_{1}, z_{1}\right)$ and perpendicular to a line with direction ratios $A, B, C$ is given as...
Solution: The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is given as...
are perpendicular, find the value of 
.Solution: It is known to us that the two lines $\frac{x-1}{3 k}=\frac{y-2}{1}=\frac{z-3}{-5} \text { and }$ $\frac{\mathrm{x}-1}{3 \mathrm{k}}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-3}{-5}$ are...
Solution: It is known to us that the angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by $\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1}...
Solution: It is known to us that, eq. of a line passing through $\left(x_{1}, y_{1}, z_{1}\right)$ and parallel to a line with direction ratios $a, b, c$ is...
Solution: The angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by $\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1}...
and 
are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are 
Solution: Let's consider $l, m, n$ be the direction cosines of the line perpendicular to each of the given lines. Therefore, $ll_{1}+m m_{1}+n n_{1}=0 \ldots(1)$ And...
Solution: Let's consider $O A$ be the line joining the origin $(0,0,0)$ and the point $A(2,1,1)$. And let $B C$ be the line joining the points $B(3,5,-1)$ and $C(4,3,-1)$ Therefore the direction...
Solution: (a) The length of perpendicular from the point $(2,3,-5)$ on the plane $x+2 y-2 z=9 \Rightarrow x+2 y-2 z-9=0$ is $\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$...
Solution: (a) The distance of the point $(0,0,0)$ from the plane $3 x-4 y+12=3 \Rightarrow$ $3 x-4 y+12 z-3=0$ is $\begin{array}{l} \frac{\left|a x_{1}+b y_{1}+c...
Solution: (a) $4 x+8 y+z-8=0$ and $y+z-4=0$ It is given that The eq. of the given planes are $4 x+8 y+z-8=0 \text { and } y+z-4=0$ It is known to us that, two planes are $\perp$ if the direction...
Solution: (a) $2 x-2 y+4 z+5=0$ and $3 x-3 y+6 z-1=0$ It is given that The eq. of the given planes are $2 x-2 y+4 z+5=0$ and $x-2 y+5=0$ It is known to us that, two planes are $\perp$ if the...
Solution: (a) $7 x+5 y+6 z+30=0$ and $3 x-y-10 z+4=0$ It is given that The eq. of the given planes are $7 x+5 y+6 z+30=0$ and $3 x-y-10 z+4=0$ Two planes are $\perp$ if the direction ratio of the...
Solution: It is given that The eq. of the given planes are $\vec{r}(2 \hat{i}+2 \hat{j}-3 \hat{k})=5 \text { and } \vec{r}(3 \hat{i}-3 \hat{j}+5 \hat{k})=5$ If $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$...
Solution: Let's say that the eq. of the plane that passes through the two-given planes $x+y+z=1$ and $2 x+3 y+4 z=5$ is $\begin{array}{l} (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\ (2 \lambda+1) x+(3...
and through the point 
Solution: Let's consider the vector eq. of the plane passing through the intersection of the planes are $\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=7...
Solution: It is given that Eq. of the plane passes through the intersection of the plane is given by $(3 x-y+2 z-4)+\lambda(x+y+z-2)=0$ and the plane passes through the points $(2,2,1)$ Therefore,...
-axis and parallel to ZOX plane.Solution: It is known to us that the equation of the plane $\mathrm{ZOX}$ is $\mathrm{y}=0$ Therefore, the equation of plane parallel to $\mathrm{ZOX}$ is of the form, $\mathrm{y}=\mathrm{a}$ As the...
Solution: It is given that The plane $2 x+y-z=5$ Let us express the equation of the plane in intercept form $x / a+y / b+z / c=1$ Where $a, b, c$ are the intercepts cut-off by the plane at $x, y$...
Solution: (a) It is given that, The points are $(1,1,-1),(6,4,-5),(-4,-2,3)$. Let, $\begin{array}{l} =\left|\begin{array}{ccc} 1 & 1 & -1 \\ 6 & 4 & -5 \\ -4 & -2 & 3...
and the normal to the plane is 
and the normal vector to the plane is 
Solution: (a) That passes through the point $(1,0,-2)$ and the normal to the plane is $\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ Let's say that the position vector of the point $(1,0,-2)$...
Solution: (a) $x+y+z=1$ Let the coordinate of the foot of $\perp \mathrm{P}$ from the origin to the given plane be $P(x, y, z)$ $x+y+z=1$ The direction ratio are $(1,1,1)$ $\begin{array}{l}...
![\overrightarrow{\mathrm{r}} \cdot[(\mathrm{s}-2 \mathrm{t}) \hat{\mathrm{i}}+(3-\mathrm{t}) \hat{\mathrm{j}}+(2 \mathrm{~s}+\mathrm{t}) \hat{\mathrm{k}}]=15](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0789422b9ad2fd168094a22f1dd6043a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let $\overrightarrow{\mathrm{r}}$ be the position vector of $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is given by $\overrightarrow{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y}...
Solution: (a) It is given that, The equation of the plane. Let $\overrightarrow{\mathrm{r}}$ be the position vector of $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is given by $\vec{r}=x...
Solution: It is given that, The vector $3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$ Vector equation of the plane with position vector $\overrightarrow{\mathrm{r}}$ is $\vec{r} \cdot...
Solution: (a) $2 x+3 y-z=5$ It is given that The eq. of the plane, $2 x+3 y-z=5 \ldots$. (1) The direction ratio of the normal $(2,3,-1)$ Using the formula,...
Solution: (a) $z=2$ It is given that The eq. of the plane, $z=2$ or $0 x+0 y+z=2 \ldots (1) .$ The direction ratio of the normal $(0,0,1)$ Using the formula, $\begin{array}{l}...
Solution: Consider the given equations $\begin{array}{l} \Rightarrow \vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \\ \vec{r}=\hat{i}-t \hat{i}+t \hat{j}-2 \hat{j}+3 \hat{k}-2 t \hat{k} \\...
and 
Solution: It is known to us that shortest distance between two lines $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and $\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$...
and 
Solution: It is known to us that the shortest distance between two lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ is given as:...
Solution: It is known to us that the shortest distance between two lines $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{1}}+\lambda \overrightarrow{\mathrm{b}_{1}}$ and...
and 
are perpendicular to each other.Solution: The equations of the given lines are $\frac{\mathrm{x}-5}{7}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}}{1}$ and $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}$ Two lines...
and 
are at right angles.Solution: The standard form of a pair of Cartesian lines is:...
and 
and 
Solution: Let's consider $\theta$ be the angle between the given lines. If $\theta$ is the acute angle between $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and...
Solution: It is given that Let's calculate the vector form: The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
Solution: Given: The origin $(0,0,0)$ and the point $(5,-2,3)$ It is known to us that The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
Write its vector form.Solution: It is given that The Cartesian equation is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \ldots \text { (1) }$ It is known to us that The Cartesian eq. of a line passing through a point...
and parallel to the line given by 
Solution: It is given that The points $(-2,4,-5)$ It is known that Now, the Cartesian equation of a line through a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and having...
and 
. is in the directionSolution: Given: Vector equation of a line that passes through a given point whose position vector is $\vec{a}$ and parallel to a given vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$ Let,...
Solution: Given that, Line passes through the point $(1,2,3)$ and is parallel to the vector. It is known to us that Vector eq. of a line that passes through a given point whose position vector is...
Solution: The points $(4,7,8),(2,3,4)$ and $(-1,-2,1),(1,2,5)$. Consider $A B$ be the line joining the points, $(4,7,8),(2,3,4)$ and $C D$ be the line through the points $(-1,-2$, 1), $(1,2,5)$. So...
Solution: Given that The points $(1,-1,2),(3,4,-2)$ and $(0,3,2),(3,5,6)$. Let's consider $A B$ be the line joining the points, $(1,-1,2)$ and $(3,4,-2)$, and $C D$ be the line through the points...
Are mutually perpendicular.Solution: Consider the direction cosines of $L_{1}, L_{2}$ and $L_{3}$ be $l_{1}, m_{1}, n_{1} ; l_{2}, m_{2}, n_{2}$ and $l_{3}, m_{3}, n_{3}$. It is known that If $\mathrm{f}_{1}, \mathrm{~m}_{1},...
Solution: Given that, The vertices are $(3,5,-4),(-1,1,2)$ and $(-5,-5,-2)$. Firstly find the direction ratios of $\mathrm{AB}$ Where, $A=(3,5,-4)$ and $B=(-1,1,2)$ Ratio of $A...
Solution: If the direction ratios of two lines segments are proportional, then the lines are collinear. It is given that $\mathrm{A}(2,3,4), \mathrm{B}(-1,-2,1), \mathrm{C}(5,8,7)$ The direction...
Solution: Given that, The direction ratios are $-18,12,-4$ Where, $a=-18, b=12, c=-4$ Consider the direction ratios of the line as $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ Direction cosines are...
Solution: Given that, Angles are equal. Let the angles be $\alpha, \beta, \mathrm{Y}$ The direction cosines of the line be I, $\mathrm{m}$ and $\mathrm{n}$ $I=\cos \alpha, m=\cos \beta \text { and }...
Solution: Let's consider the direction cosines of the line be $I, m$ and $n$. Let $\alpha=90^{\circ}, \beta=135^{\circ}$ and $\mathrm{y}=45^{\circ}$ Therefore, $I=\cos \alpha, m=\cos \beta \text {...
As per the given question, Let I = = = = = = = = =
Solution:
only (b) exhibit oxidation state of 
and (c) form 
and 
ion (d) form 
and 
ionsSolution: (b) Group 14 components have 4 valence electrons. Thus, bunch oxidation status is $+4$. In any case, the lower oxidation state turns out to be progressively steady because of the inactive...
Solution: (b) Borax is a strong base salt (NaOH) and a feeble corrosive $\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)$. In this way, it is essential thing in nature.
each.Solution: Carbon dioxide CO2 can be ready in the lab through the activity of weaken hydrochloric corrosive on calcium carbonate. Their response is as per the following: CO2 is industrially ready by...
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Solution: The balanced equations are as follow:
gives the following results. (i) Its aqueous solution is alkaline to litmus. (ii) It swells up to a glassy material 
on strong heating. (iii) When conc. 
is added to a hot solution of 
, a white crystal of an acid 
separates out Write equations for all the above reactions and identify X, 
, and 
.Solution: The salt given to litmus is antacid. $X$ is, subsequently, a salt with a solid base, and a feeble corrosive. When $X$ is warmed unnecessarily, it additionally enlarges to frame material...
is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of 
to give soluble complex (B). Compound (A) is soluble in dilute HCI to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract the metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.Solution: The given metal $X$ gives sodium hydroxide to a white accelerate, and the encourage breaks up surpassing sodium hydroxide. $X$ must, consequently, be made of aluminum. The acquired white...
(B) Write suitable equations to show their nature.Solution: $\rightarrow$ CO $=$ Neutral $\rightarrow \mathrm{B}_{2} \mathrm{O}_{3}=$ Acidic Being acidic, it responds with bases to frame salts. It responds with $\mathrm{NaOH}$ to frame sodium...
Solution: Allotropy is the presence of a component in more than one structure, having diverse actual properties however similar substance properties. Diamond's solid 3-D construction makes it a...
Solution: A tomic sweep (in pm) Aluminum Gallium In spite of the fact that Ga has more than one shell than Al, it is more modest in size than Al. This is on the grounds that the $3 \mathrm{~d}$...
Solution: Carbon ionizing enthalpy (the primary component in bunch 14 ) is exceptionally high $(1086 \mathrm{~kJ}/\mathrm{mol})$. That is normal on account of its little size. Nonetheless, there is...
solution.Solution: (a) Silicon is warmed with methyl chloride at high temperature within the sight of copper A class of organosilicon polymers called methyl-subbed chlorosilane $\mathrm{MeSiCl}_{3},...
is reacted with ammonia?Solution: (a) Borax is warmed firmly Borax goes through different changes when warmed. It is losing atoms and expands of water right away. Then, at that point, it turns into a clear fluid, which...
Solution: (a) Diborane $\mathrm{B}_{2} \mathrm{H}_{6}$ is a compound that does not have an electron. $\mathrm{B}_{2} \mathrm{H}_{6}$ just has 12 electrons $-6 \mathrm{e}^{-}$of $6 \mathrm{H}$...
(boron trifluoride) is bubbled through. Give reasons.Solution: Hydrogen fluoride is a covalent compound with an exceptionally solid intermolecular clinging to hydrogen. Along these lines, it doesn't give particles and doesn't break up aluminum...
molecule has zero dipole moment.Solution: The $B-C l$ bond is normally polar as a result of the distinction in the electronegativities of $\mathrm{Cl}$ and $B$. However the particle of $\mathrm{BCl}_{3}$ is non-polar. That is on...
and 
differ.Solution; In $\mathrm{BF}_{3}$, the length of the $\mathrm{B}-\mathrm{F}$ bond is more limited than that of the $\mathrm{B}-\mathrm{F}$ bond in $\mathrm{BF}_{4}^{-} \cdot \mathrm{BF}_{3}$ is an...
to give 
. – Lead (IV) chloride is highly unstable towards heat. – Lead is known not to form an iodide, 
.Solution: - Lead is an individual from bunch 14 of the occasional table. The two oxidation situations with bunch shows are $+2$ and $+4$. The $+2$ oxidation state turns out to be more steady when...
Solution: Diamond: 1. It has a glasslike grid 2. In precious stone, every carbon molecule is sp3 hybridized and attached to four other carbon particles through a sigma bond. 3. It has an unbending...
diamond (c) graphite?Solution: The condition of hybridization of carbon in: (a) $C O_{3}^{2-}$ c in $\mathrm{CO}_{3}^{2-}$ is sp $^{2}$ hybridized and is attached to 3 oxygen iotas. (b) Diamond Every precious stone...
and 
.Solution: For $C O_{3}^{2-}$ There are just 2 resounding designs for the bicarbonate particle. For $\mathrm{HCO}_{3}^{-}$
Solution: Amphoteric substances will be substances that display both acidic and essential characteristics. Since aluminum breaks up in the two acids and bases, it is said to have an amphoteric...
and 
. Assign the hybridisation of boron in these speciesSolution: (I) $\mathbf{B} \boldsymbol{F}_{3}$ Boron will in general shape monomeric covalent halides in view of its little size and high electronegativity. These halides of boron generally have a...
Solution: After warming orthoboric corrosive at a temperature of $370 \mathrm{~K}$ or above, it is changed over into metaboric corrosive and, upon additional warming, yields boric oxide...
Solution: Boric corrosive is a powerless monobasic corrosive which acts as a Lewis corrosive. In this way, it's anything but a protic corrosive. $\mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{HOH}...
and 
. How will they behave with water? Justify.Solution: Since it is a Lewis corrosive, $\mathrm{BCl}_{3}$ promptly goes through hydrolysis to frame boric corrosive. $\mathrm{BCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3...
Solution: The electronic arrangement of boron is $n s^{2} n p^{1}$. It contains 3 electrons in its valence shell. In this way, it can frame just 3 covalent bonds which imply that there are just 6...
as compared to 
?Solution: Thallium and boron have a place with bunch 13 of the intermittent table and $+1$ oxidation state turns out to be more steady as we drop down the gathering. Boron is more steady than...
to 
(ii) 
to 
Solution: (I) $B$ to $\mathrm{TI}$ Gathering 13 components have their electronic design of $\mathrm{ns}^{2} \mathrm{np}^{1}$ and the oxidation state displayed by these components ought to be 3 ....
According to the question, two coins are tossed once. We know, when two coins are tossed then the total number of possible outcomes are will be $2^2=4$ So, the Sample space is {HH, HT, TT, TH} ∴...
Further show: (i) which of the electrode is negatively charged? (ii) the carriers of the current in the cell. (iii) individual reaction at each electrode.Solution: The galvanic cell relating to the given redox response can be displayed as: $\mathrm{Zn}\left|Z n_{(a q)}^{2+} \| A g_{(a q)}^{+}\right| \mathrm{Ag}$ (I) Zn anode is contrarily charged on...
Arrange these metals in their increasing order of reducing power.Solution: The diminishing specialist is more grounded as the terminal potential declines. Subsequently, the expanding request of the lessening force of the given metals is as given underneath: Ag...
Solution: A metal with more grounded diminishing force uproots one more metal with more vulnerable lessening power from its answer of salt. The request for the expanding diminishing force of the...
with silver electrodes (ii) An aqueous solution with platinum electrodes (iii) A dilute solution of with platinum electrodes (iv) An aqueous solution of 
with platinum electrodes.Solution: (I) In fluid arrangement, AgNO3 ionizes to give Ag+(aq) and NO3–(aq) particles. \[AgN03\left( aq \right)\text{ }\to \text{ }Ag+\left( aq \right)\text{ }+\text{ }NO3\left( aq...
and 
(b) 
and 
(c) and (d) 
and (e) 
and 
Solution: (a) $F e_{(a q)}^{3+}$ and $I_{(a q)}^{-}$ $2 F e_{(a q)}^{3+}+2 I_{(a q)}^{-} \rightarrow 2 F e_{(a q)}^{2+}+I_{2(s)}$ Oxidation half response: $2 I_{(a q)}^{-} \rightarrow I_{2}(s)+2...
Solution: The reasonable response is as given underneath: $4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(g)}$ $4 N H_{3}=4 \times 17...
Solution: One of the responding components consistently has a component that can exist in somewhere around 3 oxidation numbers. (I) The non - metals which can show disproportionation responses are...
Solution: The redox response is as given beneath: Cl2(s)+SO2(aq)+H2O(l)→Cl(aq)-+SO4(aq)2- The oxidation half response: SO2(aq)→SO4(aq)2- Add 2 electrons to adjust the oxidation no. :...
ion is unstable in solution and undergoes disproportionation to give 
, and 
ion. Write a balanced ionic equation for the reaction.Solution: The response is as given beneath: $\mathrm{Mn}_{(a q)}^{3+} \rightarrow \mathrm{Mn}_{(a q)}^{2+}+\mathrm{MnO}_{2(s)}+\mathrm{H}_{(a q)}^{+}$ The oxidation half response: $\mathrm{Mn}_{(a...
Solution: The oxidation no. of $\mathrm{C}$ in $(C N)_{2}, C N^{-}$and $C N O^{-}$are $+3,+2$ and $+4$ separately. Let the oxidation no. of $\mathrm{C}$ be $\mathrm{y}$. $(C N)_{2}$ $2(y-3)=0$ Along...
(s) 
(b) 
(c) 
Solution:
(Basic medium) (b) 
(Acidic medium) (c) 
(Acidic medium) (d) 
(Acidic medium)Solution: (a) $M n O_{4}^{-}(a q)+I_{(a q)}^{-} \rightarrow \operatorname{MnO}_{2}(s)+I_{2}(s)$ Stage 1 The two half responses are given beneath: Oxidation half response: $I_{(a q)} \rightarrow...
(b) 
(c) ![\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}_{(l)}+2\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]_{(a q)}^{+}+3 \mathrm{OH}_{(a q)}^{-} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}_{(a q)}^{-}+2 \mathrm{Ag}_{(s)}+](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7aa1b43808014d89aabd5bf76808b357_l3.png "Rendered by QuickLaTeX.com")
(d) 
No change is observed What inference do you draw about the behavior of 
and 
from these reactions?Solution: $\mathrm{Ag}^{+}$and $C u^{2+}$ acts as oxidizing specialist in responses (I) and (ii) individually. In response (iii), $\mathrm{Ag}^{+}$oxidizes $\mathrm{C}_{6} \mathrm{H}_{5}...
What conclusion about the compound 
(of which 
is a part) can be drawn from the reaction?Solution: $X e O_{6(a q)}^{4-}+2 F_{(a q)}^{-}+6 H_{(a q)}^{+} \rightarrow X e O_{3(g)}+F_{2(g)}+3 H_{2} O_{(l)}$ The oxidation no. of Xe decreases from $+8$ in $\mathrm{XeO}_{6}^{4-}$ to $+6$ in...
Solution: $F_{2}$ can oxidize $C l^{-}$to $C l_{2}, B r^{-}$to $B r_{2}$, and $I^{-}$to $I_{2}$ as: $F_{2(a q)}+2 C l_{(s)}^{-} \rightarrow 2 F_{(a q)}^{-}+C l_{2(g)}$ $F_{2}(a q)+2 B r_{(a q)}^{-}...
Why does the same reductant, thiosulphate react differently with iodine and bromine?Solution: The normal oxidation no. of $\mathrm{S}$ in $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is $+2$. The normal oxidation no. of $\mathrm{S}$ in $S_{4} \mathrm{O}_{6}^{2-}$ is $+2.5$. The oxidation...
(b) ![H C H O_{(l)}+2\left[A g\left(N H_{3}\right)_{2}\right]_{(a q)}^{+}+3 O H_{(a q)}^{-} \rightarrow 2 A g_{(s)}+H C O O_{(a q)}^{-}+](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b8e4956dd6ca01b2ee5147606a038527_l3.png "Rendered by QuickLaTeX.com")
(c) 
(d) 
(e) 
Solution: (a) $2 \mathrm{AgBr}_{(s)}+C_{6} H_{6} O_{2}(a q) \rightarrow 2 \mathrm{Ag}_{(s)}+2 \mathrm{HBr}_{(a q)}+C_{6} \mathrm{H}_{4} O_{2}(a q)$ $\mathrm{C}_{6} \mathrm{H}_{6}...
Solution: (a) While producing benzoic corrosive from toluene, alcoholic potassium permanganate is utilized as an oxidant because of the given reasons. (I) In an impartial medium, $O H^{-}$ions are...
Solution: When there is a response between lessening specialist and oxidizing specialist, a compound is framed which has lower oxidation number if the diminishing specialist is in abundance and a...
is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?Solution: The oxidation no. of $A g$ in $A g F_{2}$ is $+2$. Be that as it may, $+2$ is entirely unsound oxidation no. of Ag. Consequently, when $A g F_{2}$ is framed, silver acknowledges an...
(b) 
Why it is more appropriate to write these reactions as : (a) 
aq 
(b) 
Also suggest a technique to investigate the path of the above (a) and (b) redox reactionsSolution: (a) Stage 1: $\mathrm{H}_{2} \mathrm{O}$ breaks to give $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$. $2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow 2...
Solution: In sulfur dioxide $\left(S O_{2}\right)$ the oxidation no. of $\mathrm{S}$ is $+4$ and the scope of oxidation no. of sulfur is from $+6$ to $-2$. Consequently, $S O_{2}$ can go about as a...
Solution: The compound where carbon has oxidation number from -4 to +4 are given below in the table:
Solution: Formulas are: (a) Mercury (II) chloride $H g C l_{2}$ (b) Nickel (II) sulphate $\mathrm{NiSO}_{4}$ (c) Tin (IV) oxide $\mathrm{SnO}_{2}$ (d) Thallium (I) sulphate $\mathrm{Tl}_{2}...
Solution: O.N. of S in H2SO5. By traditional strategy, the O.N. of S in H2SO5 is 2 (+1) + x + 5 (- 2) = 0 or x = +8 This is outlandish on the grounds that the most extreme O.N. of S can't be more...