Vector- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

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Simulator

Q. Three vectors $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{R}}$ are shown in the figure. Let S be any point on the vector $\overrightarrow{\mathrm{R}}$. The distance between the points P and S is b $|\overrightarrow{\mathrm{R}}|$. The general relation among vectors $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{S}}$ is :

$(\mathrm{A}) \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b}^{2} \overrightarrow{\mathrm{Q}}$

(B) $\overrightarrow{\mathrm{S}}=(b-1) \overrightarrow{\mathrm{P}}+b \overrightarrow{\mathrm{Q}}$

(C) $\overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$

$(\mathrm{D}) \overrightarrow{\mathrm{S}}=\left(1-\mathrm{b}^{2}\right) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$

Sol. (C)

Let vector from point P to point S be $\overrightarrow{\mathrm{c}}$

$\Rightarrow \overrightarrow{\mathrm{c}}=\mathrm{b}|\overrightarrow{\mathrm{R}}| \hat{\mathrm{R}}=\mathrm{b}|\overrightarrow{\mathrm{R}}|\left(\frac{\overrightarrow{\mathrm{R}}}{|\overrightarrow{\mathrm{R}}|}\right)=\mathrm{b} \overrightarrow{\mathrm{R}}=\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})$

from triangle rule of vector addition

$\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{S}}$

$\overrightarrow{\mathrm{P}}+\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})=\overrightarrow{\mathrm{S}}$

$\Rightarrow \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$

Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density  remains uniform throughout the volume. The rate of fractional change in density $\left(\frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}\right)$ is constant. The velocity v of any point on the surface of the expanding sphere is proportional to :

(A) $\mathrm{R}^{3}$

(B) $\frac{1}{\mathrm{R}}$

(C) R

(D) $\mathrm{R}^{2 / 3}$

Sol. (C)

Density of sphere is $\rho=\frac{\mathrm{m}}{\mathrm{v}}=\frac{3 \mathrm{m}}{4 \pi \mathrm{R}^{3}}$

$\Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}=-\frac{3}{\mathrm{R}} \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}}$

since $\Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}$ is constant

$\therefore \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}} \propto \mathrm{R}$

Velocity of any point on the circumfrence V is equal to $\frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}}$ (rate of change of radius of outer layer)

Ray Optics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

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.

Simulator

Q. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as $\left[\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} .\right]$

(A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s

[IIT-JEE 2009]

Sol. (C)

Q. A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) :

(A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39)

[IIT-JEE 2009]

Sol. (C,D)

$\mathrm{V}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \quad$ by sustituting the value of u and $\mathrm{f}$

$\mathrm{V}_{42} \Rightarrow \frac{24 \times 42}{18} \Rightarrow 56 ; \quad \mathrm{V}_{60}=\frac{24 \times 60}{36} \Rightarrow 40$

$\mathrm{v}_{48} \Rightarrow \frac{48 \times 24}{24} \Rightarrow 48 ; \mathrm{V}_{66}=\frac{66 \times 24}{42} \Rightarrow 37.71$

$\mathrm{V}_{78} \Rightarrow \frac{78 \times 24}{54} \Rightarrow 34.3$

(66, 33) ; (78, 39)

Q. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is –

(A) virtual and at a distance of 16 cm from the mirror

(B) real and at a distance of 16 cm from the mirror

(C) virtual and at a distance of 20 cm from the mirror

(D) real and at a distance of 20 cm from the mirror

[IIT-JEE 2010]

Sol. (B)

$\frac{1}{15}=\frac{1}{v}-\frac{1}{10}$

Q. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct?

(A) The ray gets totally internally reflected at face CD

(B) The ray comes out through face AD

(C) The angle between the incident ray and the emergent ray is $90^{\circ}$

(D) The angle between the incident ray and the emergent ray is $120^{\circ}$

[IIT-JEE 2010]

Sol. (A,B,C)

Q. Two transparent media of refractive indices $\mu_{1}$ and $\mu_{3}$ have a solid lens shaped transparent material of refractive index $\mu_{2}$ between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between $\mu_{1}, \mu_{2}$ and $\mu_{3}$ are given. Match them to the ray diagrams shown in Column II.

[IIT-JEE 2010]

Sol. $(\mathrm{A})-\mathrm{pr},(\mathrm{B})-\mathrm{qst},(\mathrm{C})-\mathrm{prt},(\mathrm{D})-\mathrm{qs}$

Q. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from $\mathrm{m}_{25}$ to $\mathrm{m}_{50} .$ The ratio $\frac{m_{25}}{m_{50}}$ is –

[IIT-JEE 2010]

Sol. 6

$\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}}$

Q. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from $\frac{25}{3}$ m to $\frac{50}{7}$ m in 30 seconds. What is the speed of the object in km per hour ?

[IIT-JEE 2010]

Sol. 3

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v} \Rightarrow \frac{1}{10}=\frac{1}{u}+\frac{3}{25} \Rightarrow u=-50 \mathrm{m}$

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} \Rightarrow \frac{1}{10}=\frac{1}{\mathrm{u}}+\frac{7}{50} \Rightarrow \mathrm{u}=-25 \mathrm{m}$

Speed $=\frac{25}{30} \times \frac{18}{5}=3$

Q. A large glass slab $\left(\mu=\frac{5}{3}\right)$ of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?

[IIT-JEE 2010]

Sol. 6

$r=\frac{h}{\sqrt{\mu^{2}-1}}=\frac{8}{\sqrt{\left(\frac{5}{3}\right)^{2}-1}}=6 \mathrm{cm}$

Q. A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected (R) and transmitted (T) intensities, both as function of $\theta$, are plotted. The correct sketch is –

[IIT-JEE 2011]

Sol. (C)

When $\theta=0^{\circ},$ maximum light is transmitted. At $\theta>\theta_{\mathrm{C}}$ (critical angle), no further light is transmitted

Q. Water (with refractive index = $\frac{4}{3}$) in a tank is 18 cm deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is

[IIT-JEE 2011]

Sol. 2

First refraction [ Lens-air interface]

$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$ where $\mu_{1}=1, \mathrm{u}=-24, \mu_{2}=\frac{7}{4}, \mathrm{R}=+6$

After solving v = 21

Now for second refraction [Lens-water interface]

$\frac{4 / 3}{v_{2}}-\frac{7 / 4}{21}=0 \Rightarrow v_{2}=h=16 \mathrm{cm}$

So, from bottom $18-16=2 \Rightarrow \mathrm{x}=2$

Q. A biconvex lens is formed with two thin plano-convex lenses as shown in the figure, Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be :-

(A) –280.0 cm             (B) 40.0 cm            (C) 21.5 cm             (D) 13.3 cm

[IIT-JEE 2012]

Sol. (B)

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}$

$\frac{1}{\mathrm{f}}=\left(\frac{\mu_{1}-1}{\mathrm{R}_{1}}\right)+\left(\frac{\mu_{2}-1}{\mathrm{R}_{2}}\right) \Rightarrow \frac{1}{\mathrm{f}}=\frac{5}{14}+\frac{2}{14}$

f = 20 c.m.

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-40}=\frac{1}{20}=40 \mathrm{c.m}$

Paragraph for Questions 12 and 13

Most materials have the refractive index, n>1. So, when a light ray from air enters a naturally occurring material, then by Snell’s law, $\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}}$ , it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, $n=\left(\frac{c}{v}\right)=\pm \sqrt{\varepsilon_{r} \mu_{r}}$ , where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$ are the relative permittivity and permeability of the medium respectively.

In normal materials, both $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$ are positive, implying positive n for the medium. When both $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$ are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta materials.

Q. For light incident from air on a meta-material, the appropriate ray diagram is –

[IIT-JEE 2012]

Sol. (C)

$_{1} \mu_{2}=\frac{\mu_{2}}{\mu_{1}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$

$\frac{(-)}{1}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$

sini = sin(–r)

Q. Choose the correct statement.

(A) The speed of light in the meta-material is $\mathrm{v}=\mathrm{c}|\mathrm{n}|$

(B) The speed of light in the meta-material is v = $\frac{c}{|n|}$

(C) The speed of light in the meta-material is v = c.

(D) The wavelength of the light in the meta-material $\left(\lambda_{\mathrm{m}}\right)$ is given by $\lambda_{\mathrm{m}}=\lambda_{\mathrm{air}}|\mathrm{n}|,$ where $\lambda_{\mathrm{air}}$ is the wavelength of the light in air.

[IIT-JEE 2012]

Sol. (B)

$\mu=\frac{\mathrm{c}}{\mathrm{v}} \Rightarrow \mathrm{v}=\frac{\mathrm{c}}{\mu}=\frac{\mathrm{c}}{\mathrm{n}}$

Q. A ray of light travelling in the direction $\frac{1}{2}(\hat{i}+\sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection, it travels along the direction $\frac{1}{2}(\hat{i}-\sqrt{3} \hat{j})$. The angle of incidence is :-

(A) $30^{\circ}$

(B) $45^{\circ}$

(C) $60^{\circ}$

(D) $75^{\circ}$

Sol. (A)

Here normal is along $\hat{j}$

Angle between incident ray and normal $\cos \theta=\frac{\frac{1}{2}(\hat{i}+\sqrt{3} \hat{j}) \cdot \hat{j}}{(1)(1)}=\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ}$

Q. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $\frac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is :

(A) 1 m            (B) 2 m              (C) 3 m               (D) 6 m

Sol. (C)

$\mathrm{m}=-\frac{1}{3}=\frac{\mathrm{v}}{\mathrm{u}}$

v = 8m

u = – 24 m

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \rightarrow \frac{1}{8}-\frac{1}{-24}=\frac{1}{\mathrm{f}} \Rightarrow \mathrm{f}=6 \mathrm{m}$

$\mathrm{f}=\frac{\mathrm{R}}{(\mu-1)}$

$\mu=\frac{\lambda_{\text {vaceum }}}{\lambda_{\text {medium }}}=\frac{3}{2}$

$6 \mathrm{m}=\left(\frac{\mathrm{R}}{3 / 2-1}\right) \Rightarrow \mathrm{R}=3 \mathrm{m}$

Q. A right angled prism of refractive index $\mu_{1}$ is placed in a rectangular block of refractive index $\mu_{2}$, which is surrounded by a medium of refractive index $\mu_{3}$, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between $\mu_{1}, \mu_{2},$ and $\mu_{3}$, it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’ or ‘ei’.

Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists :

Sol. (D)

(P) at prism surface ray moving towards normal so $\left(\mu_{2}>\mu_{1}\right)$at block surface ray moving away from normal so $\left(\mu_{3}<\mu_{2}\right)$

(Q) No deflection of ray on both surface; so $\mu_{1}=\mu_{2}=\mu_{3}$

(R) At prism surface ray moving away from normal so $\mu_{2}<\mu_{1}$. At block surface ray movign away from normal so µ3 < µ2 but since on total internal reflection not takes place on prism surface

$\mu_{1} \sin 45^{\circ}<\mu_{2} \sin 90^{\circ} \Rightarrow \mu_{1}<\sqrt{2} \mu_{2}$

(S) Total internal reflection takes place so $\mu_{1} \sin 45^{\circ}>\mu_{2} \sin 90^{\circ} \Rightarrow \mu_{1}>\sqrt{2} \mu_{2}$

Q. A transparent thin film of uniform thickness and refractive index $\mathrm{n}_{1}=1.4$ is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index $\mathrm{n}_{2}=1.5$, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $\mathrm{f}_{1}$ from the film, while rays of light traversing from glass to air get focused at distance $\mathrm{f}_{2}$ from the film. Then

$(\mathrm{A})\left|\mathrm{f}_{1}\right|=3 \mathrm{R}$

(B) $\left|\mathrm{f}_{1}\right|=2.8 \mathrm{R}$

(C) $\left|\mathrm{f}_{2}\right|=2 \mathrm{R}$b

$(\mathrm{D})\left|\mathrm{f}_{2}\right|=1.4 \mathrm{R}$

Sol. (A,C)

When rays are moving from air to glass,

$\frac{1.5}{\mathrm{f}_{1}}=\frac{(1.4-1)}{+\mathrm{R}}+\frac{(1.5-1.4)}{+\mathrm{R}}$

$\frac{1.5}{\mathrm{f}_{1}}=\frac{0.4}{\mathrm{R}}+\frac{0.1}{\mathrm{R}}=\frac{0.5}{\mathrm{R}}$

$\left|\mathrm{f}_{1}\right|=3 \mathrm{R}$

When rays are moving from glass to air,

$\frac{1}{\mathrm{F}_{2}}=\frac{(1-1.4)}{-\mathrm{R}}+\frac{(1.4-1.5)}{-\mathrm{R}}=\frac{0.5}{\mathrm{R}}$

$\left|\mathrm{f}_{2}\right|=2 \mathrm{R}$

Q. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is :-

(A) 1.21            (B) 1.30              (C) 1.36               (D) 1.42

Sol. (C)

From the given situation, at critical angle,

$\tan \theta=\frac{(\mathrm{d} / 2)}{\mathrm{h}}=\frac{5.77}{10}$

$\therefore \sin \theta_{\mathrm{C}} \approx \frac{1}{2}$

$\mu_{\text {denser }} \sin \theta_{\mathrm{C}}=\mu_{\text {raver }} \sin (\pi / 2)$

$\Rightarrow 2.72 \times \frac{1}{2}=\mu_{\mathrm{r}} \Rightarrow \mu_{\mathrm{r}}=1.36$

Q. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.

Sol. (B)

Q. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification $\mathrm{M}_{1}$. When the set-up is kept in a medium of refractive index 7/6 the magnification becomes $\mathrm{M}_{2}$. The magnitude $\left|\frac{M_{2}}{M_{1}}\right|$ is.

Sol. 7

For reflectionfrom concave mirror,

$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{15}=\frac{-1}{10}$

$\frac{1}{\mathrm{v}}=\frac{1}{15}-\frac{1}{10}=\frac{-1}{30}$

$\therefore \mathrm{v}=-30$

magnification $\left(\mathrm{m}_{1}\right)=-\frac{\mathrm{v}}{\mathrm{u}}=-2$

Now for refraction from lens,

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$

$\therefore$ magnification $\left(\mathrm{m}_{2}\right)=\frac{\mathrm{v}}{\mathrm{u}}=-1$

$\therefore \mathrm{M}_{1}=\mathrm{m}_{1} \mathrm{m}_{2}=2$

Now when the set-up is immersed in liquid, no effect for the image formed by mirror.

we have $\left(\mu_{\mathrm{L}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{10}$

$\Rightarrow\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{5}$

when lens is immersed in liquid,

$\frac{1}{\mathrm{f}_{\mathrm{lens}}}=\left(\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{S}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{2}{7} \times \frac{1}{5}=\frac{2}{35}$

$\therefore \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_{\mathrm{Liquid}}}$

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{2}{35}-\frac{1}{20}=\frac{8-7}{140}=\frac{1}{140}$

$\therefore$ magnification $=-\frac{140}{20}=-7$

$\therefore \mathrm{M}_{2}=2 \times 7=14$

$\therefore\left|\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}\right|=7$

Q. Two identical glass rods $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashded line) aligned. When a point source of light P is placed inside rod $\mathrm{S}_{1}$ on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside $\mathrm{S}_{2}$. The distance d is :

(A) 60 cm         (B) 70 cm           (C) 80 cm           (D) 90 cm

Sol. (B)

For first surface

$\frac{1}{V}-\frac{1.5}{-50}=\frac{1-1.5}{-10}$

V = 50 cm

for second surface

$\frac{1.5}{\infty}-\frac{1}{-(\mathrm{d}-50)}=\frac{1.5-1}{10}$

d = 70 cm

$\therefore(\mathrm{B})$

Q. A monochromatic beam of light is incident at $60^{\circ}$ on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle $\theta$(n) with the normal (see the figure). For n = $\sqrt{3}$ the value of $\theta$ is $60^{\circ}$ and $\frac{\mathrm{d} \theta}{\mathrm{dn}}=\mathrm{m}$. The value of m is.

Sol. 2

By snell’s law

$1 \sin 60=n \sin r_{1} \Rightarrow \sin r_{1}=\frac{1}{2} r_{1}=30^{\circ} \quad \ldots \ldots(i)$

By differentiating ‘w.r.t’ n

$\mathrm{O}=\sin \mathrm{r}_{1}+\mathrm{n} \cos \mathrm{r}_{1}\left(\frac{\mathrm{dr}_{1}}{\mathrm{dn}}\right)$

$=\frac{1}{2}+\sqrt{3}(\sqrt{\frac{3}{2}}) \frac{\mathrm{dr}_{1}}{\mathrm{dn}}$

$\frac{\mathrm{d} \mathrm{r}_{1}}{\mathrm{dn}}=\frac{1}{3}$ ….(ii)

By applying snell’s law

$n \sin r_{2}=1 \sin \theta$

$n \sin \left(60-r_{1}\right)=1 \sin \theta\left[\therefore A=r_{1}+r_{2}\right]$

By diffrentiating ‘w.r.t’ n

$\sin \left(60-r_{1}\right)-n \cos \left(60-r_{1}\right) \frac{d r_{1}}{d n}=\cos \theta \frac{d \theta}{d n}$$\sin \left(60-r_{1}\right)-n \cos \left(60-r_{1}\right) \frac{d r_{1}}{d n}=\cos \theta \frac{d \theta}{d n}$

By substituting value of $\mathrm{r}_{1}^{\prime}$ and $\frac{\mathrm{dr}_{1}}{\mathrm{dn}}$ from ( 1) and ( 2)

$\frac{\mathrm{d} \theta}{\mathrm{dn}}=2$

Paragraph for Question No. 23 and 24

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $\mathbf{n}_{1}$ surrounded by a medium of lower refractive index $\mathrm{n}_{2}$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $\mathbf{n}_{1}$ and $\mathrm{n}_{2}$ as shown in the figure. All rays with the angle of incidence i less than a particular value of $\dot{l}_{\mathrm{m}}$ are confined in the medium of refractive index $\mathrm{n}_{1}$. The numerical aperture (NA) of the structure is defined as sin $i_{\mathrm{m}}$.

Q. For two structures namely $\mathrm{S}_{1}$ with $\mathrm{n}_{1}=\sqrt{45} / 4$ and $\mathrm{n}_{2}=3 / 2,$ and $\mathrm{S}_{2}$ with $\mathrm{n}_{1}=8 / 5$ and $\mathrm{n}_{2}=7 / 5$ and

taking the refractive index of water to be $4 / 3$ and that of air to be $1,$ the correct option(s) is (are)

(A) NA of $\mathrm{S}_{1}$ immersed in water is the same as that of $\mathrm{S}_{2}$ immersed in liquid of refractive index $\frac{16}{3 \sqrt{15}}$

(B) NA of $\mathrm{S}_{1}$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $\mathrm{S}_{2}$ immersed in water.

(C) NA of $\mathrm{S}_{1}$ placed in air is the same as that of $\mathrm{S}_{2}$ immersed in liquid of refractive index .

(D) NA of $\mathrm{S}_{1}$ placed in air is the same as that of $\mathrm{S}_{2}$ placed in water.

Sol. (A,C)

Let the whole structure is placed in a medium of refractive index n’, then

$n^{\prime} \sin i=n_{1} \sin (90-\theta)$

$\mathrm{n}^{\prime} \sin \mathrm{i}=\mathrm{n}_{1} \cos \theta \quad \ldots(\mathrm{i})$

Here for $\mathrm{i}_{\mathrm{m}} ; \quad \theta=\mathrm{C}$ and $\sin \mathrm{C}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$

from eq. (i), $\mathrm{n}^{\prime} \sin \mathrm{i}_{\mathrm{m}}=\mathrm{n}_{1} \sqrt{\frac{1-\mathrm{n}_{2}^{2}}{\mathrm{n}_{1}^{2}}}=\sqrt{\mathrm{n}_{1}^{2}-\mathrm{n}_{2}^{2}}$

$\Rightarrow \sin i_{m}=\frac{\sqrt{n_{1}^{2}-n_{2}^{2}}}{n^{\prime}}$

Now, for $(\mathrm{A})(\mathrm{NA})_{\mathrm{s} 1}=\frac{3}{4} \sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3 \sqrt{15}}{16} \sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{3 \sqrt{15}}{16} \frac{1}{5} \sqrt{15}=\frac{9}{16}$

For (B) $\quad(\mathrm{NA})_{\mathrm{s} 1}=\frac{\sqrt{15}}{6} \times \frac{3}{4}=\frac{\sqrt{15}}{8}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3}{4}=\frac{\sqrt{15}}{5}$ Not equal

For $(\mathrm{C})(\mathrm{NA})_{\mathrm{s} 1}=1 \times \frac{3}{4}=\frac{3}{4}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{\sqrt{15}}{4} \times \frac{\sqrt{15}}{5}=\frac{15}{4 \times 5}=\frac{3}{4}$

For (D) $(\mathrm{NA})_{\mathrm{s} 1}=\frac{3}{4}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3}{4} \frac{\sqrt{15}}{5}$ Not equal

Q. If two structures of same cross-sectional area, but different numerical apertures $\mathrm{NA}_{1}$ and $\mathrm{NA}_{2}$ $\left(\mathrm{NA}_{2}<\mathrm{NA}_{1}\right)$ are joined longitudinally, the numerical aperture of the combined structure is

(A) $\frac{\mathrm{NA}_{1} \mathrm{NA}_{2}}{\mathrm{NA}_{1}+\mathrm{NA}_{2}}$

$(\mathrm{B}) \mathrm{NA}_{1}+\mathrm{NA}_{2}$

$(\mathrm{C}) \mathrm{NA}_{1}$

$(\mathrm{D}) \mathrm{NA}_{2}$

Sol. (D)

$\begin{array}{ll}{\text { It is given that }} & {\mathrm{NA}_{2}<\mathrm{NA}_{1}} \\ {\Rightarrow \mathrm{i}_{\mathrm{m} 2}<\mathrm{i}_{\mathrm{m} 1}}\end{array}$

Hence if the combination can be placed both ways i.e. 1st structure & then 2nd structure and then reversed also, then the condition of TIR is satisfied for lower $\dot{\mathbf{1}}_{\mathrm{m}}$ then it can be satisfied for all other less angler as well.

Hence $\mathrm{NA}_{2}$ will be the numerical aperture of the combined structure.

Q. A parallel beam of light is incident from air at an angle  on the side PQ of a right angled triangular prism of refractive index $\mathrm{n}=\sqrt{2}$. Light undergoes total internal reflection in the prism at the face PR when  has a minimum value of $45^{\circ}$. The angle $\theta$ of the prism is :

(A) $15^{\circ}$

(B) $22.5^{\circ}$

(C) $30^{\circ}$b

(D) $45^{\circ}$

Sol. (A)

$1 \sin 45^{\circ}=\sqrt{2} \sin \mathrm{r}_{1}$

$\mathrm{r}_{2}-\mathrm{r}_{1}=\theta$

$\theta=45^{\circ}-30^{\circ}$

$\Rightarrow \theta=15^{\circ}$

Q. A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices $\mathrm{n}_{1}$ and $\mathrm{n}_{2}\left(>\mathrm{n}_{1}\right)$, as shown in the figure. A ray of light is incident with angle $\theta_{\mathrm{i}}$ from medium 1 and emerges in medium 2 with refraction angle $\theta_{\mathrm{f}}$ with a lateral displacement . Which of the following statement(s) is(are) true ?

(A) $\ell$ is independent of $\mathrm{n}_{2}$

(B) $\ell$ is dependent on $\mathrm{n}(\mathrm{z})$

(C) $\mathrm{n}_{1} \sin \theta_{\mathrm{i}}=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \sin \theta_{\mathrm{f}}$

(D) $\mathrm{n}_{1} \sin \theta_{\mathrm{i}}=\mathrm{n}_{2} \sin \theta_{\mathrm{f}}$

Sol. (A,B,D)

Q. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?

(A) The refractive index of the lens is 2.5

(B) The radius of curvature of the convex surface is 45 cm

(C) The faint image is erect and real

(D) The focal length of the lens is 20 cm.

Sol. (A,D)

Q. A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle  = $30^{\circ}$ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are :

(A) $(25,25 \sqrt{3})$

$(\mathrm{B})\left(\frac{125}{3}, \frac{25}{\sqrt{3}}\right)$

(C) $(50-25 \sqrt{3}, 25)$

(D) (0, 0)

Sol. (A)

For lens $\mathrm{V}=\frac{(-50)(30)}{-50+30}=75$

For mirror $\mathrm{V}=\frac{\left(\frac{25 \sqrt{3}}{2}\right)(50)}{\frac{25 \sqrt{3}}{2}-50}=\frac{-50 \sqrt{3}}{4-\sqrt{3}}$

$\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}_{2}}{\mathrm{h}_{1}} \Rightarrow \mathrm{h}_{2}=-\left(\frac{\frac{-50 \sqrt{3}}{4-\sqrt{3}}}{\frac{25 \sqrt{3}}{2}}\right) \cdot \frac{25}{2}$

$\mathrm{h}_{2}=\frac{+50}{4-\sqrt{3}}$

The x coordinate of the images $=50-v \cos 30+h_{2} \cos 60 \approx 25$

The y coordinate of the images $=v \sin 30+h_{2} \sin 60 \approx 25 \sqrt{3}$

Q. For an isosceles prism of angle A and refractive index µ, it is found that the angle of minimum deviation $\delta_{\mathrm{m}}$ = A. Which of the following options is/are correct ?

(A) At minimum deviation, the incident angle $\dot{\mathbf{1}}_{1}$ and the refracting angle $\mathrm{r}_{1}$ at the first refracting surface are related by $\mathbf{r}_{1}=\left(\mathbf{i}_{1} / 2\right)$

(B) For this prism, the refractive index µ and the angle of prism A are related as $\mathrm{A}=\frac{1}{2} \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $\mathrm{i}_{1}=\sin ^{-1}[\sin \mathrm{A} \sqrt{4 \cos ^{2} \frac{\mathrm{A}}{2}-1}-\cos \mathrm{A}]$

(D) For the angle of incidence $\mathbf{i}_{1}$ = A, the ray inside the prism is parallel to the base of the prism.

Sol. (A,C,D)

i = e (for minimum deviation)

$\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}, \mathrm{r}_{1}=\mathrm{r}_{2}$

(A) $\delta_{\mathrm{m}}=2 \mathrm{i}-\mathrm{A}=\mathrm{A}$ (given)

$\Rightarrow \mathrm{i}=\mathrm{A}$

$\Rightarrow \mathrm{r}_{1}=\frac{\mathrm{A}}{2}=\frac{\mathrm{i}}{2}$

(B) $\mu=\frac{\sin (\mathrm{A})}{\sin (\mathrm{A} / 2)}=2 \cos \frac{\mathrm{A}}{2} \Rightarrow \mathrm{A}=2 \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) $\mu \sin \left(\mathrm{r}_{2}\right)=1$

$\sin \left(\mathrm{r}_{2}\right)=\frac{1}{\mu}$

$\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}$

$\mathrm{r}_{1}=\mathrm{A}-\mathrm{r}_{2}$

$=\mathrm{A}-\sin ^{-1}\left[\frac{1}{\mu}\right]$

$\sin (\mathrm{i})=\mu \sin \left(\mathrm{r}_{1}\right)$

$\mathrm{i}=\sin ^{-1}\left[\mu \sin \left[\mathrm{A}-\sin ^{-1}\left[\frac{1}{\mu}\right]\right]\right.$

$\mathrm{i}_{\mathrm{g}}=\sin ^{-1}[\sqrt{\mu^{2}-1} \sin \mathrm{A}-\cos \mathrm{A}]=\sin ^{-1}\left[\mu \sin \left(\mathrm{A}-\theta_{\mathrm{C}}\right)\right]$

(Here $\left.\mu=2 \cos \frac{\mathrm{A}}{2}\right)$

(D) Condition of min. deviation $\mathrm{i}=\mathrm{e} \& \mathrm{r}_{1}=\mathrm{r}_{2}=\frac{\mathrm{A}}{2}$

Rays will be parallel to base.

Q. A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle $\theta=30^{\circ}$. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $\mathrm{n}_{\mathrm{m}}=\mathrm{n}-\mathrm{m} \Delta \mathrm{n}$, where $\mathrm{n}_{\mathrm{m}}$ is the refractive index of the mth slab and $\Delta \mathrm{n}=0.1$ (see the figure). The ray is refracted out parallel to the interface between the (m – 1)th and mth slabs from the right side of the stack. What is the value of m ?

Sol. 8

Applying snell’s law between entry & exit surfaces,

n $\sin \theta=\mu \sin \left(\frac{\pi}{2}\right)$

Q. Sunlight of intensity $1.3 \mathrm{kW} \mathrm{m}^{-2}$ is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW m–2, at a distance 22 cm from the lens on the other side is

Sol. 130

$\frac{\mathrm{r}}{\mathrm{R}}=\frac{2}{20}=\frac{1}{10}$

$\therefore$ Ratio of area $=\frac{1}{100}$

Let energy incident on lens be E.

$\therefore$ Given $\frac{\mathrm{E}}{\mathrm{A}}=1.3$

So final, $\frac{\mathrm{E}}{\mathrm{a}}=? ?$

E = A × 1.30

Also $\frac{\mathrm{a}}{\mathrm{A}}=\frac{1}{100}$

$\therefore$ Average intensity of light at $22 \mathrm{cm}=\frac{\mathrm{E}}{\mathrm{a}}=\frac{\mathrm{A} \times 1.3}{\mathrm{a}}=100 \times 1.3=130 \mathrm{kW} / \mathrm{m}^{2}$

Q. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire ? (These figures are not to scale.) ?

Sol. (D)

Distance of point A is f/2

Let A’ is the image of A from mirror, for this image

$\frac{1}{\mathrm{v}}+\frac{1}{-\mathrm{f} / 2}=\frac{1}{-\mathrm{f}}$

$\frac{1}{\mathrm{v}}=\frac{2}{\mathrm{f}}-\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}}$

image of line AB should be perpendicular to the principle axis & image of F will form at infinity, therefor correct image diagram is

Newtons Law of Motion-JEE Advanced Previous Year Questions with Solutions

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Simulator

Q. A piece of wire is bent in the shape of a parabola y = $\mathrm{Kx}^{2}$ (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

[IIT-JEE-2009]

(A) $\frac{\mathrm{a}}{\mathrm{gk}}$            (B) $\frac{\mathrm{a}}{2 \mathrm{gk}}$            (C) $\frac{2 \mathrm{a}}{\mathrm{gk}}$                  (D) $\frac{\mathrm{a}}{4 \mathrm{gk}}$

Sol. (B)

$\operatorname{ma} \cos \theta=\operatorname{mg} \sin \theta$

$\mathrm{a}=\mathrm{g} \tan \theta$

$\frac{a}{g}=\tan \theta$

$\frac{a}{g}=2 k x$

$\frac{a}{2 g k}=x$

Polymer – JEE Advanced Previous Year Questions with Solutions

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Q. The correct functional group X and the reagent/reaction conditions Y in the following scheme are

[JEE 2011]

Sol. (A,B,C,D)

Q. On complete hydrogenation, natural rubber produces

(A) ethylene-propylene copolymer

(B) vulcanised rubber

(C) polypropylene

(D) polybutylene

Sol. (A)

Applications of Colloids | Catalysis & its Types for Class 12, IIT-JEE, NEET

## Applications of Colloids

Colloids play an important role in our daily life and industry. Some of the important applications of colloids are listed below.

Indicators like fluoresce in function by adsorption of ions on to sol particles When silver nitrate solution is run into a solution of sodium chloride containing a fluoresce in a white precipitate of AgCl is first formed. At the end point the white precipitate turns sharply pink.

## Tanning

Animal leather (skin) is soft due to the presence of very fine globules oils and fats in the’ pores of skin. These globules are of colloidal in nature. When this soft leather (skin) is placed in salt water, these globules of oils and fats coagulate and settle down in water. Now when the animal hides tare dried, they become hand. This process of making soft leather to hard leather by dipping it in salt water is called tanning usually chromium salts are used for tanning.

## Clearing action of soaps and detergents

Dust and dirt particles on clothes are colloidal in nature soaps being sodium salts of higher fatty acids coagulates them which become suspension particles called micelle and roll out due to greater volume and greater mass.

## Industrial products

Paints, inks, synthetic plastics, rubber, graphite lubricants, cement, etc are all colloidal solutions.

Colloidal solution of graphite in water is called “Aqua dag” while that in Oil is called oil dag.

The phenomenon of increase in concentration at the surface due to molecular surface force is known as adsorption.

The solid substance on the surface of which adsorption takes place is called adsorbent. Examples of absorbents are activated charcoal, Pt, Pd, Ni etc.

## Sorption

lt may be defined as the process in which both adsorption and absorption take place simultaneously.

## Absorption

When a substance is uniformly distributed throughout the body of a solid of liquid, the phenomenon is called absorption.

Adsorption is due to the fact that the surface particles of the adsorbent are in different state than the particles inside the bulk. Inside the adsorbent all the forces acting between the particles are mutually balanced but on the surface particles are not surrounded by atoms or molecules of their kind on all sides and hence they possess unbalanced or residual attractive forces.

Depending upon the nature of forces between molecules of adsorbate and adsorbent, adsorption is of two types.

1. Nature of adsorbate: Readily liquefiable gases such as HCl, etc. are more. easily adsorbed by and adsorbent than the permanent gases etc.
2. Nature of absorbent: – Activated charcoal is a better adsorbent than transitional metals.

Pressure: – Extent of physical adsorption increases as the pressure of the gas increases, till a saturation point is reached.

Freundlich gave the following relationship between extent of adsorption

Where:

and are constants, depends upon the nature of adsorbate as well as adsorbent.

(a) At low pressure: is directly proportional to the pressure

(b) At high pressure: – The extent of adsorption,

becomes independent of pressure i.e. a. PO

(c) At intermediate pressure: – will depend upon the power of

pressure which lies between 0 and 1

or (i)

On taking logarithm on both sides of

(i) We get

(ii)

Expression (i) and (ii) are called Freundlich adsorption isotherm.

Expression (ii) represents a straight line. The slope of the line will be while its intercept on the log

axis will be

Temperature: Adsorption is accompanied by evolution of heat i.e. H is negative, so the rate of adsorption should decrease with rise in temperature. It is found to be so in case of physical adsorption. The effect of temperature is represented by an adsorption isobar.

Activation of adsorbent-An adsorbent can be activated either by heating or by bringing it in finely divided state, or by making its surface rough by rubbing. For example, charcoal is activated by heating it in vacuum at

## Catalysis

The phenomena in which the rate of a reaction is altered (increased or decrease) by the presence of a substance (Catalyst) is known as catalysis.

Catalytic reactions are divided into two types.

(a) Homogeneous catalysis

(b) Heterogeneous catalysis

## Homogeneous catalysis:

When the reactants and the catalyst are in the same physical state, i.e. in the same phase, it is called homogeneous catalysis. For example.

(i) Lead chamber process: – In this process for the manufacture of sulphuric acid NO (gas) is used as a catalyst.

(ii) Inversion of cane sugar- In aqueous solution is catalysed by dilute acid (Hydrogen ions)

(b) Heterogeneous catalysis: When the catalyst and the reactants are not in the same physical state i.e. not in the same phase, it is called heterogeneous catalysis. for example.

(i) Decomposition of

(ii) Haber process for

## TYPES OF CATALYSIS

(i) Positive catalysts: –The substance which increases the rate of a reaction is known as a positive catalyst. It decreases the energy of activation for the reaction. For example:

(ii) Negative catalysts: The substance which decreases the rate of chemical reaction is called negative catalyst or inhibitor. It increases the activation energy for the reaction. For example.

(iii) Auto catalysts: –When one of the products of the reaction begins to act as a catalyst, it is called auto-catalyst for example. In the initial stages the reaction is slow but as soon as the products come into existence the the reaction rate increases

(iv) Induced catalyst:- When a chemical reaction enhances the rate of another chemical reaction it is called induced catalysis. For example:-

Soduim arsenite solution is not oxidised by air. If however, air is passed through a mixture of both of them undergo simultaneous oxidation. The oxidation of sodium sulphite, thus influences the oxidation of sodium arsenite.

Promoters: Those substance which do not themselves act as catalysts but their presence increases the activity of a catalyst are called catalytic promoters or catalyst for a catalyst. Example: In the Haber process for the synthesis of ammonia, Fe is the catalyst while molybdenum (Mo) acts as promoter.

Catalytic Poison: –The substance whose presence decreases or destroy the activity of a catalyst is called catalytic poison. For example:- Carbon monoxide or in hydrogen gas, acts as a poison for Fe catalyst in the Haber process for acts as poison for asbestos in contact process for

Inhibitors: – Those substances which retard rate of a chemical reaction are known as inhibitor. For example: glycerol or acetamide decrease the rate of decomposition of hydrogen peroxide.

General characteristics of catalysts:

(i) A catalyst remains unchanged in mass and chemical composition but change their physical state.

(ii) Only a very small amount of catalyst is sufficient to catalyse a reaction.

(iii) A catalyst does not initiate a reaction & does not controlled on chemical Rxn.

(iv) When a catalyst is a solid, it is usually more efficient where used in finely divided form.

(v) Generally, catalyst does not change the nature of products.

(vi) A catalyst does not change the equilibrium state of a reversible reaction but help to time achieve of equilibrium state. or position of equilibrium.

(vii) The catalysts are generally specific in nature.

(viii) Change rate constant of Rxn.

(ix) does not change free energy of Rxn.

THEORIES OF CATALYSIS

Intermediate compound formation theory: -This theory is explaining homogeneous catalysis mainly. According to this theory, the catalyst combines with one of the reactants to give an intermediate compound. This compound intermediately reacts with the other reactants and gives the product and regenerates the catalyst in its original form.

Thus, the reactants do not directly combine with each other, instead they react through the catalyst which provides an alternative pathway which involves lesser energy of activation.

For example: -The function of nitric oxide (NO) as a catalyst in the formation of is explained as follows.

Adsorption theory: – This theory is explaining the heterogeneous catalysis. The role of a solid catalyst in enhancing the reaction rate is explained on the basis of this theory in the following steps.

(i) The reactant molecules are absorbed on the surface of the catalyst at adjacent points. Adsorption leads to higher concentration of the adsorbed reactant on the surface of a catalyst.

(ii) As adsorption is an exothermic process, the heat of adsorption provides the necessary activation energy for the chemical reaction to proceed.

(iii) The adsorbed reactant molecules are tied on the solid sold surface of the catalyst. The bonds between the atoms of chemisorption  reactant molecules are weakened. The reactant molecules of sufficient energy combine together and with the surface of the catalyst to form surface activated complex.

This adsorbed activated complex is decomposed to form ‘products as a definite faster rate.

Catalysts in Industry: – Some of the important processes and their catalyst are given in below.

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Coagulation of Colloids | Methods | Associated Colloids for Class 12, IIT-JEE, NEET

Learn about Coagulation of Colloids, Electrophoresis, water in oil emulsion, oil in water emulsion, Associated Colloids and more. Get to know about the Colloid Protection here.

Charge on Colloidal Particles

Helmholtz Electrical Double Layer

Electrophoresis

Electro-osmosis

Coagulation of Colloids & its methods

Hardy Schulze Law

Protective Colloids

Emulsions

Associated Colloids

Gels

## Charge on colloidal particles

Colloidal particles always carry an electric charge. The mutual forces of repulsion between similarly charged particles prevent them from aggregating and settling under the action of gravity. This gives stability to the sole A list of common sols with the type of change on their particles is given ahead.

## Helmholtz Electrical Double Layer

The surface of a colloidal particle acquires a type by selective adsorption of a layer of positive ions around it. This layer attracts counter ions from the medium which form a second layer of -ve charges. The combination of the two layers of +ve and-ve charges around the sol particle was called Helmholtz double layer. According to modern view, the first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed as diffused layer.

The combination of the compact and diffused layer is referred as the stern double layer. The diffused layer is only loosely attached to the particles surface and moves in the opposite direction under an applied electric field. The potential difference between., the fixed layer and the diffused layer of opposite charge is called electro kinetic potential or zeta potential.

### Electrophoresis Meaning

If electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode, due to charge them, The movement of sol particles under an applied electric potential is called “Electrophoresis”, Depending upon the direction of movement of particles towards cathode, or anode electrophoresis can be called “cataphoresis’ or ‘Anaphoresis”.Electrophoresis provides and experimental proof to show that the colloidal particles are charged particles.

### Electro-osmosis

The medium will move in opposite direction to the dispersed phase under the influence of applied electric potential. The movement of dispersion medium under the influence of applied potential is known as ‘Electro – osmosis’.

## Coagulation of Colloids

We know that the stability of a lyophobic sol is due to the adsorption of positive or negative ions by the dispersed particles. The repulsion forces between the charged particles do not allow

them to settle. If somehow, the charge is removed there is nothing to keep the particles apart from each other. In such cases they aggregate or flocculate and settle down under the action of gravity the flocculation and settling down of the discharged sol particles is called coagulation or the precipitation can be brought about in four ways

### Methods of Coagulation

(b) By electrophoresis.

(c) By mixing two oppositely charged sols.

(d) By boiling.

### (a) By addition of electrolytes

When an electrolyte is added in excess to a sol, then the electrolyte furnishes both the type of ions in solution. The oppositely charged ions get adsorbed on the surface of colloidal particles this causes neutralization and there by the size and mass of colloidal particle increases and it becomes a suspension particle. Due to greater volume and greater mass these suspension particles settle down i.e. they coagulate. The ion responsible for neutralization of charge on the particle is called the flocculating ion,

### Hardy Schulze Law

states that the precipitating effect of an ion on dispersed phase of opposite charge increases with the valence of the ion.

The higher the valency of the flocculating ion, the greater is its precipitating power. Thus, for the precipitation of

solutions.

Similarly for precipitating sol (positive) the precipitating power of and is in the order

### FLOCCULATION VALUE

The minimum ,concentration of an electrolyte in milli moles per litre required to cause precipitation of a sol in 2 hours is called FLOCCULATION VALUE. The smaller the flocculating value, the higher will be the coagulating power of the ion.

### (b) By Electrophoresis

During electrophoresis the charged sol particles migrate towards the electrode of opposite sign. There they deposit their charge and then get coagulated (As the neutral particles can aggregate and change to suspension particles.

### (c) By mixing two oppositely charged sols-

The neutral coagulation of two sols of opposite charge can be affected by mixing them. For e.g. (positive) sol and Arsenious sulphide (negative sol) when mixed join and coagulate.

### (d) By boiling

Sols such as sulphur and silver halides disperse in water disperse in water, get coagulated when boiled due to increased collisions between sol particles and water molecules, which removes the adsorbed charged layer from the sol and therefore the sol particles settle down.

## Protective Colloids : Protection or Protective action.

Lyophobic sols are readily precipitated by small amounts of electrolytes. However, these sols often stabilized by the addition of Lyophilic sols.

The property of Lyophilic soles to prevent the precipitation or coagulation of a lyophobic salt is called protection. The Lyophilic sol used to protect a lyophobic sol from precipitation is referred to as a protective colloid. Lyophilic sols form a thin layer around lyophobic sol or around the ions furnished by electrolyte and therefore the coagulation cannot take place (as the size does not increase much). Gelatine, Albumen, Gum Arabia, Potato Starch are some of the examples of protective colloids.

The Lyophilic colloids differ in their protective power. The protective power is measured in terms of “Gold number introduced by Zsigmondy.

The number of milligrams of a hydrophilic colloid that will just prevent the precipitation of 10 of gold solon addition of 1 of 10 solution is known as gold number of that protector (Lyophilic colloid) On the onset of precipitation of the gold sol is indicated by a colour change from red to blue when the particle size just increases.

The smaller the gold number of a protective Lyophilic colloid, greater is its protection power.

Gold Number of some Lyophilic colloids

Protection Capacity

Gelatine and starch have the maximum and minimum protective powers.

## Emulsions

These are liquid-liquid colloidal systems. There are two types Emulsions

(i) Oil dispersed in water (o/w types)

(ii) Water Dispersed in oil (w/o types)

water in oil emulsion Examples
Oil in water Emulsion

In the first type water acts dispersion medium examples of this type of emulsions are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system oil acts as dispersion medium common examples of this type are butter and cream.

Emulsions of oil and water are unstable and sometimes they separate into two layers on standing. For stabilization an of an emulsion, a third component called emulsifying agent is usually added. The emulsifying agent form an inter facial film between suspended particles and the medium. The Principal agent for of emulsions are proteins, gums, soaps, etc. for w/o emulsion the principal emulsifying agents are heavy metal salts of fatty acids, long chain alcohols etc.

## Associated Colloids [Micelles]

Substances whose molecules aggregates to form particles of colloidal dimensions are called associated colloids.

The molecules of soaps and detergents are usually smaller than the colloidal particles. However, in concentrated solutions, these molecules associated and form aggregates of colloidal size. These aggregates of soaps or detergent molecules are called micelles. Soaps and detergents are strong electrolytes and gives ions when dissolved in water

The negative ions aggregate to form a micelle of colloidal size. The negative ion has a long Hydrocarbon chain and a polar groupin water it directed towards the center while the soluble polar head is on the surface in contact with water. The charge on the micelle is responsible for the stability of this system.

## Gels

A gel is a jelly like colloidal system in which a liquid is dispersed in a solid medium.

Gels may be classified into two types

(a) Elastic gels- These are those which possesses the property of elasticity. They change their shape on applying force and return to original shape when the force is removed. Gelatine, starch and soaps are examples substances which form elastic gels.

(b) Non- elastic gels- These are the gels which are rigid ego Silica gel. These are prepared by appropriate chemical action. Thus, silica gel is produced by adding concentrated Hydrochloric Acid to sodium silicate solution of the correct concentration.

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Colligative Properties | Brownian Movement for Class 12, IIT-JEE, NEET

## Colligative Properties Definition

These properties depend on the number of solute particles in solution. In case of colloidal solutions, colloidal particles are the aggregates of many ions or smaller molecules and when compared to true solutions or normal solutions, the total no of particles of solute in solution are very less and hence these solutions exhibit Colligative properties to lesser extent. Learn more about Properties of Colloidal Solution. This includes: Tyndal effect, Brownian Movement etc.

Optical Properties of Colloids with the help of Tyndall Effect Example.

Kinetic properties of Colloids through Brownian Movement.

## Properties of Colloidal Solutions

1. Heterogenous- Colloidal particles in a solution differ in sizes and are not homogenously distributed throughout the solution.

2. Visibility- Colloidal particles can not be seen with naked eyes or with the help of microscope. It is a well-known fact no particle is visible if its diameter is less than half the wavelength of light used. The visible light has greater wavelength than the size of colloidal particle.

3. Filterability- Colloidal particles pass through an ordinary filter paper but do not pass through parchment and other fine membranes.

4. Surface tension and viscosity – For Lyophobic sols, surface tension and viscosity are not very different from. From those of the medium, as there is very slight interaction between the suspended particles and the medium. On the other hand, Lyophilic sols show a high degree of solvation of the particles and therefore, the properties of the One medium are modified. Thus, the viscosity is much higher for the solution for the medium. Furthermore, the surface tension of the sol is lower than that of pure medium.

Colour- The colour hydrophobic sol depends on the wavelength of the light scattered by the dispersed particles. The wavelength of the scattered light again depends on the size and the nature of particles. For example, the colour of silver sol changes with the particle (suspended) diameter in solution.

## Optical properties

Solutions Exhibits Tyndal Effect.

When a beam of light is passed through a sol and viewed at right angles, the path of the light shows up as a hazy beam of cone ( lumited path by by bluish light). This was first observed by Faraday and later by Tyndal and in known as Tyndal effect. The bright cone of the light is called Tyndal cone. The tyndal effect-is due to the fact that the colloidal particles absorb light and scatter in all directions in space. The scatterise of light illuminates the path of the beam in the colloidal dispersion.

#### Some Examples of Tyndal Effect are

i. Blue colour of sky and sea water

ii. visibility of tails of comets

iii. Twinkling of stars

## Kinetic properties

Brownian Movement

When a sol is examined with an ultra-microscope, the suspended particles are seen as shining speaks of light. By following an individual particle, it is observed that the particle is in a state of continuous motion in zig-zag directions. The continuous rapid zig-zag motion of a colloidal particle in the dispersion medium is called “Brownian movement or motion (first observed by British botanist Robert Brown).

The Brownian movement has been explained to be due to the unbalanced bombardments of the particles by the molecules of dispersion medium

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Lyophilic and Lyophobic Colloids | Peptization for Class 12, IIT-JEE, NEET

Colloidal solutions in which the dispersed phase has considerable affinity for the dispersion phase, are called Lyophilic sols (solvent-liking). For example – dispersion of gelatin starch, gum and proteins in water. Colloidal solutions in which the dispersed phase has no affinity or attraction for the medium or for the solvent are called Lyophobic colloidal (Solvent heating) solutions. Here, Get to know about Lyophilic and Lyophobic Colloids:

Difference between Lyophilic and Lyophobic Colloids

Method of Preparation

Bredig’s Arc Method

Peptization Definition with Examples

## Methods of Preparation

Lyophilic sols may be prepared by simply warming the solid with liquid dispersion medium e.g. starch with water. On the other hand, Lyophobic sols have to be prepared by special methods. These methods fall into two categories

(a) Dispersion Methods is which larger macro sized particles are broken down to colloidal size.

(b) Condensation methods in which colloidal sized particles are built up by aggregating single ions or molecules.

This method is also known as condensation method.

## Electro-Dispersion (Bredig’s arc method)

This method is suitable for the preparation of colloidal solutions of metals like gold, silver, platinum etc, An arc is struck between the metal electrodes under, platinum etc, An arc is struck between the stabilizing agent such as a trace of KOH. The water is cooled by immersing the container in a ice bath. The intense heat of the arc vaporizes some of the metal which condenses under cold water.

Note:

1. This method is not suitable when the dispersion medium is an organic liquid as considerable charring occurs.

2. This method comprises both dispersion and condensation.

## Peptization Definition

The dispersion of a freshly precipitated material into colloidal solution by the action of an electrolyte in solution is termed Peptization. The electrolyte used is called a Peptizing agent.

#### A few examples of sols obtained by Peptization are:

(i) Freshly prepared ferric hydroxide on treatment with a small amount of ferric chloride solution once forms a dark reddish-brown solution. Ferric chloride acts as a Peptizing agent.

(ii) Freshly prepared stannic oxide on treatment with a small amount of dilute Hydrochloric acid forms a stable colloidal solution of stannic oxide.

(iii) Freshly precipitated silver chloride can be converted into a colloidal solution by adding a small amount of Hydrochloric acid.

(iv) Cadmium sulphide can be peptized with the help of hydrogen sulphide.

The process of peptization thus involves the adsorption of suitable ions (supplied by the electrolyte added particularly a common ion) and electrically charged particles than split from the precipitate as colloidal particles.

Chemical method: The chemical method involves chemical reactions in a medium in which the dispersed phase. is sparingly soluble. A condition of super-saturation is produced but the actual precipitation is avoided. Some familiar reactions used are:

(a) Double decomposition

(i) Arsenious sulphide sol: A 1 solution of arsenious oxide is prepared in hot water. The solution is cooled, filtered and is then gradually added to water saturated with hydrogen sulphide. whilst a stream of is being passed through the solution

This is continued till an intense yellow colored solution is obtained. Excess of is removed by bubbling hydrogen through the solution.

(b) Oxidation – A colloidal solution of Sulphur is obtained by passing into a solution of Sulphur dioxide.

Sulphur sol can also be obtained when is bubbled through an oxidizing agent (bromine water or Nitric acid)

(c) Reduction – Colloidal solutions of metals like gold, silver, platinum, lead etc. can be obtained when their salts solutions are act upon by reducing agents.

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Colloidal Solution Definition | Examples | Types for Class 12, IIT-JEE, NEET

# Introduction

According to colloidal solution definition, it is defined as a solution in which a material is evenly suspended in a liquid. Some of the Examples of Colloidal Solution are gelatin; muddy water, Butter, blood, Colored Glass. Know the various types of colloidal solution and difference between true solution and colloidal solution.

Colloidal Solution & its Examples

Comparison of Suspension, Colloidal Solution & True Solution

Types of Colloidal Solution

The foundation of colloid chemistry was Laid by an English scientist, Thomas Graham in 1861.

### Definition of Colloidal Solution:

A colloidal solution, generally identified as a colloidal suspension, is a mixture in which the substances are regularly suspended in a fluid. While colloidal systems can occur in any one of the three main states of matter, solid, liquid, or gas, a colloidal solution unambiguously refers to a liquid mixture.

Examples of Colloidal Solution

### Types of Colloidal solutions

A colloidal system comprises of two phases. The substance distributed as the colloidal particles is called Dispersed phase the internal phase or the discontinuous phase. Dispersion medium is defined as the second continuous phase in which the colloidal particles are dispersed. For example, for a colloidal solution of copper in water, copper particles constitute the dispersed phase and water the dispersion medium. Depending on the physical states of dispersed phase or dispersion medium, colloidal solutions are eight types

A colloidal dispersion of one gas in another is not possible since the two gases would give a homogeneous molecular structure.

We restrict our study mainly to colloidal systems which consist of a solid substance dispersed in a liquid. These are frequently referred to as sols or colloidal solutions, the colloidal solutions in water as the dispersion medium are termed hydrosols or Aqua sols. When the dispersion medium is alcohol or benzene, the sols are referred to as Alcohols and Benzenols respectively.

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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,

* 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.

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CHEMISTRY

CHEMICAL KINETICS

THEORY

It is a branch of physical chemistry deals with the “Rate of Chemical Reactions” including the effect of temperature, pressure, concentration, etc., on the rates, and the mechanism by which the reaction takes place.

Rate of chemical Reaction is defined as the change in concentration of a reactant (or a product) in a particular time interval. Average rate of reaction, Instantaneous rate of reaction.

Units of Reaction Rate are unit of concentration divided by the unit of time $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1} \text { or } \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1} \text { or so on }\right)$

Factors Affecting Reaction Rates :

(i) Concentration of reactants and

(ii) Reaction temperature

Besides these, presence of catalyst and surface area (if a reactant or a catalyst is a solid) exposure to radiation also affect the reaction rates.

Expressions or the rate :

For a general reaction $: a A+b B \longrightarrow c C+d D$

The rate of disappearance of A $=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}$; Rate of disappearance of B $=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$ ;

Rate of appearance of C = $\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}$ & Rate of appearance of D = $\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}$.

The positive sign shows that concentrations of C and D increases with time and the negative sign indicating that concentrations of A and B decrease with time.Thus the rate of general reaction.

Rate Equation and Rate constant :

An expression which relates the rate of a reaction to the concentration of the reactants is called the Rate Equation or Rate Law. Rate $\propto[\mathrm{A}]^{\mathrm{a}} \cdot[\mathrm{B}]^{\mathrm{b}}$ or Rate = Rate $=\mathrm{k}[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}$ . The constant of proportionality, k is known as the Rate Constant (specific reaction rate) and may be defined as the rate at unit concentrations of the reactants. k depends on the temperature and is independent of the initial concentrations of the reactants. At a fixed temperature, k is constant characteristic of the reaction. Larger value of k indicates fast reaction and small k indicates slow reactions.

Molecularity :

Molecularity of a reaction is defined as the numbers of particles (atoms, ions, groups or molecules) of reactants actually taking part in a single step chemical reaction.

Molecularity of a reaction is :

(i) Always a whole number (not zero) and never a fraction.

(ii) The value of molecularity of a simple or one step reaction does not exceed 3.

Order of Reaction :

It is defined as the sum of the exponents (powers) of the molar concentrations of the reactants in the experimentally determined rate equations.

If rate of reaction a $[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}[\mathrm{C}]^{\mathrm{r}}$ or Rate of reaction $=\mathrm{k}[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}[\mathrm{C}]^{\mathrm{r}}$

order of reaction = p + q + r & the order w.r.t. A, B & C are p, q & r respectively.

For a “Reaction of nth order”, the order of the reaction is n and the rate equation (or Rate law) is rate $\propto[\mathrm{A}]^{\mathrm{n}}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}$

The order of a reaction is obtained from the experimentally determined rate (and not from the stoichiometric equation) and may be zero, an integer or a fraction and never exceeds 3. In a multi-step complex reaction, the order of the reaction depends on the slowest step.

Zero order reaction :

A reaction is said to be of zero order if the rate is independent of the concentration of the reactants.

$A \longrightarrow$ products $;$ Rate $\alpha \mathrm{k}[\mathrm{A}]^{\circ}=\mathrm{k} \operatorname{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}$

Examples :

(i) $\quad \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \stackrel{\mathrm{hv}}{\longrightarrow} 2 \mathrm{HCl}(\mathrm{g})$

(ii) $\quad \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \frac{\text { hot Pt. }}{\text { Surface }} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$

(iii) $2 \mathrm{NH}_{3}$ (g) $\frac{\text { Mo or } \mathrm{W}}{\text { surface }} \mathrm{N}_{2}+3 \mathrm{H}_{2}$

(iv) $\quad 2 \mathrm{HI}(\mathrm{g}) \frac{\mathrm{Au}}{\text { surface }} \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$

Characteristics of Zero Order Reaction :

(1)Concentration of reactant decreases lineraly with time. $[\mathrm{A}]_{\mathrm{t}}=[\mathrm{A}]_{0}-\mathrm{kt}$

(2) Units of k are, mol $l^{-1} \operatorname{time}^{-1}$

(3) Time required for the completion of reaction t = $\frac{[\mathrm{A}]_{0}}{\mathrm{k}} \& \mathrm{t}_{1 / 2}=\frac{0.5[\mathrm{A}]_{0}}{\mathrm{k}}$

First order reaction :

a reaction is said to be of first order if its rate is proportinal to the concentration of one reactant only.

Characteristics Of First Order Reaction :

Examples :

(i) Radioactive disintegration is a first order reaction.

(iii) Mineral acid catalyzed hydrolysis of esters.

(iv) Decomposition of $H_{2} O_{2}$ in aqueous solution.

Second Order Reaction :

(i) When two molecules of the same reactant are involved or the concentrations of the both reactants are equal reactions $2 \mathrm{A} \longrightarrow$ products or $\mathrm{A}+\mathrm{B} \longrightarrow$ products.

Characteristics of Second Order Reaction :

(i) Unit of rate constant L $\mathrm{mol}^{-1}$ time $^{-1}$

(ii) Numerical value of k will depend upon unit of concentration.

(iii) $\left.\mathrm{t}_{1 / 2} \propto \mathrm{a}^{-1} \text { (In general } \mathrm{t}_{1 / 2} \alpha \mathrm{a}^{(1-\mathrm{n})} ; \mathrm{n}=\text { order of reactions }\right)$

(iv) 2nd order reaction conforms to first order when one of the reactant in excess.

Examples :

Side or concurent reaction :

Consecutive reaction :

Threshold Energy and Activation Energy :

For a reaction to take place the reacting molecules must colloid together, but only those collisions, in which colliding molecules possess certain minimum energy is called threshold energy $\left(\mathrm{E}_{\mathrm{T}}\right)$.

Activation Energy $\left(\mathbf{E}_{\mathbf{a}}\right):$

The extra energy needed for the reactant molecules to be able to react chemically is known as Activation energy.

Influence of Temperature on reaction rates :

Temperature coefficient :

The temperature coefficient of a chemical reaction is defined as the ratio of the reaction rates at two temperatures differing by $10^{\circ} \mathrm{C}$. Its value usually lies between 2 & 3.

Temperature coefficient $=\frac{\mathrm{k}_{\mathrm{t}+10}}{\mathrm{k}_{\mathrm{t}}}$

Arrhenius Equation :

Graphical representations are :

Methods of Determination of order of reactions :

A few methods commonly used are given below :

1. Hit & Trial Method : It is method of using integrated rate equations, where the experimental values of a, x & t are put into these equations. One which gives a constant value of k for different sets of a, x & t correspond to the order of the reaction.

2. Graphical Method :

(i) A plot of log (a – x) versus ‘t’ gives a straight lines for the First order reaction.

(ii) A plot of (a – x)– (n–1) versus ‘t’ gives a straight line any reaction of the order n (except n = 1)

3.Half Life Method : The half life of different order of reactions is given by $\mathrm{a}_{\mathrm{n}}=\left(\frac{1}{2}\right)^{\mathrm{n}} \mathrm{a}_{0}$

By experimental observation of the dependence of half life on initial concentration we can determine n, the order of reaction. N $=1+\frac{\log \mathrm{t}_{2}-\log \mathrm{t}_{1}}{\log \mathrm{a}_{1}-\log \mathrm{a}_{2}}$

4. Initial rate method. Initial rate method is used to determine the order or reaction in cases where more than one reactant is used. It involves the determination of the order of different reactants separately. A series of experiments are performed in which concentration of one particular reactant is varied whereas conc. of other reactants are kept constant. In each experiment the initial rate is determined from the plot of conc. vs. time, e.g., if conc. of A is doubled, and initial rate of reaction is also doubled, order of reaction is l.

Mechanism Of reactions :

The path way which reactants are converted into the products is called the reaction mechanism. It should be clear that experimentally determined rate expression cannot be predicted from the stiochiometry of the reaction. For example for the reaction ;

The reason is that the reaction occurs by a series of elementary steps.

The sequence of elementary processes leading to the overall stiochiometry is known as the “Mechanism of the reaction”. An in a sequence of reactions leading to the formation of products from reactants, the slowest step is the rate determining step.

The mechanism proposed for the above reaction is a two step one.

The sum of the two gives the stiochiometry & the slow step decided the rate expression.

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 Electrochemical cells

An electrochemical cell consists of two electrodes (metallic conductors) in contact with an electrolyte (an ionic conductor).

An electrode and its electrolyte comprise an Electrode Compartment. Electrochemical Cells can be classified as:

(i) Electrolytic Cells in which a nonspontaneous reaction is driven by an external source of current.

(ii) Galvanic Cells which produce electricity as a result of a spontaneous cell reaction

Note : In a galvanic cell, cathode is positive with respect to anode.In a electrolytic cell, anode is made positive with respect to cathode.

Electrolysis

The decomposition of electrolyte solution by passage of electric current, resulting into deposition of metals or liberation of gases at electrodes is known as electrolysis.

Electrolytic Cell

This cell converts electrical energy into chemical energy. The entire assembly except that of the external battery is known as the electrolytic cell

Electrodes

The metal strip at which positive current enters is called anode; anode is positively charged in electrolytic cell. On the other hand, the electrode at which current leaves is called cathode. Cathodes are negatively charged.

Electrolysis of Molten sodium chloride

$\mathrm{NaCl}(\mathrm{molten}) \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$

There are two types of electrodes used in the electroytic cell, namely attackable and non – attackable. The attackable elecrodes participitate in the electrode reaction. They are made up of reactive metals like Zn, Cu, Ag etc. In such electrodes, atom of the metal gets oxidised into the corresponding cation, which is passed into the solution. Thus, such anodes get dissolved and their mass decreases. On the other hand, non-attackable electrodes do not participiate in the electrode reaction as they made up of unreactive elements like Pt, graphite etc. Such electrodes do not dissolve and their mass remain same.

(i) First law of electrolysis :

Amount of substance deposited or liberated at an electrode is directly proportional to amount of charge passed (utilized) through the solution.

$\mathrm{w} \propto \mathrm{Q}$

W = weight liberated, Q = charge in coulomb

w = ZQ

Z = electrochemical equivalent

when Q = 1 coulomb, then w = Z

Thus, weight deposited by 1 coulomb charge is called electrochemical equivalent.

Let 1 ampere current is passed till ‘t’ seconds .

Then, Q = It w = ZIt

1 Faraday = 96500 coulomb = Charge of one mole electrons

One faraday is the charge required to liberate or deposit one gm equivalent of a substance at corresponding electrode.

Let ‘E’ is equivalent weight then ‘E’ gm will be liberated by 96500 coulomb.

$\therefore 1$ Coulomb will liberate $\frac{\mathrm{E}}{96500} \mathrm{gm} ;$ By definition, $Z=\frac{\mathrm{E}}{96500}$

$\therefore W=\frac{I t E}{96500}$

When a gas is evolved at an electrode, then above formula changes as,

$\mathrm{V}=\frac{\mathrm{ItV}_{\mathrm{e}}}{96500}$

where V = volume of liberated gas, $\mathrm{V}_{\mathrm{e}}$ = equivalent volume of gas.

Equivalent volume may be defined as:

The volume of gas liberated by 96500 coulomb at STP.

(ii) Second law of electrolysis :

When same amount of charge is passed through different electrolyte solutions connected in series then weight of substances deposited or dissolved at anode or cathode are in ratio of their equivalent weights. i.e.

$\mathrm{w}_{1} / \mathrm{w}_{2}=\mathrm{E}_{1} / \mathrm{E}_{2}$

QUALITATIVE ASPECTS OF ELECTROLYSIS

In the electrolysis process we have discussed above, we have taken molten salt as electrolyte, which contains only one cation and anion. Now, if the electrolyte taken contains more than one cation and anion (for example, aqueous solution of the ionic electrolyte), then the cation and anion that will get discharged depends on the ability of cation to get reduced and the ability of anion to get oxidised.

The ability of an ion to get oxidised or reduced depends upon the size, mass, positive charge, negative charge etc. Thus, it is not possible to predict qualitatively that which ion would be discharged first, a one factor might enhance the ability to discharge while the other factor may hamper it. This can only be predicted on the basis of quantitative value assigned based on the cumulative effect of all the factors reponsible for an ion’s ability to discharge. The value is referred as standard potential, which is determined by keeping the concentration of ion as 1 M, pressure of gas at 1 atm, and the measurement done at $25^{\circ} \mathrm{C}$. For a cation, the standard reduction potential (SRP) values are compared. The cation having higher standard reduction potential value is discharged in preference to cation with lower SRP value provided the ions are at 1 M concentration. For an anion, the standard oxidation potential (SOP) values are compared and anion having higher SOP is preferentially discharged, if the concentration is 1 M for each of the ion. The SRP values at $25^{\circ} \mathrm{C}$for some of the reduction half reactions are given in the table below.

When solution of an electroyte contains more than one type of cations and anions at concentrations different than 1 M, the discharge of an ion does not depend solely on standard potentials but also depends on the concentration of ion in the solution. This value is refered as potential, called as reduction potential for cation and oxidation potential for anion. The relation between reduction potential and standard reduction potential is given by Nernst equation, as

$\mathrm{E}_{\mathrm{RP}}=\mathrm{E}_{\mathrm{RP}}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{[\text { concentration of product }]}{[\text { concentration of reac tan } \mathrm{t}]}$

where $E_{R P}$ = Reduction potential of cation and $\mathrm{E}^{\mathrm{O}} \mathrm{RP}$= Standard reduction potential of cation.

Thus, it is possible that a cation (A+) with lower standard reduction potential getting discharged in preference to cation $\left(\mathrm{B}^{+}\right)$ having higher standard reduction potential because their concentration might be such that the reduction potential of $\mathrm{A}^{+}$ is higher than that of $\mathrm{B}^{+}$.

When two metal ions in the solution have identical values of their reduction potentials, the simultaneous deposition of both the metals will occur in the form of an alloy.

Galvanic Cell

This cell converts chemical energy into electrical energy.

Galvanic cell is made up of two half cells i.e., anodic and cathodic. The cell reaction is of redox kind. Oxidation takes place at anode and reduction at cathode. It is also known as voltaic cell. It may be represented as shown in Fig. Zinc rod immersed in $\mathrm{ZnSO}_{4}$ behaves as anode and copper rod immersed in $\mathrm{CuSO}_{4}$ behaves as cathode.

Oxidation takes place at anode:

$\mathrm{Zn}^{3 / 4} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$ (loss of electron : oxidation)

Reduction takes place at cathode:

$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\text { gain of electron } ; \text { reduction })$

Over all process:

$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}$

In galvanic cell like Daniell cell; electrons flow from anode (zinc rod) to the cathode (copper rod) through external circuit; zinc dissolves as $\mathrm{Zn}^{2+} ; \mathrm{Cu}^{2+}$ion in the cathode cell picks up two electron and become deposited at cathode.

Salt Bridge

Two electrolyte solutions in galvanic cells are seperated using salt bridge as represented in the Fig. salt bridge is a device to minimize or eliminate the liquid junction potential. Saturated solution of salt like KCI, $\mathrm{KNO}_{3}$, $\mathrm{NH}_{4} \mathrm{Cl}$ and $\mathrm{NH}_{4} \mathrm{NO}_{3}$ etc. in agar-agar gel is used in salt bridge. Salt bridge contains high concentration of ions viz. $\mathrm{K}^{+}$ and $\mathrm{NO}_{3}^{-}$at the junction with electrolyte solution. Thus, salt bridge carries whole of the current across the boundary ; more over the $\mathrm{K}^{+}$ and $\mathrm{NO}_{3}^{-}$ ions have same speed. Hence, salt bridge with uniform and same mobility of cations and anions minimize the liquid junction potential & completes the electrical circuit & permits the ions to migrate.

Representation of a cell (IUPAC conventions ): Let us illustrate the convention taking the example of Daniel cell.

(i) Anodic half cell is written on left and cathodic half cell on right hand sid

$\mathrm{Zn}(\mathrm{s})\left|\mathrm{ZnSO}_{4}(\mathrm{sol}) \| \mathrm{CuSO}_{4}(\mathrm{sol})\right| \mathrm{Cu}(\mathrm{s})$

(ii) Two half cells are separated by double vertical lines: Double vertical lines indicate salt bridge or any type of porous partition.

(iii) EMF (electromotive force) may be written on the right hand side of the cell.

(iv) Single vertical lines indicate the phase separation between electrode and electrolyte solution.

$\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}$

Concept of Electromotive force (EMF) of a Cell

Electron flows from anode to cathode in external circuit due to a pushing effect called or electromotic force (e.m.f.). E.m.f. is some times called as cell potential. Unit of e.m.f. of cell is volt.

EMF of cell may be calculated as :

$E_{\operatorname{cell}}$ reduction potential of cathode – Reduction potential of anode

Similarly, standard e.m.f. of the cell $\left(\mathrm{E}^{\circ}\right)$ may be calculated as

$\mathrm{E}_{\text {cell }}^{\circ}$ = Standard reduction potential of cathode – Standard reduction potential of anode

Sign Convention of EMF

EMF of cell should be positive other wise it will not be feasible in the given direction .

$\begin{array}{ll}{\mathrm{Zn}\left|\mathrm{ZnSO}_{4} \| \mathrm{CuSO}_{4}\right| \mathrm{Cu}} & {\mathrm{E}=+1.10 \text { volt } \quad \text { (Feasible) }} \\ {\mathrm{Cu}\left|\mathrm{CuSO}_{4} \| \mathrm{ZnSO}_{4}\right| \mathrm{Zn}} & {\mathrm{E}=-1.10 \text { volt } \quad \text { (Not Feasible) }}\end{array}$

Nernst Equation

Walter Nernst derived a relation between cell potential and concentration or Reaction quotient.

$\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{R} \mathrm{T} \ln \mathrm{Q}$ $\ldots .(1)$

where $\Delta \mathrm{G}$ and $\Delta \mathrm{G}^{\circ}$are free energy and standard free energy change; ‘Q’ is reaction quotient.

Let n, Faraday charge is taken out from a cell of e.m.f. (E) then electrical work done by the cell may be calculated as,

Work done = Charge x Potential = nFE

From thermodynamics we know that decrease in Gibbs free energy of a system is a measure of reversible or maximum obtainable work by the system if there is no work due to volume expans

$\therefore-\Delta \mathrm{G}=n \mathrm{FE}$ and $-\Delta \mathrm{G}^{\circ}=n \mathrm{FE}^{\circ}$

Thus from Eq. (i), we get $-n \mathrm{FE}=-n \mathrm{FE}^{\circ}+\mathrm{RT} \ln \mathrm{Q}$

At $25^{\circ} \mathrm{C},$ above equation may be written as $\mathbf{E}=\mathbf{E}^{\mathbf{0}}-\frac{0.0591}{\mathbf{n}} \log \mathbf{Q}$

Where ‘n’ represents number of moles of electrons involved in process.

E, E° are e.m.f. and standard e.m.f. of the cell respectively.

In general , for a redox cell reaction involving the transference of n electrons

$\mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC}+\mathrm{dD},$ the EMF can be calculated as:

$\mathrm{E}_{\mathrm{Cell}}=\mathrm{E}_{\mathrm{Cell}}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}$

Prediction and feasibility of spontaniety of a cell reaction.

Work done by the cell = nFE;

It is equivalent to decrease in free energy $\Delta \mathrm{G}=-\mathrm{nFE}$

Under standard state $\Delta \mathrm{G}^{0}=-\mathrm{nFE}^{0}$ ….. (i)

(i) From thermodynamics we know, $\Delta$G = negative for spontaneous process. Thus from eq.(i) it is clear that the EMF should be +ve for a cell process to be feasible or spontaneous.

(ii) When $\Delta \mathrm{G}$ = positive, E = negative and the cell process will be non spontaneous.

Standard free energy change of a cell may be calculated by electrode potential data.

Substituting the value of $\mathrm{E}^{0}$ (i.e., standard reduction potentialof cathode- standard reduction potential of anode) in eq. (i) we may get $\Delta \mathrm{G}^{0}$

Let us see whether the cell (Daniell) is feasible or not: i.e. whether Zinc will displace copper or not.

$\mathrm{Zn}|(\mathrm{s})| \mathrm{ZnSO}_{4}(\mathrm{sol}) \| \mathrm{CuSO}_{4}(\mathrm{sol}) | \mathrm{Cu}(\mathrm{s})$

$\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{volt} ; \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{volt}$

$\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}-\mathrm{E}_{\mathrm{zn}^{2+} / \mathrm{Zn}}^{0}$

$=0.34-(-0.76)=+1.10 \mathrm{volt}$

Since $\mathrm{E}^{0}=+\mathrm{ve}$, hence the cell will be feasible and zinc will displace copper from its salt solution. In the other words zinc will reduce copper.

8 Thermodynamic treatment of Nernst equation

Determination of equilibrium constant : We know, that

$\mathrm{E}=\mathrm{E}^{0}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Q}$ ..(i)

At equilibrium, the cell potential is zero because cell reactions are balanced, i.e. E = 0

$\therefore$ From Eq. (i), we have

$0=\mathrm{E}^{0}-\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}} \quad$ or $\quad \mathrm{K}_{\mathrm{eq}}=$ antilog $\left[\frac{n \mathrm{E}^{0}}{0.0591}\right]$

Heat of Reaction inside the cell: Let n Faraday charge flows out of a cell of e.m.f. E, then

$-\Delta \mathrm{G}=n \mathrm{FE}$ ……. (i)

Gibbs Helmholtz equation (from thermodynamics ) may be given as,

$\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ ……….. (ii)

From Eqs. (i) and (ii), we have

$-\mathrm{nFE}=\Delta \mathrm{H}+\mathrm{T}\left[\frac{\partial(-\mathrm{nFE})}{\partial \mathrm{T}}\right]_{\mathrm{p}}=\Delta \mathrm{H}-\mathrm{nFT}\left[\frac{\partial \mathrm{E}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$

$\therefore \Delta \mathrm{H}=-\mathrm{nFE}+\mathrm{nFT}\left[\frac{\partial \mathrm{E}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$

Entropy change inside the cell : We know that G = H – TS or DG = DH – TDS …(i)

where $\Delta \mathrm{G}$ = Free energy change ; DH = Enthalpy change and DS = entropy change.

According to Gibbs Helmoholtz equation,

$\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$

$\Delta \mathrm{G}=\Delta \mathrm{H}=\mathrm{T}\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$

From Eqs. (i) and (ii), we have

$-\mathrm{T} \Delta \mathrm{S}=\mathrm{T}\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ or $\quad \Delta \mathrm{S}=-\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$

$-\mathrm{T} \Delta \mathrm{S}=\mathrm{T}\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ or $\quad \Delta \mathrm{S}=-\left[\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$

where $\left[\frac{\partial \mathrm{E}}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ is called temperature coefficient of cell e.m.f.

8 Different Types of Half-Cells And Their Reduction Potential

(1) Gas-Ion Half Cell:

In such a half cell, an inert collector of electrons, platinum or graphite is in contact with gas and a solution containing a specified ion. One of the most important gas-ion half cell is the hydrogen-gas-hydrogen ion half cell. In this half cell, purified $\mathrm{H}_{2} \mathrm{gas}$ at a constant pressure is passed over a platinum electrode which is in contact with an acid solution.

$\mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightleftharpoons 1 / 2 \mathrm{H}_{2}$

$\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}=\mathrm{E}^{0}_{\mathrm{H}^{+} / \mathrm{H}_{2}}-\frac{0.0591}{1} \log \frac{\left(\mathrm{pH}_{2}\right)^{1 / 2}}{\left[\mathrm{H}^{+}\right]}$

(2) Metal-Metal Ion Half Cell:

This type of cell consist of a metal M in contact with a solution containing $M^{n+}$ ions.

$\mathrm{M}^{\mathrm{n}+}(\mathrm{aq})+\mathrm{ne}^{-} \rightleftharpoons \mathrm{M}(\mathrm{s})$

$\mathrm{E}_{\mathrm{M}^{\mathrm{n}+} / \mathrm{M}}=\mathrm{E}_{\mathrm{M}^{\mathrm{n}+} / \mathrm{M}}^{0}-\frac{0.0591}{\mathrm{n}} \log \frac{1}{\left[\mathrm{M}^{\mathrm{n}+}\right]}$

(3) Metal-Insoluble Salt – Anion Half Cell:

In this half cell, a metal coated with its insoluble salt is in contact with a solution containing the anion of the insoluble salt. eg. Silver-Silver Chloride Half Cell:

This half cell is represented as CI’AgCl/Ag. The eq

$\mathrm{AgCl}(\mathrm{s})+\mathrm{e}^{-} \rightleftharpoons \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}(\text { aq })$

potential of such cells depends upon the concentration of anions. Such cells can be used as Reference Electrode.

(4) Oxidation-reduction Half Cell:

This type of half cell is made by using an inert metal collector, usually platinum, immersed in a solution which contains two ions of the same element in different states of oxidation. eg. $\mathrm{Fe}^{2+}-\mathrm{Fe}^{3+}$ half cell.

$\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightleftharpoons \mathrm{Fe}^{2+}(\mathrm{aq})$

$\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}-\frac{0.0591}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}$

8 Concentration cell

The cells in which electrical current is produced due to transport of a substance from higherto lower concentration. Concentration gradient may arise either in electrode material or in electrolyte. Thus there are two types of concentration cell .

(i) Electrode concentration cell

(ii) Electrolyte concentration cell

Electrode Gas concentration cell :

$\mathrm{Pt}, \mathrm{H}_{2}\left(\mathrm{P}_{1}\right)\left|\mathrm{H}^{+}(\mathrm{C})\right| \mathrm{H}_{2}\left(\mathrm{P}_{2}\right), \mathrm{Pt}$

Here, hydrogen gas is bubbled at two different partial pressures at electrode dipped in the solution of same electrolyte.

Anode process: $1 / 2 \mathrm{H}_{2}\left(\mathrm{p}_{1}\right) \rightarrow \mathrm{H}^{+}(\mathrm{c})+\mathrm{e}^{-}$

Cathode process $\quad \frac{\mathrm{H}^{+}(\mathrm{c})+\mathrm{e}^{-} \rightarrow 1 / 2 \mathrm{H}_{2}\left(\mathrm{p}_{2}\right)}{1 / 2 \mathrm{H}_{2}\left(\mathrm{p}_{1}\right) \rightarrow 1 / 2 \mathrm{H}_{2}\left(\mathrm{p}_{2}\right)}$

$\therefore \quad \mathrm{E}=-\frac{2.303 \mathrm{RT}}{\mathrm{F}} \log \left[\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right]^{1 / 2}$

or $\quad \mathrm{E}=\left[\frac{2.303 \mathrm{RT}}{2 \mathrm{F}}\right] \log \left[\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right], \operatorname{At} 25^{0} \mathrm{C}, \mathrm{E}=\frac{0.059}{2 \mathrm{F}} \log \left[\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right]$

For spontanity of such cell reaction, $\mathrm{p}_{1}>\mathrm{p}_{2}$

Electrolyte concentration cells:

$\mathrm{Zn}(\mathrm{s})\left|\mathrm{ZnSO}_{4}\left(\mathrm{C}_{1}\right) \| \mathrm{ZnSO}_{4}\left(\mathrm{C}_{2}\right)\right| \mathrm{Zn}(\mathrm{s})$

In such cells, concentration gradient arise in electrolyte solutions. Cell process may be givewn a

$\mathrm{zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}\left(\mathrm{C}_{1}\right)+2 \mathrm{e} \quad$ (Anodic process)

$\frac{\mathrm{Zn}^{2+}\left(\mathrm{C}_{2}\right)+2 \mathrm{e} \rightarrow \mathrm{Zn}(\mathrm{s})}{\mathrm{Zn}^{2+}\left(\mathrm{C}_{2}\right) \rightarrow \mathrm{Zn}^{2+}\left(\mathrm{C}_{1}\right)}$ (Over all process)

$\therefore$ From Nernst equation, we have

$\mathrm{E}=0-\frac{2.303 \mathrm{RT}}{2 \mathrm{F}} \log \left[\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right]$ OR $\mathrm{E}=\frac{2.303 \mathrm{RT}}{2 \mathrm{F}} \log \left[\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\right]$

For spontanity of such cell reaction, $\mathrm{C}_{2}>\mathrm{C}_{1}$

8 CONDUCTANCE

Introduction:

Both metallic and electrolytic conductors obey Ohm’s law

i.e. V = IR

where V = Potential difference in volt; I = Current in ampere ; R = resistance in Ohm

We know, resistance is directly proportional to length of conductor and inversely proportional to cross sectional area of the conductor.

$\operatorname{R\propto} \frac{l}{\mathrm{A}} \quad$ or $\mathrm{R}=\rho \frac{l}{\mathrm{A}}$ (= Specific resistance )

Specific resistance is the resistance of a conductor having lengths of 1 cm and corss sectional area of 1cm2.

Unit of R is ohm and unit of specific resistance is ohm cm

Reciprocal of resistance is called as conductance and reciprocal of specific resistance is called as specific conductance.

$\frac{1}{\mathrm{R}}=\frac{1}{\rho} \frac{\mathrm{A}}{l}$ or $\mathrm{C}=\mathrm{K} \frac{\mathrm{A}}{l}$

where C = conductance ohm $^{-1}$; K = specific conductance ohm $^{-1} \mathrm{cm}^{-1}$

Mho and siemens are other units of conductance.

$\mathrm{K}=\frac{l}{\mathrm{A}} \mathrm{C}$

Specific conductance= Cell constant x Conductance

Specific conductance is conductance of 1 $\mathbf{C} \mathbf{M}^{3}$ of an electrolyte solution.

In case of electrolytic solution, the specific conductance is defined as the conductance of a solsution of definite concentration enclosed in a cell having two electrodes sof unit area separated by 1 cm apart.

1. Equivalent Conductance

Equivalent conductance is the conductance of an electrolyte solution containing 1 gm equivalent of electrolyte. It is densoted by $\wedge$ .

$\wedge$ = K x V

$\left(\wedge=\operatorname{ohm}^{-1} \mathrm{cm}^{-1} \mathrm{x} \mathrm{cm}^{3}=\mathrm{ohm}^{-1} \mathrm{cm}^{2}\right)$

Usually concentration of electrolyte solution is expressed as C gm equivalent per litre.

Thus, $\mathrm{V}=\frac{1000}{\mathrm{C}}$

2. Molar Conductance

Molar conductance may be defined as conductance of an electrolyte solution having 1 gm mole electrolyte in a litre. It is denoted by $\wedge_{\mathrm{m}}$ .

$\wedge_{\mathrm{m}}=\mathrm{K} \times \mathrm{V}$

Usually concentration of electrolyte solution is expressed as ‘M’ gm mole elecrtrolyte per litre.

The $\dot{\mathrm{U}}_{\mathrm{m}}$ vs $\sqrt{\mathrm{C}}$ plot of strong electrolyte being linear it can be extrapolated to zero concentration. Thus, $\dot{\mathrm{U}}_{\mathrm{m}}$ values of the solution of the test electrolyte are determined at various concentrations the concentrations should be as low as good.

$\dot{\mathrm{U}}_{\mathrm{m}}$ values are then plotted against $\sqrt{\mathrm{C}}$ when a straight line is obtained. This is the extrapolated to zero concentration. The point where the straight line intersects $\dot{\mathrm{U}}_{\mathrm{m}}$ axis is of the strong electrolyte.

However, the plot in the case weak electrolyte being non linear, shooting up suddenly at some low concentration and assuming the shape of a straight line parallel to Ùm axis. Hence extrapolation in this case is not possible. Thus, $\dot{\mathrm{U}}_{0}$of a weak electrolyte cannot be determined experimentally. It can, however, be done with the help of Kohlrausch’s law to be discussed later.

8 Kohlrausch’s Law of Independent Migration of Ions

Kohlrausch determined $\dot{\mathrm{U}}_{0}$ values of pairs of some strong electrolytes containing same cation say KF and KCl, NaF and NaCl etc., and found that the difference in $\dot{\mathrm{U}}_{0}$ values in each case remains the same:

$\wedge_{\mathrm{m}}^{0}(\mathrm{KCl})-\wedge_{\mathrm{m}}^{0}(\mathrm{KF})=\wedge_{\mathrm{m}}^{0}(\mathrm{NaCl})-\wedge_{\mathrm{m}}^{0}(\mathrm{NaF})$

He also detemined $\dot{\mathrm{U}}_{0}$ values of pairs of strong electrolytes containing same anion say KF and NaF, KCl and NaCl etc.and found that the difference in Ù0 values in each case remains the same.

$\wedge_{\mathrm{m}}^{0}(\mathrm{KF})-\wedge_{\mathrm{m}}^{0}(\mathrm{NaF})=\wedge_{\mathrm{m}}^{0}(\mathrm{KCl})-\wedge_{\mathrm{m}}^{0}(\mathrm{NaCl})$

At infinite dilution when dissociation is complete, every ion makes some definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion which with it is associated and that the molar conductance at infinite dilution for any electrolyte is given by the sum of the contribution of the two ions. Thus,

$\wedge_{\mathrm{m}}^{0}=\lambda_{+}^{0}+\lambda_{-}^{0}$

Where $\lambda_{+}^{0}$ is the contribution of the cation and $\lambda_{-}^{0}$ is the contribution of the anion towards the molar conductance at infinite dilution. These contributions are called molar ionic conductances at infinite dilution. Thus, $\lambda_{+}^{0}$ is the molar ionic conductance of cation and

$\lambda_{-}^{0}$ is the molar ionic conductnace of anion, at infinite dilution. The above equation is, however, correct only for binary electrolyte like NaCl, $\mathrm{MgSO}_{4}$ etc.

8 Application of Kohlrausch’s law :

(1) Determination of $\wedge_{\mathrm{m}}^{0}$ of a weak electrolyte:

In order to calculate $\wedge_{\mathrm{m}}^{0}$ of a weak electrolyte say CH3COOH, we determine experimentally $\wedge_{\mathrm{m}}^{0}$ values of

the following three strong electrolytes:

(a) A strong electrolyte containing same cation as in the test electrolyte, say HCl

(b) A strong electrolyte containing same anion as in the test electrolyte, say $\mathrm{CH}_{3} \mathrm{COONa}$

(c) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl.

$\wedge_{\mathrm{m}}^{0}$ of $\mathrm{CH}_{3} \mathrm{COOH}$ is then given as:

$\wedge_{\mathrm{m}}^{0}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=\wedge_{\mathrm{m}}^{0}(\mathrm{HCl})+\wedge_{\mathrm{m}}^{0}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)-\wedge_{\mathrm{m}}^{0}(\mathrm{NaCl})$

Proof:

$\wedge_{\mathrm{m}}^{0}(\mathrm{HCl})+\wedge_{\left(\mathrm{CH}_{3} \mathrm{COONa}\right)}^{0}-\wedge_{(\mathrm{NaCl})}^{0}=\lambda_{\left(\mathrm{H}^{+}\right)}^{0}+\lambda_{\left(\mathrm{CH}_{3} \mathrm{COO}^{0}\right)}^{0}=\wedge_{0\left(\mathrm{CH}_{3} \mathrm{COOH}\right)}$

(2) Determination of degree of dissociation (a) :

(3) Determination of solubility of sparingly soluble salt

The specific conductivity of a saturated solution of the test electrolyte (sparingly soluble) made in conductivity water is determined by the method as described above. From this the specific conductivity of conductivity water is deducted. The molar conductance of the saturated solution is taken to be equal to $\wedge_{\mathrm{m}}^{0}$ as the saturated solution of a sparingly soluble salt is extremely dilute.

$\wedge_{\mathrm{m}}^{0}=\frac{1000 \mathrm{\kappa}}{\mathrm{C}}$

where C is the molarity of solution and hence the solubility.

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Practicing JEE Advanced Previous Year Questions 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 last 10 years JEE Advanced papers questions with solutions.
These JEE Advanced previous Year Question for Physics plays an important role in IIT-JEE preparation. We are providing IIT-JEE Mains Previous Year Question Papers with detailed Solution.

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 Advanced 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 Physics Topic-wise Previous year question for JEE Main:

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.

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X-Rays – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Q. If $\lambda_{\mathrm{Cu}}$ is the wavelength of $\mathrm{K}_{\alpha}$ X-ray line of copper (atomic number 29) and $\lambda_{\mathrm{Mo}}$ is the wavelength of the $\mathrm{K}_{\alpha}$ X-ray line of molybdenum (atomic number 42), then the ratio is close to :-

(A) 1.99                (B) 2.14               (C) 0.50                (D) 0.48

Sol. (B)

$\sqrt{\frac{\mathrm{C}}{\lambda}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})$

$\mathrm{b}=1$

$\sqrt{\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}}=\left(\frac{Z_{\mathrm{M}_{0}}-1}{Z_{\mathrm{Cu}}-1}\right)$

$\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}=\left(\frac{41}{28}\right)^{2}=2.14$