Class 11 Rotational Motion: Question Bank For Class 11 Physics Gravitation: Question Bank For Class 11 Physics Simple Harmonic Motion: Question Bank For Class 11 Circular Motion: Question Bank For Class 11 Physics Friction: Question Bank For Class 11 Physics Newton’s Laws Of Motion: Question Bank For Class 11 Physics Conservation Of Linear Of Momentum: Question Bank For Class 11 Physics Projectile Motion: Question Bank For Class 11 Physics Unit And Dimension: Question Bank For Class 11 Physics Heat And Thermodynamics: Question Bank For Class 11 General Properties Of Matter: Question Bank For Class 11 Motion In One Dimension: Question Bank For Class 11       Class 12 Electrostatics  Moving Charges & Magnetism  Wave Optics Ray Optics  Semiconductor  Modern Physics Alternating Current  Electromagnetic Induction Capacitor  Current Electricity   
Ionic Equilibrium Short Notes for Class 11, JEE & NEET
eSaral provides chemistry short notes for JEE and NEET to help students in revising topics quickly. These notes are completely based on latest syllabus and it includes all the tips and tricks that will help you in learning chemistry better and score well. The Notes will help you to understand the important topics and remember the key points for exam point of view. You can also access detailed Notes of chemistry here. Download or View Detailed Notes for Chemistry Class 11th Download or View Detailed Notes for Chemistry Class 12th   Ionic Equilibrium Theory Fundamentals of Acids, Bases & Ionic Equilibrium Acids & Bases When dissolved in water, acids release $\mathrm{H}^{+}$ ions, base release $\mathrm{OH}^{-}$ ions. Arrhenius Theory When dissolved in water, the substances which release (i) $\mathrm{H}^{+}$ ions are called acids (ii) $\mathrm{OH}^{-}$ ions are called bases Bronsted & Lowry Concept Acids are proton donors, bases are proton acceptors Note that as per this definition, water is not necessarily the solvent. When a substance is dissolved in water, it is said to react with water e.g. $\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}$ ; HCl donates H+ to water, hence acid. $\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \quad \rightarrow \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}$ ; NH3 takes H+ from water, hence base. For the backward reaction, $N H_{4}^{+}$ donate $H^{+}$ , hence it is an acid; $O H^{-}$ accepts H+, hence it is base. $N H_{3}$ (base) & $N H_{4}^{+}$ (acid) from conjugate acid base pair. Conjugate acid and bases To get conjugate acid of a given species add $\mathrm{H}^{+}$ to it. e.g. conjugate acid of N2H4 is N2H5+. To get conjugate base of any species subtract H+ from it. e.g. Conjugate base of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{2}^{-}$. Note: Although $C I^{-}$ is conjugate base of HCl, it is not a base as an independent species. In fact,anions of all strong acid like $C I, N O_{3}^{-}, C l O_{4}^{-}$ etc. are neutral anions. Same is true for cations of strong bases like $K^{+}, N a^{+}, B a^{++}$ etc. When they are dissolved in water, they do not react with water (i.e. they do not undergo hydrolysis) and these ions do not cause any change in pH of water (others like $\left.C N^{-} d o\right)$. Some examples of : Basic Anions : $\mathrm{CH}_{3} \mathrm{COO}^{-}, \mathrm{OH}^{-}, \mathrm{CN}^{-}$ (Conjugate bases of weak acids) Acid Anions: $H S O_{3}^{-}, H S^{-}$ etc. Note that these ions are amphoteric, i.e. they can behave both as an acid and as a base. e.g. for $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}:$ Lewis Concept : Acids are substances which accept a pair of electrons to form a coordinate bond and bases are the substances which donate a pair of electrons to form a coordinate bond. Important : $\mathrm{Ca}+\mathrm{S} \rightarrow \mathrm{Ca}^{2+}+\mathrm{S}^{2-}$ is not a Lewis acidbase reaction since dative bond is not formed. Lewis Acids : As per Lewis concept, following species can acts as Lewis Acids : (i) Molecules in which central atom has incomplete octet. \text { (e.g. }\left.\mathrm{BF}_{3}, \mathrm{AlCl}_{3} \text { etc. }\right) (ii) Molecules which have a central atom with empty d orbitals \text { (e.g. }\left.\operatorname{six}_{4}, \mathrm{GeX}_{4}, \mathrm{PX}_{3}, \mathrm{TiCl}_{4} \text { etc. }\right) (iii) Simple Cations: Though all cations can be expected to be Lewis acids, \mathrm{Na}^{+}, \mathrm{Ca}^{++}, \mathrm{K}^{+} \text {etc. } show no tendency to accept electrons. However \mathrm{H}^{+}, \mathrm{Ag}^{+} etc. act as Lewis acids. (iv) Molecules having multiple bond between atoms of dissimilar electronegativity. Lewis bases are typically : (i) Neutral species having at least one lone pair of electrons. Autoprotolysis of water (or any solvent) \text { Autoprotolysis (or self-ionization) constant }\left(\mathrm{K}_{\mathrm{w}}\right)=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \text { Hence, } \mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}} \text { at all temperatures } Condition of neutrality \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right] \text {(for water as solvent) } \text { At } 25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{W}}=10^{-14} \cdot \mathrm{K}_{\mathrm{W}} increases with increase in temperature. Accordingly, the neutral point of water \left(\mathrm{pH}=7 \text { at } 25^{\circ} \mathrm{C}\right) also shifts to a value lower than 7 with increase in temperature. Important: \mathrm{K}_{\mathrm{W}}=10^{-14} s a value at \text { (i) } 25^{\circ} \mathrm{C} \text { (ii) }for water only. If the temperature changes or if some other solvent is used, autoprotolysis constant will not be same. Ionisation Constant * \quad \text { For dissociation of weak acids (eg. HCN), HCN + H_{ } 2 } \mathrm{O} \text { 1 } \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CN}^{-} \text {the equilibrium } constant expression is written as Ka =\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]} * For the Polyprotic acids \left(\mathrm{e} \cdot \mathrm{g} \cdot \mathrm{H}_{3} \mathrm{PO}_{4}\right) sucessive ionisation constants are denoted \text { by } \mathrm{K}_{1}, \mathrm{K}_{2}, \mathrm{K}_{3} \text { etc. } For \mathrm{H}_{3} \mathrm{PO}_{4}, \text { Similarly, } \mathrm{K}_{\mathrm{b}} \text { denotes basic dissociation constant for a base. } \text { Also, } \mathrm{pK}_{\mathrm{a}}=-\log _{10} \mathrm{K}_{\mathrm{a}}, \mathrm{pK}_{\mathrm{b}}=-\log _{10} \mathrm{K}_{\mathrm{b}} \text { Some Important Results: }\left[\mathrm{H}^{+}\right] \text {concentration of } Case (i) A weak acid in water Similarly for a weak base, substitute \left[\mathrm{OH}^{-}\right] \text {and } \mathrm{K}_{\mathrm{b}} instead of \left[\mathrm{H}^{+}\right] \text {and } \mathrm{K}_{\mathrm{a}} respectively in these expressions. Case (ii) (a) A weak acid and a strong acid \left[\mathrm{H}^{+}\right] is entirely due to dissociation of strong acid (b) A weak base and a strong base \left[\mathrm{H}^{+}\right] is entirely due to dissociation of strong base Neglect the contribution of weak acid/base usually. Condition for neglecting : \text { If } \mathrm{c}_{0} concentration of strong acid, c_{1} = concentration of weak acid then neglect the contribution of weak acid if \mathrm{K}_{\mathrm{a}} \leq 0.01 \mathrm{c}_{0}^{2 /} \mathrm{c}_{1} Case (iii) Two (or more) weak acids Proceed by the general method of applying two conditions (i) of electroneutrality (ii) of equilibria. The accurate treatement yields a cubic equation. Assuming that acids dissociate to \text { a negligible extent }\left[\text { i.e. } c_{0}-x \approx c_{0}\right] \quad\left[\mathrm{H}^{+}\right]=\left(\mathrm{K}_{1} \mathrm{c}_{1}+\mathrm{K}_{2} \mathrm{c}_{2}+\ldots+\mathrm{K}_{\mathrm{w}}\right)^{1 / 2} Case (iv) When dissociation of water becomes significant: \text { Dissociation of water contributes significantly to }\left[\mathrm{H}^{+}\right] \text {or }\left[\mathrm{OH}^{-}\right] \text {only when for } (i) strong acids (or bases) : 10^{-8} \mathrm{M}<\mathrm{c}_{0}<10^{-6} \mathrm{M}. Neglecting ionisation of water at 10^{-6} \mathrm{M} \text { causes } 1 \% \text { error (approvable). Below } 10^{-8} \mathrm{M}, Neglecting ionisation of water at 10^{-6} \mathrm{M} \text { causes } 1 \% \text { error (approvable). } Below 10^{-8} \mathrm{M}, contribution of acid (or base) can be neglected and pH can be taken to be practically 7. Weak acids (or bases) : \text { When } \mathrm{K}_{\mathrm{a}} \mathrm{c}_{0}<10^{-12} then consider dissociation of water as well. HYDROLYSIS * Salts of strong acids and strong bases do not undergo hydrolysis. * Salts of a strong acids and weak bases give an acidic solution. e.g. \mathrm{NH}_{4} \mathrm{Cl} when dissolved, it dissociates to give \mathrm{NH}_{4}^{+} ions and \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad 1 \quad \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \mathrm{K}_{\mathrm{h}}=\left[\mathrm{NH}_{3}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] /\left[\mathrm{NH}_{4}^{+}\right]=\mathrm{K}_{\mathrm{W}} / \mathrm{K}_{\mathrm{b}} of conjugate base of \mathrm{NH}_{4}^{+} Important! In general : \mathrm{K}_{\mathrm{a}}(\text { of an acid }) \mathrm{xK}_{\mathrm{b}} \text { (of its conjugate base) }=\mathrm{K}_{\mathrm{w}} If the degree of hydrolysis(h) is small (<<1), \quad \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}} Otherwise h = \frac{-\mathrm{K}_{\mathrm{h}}+\sqrt{\mathrm{K}_{\mathrm{h}}^{2}+4 \mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}}{2 \mathrm{c}_{0}}, \quad\left[\mathrm{H}^{+}\right]=\mathrm{c}_{0} \mathrm{h} * Salts of strong base and weak acid give a basic solution (\mathrm{pH}>7) when dissolved in water, e.g. NaCN, \mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \quad 1 \quad \mathrm{HCN}+\mathrm{OH}^{-} \quad\left[\mathrm{OH}^{-}\right]=\mathrm{c}_{0} \mathrm{h}, \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}} * Salts of weak base and weak acid Assuming degree of hydrolysis to be same for the both the ions, \mathrm{K}_{\mathrm{h}}=\mathrm{K}_{\mathrm{w}} /\left(\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}\right),\left[\mathrm{H}^{+}\right]=\left[\mathrm{K}_{\mathrm{a}} \mathrm{K}_{\mathrm{w}} / \mathrm{K}_{\mathrm{b}}\right]^{1 / 2} Note: Exact treatment of this case is difficult to solve. So use this assumption in general cases. Also, degree of anion or cation will be much higher in the case of a salt of weak acid and weak base. This is because each of them gets hydrolysed, producing \mathrm{H}^{+} \text {and } \mathrm{OH}^{-} ions. These ions combine to form water and the hydrolysis equilibrium is shifted in the forward direaction. Buffer Solutions are the solutions whose pH does not change significantly on adding a small quantity of strong base or on little dilution. These are typically made by mixing a weak acid (or base) with its conjugate base (or acid). e.g. \mathrm{CH}_{3} \mathrm{COOH} with \mathrm{CH}_{3} \mathrm{COONa}, \mathrm{NH}_{3}(\mathrm{aq}) \text { with } \mathrm{NH}_{4} \mathrm{Cl} \text { etc. } \text { If }\left.\mathrm{K}_{\mathrm{a}} \text { for acid (or } \mathrm{K}_{\mathrm{b}} \text { for base }\right) (or Kb for base) is not too high, we may write : Henderson’s Equation \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \{[\mathrm{salt}] /[\mathrm{acid}]\} for weak acid with its conjugate base. \text { or } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \{[\mathrm{salt}] /[\text { base }]\} for weak base with its conjugate acid. Important : For good buffer capacity, [salt] : [acid ratios should be as close to one as possible. In such a case, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}} (This also is the case at midpoint of titration) Buffer capacity = (no. of moles of acid (or base) added to 1L) / (change in pH) Indicators. Indicator is a substance which indicates the point of equivalence in a titration by undergoing a change in its colour. They are weak acids or weak bases. Theory of Indicators. The ionized and unionized forms of indicators have different colours. If 90 % or more of a particular form (ionised or unionised) is present, then its colour can be distinclty seen.In general, for an indicator which is weak acid, HIn l H+ + In–, the ratio of ionized to unionized form can be determined from This roughly gives the range of indicators. Ranges for some popular indicators are Table 1 : Indicators Equivalence point. The point at which exactly equivalent amounts of acid and base have been mixed. Acid Base Titration. For choosing a suitable indicator titration curves are of great help. In a titration curve, change in pH is plotted against the volume of alkali to a given acid. Four cases arise. (a) Strong acid vs strong base. The curve is almost vertical over the pH range 3.5-10. This abrupt change corresponds to equivalence point. Any indicator suitable. (b) Weak acid vs strong base. Final solution is basic 9 at equivalence point. Vertical region (not so sharp) lies in pH range 6.5-10. So, phenolphathlene is suitable. (c) Strong acid vs weak base. Final solution acidic. Vertical point in pH range 3.8-7.2. Methyl red or methyl orange suitable. (d) Weak acid vs weak base. No sharp change in pH. No suitable indicator. Note : at midpoint of titration, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}, thus by pH measurements, Ka for weak acids (or K_{b} for weak bases) can be determined. Polyprotic acids and bases. \text { Usually } \mathrm{K}_{2}, \mathrm{K}_{3} \text { etc. can be safely neglected and only } \mathrm{K}_{1} \text { plays a significant } Solubility product \left(\mathbf{K}_{\mathbf{s p}}\right). For sparingly soluble salts (eg. \mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}) an equilibrium which exists is Precipitation. Whenever the product of concentrations (raised to appropriate power) exceeds the solubility product, precipitation occurs. Common ion effects. Suppression of dissociation by adding an ion common with dissociation products. \text { e.g. } \mathrm{Ag}^{+} \text {or } \mathrm{C}_{2} \mathrm{O}_{4}^{2-} in the above example. Simultaneous solubility. While solving these problems, go as per general method i.e. (i) First apply condition of electroneutrality and (ii) Apply the equilibria conditions.
Limit – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
Q. Let $\mathrm{L}=\lim _{x \rightarrow 0} \frac{\mathrm{a}-\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}-\frac{\mathrm{x}^{2}}{4}}{\mathrm{x}^{4}}, \mathrm{a}>0 .$ If $\mathrm{L}$ is finite, then $:-$ (A) a = 2 (B) a = 1 (C) $\mathrm{L}=\frac{1}{64}$ (D) $\mathrm{L}=\frac{1}{32}$ [JEE 2009, 4]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,C) $\mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\right)^{\frac{1}{2}}-\frac{\mathrm{x}^{2}}{4} \quad \mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{2 \mathrm{a}^{2}}-\frac{1}{8} \frac{\mathrm{x}^{4}}{\mathrm{a}^{4}}\right)-\frac{\mathrm{x}^{2}}{4}$ $\mathrm{a}=2,\left(\mathrm{coefficient} \text { of } \mathrm{x}^{2}=0\right)$ $\therefore \mathrm{L}=\frac{1}{64}$

Q. If $\lim _{x \rightarrow 0}\left[1+x \ell n\left(1+b^{2}\right)\right]^{\frac{1}{x}}=2 b \sin ^{2} \theta, b>0$ and $\theta \in(-\pi, \pi],$ then the value of $\theta$ is- [JEE 2011, 3M, –1M]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. If $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4,$ then $-$ (A) a = 1, b = 4 (B) a = 1, b = –4] (C) a = 2, b = –3 (D) a = 2, b = 3 [JEE 2012, 3M, –1M]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B)

Q. Let $\alpha(\mathrm{a})$ and $\beta(\mathrm{a})$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a>-1 .$ Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are (A) $-\frac{5}{2}$ and 1 (B) $-\frac{1}{2}$ and $-1$ (C) $-\frac{7}{2}$ and 2 (D) $-\frac{9}{2}$ and 3 [JEE 2012, 3M, –1M]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B)

Q. The largest value of the non-negative integer a for which $\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$ is [JEE(Advanced)-2014, 3]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 0

Q. Let $\alpha, \beta \in \mathrm{R}$ be such that $\lim _{x \rightarrow 0} \frac{x^{2} \sin (\beta x)}{\alpha x-\sin x}=1 .$ Then $6(\alpha+\beta)$ equals [JEE(Advanced)-2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 7

Q. Let $\mathrm{f}(\mathrm{x})=\frac{1-\mathrm{x}(1+|1-\mathrm{x}|)}{|1-\mathrm{x}|} \cos \left(\frac{1}{1-\mathrm{x}}\right)$ for $\mathrm{x} \neq 1 .$ Then [JEE(Advanced)-2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,C)

Q. For any positive integer n, define $f_{\mathrm{n}}:(0, \infty) \rightarrow \square$ as $f_{\mathrm{n}}(\mathrm{x})=\sum_{\mathrm{j}=1}^{\mathrm{n}} \tan ^{-1}\left(\frac{1}{1+(\mathrm{x}+\mathrm{j})(\mathrm{x}+\mathrm{j}-1)}\right)$ for all $\mathrm{x} \in(0, \infty)$ (Here, the inverse trigonometric function $\tan ^{-1} \mathrm{x}$ assume values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$ ) Then, which of the following statement(s) is (are) TRUE? (A) $\sum_{j=1}^{5} \tan ^{2}\left(f_{j}(0)\right)=55$ (B) $\sum_{j=1}^{10}\left(1+f_{j}^{\prime}(0)\right) \sec ^{2}\left(f_{j}(0)\right)=10$ (C) For any fixed positive integer $n, \lim _{x \rightarrow \infty} \tan \left(f_{\mathrm{n}}(x)\right)=\frac{1}{n}$ (D) For any fixed positive integer $n, \operatorname{limsec}_{x \rightarrow \infty} \operatorname{ec}^{2}\left(f_{\mathrm{n}}(x)\right)=1$ [JEE(Advanced)-2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. For each positive integer $n,$ let $y_{n}=\frac{1}{n}(n+1)(n+2) \ldots(n+n)^{1 / n}$ For $x \in \square,$ let $[x]$ be the greatest integer less than or equal to $x$. If $\lim _{n \rightarrow \infty} y_{n}=L,$ then the value of $[L]$ is [JEE(Advanced)-2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 1

Matrices – JEE Main Previous Year Question with Solutions
JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
Q. Let A be the set of all 3  3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. (A) The number of matrices in A is – (A) 12            (B) 6          (C) 9                  (D) 3 (B) The number of matrices A in A for which the system of linear equations $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ has a unique solution, is – (A) less than 4 (B) at least 4 but less than 7 (C) at least 7 but less than 10 (D) at least 10 (C) The number of matrices A in A for which the system of linear equations $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ is inconsistent, is – (A) 0 (B) more than 2 (C) 2 (D) 1 [JEE 2009, 4+4+4]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. ( (A) A ,(b) B, (c) B )

Q. (A) The number of 3  3 matrices A whose entries are either 0 or 1 and for which the system $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ has exactly two distinct solutions, is (A) 0 (B) $2^{9}-1$ (C) 168 (D) 2 (B) Let $\mathrm{k}$ be a positive real number and let $\mathrm{A}=\left[\begin{array}{ccc}{2 \mathrm{k}-1} & {2 \sqrt{\mathrm{k}}} & {2 \sqrt{\mathrm{k}}} \\ {2 \sqrt{\mathrm{k}}} & {1} & {-2 \mathrm{k}} \\ {-2 \sqrt{\mathrm{k}}} & {2 \mathrm{k}} & {-1}\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}{0} & {2 \mathrm{k}-1} & {\sqrt{\mathrm{k}}} \\ {1-2 \mathrm{k}} & {0} & {2 \sqrt{\mathrm{k}}} \\ {-\sqrt{\mathrm{k}}} & {-2 \sqrt{\mathrm{k}}} & {0}\end{array}\right]$ If $\operatorname{det}(\operatorname{adj} \mathrm{A})+\operatorname{det}(\operatorname{adj} \mathrm{B})=10^{6},$ then $[\mathrm{k}]$ is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k]. (C) Let p be an odd prime number and Tp be the following set of 2  2 matrices: $\mathrm{T}_{\mathrm{p}}=\left\{\mathrm{A}=\left[\begin{array}{ll}{\mathrm{a}} & {\mathrm{b}} \\ {\mathrm{c}} & {\mathrm{a}}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1,2, \ldots \ldots, \mathrm{p}-1\}\right.$ (i) The number of $A$ in $T_{p}$ such that $A$ is either symmetric or skew symmetric or both, and det(A) divisible by p is – (A) $(\mathrm{p}-1)^{2}$ (B) 2 (p – 1) (C) $(p-1)^{2}+1$ (D) 2p –1 (ii) The number of $A$ in $T_{p}$ such that the trace of $A$ is not divisible by $p$ but det (A) is divisible by p is – [Note : The trace of a matrix is the sum of its diagonal entries.] (A) $(p-1)\left(p^{2}-p+1\right)$ (B) $\mathrm{p}^{3}-(\mathrm{p}-1)^{2}$ (C) $(p-1)^{2}$ (D) $(p-1)\left(p^{2}-2\right)$ (iii) The number of $A$ in $T_{p}$ such that $\operatorname{det}(A)$ is not divisible by $p$ is – (A) $2 p^{2}$ (B) $\mathrm{p}^{3}-5 \mathrm{p}$ (C) $\mathrm{p}^{3}-3 \mathrm{p}$ (D) $\mathrm{p}^{3}-\mathrm{p}^{2}$ [JEE 2010, 3+3+3+3+3]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. ($(a) A,(b) 4 ;(c)(i) D,(i i) C,(\text { iii }) D$) For the matrix $(6),(9),(10),(11)$ the system of linear equation is incosistent. (A) The given matrix system is a linear system in $\mathrm{x}, \mathrm{y}, \mathrm{z},$ hence it can have either a unique solution or no-solution or infinitely many solutions. It can never have exactly two distinct solutions.

Q. Let $\mathrm{M}$ and $\mathrm{N}$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $\mathrm{MN}=\mathrm{NM}$. If $\mathrm{P}^{\mathrm{T}}$ denotes the transpose of $\mathrm{P},$ then $\mathrm{M}^{2} \mathrm{N}^{2}\left(\mathrm{M}^{\mathrm{T}} \mathrm{N}\right)^{-1}\left(\mathrm{MN}^{-1}\right)^{\mathrm{T}}$ is equal to – (A) $\mathbf{M}^{2}$ (B) $-N^{2}$ (C) $-M^{2}$ (D) MN [JEE 2011, 4]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (Bonus) (Comment : Although 3 3 skew symmetric matrices can never be non-singular. Therefore the information given in question is wrong. Now if we consider only non singular skew symmetric matrices M & N, then the solution is-)

Q. Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $\left[\begin{array}{lll}{1} & {a} & {b} \\ {\omega} & {1} & {c} \\ {\omega^{2}} & {\omega} & {1}\end{array}\right],$ where each of a,b and $c$ is either $\omega$ or $\omega^{2} .$ Then the number of distinct matrices in the set $\mathrm{S}$ is- (A) 2              (B) 6              (C) 4              (D) 8 [JEE 2011, 3, (–1)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. Let M be 3  3 matrix satisfying $\mathbf{M}\left[\begin{array}{l}{0} \\ {1} \\ {0}\end{array}\right]=\left[\begin{array}{r}{-1} \\ {2} \\ {3}\end{array}\right], \mathbf{M}\left[\begin{array}{c}{1} \\ {-1} \\ {0}\end{array}\right]=\left[\begin{array}{r}{1} \\ {1} \\ {-1}\end{array}\right]$ and $\mathbf{M}\left[\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right]=\left[\begin{array}{c}{0} \\ {0} \\ {12}\end{array}\right]$ Then the sum of the diagonal entries of M is [JEE 2011, 4]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 9

Q. Let $\mathrm{P}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ be a $3 \times 3$ matrix and let $\mathrm{Q}=\left[\mathrm{b}_{\mathrm{ij}}\right],$ where $\mathrm{b}_{\mathrm{ij}}=2^{\mathrm{i}+\mathrm{j}} \mathrm{a}_{\mathrm{ij}}$ for $1 \leq \mathrm{i}, \mathrm{j} \leq 3$ If the determinant of $\mathrm{P}$ is $2,$ then the determinant of the matrix $\mathrm{Q}$ is – (A) $2^{10}$ (B) $2^{11}$ (C) $2^{12}$ $(D) 2^{13}$ [JEE 2011, 4]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. If $\mathrm{P}$ is a $3 \times 3$ matrix such that $\mathrm{P}^{\mathrm{T}}=2 \mathrm{P}+\mathrm{I},$ where $\mathrm{P}^{\mathrm{T}}$ is the transpose of $\mathrm{P}$ and $\mathrm{I}$ is the $3 \times 3 \times 3$ identity matrix, then there exists a column matrix $\mathrm{X}=\left[\begin{array}{l}{\mathrm{x}} \\ {\mathrm{y}} \\ {\mathrm{z}}\end{array}\right] \neq\left[\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right]$ such that (A) $\mathrm{PX}=\left[\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right]$ (B) PX = X (C) PX = 2X (D) PX = –X [JEE 2012, 3M, –1M]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D) $\mathrm{P}^{\mathrm{T}}=2 \mathrm{P}+\mathrm{I}$ $\Rightarrow \mathrm{P}=2 \mathrm{P}^{\mathrm{T}}+\mathrm{I}$ $\Rightarrow \mathrm{P}=2(2 \mathrm{P}+\mathrm{I})+\mathrm{I}$ $\Rightarrow P=4 P+3 I$ $\Rightarrow P=-I$ $\Rightarrow \mathrm{PX}=-\mathrm{X}$

Q. If the adjoint of a $3 \times 3$ matrix $\mathrm{P}$ is $\left[\begin{array}{ccc}{1} & {4} & {4} \\ {2} & {1} & {7} \\ {1} & {1} & {3}\end{array}\right],$ then the possible value(s) of the determinant of $\mathrm{P}$ is (are) (A) –2               (B) –1               (C) 1                     (D) 2 [JEE 2012, 4M]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,D)

Q. For 3  3 matrices M and N, which of the following statement(s) is (are) NOT correct ? (A) $\mathbf{N}^{\mathrm{T}} \mathbf{M}$ N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) MN – NM is skew symmetric for all symmetric matrices M and N (C) MN is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N [JEE-Advanced 2013, 4, (–1)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C,D) (A) $\mathrm{B}=\mathrm{N}^{\mathrm{T}} \mathrm{MN}$ $\mathrm{B}^{\mathrm{T}}=\left(\mathrm{N}^{\mathrm{T}} \mathrm{MN}\right)^{\mathrm{T}}=\mathrm{N}^{\mathrm{T}} \mathrm{M}^{\mathrm{T}} \mathrm{N}$ Now it depands on $\mathrm{M}$ if $\mathrm{M}=\mathrm{M}^{\mathrm{T}}$ So A is true $(\mathrm{B}) \mathrm{B}=(\mathrm{MN}-\mathrm{NM})$ $\mathrm{B}=(\mathrm{MN}-\mathrm{NM})^{\mathrm{T}}$ $=\mathbf{N}^{\mathrm{T}} \mathbf{M}^{\mathrm{T}}-\mathbf{M}^{\mathrm{N}} \mathbf{N}^{\mathrm{T}}$ $=\mathrm{NM}-\mathrm{MN}=-(\mathrm{B})$ Skew symmetric B is ture $(\mathrm{C}) \mathrm{B}=\mathrm{MN}$ $\mathrm{B}^{\mathrm{T}}=(\mathrm{MN})^{\mathrm{T}}$ $\mathrm{B}^{\mathrm{T}}=\mathrm{N}^{\mathrm{T}} \mathrm{M}^{\mathrm{T}}$ $\mathrm{B}^{\mathrm{T}}=\mathrm{NM} \neq \mathrm{B}$ so wrong statement (D) Obviousely wrong because adj $(\mathrm{BA})=\operatorname{adj}(\mathrm{A}) \cdot \operatorname{adJ}(\mathrm{B})$

Q. Let M be a 2  2 symmetric matrix with integer entries. Then M is invertible if (A) the first column of M is the transpose of the second row of M (B) the second row of M is the transpose of the first column of M (C) M is a diagonal matrix with nonzero entries in the main diagonal (D) the product of entries in the main diagonal of M is not the square of an integer [JEE(Advanced)-2014, 3]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C,D)

Q. Let $\mathrm{M}$ and $\mathrm{N}$ be two $3 \times 3$ matrices such that $\mathrm{MN}=\mathrm{NM}$. Further, if $\mathrm{M} \neq \mathrm{N}^{2}$ and $\mathrm{M}^{2}=$ $\mathrm{N}^{4},$ then (A) determinant of $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right)$ is 0 (B) there is a $3 \times 3$ non-zero matrix $\mathrm{U}$ such that $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \mathrm{U}$ is zero Matrix (C) determinant of $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \geq 1$ (D) for a $3 \times 3$ matrix $\mathrm{U},$ if $\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right)$ U equals the zero matrix then $\mathrm{U}$ is the zero matrix [JEE(Advanced)-2014, 3]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,B) $\mathrm{M}^{2}=\mathrm{N}^{4}$ $\mathrm{M}^{2}=\mathrm{N}^{4}=0(\therefore \mathrm{MN}=\mathrm{NM})$ $\left(\mathrm{M}+\mathrm{N}^{2}\right)\left(\mathrm{M}-\mathrm{N}^{2}\right)=0$ So $\left(\mathrm{M}+\mathrm{N}^{2}\right)=0$ $\mathrm{Now} \mathrm{M} \cdot\left(\mathrm{M}+\mathrm{N}^{2}\right)=0$ $\mathrm{M}^{2}+\mathrm{MN}^{2}=0$ $\left|\mathrm{M}^{2}+\mathrm{MN}^{2}\right|=0$ Option A is right So we know $A \cdot B=0$ when $\mathrm{B} \neq 0$ $\Rightarrow(\mathrm{A})=0$ So in $\mathrm{U}\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \mathrm{U}=0$ Because $\mathrm{U} \neq 0$ But $\left|\mathrm{M}^{2}+\mathrm{MN}^{2}\right|=0$ So option $\mathrm{B}$ is also right

Q. Let X and Y be two arbitrary, 3  3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric ? (A) $\mathrm{Y}^{3} \mathrm{Z}^{4}-\mathrm{Z}^{4} \mathrm{Y}^{3}$ (B) $\mathrm{X}^{44}+\mathrm{Y}^{44}$ (C) $\mathrm{X}^{4} \mathrm{Z}^{3}-\mathrm{Z}^{3} \mathrm{X}^{4}$ (D) $\mathrm{X}^{23}+\mathrm{Y}^{23}$ [JEE(Advanced)-2015, 4M, –2M]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C,D) $\mathrm{x}^{\mathrm{T}}=-\mathrm{x}, \mathrm{y}^{\mathrm{T}}=-\mathrm{y}, \mathrm{z}^{\mathrm{T}}=\mathrm{z}$ (A) Let $P=y^{3} z^{4}-z^{4} y^{3}$ $\mathrm{P}^{\mathrm{T}}=\left(\mathrm{y}^{3} \mathrm{z}^{4}\right)^{\mathrm{T}}-\left(\mathrm{z}^{4} \mathrm{y}^{3}\right)^{\mathrm{T}}$ $=-z^{4} y^{3}+y^{3} z^{4}=P \Rightarrow$ symmetric (B) $\quad$ Let $\mathrm{P}=\mathrm{x}^{44}+\mathrm{y}^{44}$ $\mathrm{P}^{\mathrm{T}}=\left(\mathrm{X}^{44}\right)^{\mathrm{T}}+\left(\mathrm{y}^{44}\right)^{\mathrm{T}}=\mathrm{P} \Rightarrow$ symmetric (C) Let $P=x^{4} z^{3}-z^{3} x^{4}$ $\mathrm{P}^{\mathrm{T}}=\left(\mathrm{z}^{3}\right)^{\mathrm{T}}\left(\mathrm{x}^{4}\right)^{\mathrm{T}}-\left(\mathrm{x}^{4}\right)^{\mathrm{T}}\left(\mathrm{z}^{3}\right)^{\mathrm{T}}$ $=\mathrm{z}^{3} \mathrm{x}^{4}-\mathrm{x}^{4} \mathrm{z}^{3}=-\mathrm{P} \Rightarrow$ skew symmetric (D) Let $P=x^{23}+y^{23}$ $\mathrm{P}^{\mathrm{T}}=-\mathrm{x}^{23}-\mathrm{y}^{23}=-\mathrm{P} \Rightarrow$ skew symmetric

Q. Let $P=\left[\begin{array}{ccc}{3} & {-1} & {-2} \\ {2} & {0} & {\alpha} \\ {3} & {-5} & {0}\end{array}\right],$ where $\alpha \in R,$ Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$ where $\mathrm{k} \in \mathrm{R}, \mathrm{k} \neq 0$ and $\mathrm{I}$ is the identity matrix of order $3 .$ If $\mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$ and $\operatorname{det}(\mathrm{Q})=\frac{\mathrm{k}^{2}}{2}$ then- (A)  = 0, k = 8 (B) $4 \alpha-k+8=0$ (C) $\operatorname{det}(\operatorname{Padj}(\mathrm{Q}))=2^{9}$ (D) det(Qadj(P)) = $2^{13}$ [JEE(Advanced)-2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B,C) $\mathrm{PQ}=\mathrm{kI}$ $|\mathrm{P}| \cdot|\mathrm{Q}|=\mathrm{k}^{3} \Rightarrow|\mathrm{P}|=2 \mathrm{k} \neq 0 \Rightarrow \mathrm{P}$ is an invertible matrix $\because \mathrm{PQ}=\mathrm{kI}$ $\therefore \mathrm{Q}=\mathrm{k} \mathrm{P}^{-1} \mathrm{I}$ $\therefore \mathrm{Q}=\frac{\mathrm{adj.P}}{2}$ $\because \mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$ $\therefore \frac{-(3 \alpha+4)}{2}=-\frac{k}{8} \Rightarrow k=4$ $\therefore|\mathrm{P}|=2 \mathrm{k} \Rightarrow \mathrm{k}=10+6 \alpha \ldots(\mathrm{i})$ Put value of $k$ in (i).. we get $\alpha=-1$ $\therefore 4 \alpha-k+8=0$ $\& \operatorname{det}(\mathrm{P}(\mathrm{adj} . \mathrm{Q}))=|\mathrm{P}||\operatorname{adj} . \mathrm{Q}|=2 \mathrm{k} \cdot\left(\frac{\mathrm{k}^{2}}{2}\right)^{2}=\frac{\mathrm{k}^{5}}{2}=2^{9}$

Q. Let $P=\left[\begin{array}{lll}{1} & {0} & {0} \\ {4} & {1} & {0} \\ {16} & {4} & {1}\end{array}\right]$ and $I$ be the identity matrix of order $3 .$ If $Q=\left[q_{i j}\right]$ is a matrix such that $\mathrm{P}^{50}-\mathrm{Q}=\mathrm{I},$ then $\frac{\mathrm{q}_{31}+\mathrm{q}_{32}}{\mathrm{q}_{21}}$ equals (A) 52             (B) 103               (C) 201                  (D) 205 [JEE(Advanced)-2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. Which of the following is(are) NOT the square of a 3  3 matrix with real entries ? [JEE(Advanced)-2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,B)

Q. How many $3 \times 3$ matrices $\mathrm{M}$ with entries from $\{0,1,2\}$ are there, for which the sum of the diagonal entries of $\mathrm{M}^{\mathrm{T}} \mathrm{M}$ is $5 ?$ (A) 198              (B) 126                (C) 135                 (D) 16 [JEE(Advanced)-2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. Let S be the of all column matrices $\left[\begin{array}{l}{b_{1}} \\ {b_{2}} \\ {b_{3}}\end{array}\right]$ such that $b_{1}, b_{2}, b_{3} \in \square$ and the system of equations (in real variables) $-x+2 y+5 z=b_{1}$ $2 x-4 y+3 z=b_{2}$ $x-2 y+2 z=b_{3}$ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution of each $\left[\begin{array}{l}{b_{1}} \\ {b_{2}} \\ {b_{3}}\end{array}\right] \in S ?$ (A) $\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=\mathrm{b}_{1}, 4 \mathrm{y}+5 \mathrm{z}=\mathrm{b}_{2}$ and $\mathrm{x}+2 \mathrm{y}+6 \mathrm{z}=\mathrm{b}_{3}$ (B) $x+y+3 z=b_{1}, 5 x+2 y+6 z=b_{2}$ and $-2 x-y-3 z=b_{3}$ (C) $-\mathrm{x}+2 \mathrm{y}-5 \mathrm{z}=\mathrm{b}_{1}, 2 \mathrm{x}-4 \mathrm{y}+10 \mathrm{z}=\mathrm{b}_{2}$ and $\mathrm{x}-2 \mathrm{y}+5 \mathrm{z}=\mathrm{b}_{3}$ (D) $x+2 y+5 z=b_{1}, 2 x+3 z=b_{2}$ and $x+4 y-5 z=b_{3}$ [JEE(Advanced)-2018, 4(–2)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,C,D)

Chemistry Topic-wise JEE Main Previous Year Question with Solutions
Chemistry Topic-wise JEE Main Previous Year Question with Solutions
Practicing JEE Main previous Year Question Papers will help you in many ways in your Exam preparation. It will help you to boost your confidence level. Students can check where they are lagging through practicing these previous year question papers. Here you will get the JEE-Mains previous year question papers from year  2009 to 2019 along with solutions. These JEE Main previous Year Question for Chemistry plays an important role in IIT-JEE preparation. We are providing IIT-JEE Mains Previous Year Question Papers with detailed Solution. We have tried our best to provide you last 10 years question with solutions. This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning. The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam. While preparing for the IIT-JEE exam, aspirants should be aware about the question paper structure and the format of questions to be asked in this exam. This will help to make an effective preparation strategy for the exam. JEE Main Previous Year Question Papers are the best resources to prepare for exam. This will help an individual to understand the exam pattern of JEE. This will also enhance your level of preparation. JEE aspirants must solve multiple sample papers and analyse their performances in order to recognize their strengths and weaknesses.   Here are the Chemistry Topic-wise Previous year question for JEE Main:   In the last few months of JEE Main 2020, candidates should go through the sample papers and mock tests to test their preparation level and get hands-on practice in solving questions. One of the most effective ways to prepare for the JEE Advanced, after the IIT-JEE Previous Year Papers, is to take a JEE mock test. At eSaral we provide almost all types of test series which includes topic wise tests, full length mock tests, etc.

Click Here to Download Math Topic-wise JEE Main Previous Year Question with Solutions

Click Here to Download Physics Topic-wise JEE Main Previous Year Question with Solutions

Click Here to Download JEE Advanced Previous Year Questions with solutions. 

We believe that your preparation for the JEE Mains Exam will be simple as long as an organized and consistent plan is followed. Solving the JEE Main Previous Year Papers will absolutely help you in your JEE Mains Exam preparation. As a result, no stone has been left unturned by eSaral to ensure that these IIT-JEE mock tests are similar to the actual exam.   For free study material and video tutorials, Download eSaral app
Chapter 1- 13 Mathematics Notes for Class 12 and IIT JEE
eSaral brings you detailed maths notes for class 12. eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE preparations. We hope Maths notes for class 12 will help you understand the important topics and remember the key points for the exam point of view. Below are the direct links for detailed chapter-wise12th Maths study material.   Relation and Functions Inverse Trigonometric Function Matrices Determinants Continuity and Differentiability Application of Derivatives Integrals Application of Integrals Differential Equation Vector Algebra Three Dimension Geometry Linear Programming Probability   Mathematics Notes for Class 11    Install eSaral App to learn from video tutorials by Top IITian faculties of Kota. Our eSaral application contains around 5,00,000 questions to practice for IIT JEE, NEET and Board Exams.
d & f – block – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years AIEEE/JEE Main Question
Q. In context with the transition elements, which of the following statements is incorrect? (1) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding. (2) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases. (3) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes. (4) In the highest oxidation states, the transition metal show basic character and form cationic complexes. [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect ? (1) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. (2) Ferrous oxide is more basic in nature than the ferric oxide. (3) Ferrous compounds are relatively more ionic than the corresponding ferric compounds. (4) Ferrous compounds are less volatile than the corresponding ferric compounds. [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Consider the following reaction : $\mathrm{xMnO}_{4}^{-}+\mathrm{yC}_{2} \mathrm{O}_{4}^{2-}+\mathrm{zH}^{+} \rightarrow \mathrm{xMn}^{2+}+2 \mathrm{yCO}_{2}+\frac{\mathrm{Z}}{2} \mathrm{H}_{2} \mathrm{O}$ The values of x, y and z in the reaction are respectively :- (1) 5,2 and 16 (2) 2,5 and 8 (3) 2, 5 and 16 (4) 5,2 and 8 [JEE MAIN-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Which of the following arrangements does not represent the correct order of the property stated against it ? (1) $\mathrm{V}^{2+}<\mathrm{Cr}^{2+}<\mathrm{Mn}^{2}<\mathrm{Fe}^{2+}$: paramagnetic behaviour (2) $\mathrm{Ni}^{2+}<\mathrm{Co}^{2+}<\mathrm{Fe}^{2+}<\mathrm{Mn}^{2+}$: ionic size (3)$\mathrm{Co}^{3+}<\mathrm{Fe}^{3+}<\mathrm{Cr}^{3+}<\mathrm{Sc}^{3+}$: stability in aqueous solution (4) $\mathrm{Sc}<\mathrm{Ti}<\mathrm{Cr}<$Mn : number of oxidation states [JEE MAIN-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown – red vapours of: (1) $\mathrm{CrO}_{3}$ (2) $\mathrm{Cr}_{2} \mathrm{O}_{3}$ (3) $\mathrm{CrCl}_{3}$ (4) $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$ [JEE MAIN-2013, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Explanation. $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+6 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{NaCl} \longrightarrow 2 \mathrm{KHSO}_{4}+4 \mathrm{NaHSO}_{4}+2 \mathrm{CrO}_{2} \mathrm{Cl}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

Q. The element with which of the following outer electron configuration may exhibit the largest number of oxidation states in its compounds : (1) $3 \mathrm{d}^{7} 4 \mathrm{s}^{2}$ (2) $3 \mathrm{d}^{8} 4 \mathrm{s}^{2}$ ( 3) (3) $3 \mathrm{d}^{5} 4 \mathrm{s}^{2}$ (4) (4) $3 \mathrm{d}^{6} 4 \mathrm{S}^{2}$ [JEE MAIN-2013, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Explanation $\mathrm{Mn} \longrightarrow \mathrm{Mn}^{+7}$

Q. When a small amount of $\mathrm{KMnO}_{4}$ is added to concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$a green oily compound is obtained which is highly explosive in nature. Compound may be: (1) $\mathrm{Mn}_{2} \mathrm{O}_{3}$ (2) MnSO $_{4}$ (3) $\mathrm{Mn}_{2} \mathrm{O}_{7}$ ( 4) $\mathrm{MnO}_{2}$ [JEE MAIN-2013, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Explanation

Q. Which series of reactions correctly represents chemical relations related to iron and its compound? [JEE MAIN-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. The equation which is balanced and represents the correct product (s) is: [JEE MAIN-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Which of the following is not formed when $\mathrm{H}_{2} \mathrm{S}$ reacts with acidic $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$solution ? ( 1) $\mathrm{K}_{2} \mathrm{SO}_{4}$ (2) $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ (3) $\mathrm{S}$ (4) $\mathrm{CrSO}_{4}$ [JEE MAIN-2014, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Copper becomes green when exposed to moist air for a long period. This is due to :- (1) the formation of a layer of cupric oxide on the surface of copper. (2) the formation of basic copper sulphate layer on the surface of the metal (3) the formation of a layer of cupric hydroxide on the surface of copper. (4) the formation of a layer of basic carbonate of copper on the surface of copper. [JEE MAIN-2014, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Which one of the following exhibits the largest number of oxidation states ? (1) Mn(25) (2) V(23) (3) Cr (24) (4) Ti (22) [JEE MAIN-2014, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. How many electrons are involved in the following redox reaction? $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{CO}_{2}$ (Unbalanced) (1) 3 (2) 4 (3) 5 (4) 6 [JEE MAINS-2014,Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Amongst the following, identify the species with an atom in +6 oxidation state: (1) $\left[\mathrm{MnO}_{4}\right]^{-}$ (2) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$ (3) $\mathrm{Cr}_{2} \mathrm{O}_{3}$ (4) $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$ [JEE MAIN-2014, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Match the catalysts to the correct processes :- (1) A-ii, B-iii, C-iv, D-i (2) A-iii, B-i, C-ii, D-iv (3) A-iii, B-ii, C-iv, D-i (4) A-ii, B-i, C-iv, D-iii [JEE MAIN-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Which of the following statements is false :- (1) has a Cr – O – Cr bond (2) is tetrahedral in shape (3) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a primary standard in volumetry (4) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is less soluble than $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ [JEE MAIN-2015, Online]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

src=”https://www.esaral.com/wp-content/uploads/2019/08/CB-3-31.jpg” alt=”” width=”1275″ height=”1650″ />        
Hydrogen – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator   Previous Years AIEEE/JEE Mains Questions
Q. Very pure hydrogen (99.9%) can be made by which of the following processes? (1) Reaction of salt like hydrides with water (2) Reaction of methane with steam (3) Mixing natural hydrocarbons of high molecular weight (4) Electrolysis of water [AIEEE 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Very pure hydrogen (99.9%) can be made by electrolysis of water.

Q. In which of the following reaction $\mathrm{H}_{2} \mathrm{O}_{2}$ acts as a reducing agent ? (1) (1), (3) (2) (2), (4) (3) (1), (2) (4) (3), (4) [JEE(Main) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) When $\mathrm{H}_{2} \mathrm{O}_{2}$ act as reducing agent then it evolve.

Q. Which of the following statements about $\mathrm{Na}_{2} \mathrm{O}_{2}$ is not correct ? (1) $\mathrm{Na}_{2} \mathrm{O}_{2}$ oxidises $\mathrm{Cr}^{3+}$ to $\mathrm{CrO}_{4}^{2-}$ in acid medium (2) It is diamagnetic in nature (3) It is the super oxide of sodium (4) It is a derivative of $\mathrm{H}_{2} \mathrm{O}_{2}$ [JEE(Main) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{Na}_{2} \mathrm{O}_{2}$ is peroxide of sodium

Q. Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases $\mathrm{H}_{2} \mathrm{O}_{2}$ acts as a reducing agent in acid medium ? :- (1) $\mathrm{MnO}_{4}^{-}$ (2) $\mathrm{SO}_{3}^{2-}$ (3) KI (4) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ [JEE(Main)Online-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Permanent hardness in water cannot be cured by: (1) Treatment with washing soda (2) Calgon’s method (3) Boiling (4) Ion exchange method [JEE(Main)Online-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Permanent hardness in water cannot cured by boiling of water

Q. From the following statements regarding H2O2, choose the incorrect statement : (1) It has to be stored in plastic or wax lined glass bottles in dark (2) It has to be kept away from dust (3) It can act only as an oxidizing agent (4) It decomposes on exposure to light [JEE(Main)Online-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{H}_{2} \mathrm{O}_{2}$ can act as oxidizing as well as reducing agent depend on condition.

Q. Hydrogen peroxide oxidises $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ to $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ in acidic medium but reduces $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ to $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ in alkaline medium. The other products formed are, respectively : [JEE(Main)Online-2018] (1) $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ and $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{OH}^{-}\right)$ (2) $\mathrm{H}_{2} \mathrm{O}$ and $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ (3) $\mathrm{H}_{2} \mathrm{O}$ and $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{OH}^{-}\right)$ (4) $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ and $\mathrm{H}_{2} \mathrm{O}$

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Chemical Bonding – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator   Previous Years AIEEE/JEE Mains Questions
Q. The bond dissociation energy of B–F in $\mathrm{BF}_{3}$ is 646 kJ $\mathrm{mol}^{-1}$whereas that of C–F in $\mathrm{CF}_{4}$ is 515 kJ mol–1. The correct reason for higher B–F bond dissociation energy as compared to that of C–F is :- (1) Significant $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between B and F in $\mathrm{BF}_{3}$whereas there is not possibility of such interaction between C and F in $\mathrm{CF}_{4}$. (2) Lower degree of p – p interaction between B and F in $\mathrm{BF}_{3}$ than that between C and F in $\mathrm{CF}_{4}$ (3)Smaller size of B-atom as compared to that of C-atom (4) Stronger  bond between B and F in $\mathrm{BF}_{3}$ as compared to that between C and F in $\mathrm{CF}_{4}$ [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Due to back bonding in BF3 molecule all B–F bond having partial double bond character.

Q. Using MO theory predict which of the following species has the shortest bond length ? (1) $\mathrm{o}_{2}^{-}$ (2) $\mathrm{O}_{2}^{2-}$ (3) $\mathrm{O}_{2}^{2+}$ (4) $\mathrm{O}_{2}^{+}$ [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The hybridisation of orbitals of N atom in $\mathrm{NO}_{3}^{-}, \mathrm{NO}_{2}^{+}$ and $\mathrm{NH}_{4}^{+}$ are respectively:- [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The structure of $\mathrm{IF}_{7}$ is :- (1) octahedral (2) pentagonal bipyramid (3) square pyramid (4) trigonal bipyramid [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Among the following the maximum covalent character is shown by the compound :- (1) $\mathrm{AlCl}_{3}$ (2) $\mathrm{MgCl}_{2}$ (3) FeCl $_{2}$ (4) $\mathrm{SnCl}_{2}$ [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{Al}^{+3}$ having highest polarizing power than other : having greater covalent character

Q. Which of the following has maximum number of lone pairs associated with Xe (1) $\mathrm{XeO}_{3}$ (2) $\mathrm{XeF}_{4}$ (3) XeF $_{6}$ (4) $\mathrm{XeF}_{2}$ [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The number of types of bonds between two carbon atoms in calcium carbide is :- (1) One sigma, two pi (2) One sigma, one pi (3) Two sigma, one pi (4) Two sigma, two pi [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{Ca}^{+2}(\mathrm{C} \equiv \mathrm{C})^{2-}$

Q. The molecule having smallest bond angle is :- (1) $\mathrm{PCl}_{3}$ (2) $\mathrm{NCl}_{3}$ (4) $\mathrm{SbCl}_{3}$ [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Bond angle order $\mathrm{NCl}_{3}>\mathrm{PCl}_{3}>\mathrm{AsCl}_{3}>\mathrm{SbCl}_{3}$

Q. In which of the following pairs the two species are not isostructural ? (1) $\mathrm{AIF}_{6}^{3-}$ and $\mathrm{SF}_{6}$ (3) $\mathrm{CO}_{3}^{2-}$ and $\mathrm{NO}_{3}^{-}$ (3) $\mathrm{PCl}_{4}^{+}$ and $\mathrm{SiCl}_{4}$ (4) $\mathrm{PF}_{5}$ and $\mathrm{BrF}_{5}$ [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The number of S–S bonds in $\mathrm{SO}_{3}, \mathrm{S}_{2} \mathrm{O}_{3}^{2-}, \mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{8}^{2-}$ respectively are :- (1) 1, 0, 1, 0 (2) 0, 1, 1, 0 (3) 1, 0, 0, 1 (4) 0, 1, 0, 1 [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Dipole moment is shown by :- (1) trans-2, 3-dichloro- 2-butene (2) 1, 2-dichlorobenzene (3) 1, 4-dichlorobenzene (4) trans-1, 2-dinitroethene [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Among the following species which two have trigonal bipyramidal shape ? [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (Bonus)

Q. Among the following, the species having the smallest bond is :- (1) NO (2) $\mathrm{NO}^{+}$ (3) $\mathrm{O}_{2}$ (4) $\mathrm{NO}^{-}$ [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Based on lattice energy and other considerations, which one of the following alkali metal chloride is expected to have the highest melting point ? (1) RbCl (2) LiCl (3) KCl (4) NaCl [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Highest melting is of NaCl

Q. Which of the following has the square planar structure :- (1) $\mathrm{NH}_{4}^{+}$ (2) $\mathrm{CCl}_{4}$ (3) $\mathrm{XeF}_{4}$ (4) $\mathrm{BF}_{4}^{-}$ [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Hybridisation $\mathrm{Sp}^{3} \mathrm{d}^{2}$ Shape – square planar

Q. The compound of Xenon with zero dipole moment is :- (1) $\mathrm{XeO}_{3}$ (2) $\mathrm{XeO}_{2}$ (3) $\mathrm{XeF}_{4}$ (4) $\mathrm{XeOF}_{4}$ [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Among the following the molecule with the lowest dipole moment is :- (1) $\mathrm{CHCl}_{3}$ (2) $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ (3) $\mathrm{CCl}_{4}$ (4) $\mathrm{CH}_{3} \mathrm{Cl}$ [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{CCl}_{4}<\mathrm{CHCl}_{3}<\mathrm{CH}_{2} \mathrm{Cl}_{2}<\mathrm{CH}_{3} \mathrm{Cl}(\text { Dipolar moment })$

Q. The formation of molecular complex $\mathrm{BF}_{3}-\mathrm{NH}_{3}$ results in a change in hybridisation of boron (1) from $\operatorname{sp}^{3}$ to $\mathrm{sp}^{3} \mathrm{d}$ (2) from $\operatorname{sp}^{2}$ to $\mathrm{dsp}^{2}$ (3) from $\operatorname{sp}^{3}$ to $\mathrm{sp}^{2}$ (4) from $\operatorname{sp}^{2}$ to $\mathrm{sp}^{3}$ [JEE-MAINS(Online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Among the following chloro-compound having the lowest dipole moment is :- (1) $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ (2) $\mathrm{CH}_{3} \mathrm{Cl}$ (3) (4)

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ has lowest dipole moment

Q. Which one of the following molecules is expected to exhibit diamagnetic behaviour ? (1) $\mathrm{C}_{2}$ (2) $\mathrm{N}_{2}$ (3) $\mathrm{O}_{2}$ (4) $\mathrm{S}_{2}$ [AIEEE-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\mathrm{N}_{2}$ is diamagnetic

Q. Which of the following is the wrong statement? (1) ONCl and $\mathrm{ONO}^{-}$ are not isoelectronic (2) $\mathrm{O}_{3}$ molecule is bent (3) Ozone is violet-black in solid state (4) Ozone is diamagnetic gas [JEE-maIN 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Number of electron in ONCl an $\mathrm{ONO}^{-}$ is 32 & 24 respectively.

Q. In which of the following pairs of molecules/ions, both the species are not likely to exist ? (1) $\mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{2-}$ (2) $\mathrm{H}_{2}^{-}, \mathrm{He}_{2}^{2-}$ (3) $\mathrm{H}_{2}^{2+}, \mathrm{He}_{2}$ (4) $\mathrm{H}_{2}^{-}, \mathrm{H} \mathrm{e}_{2}^{2+}$ [JEE-main 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Bond order of $\mathrm{H}_{2}^{2+} \& \mathrm{He}_{2}$ is zero i.e. these molecule do not exist.

Q. Stability of the species $\mathrm{Li}_{2}, \quad \mathrm{Li}_{2}^{-}$ and $\mathrm{Li}_{2}^{+}$ increases in the order of :- (1) $\mathrm{Li}_{2}<\mathrm{Li}_{2}^{+}<\mathrm{Li}_{2}^{-}$ (2) $\mathrm{Li}_{2}^{-}<\mathrm{Li}_{2}^{+}<\mathrm{Li}_{2}$ (3) $\mathrm{Li}_{2}<\mathrm{Li}_{2}<\mathrm{Li}_{2}^{+}$ (4) $\mathrm{Li}_{2}^{-}<\mathrm{Li}_{2}<\mathrm{Li}_{2}^{+}$ [JEE-main 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Trigonal bipyramidal geometry is shown by: (1) $\mathrm{XeO}_{3} \mathrm{F}_{2}$ (2) $\mathrm{XeOF}_{2}$ (3) $\left[\mathrm{XeF}_{8}\right]^{2-}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic ? (1) $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$ (2) $\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}$ (3) $\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{+}$ ( 4) $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}^{+}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Which one of the following molecules is polar? (1) $\mathrm{CF}_{4}$ (2) $\mathrm{SbF}_{5}$ (3) $\mathrm{IF}_{5}$ (4) $\mathrm{XeF}_{4}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Oxidation state of sulphur in anions $\mathrm{SO}_{3,}^{2-} \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ increases in the orders : (1) $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}$ (2) $\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ (3) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ (4) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{SO}_{3}^{2-}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Bond order normally gives idea of stability of a molecular species. All the molecules viz. $\mathrm{H}_{2}, \mathrm{Li}_{2}$ and $\mathrm{B}_{2}$ have the same bond order yet they are not equally stable. Their stability order is: (1) $\mathrm{Li}_{2}>\mathrm{H}_{2}>\mathrm{B}_{2}$ (2) $\mathrm{H}_{2}>\mathrm{B}_{2}>\mathrm{Li}_{2}$ (3) $\mathrm{B}_{2}>\mathrm{H}_{2}>\mathrm{Li}_{2}$ (4) $\mathrm{Li}_{2}>\mathrm{B}_{2}>\mathrm{H}_{2}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. The solubility order for alkali metal fluoride in water is : (1) $\mathrm{LiF}<\mathrm{NaF}<\mathrm{KF}<\mathrm{RbF}$ (2) $\mathrm{LiF}>\mathrm{NaF}>\mathrm{KF}>\mathrm{RbF}$ (3) $\mathrm{RbF}<\mathrm{KF}<\mathrm{NaF}<\mathrm{LiF}$ (4) $\mathrm{LiF}<\mathrm{RbF}<\mathrm{KF}<\mathrm{NaF}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Solubility order LiF < NaF < KF < RbF

Q. $\mathrm{XeO}_{4}$ molecule is tetrahedral having : (1) Two p $\pi$ -d $\pi$ bonds (2) Four p $\pi$ – d $\pi$ bonds (3) One $\mathrm{p} \pi-\mathrm{d} \pi$ bond (4) Three p $\pi$ -d $\pi$ bonds [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Bond distance in HF is 9.17 × $10^{-11}$ m. Dipole moment of HF is 6.104 × 10–30 Cm. The percent ionic character in HF will be : (electron charge = 1.60 × $10^{-19}$ C) (1) 61.0% (2) 38.0% (3) 35.5% (4) 41.5% [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The shape of $\mathrm{IF}_{6}^{-}$ is : (1) Trigonally distorted octahedron (2) Pyramidal (3) Octahedral (4) Square antiprism [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Which has trigonal bipyramidal shape ? (1) $\mathrm{XeOF}_{4}$ (2) $\mathrm{XeO}_{3}$ (3) $\mathrm{XeO}_{3} \mathrm{F}_{2}$ (4) $\mathrm{XeOF}_{2}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The catenation tendency of C, Si and Ge is in the order Ge < Si < C. The bond energies (in kJ $\mathrm{mol}^{-1}$) of C — C, Si —Si and Ge—Ge bonds are respectively : (1) 348, 260, 297 (2) 348, 297, 260 (3) 297, 348, 260 (4) 260, 297, 348 [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Catenation/Bond energy order C – C > Si – Si > Ge – Ge

Q. In which of the following sets, all the given species are isostructural ? (1) $\mathrm{BF}_{3}, \mathrm{NF}_{3}, \mathrm{PF}_{3}, \mathrm{AlF}_{3}$ (2) $\mathrm{PCl}_{3}, \mathrm{AlCl}_{3}, \mathrm{BCl}_{3}, \mathrm{SbCl}_{3}$ (3) $\mathrm{BF}_{4}^{-}, \mathrm{CCl}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{PCl}_{4}^{+}$ (4) $\mathrm{CO}_{2}, \mathrm{NO}_{2}, \mathrm{ClO}_{2}, \mathrm{SiO}_{2}$ [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{BF}_{4}^{-}, \mathrm{CC}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{PCl}_{4}^{+}$ are terahedral

Q. The internuclear distances in O —O bonds for O, $\mathrm{O}_{2}, \mathrm{O}^{-}$2 and O respectively are : (1) 1.49 Å, 1.21 Å, 1.12 Å, 1.30 Å (2) 1.30 Å, 1.49 Å, 1.12 Å, 1.21 (3) 1.12 Å, 1.21 Å, 1.30 Å, 1.49 Å (4) 1.21 Å, 1.12 Å, 1.49 Å, 1.30 Å [JEE-MAINS(Online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Which one of the following properties is not shown by NO ? (1) It combines with oxygen to form nitrogen dioxide (2) It’s bond order is 2.5 (3) It is diamagnetic in gaseous state (4) It is a neutral oxide [JEE-main 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For which of the following molecule significant $\mu \neq 0$ [JEE-main 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. The number and type of bonds in $\mathrm{C}_{2}^{2-} \mathrm{ion}$ in $\mathrm{CaC}_{2}$ are: (1) Two $\sigma$ bonds and one $\pi-$ bond (2) Two $\sigma$ bonds and two $\pi-$ bonds (3) One $\sigma$ bond and two $\pi-$ bonds (4) One $\sigma$ bond and one $\pi-$ bond [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For the compounds $\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{Br}, \mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{CH}_{3} \mathrm{F}$ (1) $\mathrm{CH}_{3} \mathrm{F}<\mathrm{CH}_{3} \mathrm{Br}<\mathrm{CH}_{3} \mathrm{Cl}<\mathrm{CH}_{3} \mathrm{I}$ (2) $\mathrm{CH}_{3} \mathrm{F}<\mathrm{CH}_{3} \mathrm{Cl}<\mathrm{CH}_{3} \mathrm{Br}<\mathrm{CH}_{3} \mathrm{I}$ (3) $\mathrm{CH}_{3} \mathrm{Cl}<\mathrm{CH}_{3} \mathrm{Br}<\mathrm{CH}_{3} \mathrm{F}<\mathrm{CH}_{3} \mathrm{I}$, the correct order of increasing C-halogen bond length is : (4) $\mathrm{CH}_{3} \mathrm{F}<\mathrm{CH}_{3} \mathrm{I}<\mathrm{CH}_{3} \mathrm{Br}<\mathrm{CH}_{3} \mathrm{Cl}$ [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Which of the following has unpaired electron(s) ? (1) $\mathrm{O}_{2}^{-}$ (2) $\mathrm{N}_{2}^{2+}$ (3) $\mathrm{O}_{2}^{2-}$ (4) $\mathrm{N}_{2}$ [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{O}_{2}^{-}$ ha one unpaired electron

Q. In allene $\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)$, the type(s) of hybridization of the carbon atoms is (are): (1) only sp $^{2}$ (2) $\mathrm{sp}^{2}$ and $\mathrm{sp}$ (3) sp and sp $^{3}$ (4) $\mathrm{sp}^{2}$ and $\mathrm{sp}^{3}$ [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated? (1) IF $_{7}:$ Pentagonal bipyramid (2) BrF $_{5}:$ Trigonal bipyramid (4) $\mathrm{BrF}_{3}:$ Planar T-shaped [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\mathrm{BrF}_{5}$ have square pyramidal shape.

Q. The correct order of bond dissociation energy among $\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{o}_{2}^{-}$ is shown in which of the following arrangements? (1) $\mathrm{N}_{2}>\mathrm{O}_{2}>\mathrm{O}_{2}^{-}$ (2) $\mathrm{O}_{2}>\mathrm{O}_{2}^{-}>\mathrm{N}_{2}$ (3) $\mathrm{N}_{2}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}$ (4) $\mathrm{O}_{2}^{-}>\mathrm{O}_{2}>\mathrm{N}_{2}$ [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Which of the following molecules has two sigma$(\sigma)$ and two $\operatorname{pi}(\pi)$ bonds :- (1) HCN (2) $\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}$ (3) $\mathrm{N}_{2} \mathrm{F}_{2}$ (4) $\mathrm{C}_{2} \mathrm{H}_{4}$ [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Which one of the following molecules is paramagnetic? (1) NO ( 2) $\mathrm{O}_{3}$ (3) $\mathrm{N}_{2}$ (4) CO [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) NO has unpaired e $^{-} \therefore$ paramagnetic is nature

Q. Amongst LiCl, RbCl, $\mathrm{BeCl}_{2}$ and $\mathrm{MgCl}_{2}$ the compounds with the greatest and the least ionic character, respectively are : (1) $\mathrm{RbCl}$ and $\mathrm{MgCl}_{2}$ (2) LiCl and RbCl (3) $\mathrm{MgCl}_{2}$ and $\mathrm{BeCl}_{2}$ (4) $\mathrm{RbCl}$ and $\mathrm{BeCl}_{2}$ [JEE-MAINS(Online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) RbCl has highest ionic & and $\mathrm{BeCl}_{2}$ is most covalent

Q. The species in which the N atom is in a state of sp hybridization is :- (1) $\mathrm{NO}_{2}$ (2) $\mathrm{NO}_{2}^{+}$ (3) $\mathrm{NO}_{2}^{-}$ (4) $\mathrm{NO}_{3}^{-}$ [JEE-MAINS 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Which of the following species is not paramagnetic :- (1)NO (2) CO (3) $\mathrm{O}_{2}$ (4) $\mathrm{B}_{2}$ [JEE-MAINS 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Total number of lone pair of electrons in $\mathrm{I}_{3}^{-} \mathrm{ion}$ is (1) 6 (2) 9 (3) 12 (4) 3 [JEE-MAINS 2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. According to molecular orbital theory, which of the following will not be a viable molecule ? (1) $\mathrm{He}_{2}^{+}$ (2) $\mathrm{H}_{2}^{-}$ (3) $\mathrm{H}_{2}^{2-}$ (4) $\mathrm{He}_{2}^{2+}$ [JEE-MAINS 2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Which of the following compounds contain(s) no covalent bond(s) ? $\mathrm{KCl}, \mathrm{PH}_{3}, \mathrm{O}_{2}, \mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{H}_{2} \mathrm{SO}_{4}$ $(1) \mathrm{KCl}, \mathrm{H}_{2} \mathrm{SO}_{4}$ (2) KC1 (3) $\mathrm{KCl}, \mathrm{B}_{2} \mathrm{H}_{6}$ (4) $\mathrm{KCl}, \mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{PH}_{3}$ [JEE-MAINS 2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Circular Motion Notes Class 11th
Rotational Motion
System of Particles and Rotational Motion Class 11 Notes for IIT JEE & NEET
System of Particles and Rotational Motion Class 11 is one of the important chapters when it comes to understanding the basics of circular motion. We all are aware of the definition of rotational motion, the center of mass and center of gravity etc.Rotational motion refers to objects that are rotating in a curved path and comprises torque, moment of inertia, angular velocity, angular displacement, angular acceleration and angular momentum. System of Particles and Rotational Motion Class 11 notes are given below: System of Particles and Rotational Motion Class 11 Physical significance of Moment of Inertia System of Particles and Rotational Motion questions with solutions transational motion and rotational motion eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. This revision series is free for all. Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41) About Saransh Gupta Sir Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”. He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams. eSaral brings you detailed Class 11th Physics study material.  eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE & NEET preparations. We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view. Get Complete Physics Notes for Physics Class 11 for easy learning and understanding. For free video lectures and complete study material, Download eSaral APP. About eSaral At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.
Work, Energy and Power
Work Energy and Power Class 11 Physics Notes for IIT JEE | NEET

This article contains Work Energy and Power Class 11 Physics Notes.

Work Energy and Power can never be ignored as it covers basic concepts which provides a smooth access to the next some topics in Mechanics. This chapter is a key to prepare for entrance examinations like IIT-JEE & NEET. This is one of the most important topic in mechanics for IIT-JEE, NEET and CBSE perspectives. The terms work, energy and power have specific meanings in Physics. Work is done when a force produces motion. To do work, energy is required. This energy we acquire from the food which we eat and if work is done by machine, then energy is supplied by fuels or by current.

There are various forms of energy – kinetic energy, potential energy, heat energy, chemical energy, electrical energy, nuclear energy etc. Work Energy and Power Class 11 Physics notes  are given here:

  work energy and power class 11 notes Click Here to Study Newton Laws of Motion Class 11th To watch free video tutorials on Physics topics by Saransh Sir, AIR-41 Install eSaral App   eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. This revision series is free for all. Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41) About Saransh Gupta Sir Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”. He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams. eSaral brings you detailed Class 11th Physics study material.  eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE & NEET preparations. We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view. Get Complete Physics Notes for Physics Class 11 for easy learning and understanding. For free video lectures and complete study material, Download eSaral APP. About eSaral At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.
Score-good-in-JEE Mains-Exam
Tips to Score Maximum Marks in JEE Mains Exam

To get into IIT is the dream of every JEE Aspirant. Most of them started preparing for JEE from foundation classes while some realises their interest and inclination towards engineering after class 10th. Now for the aspirants who are preparing for 2020, few months left for JEE-Mains 2019. Many students started worrying on how to score good marks in JEE mains exam. During this time of preparation, aspirants have many more questions in their mind, Such as:

  • How to improve marks during mock test
  • How to score above 300 marks in JEE mains
  • How to attempt questions to score well in JEE mains
  • Where to practice JEE mains mock Test.
  • Where to get online test series for JEE Mains.

If you are seeking answer for any of the question above, then this article is for you!!

From last year, NTA (National Testing Agency) is conducting JEE Exam. From 2019 JEE mains exam held twice in year. In January and April.

Lot of hard work and practice is required to get a good score in JEE.

To score well in JEE mains, focus more on your learning & basic understanding of the concepts of Physics, Chemistry, and Mathematics. Click to watch Exam Pattern of JEE mains.

How to Study Physics for JEE Mains

Start with the basic concepts of Physics. Then memorize all the important Physics formulas for JEE Mains.

Practice makes everyone perfect so, practice questions more and more. Also practice previous year’s questions to make an idea on the question patterns that will appear in JEE mains. By practicing previous year questions, you will also get to know the most important chapters in JEE mains Physics. This will also help in strengthening your concepts and will make you familiar with a variety of questions asked in the exam.

Practice Mock Test on regular basis to improve score.

Don’t be overconfident. Sometimes it make mistakes if the question seems easy. Mostly mistakes are made in unit system where student ignore the difference between CGS and MKS systems. So, be easy-going and have the presence of mind while solving the problems.

How to Study Chemistry for JEE Mains

Chemistry comprises of three sections i.e., Physical Chemistry, Inorganic Chemistry and Organic Chemistry. The numerical questions are asked only from Physical Chemistry. Chemistry is a scoring subject in JEE mains Exam as the difficulty level of the questions in Chemistry is easier than the other two subjects i.e., Physics and Mathematics. That’s why, Chemistry is the easiest part of the JEE Main Examination for many students.

Learn all important formulas in Physical Chemistry which will help you to solve numerical problems in less time in JEE Main Examination 2020.

Don’t be overconfident. Chemistry is the easiest part of the JEE Examination. Therefore, some students did not give sufficient time to Chemistry subject. As a result, they score less marks in JEE Main Examination. They think that they can learn the complete subject within a month which is not possible.

How to study Mathematics for JEE Mains

For most of the aspirants, Math is one of the toughest subject among all three. Mostly, performance of JEE mains aspirants are affected by fear of mathematics.

To cope up with the fear of math, there is only one solution. Practice! Practice! And Practice! You can also make an actionable study plan to conquer the JEE Mains Mathematics Preparation.

An Ideal rule of solving mathematics problem is that never start with a challenging problem. If you can’t solve few basic problems, it will make you nervous & hence it will distress your complete performance in JEE Mains paper.

Solve the easier mathematical problems first & then progress towards the challenging ones.

The more you practice Mathematics, the less you make mistakes while solving problems and hence you will get good score in the JEE Mains.

Take mock tests and practice previous year questions to get first real look and feel of the examination. JEE Mains mock tests provide good practice and are a guide to judge your abilities. You can get mock test at eSaral app. eSaral also provides Topic-wise tests and offers more than 5 lakh questions to practice. For more details Click here.

It has been observed from the toppers talk that mock tests during exam preparation helped them a lot in managing time as well as scoring well in JEE mains exam paper. Hence Mock tests plays an important role in scoring good marks.

Finally, I would like to say have faith in yourself and be confident on your learning and preparation for JEE exam. Focus on every subject individually to score well.

Hope you liked reading the above post. Please share it with your friends to help them in getting the relevant information. Stay motivated and keep learning from eSaral!!!

Avoid-Distraction-While-studying
How to Avoid Distractions While studying?
Distraction is everywhere around us. Here we will get to know about How to avoid distractions while studying and how to focus on our goal. Let’s think about the time when you were “lost” in something you enjoy:  music, a good game, a sport, or a movie. It is not surprising to know that students spending more time unreasonably in unwanted things, that cause distraction rather than pursuing their careers. As per the survey it has been observed that an average person checks/unlocks her cell phone 110 times in a day. Those who falls under the category of addiction, they checks their phone up to 900 times in a day. Even most of the phone addicts experiences false vibrations. Smart phones not only causes distraction in studies but also affects health and may cause long-term, incurable Side effects. Nearby 63% of smartphone users of age 16-25 fall asleep with a cell phones.  Instead of making us more connected, smartphones could be making us more isolated. From age 14-22, around 52% of the students spending more than 4 hours a day on electronic gadgets and get improper sleep, that is causing long term high B.P. and weigh gain. In which 27% of them are at high risk of depression. 86% of smartphone users will check their device while speaking with friends and family. Research study has shown that most of the student are facing problems in concentration. While studying and they blame out to the outside distractions for their problem. But the main root of distraction is lack of passion, Interest and determination. It is because knowingly or unknowingly, students are indulging themselves in unwanted thoughts. Distractions that hinders their output and hamper their careers in the long run. For IIT-JEE or NEET aspirant, you must avoid distractions during your preparation. Because it is a golden period of your life. Your hard work towards your goal will decide better future for you. So, stay focused and avoid unwanted thoughts. To crack IIT-JEE and NEET, an aspirant need to be completely focus on your preparation and avoid distractions while studying. There are majorly two main causes of distraction among students:  
  1. Mobile Phones
  2. Unwanted Gossips
The major cause of distraction considered among students nowadays is ‘Smart Phone’. According to research, it has been observed that Mobile phone is as addictive as Alcohol& tobacco. But in actual Phone is not an addiction. The things inside the phones are addictive such as social networking sites & applications as such as Instagram, Facebook, What’s app.
The people around you are also responsible in your success. So, choose your friends wisely. Choose the peoples that helps you in achieving your goal. Don’t involve in irrelevant & unwanted gossips. Try to escape your time from unwanted things and be focused in your exam preparation and get rid of any fuss to avoid distraction while studying. Follow some steps to avoid mobile phone while studying You can delete the addictive apps from your phone. If that’s too radical, move them off the main screen or into a folder that’s out of sight. You can also put your mobile phone on grey scale so you aren’t as enchanted by the colorful graphics. Turn off notifications & delete unnecessary apps from mobile phone to reduce distraction while studying. While studying Keep your mobile data off. This will help you to keep away the unwanted notification. This will also helps in avoiding unwanted thoughts while studying. But Mobile phone became necessary part of our life. So firstly we must keep in mind that Mobile phones must be used to increase productivity not to waste time. The main thing behind this is that you’re controlling your phone or the phone is controlling you. That makes the main difference.   You can also do one thing to stop getting distracted while studying is- Setting up reminder for every 30 minutes Generally we do not think of getting distracted, it occurs automatically right? You cannot learn to focus in One day night! So, this will help you to check after every 30 minutes or accordingly that you are studying in actual or just staring at the textbook and thinking about some irrelevant situation. For students who also want to increase their study outputs, there is an amazing video by Saransh Gupta Sir. He has revealed some wonderful tips to increase study hours and also to increase study output. Eventually this will also help you to avoid distractions while studying. Watch How to increase self-study output. Saransh Gupta Sir is the founder of eSaral and secured AIR-41 in IIT-JEE 2006. He has encouraged and motivated thousands of students during their IIT-JEE & NEET preparation. To watch out Physics tutorials for free by Saransh Sir, Download eSaral for IIT-JEE & NEET  preparation. With eSaral, prepare for IIT-JEE & NEET at your place without getting distracted. eSaral provides a tab consisting all the study material, video lectures, doubt clearing session, and more related to your preparation. eSaral Tab comprises an amazing feature, that a student can access only his study material in that tab. The tab is totally locked to use other sites such as Facebook, Instagram, etc. That will help him to get rid of distractions while studying. Click Here for more Information.      
How to Wake Up Early in the Morning for study
How to wake up early in the morning for study
We all want to wake up early, but most of us end up failing in this desire. Do you also have the same desire and you want to wake up early in the morning, but you do not know how? As you know that the time before the dawn covers most hours of the day. Every morning there are two opportunities for you: Continuing sleep with dreams. OR Wake up early to chase your dreams. The choice is yours only. Waking up early in the morning will help you in maintaining the sleep cycle. Thus, improves health. But also a good night’s sleep is extremely important for students to produce improved study outputs. Some other benefits to wake up early in the morning for students are:
  • Better concentration power.
  • The stress level is minimum.
  • Better Study Output.
  • Improves mental health.
  • Improves brain efficiency.
  • Less traffic, more peace.
But most of the students are concerned about on how to wake up in the early morning for study. Because they may be probably making some mistakes that are making difficult to wake up early for them. Such as:
  • Not having a regular sleep schedule.
  • Sleeping late in the night.
  • Drinking Caffeine during study hours.
  • Taking long naps instead of short naps.
  • Eating late at night.
  • Hitting snooze.
Saransh Gupta Sir, founder of eSaral has revealed some tips to wake up early in the morning. He mentored many students and the tips are proved to be useful in getting up early. So Guys, watch out the Video on How to wake up early in the morning by Saransh Sir.
He also talked about productive time for study in his previous video, where he talked about the benefits of early morning study for IIT JEE & NEET aspirants. If you haven’t watched the video, then watch the video on Early Morning or Late Night Study! Research study has also shown that morning time is the best time to study and memorize. So wake up early to make your dream come true. Here are some tips to help you in the fight & wake up early in the morning:
  • Get excited to wake up early: find a strong purpose to get up early morning.
  • Don’t put the alarm on the bed or near to the bed. Keep it at distance.
  • You can also find a wake-up partner to help you to wake up early in the morning.
  • Sleep at the right time to get proper sleep of 6-7 hours.
  • Do stretch your body after waking up. This will help you in activating blood circulation and also helps in reducing laziness.
  • Give 10-15 minutes to meditation or exercise to maintain the oxygen level.
  • Hydrate your body by drinking 2-3 glasses of water just after waking up.
So Guys, get a head start on the rest of the world by skipping the snooze button and outworking your competition. Because life has no remote, get up and change it yourself! For more videos related to Boards, IIT JEE & NEET preparation. Click Here.