datavorous/entrance-exam-dataset huggingface dataset:
A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?

Options:
A Linear velocity
B Angular velocity
C Angular momentum
D Potential energy

The correct answer is:
Angular momentum

Solution:
When comet orbits around Sun in an elliptical orbit, it is under action of a central force and its angular momentum remains constant.

Topic: Physics, Gravitation, NEET
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Options:

None

Solution:
(a) The linear speed of the comet is variable in accordance with Kepler's second law. When comet is near the sun, its speed is higher. When the comet is far away from the sun, its speed is very less.
(b) Angular speed also varies slightly.
(c) Comet has constant angular momentum.
(d) Kinetic energy does not remain constant.
(e) Potential energy varies along the path.
(f) Total energy throughout the orbit remains constant.

Topic: Physics, Gravitation
A comet orbits the sun in a highly elliptical orbit. Which of the following quantities remain constant in a comet?

Options:
A Linear Speed
B Angular Speed
C Angular Momentum
D Kinetic Energy

The correct answer is:
Angular Momentum

Solution:
Correct Option is : (C)
Angular Momentum

Topic: Chemistry, Gravitation, NEET
A comet revolves around the sun in an elliptical orbit. When it is closest to the sun at a distance d, its corresponding kinetic energy is k0. If it is farthest from the sun at distance 3d then the corresponding kinetic energy will be

Options:
A k09
B 8k09
C k04
D 4k09

The correct answer is:
k09

Solution:
v1r1=v2r2= constant
â´ v0d=3dv1
â v1=v03
â´ KE=12mv029
â KE=k09

Topic: Physics, Gravitation, JEE Main
A commercial sample of hydrogen peroxide marked as \( 100 \) volume \( \mathrm{H}_{2} \mathrm{O}_{2} \), it means that-

Options:
A \( 1 \mathrm{ml} \) of \( \mathrm{H}_{2} \mathrm{O}_{2} \) will give \( 100 \mathrm{ml} \mathrm{O}_{2} \) at STP
B \( 1 \mathrm{~L} \) of \( \mathrm{H}_{2} \mathrm{O}_{2} \) will give \( 100 \mathrm{ml} \mathrm{O}_{2} \) at STP
C \( 1 \mathrm{~L} \) of \( \mathrm{H}_{2} \mathrm{O}_{2} \) will give \( 22.4 \mathrm{~L} \) at STP
D \( 1 \mathrm{ml} \) of \( \mathrm{H}_{2} \mathrm{O}_{2} \) will give \( 1 \) mole of \( \mathrm{O}_{2} \) at STP

The correct answer is:
\( 1 \mathrm{ml} \) of \( \mathrm{H}_{2} \mathrm{O}_{2} \) will give \( 100 \mathrm{ml} \mathrm{O}_{2} \) at STP

Solution:
If a commercial sample of hydrogen peroxide marked as 100 volume H2O2 that means at STP, 100 ml of oxygen is liberated by 1 ml of this solution on heating.

Topic: Chemistry, Hydrogen, JEE Main
A committee consisting of atleast three members is to be formed from a group of 6 boys and 6 girls such that it always has a boy and a girl. The number of ways to form such a committee is equal to

Options:
A 212-27-13
B 211-26-13
C 211-27-35
D 212-27-35

The correct answer is:
212-27-35

Solution:
Groups having less than three members =12C0+12C1+12C2

Groups having only boys = groups having only girls =6C3+6C4+6C5+6C6
Total number of groups =12C0+12C1+12C2+…+12C10+12C11+12C12=212
Hence, required number of groups =212−C012+C112+C212−26C3+6C4+6C5+6C6

=212−12C0+12C1+12C2+6C3−6C3+6C4+6C2+6C5+6C1+6C6+6C0
=212-1+12+66+20-26=212-27-35

Topic: Mathematics, Permutation Combination, JEE Main
A committee of 4 persons is to be formed from 2 ladies, 2 old men and 4 young men such that it includes at least 1 lady, at least 1 old man and at most 2 young men. Then the total number of ways in which this committee can be formed is :

Options:
A
40

B
41

C
16

D
32

The correct answer is:
41

Solution:
Required number of ways
$$
\begin{aligned}
& ={ }^2 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1 \times{ }^2 \mathrm{C}_2+{ }^2 \mathrm{C}_1 \times{ }^2 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1+{ }^2 \mathrm{C}_2 \times \\
& { }^2 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1+{ }^2 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2 \times{ }^4 \mathrm{C}_0 \\
& =2 \times 2 \times \frac{4 \times 3}{2} \\
& \quad \quad \quad 2 \times 1 \times 4+1 \times 2 \times 4+1 \times 1 \times 1
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2013 (09 Apr Online)
A committee of 5 is to be formed out of 6 men and 4 ladies. The number of ways this can be done, when at most 2 ladies are include, is

Options:
A 240
B 186
C 60
D 120

The correct answer is:
186

Solution:
The committee can be formed in following ways:
( 5 men), (4 men, 1 lady) , (3 men, 2 ladies)
$\begin{aligned}
& \therefore \text { Number of was }=\left({ }^6 \mathrm{C}_5\right)+\left({ }^6 \mathrm{C}_4 \times{ }^4 \mathrm{C}_1\right)+\left({ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2\right) \\
& =(6)+(15 \times 4)+(20 \times 6)=6+60+120=186
\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
A committee of 5 persons is to be randomly selected from a group of 5 men and 4 women and a chairperson will be randomly selected from the committee. The probability that the committee will have exactly 2 women and 3 men and the chairperson will be a man is p, then 1p is equal to

Options:

The correct answer is:
3.5

Solution:
Total cases =C95ÃC51 (â´ there are 5 ways to choose a chairperson)
Favourable cases =3 men can be chosen in C53 ways, 2 women can be chosen in C42 ways and there are 3 ways to choose the chairperson
Total number of favourable ways =C53ÃC42Ã3
Required probability = C 5 3 × C 4 2 ×3 C 9 5 ×5
=10Ã6Ã3126Ã5=27

Topic: Mathematics, Probability, JEE Main
A committee of 11 member is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then:

Options:
A m=n=68
B n= m –8
C m=n=78
D m+n=68

The correct answer is:
m=n=78

Solution:
m=(6M, 5F)+(7M, 4F)+(8M, 3F)

=C6 8×C5 5+C7 8×C4 5+C8 8×C3 5

=28×1+8×5+1×10=78

n=(3F, 8M)+(4F, 7M)+(5F, 6M)

=C3 5×C8 8+C4 5×C7 8+C5 5×C6 8

=10×1+5×8+1×28=78

Hence, m=n=78.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A committee of 12 members is to be formed from 9 women and 8 men. The number of committees in which the women are in majority is

Options:
A $2720$
B $2702$
C $2270$
D $2278$

The correct answer is:
$2702$

Solution:
A committee of 12 members is to be formed when women are in majority.
Case I 9 women and 3 men
$\begin{aligned} \therefore \text { Number of ways }={ }^9 C_9 \times{ }^8 C_3 & =1 \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \\ & =56\end{aligned}$
Case II 8 women and 4 men
$\therefore$ Number of ways $={ }^9 C_8 \times{ }^8 C_4$
$\begin{aligned} & =9 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \\ & =630\end{aligned}$
Case III 7 women and 5 men
$\therefore$ Number of ways $={ }^9 C_7 \times{ }^8 C_5$
$\begin{aligned} & =\frac{9 \times 8}{2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \\ & =36 \times 56=2016\end{aligned}$
$\begin{aligned} \therefore \text { Required number of ways } & =56+630+2016 \\ & =2702\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
A committee of 12 members is to be formed from 9 women and 8 men. The number of ways of forming the committee with atleast 5 women is

Options:
A 1008
B 2702
C 6062
D 2352

The correct answer is:
6062

Solution:
The number of ways of forming the committee with atleast 8 women is
C 9 5 × C 8 7 + C 9 6 × C 8 6 + C 9 7 × C 8 5 + C 9 8 × C 8 4 + C 9 9 × C 8 3
=9Ã8Ã7Ã64Ã3Ã2Ã8+9Ã8Ã73Ã2Ã8Ã72+
9Ã82Ã8Ã7Ã63Ã2+9Ã8Ã7Ã6Ã54Ã3Ã2+1Ã8Ã7Ã63Ã2
=1008+2352+2016+630+56
=6062

Topic: Mathematics, Permutation Combination, JEE Main
A committee of 12 members is to be formed from 9 women and 8 men. The number of ways of forming the committee with women in majority is

Options:
A 1008
B 2702
C 6062
D 2352

The correct answer is:
2702

Solution:
The number of ways of forming the committee with women in majority
= C 9 7 × C 8 5 + C 9 8 × C 8 4 + C 9 9 × C 8 3
=9Ã82Ã8Ã7Ã63Ã2+9Ã8Ã7Ã6Ã54Ã3Ã2+1Ã8Ã7Ã63Ã2
=2016+630+56=2702

Topic: Mathematics, Permutation Combination, JEE Main
A committee of ten is to be formed from eight teachers and twelve students of whom four are girls. The number of committees which contains atleast four of either group (teachers and students) and atleast two girls and atleast two boys is equal to m. The number of divisors of m is n, then n6 is equal to

Options:

The correct answer is:
5

Solution:
Teachers 8	Students 12	8B+4G

4	6	2B4G,3B3G,4B2G

5	5	2B3G,3B2G

6	4	2B,2G

Required number of ways =m
=C84C82C44+C83C43+C84C42+C85C82C43+C83C42+C86C82C42
=7028+56×4+70×6+5628×4+56×6+2828×6=47040+25088+4704
⇒m=76832
m= 2 5 ⋅ 7 4
Number of divisions n=5+14+1=30

So, n6=5

Topic: Mathematics, Permutation Combination, JEE Main
A common characteristic feature of plant sieve tube cells and most of mammalian erythrocytes is

Options:
A
absence of mitochondria

B
presence of cell wall

C
presence of haemoglobin

D
absence of nucleus

The correct answer is:
absence of nucleus

Solution:
The mammalian RBCs lose their nuclei at maturity and live for a few months only. The food conducting phloem cells known as sieve tubes, of flowering plants also lose their nuclei at maturity but they remain functional for many years.

Topic: Biology, Cell - The Unit of Life
A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its dc current gain is 250, RC=1 kΩ and VCC=10 V . The minimum base current for VCE to reach saturation is

Options:
A 10Â Î¼A
B 40 Î¼A
C 7Â Î¼A
D 100Â Î¼A

The correct answer is:
40 Î¼A

Solution:
Given, β=250
RC=1000 Ω
VCC=10 V

At saturation, VCE=0
⇒ie=10 V1000 Ω=10 mA
Current gain factor, β=ieiB
⇒iB=ieβ=10 mA250
=40 μA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is

Options:
A 500
B 1000
C 1250
D 100

The correct answer is:
1250

Solution:
AC power gain = Change in output powerChange in input power
= ∆Vc×∆ic∆Vi×∆ib=∆Vc∆Vi×∆ic∆ib=AV×βAC
where AV is voltage gain and βAC is AC current gain.
Also,
AV=βAC×resistance gain =RoRi
Given, AV=50, Ro=200 Ω, Ri=100 Ω
Hence, 50 = βAC×200100
∴ βAC=25
Now, AC power gain = AV×βAC = 50 × 25 = 1250

Topic: Physics, Semiconductors, JEE Main
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is

Options:
A 500
B 1000
C 1250
D 100

The correct answer is:
1250

Solution:
AC power gain = Change in output powerChange in input power
= âVc×âicâVi×âib=âVcâVi×âicâib=AV×βAC
where AV is voltage gain and βAC is AC current gain.
Also,
AV=βAC×resistance gain =RoRi
Given, AV=50, Ro=200 Ω, Ri=100 Ω
Hence, 50 = βAC×200100
∴ βAC=25
Now, AC power gain = AV×βAC = 50 × 25 = 1250

Topic: Physics, Semiconductors, JEE Main
A common emitter amplifier has a voltage gain of 50 , an input impedance of $100 \Omega$ and an output impedance of $200 \Omega$. The power gain the amplifier is

Options:
A 500
B 1000
C 1250
D 50

The correct answer is:
1250

Solution:
Voltage gain $=\beta \times$ impedance gain
$$
\Rightarrow \quad 50=\beta \times \frac{200}{100} \Rightarrow \beta=25
$$
Also, power gain $=\beta^2 \times$ impedance gain
$$
=(25)^2 \times \frac{200}{100}=1250
$$

Topic: Physics, Semiconductors, JEE Main
A common emitter amplifier is designed with NPN transistor $(\alpha=0.99)$. The input impedance is $1 \mathrm{~K} \Omega$ and load is $10 \mathrm{~K} \Omega$. The voltage gain will be

Options:
A 9.9
B 99
C 990
D 9900

The correct answer is:
990

Solution:
Voltage gain $=\beta \times$ Resistance gain
$\beta=\frac{\alpha}{1-\alpha}=\frac{0.99}{(1-0.99)}=99$
Resistance gain $=\frac{10 \times 10^3}{10^3}=10$
$\Rightarrow$ Voltage gain $=99 \times 10=990$.

Topic: Physics, Semiconductors, JEE Main
A common emitter transistor amplifier has a current gain of 50 . If the load resistance is $4 \mathrm{k} \Omega$ and input resistance is $500 \Omega$, the voltage gain of the amplifier is

Options:
A $100$
B $200$
C $300$
D $400$

The correct answer is:
$400$

Solution:
$$
\begin{aligned}
\beta=50, R_L=4 \mathrm{k} \Omega & =4000 \Omega \\
R_i & =500 \Omega \\
\beta & =\frac{I_c}{I_b}
\end{aligned}
$$
Voltage gain of amplifier
$$
\begin{aligned}
& =\frac{\text { output voltage }}{\text { input voltage }}=\frac{I_c \times R_L}{I_b \times R_i} \\
& =\beta\left(\frac{R_L}{R_i}\right)=\frac{50 \times 4000}{500}=400
\end{aligned}
$$

Topic: Physics, Semiconductors, JEE Main
A common tangent to the conics $x^2=6 y$ and $2 x^2-4 y^2=9$ is:

Options:
A
$x-y=\frac{3}{2}$

B
$x+y=1$

C
$x+y=\frac{9}{2}$

D
$x-y=1$

The correct answer is:
$x-y=\frac{3}{2}$

Solution:
$$
\begin{aligned}
&x^2-6 y \\
&2 x^2-4 y^2=9
\end{aligned}
$$
Consider the line,
$$
x-y=\frac{3}{2}
$$
On solving (i) and (iii), we get only
$$
x=3, y=\frac{3}{2}
$$
Hence $\left(3, \frac{3}{2}\right)$ is the point of contact of conic (i), and line (iii)
On solving (ii) and (iii), we get only $x=3$,
$$
y=\frac{3}{2}
$$
Hence $\left(3, \frac{3}{2}\right)$ is also the point of contact of conic (ii) and line (iii).
Hence line (iii) is the common tangent to both the given conics.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2013 (25 Apr Online)
A common transistor radio set requires 12 V D.C. for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V A.C. on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are

Options:

The correct answer is:
440

Solution:
NPNS=VPVS

NP24=22012

NP=220×2412

NP=440

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A company dissolves 'x' amount of CO2 at 298 K in 1 litre of water to prepare soda water. X=×10-3 g. (nearest integer)

(Given: partial pressure of CO2 at 298 K=0.835 bar. Henry's law constant for CO2 at 298 K=1.67 kbar. Atomic mass of H,C and O is 1,12, and 6 g mol-1, respectively)

Options:

The correct answer is:
1221

Solution:
Using Henry’s law
Pgas = KH .Xgas

10-3X0.835=1.67nCO2nCO2+55.510-3X0.5=nCO2nCO2+55.5nCO255.5=0.5 X10-3nCO2=27.75X10-3WCO2=27.75X10-3X44=1221X10-3g

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places BQ:BP is:

Options:
A 3:4
B 4:3
C 3:2
D 2:3

The correct answer is:
3:4

Solution:
At point P the needle oscillates 20 times per minute. Therefore, time period TP=6020= 3 s and at point Q, TQ=6010=6 s.

The time period of oscillation magnetometer is given by,

T=2πIBHM

⇒TP=3=2πIBPcos30°M

And TQ=6 s=2πIBQcos60°M

Taking ratio,

36=1 BP32×BQ2

⇒34=BQBP

⇒BQ:BP=3:4

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A compass needle oscillates 20 times per minute at a place where the dip is 45° and the magnetic field is B1. The same needle oscillates 30 times per minute at a place where the dip is 30° and magnetic field is B2. Then B1:B2 is

Options:
A 93:42
B 42:93
C 33:22
D 22:33

The correct answer is:
22:33

Solution:
The frequency of oscillations of a magnetic needle in earth’s magnetic field is given as f=12πμBHI, where, horizontal component of earth’s magnetic field BH=Bcosθ, μ is the magnetic moment of the magnetic needle and I is moment of inertia of the magnetic needle about the axis about which it oscillates. Here, θ is angle of dip.

As the same magnetic needle is used, so the value of magnetic moment and the moment of inertia will remain the same at the two places.

Now, f∝Bcosθ.

Then the ratio, f1f2=2030=B1cos45°B2cos30°

⇒400900=B1×12B2×32⇒B1B2=23×23×2⇒B1:B2=22:33

Topic: Physics, Magnetic Properties of Matter, JEE Main
A compass needle oscillates 20 times per minute at a place where the dip is 45° and the magnetic field is B1. The same needle oscillates 30 times per minute at a place where the dip is 30° and magnetic field is B2. Then B1:B2 is

Options:
A 93:42
B 42:93
C 33:22
D 22:33

The correct answer is:
22:33

Solution:
The frequency of oscillations of a magnetic needle in earth’s magnetic field is given as f=12πμBHI, where, horizontal component of earth’s magnetic field BH=Bcosθ, μ is the magnetic moment of the magnetic needle and I is moment of inertia of the magnetic needle about the axis about which it oscillates. Here, θ is angle of dip.

As the same magnetic needle is used, so the value of magnetic moment and the moment of inertia will remain the same at the two places.

Now, f∝Bcosθ.

Then the ratio, f1f2=2030=B1cos45°B2cos30°

⇒400900=B1×12B2×32⇒B1B2=23×23×2⇒B1:B2=22:33

Topic: Physics, Magnetic Properties of Matter, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 1)
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It

Options:
A will become rigid showing no movement
B will stay in any position
C will stay in north-south direction only
D will stay in east-west direction only

The correct answer is:
will stay in any position

Solution:
It will stay in any position at geomagnetic north and south poles.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A competitive inhibitor of succinic dehydrogenase is

Options:
A Malate
B Malonate
C Oxaloacetate
D $\alpha_{\text {- ketoglutarate }}$

The correct answer is:
Malonate

Solution:
Malonate competes with succinate for active sites of succinate dehydrogenase.

Topic: Biology, Biomolecules (B), NEET, NEET 2008 (Mains)
A complex compound in which the oxidation number of a metal is zero is

Options:
A $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
B $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
C $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$
D $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}$

The correct answer is:
$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$

Solution:
CO is a neutral ligand, so the oxidation state of metal in metal carbonyls is always zero.
$$
\begin{array}{r}
{\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]} \\
\mathrm{x}+(0 \times 4)=0 \\
\mathrm{x}=0
\end{array}
$$

Topic: Chemistry, Coordination Compounds, JEE Main
A complex compound of $\mathrm{Co}^{3+}$ with molecular formula $\mathrm{CoCl}_{\mathrm{x}} \cdot \mathrm{yNH}_3$ gives a total of 3 ions when dissolved in water. How many $\mathrm{Cl}^{-}$ions satisfy both primary and secondary valencies in this complex?

Options:
A 3
B 1
C 4
D zero

The correct answer is:
1

Solution:
As cobalt is present as $\mathrm{Co}^{3+}$ and co-ordination number of cobalt is 6 , the molecular formula of compound should be $\mathrm{CoCl}_3 \cdot \mathrm{yNH}_3$. Now, as it gives a total of 3 ions when dissolved in water, its structural formula must be $\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2$
$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2 \rightleftharpoons\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}+2 \mathrm{Cl}^{-}$
Thus only one $\mathrm{Cl}^{-}$ion is satisfying both primary and secondary valency of $\mathrm{Co}^{3+}$ in this compound.

Topic: Chemistry, Coordination Compounds, JEE Main
A complex is represented as CoCl3.xH2O. Its 0.1 m solution in aqueous medium shows ÎTf= 0.558 K. Kf for water is 1.86 kg molâ1 K and assuming 100% ionisation of the complex having coordination number 6 . The formula of the complex is

Options:
A CoH2O4ClCl2
B CoH2O4Cl2Cl
C CoH2O6Cl3
D CoH2O5ClCl2

The correct answer is:
CoH2O5ClCl2

Solution:
ΔTf=i×Kf×m
0.558=i×1.86×0.1⇒i=3
So, the total no of ions produce by complex per mole should be 3
CoH2O5ClCl2→CoH2O5Cl2++2Cl-

Topic: Chemistry, Solutions, JEE Main
A complex number z is said to be unimodular if z=1. Let z1 and z2 are complex numbers such that z1-2z22-z1z2 is unimodular and z2 is not unimodular, then the point z1 lies on a

Options:
A circle of radius 2
B straight line parallel to x-axis
C straight line parallel to y-axis
D circle of radius 2

The correct answer is:
circle of radius 2

Solution:
Given, z1-2z22-z1z¯2 is unimodular.

⇒ z1-2z22-z1z¯2=1

⇒ z1-2z2=2-z1z¯2
Squaring both the sides, we get,
z1-2z22=2-z1z22
⇒ z1-2z2z¯1-2z¯2=2-z1z¯22-z¯1z2
∵ z2=zz¯
⇒ z1z¯1−2z1z¯2−2z¯1z2+4z¯2z¯2
=4-2z¯1 z2-2z1 z¯2+z1 z¯1 z2 z¯2
⇒ z12+4z22=4+z12 z22
⇒ z12-4+4z22-z12 z22=0
⇒ z12-41-z22=0
⇒ z1=2 or z2=1

Given, z2 is not unimodular
∴ z1=2
∴ Point z1 lies on a circle of radius 2.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2015 (04 Apr)
A complex number z is said to be unimodular if |z|=1. Let, z1 and z2 are complex numbers such that z 1 - 2z 2 2 - z 1 z Â¯ 2 is unimodular and z2 is not unimodular, then the point z1 lies on a

Options:
A circle of radius 2
B straight line parallel to x-axis
C straight line parallel to y-axis
D circle of radius 2

The correct answer is:
circle of radius 2

Solution:
Given, z 1 - 2z 2 2 - z 1 z ¯ 2 is unimodular
⇒ z 1 - 2 z 2 2 - z 1 z ¯ 2 = 1
⇒ z 1 - 2 z 2 = 2 - z 1 z ¯ 2
Squaring both the sides, we get,
z 1 - 2z 2 2 = 2 - z 1 z ¯ 2 2
⇒ z 1 - 2 z 2 z ¯ 1 - 2 z ¯ 2 = 2 - z 1 z ¯ 2 2 - z ¯ 1 z 2
∵ z 2 = z z ¯
⇒ z 1 z ¯ 1 - 2 z 1 z ¯ 2 - 2 z ¯ 1 z 2 + 4 z 2 z ¯ 2
= 4 - 2 z ¯ 1 z 2 - 2 z 1 z ¯ 2 + z 1 z ¯ 1 z 2 z ¯ 2
⇒ z 1 2 + 4 z 2 2 = 4 + z 1 2 z 2 2
⇒ z 1 2 - 4 + 4 z 2 2 - z 1 2 z 2 2 = 0
⇒ z 1 2 - 4 1 - z 2 2 = 0
⇒ z1=2 or z2=1

Given, z2 is not unimodular
∴ z 1 = 2
∴ Point z1 lies on a circle of radius 2.

Topic: Mathematics, Complex Number, JEE Main
A complex number \( z \) is said to be unimodular, if \( |z|=1 \) Suppose \( z_{1} \) and \( z_{2} \) are complex numbers such that \( \frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}} \) is unimodular and \( z_{2} \) is not unimodular. Then, the point \( z_{1} \) lies on a

Options:
A straight line parallel to \( X \)-axis
B straight line parallel to \( Y \)-axis
C circle of radius \( 2 \)
D circle of radius \( \sqrt{2} \).

The correct answer is:
circle of radius \( 2 \)

Solution:
Given, z2 is not unimodular, i.e., z2≠1

and z1-2z22-z1z¯2 is unimodular.

⇒z1-2z22-z1z¯2=1⇒z1-2z22=2-z1z¯22

⇒z1-2z2z¯1-2z¯2=2-z1z¯22-z¯1z2

∵zz¯=|z|2

⇒z12+4z22-2z¯1z2-2z1z¯2

=4+z12z22-2z¯1z2-2z1z¯2

⇒z22-1z12-4=0

∵z2≠1

∴z1=2

Let z1=x+iy⇒x2+y2=22

∴Point z1 lies on a circle of radius 2

Topic: Mathematics, Complex Number, JEE Main
A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a strandarized acid as follows :
CoNH3xClyaq+HClâNH4+aq+Coy++aq+Clâaq A 1.8 g complex required 20.00 mL 1.54 M HCl to reach the equivalence point. Also, if the reaction mixture at equivalence point is treated with excess of AgNO3 solution, 7.735 g of AgCl precipitate was produced. What is the formula of this complex?
[Given: atomic weight of CO = 59 gmol-1]

Options:
A Co NH 3 4 Cl 3
B Co NH 4 4 Cl 3
C Co NH 3 3 Cl 4
D Co NH 4 3 Cl 4

The correct answer is:
Co NH 3 4 Cl 3

Solution:
Co(NH3)xCly1.8M+xHCL1.8MâxNH4+aq+Coy++(x+y)Cl-1.8(x+r)M
1 . 8 x M = 2 0 Ã 1 . 5 4 Ã 1 0 - 3 = 3 0 . 8 Ã 1 0 - 3 ........(I)
1 Â· 8 x + y M = 7 . 7 3 5 1 4 3 . 5 = 5 3 . 9 Ã 1 0 - 3 ........(II)
II I â 5 3 . 9 Ã 1 0 - 3 3 0 . 8 Ã 1 0 - 3 = 1 . 7 5 = x + y x = 1 + y x
y x = 0 . 7 5
Substituting in (I) â
1 . 8 x 5 9 + 1 7 x + 3 5 . 5 Ã 0 . 7 5 x = 3 0 . 8 Ã 1 0 - 3
â´ x = 3.98 = 4 â´ y = 3

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A complex of iron contains \( 45.6 \% \) iron by mass. Find out the number of iron atoms in \( 5.00 \mathrm{~g} \) of this complex number of iron atoms \(=x^{*} 10^{\wedge} 22\) find the \(x\)

Options:

The correct answer is:
2.4

Solution:
Given,

Percentage by mass of Fe in the compound =45.6%

Mass of Fe in 5 g of complex =45.6×5100=2.28 g

Number of atoms = Number of moles × Avogadro number

But,

Number of moles = Given mass of FeMolar mass of Fe=2.2856=0.04

Now,

Number of atoms =0.04×6.022×1023=2.4×1022

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main