datavorous/entrance-exam-dataset huggingface dataset:
The driver of a car traveling with speed 30 $\mathrm{m} / \mathrm{sec}$ towards a hill sounds a horn of frequency $600 \mathrm{~Hz}$. If the velocity of sound in air is 330 $\mathrm{m} / \mathrm{s}$, the frequency of reflected sound as heard by driver is :

Options:
A $500 \mathrm{~Hz}$
B $550 \mathrm{~Hz}$
C $555.5 \mathrm{~Hz}$
D $720 \mathrm{~Hz}$

The correct answer is:
$720 \mathrm{~Hz}$

Solution:
$$
\text { } \begin{aligned}
\mathrm{n}^{\prime} & =v\left(\frac{\mathrm{c}+\mathrm{u}}{\mathrm{c}-\mathrm{u}}\right) \\
& =600\left(\frac{330+30}{330-30}\right)=720 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2009 (Mains)
The driver of bus approaching a big wall notices that the frequency of his bus's horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is 330 ms–1:

Options:
A 91 kmh-1
B 81 kmh-1
C 61 kmh-1
D 71 kmh-1

The correct answer is:
91 kmh-1

Solution:
Frequency appeared at wall

fw=330330-v.f ......(1)

f'=330+v330·fw=330+v330-vf

490=330+v330-v·420

v=330×791≈25.38m/s=91Km/s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The $\mathrm{E}_{\mathrm{M}^{+3} / \mathrm{M}^{2+}}^{\circ}$ values for $\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}$ and Co are $-0.41,+1.57,+0.77$ and $+1.97 \mathrm{~V}$ respectively. For which one of these metals the change in oxidation state form $+2$ to $+3$ is easiest?

Options:
A
$\mathrm{Cr}$

B
Co

C
$\mathrm{Fe}$

D
Mn

The correct answer is:
$\mathrm{Cr}$

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2004
The ∂E∂TP of different types of half cells are as follows :
ABCD1×10-42×10-40.1×10-40.2×10-4

(Where E is the electromotive force). Which of the above half cells would be preferred to be used as reference electrode

Options:
A A
B B
C C
D D

The correct answer is:
C

Solution:
Metal Which have lower value of dEdTp is best used for standard or reference half Electrode because if can be used for wider range of temperature Without much derivation from standard Value.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The EAN of central metal atom in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is
[Atomic number of $\mathrm{Co}=27$ ]

Options:
A 34
B 35
C 36
D 32

The correct answer is:
36

Solution:
EAN of central atom in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ $=Z-(\mathrm{ON})+2 \times(\mathrm{CN})$
where, $Z=$ atomic number
$\mathrm{ON}=$ oxidation number
$\mathrm{CN}=$ coordination number
ON of Co $\Rightarrow x+6 \times 0-3=0$
$\therefore \quad x=3$
$\mathrm{CN}=6$
$\therefore \quad \mathrm{EAN}=27-3+2 \times 6=36$

Topic: Chemistry, Coordination Compounds, JEE Main
The EAN of iron in potassium ferricyanide is

Options:
A 18
B 54
C 35
D 23

The correct answer is:
35

Solution:
EAN of a central metal ion=(atomic no. of central atom) - oxidation state + no. of ligands $\times 2=26-3+(6 \times 2)=23+12=35$

Topic: Chemistry, Coordination Compounds, JEE Main
The EAN of metal atoms in Fe CO 2 (NO + ) 2 is 'x' and EAN of metal atoms in Co 2 (CO) 8 is 'y'. The value of xâ+ây is

Options:

The correct answer is:
72

Solution:
Oxidation state of iron in Fe CO 2 NO + 2  = x + 2 0  + 2 = 0; x = −2
So, EAN =26--2+8=36=x
Oxidation state of cobalt in Co2CO8=2x+8-0= or x=0
So, EAN =27+1+8=36=y
So, x+y=72

Topic: Chemistry, Coordination Compounds, JEE Main
The EAN of Zn in [Zn(NH3)4]2+ is

Options:

The correct answer is:
36

Solution:
Z= atomic number of (Zn) metal =30
X= number of electrons lost during oxidation of metal to metal ion =2
Y= number of electrons donated by ligands =(4Ã2)=8
EAN=Z-X+Y=30-2+8=36

Topic: Chemistry, Coordination Compounds, JEE Main
The earliest attempt to the scientific basis for classification is made by

Options:
A Linnaeus
B Whittaker
C Aristotle
D Theophrastus

The correct answer is:
Aristotle

Solution:
Aristotle was a Greek philosopher who classified the living organisms on a scientific basis for the first time. He classified the plants as herbs, shrubs and trees and animals were categorized into two groups, animals with red blood and animals without red blood. The need for a higher and a better classification was always felt. Linnaeus was responsible for the two kingdom classification (Plantae and Animalia) and R.H.Whittaker introduced the five-kingdom classification whose main basis was the mode of nutrition. Theophrastus was a successor to Aristotle and was known for classifying plants and how they reproduced.

Topic: Biology, The Living World, NEET
The earth $E$ moves in an elliptical orbit with the sun $S$ at one of the foci as shown in figure. Its speed of motion will be maximum at the point

Options:
A C
B A
C B
D D

The correct answer is:
A

Solution:
The correct answer is A
According to the Kepler's Second Law, a line between the sun and the planet sweeps equal areas in equal times. Thus, the speed of the planet increases as it nears the sun and decreases as it recedes from the sun. Therefore the speed of the planet will be maximum at point $A$ and minimum at point B

Topic: Physics, Gravitation, JEE Main
The earth is considered as a short magnet with its centre coinciding with the geometric centre of earth. The angle of $\operatorname{dip} \phi$ related to the magnetic latitude $\alpha \lambda$ as

Options:
A $\tan \phi=\frac{1}{2 \tan \alpha}$
B $\tan \lambda=2 \tan \phi$
C $\tan \lambda=2 \tan \phi$
D $\tan \phi=2 \tan \lambda$

The correct answer is:
$\tan \phi=2 \tan \lambda$

Solution:
For a dipole at position $(\mathrm{R}, \mathrm{Q})$
$\mathrm{B}_{\mathrm{R}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \mathrm{M} \cos \theta}{\mathrm{R}^{3}}$ ...(i)
and $\mathrm{B}_{\mathrm{Q}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\mathrm{M} \sin \theta}{\mathrm{R}^{3}}$ ...(ii)
Also $\tan \phi=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}=-\frac{\mathrm{B}_{\mathrm{R}}}{\mathrm{B}_{\mathrm{Q}}}$ ...(iii)

Dividing eq. (i) by (ii)
$\frac{\mathrm{B}_{\mathrm{R}}}{\mathrm{B}_{\mathrm{Q}}}=\frac{2 \cos \theta}{\sin \theta}=2 \cot \theta$
From eq. (iii) and (iv)
$\tan \phi=-2 \cot \theta$
From figure, $\theta=90^{\circ}+\lambda$
$\therefore \quad \tan \phi=-2 \cot (90+\lambda)$
$\tan \phi=2 \tan \lambda$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The earth is moving towards a fixed star with a velocity of 30 km s-1. An observer on the earth observes a shift of 0.58 Å in the wavelength of light coming from the star. What is the actual wavelength of light (in μm ) emitted by the star?

Options:

The correct answer is:
0.58

Solution:
Using, Δλ=vcλ
Here v=30 km s-1=3×104 m s-1
c=3×108 m s-1
Δλ=0.58×10-10m
∴ λ=cvΔλ
=3×1083×104×0.58×10-10 m
=0.58×10-6 m
=0.58 μm

Topic: Physics, Wave Optics, JEE Main
The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by

Options:
A
Rayleigh Jeans law

B
Planck's law of radiation

C
Stefan's law of radiation

D
Wien's law

The correct answer is:
Wien's law

Solution:
No Solution Available

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2003
The earth receives on its surface radiation from the sun at the rate of 1400 W m-2. The distance of the centre of the sun from the surface of the earth is 1.5 Ã1011 m and the radius of the sun is 7.0Ã108 m. Treating sun as a block body, it follows from the above data that its surface temperature is

Options:
A 5801 K

B 106 K

C 50.1 K

D 5801 °C

The correct answer is:
5801 K

Solution:
According to Stefan's law,
dQdT=σT4
The total surface area of the sun = 4 π R s 2
Therefore, the total energy radiated per second by the sun per unit solid angle
= - σ T 4 × 4 π R s 2 4 π = σ T 4 R s 2
Let Res be the distance of the earth from the sun. Hence, the intensity of radiation of the earth
l⁡ = σ T 4 R s 2 / R es 2
∴ 1 4 0 0 = 5.6 × 1 0 - 8 T 4 7.0 × 1 0 8 1.5 × 1 0 1 1 2
T = 5801 K

Topic: Physics, Thermal Properties of Matter, JEE Main
The earth's magnetic field at the equator is approximately \( 0.4 \mathrm{G} \). Then the earth's dipole moment is

Options:
A \( 3.7 \times 10^{23} \mathrm{~A} \mathrm{~m}^{2} \)
B \( \sqrt{2} \mathrm{~A} \mathrm{~m}^{2} \)
C \( 1.5 \times 10^{23} \mathrm{~A} \mathrm{~m}^{2} \)
D \( 1.05 \times 10^{23} \mathrm{~A} \mathrm{~m}^{2} \)

The correct answer is:
\( 1.05 \times 10^{23} \mathrm{~A} \mathrm{~m}^{2} \)

Solution:
Given, earth's magnetic field at equator Beq=0.4 G=0.4×10-4 T.

Radius of the earth is R=6400 km=6.4×106 m.

Let the magnetic moment of the earth be 'M'.

Magnetic field at a point on equator is given by

Beq=μ04πMR3

⇒M=BeqR3μ04π

⇒M=0.4×10-4×6.4×106310-7

⇒M=1.05×1023 A m2.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The earth's magnetic field lines resemble that of a dipole at the centre of the earth. If the magnetic moment of this dipole is close to $8 \times 10^{22} \mathrm{Am}^2$, the value of earth's magnetic field near the equator is close to (radius of the earth $\left.=6.4 \times 10^6 \mathrm{~m}\right)$

Options:
A
$0.6$ Gauss

B
$1.2$ Gauss

C
$1.8$ Gauss

D
$0.32$ Gauss

The correct answer is:
$0.6$ Gauss

Solution:
Given $M=8 \times 10^{22} \mathrm{Am}^2$
$$
\mathrm{d}=\mathrm{R}_{\mathrm{e}}=6.4 \times 10^6 \mathrm{~m}
$$
Earth's magnetic field, $\mathrm{B}=\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{M}}{\mathrm{d}^3}$
$$
\begin{aligned}
&=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 8 \times 10^{22}}{\left(6.4 \times 10^6\right)^3} \\
&\cong 0.6 \text { Gauss }
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2013 (25 Apr Online)
The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

Options:
A 3° < 2° < 1°
B 3Â° > 2Â° > 1Â°
C 3° < 2° > 1°
D 3° > 2° < 1°

The correct answer is:
3Â° > 2Â° > 1Â°

Solution:
Such dehydrohalogenation follow E2 mechanism. The driving force of such reaction is the stability of alkene produced. Since, tertiary alkyl halide can give more substituted alkene, it reacts fastest followed by secondary and primary i.e., 3Â°>2Â°>1Â°.

Topic: Chemistry, General Organic Chemistry, JEE Main
The ease of dehydrohalogenation of alkyl halide with alcoholic \(\mathrm{KOH}\) is

Options:
A \(3^{\circ} < 2^{\circ} < 1^{\circ}\)
B \(3^{\circ} > 2^{\circ} > 1^{\circ}\)
C \(3^{\circ} < 2^{\circ} > 1^{\circ}\)
D \(3^{\circ} > 2^{\circ} < 1^{\circ}\)

The correct answer is:
\(3^{\circ} > 2^{\circ} > 1^{\circ}\)

Solution:
Hints : Such dehydrohalogenation follows \(\mathrm{E}_2\) mechanism. The driving force of such reactions is the stability of alkene produced. Since tetriary alkyl halide can give more substituted alkene, it reacts fastest followed by secondary and primary i.e. \(3^{\circ} > 2^{\circ} > 1^{\circ}\).

Topic: Chemistry, Hydrocarbons, JEE Main
The eccentric angle of a point on the ellipse x2+3y2=6 lying at a distance of 2 units from its centre is

Options:
A π6
B π4
C π3
D π2

The correct answer is:
π4

Solution:
The given equation of ellipse can be written in the form x262+y222=1.

Now let the eccentric angle of the point be θ, then its co-ordinates are 6cosθ,2sinθ.

Since the distance of the point from the centre is 2 units

∴ 6cosθ-02+2sinθ-02=4

⇒6cos2θ+21-cos2θ=4⇒4cos2θ=2

⇒ cosθ=±12 ∴ θ=π4 or 3π4

Topic: Mathematics, Ellipse, JEE Main
The eccentricity of a rectangular hyperbola is

Options:
A 2
B $2 \sqrt{2}$
C 1
D $\sqrt{2}$

The correct answer is:
$\sqrt{2}$

Solution:
$\lim _{x \rightarrow \pi} \frac{(1+\cos x)-\sin x}{(1+\cos x)+\sin x}$
$=\lim _{x \rightarrow \pi} \frac{2 \cos ^{2} x / 2-2(\sin x / 2) \cos x / 2}{2 \cos ^{2} x / 2+2(\sin x / 2) \cos x / 2}$
$=\lim _{x \rightarrow \pi} \frac{2 \cos ^{2} \frac{x}{2}\left(1-\tan \frac{x}{2}\right)}{2 \cos ^{2} \frac{x}{2}\left(x+\tan \frac{x}{2}\right)}$
$=\lim _{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=-1$
$\therefore f(\pi)=\lim _{x \rightarrow \pi} f(x)=-1$

Topic: Mathematics, Hyperbola, JEE Main
The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points 4,-1 and -2,2 is

Options:
A 32
B 34
C 25
D 12

The correct answer is:
32

Solution:
Equation of ellipse whose centre at 0,0 is

x2a2+y2b2=1

It passes through (4,−1)

⇒16a2+1b2=1

⇒16b2+a2=a2b2 ...i

Again it passes through (−2,2)

⇒4a2+4b2=1

⇒4b2+4a2=a2b2 ...ii

From the equation i and ii, we get,

16b2+a2=4a2+4b2

⇒a2=4b2

Now, we have, e2=1-b2a2=1-14=34

⇒e=32.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2017 (09 Apr Online)
The eccentricity of an ellipse whose centre is at the origin is 12 . If one of its directrices is x=-4 , then the equation of the normal to it at 1,32 is:

Options:
A 2y-x=2
B 4x-2y=1
C 4x+2y=7
D x+2y=4

The correct answer is:
4x-2y=1

Solution:
Given, e=12 and ae=4
⇒a=2,

Also, b2=a21-e2
⇒b2=3

Equation of ellipse is x24+y23=1
We know that, equation of normal to the ellipse x2a2+y2b2=1 at x1, y1 is a2xx1-b2yy1=a2-b2
So, equation of normal at 1, 32 is 4x1-3y3/2=4-3
⇒4x-2y=1

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2017 (02 Apr)
The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is $x=$ 4 , then the equation of the ellipse is

Options:
A
$3 x^2+4 y^2=1$

B
$3 x^2+4 y^2=12$

C
$4 x^2+3 y^2=12$

D
$4 x^2+3 y^2=1$

The correct answer is:
$3 x^2+4 y^2=12$

Solution:
Equation of directrix is $x=a / e=4 \Rightarrow a=2$
$$
b^2=a^2\left(1-e^2\right) \Rightarrow b^2=3
$$
Hence equation of ellipse is $3 x^2+4 y^2=12$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2004
The eccentricity of an ellipse, with its centre at the origin, is $1 / 2$. If one of the directrices is $\mathrm{x}=4$ then the equation of the ellipse is:

Options:
A $4 x^{2}+3 y^{2}=1$
B $3 x^{2}+4 y^{2}=12$
C $4 x^{2}+3 y^{2}=12$
D $3 x^{2}+4 y^{2}=1$

The correct answer is:
$3 x^{2}+4 y^{2}=12$

Solution:
$\mathrm{e}=\frac{1}{2} .$ Directrix $, \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4$

$$

\therefore \mathrm{a}=4 \times \frac{1}{2}=2 \quad \therefore \mathrm{b}=2 \sqrt{1-\frac{1}{4}}=\sqrt{3}

$$

Equation of ellispe is

$$

\frac{x^{2}}{4}+\frac{y^{2}}{3}=1 \Rightarrow 3 x^{2}+4 y^{2}=12

$$

Topic: Mathematics, Ellipse, JEE Main
The eccentricity of curve $x^2-y^2=1$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{\sqrt{2}}$
C $2$
D $\sqrt{2}$

The correct answer is:
$\sqrt{2}$

Solution:
Since it is a rectangular hyperbola, therefore eccentricity $e=\sqrt{2}$.

Topic: Mathematics, Hyperbola, JEE Main
The eccentricity of earth's orbit is 0.0167 . The ratio of its maximum speed in its orbit to its minimum speed is

Options:
A $2.507$
B $1.033$
C $8.324$
D $1.000$

The correct answer is:
$1.033$

Solution:
$\frac{v_{\text {max }}}{v_{\text {min }}}=\frac{1+e}{1-e}=\frac{1+0.0167}{1-0.0167}=1.033$

Topic: Physics, Gravitation, JEE Main
The eccentricity of ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

Options:
A $\frac{7}{16}$
B $\frac{5}{4}$
C $\frac{\sqrt{7}}{4}$
D $\frac{\sqrt{7}}{2}$

The correct answer is:
$\frac{\sqrt{7}}{4}$

Solution:
Given ellipse is
$\frac{x^2}{16}+\frac{y^2}{9}=1$
Here, $a^2=16, b^2=9$
Now, $b^2=a^2\left(1-e^2\right)$
$\begin{array}{rlrl} & 9 & =16\left(1-e^2\right) \\ \Rightarrow & & 1-e^2 & =\frac{9}{16} \\ \Rightarrow & e^2 & =1-\frac{9}{16}=\frac{7}{16} \\ \Rightarrow & e & =\frac{\sqrt{7}}{4}\end{array}$

Topic: Mathematics, Ellipse, JEE Main
The eccentricity of the conic $36 x^2+144 y^2-36 x-96 y-119=0$ is

Options:
A $\frac{\sqrt{3}}{2}$
B $\frac{1}{2}$
C $\frac{\sqrt{3}}{4}$
D $\frac{1}{\sqrt{3}}$

The correct answer is:
$\frac{\sqrt{3}}{2}$

Solution:
Given equation of conic is
$$
\begin{aligned}
& 36 x^2+144 y^2-36 x-96 y-119=0 \\
& \Rightarrow \quad 36\left(x^2-x\right)+144\left(y^2-\frac{2}{3}\right)=119 \\
& \Rightarrow \quad 36\left(x^2-x+\frac{1}{4}\right)+144\left(y^2-\frac{2}{3}+\frac{1}{9}\right) \\
& =119+9+16 \\
& \Rightarrow \quad 36\left(x-\frac{1}{2}\right)^2+144\left(y-\frac{1}{3}\right)^2=114 \\
& \Rightarrow \quad \frac{\left(x-\frac{1}{2}\right)^2}{4}+\frac{\left(y-\frac{1}{3}\right)^2}{1}=1 \\
&
\end{aligned}
$$
This is the equation of ellipse.
Here,
$$
\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
$$

Topic: Mathematics, Ellipse, JEE Main
The eccentricity of the conjugate hyperbola of the hyperbola $x^2-3 y^2=1$, is

Options:
A $2$
B $\frac{2}{\sqrt{3}}$
C $4$
D $\frac{4}{3}$

The correct answer is:
$2$

Solution:
Eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $e=\sqrt{\frac{a^2+b^2}{a^2}}$
Eccentricity of conjugate hyperbola,
$e^{\prime}=\sqrt{\frac{a^2+b^2}{b^2}}$
Write the given equation in standard form,
$\begin{aligned}
& \frac{x^2}{1}-\frac{y^2}{1 / 3}=1 \Rightarrow a^2=1, b^2=\frac{1}{3} \\
& \quad e^{\prime}=\sqrt{\frac{1+1 / 3}{1 / 3}}=\sqrt{4}=2 .
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The eccentricity of the hyperbola \(4 x^2-9 y^2=36\) is

Options:
A \(\frac{\sqrt{11}}{3}\)
B \(\frac{\sqrt{15}}{3}\)
C \(\frac{\sqrt{13}}{3}\)
D \(\frac{\sqrt{14}}{3}\)

The correct answer is:
\(\frac{\sqrt{13}}{3}\)

Solution:
Hints: \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
\(\begin{aligned}
& a=3, b=2 \\
& \therefore e=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}
\end{aligned}\)

Topic: Mathematics, Hyperbola, JEE Main
The eccentricity of the hyperbola can never be equal to

Options:
A $\sqrt{\frac{9}{5}}$
B $2 \sqrt{\frac{1}{9}}$
C $3 \sqrt{\frac{1}{8}}$
D $2$

The correct answer is:
$2 \sqrt{\frac{1}{9}}$

Solution:
Since $e\gt1$ always for hyperbola and $\frac{2}{3} \lt 1$

Topic: Mathematics, Hyperbola, JEE Main
The eccentricity of the hyperbola whose length of its conjugate axis is equal to half of the distance between its foci, is

Options:
A 23
B 3
C 43
D 43

The correct answer is:
23

Solution:
For any standard hyperbola we know that

eccentricity, e=1+b2a2, conjugate axis =2b and distance between the foci =2ae.

Given, 2b=122ae.

⇒ ba=e2

Now, e= 1+b2a2= 1+e24

⇒ e2=1+e24

⇒ 34e2=1

⇒ e2=43

⇒ e=23

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2016 (03 Apr)
The edge length of a face centered cubic cell of an ionic substance is $508 \mathrm{pm}$. If the radius of the cation is $110 \mathrm{pm}$, the radius of the anion is

Options:
A
$288 \mathrm{pm}$

B
$398 \mathrm{pm}$

C
$618 \mathrm{pm}$

D
$144 \mathrm{pm}$

The correct answer is:
$144 \mathrm{pm}$

Solution:
For an ionic substance in FCC arrangement,
$$
\begin{aligned}
& 2\left(r^{+}+r^{-}\right)=\text {edge length } \\
& 2\left(110+r^{-}\right)=508 \\
& r^{-}=144 \mathrm{pm}
\end{aligned}
$$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2010
The edge length of sodium chloride unit cell is 564 pm. If the size of Cl- ion is 181 pm. The size of Na+ ion will be

Options:
A 101 pm
B 181 pm
C 410 pm
D 202 pm

The correct answer is:
101 pm

Solution:
2r++r-=a
2rNa+rCl-=564 pm
r Na + = 564 2 −181=101 pm

Topic: Chemistry, Solid State, JEE Main
The edge length of unit cell of a metal (molecular weight \( 24 \) ) having cubic structure is \( 4.53 \mathrm{~A} \). If the density of metal is
\( 1.74 \mathrm{~g} / \mathrm{cm}^{3} \), What is the radius of metal? (Avogadro Number \( =6 \times 10^{23} \) )

Options:
A \( 180 \mathrm{pm} \)
B \( 160 \mathrm{pm} \)
C \( 140 \mathrm{pm} \)
D \( 190 \mathrm{pm} \)

The correct answer is:
\( 160 \mathrm{pm} \)

Solution:
d=n×Atomic weighta3×Av. number
⇒ 1.74 = n × 24 g/mole 6 × 1 0 2 3 × 4.53 × 1 0 - 8 3
n = 1.74 × 6 × 9.296 2 4 = 4
(n is number of atoms per unit cell)
∴ n = 4 and metal crystallises in fcc.

Then,

r=a22=4.2322=1.6 A° = 160 pm.

Topic: Chemistry, Solid State, JEE Main
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Options:
A 3.92×10-7 m

B 5.98Ã10-7 m
C 2.72Ã10-7 m
D 4.82Ã10-7 m

The correct answer is:
3.92×10-7 m

Solution:
Given, side of a cube (l) = 10 cm = 0.1 m
âArea its each face (A) = l2 = (0.1)2 = 0.01 m2
Load (m) = 100 kg
Tangential force acting on one face of the cube,
âF = mg = 100 Ã 9.8 = 980 N
Shear stress acting on this face = F A
= 980 0.01 N/m 2
= 9.8 Ã 104 N/m2
Shear modulus of aluminium (Î·)) = 25 GPa
= 25 Ã 109 N/m2
Shear modulus Î· = Shearing stress Shearing strain
or shearing strain Î y L = Shearing stress Shear modulus
or Î y = Shearing stress Shear modulus Ã L
= 9.8 Ã 10 4 25 Ã 10 9 Ã 0.1
= 0.392 Ã 10-5 m
= 3.92 Ã 10-7 m

Topic: Physics, Mechanical Properties of Solids, JEE Main
The effect of addition of helium gas to the following reaction in equilibrium state at constant volume, is :
PCI5( g)⇌PCl3( g)+Cl2( g)

Options:
A the equilibrium will shift in the forward direction and more of Cl2 and PCl3 gases will be produced.
B the equilibrium will go backward due to suppression of dissociation of PCl5.
C helium will deactivate PCl5 and reaction will stop.
D addition of helium will not affect the equilibrium.

The correct answer is:
addition of helium will not affect the equilibrium.

Solution:
PCI5( g)⇌PCl3( g)+Cl2( g)
In the above reaction vessel, when helium(an inert gas) is added following cases can happen for isobaric and isochoric processes respectively:

Case 1 : At constant P - volume will increase so reaction will shift in forward direction then answer will be A

Case 2 : At constant volume no change in active mass so reaction will not shift in any direction then answer will be D.
Thus, at constant volume, addition of helium will not have any impact on the equilibrium.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The effect of increase in temperature on the number of electrons in conduction band (ne) and resistance of a semiconductor will be as:

Options:
A Both ne and resistance decrease
B Both ne and resistance increase
C ne increases, resistance decreases
D ne decreases, resistance increases

The correct answer is:
ne increases, resistance decreases

Solution:
In case of semiconductors as the temperature increases the electrons in the valence band get excited and jump into the conduction band and hence the conductance increases resulting in the decrease of resistance. Thus, as temperature increases, ne increases and resistance decreases.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The effect of lanthanoid contraction in the lanthanoid series of elements by and large means

Options:
A increase in atomic radii and decrease in ionic radii,
B decrease in atomic radii and increase in ionic radii.
C increase in both atomic and ionic radii.
D decrease in both atomic and ionic radii.

The correct answer is:
decrease in both atomic and ionic radii.

Solution:
In lanthanide series, the presence of additional 14 electrons and protons increases the effective nuclear charge. Hence, both atomic and ionic radii decrease.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The effect on the interference fringes if the monochromatic source is replaced by a source of white light would be:

Options:
A Interference pattern disappears
B Fringe separation remains fixed
C Fringe closest to the central fringe is red.
D Fringe separation increases

The correct answer is:
Fringe closest to the central fringe is red.

Solution:
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at same position. Therefore, the central fringe is white.
For a point P for which:
S 2 P - S 1 P= Î» b /2
Î»bâ4000 Aâ represents the wavelength of blue colour, the blue colour will be absent and the fringe will appear red in colour. Slightly farther away where,
S 2 Q - S 1 Q= Î» b = Î» r /2
Where Î»râ8000 Aâ represents the wavelength of red colour, the fringe will be predominantly blue.

Topic: Physics, Wave Optics, JEE Main