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.

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Previous Years JEE Advanced Questions

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

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

[JEE Advanced – 2017]

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

[JEE Advanced – 2017]

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.

 

 

Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions 

 

Download eSaral app  for free study material and video tutorials.

Simulator

Previous Years JEE Advanced Questions

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) $75^{\circ}$

[JEE-Advance-2013]

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

[JEE-Advance-2013]

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 :

[JEE-Advance-2013]

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

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

[JEE-Advance-2014]

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

[JEE-Advance-2014]

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.

[JEE-Advance-2014]

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.

[JEE-Advance-2015]

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

[JEE-Advance-2015]

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.

[JEE-Advance-2015]

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.

[JEE-Advance-2015]

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

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

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

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

[JEE-Advance-2015]

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) $45^{\circ}$

[JEE-Advance-2016]

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

[JEE-Advance-2016]

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.

[JEE-Advance-2016]

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)

[JEE-Advance-2016]

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.

[JEE-Advance-2017]

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 ?

[JEE-Advance-2017]

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

[JEE-Advance-2018]

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.) ?

[JEE-Advance-2018]

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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Previous Years JEE Advanced Questions

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)

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

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

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

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

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


Maxima-Minima – JEE Main Previous Year Question with Solutions

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

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Mathematics

Tangent & Normal – JEE Main Previous Year Question with Solutions

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

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Mathematics

Ray Optics – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

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Communication System – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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Previous Years AIEEE/JEE Mains Questions

Q. This questions has Statement-1 and Statement-2. Of the four choice given after the statements, choose the one that best describes the two statements.

Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.

Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season.

(1) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of statement-1

(2) Statement-1 is true, Statement-2 is true, Statrment-2 is not the correct explanation of Statement-1

(3) Statement-1 is false, Statement-2 is true

(4) Statement-1 is true, Statement-2 is false

[AIEEE-2011]

Sol. (2)


Q. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance up to which it can detect object located on the surface of the earth (Radius of earth $\left.=6.4 \times 10^{6} \mathrm{m}\right)$ is :

(1) 40 km (2) 64 (3) 80 km (4) 16km

[AIEEE-2012]

Sol. (3)

$\mathrm{d}=\sqrt{2 \mathrm{Rh}}=\sqrt{2 \times 6400 \times 0.5}=80 \mathrm{km}$


Q. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

(1) 10.62kHz (2) 5.31 MHz (3) 5.31 kHz (4) 10.62MHz

[JEE Main-2013]

Sol. (3)

$\mathrm{f}_{\mathrm{max}} \leq \frac{\sqrt{\frac{1}{\mathrm{m}^{2}}-1}}{2 \pi \mathrm{RC}}$

$\mathrm{f}_{\max } \leq \frac{8}{6 \times 2 \pi \times 100 \times 10^{3} \times 250 \times 10^{-12}}$

$\mathrm{f}_{\max } \leq \frac{8 \times 10^{6}}{12 \pi \times 25}$

$\mathrm{f}_{\max } \leq 8.4925 \mathrm{kHz}$


Q. A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are –

(1) 2005 kHz, 2000 kHz and 1995 kHz

(2) 2000 kHz and 1995 kHz

(3) 2 MHz only

(4) 2005 kHz and 1995 kHz

[JEE Main-2015]

Sol. (1)

Frequency present after modulation

$\mathrm{f}_{\mathrm{c}}, \mathrm{f}_{\mathrm{c}} \pm \mathrm{f}_{\mathrm{s}}$

$\Rightarrow 2000 \mathrm{KHz}, 2005 \mathrm{KHz}$ and $1995 \mathrm{KHz}$


Q. Choose the correct statement :

(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.

(2) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(3) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

[JEE Main-2016]

Sol. (2)


Q. In amplitude modulation, sinusoidal carrier frequency used is denoted by $\omega_{\mathrm{c}}$ and the signal frequency is denoted by $\omega_{\mathrm{m}}$. The bandwidth $\left(\Delta \omega_{\mathrm{m}}\right)$ of the signal is such that $\Delta \omega_{\mathrm{m}}<<\omega_{\mathrm{c}}$. Which of the following frequencies is not contained in the modulated wave ?

[JEE Main-2017]

Sol. (3)

Refer NCERT Page No. 526 Three frequencies are contained $\omega_{\mathrm{m}}+\omega_{\mathrm{c}}, \omega_{\mathrm{c}}-\omega_{\mathrm{m}} \& \omega_{\mathrm{c}}$


Q. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?

[JEE Main-2018]

Sol. (2)

Since the carrier frequency is distributed as band width frequency, so 10% of 10 GHz = n × 5 kHz

where n = no of channels $\frac{10}{100} \times 10 \times 10^{9}=n \times 5 \times 10^{3} \mathrm{n}=2 \times 10^{5}$ telephonic channels


Kinematics 2D – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

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Previous Years AIEEE/JEE Mains Questions

Q. A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$ where K is a constant. The general equation for its path is :

(1) $\mathrm{y}^{2}=\mathrm{x}^{2}+$ constant

(2) $\mathrm{y}=\mathrm{x}^{2}+$ constant

(3) $\mathrm{y}^{2}=\mathrm{x}+$ constant

(4) xy = constant

[AIEEE – 2010]

Sol. (1)

$\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$

$\overrightarrow{\mathrm{v}}=\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} \hat{\mathrm{i}}+\frac{\mathrm{dy}}{\mathrm{dt}} \hat{\mathrm{j}}$

$\frac{d x}{d t}=k y \& \frac{d y}{d t}=k x$

$\Rightarrow \frac{d x}{d y}=\frac{y}{x}$

$\Rightarrow \mathrm{y}^{2}=\mathrm{x}^{2}+$ constant


Q. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :-

(1) $\frac{\pi}{2} \frac{v^{4}}{g^{2}}$

(2) $\pi \frac{\mathrm{v}^{2}}{\mathrm{g}^{2}}$

( 3)$\pi \frac{\mathrm{v}^{2}}{\mathrm{g}}$

( 4)$\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$

[AIEEE – 2011]

Sol. (4)

$\mathrm{r}=\mathrm{R}_{\max }=\frac{\mathrm{v}^{2}}{\mathrm{g}}$

area $=\pi \mathrm{r}^{2}$

$=\pi\left(\frac{\mathrm{v}^{2}}{\mathrm{g}}\right)^{2}$

$=\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$


Q. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force $\mathrm{F}(\mathrm{t})=\mathrm{F}_{0} \mathrm{e}^{-\mathrm{bt}}$ in the x direction. Its speed v(t) is depicted by which of the following curves ?

[AIEEE – 2012]

Sol. (3)

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}}$

$\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \int_{0}^{\mathrm{t}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$

$\mathrm{v}=\frac{-\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[\mathrm{e}^{-\mathrm{bt}}-1\right]$

$=\frac{\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[1-\mathrm{e}^{-\mathrm{bt}}\right]$


Q. A projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) m / s$ where $\hat{\mathbf{i}}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is :

(1) $\mathrm{y}=\mathrm{x}-5 \mathrm{x}^{2}$

(2) $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

(3) $4 y=2 x-5 x^{2}$

(4) $4 y=2 x-25 x^{2}$

[AIEEE – 2013]

Sol. (2)

$\mathrm{u}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}$

$\mathrm{u}_{\mathrm{x}}=1$

$\mathrm{u}_{\mathrm{y}}=2$

$\tan \theta=\frac{2}{1}$

$\mathrm{y}=\mathrm{x} \tan \theta-\frac{1}{2} \mathrm{g} \frac{\mathrm{x}^{2}}{\mathrm{u}^{2} \cos ^{2} \theta}$

$\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$


Q. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take

g = 10 $\mathrm{M} / \mathrm{S}^{2}$) (The figure are schematic and not drawn to scale)

[JEE Mains – 2015]

Sol. (1)

$\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

For particle 2

$-240=40 \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$

$5 t^{2}-40 t-240=0$

$\mathrm{t}_{2}=12 \mathrm{sec}$

for $0<\mathrm{t}<8$ sec $\rightarrow \mathrm{a}_{\text {rel }}=0$

straight line x-t graph

for $8<\mathrm{t}<12$ sec $\rightarrow \mathrm{a}_{\mathrm{rel}}=-\mathrm{g}$

downward parabola

for $\mathrm{t}>12 \mathrm{sec} \rightarrow$ Both particles comes to rest


Kinematics 2D – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their 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.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

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Previous Years JEE Advanced Questions

Q. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in $\mathrm{m} / \mathrm{s}^{2}$, is –

[IIT-JEE 2011]

Sol. 5


Q. A rocket is moving in a gravity free space with a constant acceleration of 2 $\mathrm{ms}^{-2}$along + x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 $\mathrm{ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 $\mathrm{ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is –

[JEE Advanced 2014]

Sol. 8 or 2

Assuming open chamber

$\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$

$S_{\text {relative }}=4 \mathrm{m}$

time $=\frac{4}{0.5}=8 \mathrm{s}$

Alternate

Assuming closed chamber

In the frame of chamber :

Maximum displacement of ball A from its left end is $\frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m}$

This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be very close to the left end.

Hence, time taken by ball B to reach left end will be given by

$\mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}$

$4=(0.2)(t)+\frac{1}{2}(2)(t)^{2}$

Solving this, we get

$\mathrm{t} \approx 2 \mathrm{s}$


Q. Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $\mathrm{ms}^{-1}$. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = $t_{0}$, A just escapes being hit by B, $t_{0}$ in seconds is

[JEE Advanced 2014]

Sol. 5

As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A

$\mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3}$

$\mathrm{V}_{\mathrm{B}}=200$

If A is observer A remains stationary therefore

$\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$


Q. A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is

[JEE Advanced 2018]

Sol. 30

$\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} 45}{2 \mathrm{g}}=120$

$\Rightarrow \frac{\mathrm{u}^{2}}{4 \mathrm{g}}=120$ ….(i)

when half of kinetic energy is lost $\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$

$\mathrm{H}_{2}=\frac{\left(\frac{\mathrm{u}}{\sqrt{2}}\right)^{2} \sin ^{2} 30}{2 \mathrm{g}}=\frac{\mathrm{u}^{2}}{16 \mathrm{g}}$

from (i) & (ii)

$\mathrm{H}_{2}=\frac{\mathrm{H}_{1}}{4}=30 \mathrm{m}$ on 30.00


Kinematics 1D- JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

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Simulator

Previous Years AIEEE/JEE Mains Questions

Q. A particle has an initial velocity of $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}$. Its speed after 10s is :-

(1) 7 units

(2) 8.5 units

(3) 10 units

(4) $7 \sqrt{2}$ units

[AIEEE-2009]

Sol. (4)

v = u + at

u = 3i + 4j + (0.4 i + 0.3 j) × 10

= 3i + 4j + 4i + 3i

u = 7i + 7j

$|\overrightarrow{\mathrm{u}}|=\sqrt{7^{2}+7^{2}}$

$=7 \sqrt{2}$


Q. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :-

(1) 4 s (2) 8 s (3) 1 s (4) 2 s

[AIEEE-2011]

Sol. (4)

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$

$\int \mathrm{d} \mathrm{v} \times \mathrm{v}^{-1 / 2}=\int-2.5 \mathrm{dt}$

$\frac{\mathrm{v}^{\frac{1}{2}}}{\frac{1}{2}}=-2.5 \mathrm{t}$

$2 \sqrt{\mathrm{v}}=-2.5 \mathrm{t}+\mathrm{c}$

$2 \sqrt{6.25}=-2.5 \mathrm{t}+\mathrm{c}$

2 × 0.25 = c

5 = c

$2 \sqrt{\mathrm{v}}=-\mathrm{kt}+5$

0 = –kt + 5

kt = 5

$\mathrm{t}=\frac{5 \mathrm{sec}}{2.5}=2 \mathrm{sec}$


Q. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is :

(1) $2 \mathrm{g} \mathrm{H}=\mathrm{nu}^{2}(\mathrm{n}-2)$

(2) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2) \mathrm{u}^{2}$

(3) $2 \mathrm{g} \mathrm{H}=\mathrm{n}^{2} \mathrm{u}^{2}$

(4) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2)^{2} \mathrm{u}^{2}$

[JEE-MAIN-2014]

Sol. (1)

Time to reach highest point $=\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$

time to reach ground = nt

$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

$-\mathrm{H}=\mathrm{u}(\mathrm{nt})-\frac{1}{2} \mathrm{g}(\mathrm{nt})^{2}$

$\Rightarrow 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2)$


Q. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ?

[JEE Mains – 2017]

Sol. (1)

Velocity at any time t is

given by

v = u + at

v = v0 + (–g)t

$\mathrm{v}=\mathrm{v}_{0}-\mathrm{gt}$

$\Rightarrow$ straight line with negative slope


Q. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

[JEE Mains – 2018]

Sol. (1)

In this question option (2) and (4) are the corresponding position – time graph and velocity –position graph of option (3) and its distance – time graph is given as

Hence incorrect graph is option (1)


Quadratic Equation – JEE Main Previous Year Question with Solutions

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

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Simulator

Previous Years AIEEE/JEE Mains Questions

Q. If the roots of the equation $b x^{2}+c x+a=0$ be imaginary, then for all real values of x, the expression $3 b^{2} x^{2}+6 b c x+2 c^{2}$ is :-

(1) Greater than –4ab (2) Less than –4ab (3) Greater than 4ab (4) Less than 4ab

[AIEEE-2010]

Sol. (1)

since the roots of $\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{a}=0$ are imaginary

$\because c^{2}-4 a b<0 \Rightarrow c^{2}<4 a b$

for exp. $3 b^{2} x^{2}+6 b c x+2 c^{2}$

Min. value $=\frac{-\mathrm{D}}{4 \mathrm{a}}=\frac{\left(36 \mathrm{b}^{2} \mathrm{c}^{2}-24 \mathrm{b}^{2} \mathrm{c}^{2}\right)}{12 \mathrm{b}^{2}}=\frac{-12 \mathrm{b}^{2} \mathrm{c}^{2}}{12 \mathrm{b}^{2}}=-\mathrm{c}^{2}$

$\therefore-\mathrm{c}^{2}>-4 \mathrm{ab}$

So exp. is greater than (– 4ab)


Q. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+1=0,$ then $\alpha^{2009}+\beta^{2009}=$

(1) –2 (2) –1 (3) 1 (4) 2

[AIEEE-2010]

Sol. (3)

Roots of equation $\mathrm{x}^{2}-\mathrm{x}+1=0 \mathrm{ar}$

$\alpha=-\omega, \quad \beta=-\omega^{2}$

$\alpha^{2009}+\beta^{2009}=(-\omega)^{2009}+\left(-\omega^{2}\right)^{2009}$c

$=-\left(\omega^{2}+\omega\right)=1$


Q. Let for $a \neq a_{1} \neq 0, f(x)=a x^{2}+b x+c, g(x)=a_{1} x^{2}+b_{1} x+c_{1}$ and $p(x)=f(x)-g(x)$

If $\mathrm{p}(\mathrm{x})=0$ only for $\mathrm{x}=-1$ and $\mathrm{p}(-2)=2,$ then the value of $\mathrm{p}(2)$ is:

(1) 18 (2) 3 (3) 9 (4) 6

[AIEEE-2011]

Sol. (1)

$\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}+1)^{2}$

$\mathrm{P}(-2)=2=\mathrm{k}(-1)^{2}$

$\Rightarrow \mathrm{k}=2$

$\therefore \mathrm{P}(\mathrm{x})=2(\mathrm{x}+1)^{2}$

$\Rightarrow \mathrm{P}(2)=18$

Aliter :

$\mathrm{P}(\mathrm{x})=\left(\mathrm{a}-\mathrm{a}_{1}\right) \mathrm{x}^{2}+\left(\mathrm{b}-\mathrm{b}_{1}\right) \mathrm{x}+\left(\mathrm{c}-\mathrm{c}_{1}\right)=0$

only x = –1

P(x) = 0

So roots are equal.

it means D = 0

$\left(b-b_{1}\right)^{2}=4\left(a-a_{1}\right)\left(c-c_{1}\right) \Rightarrow q^{2}=4 p r$

Let

$\mathrm{a}-\mathrm{a}_{1}=\mathrm{p}$

$\mathrm{b}-\mathrm{b}_{1}=\mathrm{q}$

$\mathrm{c}-\mathrm{c}_{1}=\mathrm{r}$

P(–1) = 0

p – q + r = 0 …… (1)

4p – 2q + r = 2 ….. (2)

4p + 2q + r = ?

4p + 2q + r = ?

$(p+r)^{2}-4 p r=0$

$(p-r)^{2}=0$                       

from eq. (1) q = 2r

So from eq. (2) 4r – 4r + r = 2

r = 2

So 4p + 2q + r = 4r + 4r + r = 9r = 18


Q. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are:

(1) –4, –3 (2) 6, 1 (3) 4, 3 (4) –6, –1

[AIEEE-2011]

Sol. (2)


Q. The equation $\mathrm{e}^{\mathrm{sinx}}-\mathrm{e}^{-\mathrm{sinx}}-4=0$ has :

(1) exactly four real roots.

(2) infinite number of real roots.

(3) no real roots.

(4) exactly one real root.

[AIEEE-2012]

Sol. (3)


Q. If the equations $\mathrm{x}^{2}+2 \mathrm{x}+3=0$ and $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$, have a common root, then a : b : c is :

(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2

[JEE-MAIN-2013]

Sol. (1)

$\mathrm{x}^{2}+2 \mathrm{x}+3=0$

$\mathrm{D}<0$

$\therefore \quad \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 \mathrm{has}$

both roots common

$\therefore \quad a: b: c=1: 2: 3$


Q. Let  and $\beta$ be the roots of equation $\mathrm{x}^{2}-6 \mathrm{x}-2=0 .$ If $\mathrm{a}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}},$ for $\mathrm{n} \geq$ 1, then the value of $\frac{\mathrm{a}_{10}-2 \mathrm{a}_{8}}{2 \mathrm{a}_{9}}$ is equal to :

(1) 3 (2) – 3 (3) 6 (4) – 6

[JEE-MAIN-2015]

Sol. (1)


Q. The sum of all real values of x satisfying the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is :

(1) 5 (2) 3 (3) –4 (4) 6

[JEE-MAIN-2016]

Sol. (2)

$\mathrm{x}^{2}-5 \mathrm{x}+5=1 \Rightarrow \mathrm{x}=1,4$

$x^{2}-5 x+5=-1 \Rightarrow x=2,3$

but 3 is rejected

$\mathrm{x}^{2}+4 \mathrm{x}-60=0 \Rightarrow \mathrm{x}=-10,6$

Sum = 3


Q. If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to-

If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to-

[JEE-MAIN-2018]

Sol. (2)


Definite Integration – JEE Main Previous Year Question with Solutions

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

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Q. $\int_{0}^{\pi}[\cot x] d x,$ where $[.]$ denotes the greatest integer function, is equal to –

(1) –1 $         

(2)-\frac{\pi}{2}$           

(3) $\frac{\pi}{2}$           

(4) 1

[AIEEE-2009] 

Sol. (2)


Q. Let p(x) be a function defined on R such that p'(x) = p'(1 – x), for all x $\in$0, 1], p(0) = 1 and p(1) = 41. Then $\int_{0}^{1}$ p(x) dx equals :-

(1) $\sqrt{41}$

(2) 21

(3) 41

(4) 42

[AIEEE-2010]

Sol. (2)


Q. The value of $\int_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x$ is :-

(1) $\frac{\pi}{2} \log 2$

(2) $\log 2$

(3) $\pi \log 2$

(4) $\frac{\pi}{8} \log 2$

[AIEEE-2011]

Sol. (3)


Q. Let [.] denote the greatest integet function then the value of $\int_{0}^{1.5} \mathrm{x}\left[\mathrm{x}^{2}\right] \mathrm{dx}$ is :-

( 1)$\frac{5}{4}$ (2) 0 (3) $\frac{3}{2}$ (4) $\frac{3}{4}$

[AIEEE-2011]

Sol. (4)


Q. If $\mathrm{g}(\mathrm{x})=\int_{0}^{\mathrm{x}} \cos 4 \mathrm{t} \mathrm{dt},$ then $\mathrm{g}(\mathrm{x}+\pi)$ equals :

(1) $\mathrm{g}(\mathrm{X}) \cdot \mathrm{g}(\pi)$

(2) $\frac{\mathrm{g}(\mathrm{x})}{\mathrm{g}(\pi)}$

(3) $\mathrm{g}(\mathrm{x})+\mathrm{g}(\pi)$

(4) $\mathrm{g}(\mathrm{x})-\mathrm{g}(\pi)$

[AIEEE-2012]

Sol. (3,4)


Q. Statement-I : The value of the integral $\int_{\pi / 6}^{\pi / 3} \frac{\mathrm{dx}}{1+\sqrt{\tan \mathrm{x}}}$ is equal to $\frac{\pi}{6}$

Statement-II : $\int_{a}^{b} f(x) d x-\int_{a}^{b} f(a+b-x) d x$

(1) Statement-I is true, Statement-II is true; Statement-II is a correct explanation for Statement- I.

(2) Statement-I is true, Statement-II is true; Statement-II is not a correct explanation for Statement-I.

(3) Statement-I is true, Statement-II is false.

(4) Statement-I is false, Statement-II is true.

[JEE-MAIN-2013]

Sol. (4)


Q. The integral $\int_{0}^{\pi} \sqrt{1+4 \sin ^{2} \frac{x}{2}-4 \sin \frac{x}{2}} d x$ equals :

(1) $\pi-4$

(2) $\frac{2 \pi}{3}-4-4 \sqrt{3}$

(3) $4 \sqrt{3}-4$

(4) $4 \sqrt{3}-4-\frac{\pi}{3}$

[JEE-MAIN-2014]

Sol. (4)


Q. The integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x$ is equal to :

(1) 1 (2) 6 (3) 2 (4) 4

[JEE-MAIN-2015]

Sol. (1)


Q. The integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{dx}}{1+\cos x}$ is equal to :-

(1) –1 (2) –2 (3) 2 (4) 4

[JEE-MAIN-2017]

Sol. (3)


Q. The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+2^{x}} d x$ is :

( 1)$\frac{\pi}{2}$

(2) $4 \pi$

(3) $\frac{\pi}{4}$

(4) $\frac{\pi}{8}$

[JEE-MAIN-2018]

Sol. (3)


 

Probability – JEE Main Previous Year Question with Solutions

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

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

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Mathematics

3D Geometry – JEE Main Previous Year Question with Solutions

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

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Download eSaral app for free study material and video tutorials.

Q. Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3 y-\alpha z+\beta=0 .$ Then $(\alpha, \beta)$ equals

(1) (5, – 15)             (2) (–5, 5)             (3) (6, –17)               (4) (–6, 7)

[AIEEE-2009]

Sol. (4)

Line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2} \ldots .(1)$

Plane $\mathrm{x}+3 \mathrm{y}-\alpha \mathrm{z}+\beta=0 \quad \ldots .(2)$

Point $(2,1,-2)$ put in ( 2)

$2+3+2 \alpha+\beta=0 \Rightarrow 2 \alpha+\beta=-5$

$\mathrm{Now} \mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0$

$3-15-2 \alpha=0 \Rightarrow-12-2 \alpha=0 \Rightarrow \alpha=-6$

$-12+\beta=-5 \Rightarrow \beta=7$

$\therefore \alpha=-6, \beta=7$


Q. The projections of a vector on the three coordinate axis are 6, –3, 2 respectively. The direction cosines of the vector are :-

(1) $\frac{6}{7}, \frac{-3}{7}, \frac{2}{7}$

(2) $\frac{-6}{7}, \frac{-3}{7}, \frac{2}{7}$

(3) 6, –3, 2

(4) $\frac{6}{5}, \frac{-3}{5}, \frac{2}{5}$

[AIEEE-2009]

Sol. (1)


Q. Statement–1 : The point A(3,1,6) is the mirror image of the point B(1, 3, 4) in the plane

x – y + z = 5.

Statement–2 : The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4).

(1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement– 1.

(2) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for statement– 1.

(3) Statement–1 is true,n Statement–2 is false.

(4) Statement–1 is false, Statement–2 is true.

[AIEEE-2010]

Sol. (2)

Mirror image of $\mathrm{B}(1,3,4)$ in plane $\mathrm{x}-\mathrm{y}+\mathrm{z}=5$

$\frac{x-1}{1}=\frac{y-3}{-1}=\frac{z-4}{1}=-2 \frac{(1-3+4-5)}{1+1+1}=2$

$\Rightarrow x=3, y=1, z=6$

$\therefore$ mirror image of $\mathrm{B}(1,3,4)$

is $\mathrm{A}(3,1,6)$

statement-1 is correct

statement-2 is true but it is not the correct explanation.

because it is not neccessary that to bisect the line joining (1,3,4) and (3,1,6) plane is always perpendicular to line AB


Q. If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1}(\sqrt{\frac{5}{14}})$, then $\lambda$ equals:-

( 1)$\frac{2}{5}$

( 2)$\frac{5}{3}$

( 3)$\frac{2}{3}$

( 4)$\frac{3}{2}$

[AIEEE-2011]

Sol. (3)

$\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda}$ equation of line

equation of plane $x+2 y+3 z=4$

$\sin \theta=\frac{1+4+3 \lambda}{\sqrt{14} \sqrt{1+4}+\lambda^{2}}$

$\Rightarrow \lambda=\frac{2}{3}$


Q. Statement-1: The point $\mathrm{A}(1,0,7)$ is the mirror image of the point $\mathrm{B}(1,6,3)$ in the line:

\[ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \]

Statement-2: The line : $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ bisects the line segment joining $A(1,0,7)$ and B(1, 6, 3).

(1) Statement-1 is true, Statement-2 is false.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

[AIEEE-2011]

Sol. (4)

$1(1-1)+2(0-6)+3(7-3)$

$=0-12+12=0$

mid point $\mathrm{AB}(1,3,5)$

lies on $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$


Q. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along a straight line x = y = z is :

(1) $3 \sqrt{5}$

(2) $10 \sqrt{3}$

(3) $5 \sqrt{3}$

(4) $3 \sqrt{10}$

[AIEEE-2011]

Sol. (2)

Distance $\mathrm{PM}=\sqrt{100+100+100}=10 \sqrt{3}$


Q. An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is :

(1) x – 2y + 2z + 5 = 0

(2) x – 2y + 2z – 3 = 0

(3) x – 2y + 2z + 1 = 0

(4) x – 2y + 2z – 1 = 0

[AIEEE-2012]

Sol. (2)

Equation of plane parallel to

$x-2 y+2 z-5=0$

is $x-2 y+2 z=k$

Distance of plane from origen $=1$

$\left|\frac{k}{\sqrt{9}}\right|=1$

$\mathrm{k}=\pm 3$

Equation of required plane is

$x-2 y+2 z \pm 3=0$


Q. If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k$ is equal to :

(1) 0

(2) – 1

( 3)$\frac{2}{9}$

( 4)$\frac{9}{2}$

[AIEEE-2012]

Sol. (4)

$\left|\begin{array}{ccc}{3-1} & {\mathrm{K}+1} & {0-1} \\ {2} & {3} & {4} \\ {1} & {2} & {1}\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}{2} & {\mathrm{K}+1} & {-1} \\ {2} & {3} & {4} \\ {1} & {2} & {1}\end{array}\right|=0$

$\Rightarrow 2 \mathrm{K}-9=0 \Rightarrow \mathrm{K}=\frac{9}{2}$


Q. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is :-

( 1)$\frac{3}{2}$

( 2)$\frac{5}{2}$

(3) $\frac{7}{2}$

( 4)$\frac{9}{2}$

[JEE-MAIN 2013]

Sol. (3)

$4 x+2 y+4 z+5=0$

$4 x+2 y+4 z-16=0$

$\Rightarrow \quad d=\left|\frac{21}{\sqrt{36}}\right|=\frac{7}{2}$


Q. If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then $k$ can have :

(1) any value

(2) exactly one value

(3) exactly two values

(4) exactly three values.

[JEE-MAIN 2013]

Sol. (3)

$\Rightarrow(\bar{a}-\bar{b}) \cdot(\bar{c} \times \bar{d})=0$

$\Rightarrow\left|\begin{array}{ccc}{1} & {-1} & {-1} \\ {1} & {1} & {-k} \\ {k} & {2} & {1}\end{array}\right|=0$

$\Rightarrow \quad(1+2 k)+\left(1+k^{2}\right)-(2-k)=0$

$\Rightarrow \quad \mathrm{k}^{2}+3 \mathrm{k}=0<\begin{array}{c}{0} \\ {-3}\end{array}$


Q. A vector $\overrightarrow{\mathrm{n}}$ is inclined to x-axis at $45^{\circ},$ to $\mathrm{y}$ -axis at $60^{\circ}$ and at an acute angle to z-axis. If $\overrightarrow{\mathrm{n}}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1),$ then the equation of the plane is :

(1) $\sqrt{2} x-y-z=2$

(2) $\sqrt{2} x+y+z=2$

(3) $3 \sqrt{2} x-4 y-3 z=7$

(4) $4 \sqrt{2} x+7 y+z=2$

[JEE-MAIN Online 2013]

Sol. (2)

Let $\alpha, \beta, \gamma$ v be direction cosine of vector $\vec{n}$

$\alpha^{2}+\beta^{2}+\gamma^{2}=1$

$\frac{1}{2}+\frac{1}{4}+\gamma^{2}=1$

$\gamma=\frac{1}{2} \quad\left\{-\frac{1}{2} \text { rejected }\right\}$

$\overline{\mathrm{n}}=\frac{\hat{\mathrm{i}}}{\sqrt{2}}+\frac{\hat{\mathrm{j}}}{2}+\frac{\hat{\mathrm{k}}}{2}=\sqrt{2 \hat{\mathrm{i}}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overline{\mathrm{a}}=\sqrt{2} \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$d=\bar{a} \cdot \bar{n}=2-1+1=2$

$\overline{\mathrm{r}} \cdot(\sqrt{2} \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2$

$\sqrt{2} x+y+z=2$


Q. The acute angle between two lines such that the direction cosines $\ell, \mathrm{m}, \mathrm{n}$ of each of them satisfy the equations $\ell+\mathrm{m}+\mathrm{n}=0$ and $\ell^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=0$ is :-

(1) $30^{\circ}$ (2) $45^{\circ}$ (3) $60^{\circ}$ (4) $15^{\circ}$

[JEE-MAIN Online 2013]

Sol. (3)

$\begin{array}{rl}{1+\mathrm{m}+\mathrm{n}} & {=0 ; \ell^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=0} \\ {\mathrm{a} \ell=-(\mathrm{m}+\mathrm{n})} & {} \\ {(\mathrm{m}+\mathrm{n})^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}} & {=0} \\ {(\mathrm{m}+\mathrm{n})[\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}]} & {=0} \\ {\mathrm{m}+\mathrm{n}=0} & {\mathrm{m}=0} \\ {\mathrm{m}=-\mathrm{n}}\end{array}$

Dr’s of line $1 \Rightarrow 0,-\mathrm{n}, \mathrm{n}$ or $0,-1,1$

Dr’s of line $2 \Rightarrow-\mathrm{n}, 0, \mathrm{n}$ or $-1,0,1$

$\cos \theta=\frac{1}{\sqrt{2} \sqrt{2}}$

$\theta=60^{\circ}$


Q. Let Q be the foot of perpendicular from the origin to the plane 4x – 3y + z + 13 = 0 and R be a point (–1, 1, –6) on the plane. Then length QR is :

(1) $3 \sqrt{\frac{7}{2}}$

(2) $\sqrt{14}$

(3) $\sqrt{\frac{19}{2}}$

(4) $\frac{3}{\sqrt{2}}$

[JEE-MAIN Online 2013]

Sol. (1)


Q. If the projections of a line segment on thex, y and z-axes in 3-dimensional space are 2, 3 and 6 respectively, then the length ofthe line segment is :

(1) 7             (2) 9           (3) 12               (4) 6

[JEE-MAIN Online 2013]

Sol. (1)

i.e. $[\mathrm{OM}=2, \mathrm{ON}=3, \mathrm{OQ}=6]$

$\therefore \mathrm{P}(2,3,6)$

$\begin{aligned} \therefore \mathrm{OP} &=\sqrt{2^{2}+3^{2}+6^{2}} \\ &=\sqrt{49}=7 \end{aligned}$


Q. If two lines $L_{1}$ and $L_{2}$ in space, are defined by $\mathrm{L}_{1}=\{\mathrm{x}=\sqrt{\lambda} \mathrm{y}+(\sqrt{\lambda}-1)$

$\mathrm{z}=(\sqrt{\lambda}-1) \mathrm{y}+\sqrt{\lambda}\}$ and

$\mathrm{L}_{2}=\{\mathrm{x}=\sqrt{\mu} \mathrm{y}+(1-\sqrt{\mu})$

$\mathrm{z}=(1-\sqrt{\mu}) \mathrm{y}+\sqrt{\mu}\}$

then $\mathrm{L}_{1}$ is perpendicular to $\mathrm{L}_{2},$ for all non-negative reals $\lambda$ and $\mu,$ such that :

(1) $\lambda=\mu$

(2) $\lambda \neq \mu$

(3) $\sqrt{\lambda}+\sqrt{\mu}=1$

(4) $\lambda+\mu=0$

[JEE-MAIN Online 2013]

Sol. (1,4)

$\mathrm{L}_{1}=\{\mathrm{x}=\sqrt{\lambda} \mathrm{y}+(\sqrt{\lambda}-1), \mathrm{z}=(\sqrt{\lambda}-1) \mathrm{y}+\sqrt{\lambda}\}$

$\therefore \mathrm{L}_{1}: \frac{\mathrm{x}-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{Z}-\sqrt{\lambda}}{(\sqrt{\lambda}-1)} \quad \& \mathrm{L}_{2}:\{\mathrm{x}=\sqrt{\mu} \mathrm{y}+(1-\sqrt{\mu}), \mathrm{z}=(1-\sqrt{\mu}) \mathrm{y}+\sqrt{\mu}\}$

$\therefore \mathrm{L}_{2}: \frac{\mathrm{x}-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\sqrt{\mu}}{(1-\sqrt{\mu})}$

$\therefore \mathrm{L}_{1} \perp^{r} \mathrm{L}_{2} \Rightarrow(\sqrt{\lambda})(\sqrt{\mu})+1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0$

$\Rightarrow \sqrt{\lambda} \sqrt{\mu}+1-\sqrt{\lambda} \sqrt{\mu}+\sqrt{\lambda}+\sqrt{\mu}-1=0$

$\Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \lambda=\mu=0$


Q. The equation of a plane through the line of intersection of the planes x + 2y = 3, y – 2z+ 1 = 0, and perpendicular to the first plane is :

(1) 2x – y + 7z = 11

(2) 2x – y + 10 z = 11

(3) 2x – y – 9z = 10

(4) 2x – y – 10z = 9

[JEE-MAIN Online 2013]

Sol. (2)

$(x+2 y-3)+\lambda(y-2 z+1)=0$

$(1)(1)+(2+\lambda) 2+(-2 \lambda) \times 0=0$

$1+4+2 \lambda=0$

$\lambda=-\frac{5}{2}$

$(x+2 y-3)-\frac{5}{2}(y-2 z+1)=0$

$2 \mathrm{x}-\mathrm{y}+10 \mathrm{z}-11=0$


Q. Let ABC be a triangle with vertices at points $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ in three dimensional space. If the median through A is equally inclined with the axes, then $(\lambda, \mu$.) is equal to :

(1) (10, 7)                  (2) (7.5)               (3) (7, 10)                    (4) (5,7)

[JEE-MAIN Online 2013]

Sol. (3)


Q. The angle between the lines whose direction cosines satisfy the equations $\ell+\mathrm{m}+\mathrm{n}=0$ and $\ell^{2}=\mathrm{m}^{2}+\mathrm{n}^{2}$ is

( 1)$\frac{\pi}{3}$

( 2)$\frac{\pi}{4}$

( 3)$\frac{\pi}{6}$

(4) $\frac{\pi}{2}$

[JEE-MAIN 2014]

Sol. (1)


Q. The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2 x-y+z+3=0$ is the line:

(1) $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

(2) $\frac{x+3}{-3}=\frac{y-5}{-1}=\frac{z+2}{5}$

(3) $\frac{x-3}{3}=\frac{y+5}{1}=\frac{z-2}{-5}$

(4) $\frac{x-3}{-3}=\frac{y+5}{-1}=\frac{z-2}{5}$

[JEE-MAIN 2014]

Sol. (1)

$\mathrm{L}: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-5}$

$P: 2 x-y+z+3=0$

It can be observed given line is parallel to given plane.

Image of $(1,3,4)$ in given plane can be calculated as

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\frac{-2(2 \times 1-3+4+3)}{6}=-2$

$\Rightarrow x=-3 ; y=5 ; z=2$


Q. The equation of the plane containing the line 2x – 5y + z = 3 ; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is :

(1)x + 3y + 6z = 7

(2) 2x + 6y + 12z = – 13

(3) 2x + 6y + 12z = 13

(4) x + 3y + 6z = – 7

[JEE(Main)-2015]

Sol. (1)

Let equation of plane parallel to $\mathrm{x}+3 \mathrm{y}+6 \mathrm{z}=1$ is $\mathrm{x}+3 \mathrm{y}+6 \mathrm{z}=\lambda$ point on line of intersection is $(4,1,0),$ it also lie on plane $x+3 y+6 z=\lambda$ so required plane is $x+3 y+6 z=7$


Q. The distance of the point (1, 0, 2) from the point of intersection of the line $\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}$ and the plane $\mathrm{x}-\mathrm{y}+\mathrm{z}=16,$ is :

(1) $3 \sqrt{21}$

(2) 13

(3) $2 \sqrt{14}$

(4) 8

[JEE(Main)-2015]

Sol. (2)


Q. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :

(1) $\frac{20}{3}$

(2) $3 \sqrt{10}$

(3) $10 \sqrt{3}$

( 4)$\frac{10}{\sqrt{3}}$

[JEE(Main)-2016]

Sol. (3)

Equation of line parallel to $\mathrm{x}=\mathrm{y}=\mathrm{z}$ through $(1,-5,9)$ is $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+5}{1}=\frac{\mathrm{z}-9}{1}=\lambda$

If $\mathrm{P}(\lambda+1, \lambda-5, \lambda+9)$ be point of intesection of line and plane.

$\Rightarrow$ Required distance $=10 \sqrt{3}$


Q. If the line, $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane, $1 x+m y-z=9,$ then $1^{2}+m^{2}$ is equal to :-

(1) 2               (2) 26              (3) 18                  (4) 5

[JEE(Main)-2016]

Sol. (1)

Given line

\[ \frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3} \]

and Given plane is $\ell \mathrm{x}+\mathrm{my}-\mathrm{z}=9$

Now, it is given that line lies on plane $\therefore 2 \ell-\mathrm{m}-3=0 \Rightarrow 2 \ell-\mathrm{m}=3$

Also, $(3,-2,-4)$ lies on plane

\[ 3 \ell-2 m=5 \]

Solving ( 1) and $(2),$ we get

\[ \ell=1, \mathrm{m}=-1 \]

$\therefore \ell^{2}+\mathrm{m}^{2}=2$


Q. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to line, $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}$ is $\mathrm{Q},$ then $\mathrm{PQ}$ is equal to :-

(1) $6 \sqrt{5}$

(2) $3 \sqrt{5}$

(3) $2 \sqrt{42}$

(4) $\sqrt{42}$

[JEE(Main)-2017]

Sol. (3)

Line $\mathrm{PQ} ; \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+2}{4}=\frac{\mathrm{z}-3}{5}$

Let $\mathrm{F}(\lambda+1,4 \lambda-2,5 \lambda+3)$

                         

F lies on the plane

$2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$

$\Rightarrow-6 \lambda+6=0 \Rightarrow \lambda=1$

$\mathrm{F}(2,2,8)$

$\mathrm{PQ}=2 \quad \mathrm{PF}=2 \sqrt{42}$


Q. The distantce of the point (1, 3, –7) from the plane passing through the point (1, 1, –1), havingnormal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1},$ is :

(1) $\frac{10}{\sqrt{74}}$

(2) $\frac{20}{\sqrt{74}}$

(3) $\frac{10}{\sqrt{83}}$

(4) $\frac{5}{\sqrt{83}}$

[JEE(Main)-2017]

Sol. (1)

Normal vector

$\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {1} & {-2} & {3} \\ {2} & {-1} & {-1}\end{array}\right|=5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

So plane is 5(x – 1) + 7(y + 1) + 3(z + 1) = 0

$\Rightarrow 5 x+7 y+3 z+5=0$

Distance $\frac{5+21-21+5}{\sqrt{25+49+9}}=\frac{10}{\sqrt{83}}$


Q. The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is :

( 1)$\frac{2}{3}$

(2) $\frac{1}{3}$

(3) $\sqrt{\frac{2}{3}}$

(4) $\frac{2}{\sqrt{3}}$

[JEE(Main)-2018]

Sol. (3)


Q. If $\mathrm{L}_{1}$ is the line of intersection of the planes $2 \mathrm{x}-2 \mathrm{y}+3 \mathrm{z}-2=0, \mathrm{x}-\mathrm{y}+\mathrm{z}+1=0$ and $\mathrm{L}_{2}$ is the line of intersection of the planes $\mathrm{x}+2 \mathrm{y}-\mathrm{z}-3=0,3 \mathrm{x}-\mathrm{y}+2 \mathrm{z}-1=0,$ then the distance of the origin from the plane, containing the lines $L_{1}$ and $L_{2}$ is $z$

(1) $\frac{1}{3 \sqrt{2}}$

(2) $\frac{1}{2 \sqrt{2}}$

(3) $\frac{1}{\sqrt{2}}$

( 4)$\frac{1}{4 \sqrt{2}}$

[JEE(Main)-2018]

Sol. (1)

Plane passes through line of intersectuion of first two planes is $(2 x-2 y+3 z-2)+\lambda(x-y+z+1)=0$

$\mathrm{x}(\lambda+2)-\mathrm{y}(2+\lambda)+\mathrm{z}(\lambda+3)+(\lambda-2)=0 \ldots \ldots(1)$

is having infinite number of solution with x + 2y – z – 3 = 0 and 3x – y + 2z – 1 = 0 then

$\left|\begin{array}{ccc}{(\lambda+2)} & {-(\lambda+2)} & {(\lambda+3)} \\ {1} & {2} & {-1} \\ {3} & {-1} & {2}\end{array}\right|=0$

Solving $\lambda=5$ $7 x-7 y+8 z+3=0$

perpendicular distance from $(0,0,0)$

is $\frac{3}{\sqrt{162}}=\frac{1}{3 \sqrt{2}}$


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Previous Years JEE Advanced Questions

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

[JEE – ADV. 2016]

Sol. (A)


JEE Advanced Previous Year Papers
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Physics JEE Advanced Previous Year Questions with Solutions
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Previous Years AIEEE/JEE Main Questions

Q. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $\mathrm{U}(\mathrm{x})=\frac{\mathrm{a}}{\mathrm{x}^{12}}-\frac{\mathrm{b}}{\mathrm{x}^{6}}$ , where a and b are constant and x is the distance between the atoms. if the dissociation energy of the molecule is $\left[\mathrm{U}(\mathrm{x}=\infty)-\mathrm{U}_{\text {at equilibrium }}\right], \mathrm{D}$ is :

( 1)$\frac{b^{2}}{6 a}$

( 2)$\frac{\mathrm{b}^{2}}{2 \mathrm{a}}$

(3) $\frac{b^{2}}{12 a}$

(4) $\frac{\mathrm{b}^{2}}{4 \mathrm{a}}$

[AIEEE-2010]

Sol. (4)

$\mathrm{U}(\mathrm{x}=\infty)=0,$ at equilibrium

$\frac{\mathrm{d} \mathrm{U}}{\mathrm{dx}}=0 ;-\frac{12 \mathrm{a}}{\mathrm{x}^{13}}+\frac{6 \mathrm{b}}{\mathrm{x}^{7}}=0 \Rightarrow \mathrm{x}^{6}=\frac{2 \mathrm{a}}{\mathrm{b}}$


Q. At time t = 0s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time ‘t’, the net force acting on it must be proportional to :-

(1) $\sqrt{t}$

(2) constant

(3) t

(4) $\frac{1}{\sqrt{t}}$

[AIEEE-2011]

Sol. (4)

$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{kt}$

$\mathrm{v}=\sqrt{\frac{2 \mathrm{kt}}{\mathrm{m}}}$

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{mt}}}$

$\mathrm{F}=\mathrm{m} \cdot \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$

$\mathrm{F} \propto \frac{1}{\sqrt{\mathrm{t}}}$


Q. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

If two springs $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of force constants $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$, respectively, are stretched by the same force, it is found that more work is done on spring $\mathrm{S}_{1}$ than on spring $\mathrm{S}_{2}$.

Statement-1: If stretched by the same amount, work done on $\mathrm{S}_{1}$, will be more than that on $\mathrm{S}_{2}$

Statement-2 : $\mathrm{k}_{1}$ < $\mathrm{k}_{2}$

(1) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is false

(4) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of statement-1.

[AIEEE-2012]

Sol. (2)

Given same force $\mathrm{F}=\mathrm{k}_{1} \mathrm{x}_{1}=\mathrm{k}_{2} \mathrm{x}_{2}$

$\Rightarrow \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}$

$\mathrm{W}_{1}=\frac{1}{2} \mathrm{k}_{1} \mathrm{x}_{1}^{2} \& \mathrm{W}_{2}=\frac{1}{2} \mathrm{k}_{2} \mathrm{x}_{2}^{2}$

As $\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}>1$ so $\frac{\frac{1}{2} \mathrm{k}_{1} \mathrm{x}_{1}^{2}}{\frac{1}{2} \mathrm{k}_{2} \mathrm{x}_{2}^{2}}>1$

$\Rightarrow \frac{\mathrm{Fx}_{1}}{\mathrm{Fx}_{2}}>1 \Rightarrow \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}>1$

$\therefore \mathrm{k}_{2}>\mathrm{k}_{1}$ statement 2 is true

OR

if $\mathrm{x}_{1}=\mathrm{x}_{2}=\mathrm{x}$

$\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}=\frac{\frac{1}{2} \mathrm{K}_{1} \mathrm{x}^{2}}{\frac{1}{2} \mathrm{K}_{2} \mathrm{x}^{2}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}$

$\therefore \frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}<1$

$\therefore \mathrm{W}_{1}<\mathrm{W}_{2}$

statement 1 is false


Q. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude $\mathrm{F}=\mathrm{ax}+\mathrm{bx}^{2}$ where a and b are constants. The work done in stretching the unstretched rubber-band by L is:-

(1) $\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$

(2) $\frac{1}{2}\left(\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{b} \mathrm{L}^{3}}{3}\right)$

(3) aL $^{2}+\mathrm{bL}^{3}$

(4) $\frac{1}{2}\left(\mathrm{aL}^{2}+\mathrm{bL}^{3}\right)$

[JEE-Mains-2014]

Sol. (1)

Work done $=\int_{0}^{\mathrm{L}} \mathrm{Fdx}$

$=\int_{0}^{\mathrm{L}}\left(\mathrm{ax}+\mathrm{bx}^{2}\right) \mathrm{d} \mathrm{x}$

$=\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$


Q. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7}$ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 $\mathrm{ms}^{-2}$ :-

(1) $12.89 \times 10^{-3} \mathrm{kg}$

(2) $2.45 \times 10^{-3} \mathrm{kg}$

(3) $6.45 \times 10^{-3} \mathrm{kg}$

(4) $9.89 \times 10^{-3} \mathrm{kg}$

[JEE-Mains-2016]

Sol. (1)

Work done against gravity = (mgh) 1000

in lifting 1000 times

$=10 \times 9.8 \times 10^{3}$

$=9.8 \times 10^{4}$ Joule

$20 \%$ efficiency is to converts fat into energy.

$\left[20 \% \text { of } 3.8 \times 10^{7} \mathrm{J}\right] \times(\mathrm{m})=9.8 \times 10^{4}$

(Where m is mass)

$\mathrm{m}=12.89 \times 10^{-3} \mathrm{kg}$


Q. A point particle of mass, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals . The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and PR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction $\mu$ and the distance x(=QR) are, respecitvely close to :

[JEE-Mains-2016]

Sol. (4)


Q. A body of mass m $=10^{-2}$ kg is moving in a medium and experiences a frictional force F = $-\mathrm{K} \mathrm{V}^{2}$. Its intial speed is $\mathrm{v}_{0}$ = 10 $\mathrm{ms}^{-1}$. If, after 10 s, its energy is , the value of k will be :-

(1) $10^{-4} \mathrm{kg} \mathrm{m}^{-1}$

(2) $10^{-1} \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}$

(3) $10^{-3} \mathrm{kg} \mathrm{m}^{-1}$

(4) $10^{-3} \mathrm{kg} \mathrm{s}^{-1}$

[JEE MAINS-2017]

Sol. (1)

$\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{8} \mathrm{mv}_{0}^{2}$

$\mathrm{v}_{\mathrm{f}}=\frac{\mathrm{v}_{0}}{2}=5 \mathrm{m} / \mathrm{s}$

$\left(10^{-2}\right) \frac{d V}{d t}=-k v^{2}$

$\int_{10}^{5} \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-100 \mathrm{k} \int_{0}^{10} \mathrm{dt}$

$\frac{1}{5}-\frac{1}{10}=100 \mathrm{k}(10)$

$\mathrm{k}=10^{-4}$


Q. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :

(1) 9 J            (2) 18 J              (3) 4.5 J            (4) 22 J

[JEE MAINS-2017]

Sol. (3)

F = 6t = ma

$\Rightarrow \mathrm{a}=6 \mathrm{t}$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}$

$\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\int_{0}^{1} 6 \mathrm{t} \mathrm{dt}$

$\mathrm{v}=\left(3 \mathrm{t}^{2}\right)_{0}^{1}=3 \mathrm{m} / \mathrm{s}$

From work energy theorem

$\mathrm{W}_{\mathrm{F}}=\Delta \mathrm{K.E}=\frac{1}{2} \mathrm{m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$

$=\frac{1}{2}(1)(9-0)=4.5 \mathrm{J}$


Q. A particle is moving in a circular path of radius a under the action of an attractive potential $\mathrm{U}=-\frac{\mathrm{k}}{2 \mathrm{r}^{2}} .$ Its total energy is :-.

(1) $\frac{\mathrm{k}}{2 \mathrm{a}^{2}}$

(2) Zero

(3) $-\frac{3}{2} \frac{\mathrm{k}}{\mathrm{a}^{2}}$

$(4)-\frac{\mathrm{k}}{4 \mathrm{a}^{2}}$

[JEE MAINS-2018]

Sol. (2)

$\mathrm{F}=-\frac{\partial \mathrm{u}}{\partial \mathrm{r}}=\frac{\mathrm{K}}{\mathrm{r}^{3}}$

Since it is performing circular motion

$\mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{K}}{\mathrm{r}^{3}}$

$\mathrm{mv}^{2}=\frac{\mathrm{K}}{\mathrm{r}^{2}}$

$\Rightarrow \mathrm{K.E.}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{K}}{2 \mathrm{r}^{2}}$

Total energy = P.E. + K.E.

$=-\frac{\mathrm{K}}{2 \mathrm{r}^{2}}+\frac{\mathrm{K}}{2 \mathrm{r}^{2}}=\mathrm{Zero}$


Radioactivity – JEE Main Previous Year Questions with Solutions

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Simulator

Previous Years AIEEE/JEE Mains Questions

Q. The half life of a radioactive substance is 20 minutes. The approximate time interval $\left(\mathfrak{t}_{2}-\mathfrak{t}_{1}\right)$ between the time $t_{2}$ when $\frac{2}{3}$ of it has decayed and time $t_{1}$ when $\frac{1}{3}$ of it had decayed is :-

(1) 20 min (2) 28 min (3) 7 min (4) 14 min

[AIEEE – 2011]

Sol. (1)

\because \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{t / T} \quad \therefore \frac{1}{3}=\left[\frac{1}{2}\right]^{\frac{t_{2}}{T}} \& \frac{2}{3}=\left[\frac{1}{2}\right]^{\frac{t_{1}}{T}}

\Rightarrow \frac{1}{2}=\left[\frac{1}{2}\right]^{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \frac{1}{\mathrm{T}}} \Rightarrow 1=\frac{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}{\mathrm{T}} \Rightarrow \mathrm{T}=\mathrm{t}_{2}-\mathrm{t}_{1}

\Rightarrow \mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{min}


Q. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes , the ratio of decayed numbers of A and B nuclei will be :-

(1) 5 : 4 (2) 1 : 16 (3) 4 : 1 (4) 1 : 4

[JEE-Mains – 2016]

Sol. (1)

\mathrm{t}=80 \mathrm{min}=4 \mathrm{T}_{\mathrm{A}}=2 \mathrm{T}_{\mathrm{B}}

\therefore \text { no. of nuclei of } \mathrm{A} \text { decayed }=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{4}}=\frac{15 \mathrm{N}_{0}}{16}

\therefore \text { no. of nuclei of } \mathrm{B} \text { decayed }=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{2}}=\frac{3 \mathrm{N}_{0}}{4}

required ratio $=\frac{5}{4}$


Q. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

(1) $\mathrm{t}=\mathrm{T} \log (1.3)$

$(2) \mathrm{t}=\frac{\mathrm{T}}{\log (1.3)}$

(3) $\mathrm{t}=\frac{\mathrm{T}}{2} \frac{\log 2}{\log 1.3}$

(4) \mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2}

[JEE-Mains – 2017]

Sol. (4)

At time t

$\frac{\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}=.3 \Rightarrow \mathrm{N}_{\mathrm{B}}=.3 \mathrm{N}_{\mathrm{A}}$

also let initially there are total $\mathrm{N}_{0}$ number of nuclei

$\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}$

$\mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{0}}{1.3}$

Also as we know

$\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\frac{\mathrm{N}_{0}}{1.3}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\frac{1}{1.3}=\mathrm{e}^{-\lambda t} \Rightarrow \ell \mathrm{n}(1.3)=\lambda \mathrm{t}$ or $\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\lambda}$

$\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\frac{\ell \mathrm{n}(2)}{\mathrm{T}}}=\frac{\ell \mathrm{n}(1.3)}{\ell \mathrm{n}(2)} \mathrm{T}$