datavorous/entrance-exam-dataset huggingface dataset:
In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :

Options:
A 16
B 56
C 13
D 23

The correct answer is:
16

Solution:
Let A& B be the sets for NCC and NSS respectively.

Given nA=40, nB=30, nA∩B=20

∴n A∪B=nA+nB-nA∩B

=40+30-20=50

∴P A∪B=5060=56

∴P A∪B′=1-P A∪B

=1-56=16

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (12 Jan Shift 2)
In a class of 60 students, 45 students like music, 50 students like dancing, 5 students like neither. Then the number of students in the class who like both music and dancing is

Options:
A 35
B 40
C 50
D 5

The correct answer is:
40

Solution:
Let ' $x$ ' be the number of students wholikes both music
and dance. 5 students likes neither music nor dancing. Hence, total number of remaining students $=60-5=55$
Nowfrom Venn diagram, $45-x+x+$
$\therefore 95-x=55$
$\therefore x=95-55=40$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?

Options:
A
73

B
65

C
68

D
74

The correct answer is:
65

Solution:
Total student $=100 ;$ for 70 stds $75 \times 70=5250 \Rightarrow 7200-5250=1950$
$$
\text { Average of girls }=\frac{1950}{30}=65
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2002
In a class of 125 students 70 passed in Mathematics, 55 in Statistics and 30 in both. The probability that a student selected at random from the class has passed in only one subject is

Options:
A $\frac{13}{25}$
B $\frac{3}{25}$
C $\frac{17}{25}$
D $\frac{8}{25}$

The correct answer is:
$\frac{13}{25}$

Solution:
Consider the following events :
$A=A$ student is passed in Mathematics,
$B=$ A student is passed in Statistics.
Then, $P(A)=\frac{70}{125}, \quad P(B)=\frac{55}{125}$, $P(A \cap B)=\frac{30}{125}$.
Then Required probability is $P(A \cap \bar{B})+P(\bar{A} \cap B)=P(A)+P(B)-2 P(A \cap B)$
$=\frac{70}{125}+\frac{55}{125}-\frac{60}{125}=\frac{65}{125}=\frac{13}{25}$.

Topic: Mathematics, Probability, JEE Main
In a class of 125 students 70 passed in Mathematics, 55 passed in Statistics and 30 passed in both. What is the probability that a student selected at random from the class has passed in only one subject?

Options:
A $13 / 25$
B $3 / 25$
C $17 / 25$
D $8 / 25$

The correct answer is:
$13 / 25$

Solution:
Required probability
$=\frac{40}{125}+\frac{25}{125}=\frac{65}{125}=\frac{13}{25}$

Topic: Mathematics, Probability, JEE Main
In a class of 140 students numbered 1 to140 , all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is:

Options:
A 42
B 1
C 38
D 102

The correct answer is:
38

Solution:
From the given 140 students, all even numbered students opted Mathematics M, hence, the number of students opted Mathematics is M≡1402=70, where x denotes the greatest integer function of x.

Similarly, the students who opted Physics are P≡1403=46

And, the students who opted Chemistry are C≡1405=28

Similarly, the students who opted Mathematics and Physics are the numbers which are divisible by both 2 & 3 i.e. 6, and their number is 1406=23

The students who opted Mathematics and Chemistry are the numbers which are divisible by both 2 & 5 i.e. 10, and their number is 14010=14

The students who opted Physics and Chemistry are the numbers which are divisible by both 3 & 5 i.e. 15, and their number is 14015=9

And, the students who opted all the three subjects are the numbers which are divisible by 2, 3 & 5 i.e. 30, and their number is 14030=4

Now, using
nP∪C∪M=nP+nC+nM-nP∩C-nC∩M-nM∩P+nP∩M∩C

=46+28+70-9-14-23+4

=102

Thus, 102 students have opted for atleast one of the three subjects.

Hence, number of students who did not opt for any of the three courses =140-102=38.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2019 (10 Jan Shift 1)
In a class tournament where the participants were to play one game with one another, two of the class players fell ill, having played 3 games each. If the total number of games played is 27, then the number of participants at the beginning was

Options:

The correct answer is:
9

Solution:
Suppose, the two players did not play at all such that the remaining n-2 players played n-2C2 matches.
Since, these two players played 3 matches each, hence the total number of matches is n-2C2+3+3=27⇒n-2C2=21
n-2C2=21⇒n-2n-32=7×62
⇒n-2=7⇒n=9.

Topic: Mathematics, Permutation Combination, JEE Main
In a classroom, one-fifth of the boys leave the class and the ratio of the remaining boys to girls is $2: 3$. If further 44 girls leave the class, the ratio of boys to girls is $5: 2$. How many more boys should leave the class so that the number of boys equals that of girls?

Options:
A 16
B 24
C 30
D 36

The correct answer is:
24

Solution:
Let no of Boys $=x$ $\&$ Let no of girls $=y$
Given $\frac{\left(\frac{4 x}{5}\right)}{y}=\frac{2}{3}$
$\frac{2 x}{5 y}=\frac{1}{3}$
$y=\frac{6 x}{5}$ .........(i)
Also, $\frac{\left(\frac{4 x}{5}\right)}{y-44}=\frac{5}{2}$
$8 x=25(y-44)$
$8 x=25\left(\frac{6 x}{5}-44\right)$ (use 1)
$x=50$
$y=60$

Topic: Mathematics, Basic of Mathematics, KVPY, KVPY 2017 (19 Nov SA)
In a closed circuit, the current I(in ampere) at an instant of time t(in second) is given by I=4-0.08t. The number of electrons flowing in 50s through the cross-section of the conductor is

Options:
A 1.25×1019
B 6.25×1020
C 5.25×1019
D 2.55×1020

The correct answer is:
6.25×1020

Solution:
I=4-0.08t A
Or dqdt=4-0.08t A
or q=∫0504-0.08tdt C
or Ne=4t-0.08t22050=100 C
Where N is number of electrons.
or N=100e=1001.6×10-19
=6.25×1020

Topic: Physics, Current Electricity, JEE Main
In a closed flask of 5 litrer, $1.0 \mathrm{~g}$ of $\mathrm{H}_2$ is heated from $300 \mathrm{to} 600 \mathrm{~K}$. which statement is not correct

Options:
A Pressure of the gas increases
B The rate of collision increases
C The number of moles of gas increases
D The energy of gaseous molecules increases

The correct answer is:
The number of moles of gas increases

Solution:
At constant $V$ of a definite mass
$$
\frac{P_1}{T_1}=\frac{P_2}{T_2} \therefore \frac{P_1}{P_2}=\frac{300}{600}=\frac{1}{2}
$$
i.e. pressure increases and on increasing temperature energy of molecules increases so the rate of collisions also increases and number of moles remains constant because there is neither addition nor removal of gas in the occurring.

Topic: Chemistry, States of Matter, JEE Main
In a closed flask of 5 litres, 1.0 g of H 2 is heated from 300 to 600 K. Which statement is not correct?

Options:
A The number of moles of gas increases
B The rate of collision increases
C The energy of gaseous molecules increases
D Pressure of the gas increases

The correct answer is:
The number of moles of gas increases

Solution:
In a closed vessel for a fixed mass of gas (H2). The no. of moles remains constant.

Topic: Chemistry, States of Matter, NEET
In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm2, the amount of water poured in the organ tube is ________g. (Take speed of sound in air is 330 m s-1 )

Options:

The correct answer is:
400

Solution:
Fundamental frequency for closed organ pipe in the first part:

V4l1=30⇒l1=114 m

Similarly, for second part:

V4l2=110⇒l2=34 m

Therefore, change in length Δl=2 m

Change in volume =AΔl=400 cm3

M=400 g;∵ρ=1 g cm-3

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (30 Jan Shift 1)
In a closed tube whose length can be altered by sliding one of the halves relative to other is shown in the figure. In the initial condition the detector detects minimum intensity. One of the tubes is then displaced by 5 cm, during this displacement detector detects maximum intensity for 10 times and then a minimum intensity when the displacement is complete. What is the wavelength of sound?

Options:
A 10/9 cm
B 1 cm
C 1/2 cm
D 5/9 cm

The correct answer is:
1 cm

Solution:
Path difference introduced due to displacement of tube is given by
2x=10 cm
Now due to one wavelength change, maxima/minima will be attained once hence for 10 maxima's

ÎP=10Î»=10 cm
So, Î»=1 cm

Topic: Physics, Waves and Sound, JEE Main
In a closed vesselat STP, 50 L of $\mathrm{CH}_{4}$ is ignited with $750 \mathrm{~L}$ of air (containing $\left.20 \% \mathrm{O}_{2}\right)$. The number of moles of $\mathrm{O}_{2}$ remaining in the vessel on cooling to room temperature is closest to

Options:
A $5.8$
B $2.2$
C $4.5$
D $6.7$

The correct answer is:
$2.2$

Solution:
$\begin{array}{l}
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\
50 \mathrm{~L} \quad 150 \mathrm{~L}
\end{array}$
$\mathrm{CH}_{4}$ is limiting reagent
Moles of $\mathrm{O}_{2}$ remaining $=\frac{50}{22.4}=2.2$ moles

Topic: Chemistry, Some Basic Concepts of Chemistry, KVPY, KVPY 2017 (5 Nov SA)
In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :

Options:
A outside the cable
B inside the outer conductor
C inside the inner conductor
D in between the two conductors

The correct answer is:
outside the cable

Solution:
$\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_0 \mathrm{i}_{\mathrm{enc}}=0$
$\therefore \mathrm{B}=0$ outside the cable

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (05 Apr Shift 1)
In a coil of resistance 8 Ω, the magnetic flux due to an external magnetic field varies with time as ϕ=239-t2. The value of total heat produced in the coil, till the flux becomes zero, will be _____ J.

Options:

The correct answer is:
2

Solution:
It is given that magnetic flux ϕ=239-t2

For magnetic flux to be zero 0=239-t2⇒t=3 s

According to Faraday's law of electromagnetic induction,

ε=-dϕdt=-230-2t=4t3

Heat produced from 0 to 3 s,

H=∫03ε2Rdt=∫0316t29×8dt=2 J

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (26 Jul Shift 2)
In a coil of resistance $10 \Omega$, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in weber is

Options:
A 8
B 2
C 6
D 4

The correct answer is:
2

Solution:
$$
\begin{aligned}
I & =\left|\frac{1}{R} \frac{d \phi}{d t}\right| \\
|d \phi| & =|I R d t| \\
d \phi & =(\text { area of triangle }) \times R \\
& =\left(\frac{1}{2} \times 4 \times 0.1\right) \times 10=2 \mathrm{~Wb}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2012 (Mains)
In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is:

Options:
A 275 Wb
B 200 Wb
C 225 Wb
D 250 Wb

The correct answer is:
250 Wb

Solution:
Emf=iR⇒-dϕdt=iR
∫-dϕ=∫(idt)R
q=ΔϕR= Area under i-t graph
⇒∆ϕ=12×10×0.5×100=250 Wb

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2017 (02 Apr)
In a coil, the current changes from $-2 \mathrm{~A}$ to $+2 \mathrm{~A}$ in $0.2 \mathrm{~s}$ and induces an emf of $0.1 \mathrm{~V}$. The self inductance of the coil is :

Options:
A $4 \mathrm{mH}$
B $1 \mathrm{mH}$
C $5 \mathrm{mH}$
D $2.5 \mathrm{mH}$

The correct answer is:
$5 \mathrm{mH}$

Solution:
$(E m f)_{\text {induced }}=-L \frac{d i}{d t}$
In magnitude form,
$\begin{aligned}
& \left|\operatorname{Emf}_{\text {ind }}\right|=\left|(-) L \frac{\mathrm{di}}{\mathrm{dt}}\right| \\
& \Rightarrow 0.1=\frac{(\mathrm{L})[+2-(-2)]}{0.2} \\
& \Rightarrow \mathrm{L}=\frac{0.1 \times 0.2}{4}=5 \mathrm{mH}
\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (06 Apr Shift 2)
In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is

Options:
A v02
B v04
C 2 v0
D v02

The correct answer is:
2 v0

Solution:
KE2=32mv02. Conservation of momentum, v 1 + v 2 = v 0 . Conservation of energy, 12mv12+12mv22=12 × 32mv02
⇒v12+v22=32v02⇒v12+v22+2v1v2=v02⇒32v02+2v1v2=v02
⇒2v1v2=−12v02⇒(v1−v2)2=32v02+12v02 =2 v 0 2
⇒v1−v2=2v0.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2018 (08 Apr)
In a common base amplifier, the phase difference between the input signal voltage and output voltage is

Options:
A
$\frac{\pi}{4}$

B
$\pi$

C
$0$

D
$\frac{\pi}{2}$

The correct answer is:
$0$

Solution:
No phase difference between input and output signal

Topic: Physics, Semiconductors, JEE Main, JEE Main 2005
In a common base mode of a transistor, $\mathrm{t}$ collector current is $5.488 \mathrm{~mA}$ for an emit current of $5.60$ $\mathrm{mA}$. The value of the base current amplification factor $(\beta)$ will be

Options:
A
48

B
49

C
50

D
51

The correct answer is:
49

Solution:
$I_b=I_e-I_c$
$\beta=\frac{I_c}{I_b}=49$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2006
In a common base transistor circuit, the current gain is 0.98. On changing the emitter current by 5.00 mA, the change in collector current is

Options:
A 0.196 mA
B 2.45Â mA
C 4.9 mA
D 5.1Â mA

The correct answer is:
4.9 mA

Solution:
Given, Î±=0.98
â´ Î±=âIcâIe
0.98=âIc5
â âIc=4.9 mA

Topic: Physics, Semiconductors, JEE Main
In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be:

Options:
A 180°
B 45°
C 90°
D 135°

The correct answer is:
180°

Solution:
In a common emitter amplifier circuit, the input voltage must be forward biased and output voltage must be reversed biased.

So the input voltage must be out of phase of output voltage.

Therefore, the phase difference between the input and output voltage is 180°.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2017 (02 Apr)
In a common emitter amplifier, the input AC signal to be amplified is applied across

Options:
A Forward biased emitter-base junction
B Reverse biased collector-base junction
C Reverse biased emitter-base junction
D Forward biased collector-base junction

The correct answer is:
Forward biased emitter-base junction

Solution:
The circuit diagram for CE amplifier is given below
Input signal is applied across base-emitter junction.

Topic: Physics, Semiconductors, JEE Main
In a common emitter configuration, a transistor has $\beta=50$ and input resistance $1 \mathrm{k} \Omega$. If the peak value of a.c. input is $0.01 \mathrm{~V}$ then the peak value of collector current is

Options:
A $0.01 \mu \mathrm{A}$
B $0.25 \mu \mathrm{A}$
C $100 \mu \mathrm{A}$
D $500 \mu \mathrm{A}$

The correct answer is:
$500 \mu \mathrm{A}$

Solution:
$$
\text { Hints: } \beta=50 \Rightarrow \beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \Rightarrow \Delta \mathrm{I}_{\mathrm{C}}=\beta \times \Delta \mathrm{I}_{\mathrm{B}}
$$
$$
\begin{aligned}
& \Delta \mathrm{I}_{\mathrm{B}}=\frac{0.01}{10^3}=10^{-2} \times 10^{-3}=10^{-5} \\
& \Delta \mathrm{I}_{\mathrm{C}}=50 \times 10^{-5}=500 \times 10^{-6}=500 \mu \mathrm{A}
\end{aligned}
$$

Topic: Physics, Semiconductors, JEE Main
In a common emitter configuration with suitable bias, it is given than $\mathrm{R}_{\mathrm{L}}$ is the load resistance and $R_{B E}$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:
[ $\beta$ is current gain, $I_B, I_C, I_E$ are respectively base, collector and emitter currents]

Options:
A
$\beta \frac{R_L}{R_{B E}}, \frac{\Delta I_E}{\Delta I_B}, \beta^2 \frac{R_L}{R_{B E}}$

B
$\beta^2 \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_B}, \beta \frac{R_L}{R_{B E}}$

C
$\beta^2 \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_E}, \beta^2 \frac{R_L}{R_{B E}}$

D
$\beta \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_B}, \beta^2 \frac{R_L}{R_{B E}}$

The correct answer is:
$\beta \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_B}, \beta^2 \frac{R_L}{R_{B E}}$

Solution:
Current gain $\beta=\frac{\Delta I_C}{I_B}$
Voltage gain $\mathrm{A}_{\mathrm{v}}=$ Current gain $\times$ Resistance gain $=\beta \frac{R_L}{R_{B E}}$
Power gain $A_p=(\text { Current gain })^2 \times$ Resistance gain
$$
=\beta^2 \frac{R_L}{R_{B E}}
$$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In a common emitter configuration with suitable bias, it is given that RL is the load resistance and RBE is small-signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by: β is current gain, IB, IC and IE are respectively base, collector, and emitter currents.

Options:
A βRLRBE,ΔICΔIB,β2RLRBE
B βRLRBE,ΔIEΔIB,β2RLRBE
C β2RLRBE,ΔICΔIB,βRLRBE
D β2RLRBE,ΔICΔIE,β2RLRBE

The correct answer is:
βRLRBE,ΔICΔIB,β2RLRBE

Solution:
For a common emitter configuration, current gain, β=ΔICΔIB. Voltage gain is the ratio of output voltage to the input voltage. It is given as, AV=ΔVCE RBEΔIB=βRLRBE (Since from Ohm's law, RL=∆VCE∆IB). Power gain AP=βAV=β2RLRBE.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2018 (15 Apr)
In a common emitter NPN transistor, 1010 electrons enter the emitter in 10-6 s when it is connected to the battery. About 5% of electrons recombine with holes in the base. The current gain of the transistor is
e=1.6Ã10-19 C

Options:
A 0.98
B 19
C 49
D 0.95

The correct answer is:
19

Solution:
Current gain of common emitter Î²=iCiB
Î²=95% of iE5% of iE
Î²=955=19

Topic: Physics, Semiconductors, JEE Main
In a common emitter transistor amplifier, signal voltage across the collector resistance of 2 kÎ© is 2 volt. What should be the value of resistance R B in series with V BB voltage supply of 2 volt in the input circuit, if the D.C. base current is 10 times the signal current in R B ?

( Given: base emitter voltage VBE=0.6 V, current amplification factor Î² ac â Î² dc =100 )

Options:
A 7 kΩ
B 26 kΩ
C 10 kΩ
D 14 kΩ

The correct answer is:
14 kΩ

Solution:
icAC=Vout ACRC=2 volt 2 kΩ=1 mA

iBAC=icACβac=1 mA100=0.01 mA

iBDC=10iBAC=0.1 mA=10-4 A

RB=ViB=2-0.610-4=14 kΩ

Topic: Physics, Semiconductors, JEE Main
In a common emitter transistor amplifier, signal voltage across the collector resistance of 2 kÎ© is 2 volt. What should be the value of resistance R B in series with V BB voltage supply of 2 volt in the input circuit, if the D.C. base current is 10 times the signal current in R B ?

( given base emitter voltage V BE =0.6 volt, current amplification factor Î² ac â Î² dc =100 )

Options:
A 7 kΩ
B 26 kΩ
C 10 kΩ
D 14 kΩ

The correct answer is:
14 kΩ

Solution:
icAC=Vout ACRC=2 volt 2 kΩ=1 mA

iBAC=icACβac=1 mA100=0.01 mA

iBDC=10iBAC=0.1 mA=10-4 A

RB=ViB=2-0.610-4=14 kΩ

Topic: Physics, Semiconductors, JEE Main
In a common-emitter configuration of a transistor, the base current IE=2 Î¼A, Î±=0.9 then the value of IC is

Options:
A 3.0 μA

B 2.25 μA

C 4.9 μA

D 1.8 Î¼A

The correct answer is:
1.8 Î¼A

Solution:
We know that current gain is given by,
α=IcIb
∴ Ic=α×Ib=0.9×2=1.8 μA

Topic: Physics, Semiconductors, JEE Main
In a common-emitter transistor amplifier, the output resistance is 500 kÎ© and the current gain Î²=49. If the power gain of the amplifier is 5Ã106, the input resistance is

Options:
A 165 Ω
B 198 Ω
C 225 Ω
D 240 Ω

The correct answer is:
240 Ω

Solution:
Given, Ro= 500 kΩ, β =49 and P = 5×106
We have P=β 2 RoRi
5×106=492×500×103Ri
Ri=49×49×500×1035×106=240 Ω

Topic: Physics, Semiconductors, JEE Main
In a communication system operating at wavelength \( 800 \mathrm{~nm} \), only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting \( T V \) signals of band width \( 6 \mathrm{MHz} \) are (Take velocity of light \( c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, h=6.6 \times 10^{-34} J_{-s} \) )

Options:
A \( 6.25 \times 10^{5} \)
B \( 4.87 \times 10^{5} \)
C \( 3.75 \times 10^{6} \)
D \( 3.86 \times 10^{6} \)

The correct answer is:
\( 6.25 \times 10^{5} \)

Solution:
Available signal width =1100×cλ
=1100×3×108800×10-9 Hz=3.75×1012 Hz
Number of available channels =3.75×10126×106
=6.25×105

Topic: Physics, Communication System, JEE Main
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c=3×108 m/s, h=6.6×10-34 J-s )

Options:
A 6.25×105
B 4.87×105
C 3.75×106
D 3.86×106

The correct answer is:
6.25×105

Solution:
Available signal width =1100×cλ
=1100×3×108800×10-9 Hz=3.75×1012 Hz
Number of available channels =3.75×10126×106
=6.25×105

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (09 Jan Shift 2)
In a complexometric titration of metal ion with ligand M (Metal ion)+L(Ligand)→C (Complex) end point is estimated spectrophotometrically (through light absorption). If 'M' and 'C' do not absorb light and only 'L' absorbs then the titration plot between absorbed light (A) versus volume of ligand 'L' (V) would look like:

Options:
A
B
C
D

The correct answer is:

Solution:
Initially ligand consumed by metal due to formation of complex. So absorbed light (A) remain constant, after complex formation is completed, extra volume of ligand solution increases ligand concentration and also increases absorbed light.

Topic: Chemistry, Coordination Compounds, JEE Main
In a complexometric titration of metal ion with ligand M (Metal ion)+L(Ligand)→C (Complex) end point is estimated spectrophotometrically (through light absorption). If 'M' and 'C' do not absorb light and only 'L' absorbs then the titration plot between absorbed light (A) versus volume of ligand 'L' (V) would look like:

Options:
A
B
C
D

The correct answer is:

Solution:
Initially ligand consumed by metal due to formation of complex. So absorbed light (A) remain constant, after complex formation is completed, extra volume of ligand solution increases ligand concentration and also increases absorbed light.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (16 Apr Online)
In a compound atoms of element $Y$ from ccp lattice and those of element $X$ occupy $2 / 3^{\text {rd }}$ of tetrahedral voids. The formula of the compound will be

Options:
A
$X_4 Y_3$

B
$X_2 Y_3$

C
$X_2 Y$

D
$\mathrm{X}_3 \mathrm{Y}_4$

The correct answer is:
$X_4 Y_3$

Solution:
No. of atoms of $Y=4$
No. of atoms of $X=\frac{2}{3} \times 8$
Formula of compound will be $X_4 Y_3$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2008
In a compound C, H and N atoms are present in 9 : 1 : 35 by weight. Molecular weight of
compound is 108. Molecular formula of compound is

Options:
A
$\mathrm{C}_2 \mathrm{H}_6 \mathrm{N}_2$

B
$\mathrm{C}_3 \mathrm{H}_4 \mathrm{N}$

C
$\mathrm{C}_6 \mathrm{H}_8 \mathrm{N}_2$

D
$\mathrm{C}_9 \mathrm{H}_{12} \mathrm{N}_3$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_8 \mathrm{N}_2$

Solution:
According to molecular weight given

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2002
In a compound microscope, if the objective produces an image $I_0$ and the eye piece produces an image $I_e$, then

Options:
A $I_o$ is virtual but $I_e$ is real
B $I_o$ is real but $I_e$ is virtual
C $I_o$ and $I_e$ are both real
D $I_o$ and $I_e$ are both virtual

The correct answer is:
$I_o$ is real but $I_e$ is virtual

Solution:
The ray diagram of a compound microscope is shown below.
From the ray diagram, we can see that, the objective lens produces a real image while the eyepiece produces a virtual image.
Therefore, $I_o$ is real, but $I_e$ is virtual.
Hence, option (2) is the correct answer.

Topic: Physics, Ray Optics, JEE Main