datavorous/entrance-exam-dataset huggingface dataset:
In a graph of population, with time on X-axis and population on Y-axis is plotted. A parallel line to X-axis shows:

Options:
A Natality equal to mortality
B Natality decreases, mortality increases
C Natality constant, mortality increases
D Natality increases, mortality decreases

The correct answer is:
Natality equal to mortality

Solution:
Natality rate, also called birth rate for a population. It is a measure of number of births occurring in a population in a particular time period. It is denoted by letter B.
Mortality rate is the measure of number deaths in a particular population for a particular time period. It is denoted by letter D.
As per the graph, the population is maintained where in the population count is remains stable. Such scenario occurs when the birth rate is same as that of the death rate. Thus, natality is equal to mortality.

Topic: Biology, Organisms and Populations, NEET
In a gravitational field, at a point where the gravitational potential is zero

Options:
A The gravitational field is necessarily zero
B The gravitational field is not necessarily zero
C Nothing can be said definitely about the gravitational field
D None of these

The correct answer is:
The gravitational field is necessarily zero

Solution:
$I=\frac{-d V}{d x}$
If $V=0$ then gravitational field is necessarily zero.

Topic: Physics, Gravitation, JEE Main
In a group of representative metals like alkali metals, what happens to reactivity as you move down the group?

Options:
A Reactivity decreases
B Reactivity remains constant
C Reactivity increases
D Reactivity becomes unpredictable

The correct answer is:
Reactivity increases

Solution:
Within a group of representative metals, such as alkali metals, reactivity increases as you move down the group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
In a half wave rectifier, the AC input source of frequency $50 \mathrm{~Hz}$ is used. The fundamental frequency of the output is

Options:
A $50 \mathrm{~Hz}$
B $150 \mathrm{~Hz}$
C $200 \mathrm{~Hz}$
D $75 \mathrm{~Hz}$

The correct answer is:
$50 \mathrm{~Hz}$

Solution:
In half wave rectifier frequency remain unchanged.
$\therefore$ Fundamental frequency of the output is $50 \mathrm{~Hz}$.

Topic: Physics, Semiconductors, TS EAMCET, TS EAMCET 2015
In a harmonic progression t1,t2,t3,â¦â¦â¦.., it is given that t5=20 and t6=50. If Sn denotes the sum of first n terms of this, then the value of n for which Sn is maximum is

Options:
A 6
B 7
C 9
D 10

The correct answer is:
6

Solution:
Let the first term be a and the common difference be d, then

20=1a+4d and 50=1a+5d
On solving, we get,

d=-3100 and a=17100
Hence, tr=10020-3r>0⇒r<203⇒rmax=6
So, sum is maximum for n=6, as after this terms are negative

Topic: Mathematics, Sequences and Series, JEE Main
In a head on elastic collision of a very heavy body moving at \( v \) with a light body at rest, velocity of heavy body after collision is

Options:
A v
B \( 2 v \)
C Zero
D \( \frac{v}{2} \)

The correct answer is:
v

Solution:
In a perfectly elastic collision the relative velocity remains unchanged in magnitude but reserved in direction. Therefore , velocity of heavy body after collision is v.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In a homogeneous catalysis

Options:
A The catalyst and the reactants should be gases
B The catalyst and the reactants should form a single phase
C Catalyst and the reactants are all solids
D The catalyst and the reactions are all liquids

The correct answer is:
The catalyst and the reactants should form a single phase

Solution:
Correct option is $\mathrm{2}$. The catalyst and the reactants should from a single phase It is known as homogeneous catalysis if the state of catalyst and the reactants are same.

Topic: Chemistry, Surface Chemistry, JEE Main
In a horizontal tube the water pressure changes by $1500 \mathrm{Nm}^{-2}$ between $A$ and $B$ as shown in figure below. The cross-sectional areas at $A$ and $B$ of the tube are $40 \mathrm{~cm}^2$ and $20 \mathrm{~cm}^2$, respectively. Find the rate of flow of water through the tube.

Options:
A $1000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$
B $2000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$
C $4000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$
D $6000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$

The correct answer is:
$4000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$

Solution:
Given that, pressure difference
By using Bernoulli's equation,
$$
p_A+\frac{1}{2} \rho v_A^2=p_B+\frac{1}{2} \rho v_B^2
$$
Density of water, $\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Cross-section areas at points $A$ and $B$,
$$
\begin{aligned}
& a_A=40 \mathrm{~cm}^2=40 \times 10^{-4} \mathrm{~m}^2 \\
& a_B=20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
By equation of continuity, rate of flow of water through the tube
$$
\frac{\Delta V}{\Delta t}=a_A v_A=a_B v_B \Rightarrow \frac{v_A}{v_B}=\frac{a_B}{a_A}=\frac{1}{2}
$$
Substituting the values from Eqs. (i) and (iii), in Eq. (ii), we get
$$
\begin{aligned}
1500 & =\frac{1}{2} \times 10^3\left[\left(2 v_A\right)^2-v_A^2\right] \\
1500 & =500\left(4 v_A^2-v_A^2\right) \\
v_A^2 & =1 \Rightarrow v_A=1 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$\therefore$ Rate of flow of water,
$$
\begin{aligned}
\frac{\Delta V}{\Delta t} & =a_A v_A=40 \times 10^{-4} \times 1 \mathrm{~m}^3 / \mathrm{s} \\
& =4000 \mathrm{~cm}^3 / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In a H.P., $p^{\text {th }}$ term is $q$ and the $g^{\text {th }}$ term is $p$. Then $p q^{\text {th }}$ term is

Options:
A $0$
B $1$
C $p q$
D $p q(p+q)$

The correct answer is:
$1$

Solution:
Let $a$ be the first term and $d$ be the common difference of the corresponding A.P.
$p^{\text {th }}$ term of A.P.$\left(T_p\right)=a+(p-1) d=\frac{1}{q}$\ldots(i)
$q^{\text {th }}$ term of A.P.$\left(T_q\right)=a+(q-1) d=\frac{1}{p}$\ldots(ii)
From (i) - (ii),$(p-q) d=\frac{1}{q}-\frac{1}{p}$ $=\frac{p-q}{p q} \Rightarrow d=\frac{1}{p q}$
From (i) $a+(p-1) \frac{1}{p q}=\frac{1}{q} \Rightarrow a=\frac{1}{p q}$
$\therefore \quad T_{p q}=a+(p q-1) d=$ $\frac{1}{p q}+(p q-1) \frac{1}{p q}=1$
Therefore $p q^{\text {th }}$ term is 1.

Topic: Mathematics, Sequences and Series, JEE Main
In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is

Options:
A
1.51 eV

B
3.4 eV

C
6.04 eV

D
13.6 eV

The correct answer is:
1.51 eV

Solution:
$2^{\text {nd }}$ excited state will be the 3 rd energy level
$$
\mathrm{E}_{\mathrm{n}}=\frac{13.6}{\mathrm{n}^2} \mathrm{eV} \text { or } \mathrm{E}=\frac{13.6}{9} \mathrm{eV}=1.51 \mathrm{eV}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2002
In a hydrogen atom, the electron is making $6.6 \times 10^{15} \mathrm{revs}^{-1}$ around the nucleus in an orbit of radius $0.528 \mathrm{~A}$. The magnetic moment $\left(\mathrm{A}-\mathrm{m}^{2}\right)$ will be

Options:
A $1 \times 10^{-15}$
B $1 \times 10^{-10}$
C $1 \times 10^{-23}$
D $1 \times 10^{-27}$

The correct answer is:
$1 \times 10^{-23}$

Solution:
Current, $I=6.6 \times 10^{15} \times 1.6 \times 10^{-19}$
$=10.5 \times 10^{-4} \mathrm{~A} .$
Area $A=\pi R^{2}=3.142 \times(0.528)^{2} \times 10^{-20} \mathrm{~m}^{2}$
So, Magnetic moment $M=I A=10.5 \times 10^{-4} \times 3.142$
$\begin{aligned} \times(0.528)^{2} \times 10^{-20} & \\=& 10 \times 10^{-24}=10^{-23} \text { units } \end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
In a hydrogen atom the electron makes a transition from n + 1th level to the nth level. If n>>1, the frequency of radiation emitted is proportional to :

Options:
A 1n
B 1n3
C 1n2
D 1n4

The correct answer is:
1n3

Solution:
En=-Rhcn2
En+1=-Rhc(n+1)2
ΔE=En+1-En
hv=Rhc1n2-1(n+1)2
v=R.cn+12-n2n2n+12
v=R.c1+2nn2n+12

n >> 1 ⇒v=R.c2nn2×n2=2RCn3

v∞1n3

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2020 (02 Sep Shift 2)
In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number n - 1 . If n >> 1 , the frequency of radiation emitted is proportional to :

Options:
A 1 n 3 / 2
B 1 n 3
C 1 n
D 1 n 2

The correct answer is:
1 n 3

Solution:
Δ E = h ν
ν = Δ E h = k 1 n - 1 2 - 1 n 2 = k 2 n n 2 n - 1 2
≈ 2 k n 3 ∝ 1 n 3

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2013 (07 Apr)
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $L$. If an electron jumps from $\mathrm{N}$ -shell to the $\mathrm{L}$ -shell, the wavelength of emitted radiation will be:

Options:
A $\frac{27}{20} \lambda$
B $\frac{16}{25} \lambda$
C $\frac{25}{16} \lambda$
D $\frac{20}{27} \lambda$

The correct answer is:
$\frac{20}{27} \lambda$

Solution:
When electron jumps from $\mathrm{M} \rightarrow \mathrm{L}$ shell

$\frac{1}{\lambda}=\mathrm{K}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{K} \times 5}{36}$

When eletron jumps from $\mathrm{N} \rightarrow \mathrm{L}$ shell

$\frac{1}{\lambda^{\prime}}=K\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{K \times 3}{16}$

solving equation (i) and (ii) we get

$\lambda^{\prime}=\frac{20}{27} \lambda$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (11 Jan Shift 2)
In a hydrogen spectrum λ be the wavelength of first transition line of Lyman series. The wavelength difference will be "aλ" between the wavelength of 3rd transition line of Paschen series and that of 2nd transition line of Balmer Series where a= _____.

Options:

The correct answer is:
5

Solution:
Using Rydberg formula,

1λ=R1nf2-1ni2, where, R is Rydberg constant and n is quantum number.

For first line of Lyman series

1λ=R1-14=R34

⇒λ=43R ...1

For 3rd line of Paschen series

1λ3=R132-162=R9×34

For 2nd line of Balmer series

1λ2=R122-142=R4×34

Thus, aλ=λ3-λ2=12R-163R=203R

Putting equation 1

a43R=203R⇒a=5

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
In a hydrogen-like species, the net force acting on a revolving electron in an orbit as given by the Bohr model is proportional to [nâprincipal quantum number, Zâ atomic number]

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

The correct answer is:
Z 3 n 4

Solution:
The net force on a revolving electron is

F= m v 2 r

F∝vn2rn
The velocity of the electron in the nth shell

vn∝Zn
and the radius of the nth shell

rn∝n2Z

⇒F∝Z3n4

Topic: Physics, Atomic Physics, JEE Main
In a hypothetical fission reaction
${ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R$
The identity of emitted particles ( $\mathrm{R})$ is :

Options:
A Electron
B Neutron
C $\gamma$-radiations
D Proton

The correct answer is:
Neutron

Solution:
$\begin{aligned}
& \mathrm{Z} \text { in LHS }=92 \\
& \mathrm{Z} \text { in } \mathrm{RHS}=56+36=92 \\
& \mathrm{~A} \text { in } \mathrm{LHS}=236 \\
& \mathrm{~A} \text { in } \mathrm{RHS}=141+92=233
\end{aligned}$
So 3 neutrons are released.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (08 Apr Shift 2)
In a hypothetical hydrogen like atom, if transition of electron from n=4 to n=3 produces visible light, then the possible transition to obtain infrared radiation is

Options:
A n=5 to n=3
B n=4 to n=2
C n=3 to n=1
D n=5 to n=4

The correct answer is:
n=5 to n=4

Solution:
ΔE should be less than E4–E3
⇒n1=5, n2=4

Topic: Physics, Atomic Physics, JEE Main
In a hypothetical solid, C atoms are found to form cubical close-packed lattice. A atoms occupy all tetrahedral voids and B atoms occupy all octahedral voids.

A and B atoms are of appropriate size, so that there is no distortion in the CCP lattice of C atoms. Now, if a plane as shown in the following figure is cut, then the cross section of this plane will look like

Options:
A
B
C
D

The correct answer is:

Solution:
From figure, it is clear that 4 corners and 2 face centers lie on the shaded plane. Therefore, there will be six C atoms, and atoms (marked A) in TVs do not touch other.
Fig (i) is not possible; four atom marked C.
Fig. (ii) is not possible, atoms A in TVs are not shown in figure.
Fig. (iii) is possible, since atoms A in TVs are not touching each other. There are four atoms A on two body diagonals contained in shades plane.
Fig (iv) is not possible, since atoms A in TVs are touching each other.

Topic: Chemistry, Solid State, JEE Main
In a $L-C-R$ circuit, the capacitance is made one-fourth, then what should be change in inductance, so that the circuit remains in resonance?

Options:
A 8 times
B $1 / 4$ times
C 2 times
D 4 times

The correct answer is:
4 times

Solution:
Resonance frequency $f=\frac{1}{2 \pi \sqrt{L C}}$
For the circuit to remain in resonance, $\sqrt{L C}=$ constant
$\begin{aligned} or \quad L_1 C_1 & =L_2 C_2 \\ or \quad L C & =L_2 \frac{C}{4} \\ \therefore L_2 & =4 L\end{aligned}$
$\therefore$ Inductance should be change to four times.

Topic: Physics, Alternating Current, JIPMER, JIPMER 2006
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:

Options:
A 8 A
B 10 A
C 12 A
D 14 A

The correct answer is:
12 A

Solution:
The power generated by each electrical device is calculated below

Item	Number	Power Consumed

40 W bulb	15	40×15=600 Watt

100 W bulb	5	100×5=500 Watt

80 W fan	5	80×5=400 Watt

1000 W heater	1	1000 Watt

The total power consumed=2500 Watt

So the minimum current capacity

i=PV=2500220=12511=11.36≅12 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2014 (06 Apr)
In a LCR circuit capacitance is changed from $C$ to $2 C$. For the resonant frequency to remain unchanged, the inductance should be changed from $L$ to

Options:
A
$4 \mathrm{~L}$

B
$2 \mathrm{~L}$

C
L/2

D
L/4

The correct answer is:
L/2

Solution:
$$
\omega_{\text {res }}=\frac{1}{\sqrt{L C}}
$$
if $\omega_{\text {res }}$ is to remain same, the product LC should also not change.
$$
\begin{aligned}
& \Rightarrow \mathrm{LC}=\mathrm{L}^{\prime} \mathrm{C}^{\prime} \\
& \Rightarrow \mathrm{LC}=\mathrm{L}^{\prime} 2 \mathrm{C} \\
& \Rightarrow \mathrm{L}^{\prime}=\mathrm{L}^2 2
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2004
In a lime kiln, to get higher yield of $\mathrm{CO}_{2}$, the measure that can be taken is

Options:
A to remove $\mathrm{CaO}$
B to add more $\mathrm{CaCO}_{3}$
C to maintain high temperature
D to pump out $\mathrm{CO}_{2}$

The correct answer is:
to pump out $\mathrm{CO}_{2}$

Solution:
In a lime kiln, to get higher yield of $\mathrm{CO}_{2}$, the measure that can be taken is to pump out $\mathrm{CO}_{2}$. This will reduce the pressure of $\mathrm{CO}_{2}$ and reaction will shift in forward direction.
$$
\mathrm{CaCO}_{3} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}
$$

Topic: Chemistry, Chemical Equilibrium, COMEDK, COMEDK 2013
In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be: (Radius of the Earth =6.4×106 m)

Options:
A 20 m
B 51 m
C 40 m
D 32 m

The correct answer is:
32 m

Solution:
Range, d=2 RhT+2 Rhr
⇒50=2×6400×hT+2×6400×70×10-3
⇒2×6400×hT=50-814
hT=50-81422×6400=0.031458 km
hT≈32 m

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (08 Apr Shift 2)
In a line of sight radio communication, a distance of about \( 50 \mathrm{~km} \) is kept between the transmitting and receiving antennas. If the height of the receiving antenna is \( 70 \mathrm{~m} \), then the minimum height of the transmitting antenna should be: (Radius of the Earth \( =6.4 \times 10^{6} \mathrm{~m} \) )

Options:
A \( 20 \mathrm{~m} \)
B \( 51 \mathrm{~m} \)
C \( 40 \mathrm{~m} \)
D \( 32 \mathrm{~m} \)

The correct answer is:
\( 32 \mathrm{~m} \)

Solution:
Range, d=2 RhT+2 Rhr
⇒50=2×6400×hT+2×6400×70×10-3
⇒2×6400×hT=50-814
hT=50-81422×6400=0.031458 km
hT≈32 m

Topic: Physics, Communication System, JEE Main
In a linear Simple Harmonic Motion (SHM)

(A) Restoring force is directly proportional to the displacement.

(B) The acceleration and displacement are opposite in direction.

(C) The velocity is maximum at mean position.

(D) The acceleration is minimum at extreme points.

Choose the correct answer from the options given below:

Options:
A C and D only
B A, C and D only
C A, B and C only
D A, B and D only

The correct answer is:
A, B and C only

Solution:
The equations of SHM is given by

F=–kx

⇒a=–ω2x

Velocity is given by, v=ωA2-x2.

Clearly, velocity is maximum at mean position and acceleration is maximum at extreme positions.

Hence, this is the correct option.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (15 Apr Shift 1)
In a linear tetrapeptide (Constituted with different amino acids), (number of amino acids) - (number of peptide bonds) is

Options:

The correct answer is:
1

Solution:
A tetrapeptide is a peptide, classified as an oligopeptide, since it only consists of four amino acids joined by peptide bonds.

A peptide bond, also referred to as an amide bond, is formed between the α-nitrogen atom of one amino acid and the carbonyl carbon of a second.

Example of tetrapeptide is: Tuftsin (L-threonyl)-(L-lysyl)-(L-prolyl)-(L-arginine) is a peptide related primarily to the immune system function

As we can see in this example

No. of Amino Acids =4

No. of Peptide bonds =3

Hence, (number of amino acids) - (number of peptide bonds) =1

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (29 Jul Shift 1)
In a liquid a square wire frame of size \( L \) is dipped. On taking out a membrane is formed. \( T \) is the surface tension of liquid, then force acting on a frame will be

Options:
A \( 2 T L \)
B \( 4 T L \)
C \( 8 T L \)
D \( 16 T L \)

The correct answer is:
\( 8 T L \)

Solution:
Surface tension makes fluid films minimize their surface areas. Since the square frame is a closed polygon, the net vector force acting on it due to the fluid film will be zero, but this force will pull on the frame and try to collapse it. This force has thus got a nature to try to deform the frame. So we need to find what this total deforming force is.

Since a membrane has two sides, there will be two pulling forces on each arm of the frame, one due to each of the sides. Now, the pulling force due to surface tension of one side of the film is given by TL.

Since the frame has 4 arms and each arm is pulled by two sides of the membrane, the total deforming force is 8TL.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In a lottery, 16 tickets are sold and 4 prizes are awarded. If a person buys 4 tickets,what is the probability of his winning a prize?

Options:
A $\frac{4}{16^{4}}$
B $\frac{175}{256}$
C $\frac{1}{4}$
D $\frac{81}{256}$

The correct answer is:
$\frac{1}{4}$

Solution:
$16$ tickets are sold and 4 prizes are awarded. A person
buys 4 tickets, then required probability $=\frac{4}{16}=\frac{1}{4}$

Topic: Mathematics, Probability, JEE Main
In a magnetic field of 0.05 T, area of a coil changes from 101 cm 2 to 100 cm 2 without changing the resistance which is 2 Î©. The amount of charge (in Î¼C) that flows during this period is

Options:

The correct answer is:
2.5

Solution:
B=0.05 T
A1=101  cm2=101×10−4  m2
A2=100  cm2=100×10−4  m2
R=2  Ω
Amount of charge flown is given by,
q=B(ΔA)R
=0.05×101×10-4-100×10-42
=0.05×1×10-42
=2.5×10-6 C

Topic: Physics, Magnetic Effects of Current, JEE Main
In a magnetic field of 0.05 T, area of a coil changes from 101 cm 2 to 100 cm 2 without changing the resistance which is 2 Î©. The amount of charge that flows during this period is

Options:
A 2 Â· 5 Ã 10 â6 C
B 2× 10 −6 C
C 10 −6 C
D 8× 10 −6 C

The correct answer is:
2 Â· 5 Ã 10 â6 C

Solution:
B=0.05 T
A1=101  cm2=101×10−4  m2
A2=100  cm2=100×10−4  m2
R=2  Ω
Amount of charge flown is given by,
q=B(Δ A)R
=0.05×101×10-4-100×10-42
=0.05×1×10-42
=2.5×10-6 C

Topic: Physics, Electromagnetic Induction, JEE Main
In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A) of the building should be

Options:
A 20 A
B 15 A
C 10 A
D 25 A

The correct answer is:
20 A

Solution:
Total power is

(15Ã45)+(15Ã100)+(15Ã10)+(2Ã1000)=4325 W
So current is =4325220=19.66 A

Ans is 20 A

Topic: Physics, Current Electricity, NEET
In a mechanical refrigerator, the low temperature coils are at a temperature of $-23^{\circ} \mathrm{C}$ and the compressed gas in the condenser has a temperature of $27^{\circ} \mathrm{C}$. The theoretical coefficient of performance is

Options:
A $5$
B $8$
C $6$
D $6.5$

The correct answer is:
$5$

Solution:
Coefficient of performance $K=\frac{T_2}{T_1-T_2}$
$=\frac{(273-23)}{(273+27)-(273-23)}=\frac{250}{300-250}=\frac{250}{20}=5$

Topic: Physics, Thermodynamics, JEE Main
In a medium the speed of light wave decreases to 0.2 times to its speed in free space. The ratio of relative permittivity to the refractive index of the medium is x:1. The value of x is ______.

(Given speed of light in free space =3×108 m s-1 and for the given medium μr=1)

Options:

The correct answer is:
5

Solution:
Speed of light in medium is related to speed of light in vacuum c and refractive index μ by v=cμ.

Then, we have μ=cv=c0.2c

⇒μ=5

Refractive index in terms of relative permeability μrand relative permittivityεr of the medium is stated as μ=ϵr μr

⇒ϵr=μ2μr

∴ ϵrμ=μμr=5

Hence, the value of x=5.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
In a meeting $60 \%$ of the members favour and $40 \%$ oppose a certain proposal. A member is selected at random and we take $\mathrm{X}=0$ if the opposed and $\mathrm{X}=1$ if he is in favour, then Var $\mathrm{X}=$

Options:
A 0.36
B 0.24
C 0.6
D 0.06

The correct answer is:
0.24

Solution:
$$
\text { From the given data, we write }
$$
$$
\text { Variance }=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=0.6-(0.6)^2=0.6-0.36=0.24
$$

Topic: Mathematics, Statistics, JEE Main
In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor

Options:
A Drift from higher potential to lower potential
B Move with the uniform velocity throughout from lower potential to higher potential
C Move in the straight line paths in the same direction
D Move in the curved paths from lower potential to higher potential

The correct answer is:
Move in the curved paths from lower potential to higher potential

Solution:
The electrons inside the conductor move with an average velocity due to the applied field in a direction opposite to the field, this velocity is known as drift velocity. Electric field are applied from high to low potential region of the conductor. Also, electrons are negatively charged.

Electron drifts from lower potential to higher potential on curved path.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (10 Apr Shift 2)
In a meter bridge as shown in the figure, it is given that resistance Y=12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by jockey J). After interchanging the resistances X and Y a new balance point is found at a distance l2 from end A. What are the values of X (in Ω) and l2?

Options:
A 19.15 Ω and 39.5 cm
B 8.16 Ω and 60.5 cm
C 8.16 Ω and 39.5 cm
D 19.15 Ω and 60.5 cm

The correct answer is:
8.16 Ω and 60.5 cm

Solution:
For a balanced meter bridge, (y)×​39.5=x×100-39.5,
x=12.5×39.560.5=8.16 Ω. When x and y are interchanged l 1 and ( 100− l 1 ) will also interchange, l2=60.5 cm.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (15 Apr)