datavorous/entrance-exam-dataset huggingface dataset:
A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A=100 g mol-1 ; B=200 g mol-1 ; C=10,000 g mol-1]

Options:
A B>C>A
B C>B>A
C A>B>C
D A>C>B

The correct answer is:
A>B>C

Solution:
Number of moles = wtgmMW=10MW

Relative lowering in vapour pressure depends on no. of mole of solute greater the no. of moleof solute greater in RLVP and smaller will be vapour pressure.

So order of vapour pressure is A>B>C.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A set $\mathrm{S}$ contains 7 elements. A non-empty subset $A$ of $S$ and an element $x$ of $S$ are chosen at random. Then the probability that $\mathrm{x} \in \mathrm{A}$ is:

Options:
A
$\frac{1}{2}$

B
$\frac{64}{127}$

C
$\frac{63}{128}$

D
$\frac{31}{128}$

The correct answer is:
$\frac{64}{127}$

Solution:
Let $\mathrm{S}=\left\{x_1, x_2, x_3, x_4, x_5, x_6, x_7\right\}$
Let the chosen element be $\mathrm{x}_{\mathrm{i}}$.
Total number of subsets of $\mathrm{S}=2^7=128$
No. of non-empty subsets of S $=128-1$ $=127$
We need to find number of those subsets that contains $x_i$.
\begin{array}{|l|l|l|l|l|l|l|}
\hline 2 & 2 & 2 & 2 & 1 & 2 & 2 \\
\hline
\end{array}
$x_1 x_2$-—- $x_i=-x_7$
For those subsets containing $x_i$, each element has 2 choices.
i.e., (included or not included) in subset, However as the subset must contain $x_i$, $x_i$ has only one choice. (included one)
So, total no. of subsets containing $x_i=2 \times 2 \times 2 \times 2 \times 1 \times 2 \times 2=64$
Required prob
$$
=\frac{\text { No. of subsets containing } x_i}{\text { Total no. of non-empty subsets }}
$$
$$
=\frac{64}{127}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2014 (11 Apr Online)
A seven digit number is formed using digits 3,3,4,4,4,5,5. The probability, that number so formed is divisible by 2, is

Options:
A 47
B 37
C 17
D 67

The correct answer is:
37

Solution:
Digits =3,3,4,4,4,5,5

Total 7 digit numbers =7!2!2!3!

Number of 7 digit number divisible by 2

⇒last digit =4

Now 7 digit numbers which are divisible by 2

=6!2!2!2!
Required probability =6!2!2!2!7!3!2!2!=37

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Options:
A
decreases

B
remains unchanged

C
becomes infinite

D
increases

The correct answer is:
remains unchanged

Solution:
No Solution Available

Topic: Physics, Capacitance, JEE Main, JEE Main 2003
A shell bursts on contact with the ground and pieces from it fly in all directions with a velocity equal to 60 m s-1. If it is known that a man who is 180 m away is in danger for a time interval of x seconds, then what is the value of x? [Neglect the height of the man]

Options:

The correct answer is:
72

Solution:
Given, u=60 m s-1, R=180 m
∴ R=u2gsin⁡2θ
180=60210sin⁡2θ
⇒ sin⁡2θ=12
⇒ θ=15o
As there are always two directions of projections θ and 90o-θ for the given range, hence two directions of projection, i.e. 15o and 90o-15o, i.e. 15o and 75o. Let T1 and T2 be the times of flight in the two cases, where
T1=2usin⁡15og and T2=2usin⁡75og
The man is in danger for a time
T2-T1=2ugsin⁡75o-sin⁡15o
=2ug2cos⁡45osin⁡30o
using identity, sin⁡C-sin⁡D=2 cos⁡C+D2sin⁡C-D2
=2×6010×2×12×12
=72 s

Topic: Physics, Motion In Two Dimensions, JEE Main
A shell in air explodes into n equal fragments, k of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

Options:
A g
B n-kg
C n-kgk
D n-kng

The correct answer is:
n-kng

Solution:
acm=ΣmiaiΣmi
Till all the particle remains in air center of mass remains stationary. When k particles hit the ground then center of mass shifts with accelaration :
acm=n-kng

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A shell in flight explodes into n equal fragments, k of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

Options:
A G
B n-kg
C n-kgk
D n-kng

The correct answer is:
n-kng

Solution:
acm=ΣmiaiΣmi
Till all the particle remains in air center of mass remains stationary. When k particles hit the ground then center of mass shifts with accelaration :
acm=n-kng

Topic: Physics, Center of Mass Momentum and Collision, NEET
A shell initially at rest explodes into two pieces of equal mass, then the two pieces will

Options:
A Be at rest
B Move with different velocities in different directions
C Move with the same velocity in opposite directions
D Move with the same velocity in same direction

The correct answer is:
Move with the same velocity in opposite directions

Solution:
According to law of conservation of linear momentum both pieces should possess equal momentum after explosion. As their masses are equal therefore they will possess equal speed in opposite direction.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is:

Options:
A R/2g
B R/g
C 2R/g
D R/4g

The correct answer is:
2R/g

Solution:
We know for complementary angle range is same for same projection speed.
t1=2usin(θ)g ….. (i)
t2=2usin(90-θ)g ….. (ii)
R=2u2sinθcosθg .… (iii)
From equation (i) and (ii)
t1t2=4 U2sinθ cosθg2
(Using the value of R from equation iii)
t1t2=2Rg

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A shell of mass $5 M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M, 2 M$ and $2 M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes 2v and $v$ respectively. The third fragment will

Options:
A move with a velocily $v$ in a direction perpendicular To the other two
B move with a velocity $2 v$ in the direction of velocity of the first fragment
C be at rest
D move with velocity $v$ in the direction of velocity of the second fragment

The correct answer is:
be at rest

Solution:
Let the velocity of third fragment is $v'$. Then by the conservation of linear momentum
$$
\begin{aligned}
& 5(0)=M \times 2 \mathrm{v}-2 M \times \mathrm{v}+2 \mathrm{M} \times \mathrm{v}^{\prime} \\
\Rightarrow \quad & \mathrm{v}^{\prime}=0
\end{aligned}
$$
i.e., the third fragment will be at rest.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A shell of mass $200 \mathrm{gm}$ is ejected from a gun of mass $4 \mathrm{~kg}$ by an explosion that generates $1.05 \mathrm{~kJ}$ of energy. The initial velocity of the shell is

Options:
A $120 \mathrm{~ms}^{-1}$
B $100 \mathrm{~ms}^{-1}$
C $80 \mathrm{~ms}^{-1}$
D $40 \mathrm{~ms}^{-1}$

The correct answer is:
$100 \mathrm{~ms}^{-1}$

Solution:
Let the initial velocity of the shell be $\mathrm{v}$, then velocity of gun is $\left(\frac{0.2 v}{4}\right)$. (from
conservation of linar moment)
Now $1.05 \times 10^3=\frac{1}{2}(0.2) v^2+\frac{1}{2}(4)\left(\frac{0.2 v}{4}\right)^2 \Rightarrow v=100 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2008 (Mains)
A shell of mass $m$ moving with velocity $v$ suddenly breaks into two pieces. If one of those parts having mass $m / 6$ remains stationary, find the velocity of the other part.

Options:
A $\frac{6 v}{5}$
B $2 v$
C $\frac{3 v}{4}$
D $\frac{4 v}{3}$

The correct answer is:
$\frac{6 v}{5}$

Solution:
Given that, mass of shell $=m$
Mass of lst part, $m_1=\frac{m}{6}$
Then, mass of 2nd part, $m_2=m-\frac{m}{6}=\frac{5 m}{6}$
Velocity of lst part, $v_1=0$
Velocity of shell $=v$
Let velocity of 2nd part $=v_2$
By using law of conservation of linear momentum,
Initial linear momentum $=$ final linear momentum.
$m v=m_1 v_1+m_2 v_2$
Substituting the values, we get
$m v=\frac{m}{6} \times 0+\frac{5 m}{6} \times v_2$
$\Rightarrow \quad v_2=\frac{6}{5} v$
Hence, the velocity of 2 nd part is $\frac{6}{5} v$ in the same direction as that of shell.

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 2)
A shopkeeper has \( 10 \) copies each of nine different books, then the number of ways in which atleast one book can be selected is

Options:
A \( 9^{11}-1 \)
B \( 10^{10}-1 \)
C \( 11^{9}-1 \)
D \( 10^{9} \)

The correct answer is:
\( 11^{9}-1 \)

Solution:
B 1 ï¸¸ 10 B 2 ï¸¸ 10 B 3 ï¸¸ 10 â¦â¦.. B 9 ï¸¸ 10
1st kind of book can be selected in 10+1 ways.
Similarly each kind of book can be selected in 10+1 ways.
So, selection of atleast one book =10+110+1â¦â¦â¦..10+19 times-1
=119-1

Topic: Mathematics, Permutation Combination, JEE Main
A shopkeeper has 10 copies each of nine different books, then the number of ways in which atleast one book can be selected is

Options:
A 911-1
B 1010-1
C 119-1
D 109

The correct answer is:
119-1

Solution:
B 1 ï¸¸ 10 B 2 ï¸¸ 10 B 3 ï¸¸ 10 â¦â¦.. B 9 ï¸¸ 10
1st kind of book can be selected in 10+1 ways.
Similarly each kind of book can be selected in 10+1 ways.
So, selection of atleast one book =10+110+1â¦â¦â¦..10+19 times-1
=119-1

Topic: Mathematics, Permutation Combination, JEE Main
A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in Am2 is close to:
(Given μ04π=10-7 in SI units and BH= Horizontal component of earth's magnetic field =3.6×10-5 Tesla.)

Options:
A 4.9
B 14.6
C 19.4
D 9.7

The correct answer is:
9.7

Solution:
Horizontal component of earth's magnetic field is along geographical south its North.
M→= Magnetic dipole moment.
At neutral point on the equatorial line the magnetic field B′ cancels BH
∴B′=BH
⇒μ04πMx3=BH
⇒10-7×M301003=3.6×10-5
⇒10-7×103×M27=3.6×10-5
M=27×3.6×10-510-4
M=27×3.6×10-1
=9.72
M≈9.7

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2015 (11 Apr Online)
A short bar magnet of magnetic moment $0.4 \mathrm{~J} \mathrm{~T}^{-1}$ is place in a uniform magnetic field of $0.16 \mathrm{~T}$. The magnet is stable equilibrium when the potential energy is

Options:
A $-0.64 \mathrm{~J}$
B zero
C $-0.082 \mathrm{~J}$
D $0.064 \mathrm{~J}$

The correct answer is:
$0.064 \mathrm{~J}$

Solution:
We know that
$U=-M B \cos \theta$
For stable equilibrium $\theta=0^{\circ}$
$\begin{aligned}
U & =-M B \\
& =-(0.4)(0.16)=-0.064 \mathrm{~J}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2011 (Mains)
A short bar magnet placed in a horizontal plane has its axis aligned along north-south direction. Null points are found on the axis of the magnet at $20 \mathrm{~cm}$ from the centre of magnet. The earth's magnetic field at the place is $B$ and angle of dip is $0^{\circ}$. If the total magnetic field on the normal bisector of the magnet $20 \mathrm{~cm}$ from the centre of the magnet is $0.6 \mathrm{G}$, then the magnitude of $B$ is.

Options:
A $0.2 \mathrm{G}$
B $0.4 \mathrm{G}$
C $1.2 \mathrm{G}$
D $0.3 \mathrm{G}$

The correct answer is:
$0.4 \mathrm{G}$

Solution:
At null point (and along the axis), earth's magnetic field and bar's magnetic field are opposite in direction.
On equitorial line, bar's mågnetic field is opposite in idirection to its field on the axis. Hence, on equatorial line, the two fields add up As null points are on the axis of the bar magnet, therefore,
$B_1=\frac{\mu_0}{4 \pi} \frac{2 M}{d^3}=H$
on the equitorial line of magnet at same distance (d), field due to the magnet is
$B_2=\frac{\mu_0}{4 \pi} \frac{M}{d^3}=\frac{B_1}{2}=\frac{H}{2}$
Therefore, total magnetic field at this point on equitorial line is
$B=B_2+H=\frac{3 H}{2}=\frac{3}{2} \times H$
Since, given $B=0.69$
$0.6=\frac{3}{2} \times H \Rightarrow H=\frac{0.6 \times 2}{3}=0.4 \mathrm{G}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A short magnetic needle is pivoted in a uniform magnetic field of induction $1 \mathrm{~T}$. Now, simultaneously another magnetic field of induction $\sqrt{3} \mathrm{~T}$ is applied at right angles to the first field; the needle deflects through an angle $\theta$ whose value is

Options:
A $30^{\circ}$
B $45^{\circ}$
C $90^{\circ}$
D $60^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
In balance condition $B_2=B_1 \tan \theta$
or $\quad \tan \theta=\frac{\sqrt{3}}{1}=\sqrt{3}$
$\theta=60^{\circ}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A short peptide on complete hydrolysis produces 3 moles of glycine G, two moles of leucine L and two moles of valine V per mole of peptide. The number of peptide linkages in it are _____ .

Options:

The correct answer is:
6

Solution:
Peptide linkage is also known as peptide bond. It is an amide formed between −COOH and −NH2 group by elimination of a water molecule. It is represented as
-CO-NH-.

Number of peptide linkages =(Total amino acids -1)

Now, given that short peptide on hydrolysis give seven amino acids.
=7-1=6

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

Options:
A
blue

B
green

C
violet

D
red

The correct answer is:
red

Solution:
As we know that when light ray goes from one medium to other medium, the frequency of light remains unchanged. And, $\quad \mathrm{c}=v \lambda$
So, $\mathrm{c} \propto \lambda$ the light of red colour is of highest wavelength and therefore of highest speed. Thus, after travelling through the slab, the red colour emerge first,

Topic: Physics, Ray Optics
A short solenoid (length l and radius r with n turns per unit length) lies well inside and on the axis of a very long, coaxial solenoid (length L, radius R and N turns per unit length, with R>r). Current I follows in the short solenoid. Choose the correct statement.

Options:
A There is uniform magnetic field μ0nI in the long solenoid.
B Mutual inductance of the solenoids is πμ0r2nNl.
C Flux through outer solenoid due to current I in the inner solenoid is proportional to the ratio R/r.
D Mutual inductance of the solenoids is πμ0rRnNlL/rR1/2.

The correct answer is:
Mutual inductance of the solenoids is πμ0r2nNl.

Solution:
We use reciprocatory theorem to calculate mutüal inductance of the system of solenoids.

Let outer solenoid carries a current I1. Flux linked with small solenoid 2 is proportional to the current in large solenoid 1.

⇒ϕ21∝I1⇒nlμ0NI1πr2∝I1

Proportionality constant is the coefficient of mutual inductance M of system.

Note that M12=M21 in all cases.

So, n/μ0Nl1πr2=Ml1 or M=πr0r2nNI

Topic: Physics, Electromagnetic Induction, JEE Main
A short straight object of height 100 cm lies before the central axis of a spherical mirror whose focal length has absolute value f=40 cm. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information:

Options:
A Image is real, same side of convex mirror.
B Image is virtual, opposite side of concave mirror.
C Image is real, same side of concave mirror.
D Image is virtual, opposite side of convex mirror.

The correct answer is:
Image is virtual, opposite side of convex mirror.

Solution:
Since orientation is same image is virtual. Since image is smaller the mirror has to be convex

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A short straight object of length $l$ lies along the central axis of a spherical concave mirror, at a distance $X$ from the mirror. The focal length of the mirror is $F$. If the length of the image in the mirror is $l^{\prime}$, then ratio $\left(\frac{l^{\prime}}{l}\right)$ is (assume, $l< < $ and $l< < F$ )

Options:
A $\frac{F-X}{F}$
B $\left(\frac{F-X}{F}\right)^2$
C $\left(\frac{F}{F-X}\right)^2$
D $\frac{F}{X}$

The correct answer is:
$\left(\frac{F}{F-X}\right)^2$

Solution:
The given situation is shown in the following figure.
$\begin{aligned}
& A B \rightarrow \text { object, } A^{\prime} B^{\prime} \rightarrow \text { image } \\
& \qquad A B=l, A^{\prime} B^{\prime}=l^{\prime}, u=-X,
\end{aligned}$
Hence, by mirror formula,
$\begin{aligned}
\frac{1}{-F} & =\frac{1}{v}+\frac{1}{(-X)} \quad\left[\because \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\right] \\
\Rightarrow \quad \frac{1}{v} & =\frac{1}{x}-\frac{1}{F} \Rightarrow \frac{1}{v}=\frac{F-X}{F X}
\end{aligned}$
Longitudinal magnification for small object is given as
$\begin{aligned}
M & =\frac{l^{\prime}}{l}=-\frac{v^2}{u^2}=-\frac{\left(\frac{F X}{F-X}\right)^2}{(-X)^2} \\
& =-\frac{f^2 X^2}{(F-X)^2 X^2}=-\left(\frac{F}{F-X}\right)^2
\end{aligned}$
Magnitude of longitudinal magnification,
$\frac{l^{\prime}}{l}=\left(\frac{F}{F-X}\right)^2$

Topic: Physics, Ray Optics, JEE Main
A shunt of resistance $1 ~\Omega$ is connected across a galvanometer of $120 ~\Omega$ resistance. A current of $5.5$ ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly :

Options:
A
$5.5$ ampere

B
$0.5$ ampere

C
$0.004$ ampere

D
$0.045$ ampere

The correct answer is:
$0.045$ ampere

Solution:
The current that will given full scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when shunt is connected i.e. $\mathrm{I}_{\mathrm{g}}$
As $I_g=\frac{I S}{G+S}$
$=\frac{5.5 \times 1}{120+1}=0.045$ ampere

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (09 Apr Online)
A signal is to be transmitted through a wave of wavelength λ, using a linear antenna. The length l of the antenna and effective power radiated Peff will be given, respectively, as-
(K is a constant of proportionality)

Options:
A λ8, Peff=Klλ
B λ16, Peff=Klλ3
C λ5, Peff=Klλ12
D λ, Peff=Klλ2

The correct answer is:
λ, Peff=Klλ2

Solution:
Length of antenna = comparable to λ
Power P∝ ( l λ ) 2 .

Topic: Physics, Communication System, JEE Main, JEE Main 2017 (09 Apr Online)
A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5 dB per km and cable length is 20 km. The power received at the receiver is 10-x W. The value of x is _________. [Gain in dB=10 log10POPi]

Options:

The correct answer is:
8

Solution:
Sound level decreases by 5 dB every km so, sound level decreased in 20 km=100 dB

β2-β1=10 log10I2I1

-100=10 log10I2I1⇒I1I2=1010

I2=10-10I1⇒P2=10-10P1=10-8 W

x=8

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:

Options:
A 2000 kHz and 1995 kHz
B 2 MHz only
C 2005 kHz and 1995 kHz
D 2005 kHz, 2000 kHz and 1995 kHz

The correct answer is:
2005 kHz, 2000 kHz and 1995 kHz

Solution:
Frequency of signal wave =5 kHz=f. Carrier wave frequency =2MHz
=2000kHz=fc. The resultant signal maximum frequency=f+fc
= 5 + 2 0 0 0 kHz
= 2 0 0 5 kHz
Resultant signal minimum frequency
=fc-f
= 2 0 0 0 - 5 kHz
= 1995 kHz

Topic: Physics, Communication System, JEE Main, JEE Main 2015 (04 Apr)
A signal of 100THz frequency can be transmitted with maximum efficiency by

Options:
A Coaxial cable
B Optical fibre
C Twisted pair of copper wires
D Water

The correct answer is:
Optical fibre

Solution:
Fiber optics, or optical fiber, refers to the technology that transmits information as light pulses along a glass or plastic fiber. Fiber optics have high efficiency of data transmission.

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.

Options:
A Modulation index =5, side frequency bands are at 1400 kHz and 1000 kHz.
B Modulation index=5, side frequency bands are at 21.2 kHz and 18.8 kHz.
C Modulation index =0.2, side frequency bands are at 1220 kHz and 1180 kHz.
D Modulation index =5, side frequency bands are at 24.2 kHz and 18.8 kHz.

The correct answer is:
Modulation index =0.2, side frequency bands are at 1220 kHz and 1180 kHz.

Solution:
Modulation index defines the modulated signal's ratio of maximum to minimum voltage

=VmVc=525=0.2
Carrier Frequency 1.2×103 kHz
fc=1200 kHz.
Lower side-band frequency, f1=1200-20=1180 KHz.
Upper band frequency, f2 =1200+20=1220 KHz .

Topic: Physics, Communication System, JEE Main, JEE Main 2017 (08 Apr Online)
A signal wave of frequency 12 kHz is modulated with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively

Options:
A 2512 kHz and 2508 kHz
B 2522 kHz and 2488 kHz
C 2522 kHz and 2498 kHz
D 2512 kHz and 2488 kHz

The correct answer is:
2522 kHz and 2498 kHz

Solution:
Upper side band (USB) frequency =fc+fm
=2.51MHz+12kHz=2510+12 kHz=2522 kHz
Lower side band (LSB) frequency =fc-fm
=2.51MHz-12kHz=2510-12 kHz=2498 kHz

Topic: Physics, Communication System, JEE Main
A significant amount of progesterone is produced by the

Options:
A corpus luteum
B testis
C placenta
D Both A and C

The correct answer is:
Both A and C

Solution:
Progesterone is mainly the female sex hormone responsible for maintaining the endometrium of the uterus alongwithestrogen.It is also responsible for secondary sexual characters in females like development of mammary glands. It is secreted by corpus luteum. Corpus luteum forms after ovulation. This secretes progesterone so that after ovulation if the sperm enters the fertilised egg can easily get implanted in the uterus. Once the implantation is completed it is very important to maintain the endometrium till parturition. Thus progesterone is secreted by placenta too.

Topic: Biology, Human Reproduction, NEET
A silicon specimen is made into a p-type semiconductor by doping, on an average, one indium atom per 5×107 silicon atoms. If the number density of atoms in the silicon specimen is 5×1028 atoms m-3, then the number of acceptor atoms in silicon per cubic centimetre will be

Options:
A 2.5×1030 atom cm-3
B 2.5×1035 atom cm-3
C 1×1013 atom cm-3
D 1×1015 atom cm-3

The correct answer is:
1×1015 atom cm-3

Solution:
Number density of atoms in silicon specimen
= 5×1028 atoms-m-3=5×1022 atoms cm-3.
Since, 1 atom of indium is doped in 5×107 silicon atoms, so total number of indium atoms doped per cm3 of silicon will be
n=5×10225×107=1015 atom-cm-3.

Topic: Physics, Semiconductors, JEE Main
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012 s-1 . What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver,=108 g mol-1 and Avogadro number =6.02×1023)

Options:
A 5.5 N m-1
B 6.4 N m-1
C 7.1 N m-1
D 2.2 N m-1

The correct answer is:
7.1 N m-1

Solution:
Given, molecular weight of silver M=108 g and Avogadro's numberNA=6.02×1023,
frequency f=1012 s.

Mass of single silver atom will be given by,

m=MN=1086.02×1023=1.79×10-22 g=1.79×10-25 kg

The bonds between the atoms can be considered as spring and then using the spring analog, the time period of the oscillation of the atoms is given by,

T=1f=2πmk; where k is spring constant.

11012=2π1.79×10-22×10-3k

k=7.1 N m-1.

Topic: Physics, Oscillations, JEE Main, JEE Main 2018 (08 Apr)
A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. UV light of wavelength 200 nm is incident on the ball for some time during which total energy of 1Ã10â7 J falls on the surface. Assuming that on an average, one out of 103 photons incident on the ball is able to eject an electron, then the potential of the sphere will be

Options:
A 1 V
B 2 V
C 3 V
D zero

The correct answer is:
3 V

Solution:
The total number of photons striking the surface of the ball is

np=Eλhc=1 × 10-7×200×10-96.6×10-34×3×108=1×1011

The total number of electrons ejected is

ne=1011103=108

∴V=q4πε0r=108×1.6×10-19×9×1094.8×10-2=3 V

Topic: Physics, Dual Nature of Matter, JEE Main