datavorous/entrance-exam-dataset huggingface dataset:
A ray of light is incident on a $60^{\circ}$ prism at the minimum deviation position. The angle of refraction at the first face (ie, incident face) of the prism is

Options:
A zero
B $30^{\circ}$
C $45^{\circ}$
D $60^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
The refracting angle of prism
$A=r_1+\mathrm{r}_2$
For minimum deviation
$\begin{array}{l}
\mathrm{r}_1 =\mathrm{r}_2=\mathrm{r} \\
\therefore \mathrm{A} =2 \mathrm{r} \\
\text {or } \mathrm{r} =\frac{\mathrm{A}}{2}=\frac{60^{\circ}}{2}=30^{\circ}
\end{array}$

Topic: Physics, Ray Optics, NEET, NEET 2010 (Mains)
A ray of light is incident on a surface of glass slab at an angle 45Â°. If the lateral shift produced per unit thickness is13 m, then the angle of refraction produced at the incident surface is

Options:
A tan-1⁡32
B tan-1⁡1-23
C sin-1⁡1-23
D tan-1⁡23-1

The correct answer is:
tan-1⁡1-23

Solution:
Here, angle of incidence i=45Â°
Lateral shiftdThickness of glass slab t=13
Lateral shift d=tsinâ¡Î´cosâ¡r=tsinâ¡(i-r)cosâ¡r
âdt=sinâ¡(i-r)cosâ¡r
or dt=sinâ¡icosâ¡r-cosâ¡isinâ¡rcosâ¡r
or dt=sinâ¡45Â°cosâ¡r-cosâ¡45Â°sinâ¡rcosâ¡r=cosâ¡r-sinâ¡r2cosâ¡r
or dt=121-tanâ¡r
or 13=121-tanâ¡r
or tanâ¡r=1-23
or r=tan-1â¡1-23

Topic: Physics, Ray Optics, JEE Main
A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster's angle $\phi$. If $\mu$ represents the refractive index of glass with respect to air, then the angle between reflected and refracted rays is

Options:
A $90+\phi$
B $\sin ^{-1}(\mu \cos \phi)$
C $90^{\circ}$
D $90^{\circ}-\sin ^{-1}(\sin \phi / \mu)$

The correct answer is:
$90^{\circ}$

Solution:
At polarizing angle, the reflected and refracted rays are mutually perpendicular to each other.

Topic: Physics, Wave Optics, JEE Main
A ray of light moving parallel to the $x$-axis gets reflected from a parabolic mirror whose equation is $(y-2)^2=4(x+1)$. After reflection, the ray must pass through the point

Options:
A $(-2,0)$
B $(-1,2)$
C $(0,2)$
D $(2,0)$

The correct answer is:
$(0,2)$

Solution:
Given parabola is $(y-2)^2=4(x+1)$
$\Rightarrow \mathrm{Y}^2=4 \mathrm{aX}$ where $\mathrm{X}=\mathrm{x}+1, \mathrm{Y}=\mathrm{Y}-2, \mathrm{a}=1$
We know focus of parabola $Y^2=4 \mathrm{aX}$ is $(\mathrm{X}=-\mathrm{a}, \mathrm{Y}=0)$
$\Rightarrow(\mathrm{x}+1=1, \mathrm{y}-2=0)=(0, 2)$
Thus focus of the given parabola is $(0,2)$
We know every reflected ray from parabolic mirror passes through its focus.
Hence option '3' is correct choice.

Topic: Mathematics, Parabola, JEE Main
A ray of light moving parallel to $\mathrm{x}$-axis gets reflected from the parabolic mirror whose equation is given by $y^2-10 y-8 x+41=0$ After reflection, the ray must passes through the point

Options:
A $(-4,-5)$
B $(-4,5)$
C $(4,5)$
D $(4,-5)$

The correct answer is:
$(4,5)$

Solution:
As $y^2-10 y-8 x+41=0$
$\begin{aligned} & \Rightarrow(\mathrm{y}-5)^2=8(\mathrm{x}-2) \\ & \Rightarrow \mathrm{Y}^2=8 \mathrm{X}=4(2) \mathrm{X}\end{aligned}$
Where $Y=y-5, X=x-2$
$\Rightarrow x-2=2, y-5=0$
$\Rightarrow x-2=2, y-5=0$
$\Rightarrow x=4, y=5$
$\therefore$ After reflection, the ray must pass through the focus $(4,5)$ Hence choice (3) is correct answer

Topic: Mathematics, Parabola, JEE Main
A ray of light of intensity $\mathrm{I}$ is incident on a parallel glass slab at point $\mathrm{A}$ as shown in diagram. It undergoes partial reflection and refraction. At each reflection, $25 \%$ of incident energy is reflected. The rays $\mathrm{AB}$ and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ undergo interference. The ratio of $\mathrm{I}_{\max }$ and $\mathrm{I}_{\min }$ is :

Options:
A
$49: 1$

B
$7: 1$

C
$4: 1$

D
$8: 1$

The correct answer is:
$49: 1$

Solution:
From figure $I_1=\frac{I}{4}$ and $I_2=\frac{9 I}{64}$
$\Rightarrow \frac{I_2}{I_1}=\frac{9}{16}$
By using $\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{\frac{I_2}{I_1}}+1}{\sqrt{\frac{I_2}{I_1}}-1}\right)^2$
$=\left(\frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1}\right)^2=\frac{49}{1}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (09 Apr Online)
A ray of light of intensity I is incident on a parallel glass-slab at the point A and it undergoes partial reflection and refraction as shown in the figure. At each reflection, 25% of the incident energy is reflected and the rest is transmitted. If the rays AB and A′B′ undergo interference, then the ratio ImaxImin is

Options:
A 4Â :1
B 8Â :1
C 7Â :1
D 49 :1

The correct answer is:
49 :1

Solution:
From figure I1=I4 and I2=9I64⇒I2I1=916
By using ImaxImin= Iâ¡ 2 Iâ¡ 1 + 1 Iâ¡ 2 Iâ¡ 1 - 1 2 = 9 1 6 + 1 9 1 6 - 1 2 = 4 9 1

Topic: Physics, Wave Optics, JEE Main
A ray of light passes four transparent glass slabs having refractive indices μ1, μ2, μ3 and μ4 as shown in the figure. If the emergent ray CD is parallel to the incident ray AB, then

Options:
A μ1=μ2
B μ2=μ3
C μ3=μ4
D μ4=μ1

The correct answer is:
μ4=μ1

Solution:
Applying Snell's law at B and C,

Î¼ sini=constant or Î¼1siniB=Î¼4 sub ic

But since A B||C D

â´ iB=iC

or Î¼1=Î¼4

The Snell's law, which the students read in their plus two syllabus Î¼=sinisinrcomes from Î¼sini=constant

Î¼1=Î¼1sini1=Î¼2sini2

Î¼2=Î¼ or sini=Î¼sinr

or Î¼=sinisinr

Topic: Physics, Wave Optics, JEE Main
A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angle of reflection and refraction are respectively r and r'. The critical angle is given by,

Options:
A sin-1(cotr)
B tan-1(sini)
C sin-1tanr'
D sin-1(tanr)

The correct answer is:
sin-1(tanr)

Solution:
r+r'+90°=180°⇒r'=90-r=90-i

n1sini=n2sinr'=n2sin(90-i)
n1sini=n2cosi⇒tani=n2n1
Now sinC=n2n1=tani
⇒C=sin-1(tani)=sin-1(tanr)

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
A ray of light passes from a medium \( A \) having refractive index \( 1.6 \) to the medium \( B \) having refractive index \( 1.5 \). The value
of critical angle of medium \( A \) is

Options:
A \( \sin ^{-1}\left(\frac{16}{15}\right) \)
B \( \sin ^{-1} \sqrt{\frac{16}{15}} \)
C \( \sin ^{-1}\left(\frac{1}{2}\right) \)
D \( \sin ^{-1}\left(\frac{15}{16}\right) \)

The correct answer is:
\( \sin ^{-1}\left(\frac{15}{16}\right) \)

Solution:
As the law of refraction we have,

sinicsin90=1.51.6,

where ic is the critical angle.

sinic=1516

ic=sin-11516

Topic: Physics, Ray Optics, JEE Main
A ray of light passes from vacuum into a medium of refractive index \( \mu \), the angle of incidence is found to be twice the angle of refraction. Then the angle of incidence is

Options:
A \( \cos ^{-1}\left(\frac{\mu}{2}\right) \)
B \( 2 \cos ^{-1}\left(\frac{\mu}{2}\right) \)
C \( 2 \sin ^{-1} \mu \)
D \( 2 \sin ^{-1}\left(\frac{\mu}{2}\right) \)

The correct answer is:
\( 2 \cos ^{-1}\left(\frac{\mu}{2}\right) \)

Solution:
Let i be the incident angle and r be the refracted angle.

Given,

i=2rr=i2

The refractive index of vacuum is 1 and that another medium is μ. From Snell's law.

1 sin i=μ sin r

sin i=μ sin i2

2 sin i2cosi2=μ sin i2

2 cosi2-μ=0i=2cos-1μ2

The angle of incidence is

∴ i=2cos-1μ2

Topic: Physics, Ray Optics, JEE Main
A ray of light passes through four transparent media with refractive indices $\mu_1, \mu_2 \mu_3$, and $\mu_4$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$, we must have

Options:
A $\mu_1=\mu_2$
B $\mu_2=\mu_3$
C $\mu_3=\mu_4$
D $\mu_4=\mu_1$

The correct answer is:
$\mu_4=\mu_1$

Solution:
The correct option is $\mathbf{4} \mu_1=\mu_4$
Here, incident ray $A B$ is parallel to emergent ray $C D$
$\therefore \theta_1=\theta_4$
Since, incident ray $A B$ is parallel to emergent ray $C D$, so we can use
$\mu \sin \theta=$ constant
So, for first and last media,
$\begin{aligned} & \mu_1 \sin \theta_1=\mu_4 \sin \theta_4 \\ & \Rightarrow \mu_1=\mu_4 \quad\left(\because \theta_1=\theta_4\right)\end{aligned}$
Hence, option (4) is correct.

Topic: Physics, Ray Optics, JEE Main
A ray of light passing through a prism (μ=3) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is (in degrees).

Options:

The correct answer is:
60

Solution:
At minimum deviation, r1=r2=A2

Also given i=2r1=A
Now 1·sini=3sinr1
1sinA=3sinA2

⇒ 2sinA2cosA2=3sinA2
⇒cosA2=32⇒A2=30°
⇒A=60°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
A ray of light propagates from glass (refractive index $=3 / 2$ ) to water (refractive index $=4 / 3$ ). The value of the critical angle

Options:
A $\sin ^{-1}(1 / 2)$
B $\sin ^{-1}\left(\frac{\sqrt{8}}{9}\right)$
C $\sin ^{-1}(8 / 9)$
D $\sin ^{-1}(5 / 7)$

The correct answer is:
$\sin ^{-1}(8 / 9)$

Solution:
$C=\sin ^{-1}\left(\frac{1}{{ }_w \mu_g}\right)=\sin ^{-1}\left(\frac{\mu_w}{\mu_g}\right)=\sin ^{-1}\left(\frac{8}{9}\right)$

Topic: Physics, Ray Optics, JEE Main
A ray of light R1 is incident on a glass slab at an angle equal to critical angle of glass-air system. The refracted ray R2 undergoes partial reflection and refraction at other surface of slab. The angle between the reflected ray R3 and refracted ray R4 is

Options:
A 45°
B 135°
C 105Â°
D 95°

The correct answer is:
105°

Solution:
sin⁡45°sin⁡θ=2
θ=30°
α=180°-30°+45°=105°

Topic: Physics, Experimental Physics, JEE Main
A ray of light through the point A1,2,3 strikes the plane x+y+z=12 at a point B and on reflection passes through the point C3,5,9. If the equation of a plane containing the incident ray and the reflected ray is P=0 and the distance of P=0 from 0,0,0 is Î» units, then the value of 13Î»2 is equal to

Options:
A 1
B 2
C 4
D 6

The correct answer is:
2

Solution:
A plane which contains the incident ray and the reflected ray must contain the points A,B,C and should be perpendicular to the given plane.
So, equation of the plane is x-1y-2z-3111236=0
âx-13-y-24+z-33-2=0
â3x-3-4y+8+z-3=0
â3x-4y+z+2=0
Hence, the distance of the plane from 0,0,0 is 0-0+0+29+16+1=226 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A ray of light travelling in a denser medium strikes the boundary of a rarer medium. The ray gets partially reflected and refracted. The angle of reflection is r and that of refraction is r'. If the reflected and refracted rays make an angle of 90Â° with each other then the critical angle for the pair of media is

Options:
A sin-1 (tan r)
B tan-1 (sin r)
C sin-1 (tan r')
D tan-1 (sin r')

The correct answer is:
sin-1 (tan r)

Solution:
D μ R = sin r sin r ′ -----(1)
At critical angle D μ R = sin C sin 90 o -----(2)
from equation (1) & (2)
sin C = sin r sin 90 - r ⇒ sin r = tan r ⇒ C = sin -1 tan r

Topic: Physics, Ray Optics, JEE Main
A ray of light, travelling in medium A, is incident on plane interface of two media A and B and gets refracted into medium B. The angle of incidence is i and that of refraction is r. Graph between sin(i) and sin(r) is as shown in diagram. The correct statement among the following is

Options:
A Speed of light in medium B is three fourth of that in medium A
B Total internal reflection can take place
C Refraction index of medium A is greater than that of medium B
D None of these

The correct answer is:
Speed of light in medium B is three fourth of that in medium A

Solution:
sin i sin r = cot 3 7 = 4 3 = V A V B
as going from rarer to denser no TIR can take place
μ B μ A > 1 ⇒ μ B > μ A

Topic: Physics, Ray Optics, JEE Main
A ray of light travelling in water is incident on a glass plate immersed in it. When the angle of incidence is 51o, the reflected ray is totally polarized. Find out the refractive index of glass.

[The refractive index of water is 1.3 and tan 51o=1.235]

Options:
A 1.605
B 1.305
C 1.33
D 1.805

The correct answer is:
1.605

Solution:
μgμw=sin⁡isin⁡r=tan⁡i ; for total plane polarization.
{r=90-i}
μg=μw×tan⁡51
=1.605

Topic: Physics, Wave Optics, JEE Main
A ray of sunlight enters a spherical water droplet (n =43) at an angle of incidence 53Â° measured with respect to the normal to the surface. It is reflected from the back surface of the droplet and re-enters into the air. The angle between the incoming and outgoing ray is [Take sin 53Â° = 0.8]

Options:
A 15°

B 34°

C 138°

D 30°

The correct answer is:
138°

Solution:
Applying law of refraction⇒1×sin 53°=43×sin r
45=43sin r⇒sin r=35⇒sin r=0.6⇒r=37°
δ1=53°-37°⇒δ1=16°C.W.
δ2=180-2×37°⇒δ2=106°C.W.
δ3=53°-37°⇒δ3=16°C.W.
δnet=δ1+δ2+δ3=138°

Topic: Physics, Ray Optics, JEE Main
A reactant A forms two products
A → k 1 B; Activation energy Ea1
A → k 2 C; Activation energy Ea2
If Ea2=2Ea1, then k1 and k2 will be related as

Options:
A k 2  =  k 1 e − E a 1 RT
B k2=k1eEa2RT
C k1=Ak2eEa1RT
D k1=2k2eEa2RT

The correct answer is:
k 2  =  k 1 e − E a 1 RT

Solution:
k2=Ae-Ea2RT,k1=Ae-Ea1RT
So, k 2 k 1 â=â e â 2E a 1 +E a 1 RT
Since, Ea2=2Ea1
k 2  =  k 1 e − E a 1 RT

Topic: Chemistry, Chemical Kinetics, JEE Main
A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following:

Options:
A ∆H is negative while ∆S is positive
B Both ∆H and ∆S are negative
C ∆H is positive while ∆S is negative
D Both ∆H and ∆S are positive

The correct answer is:
Both ∆H and ∆S are positive

Solution:
∆G = ∆H – T∆S (For Spontaneity) (∆G < 0)
So If ∆H > 0 and ∆S > 0 for spontaneity then the process is feasible at high temperature and non-feasible at low temperature.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2016 (09 Apr Online)
A reaction between a carbonyl compound and a Grignard reagent is termed as

Options:
A nucleophilic addition
B electrophilic addition
C $\mathrm{S}_{\mathrm{N}} \mathrm{l}$ reaction
D $\mathrm{S}_{\mathrm{N}}$ 2 reaction

The correct answer is:
nucleophilic addition

Solution:
Addition of Grignard reagents to carbonyl compounds is an example of nucleophilic addition reactions.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A reaction between substance A and B is represented as

A+B→  C

Observations on the rate of this reaction are obtained as

Order of reaction will respect to A and B respectively are :-

Options:
A 1, 2
B 1, 1
C 2, 1
D 2, 2

The correct answer is:
1, 2

Solution:
r=k[A]x[B]y
5âÃâ 10 â3 â=âk(0 .1) 2 â (1) y ...(1)
2âÃâ 10 â2 â=âk(0 .1) 2 â (2) y ...(2)
equation 1Ã·2
0.1 0.1 x â 1 2 y â=â 5Ã10 â3 2Ã10 â2
12y=14
y=2
2.5 ×  10 −3  = k(0 .05) x   (1) y ...(3)
equation 1Ã·3
0.1 0.05 x â 1 1 y â=â 5Ã10 â3 2 .5Ã10 â3
10050x=2
x=1
Order of reaction will respect to A and B respectively are 1 and 2

Topic: Chemistry, Chemical Kinetics, JEE Main
A reaction has both \( \Delta \mathrm{H} \) and \( \Delta \mathrm{S} \) - ve. The rate of reaction

Options:
A increases with increase in temperature
B increases with decrease in temperature
C remains unaffected by change in temperature
D cannot be predicted for change in temperature

The correct answer is:
increases with decrease in temperature

Solution:
The reaction is thermodynamic ally favored when \( \Delta G < 0 \)
\[
\begin{array}{l}
\Delta G=\Delta H-T \Delta S \\
\Delta G=-v e-T(-v e) \\
\Delta G=-v e+T(+v e)
\end{array}
\]
Therefore, on increasing the temperature \( \Delta \mathrm{G} \) becomes positive (+ve). Hence, the rate of reaction increases with
decrease in temperature.

Topic: Chemistry, Chemistry in Everyday Life, KCET, KCET 2017
A reaction involving two different reactants can never be

Options:
A
Unimolecular reaction

B
First order reaction

C
second order reaction

D
Bimolecular reaction

The correct answer is:
Unimolecular reaction

Solution:
No Solution Available

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2005
A reaction is 50% completed in 2 hours and 75% completed in 4 hours. The order of the reaction is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
A reaction is 50% completed in 2 hours and 75% completed in 4 hours
t 3/4 =2 t 1/2 for first-order reaction

Topic: Chemistry, Chemical Kinetics, JEE Main
A reaction is $A+B \rightarrow C+D$. Initially we start with equal concentration of $A$ and $B$. At equilibrium we find the moles of $C$ is two times of $A$. What is the equilibrium constant of the reaction.

Options:
A 4
B 2
C $1 / 4$
D $1 / 2$

The correct answer is:
4

Solution:
$A+B \rightleftharpoons C+D$
at equilibrium $a a \quad 2 \alpha 2 a$
$K=\frac{2 a \times 2 a}{a \times a}=4$

Topic: Chemistry, Chemical Equilibrium, JEE Main
A reaction is carried out at 600 K . If the same reaction is carried out in the presence of catalyst at the same rate and same frequency factor, the temperature required is 500 K . What is the activation energy of the reaction, if the catalyst lowers the activation energy barrier by 20 kJ/mol?

Options:
A 100 kJ/mol
B 120 kJ/mol
C 80
D None of these

The correct answer is:
120 kJ/mol

Solution:
k=Ae-Ea/RÃ600
k â² = Ae â Eaâ20 RÃ500
k= k â²
EaRÃ600=Ea-20RÃ500
Ea6=Ea-205
56Ea=Ea-20
20=Ea-56Ea
Ea=20Ã6
Ea=120 kJ/mol

Topic: Chemistry, Chemical Kinetics, JEE Main
A reaction is first order in A and second order in B : How is the rate affected when the concentrations of both A and B are doubled ?

Options:

The correct answer is:
8

Solution:
Rate = k[2A] [2B]2 = 8k[A] [B]2 Rate of reaction becomes 8 times

Topic: Chemistry, Chemical Kinetics, JEE Main
A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n×10-1, when n=_____ (Round off to the Nearest Integer).
Given : Atomic masses: C: 12.0 u, H:1.0 u, N:14.0 u, Br : 80.0 u

Options:

The correct answer is:
3

Solution:
no of moles=3 ×10-1 mol

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet-coloured) and B (green-coloured). A can show optical activity, but, B is optically inactive. What type of isomers do A and B represent?

Options:
A Coordination isomers
B Linkage isomers
C Ionisation isomers
D Geometrical isomers

The correct answer is:
Geometrical isomers

Solution:
The coordination number of cobalt is 6.

The mole ratio of CoCl3 and en is 1 : 2.

Hence,

CoCl3+2 en→Coen2Cl2 Cl

The complex formed shows geometrical isomerism.

The complex exists as cis and trans isomers among which the trans isomer is optically inactive.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \mathrm{~min}^{-1}$. If we start with $[A]=0.5 \mathrm{~mol} l^{-1}$, when would $[A]$ reach the value $0.05 \mathrm{~mol} l^{-1}$

Options:
A $0.384 \mathrm{~min}$
B $0.15 \mathrm{~min}$
C $3 \mathrm{~min}$
D $3.84 \mathrm{~min}$

The correct answer is:
$0.384 \mathrm{~min}$

Solution:
We know that for first order kinetics
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
$(a-x)=0.05 \mathrm{~mol} \mathrm{L}$
$6=\frac{2.303}{t} \log \frac{0.5}{0.05}$
or $t=\frac{2.303}{6} \log \frac{0.5}{0.05}=\frac{2.303}{6}=0.384 \mathrm{~min}$

Topic: Chemistry, Chemical Kinetics, JEE Main
A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will

Options:
A
remain unchanged

B
triple

C
increase by a factor of 4

D
double

The correct answer is:
increase by a factor of 4

Solution:
$\mathrm{R} \propto[\mathrm{W}]^2$
$\mathrm{R}^{\prime} \propto[2 \mathrm{CO}]^2$
$\mathrm{R} \propto 4[\mathrm{~W}]^2$
$\mathrm{R} \propto 4 \mathrm{M}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2006
A reagent which gives brilliant red precipitate with Nickel ions in basic medium is

Options:
A sodium nitroprusside
B neutral FeCl3
C meta-dinitrobenzene
D dimethyl glyoxime

The correct answer is:
dimethyl glyoxime

Solution:
The reagent that gives a brilliant red precipitate with nickel ions in basic medium is dimethylglyoxime (DMG) or its sodium salt, sodium dimethylglyoximate. This reaction is commonly used in qualitative analysis to detect the presence of nickel ions in a solution. When DMG is added to a basic solution containing nickel ions, a red precipitate of nickel dimethylglyoximate is formed. This reaction is often used as a confirmatory test for the presence of nickel ions.

Ni2++2dmg-→Ni(dmg)2

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A real gas most closely approaches the behaviour of an ideal gas at

Options:
A $15 \mathrm{~atm}$ and $200 \mathrm{~K}$
B $1 \mathrm{~atm}$ and $273 \mathrm{~K}$
C $0.5 \mathrm{~atm}$ and $500 \mathrm{~K}$
D $15 \mathrm{otm}$ and $500 \mathrm{~K}$

The correct answer is:
$0.5 \mathrm{~atm}$ and $500 \mathrm{~K}$

Solution:
$0.5 \mathrm{~atm}$ and $500 \mathrm{~K}$

Topic: Chemistry, States of Matter, JEE Main
A real image is formed by a convex lens. If we put a concave lens in contact with it, the combination again forms a real image. The new image

Options:
A is closer to the lens system
B is farther from the lens system
C is at the original position
D may be anywhere depending on the focal length of the concave lens.

The correct answer is:
is farther from the lens system

Solution:
When concave lens is put in contact with convex lens, real/image is formed i.e., equivalent lens is convex. The focal length of equivalent lens F is greater than focal length of convex lens.
1v-1-u=1f
⇒1v=1f-1u=u-ffu
v=ufu-f=uuf-1
As f increases, uf decreases
uf-1 also decreases
Thus, v increases.

Topic: Physics, Ray Optics, JEE Main
A real value of a, for which the sum of the roots of the equation x2-2ax+2a-1=0 is equal to the sum of the square of its roots, is

Options:
A 12
B 32
C 52
D 2

The correct answer is:
12

Solution:
Given, Î±+Î²=Î±2+Î²2
Also, Î±+Î²=2a and Î±Î²=2a-1
Î±+Î²=Î±+Î²2-2Î±Î²
â2a=2a2-22a-1
â2a=4a2-4a+2
â4a2-6a+2=0
â2a2-3a+1=0
âa=1,12

Topic: Mathematics, Quadratic Equation, JEE Main
A real valued function $f(x)$ satisfies the functional equation $f(x-y)=f(x) f(y)-f(a-x)$ $f(a+y)$ where $a$ is a given constant and $f(0)=1, f(2 a-x)$ is equal to

Options:
A
$-f(x)$

B
$f(x)$

C
$f(a)+f(a-x)$

D
$f(-x)$

The correct answer is:
$-f(x)$

Solution:
$f(a-(x-a))=f(a) f(x-a)-f(0) f(x)$
$=-f(x)\left[\because x=0, y=0, f(0)=f^2(0)-f^2(a) \Rightarrow f^2(a)=0 \Rightarrow f(a)=0\right]$.

Topic: Mathematics, Functions, JEE Main, JEE Main 2005
A realistic graph depicting the variation of the reciprocal of input resistance in an input characteristics measurement in a common emitter transistor configuration is :

Options:
A

B

C

D

The correct answer is:

Solution:
When V BE exceeds barrier potential current starts following and gets saturated after some values

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (10 Apr Online)
A rectangle \(A B C D\) has its side parallel to the line \(y=2 x\) and vertices \(A, B, D\) are on lines \(y=1, x=1\) and \(x=-1\) respectively. The coordinate of \(C\) can be

Options:
A \((3,8)\)
B \((-3,8)\)
C \((-3,-1)\)
D None of these

The correct answers are:
None of these

Solution:
Hint: Case I
line \(y=2 x\ldots(1)\)
\(A D\) parallel to \(y=2 x\)
\(\Rightarrow \mathrm{M}_{\mathrm{AD}}=2 ~\&~ \mathrm{M}_{\mathrm{AB}}=-\frac{1}{2}\)
for \(a=3\)
\(A(3,1), B(1,2), C(-3,-6), D(-1,-7)\)
\(\begin{aligned} & \text { for } a=-3 \\ & C(3,3), A(-3,1), B(1,-1), D(-1,5)\end{aligned}\)
Case II AD perpendicular to line (1)
\(\Rightarrow M_{A D}=\frac{1}{2}, M_{A B}=2 \Rightarrow b=3-2 a \Rightarrow d=\frac{a+1}{2}\)
for \(a=3, C(-3,-2)\)
for \(a=-3, C(3,7)\)
\({ }^*\) In both cases if abscissa of \(C: 3\) or -3
then co-ordinates of \(C\) can be \((3,3)\) or \((3,7)\) or \((-3,-6)\) or \((-3,-2)\)

Topic: Mathematics, Straight Lines, JEE Main