datavorous/entrance-exam-dataset huggingface dataset:
The average marks of boys in a class is $52$ and that of girls is $42$. The average marks of boys and girls combined is $50$. The percentage of boys in the class is

Options:
A
$40$

B
$20$

C
$80$

D
$60$

The correct answer is:
$80$

Solution:
$52 x+42 y=50(x+y)$
$2 x=8 y$
$\Rightarrow \frac{x}{y}=\frac{4}{1}$ and $\frac{x}{x+y}=\frac{4}{5}$
$\therefore \%$ of boys $=80$.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2007
The average mass of rain drops is $3.0 \times 10^{-5} \mathrm{~kg}$ and their avarage terminal velocity is $9 \mathrm{~m} / \mathrm{s}$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives $100 \mathrm{~cm}$ of rain in a year.

Options:
A
$3.5 \times 10^5 \mathrm{~J}$

B
$4.05 \times 10^4 \mathrm{~J}$

C
$3.0 \times 10^5 \mathrm{~J}$

D
$9.0 \times 10^4 \mathrm{~J}$

The correct answer is:
$4.05 \times 10^4 \mathrm{~J}$

Solution:
Total volume of rain drops, received $100 \mathrm{~cm}$ in a year by area $1 \mathrm{~m}^2$
$$
=1 \mathrm{~m}^2 \times \frac{100}{100} \mathrm{~m}=1 \mathrm{~m}^3
$$
As we know, density of water,
$$
d=10^3 \mathrm{~kg} / \mathrm{m}^3
$$
Therefore, mass of this volume of water $M=d \times v=10^3 \times 1=10^3 \mathrm{~kg}$
Average terminal velocity of rain drop $v=9 \mathrm{~m} / \mathrm{s}$ (given)
Therefore, energy transferred by rain,
$$
\begin{aligned}
&E=\frac{1}{2} m v^2 \\
&=\frac{1}{2} \times 10^3 \times(9)^2 \\
&=\frac{1}{2} \times 10^3 \times 81=4.05 \times 10^4 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (11 Apr Online)
The average molar mass of chlorine is 35.5 g mol-1. The ratio of Cl35 to Cl37 in naturally occurring chlorine is close to :

Options:
A 4 : 1
B 3 : 1
C 2 : 1
D 1 : 1

The correct answer is:
3 : 1

Solution:
35CI37CIMolar ratiox1−x

Mavg.=35×x+371−x=35.5

35x+37−37x=35.5

2x=1.5

x=34

So, ratio of 35CI:37CI=3:1

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The average of five consecutive odd numbers is 61. Then the difference between the highest and lowest numbers is

Options:
A 2
B 5
C 8
D Cannot be determined

The correct answer is:
8

Solution:
Let, the number are k,âk+2,ââk+4, âk+6, âk+8
so, the difference between highest & lowest = 8

Topic: Mathematics, Statistics, JEE Main
The average power dissipated in a pure inductor of inductance $L$ when an ac current is passing through it, is

Options:
A $\frac{1}{2} L I^2$
B $\frac{1}{4} L I^2$
C $2 L i^2$
D Zero

The correct answer is:
Zero

Solution:
The average power dissipated in a circuit consisting only of inductors or capacitors is zero.
$\text { power }=1 / T \int v(t) i(t) d t$
In sinusoidal wave, $v$ and $I$ differ by $90^{\circ}$
$P_a=\omega 2 \pi \int A \omega \sin (\omega t) \cos (\omega t) d t$
where $A$ is a constant depending on the value of the inductance $L$ or capacitance $C$ of the component.
$P=A \omega^* / 2 \pi\left[\int 2 \pi / \sin (2 \omega t) d t\right]$
period of $\sin (2 \omega t)$ is $T^{\prime}=\pi / \omega$
interval of integration of $\mathrm{P}$ is exactly two periods of $\sin (2 \omega t)$
so finally:
$P=A \omega^z / 2 \pi \int \sin (2 \omega t) d t=0\left[\text { integral from o to } 2 T^{\prime}\right]$

Topic: Physics, Alternating Current, JEE Main
The average rms speed of molecules in a sample of oxygen gas at 300 K is 484 m s-1, respectively. The corresponding value at 600 K is nearly (assuming ideal gas behaviour)

Options:
A 968 m s-1
B 986 m s-1
C 648 m s-1
D 684 m s-1

The correct answer is:
684 m s-1

Solution:
The formula for rms velocity is
C rms = 3 KT m
∴ C rms 6 0 0 K C rms 3 0 0 K = 6 0 0 3 0 0
⇒ Crms600K=2×484=684 m s-1

Topic: Physics, Thermodynamics, JEE Main
The average S-F bond energy in kJ mol-1 of SF6 is _. (Rounded off to the nearest integer)

[Given : The values of standard enthalpy of formation of SF6g, Sg and Fg are -1100, 275 and 80 kJ mol-1 respectively.]

Options:

The correct answer is:
309

Solution:
SF6g→Sg+6 Fg

If ∈- bond enthalpy

ΔrH=6×ϵS-F

=ΔfHS, g+6×ΔfHF, g-ΔfHSF6, g

=275+6×80--1100

=1855 kJ

ϵS-F=18556=309.16 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (26 Feb Shift 2)
The average speed $v$ and r.m.s. speed $\bar{v}$ of the molecules are related as

Options:
A $\bar{v}=0.92 v$
B $\bar{v}^2=0.29 v^2$
C $\bar{v}=v$
D $v=0.92 \bar{v}$

The correct answer is:
$v=0.92 \bar{v}$

Solution:
$v_{r m s}=\sqrt{\frac{3 R T}{M}}$ and $\quad v_{\mathrm{av}}=\sqrt{\frac{8 R T}{\pi M}}$
$\Rightarrow v_{a v}=\sqrt{\frac{8}{3 \pi}} v_{r m s} \Rightarrow v=0.92 \bar{v}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The average translational kinetic energy of \(\mathrm{O}_2\) (molar mass 32 ) molecules at a particular temperature is \(0.048 \mathrm{eV}\). The translational kinetic energy of \(\mathrm{N}_2\) (molar mass 28) molecules in \(\mathrm{eV}\) at the same temperature is?

Options:
A \(0.0015\)
B \(0.048\)
C \(0.003\)
D \(0.768\)

The correct answer is:
\(0.048\)

Solution:
Given: Molar mass of oxygen \(=32\)
Molar mass of nitrogen \(=28\)
Kinetic energy of oxygen molecules \(=0.048 \mathrm{eV}\)
Solution: The average translational energy for any gas is given by,
\(\mathrm{E}=\frac{3}{2} \mathrm{KT}\)
where, \(\mathrm{K}\) is Boltzmann constant and \(\mathrm{T}\) is the temperature.
From the above equation we conclude that energy is directly proportional to the temperature.
As, according to the question temperature is same for the nitrogen gas, So the kinetic energy for nitrogen is also same as that of oxygen. i.e. \(0.048 \mathrm{eV}\)

Topic: Physics, Kinetic Theory of Gases, JEE Main
The average translational kinetic energy of N2 gas molecules at _________C∘ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt.

(Given kB=1.38×10-23 J K-1 (Fill the nearest integer).

Options:

The correct answer is:
500

Solution:
Given

Translation K.E. of N2=K.E of electron

32kT=eV

32×1.38×10-23 T=1.6×10-19×0.1 ⇒T=773 K

T=773-273=500°C

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The average value of a.c. voltage is given by $\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin (\omega \mathrm{t})$ over time interval $t=0$ to $=\frac{\pi}{\omega}$ is

Options:
A $\frac{\mathrm{v}_{\mathrm{m}}}{\pi}$
B $0$
C $\mathrm{V}_{\mathrm{m}}$
D $\frac{2 \mathrm{~V}_{\mathrm{m}}}{\pi}$

The correct answer is:
$\frac{2 \mathrm{~V}_{\mathrm{m}}}{\pi}$

Solution:
What is the average value of ac voltage $\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$ over the
time interval $t=0$ to $t=\frac{\pi}{\omega}$
$\begin{aligned} & \mathrm{V}_{\mathrm{av}}=\frac{\int_0^{\frac{\pi}{\omega}} \mathrm{Vdt}}{\int_0^{\frac{\pi}{\omega}} \mathrm{dt}}=\frac{\int_0^{\frac{\pi}{\omega}} \mathrm{V}_0 \sin \omega t \mathrm{dt}}{[\mathrm{t}]_0^{\pi / \omega}} \\ & =\frac{\mathrm{V}_0\left\{-\frac{\cos \omega \mathrm{t}}{\omega}\right\}_0^\omega}{\omega} \\ & =-\frac{\mathrm{V}_0}{\pi}[\cos \pi-\cos 0]\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
The average weight of students in a class of 32 students is 40 kg. If the weight of the teacher be included, the average rises by 13 kg, then the weight of the teacher is

Options:
A 40.5 kg
B 50 kg
C 41 kg
D 51 kg

The correct answer is:
51 kg

Solution:
Let, the weight of the teacher be w kg, then

40+13=32×40+w32+1
⇒33×40+33×13=32×40+w⇒w=51
∴ Weight of the teacher =51 kg

Topic: Mathematics, Statistics, JEE Main
The average weight of students in a class of 35 students is 40 kg. If the weight of the teacher is included, then the average rises by 13 kg. The weight of the teacher is

Options:
A 40.5 kg
B 50 kg
C 41 kg
D 52 kg

The correct answer is:
52 kg

Solution:
Let, the weight of the teacher be w kg, then
40+13=35Ã40+w35+1
â36Ã40+36Ã13=35Ã40+wâw=52
â´ Weight of the teacher =52 kg.

Topic: Mathematics, Statistics, JEE Main
The average weight of the students in a class of 39 students is 40kg. If the weight of the teacher is to be included, then the average rises by 14kg. The weight of the teacher is

Options:
A 40.5kg
B 50kg
C 41kg
D 51kg

The correct answer is:
50kg

Solution:
Let the weight of the teacher be w kg, then

40+14=39×40+w39+1
⇒40×40+40×14=39×40+w⇒w=50
∴ Weight of the teacher =50kg

Topic: Mathematics, Statistics, JEE Main
The axis of the parabola $y^{2}+2 x=0$ is

Options:
A $x=0$
B $y=0$
C $x=2$
D $y=2$

The correct answer is:
$y=0$

Solution:
$y^{2}+2 x=0 \Rightarrow y^{2}=-2 x$, which is in the form $y^{2}=-4 a x .$ Therefore axis of parabola is $x$ -axis and its equation is $\mathrm{y}=0$.

Topic: Mathematics, Parabola, JEE Main
The Azimuthal quantum number for the valence electrons of Ga+ ion is (Atomic number of Ga=31)

Options:

The correct answer is:
0

Solution:
Ga+:1s2 2s2 2p6 3s2 3p6 3d10 4s2
The azimuthal quantum number for the valence electrons (4s-subshell) of Ga+ ion is zero(0).

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns cm-1. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is,

Options:
A 2 mA
B 20 μA
C 1 mA
D 40 μA

The correct answer is:
1 mA

Solution:
The coercivity of ferromagnet, H=100 A m-1, n×I=100
⇒I=100105=1 mA.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2018 (15 Apr)
The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns cm-1. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is

Options:
A 2 mA
B 20 μA
C 1 mA
D 40 μA

The correct answer is:
1 mA

Solution:
Coercivity of ferromagnet, H=100 A/m.
n×I=100
I=100105=1 mA

Topic: Physics, Magnetic Properties of Matter, JEE Main
The back emf in a DC motor is maximum when,

Options:
A the motor has picked up maximum speed
B the motor has just started moving
C the speed of motor is still on increase
D the motor has just been switched off

The correct answer is:
the motor has picked up maximum speed

Solution:
The value of back emf, $e \propto W$. So, back emf is maximum when the motor has picked up the maximum speed.

Topic: Physics, Electromagnetic Induction, JEE Main
The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

Options:
A 1 henry
B 4 henry
C $10^{-3}$ henry
D $4 \times 10^{-3}$ henry

The correct answer is:
$4 \times 10^{-3}$ henry

Solution:
$\mathrm{e}=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$
But $\mathrm{e}=4 \mathrm{~V}$ and $\frac{\mathrm{di}}{\mathrm{dt}}=\frac{0-1}{10^{-3}}=-1 / 10^{-3}$
$\therefore \frac{-1}{10^{-3}}(-\mathrm{L})=4 \Rightarrow \mathrm{L}=4 \times 10^{-3}$ henry

Topic: Physics, Electromagnetic Induction, JEE Main
The back e.m.f. induced in a coil, when current changes from 1 ampere to zero in one millisecond, is 4 volts, the self inductance

Options:
A $1 \mathrm{H}$
B $4 \mathrm{H}$
C $10^{-3} \mathrm{H}$
D $4 \times 10^{-3} \mathrm{H}$

The correct answer is:
$4 \times 10^{-3} \mathrm{H}$

Solution:
$e=-L \frac{d i}{d t}$ but $e=4 V$ and $\frac{d i}{d t}=\frac{0-1}{10^{-3}}=-1 / 10^{-3}$
$\therefore \frac{-1}{10^{-3}}(-L)=4 \Rightarrow L=4 \times 10^{-3}$ henry

Topic: Physics, Electromagnetic Induction, JEE Main
The bacterium Bacillus thuringiensis is widely used in contemporary biology as -

Options:
A Source of industrial enzyme
B Indicator of water pollution
C Insecticide
D Agent for production of dairy products

The correct answer is:
Insecticide

Solution:
No Solution Available

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2009 (Mains)
The balancing length for a cell is 560cm in a potentiometer experiment. When an external resistance of 10Ω is connected in parallel to the cell, the balancing length changes by 60cm . If the internal resistance of the cell is n10Ω , where n is an integer then value of n is ________

Options:

The correct answer is:
12

Solution:
Let the emf of cell is ε internal resistance is 'r' and potential gradient is x .
Only cell connected:
E=560x -----1
After connecting the resistor
E×1010+r=500x ------2
From 1 and 2
560x×1010+r=500x
56=50+5r
r=65=1.2Ω
so,n=12

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The band gap of germanium at room temperature is

Options:
A $2.3 \mathrm{eV}$
B $0.7 \mathrm{eV}$
C $1.1 \mathrm{eV}$
D $3.4 \mathrm{eV}$

The correct answer is:
$0.7 \mathrm{eV}$

Solution:
Concept:
The energy difference between the valence band and conduction band is called the forbidden energy gap.
$\begin{array}{|c|c|c|}
\hline \text{Semiconductor} & \text{Bandgap } (0K) & \text{Bandgap } (300K) \\
\hline \mathrm{Si} & 1.21 \mathrm{eV} & 1.1 \mathrm{eV} \\
\hline \mathrm{Ge} & 0.745 \mathrm{eV} & 0.67 \mathrm{eV} \\
\hline \mathrm{GaAs} & 1.52 \mathrm{eV} & 1.47 \mathrm{eV} \\
\hline
\end{array}$

Topic: Physics, Semiconductors, JEE Main
The base of an equilateral triangle is along the line given by $3 x+4 y=9$. If a vertex of the triangle is $(1,2)$, then the length of a side of the triangle is:

Options:
A
$\frac{2 \sqrt{3}}{15}$

B
$\frac{4 \sqrt{3}}{15}$

C
$\frac{4 \sqrt{3}}{5}$

D
$\frac{2 \sqrt{3}}{5}$

The correct answer is:
$\frac{4 \sqrt{3}}{15}$

Solution:
Shortest distance of a point $\left(x_1, y_1\right)$ from line $a x+b y=c$ is $d=\left|\frac{a x_1+b y_1-c}{\sqrt{a^2+b^2}}\right|$ Now shortest distance of $\mathrm{P}(1,2)$ from $3 x$ $+4 y=9$ is
$$
\mathrm{PC}=d=\left|\frac{3(1)+4(2)-9}{\sqrt{3^2+4^2}}\right|=\frac{2}{5}
$$
Given that $\triangle \mathrm{APB}$ is an equilateral triangle Let ' $a$ ' be its side
then $\mathrm{PB}=a, \mathrm{CB}=\frac{a}{2}$
Now, In $\Delta \mathrm{PCB},(\mathrm{PB})^2=(\mathrm{PC})^2+(\mathrm{CB})^2$
(By Pythagoras theorem)
$$
a^2=\left(\frac{2}{5}\right)^2+\frac{a^2}{4}
$$
$a^2-\frac{a^4}{4}=\frac{4}{25} \Rightarrow \frac{3 a^2}{4}=\frac{4}{25}$
$a^2=\frac{16}{75} \Rightarrow a=\sqrt{\frac{16}{75}}=\frac{4}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4 \sqrt{3}}{15}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (11 Apr Online)
The basic character of hydrides of the $V$-group elements decreases in the order

Options:
A $\mathrm{SbH}_3 \gt \mathrm{PH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{NH}_3$
B $\mathrm{NH}_3 \gt \mathrm{SbH}_3 \gt \mathrm{PH}_3 \gt \mathrm{AsH}_3$
C $\mathrm{NH}_3 \gt \mathrm{PH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{SbH}_3$
D $\mathrm{SbH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{PH}_3 \gt \mathrm{NH}_3$

The correct answer is:
$\mathrm{NH}_3 \gt \mathrm{PH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{SbH}_3$

Solution:
$\mathrm{NH}_3 \gt \mathrm{PH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{SbH}_3$
On moving down the group atomic size increases and availability of lone pair decreases Hence, basic character decreases.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The basic character of the transition metal monoxides follows the order:
(Atomic nos. $\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Fe}$ $=26$ )

Options:
A $\mathrm{VO}>\mathrm{CrO}>\mathrm{TiO}>\mathrm{FeO}$
B $\mathrm{CrO}>\mathrm{VO}>\mathrm{FeO}>\mathrm{TiO}$
C $\mathrm{TiO}>\mathrm{FeO}>\mathrm{VO}>\mathrm{CrO}$
D $\mathrm{TiO}>\mathrm{VO}>\mathrm{CrO}>\mathrm{FeO}$

The correct answer is:
$\mathrm{TiO}>\mathrm{VO}>\mathrm{CrO}>\mathrm{FeO}$

Solution:
As we move from left side to right side in the first transition series atomic number of elements increases and no. of electrons also increases. Size of elements on moving left to right decreases because effective nuclear charge of nucleus on the outermost shell electron increases. It means electronegativity of atoms increases from left to right, then the power of donation of electrons decreases which means the basicity of that element decreases.
Caution
Students get confused on the order of size of elements in the series, because order of size of elements is not regular in the periods and groups. In periods of increasing atomic number size decreases and on the groups with the increase of atomic number size also increases.

Topic: Chemistry, d and f Block Elements, JEE Main
The basic character of the transition metal monoxides follows the order
(Atomic numbers Ti = 22, V = 23, Cr = 24, Fe = 26)

Options:
A VO>CrO>TiO>FeO
B CrO>VO>FeO>TiO
C TiO>FeO>VO>CrO
D TiO>VO>CrO>FeO

The correct answer is:
TiO>VO>CrO>FeO

Solution:
Oxides of transition metals in low oxidation states +2 and +3 (MO, M3O4 and M2O3 ) are generally basic except Cr2O3 which is amphoteric in character.

Topic: Chemistry, d and f Block Elements, JEE Main
The basic ligand is

Options:
A $\mathrm{NH}_3$
B $\mathrm{CN}$
C $F$
D All

The correct answer is:
All

Solution:
All

Topic: Chemistry, Coordination Compounds, JEE Main
The basic structural unit of feldspar, zeolites mica, and asbestos is:

Options:
A SiO44-
B
C SiO32-
D SiO2

The correct answer is:
SiO44-

Solution:
Feldspars are a group of rock-forming tectosilicate minerals.
Mica minerals are phyllosilicate compounds.
Zeolites are a large group of minerals consisting of hydrated aluminosilicates of Sodium, Potassium, Calcium and Barium.
Thus, it is evident that Silicates are the basic component in Feldspars, Mica, Zeolites and Asbestos and have SiO44- basic unit.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (12 Apr Shift 1)
The basic structural unit of silicates is

Options:
A $\mathrm{SiO}^{-}$
B $\mathrm{SiO}_4^{4-}$
C $\mathrm{SiO}_3^{2-}$
D $\mathrm{SiO}_4^{2-}$

The correct answer is:
$\mathrm{SiO}_4^{4-}$

Solution:
The basic building unit of all silicates is the tetrahedral $\mathrm{SiO}_4^{4-}$. It is represented as

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The basic structure of the cell membrane

Options:
A is a lipid bilayer that serves as a barrier to water-soluble molecules.
B is a lipid bilayer that serves as a barrier to lipid-soluble molecules.
C is best described as lipids dispersed within a protein bilayer.
D ​is a protein bilayer that serves as a barrier to lipid-soluble molecules.

The correct answer is:
is a lipid bilayer that serves as a barrier to water-soluble molecules.

Solution:
Plasma membrane or the cell membrane separates the inside environment of cell from the outside environment. It protects the cell from the exterior environment. The cell membrane is mainly made of proteins and phospholipids. The phospholipids are arranged in a bilayer pattern with the hydrophobic tail on the inside and the hydrophilic head towards the outside. This protects the tail from aqueous environment and serves as a barrier to water-soluble molecule. The proteins are found on the surface of the membrane and also embedded within the membrane.

Topic: Biology, Cell - The Unit of Life, NEET
The beam of light has three wavelengths $4144 Ã, 4972 Ã$ and $6216 Ã$ with a total intensity of $3.6 \times 10^{-3} \mathrm{Wm}^{2}$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1 \mathrm{~cm}^{2}$ of a clean metallic surface of work function $2.3 \mathrm{eV}$. Assume that there is $\mathrm{n} 0$ loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \mathrm{~s}$

Options:
A $2 \times 10^{9}$
B $1.075 \times 10^{12}$
C $9 \times 10^{8}$
D $3.75 \times 10^{6}$

The correct answer is:
$1.075 \times 10^{12}$

Solution:
As we know, threshold wavelength

$$

\begin{array}{c}

\left(\lambda_{0}\right)=\frac{\mathrm{hc}}{\phi} \\

\Rightarrow \lambda_{0}=\frac{\left(6.63 \times 10^{-34}\right) \times 3 \times 10^{8}}{2.3 \times\left(1.6 \times 10^{-19}\right)}=5.404 \times 10^{-7} \mathrm{~m}

\end{array}

$$

$\Rightarrow \lambda_{0}=5404 Ã…$

Hence, wavelength $4144 Ã…$ and $4972 Ã…$ will emit electron from the metal surface. For each wavelength energy incident on the surface per unit time = intensity of each $\times$ area of the surface wavelength

$$

=\frac{3.6 \times 10^{-3}}{3} \times(\mathrm{lcm})^{2}=1.2 \times 10^{-7} \text {joule }

$$

Therefore, energy incident on the surface for each wavelength in $2 \mathrm{~s}$

$$

\mathrm{E}=\left(1.2 \times 10^{-7}\right) \times 2=2.4 \times 10^{-7} \mathrm{~J}

$$

Number of photons $\mathrm{n}_{1}$ due to wavelength $4144 Ã…$

$$

n_{1}=\frac{\left(2.4 \times 10^{-7}\right)\left(4144 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=0.5 \times 10^{12}

$$

Number of photons $\mathrm{n}_{2}$ due to the wavelength $4972 Ã…$

$$

\mathrm{n}_{2}=\frac{\left(2.4 \times 10^{-7}\right)\left(4972 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=0.572 \times 10^{12}

$$

Therefore total number of photoelectrons liberated in $2 \mathrm{~s}$

$$

\begin{aligned}

\mathrm{N} &=\mathrm{n}_{1}+\mathrm{n}_{2} \\

&=0.5 \times 10^{12}+0.575 \times 10^{12} \\

&=1.075 \times 10^{12}

\end{aligned}

$$

Topic: Physics, Dual Nature of Matter, JEE Main
The below plot shows the variation of binding energy per nucleon Eb, against the nuclear mass M. The different points shown A, B, C, D and E correspond to different nuclei. In which of the following reactions, the energy released Q will be positive?

Options:
A A + B â C + Q
B C â A + B + Q
C D + E â F + Q
D F â D + E + Q ,

The correct answer is:
F â D + E + Q ,

Solution:
If the binding energy of product nuclei is greater than that of reactants nuclei then energy is released.

Topic: Physics, Nuclear Physics, JEE Main
The below representation is for the diagnosis of __________ fever?

Options:
A

Typhus

B

Malarial

C

Enteric

D Dengue

The correct answer is:

Enteric

Solution:
Salmonella typhi is a pathogenic bacterium that causes typhoid or enteric fever in human beings. Poor sanitation, contaminated water and infected milk are some of the main factors responsible for typhoid.

Diagnosis is made by any blood, bone marrow, or stool cultures and with the Widal test.

Sustained high fever ( 39Â°âtoâ40Â°C ), weakness, stomach pain, constipation, headache and loss of appetite are some of the common symptoms of this disease.

Patients should be kept on a liquid diet of orange, barley juice and milk. Orange juice, especially, hastens recovery as it increases energy, promotes body immunity and increases urinary output.

Topic: Biology, Human Health and Diseases, NEET
The bending of beam of light around corners of obstacles is called

Options:
A Reflection
B Diffraction
C Refraction
D Interference

The correct answer is:
Diffraction

Solution:
Diffraction

Topic: Physics, Wave Optics, JEE Main
The best coagulant for the precipitation of \( \mathrm{Fe}(\mathrm{OH})_{3} \) is

Options:
A \( \mathrm{Na}_{3} \mathrm{PO}_{4} \)
B \( \mathrm{NaNO}_{3} \)
C \( \mathrm{Na}_{2} \mathrm{SO}_{4} \)
D \( \mathrm{NaCl} \)

The correct answer is:
\( \mathrm{Na}_{3} \mathrm{PO}_{4} \)

Solution:
If the ion layer is gathered close to the colloidal particles, the overall charge is effectively neutralized and two colloidal particles can approach close enough to aggregate and precipitate out.
Different electrolytes have different coagulation values. Smaller the coagulation value of the electrolyte, larger is its coagulating power. According to Hardy and Schulze, coagulation of colloids by electrolytes is governed by two factors, namely

Ions carrying a charge opposite to that of the colloidal particles are effective in bringing about coagulation.

Coagulation power of an electrolyte is directly proportional to the valency of its ions.

For positively charged colloidal particles, PO43- is much more effective than Cl- to bring about coagulation.
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PO 4 3 - has maximum coagulating power for Fe(OH)3, a positively charged sol.

Topic: Chemistry, Surface Chemistry, JEE Main
The best electrolyte for coagulating As2S3 sol is

Options:
A NaCl
B CuSO4
C AlNO33
D ThSO42

The correct answer is:
ThSO42

Solution:
As2S3 sol has negatively charged sol particles and therefore, positive ions are effective in causing the coagulating of As2S3 sol.
Coagulating efficiencies of electrolytes increase with the increase in the valence of effective ion. Therefore, the best electrolyte for coagulating As2S3 sol is ThSO42 .
Na+<Cu2+<Al3+<Th4+

Topic: Chemistry, Surface Chemistry, JEE Main