datavorous/entrance-exam-dataset huggingface dataset:
If z is a non-real complex number, then the minimum value of Im z5Im z5 is (Where Im z = Imaginary part of z)

Options:
A -2
B -4
C -5
D -1

The correct answer is:
-4

Solution:
Let, z=r(cosθ+isinθ)
z 5 = r 5 (cos5θ+isin5θ)
Imz5=r5sin5θ
and Im z5=r5sin5θ
Im (z5)Im z5= sin5θsin5θ=A (Let)
⇒dAdθ=5sin5θcos5θ-5sin5θsin4θcosθsin5θ2
⇒dAdθ=5sin4θsinθcos5θ-sin5θcosθsin10θ
⇒ dA dθ = 5 sin 6 θ sin −4θ =0
⇒θ=π4
The minimum value of A will be at θ=π4.
⇒A=sin5π4sinπ45
=-12125
=-24=-4

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2015 (11 Apr Online)
If $z$ is any complex number satisfying $|z-3-2 i| \leq 2$, then the maximum value of $|2 z-6+5 i|$ is

Options:

The correct answer is:
5

Solution:
Given, $|z-3-2 i| \leq 2$
To find minimum of $|2 z-6+5 i|$ or $\quad 2\left|z-3+\frac{5}{2} i\right|$
Using triangle inequality, i.e. ||$z_1|-| z_2|| \leq\left|z_1+z_2\right|$ $\therefore\left|z-3+\frac{5}{2} i\right|$ $=\left|z-3-2 i+2 i+\frac{5}{2} i\right|$ $=\left|(z-3-2 i)+\frac{9}{2} i\right| \geq|| z-3-2 i\left|-\frac{9}{2}\right|$ $\geq\left|2-\frac{9}{2}\right| \geq \frac{5}{2} \Rightarrow\left|z-3+\frac{5}{2} i\right| \geq \frac{5}{2}$ or $\quad|2 z-6+5 i| \geq 5$

Topic: Mathematics, Complex Number, JEE Main
If $z$ is complex number such that $\left|z-\frac{4}{z}\right|=2$, then the greatest value of $|z|$ is

Options:
A $1+\sqrt{2}$
B $\sqrt{2}$
C $\sqrt{3}+1$
D $1+\sqrt{5}$

The correct answer is:
$1+\sqrt{5}$

Solution:
Given, $\left|z-\frac{4}{z}\right|=2$
$\begin{array}{cc}\therefore & \quad|z|=\left|z-\frac{4}{z}+\frac{4}{z}\right| \\ & \leq\left|z-\frac{4}{z}\right|+\left|\frac{4}{z}\right| \\ \Rightarrow & |z| \leq 2+\frac{4}{|z|} \\ \Rightarrow & |z|^2-2|z| \leq 4 \\ \Rightarrow & \left(|z|^2-1\right)^2 \leq 5 \\ \Rightarrow & (|z|-1) \leq \sqrt{5} \\ \Rightarrow & |z| \leq \sqrt{5}+1\end{array}$

Topic: Mathematics, Complex Number, JEE Main
If Z is the compressibility factor, then Van der Waal's equation at low pressure can be written as:

Options:
A Z = 1 + RT Pb
B Z = 1 - a VRT
C Z = 1 - Pb RT
D Z = 1 + Pb RT

The correct answer is:
Z = 1 - a VRT

Solution:
Van der Waal's Equation of state for one mole
P + a V 2 V - b = RT
At low pressure, the characteristic constant (a) becomes significant whereas (b) becomes negligible. So (b) is ignored.
P + a V 2 V = RT
PV + a V = RT
PV RT = RT RT - a VRT = 1 - a VRT
At low pressure for most gases z < 1.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2014 (06 Apr)
If ' $Z$ ' is the number of atoms in the unit cell that represents the closest packing sequence $---A B C A B C---$, the number of tetrahedral voids in the unit cell is equal to

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

The correct answer is:
$2 Z$

Solution:
Number of tetrahedral voids in the unit cell $=2 \times$ number of atoms $=2 Z$.

Topic: Chemistry, Solid State, JEE Main
If $z=r e^{i \theta}$, then $\left|e^{i z}\right|=$

Options:
A $e^{r \sin \theta}$
B $e^{-r \sin \theta}$
C $e^{-r \cos \theta}$
D $e^{r \cos \theta}$

The correct answer is:
$e^{-r \sin \theta}$

Solution:
$\begin{aligned} & \text {If } z=r e^{i \theta}=r(\cos \theta+i \sin \theta) \\ & \Rightarrow i z=i r(\cos \theta+i \sin \theta)=-r \sin \theta+i r \cos \theta \\ & \text {or } e^{i z}=e^{(-r \sin \theta+i r \cos \theta)}=e^{-r \sin \theta} e^{r i \cos \theta} \\ & \text {or }\left|e^{i z}\right|=e^{-r \sin \theta} \| e^{r i \cos \theta}\left|=e^{-r \sin \theta}\right| e^{i r \cos \theta} \mid \\ & =e^{-r \sin \theta}\left[\left\{\cos ^2(r \cos \theta)+\sin ^2(r \cos \theta)\right\}\right]^{1 / 2}=e^{-r \sin \theta}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $Z_{r}=\sin \frac{2 \pi}{11}-i \cos \frac{2 \pi}{11},$ then $\sum_{r=0}^{10} Z_{r}$ is equal to

Options:
A -1
B 0
C i
D $-i$

The correct answer is:
0

Solution:
We have, $Z,=\sin \frac{2 \pi r}{11}-i \cos \frac{2 \pi}{11}$
$$
=-i\left(\cos \frac{2 \pi r}{11}+i \sin \frac{2 \pi r}{11}\right)
$$
$\therefore$
$$
\begin{aligned}
&=-i e^{\frac{i 2 \pi}{11}} \\
\sum_{r=0}^{10} Z_{r} &=-i \sum_{r=0}^{10} e^{\frac{i 2 \pi}{11}}
\end{aligned}
$$
$$
=-i \times 0=0
$$

Topic: Mathematics, Complex Number, JEE Main
If $z$ satisfies the equation $|z|-z=1+2 i$, then $z$ is equal to

Options:
A $\frac{3}{2}+2 i$
B $\frac{3}{2}-2 i$
C $2-\frac{3}{2} i$
D $2+\frac{3}{2} i$

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

Solution:
Given : $|z|-z=1+2 i$
If $z=x+i y$, then th is equation reduces to
$$
\begin{array}{l}
|x+i y|-(x+i y)=1+2 i \\
\Rightarrow\left(\sqrt{x^{2}+y^{2}}-x\right)+(-i y)=1+2 i
\end{array}
$$
On comparing real and imaginary parts of both sides of this equation, we get
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}}-x=1 \\
\Rightarrow \sqrt{x^{2}+y^{2}}=1+x \Rightarrow x^{2}+y^{2}=(1+x)^{2} \\
\Rightarrow x^{2}+y^{2}=1+x^{2}+2 x \\
\Rightarrow y^{2}=1+2 x...(i) \\
\text { and }-y=2 \\
\Rightarrow y=-2
\end{array}
$$
Putting this value in eq. (i), we get
$$
\begin{array}{l}
(-2)^{2}=1+2 x \\
\Rightarrow 2 x=3 \Rightarrow x=\frac{3}{2} \\
\therefore z=x+i y=\frac{3}{2}-2 i
\end{array}
$$

Topic: Mathematics, Parabola, JEE Main
If $z=\sec (y-a x)+\tan (y+a x)$, $\frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}$ is equal to

Options:
A $0$
B $-z$
C $z$
D $2 x$

The correct answer is:
$0$

Solution:
We have,
$$
\begin{aligned}
& z=\sec (y-a x)+\tan (y+a x) \\
& \frac{\partial z}{\partial x}=-a \sec (y-a x) \tan (y-a x)+a \sec ^2(y+a x) \\
& \frac{\partial^2 z}{\partial x^2}=-a\left[-a \sec (y-a x) \tan ^2(y-a x)\right. \\
& \left.\sec ^2(y+a x) \tan (y+a x) \quad-a \sec ^3(y-a x)\right]+2 a^2 \\
& \operatorname{Similarly}^2 \\
& \frac{\partial^2 z}{\partial y^2}=\sec (y-a x) \tan ^2(y-a x) \\
& \quad+\sec ^3(y-a x)+2 \sec ^2(y+a x) \tan (y+a x) \\
& \therefore \quad \frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}=0
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $z=x-i y$ and $z^{\frac{1}{3}}=p+i q$, then $\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^2+q^2\right)}$ is equal to

Options:
A
1

B
$-2$

C
2

D
$-1$

The correct answer is:
$-2$

Solution:
$z=(p+i q)^3=p\left(p^2-3 q^2\right)-i q\left(q^2-3 p^2\right)$
$\Rightarrow \frac{x}{p}=p^2-3 q^2 \& \frac{y}{q}=q^2-3 p^2 \Rightarrow \frac{\frac{x}{p}+\frac{y}{q}}{\left(p^2+q^2\right)}=-2$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2004
If $z=x+i y$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$, then the locus of $z$ is

Options:
A $x$-axis
B $y$-axis
C $y=x$
D $2y=x$

The correct answer is:
$x$-axis

Solution:
We have,
$$
\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2
$$
$$
\begin{array}{rlrl}
\Rightarrow & \left|x+i y+\frac{i}{2}\right|^2 & =\left|x+i y-\frac{i}{2}\right|^2 \\
\Rightarrow & \left|x+i\left(y+\frac{1}{2}\right)\right|^2 & =\left|x+i\left(y+\frac{1}{2}\right)\right|^2 \\
\Rightarrow & x^2+\left(y+\frac{1}{2}\right)^2 & =x^2+\left(y-\frac{1}{2}\right)^2 \\
\Rightarrow & x^2+y^2+\frac{1}{4}+y & =x^2+y^2+\frac{1}{4}-y \\
\Rightarrow & & 2 y & =0 \\
\Rightarrow & y & =0
\end{array}
$$
Locus of $z$ is $x$-axis.

Topic: Mathematics, Complex Number, JEE Main
If z=x+iy satisfies z-2=0 and z-i-z+5i=0, then

Options:
A x+2y-4=0
B x2+y-4=0
C x+2y+4=0
D x2-y+3=0

The correct answer is:
x+2y+4=0

Solution:
Given, z=x+iy

Also, z=2

So, x2+y2=4 ...1

Now, z-i=z+5i

⇒z-i2=z+5i2

⇒x2+y-12=x2+y+52

⇒y=-2 ...2

So, x=0 by solving equation 1 & 2

Hence, only x+2y+4=0 is true, as point only satisfy this equation.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (26 Jul Shift 2)
If z=x+iy,xy≠0, satisfies the equation z2+iz¯=0, then z2 is equal to :

Options:
A 9
B 1
C 4
D 14

The correct answer is:
1

Solution:
Given: z2+iz=0

⇒z2=-iz

Now taking z=x+iy we get,

⇒x2-y2+2ixy=-ix-y

Now on comparing both side we get,

⇒x2-y2=-y & 2xy=-x

⇒x2-y2=-y & x2y+1=0

⇒x2--122=--12 & y=-12

⇒x2=34 & y=-12

⇒x=±32 & y=-12

Hence, z=±32 +i-12

⇒z2=34 -14±2i3212

⇒z2=12 ±i32

⇒z2=14+342=1

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If z=x+iy, âx,yâR, i2=-1, xyâ 0 and z=2, then the imaginary part of z+2z-2 cannot be

Options:
A 1
B 3
C 2
D 4

The correct answer is:
1

Solution:
Given, z=2 ⇒x2+y2=4 & xy≠0

Let, z=x+iy=2cos⁡θ+isin⁡θ
So, z+2z-2=x+2+iyx-2+iy

=x2+y2-4+iyx-2-yx+2x-22+y2=-4y8-4xi
So, the imaginary part =-4y8-4x=yx-2

=-2sin⁡θ2-2cos⁡θ=-cot⁡θ2, θ∈0,2π
Now, θ≠π2,π,3π2⇒cot⁡θ2≠-1,0,1

Topic: Mathematics, Complex Number, JEE Main
If $z=x$ - iy and $z^{1 / 3}=p+$ iq $(x, y, p, q \in \mathbb{R})$, then $\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^2+q^2\right)}$ is equal to

Options:
A 2
B -1
C 1
D $-2$

The correct answer is:
$-2$

Solution:
$z=p^3-3 p q^2+i\left(3 p^2 q-q^3\right)=x-i y$
$\begin{aligned}
&x=p\left(p^2-3 q^2\right) \\
&y=-q\left(3 p^2-q^2\right) \\
&\therefore \frac{x}{p}+\frac{y}{q}=p^2-3 q^2-3 p^2+q^2=-2\left(p^2+q^2\right)
\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\left|z-\frac{4}{z}\right|=2$, then the maximum value of $|z|$ is equal to

Options:
A
$\sqrt{3}+1$

B
$\sqrt{5}+1$

C
2

D
$2+\sqrt{2}$

The correct answer is:
$\sqrt{5}+1$

Solution:
$$
\begin{aligned}
& |z|=\left|\left(z-\frac{4}{z}\right)+\frac{4}{z}\right| \Rightarrow|z|=\left|z-\frac{4}{z}+\frac{4}{z}\right| \\
& \Rightarrow|z| \leq\left|z-\frac{4}{z}\right|+\frac{4}{|z|} \Rightarrow|z| \leq 2+\frac{4}{|z|}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow|z|^2-2|z|-4 \leq 0 \\
& (|z|-(\sqrt{5}+1))(|z|-(1-\sqrt{5})) \leq 0 \Rightarrow 1-\sqrt{5} \leq|z| \leq \sqrt{5}+1
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2009
If z - αz + αα∈R is a purely imaginary number and z=2, then a value of α is :

Options:
A 1
B 12
C 2
D 2

The correct answer is:
2

Solution:
If a complex number is purely imaginary, then it must be equal to negative of its conjugate.

⇒z-αz+α=-z-αz+α⇒z-αz+α=-z--αz-+α

⇒z-αz-+α=z+αz--α

⇒ zz-+αz-αz--α2=-zz--αz+αz--α2

⇒z2=α2

⇒α2=4 ∵z=2

⇒α=±2

Topic: Mathematics, Complex Number, JEE Main
If z - αz + αα∈R is a purely imaginary number and z=2, then a value of α is :

Options:
A 1
B 12
C 2
D 2

The correct answer is:
2

Solution:
If a complex number is purely imaginary, then it must be equal to negative of its conjugate.

⇒z-αz+α=-z-αz+α⇒z-αz+α=-z--αz-+α

⇒z-αz-+α=z+αz--α

⇒ zz-+αz-αz--α2=-zz--αz+αz--α2

⇒z2=α2

⇒α2=4 ∵z=2

⇒α=±2

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (12 Jan Shift 1)
If \( |z-2|=\min \{|z-1|,|z-5|\} \), where \( z \) is a complex number, then-

Options:
A \( R e=\frac{3}{2} \)
B \( R e=\frac{7}{2} \)
C \( R e \in\left\{\frac{3}{2}, \frac{7}{2}\right\} \)
D None of these

The correct answer is:
\( R e \in\left\{\frac{3}{2}, \frac{7}{2}\right\} \)

Solution:
Given, z–2=minz–1,z–5 ⋯i

case 1. when z-1<z-5
i.e., z-2=z-1 …ii

so, here it is clear that z lies on perpendicular bisector of line joining oints 1,0,2,0 on real axis.,
⇒ Rez=32 which satisfy z-1<z-5.

again case 2 when z-5<z-1
⇒z-2=z-5

here z lies on real line which perpendicular bisector of line joinig points 2,0,5,0
⇒ Rez=72 which satisfy z-5<z-1 .

Hence option 3 is correct.

Topic: Mathematics, Complex Number, JEE Main
If \( |z+4| \leq 3 \), then the maximum value of \( |z+1| \) is

Options:
A \( 4 \)
B \( 10 \)
C \( 6 \)
D \( 0 \)

The correct answer is:
\( 6 \)

Solution:
From the argand diagram, maximum value of z+1 is 6.

Aliter

z+1=z+4-3≤z+4+-3≤6

Thus, maximum value of z+1 is 6.

Topic: Mathematics, Complex Number, JEE Main
If \(z=\left(\begin{array}{ccc}1 & 1+2 i & -5 i \\ 1-2 i & -3 & 5+3 i \\ 5 i & 5-3 i & 7\end{array}\right)\) then \((i=\sqrt{-1})\)

Options:
A \(z\) is purely real
B \(z\) is purely imaginary
C \(z+\bar{z}=0\)
D \((z-\bar{z})\) i is purely imaginary

The correct answer is:
\(z\) is purely real

Solution:
\(\begin{aligned}
& \text { Hints : } z=\left|\begin{array}{ccc}
1 & 1+2 i & -5 i \\
1-2 i & -3 & 5+3 i \\
5 i & 5-3 i & 7
\end{array}\right|=1(-21-64)-((1-2 i)(7(1+2 i)+5 i(5-3 i)))+5 i((1+2 i)(5+3 i)-15 i) \\
& \text { = Real } \\
&
\end{aligned}\)

Topic: Mathematics, Determinants, JEE Main
If \( z=r e^{i \theta} \), then \( \left|e^{i z}\right| \) equals

Options:
A \( e^{r \sin \theta} \)
B \( e^{-r \sin \theta} \)
C \( e^{-r \cos \theta} \)
D \( e^{r \cos \theta} \)

The correct answer is:
\( e^{-r \sin \theta} \)

Solution:
If z=reiθ=rcosθ+isinθ

⇒iz=ircosθ+isinθ

=-rsinθ+ir cosθ

⇒eiz=e-rsinθ+ircosθ

=e-rsinθeircosθ

⇒eiz=e-rsinθericosθ

⇒eiz=e-rsinθcosrcosθ+isinrcosθ

⇒eiz=e-rsinθcos2rcosθ+sin2rcosθ12

=e-rsinθ.

Topic: Mathematics, Complex Number, JEE Main
IfM. D. is 12, the value of S.D. will be

Options:
A 15
B 12
C 24
D None of these

The correct answer is:
15

Solution:
We know that Q.D $=\frac{5}{6} \times M \cdot D .=\frac{5}{6} \times 12=10$

$\therefore \mathrm{S} . \mathrm{D}=\frac{3}{2} \times Q \cdot D .=\frac{3}{2} \times 10 \Rightarrow S . D .=15$

Topic: Mathematics, Statistics, JEE Main
IfP(E) denotes the probability of an event $\mathrm{E}$, then $\mathrm{E}$ is called certain event if:

Options:
A $\quad \mathrm{P}(\mathrm{E})=0$
B $\mathrm{P}(\mathrm{E})=1$
C $\mathrm{P}(\mathrm{E})$ is either 0 or 1
D $\mathrm{P}(\mathrm{E})=1 / 2$

The correct answer is:
$\mathrm{P}(\mathrm{E})=1$

Solution:
Prob. (certain event) $=1$ $\Rightarrow \mathrm{P}(\mathrm{E})=1$

Topic: Mathematics, Probability, JEE Main
Illustrate with examples the limitations of Willamson synthesis for the preparation of certain types of ethers.

Options:

None

Solution:
Williamson's synthesis is a versatile method for the synthesis of both symmetrical and unsymmetrical ethers. However, for the synthesis of unsymmetrical ethers, a proper choice of reactants is necessary. Since Williamson's synthesis occurs by $\mathrm{S}_{\mathrm{N}} 2$ mechanism and primary alkyl halides are most reactive in $\mathrm{S}_{\mathrm{N}} 2$ reaction, therefore, best yields of unsymmetrical ethers are obtained when the alkyl halides are primary and the alkoxide may be primary, secondary or tertiary. For example, tertbutylethyl ether is prepared by treating ethyl bromide with sodium tert-butoxide.
The above ether cannot be prepared by treating sodium ethoxide with tert-butyl chloride or bromide since under these condition an alkene, i.e., isobutylene is the main product.
Aryl and vinyl halides cannot be used as substrates because they are less reactive in nucleophilic substitution.

Topic: Chemistry, Alcohols Phenols and Ethers
Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 253m to 507m in 30 s. If the object is moving uniformly, then its speed is

Options:
A 3 km h-1
B 4 km h-1
C 5 km h-1
D 6 km h-1

The correct answer is:
3 km h-1

Solution:
Using mirror formula,
Or 1u1= 325 - 110
Or u1 = 50 m
And
â´1u2 = 750 - 110
Or u2 = 25 m
Speed of object =u1- u2time
=2530 m s-1
= 3 km h-1

Topic: Physics, Ray Optics, JEE Main
Image of line x-23=y-11=z-1-4 in the plane x+y+z=7 is

Options:
A x-41=y-31=z-31
B x-31=y-41=z-31
C x-43=y-31=z-3-4
D x-33=y-41=z-3-4

The correct answer is:
x-43=y-31=z-3-4

Solution:
The line is perpendicular to normal, hence the line is parallel to plane, so the reflected line has the same direction.
Now, image of 2,1,1 in the plane is Î±,Î²,Î³ say, then
Î±-21=Î²-11=Î³-11=-24-71+1+1=2
So, the point is 4,3,3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Image point of $(1,3,4)$ in the plane $2 x-y+z+3=0$ is

Options:
A $(-3,5,2)$
B $(3,5,-2)$
C $(3,-5,3)$
D None of these

The correct answer is:
$(-3,5,2)$

Solution:
Let $Q$ be image of the point $P(1,3,4)$ in the given plane, then $P Q$ is normal to the plane. The d.r.'s of $P Q$ are $2,-1,1$
Since $P Q$ passes through $(1,3,4)$ and has d.r's $2,1,-1$;
therefore, equation of plane is $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r$
$\therefore \quad x=2 r+1, y=-r+3, z=r+4$
$\therefore \quad x=2 r+1, y=-r+3, z=r+4$
So, co-ordinates of $Q$ be
$(2 r+1,-r+3, r+4)$
Let $R$ be the mid point of $P Q$, then co-ordiantes of $R$ are $\left(\frac{2 r+1+1}{2}, \frac{-r+3+3}{2}, \frac{r+4+4}{2}\right)$
i.e., $\left(r+1, \frac{-r+6}{2}, \frac{r+8}{2}\right)$
Since $R$ lies on the plane
$\therefore 2(r+1)-\left(\frac{-r+6}{2}\right)+\left(\frac{r+8}{2}\right)+3=0 \Rightarrow r=-2$
So, co-ordinates of $Q$ are $(-3,5,2)$.
Trick : From option (a), mid point of $(-3,5,2)$ and $(1,3,4)$ satisfies the equation of plane $2 x-y+z+3=0$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as that of the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength Î», (given in terms of Rydberg constant R for hydrogen atom) equal to

Options:
A 95R
B 365R
C 185R
D 4R

The correct answer is:
185R

Solution:
In hydrogen atom, En=Rhcn2
Also, Enâm, where m is the mass of the electron. Here, the electron has been replaced by a particle, whose mass is double the mass of an electron. Therefore, for this hypothetical atom, energy in nth orbit will be given by
En=-2Rhcn2
The longest wavelength (or minimum energy) photon will correspond to the transition of the particle from n=3 to n=2
â hcÎ»max=E3-E2=2Rhc122-132=2RhcÃ536
â´ Î»max=hc518Rhc=185R

Topic: Physics, Atomic Physics, JEE Main
Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength X (given in terms of the Rydberg constant R for the hydrogen atom) is equal to

Options:
A 95R

B 365R

C 185R

D 4R

The correct answer is:
185R

Solution:
In hydrogen atom E n -Rhcn2
where m is the mass of the electron. Here, the electron has been replaced by a particle whose mass is double of an electron. Therefore, energy in nth orbit for this hypothetical atom will be given by
En=-2Rhcn2
The longest wavelength λmax (or minimum energy) photon will correspond to the transition of the particle from n = 3 to n = 2

hcλmax=E3-E2=2Rhc122-132
This gives, λ max=18 5R

Topic: Physics, Atomic Physics, JEE Main
Imagine an electron moving in the potential field of a proton given by V = V 0 ln r r 0 where r0 is a positive constant and r is the distance between the electron and proton. Assuming Bohr's model to be applicable, the variation of rn with n (principal quantum number) is

Options:
A r n â n
B r n ∝ 1 n
C r n ∝ n 2
D r n ∝ 1 n 2

The correct answer is:
r n ∝ n

Solution:
U = eV = eV 0 ln r r 0
F = - dU dr = eV 0 r
This force provide the necessary centripetal force.
Hence
mv 2 r = eV 0 r
or v = eV 0 m ...(1)
Moreover,
mvr = nh 2 Ï ...(2)
Dividing Eq. (2) by Eq. (1), we have
mr = nh 2 Ï m eV 0 or r n â n

Topic: Physics, Atomic Physics, JEE Main
Imagine removing one electron from $\mathrm{He}^4$ and $\mathrm{He}^3$. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why?

Options:

None

Solution:
We know that, If we remove one electron from the isotopes of $\mathrm{He}^4$ and $\mathrm{He}^3$, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass. Also after removing one electron from $\mathrm{He}^4$ and $\mathrm{He}^3$ atoms will have one electron as in hydrogen atoms, and the nucleus is much heavier then $\mathrm{H}$ atom. So, stability can remain and these atoms are very close to the H atom so the energy levels are as of Hydrogen and these atom will be very close.

Topic: Physics, Atomic Physics
Imagine that a reactor converts all the given mass into energy and that it operates at a power level of 109 Watt . The mass of the fuel consumed per hour, in the reactor, will be:
(velocity of light, c is 3×108 m s-1)

Options:
A 6.6Ã10-5 g
B 0.96Â g
C 4Ã10-2 g
D 0.8Â g

The correct answer is:
4Ã10-2 g

Solution:
Power of reactor, P=EΔt=Δmc2Δt(From mass-energy equivalence equation)

Given in the problem power level =109 Watt .
∆m∆t=Pc2 =1093×1082
⇒ ∆m∆t=10-79 kg s-1
The mass of fuel consumed per hour=10-79×1000 g3600 h-1=4×10-2 g h-1.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2017 (09 Apr Online)
Imagine that the electron in a hydrogen atom is replaced by a muon μ. The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be:-

Options:
A 13.6eV
B 2815.2eV
C 331.2eV
D 27.2eV

The correct answer is:
2815.2eV

Solution:
E∝1r⇒r∝1 m⇒E∝m. Ionization potential =13.6×MassμeVMasse
=13.6×207eV=2815.2eV.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Imine formation using an aldehyde/ketone and primary amine is acid-catalyzed, yet the rate drops below pH 4.5. Why does the rate drop below this pH?

Options:
A The carbinolamine intermediate is stable at low pH
B The imine product is hydrolyzed at low pH
C Protonation of the amine decreases its nucleophilicity
D The amine is hydrolyzed at low pH

The correct answer is:
Protonation of the amine decreases its nucleophilicity

Solution:
Imine formation:
The reaction is possible pH 4 to 5 only. Hence, the reaction is pH dependent.
At very low pH (high conc. of H+ ): ammonia derivative will form their respective salts due to their basic nature and at the same time will lose their nucleophilic nature.
At very high pH (high conc. of OH- ): carbonyl group will not be able to undergo sufficient protonation.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Impairment of adrenal cortex causes decrease concentration of _______ in blood.

Options:
A sodium
B calcium
C ACTH
D water

The correct answer is:
sodium

Solution:
Adrenal glands are present over the kidneys. These consist of the outer part known as adrenal cortex and inner part adrenal medulla. Adrenal cortex is further divided into three zones and each zone secretes certain hormones. The outermost zone is zona glomerulosa that secretes mineralocorticoids, the middle zone is zona fasciculata that secretes glucocorticoids and innermost zone is zona reticularis that secretes sex corticoids.
Aldosterone is the most important mineralocorticoid secreted. This hormone is responsible for maintenance of sodium levels in the blood. The main function is to reabsorb sodium so that its excretion is decreased. Thus if there is impairment of adrenal cortex the reabsorption will not occur and its excretion will increase and this will lower the level of sodium in blood.

Topic: Biology, Chemical Coordination and Integration, NEET
Importance of day length in flowering of plants was first shown in

Options:
A Petunia
B Lemna
C Tobacco
D Cotton

The correct answer is:
Tobacco

Solution:
Discovered by Garnar \& Allard in a maryland mammoth variety of tobacco.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2008 (Mains)
Important site for formation of glycoproteins and glycolipids is

Options:
A Golgi apparatus
B lysosome
C plastid
D vacuole

The correct answer is:
Golgi apparatus

Solution:
The Golgi apparatus principally performs the function of packaging materials. Golgi apparatus is the main site of formation of glycoproteins and glycolipids.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2011 (Screening)
Impossible orbital among the following is

Options:
A $3 f$
B $2 p$
C $4 d$
D $2 s$

The correct answer is:
$3 f$

Solution:
According to Bohr - Burry's scheme, Maximum number of $\mathrm{e}^{-}$in 3 rd orbital $=3 \mathrm{n}^{2}$
$$
=3(3)^{2}=18
$$
and maximum number of $\mathrm{e}^{-}$in
s-subshell $=2$
p-subshell $=6$
d-subshell $=10$
f-subshell $=14$
Therefore, $18 \mathrm{e}^{-}$present in 3rd orbital enter in s, $\mathrm{p}$ and $\mathrm{d}$ subshell. Thereafter, there is no $\mathrm{e}^{-}$ available for f-subshell. That's why $3 \mathrm{f}$ subshell is not possible.

Topic: Chemistry, Structure of Atom, JEE Main