datavorous/entrance-exam-dataset huggingface dataset:
There are 17 cricket players, out of which 5 players can bowl. In how many ways can a team of 11 players be selected so as to include 3 bowlers?

Options:
A $\mathrm{C}(17,11)$
B $\mathrm{C}(12,8)$
C $\mathrm{C}(17,5) \times(5,3)$
D $\mathrm{C}(5,3) \times \mathrm{C}(12,8)$

The correct answer is:
$\mathrm{C}(5,3) \times \mathrm{C}(12,8)$

Solution:
3 bowlers are selected among 5 bowlers in ${ }^{5} \mathrm{C}_{3}$ ways. Remaining 8 player's are selected from 12 player's in ${ }^{12} \mathrm{C}_{8}$ ways.
$\therefore$ total number of ways $={ }^{12} \mathrm{C}_{8} \times{ }^{5} \mathrm{C}_{3}$.

Topic: Mathematics, Permutation Combination, NDA, NDA 2018 (Phase 2)
There are 100 students in a class. In an examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failed in Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then, the number of students failing in all the three subjects

Options:
A is 12
B is 4
C is 2
D cannot be determined from the given information

The correct answer is:
is 2

Solution:
According to question,
$\begin{array}{c}
a+b+c+d+e+f+g+h=100 \\
h=1 \\
a+b+d+e=50
\end{array}$
Venn diagram shows failed students $b+c+d+f=45$
$\begin{array}{l}
d+e+f+g=40 \\
b+e+f=32
\end{array}$
On solving, the above equations, we get $d=2$

Topic: Mathematics, Sets and Relations, JEE Main
There are eight rooms on the first floor of a hotel, with four rooms on each side of the corridor, symmetrically situated (that is each room is exactly opposite to one other room). Four guests have to be accommodated in four of the eight rooms (that is, one in each) such that no two guests are in adjacent rooms or in opposite rooms. If N is the number of ways in which guests can be accommodated. Then the value of N6 is

Options:

The correct answer is:
8

Solution:
Clearly guests will stay either in '✓' or in '×'
Therefore, number of required ways =2×4!=48 .

Topic: Mathematics, Permutation Combination, JEE Main
There are fifty persons among whom 2 are brothers. The number of ways they can be arranged in a circle, if there is exactly one person between the two brothers, is

Options:
A 2×48!
B 12
C 360
D 7×8!

The correct answer is:
2×48!

Solution:
Consider three persons as one unit. This unit and remaining 47 persons (total 48) can be arranged in a circle in 47! ways.
In this unit, the two brothers can be interchanged in 2 ways.
The person between the two brothers can be any of the remaining 48 persons who can be selected in C 48 1 ways.
Hence, the required number of ways = C 48 1 ×47!×2=2×48!

Topic: Mathematics, Permutation Combination, JEE Main
There are following the three solvents \( \mathrm{X}, \mathrm{Y} \& \mathrm{Z} \). They have identical molar masses. Match their boiling point with their \( \mathrm{k}_{\mathrm{b}} \)
values:
\( \begin{array}{ccc}\text { Solvent } & \text { B. Pt. } & \mathrm{K}_{\mathrm{b}} \\ \mathrm{X} & 100^{\circ} \mathrm{C} & 0.92 \\ \mathrm{Y} & 27^{\circ} \mathrm{C} & 0.63 \\ \mathrm{Z} & 283^{\circ} \mathrm{C} & 0.53\end{array} \)
Among the following, which is the correct relation between the boiling point and \( \mathrm{K}_{\mathrm{b}} \) values for \( \mathrm{X}, \mathrm{Y}, \mathrm{Z} \) respectively?

Options:
A \( 100^{\circ} \mathrm{C}-0.92,27^{\circ} \mathrm{C}-0.63,283^{\circ} \mathrm{C}-0.53 \)
B \( 100^{\circ} \mathrm{C}-0.53,27^{\circ} \mathrm{C}-0.63,283^{\circ} \mathrm{C}-0.92 \)
C \( 100^{\circ} \mathrm{C}-0.63,27^{\circ} \mathrm{C}-0.53,283^{\circ} \mathrm{C}-0.92 \)
D \( 100^{\circ} \mathrm{C}-0.53,27^{\circ} \mathrm{C}-0.92,283^{\circ} \mathrm{C}-0.63 \)

The correct answer is:
\( 100^{\circ} \mathrm{C}-0.63,27^{\circ} \mathrm{C}-0.53,283^{\circ} \mathrm{C}-0.92 \)

Solution:
K b = MRT b ∘ 2 1 0 0 0 Δ H vap
K b ∝ M T b ∘ Δ H vap T b ∘
K b ∝ T b ∘ Δ S vap ∘
Δ S ∘ will be same for three liquids changing to vapours for 1 mole
∴ K b ∝ T b ∘

Topic: Chemistry, Solutions, JEE Main
There are four chairs with two chairs in each row. In how many ways can four persons be seated on the chairs, so that no chair remains unoccupied?

Options:
A 6
B 12
C 24
D 48

The correct answer is:
24

Solution:
First chair can be occupied in 4 ways and second chair can be occupied in 3 ways, third chair can be occupied in 2 ways and last chair can be occupied in one ways only. So total number of ways $=4 \times 3 \times 2 \times 1=24$

Topic: Mathematics, Permutation Combination, NDA, NDA 2006 (Phase 1)
There are four complexes of Ni. Select the complexes/es which will be attracted by magnetic field:
(I) NiCN42- (II) NiCl42- (III) NiCO4 (IV) NiNH362+

Options:
A I only
B II only
C II, III and IV
D II and IV

The correct answer is:
II and IV

Solution:
Only NiCl42- and NiNH362+ are paramagnetic so will be attracted in magnetic field.

Complex	O.S	HBZ	Geometry	Unpaired	Magnetic property
NiCl42-	+2	sp3	Tetrahadral	2	Paramagnetic
NiCN42-	+2	dsp2	Square planar	0	Diamagnetic
NiCO4	0	sp3	Tetrahadral	0	Diamagnetic
NiNH362+	+2	sp3d2	Octahedral	2	Paramagnetic

Topic: Chemistry, Coordination Compounds, JEE Main
There are four lightweight rod samples A, B, C, D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted:

(i) A is feebly repelled
(ii) B is feebly attracted
(iii) C is strongly attracted
(iv) D remains unaffected

Which one of the following is true?

Options:
A B is of a paramagnetic material
B C is of a diamagnetic material
C D is of a ferromagnetic material
D A is of a nonmagnetic material

The correct answer is:
B is of a paramagnetic material

Solution:
A→ diamagnetic
B→ paramagnetic
C→ Ferromagnetic
D→ Non - magnetic

Topic: Physics, Magnetic Properties of Matter, NEET
There are four light-weight-rod samples $A, B, C$, $D$ separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted
(i) $A$ is feebly repelled
(ii) $B$ is feebly attracted
(iii) $C$ is strongly attracted
(iv) $D$ remains unaffected
Which one of the following is true?

Options:
A $C$ is of a diamagnetic material
B $D$ is of a ferromagnetic material
C $A$ is of a non-magnetic material
D $B$ is of a paramagnetic material

The correct answer is:
$B$ is of a paramagnetic material

Solution:
Paramagnetic materials will be feebly attract, dimagnetic material will be feebly repelled and Ferromagnetic material will be strongly attracted.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2011 (Screening)
There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6 . If the first and the last numbers are equal then two other numbers are

Options:
A -2,4
B -4,2
C 2,6
D none

The correct answer is:
-4,2

Solution:
Let the last three numbers in A.P. be a, a +6 , a $+12,$ then the first term is alsoa +12 . But $a+12, a, a+6$ are in G.P.

$\therefore \mathrm{a}^{2}=(\mathrm{a}+12)(\mathrm{a}+6) \Rightarrow \mathrm{a}^{2}=\mathrm{a}^{2}+18 \mathrm{a}+72$

$\therefore \mathrm{a}=-4$

$\therefore$ The numbers are 8,-4,2,8.

Topic: Mathematics, Sequences and Series, JEE Main
There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :

Options:
A 11
B 12
C 7
D 9

The correct answer is:
12

Solution:
Given, number of men is m and number of women is 2.

Since each participants competes two games with each other, Hence count of men Vs men total number of game plays =2 mC2.

count of men Vs women total number of game plays =4m.

As per given condition in question,

2 mC2=84+4 m

⇒m2-5 m-84=0

⇒m2-12 m+7 m-84=0

⇒m-12m+7=0.

⇒m=12, -7

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (12 Jan Shift 2)
There are \(n\) elastic balls placed on a smooth horizontal plane. The masses of the balls are \(m, \frac{m}{2}, \frac{m}{2^2}, \ldots . \frac{m}{2^{n-1}}\) respectively. If the first ball hits the second ball with velocity \(\mathrm{v}_0\), then the velocity of the \(\mathrm{n}^{\text {th }}\) ball will be,

Options:
A \(\frac{4}{3} v_0\)
B \(\left(\frac{4}{3}\right)^n v_0\)
C \(\left(\frac{4}{3}\right)^{n-1} v_0\)
D \(v_0\)

The correct answer is:
\(\left(\frac{4}{3}\right)^{n-1} v_0\)

Solution:
Hint : 1st Collision
\(\mathrm{V}_1=\frac{2 \times \mathrm{m}}{\frac{\mathrm{m}}{2}+\mathrm{m}} \mathrm{V}_0=\frac{4}{3} \mathrm{~V}_0\)
2nd Collision
\(V_2=\frac{2 \times \frac{m}{2}}{\frac{m}{2}+\frac{m}{2^2}} \times V_0=\left(\frac{4}{3}\right)^2 V_0\)
3rd Collision
\(\left(\frac{4}{3}\right)^3 \mathrm{~V}_0\)
\(\ldots(n-1)\) collision, \(\therefore V_{n-1}=\left(\frac{4}{3}\right)^{n-1} V_0\)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
There are n sets of observations given as 1, 2,3, 4,5,6, 7,8,9,10,..... The mean of the 13th set of observations is equal to

Options:
A 70
B 80
C 75
D 85

The correct answer is:
85

Solution:
In the nth set of observation, the total observations are ânâ
and the first observation of nth set is
1+2+3+....+n-1+1=n-1n2+1=a
Sum of all the n observations in the nth set
=n22a+n-1d
=n22n-1n2+1+n-1
=n2n2-n+2+n-1=nn2+12
Mean =nn2+12n=n2+12
Hence, for n=13

Mean =132+12=1702=85

Topic: Mathematics, Sequences and Series, JEE Main
There are $n$ similar conductors each of resistance $R$. The resultant resistance comes out to be $x$ when connected in parallel. If they are connected in series, the resistance comes out to be

Options:
A $x / n^2$
B $n^2 x$
C $x / n$
D $n x$

The correct answer is:
$n^2 x$

Solution:
In parallel, $x=\frac{R}{n} \quad R=n x$
In series, $R+R+R \ldots n$ times $=n R=n(n x)=$ $n^2 x$

Topic: Physics, Current Electricity, JEE Main
There are six periods in each working day of a school. The number of ways in which 5 subjects can be arranged such that each subject is allotted at least one period and no period remains vacant is

Options:
A 210
B 1800
C 360
D 120

The correct answer is:
1800

Solution:
Select one subject which is to be repeated in C15 ways
Select two periods in which the selected subject will be taught in C26 ways
Remaining 4 subjects will be taught in 4 periods in 4! ways.
⇒ Required number of ways =C51×C62×4!=5×15×24=1800

Topic: Mathematics, Permutation Combination, JEE Main
There are ten boys B1, B2,….,B10 and five girls G1,G2,….G5 in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B1 and B2 together should not be the members of a group, is _____.

Options:

The correct answer is:
1120

Solution:
Given, total number of boys is 10 and total number of girls is 5

So total number of selections =C310·C35=1200

No. of selections in which B1 & B2 both are there in group =C18·C35=80

Required no. of selections =1200-80=1120

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (26 Jun Shift 1)
There are ten seats out of which four are to be occupied. The number of ways of arranging four persons on these four seats such that each person has exactly one neighbour is

Options:
A 4P2×7P2
B 4P2×7C2
C 4C2×7C2
D 4C2×7P2

The correct answer is:
4P2×7P2

Solution:
Group of two persons can be selected in 4C2 ways. Let P1P2 are together and P3P4 are together. Let x1 be the number of seats vacant at the left end, x2 seats are between two pairs and x3 seats are vacant at the right end.
We have, x1+x2+x3=6, where x1, x3â¥0 and x2â¥1
âx1+x2+1+x3=6, where x1, x2, x3â¥0
âx1+x2+x3=5
Number of ways of selecting seats =n+r-1Cr-1=3+5-1C3-1=7C2
Persons can interchange their seats in 2!Ã2! ways
â Required ways =4C2Ã7C2Ã2!Ã2!=4P2Ã7P2

Topic: Mathematics, Permutation Combination, JEE Main
There are three bags $X, Y$ and $Z$. Bag $X$ contains 5 one-rupee coins and 4 five-rupee coins; Bag $Y$ contains 4 one-rupee coins and 5 five-rupee coins and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag Y, is :

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{5}{12}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{3}$

Solution:

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (08 Apr Shift 2)
There are three basic steps in creating genetically modified organisms. Arrange these steps in the correct sequence.

a. Introduction of the identified DNA into the host
b. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny
c. Identification of DNA with desirable genes

Options:
A a → b → c
B b → c → a
C c → b → a
D c → a → b

The correct answer is:
c → a → b

Solution:
To modify any organism genetically, desired gene of interest is identified and isolated; the gene is then inserted into the desired host using the suitable method at such a position that the gene multiplies into the host and is transferred to all the progeny cells.

Topic: Biology, Biotechnology - Principles and Processes, NEET
There are three coplanar parallel lines. If any \( p \) points are taken on each of the lines, the maximum number of triangles with vertices at these points are

Options:
A \( 3 p^{2}(p-1)+1 \)
B \( 3 p^{2}(p-1) \)
C \( p^{2}(4 p-3) \)
D \( p^{3} \)

The correct answer is:
\( p^{2}(4 p-3) \)

Solution:
We can either select the vertices of the triangle all on three different lines or two vertices on one line and one on another line.

The number of triangles with vertices on different lines

=C1p×C1p×C1p=p3.

The number of triangles with two vertices on one line and the third vertex on any one of the other two lines

=C13C2P×C12p=6p·pp-12.

So, the required number of triangles

=p3+3p2p-1=p24p-3.

Topic: Mathematics, Permutation Combination, JEE Main
There are three main well known allotropes of phosphorous in which one form is interconvertable to other as
\( [\mathrm{X}],[\mathrm{Y}] \) and \( [\mathrm{Z}] \) represent the

Options:
A Red \( \mathrm{P} \), white \( \mathrm{P} \) and black \( \mathrm{P} \) respectively
B White P, red P, and black P respectively
C Black P, red P and white P respectively
D Red P, black P and white P respectively

The correct answer is:
White P, red P, and black P respectively

Solution:
This problem is based on the preparation and interconversion of different types of phosphorous. Students must be familiar with condition of these interconversions.
Synthesis of red and black phosphorus
White phosphorus is converted to red phosphorus on heating at 573 K. Red phosphorous so formed is heated under high pressure to produces black P.
White P → Inert atmosphere 73 K Red P → High p Δ Black P

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
There are two balls in an urn. Each ball can be either white or black. If a white ball is put into the urn and there after a ball is drawn at random from the urn, then the probability that it is white is

Options:
A
$\frac{1}{4}$

B
$\frac{2}{3}$

C
$\frac{1}{5}$

D
$\frac{1}{3}$

The correct answer is:
$\frac{2}{3}$

Solution:
Total possible event when one ball is taken out $={ }^3 \mathrm{C}_1$ Let $E$ : The event of 1 white ball coming out
No. of ways to 1 white ball coming out $={ }^2 \mathrm{C}_1$
$$
\therefore P(E)=\frac{{ }^2 C_1}{{ }^3 C_1}=\frac{2}{3}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2012 (26 May Online)
There are two bodies of masses $100 \mathrm{~kg}$ and $10000 \mathrm{~kg}$ separated by a distance of $1 \mathrm{~m}$. At what distance from the smaller body, the intensity of gravitational field will be zero

Options:
A $\frac{1}{9} m$
B $\frac{1}{10} m$
C $\frac{1}{11} m$
D $\frac{10}{11} m$

The correct answer is:
$\frac{1}{11} m$

Solution:
$\frac{G \times 100}{x^2}=\frac{G \times 10000}{(1-x)^2} \Rightarrow \frac{10}{x}=\frac{100}{1-x} \Rightarrow x=\frac{1}{11} m$

Topic: Physics, Gravitation, JEE Main
There are two children in a family. The probability that both of them are boys is

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D None of these

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

Solution:
To be both boys the probability $=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$

Topic: Mathematics, Probability, JEE Main
There are two force vectors, one of $5 \mathrm{~N}$ and other of $1.2 \mathrm{~N}$ at what angle the two vectors be added to get resultant vector of $1.7 \mathrm{~N}, 7 \mathrm{~N}$ and $1.3 \mathrm{~N}$ respectively

Options:
A $0^{\circ}, 180^{\circ}$ and $90^{\circ}$
B $0^{\circ}, 90^{\circ}$ and $180^{\circ}$
C $0^{\circ}, 90^{\circ}$ and $90^{\circ}$
D $180^{\circ}, 0^{\circ}$ and $90^{\circ}$

The correct answer is:
$0^{\circ}, 180^{\circ}$ and $90^{\circ}$

Solution:
For 1.7 N both the vector should be parallel i.e. angle between them should be zero.
For $7 \mathrm{~N}$ both the vectors should be antiparallel i.e angle between them should be $180^{\circ}$
For $1.3 \mathrm{~N}$ both the vectors should be perpendicular to each other i.e. angle between them should be $90^{\circ}$

Topic: Physics, Mathematics in Physics, JEE Main
There are two identical chambers, completely thermally insulated from surroundings. Both chambers have a partition wall dividing the chambers in two compartments. Compartment $1$ is filled with an ideal gas and Compartment $3$ is filled with a real gas. Compartments $2$ and $4$ are vacuum. A small hole (orifice) is made in the partition walls and the gases are allowed to expand in vacuum.
$\mathrm{Statement-1:}$ No change in the temperature of the gas takes place when ideal gas expands in vacuum. However, the temperature of real gas goes down (cooling) when it expands in vacuum.
$\mathrm{Statement-2:}$ The internal energy of an ideal gas is only kinetic. The internal energy of a real gas is kinetic as well as potential.

Options:
A
Statement-1 is false and Statement-2 is true.

B
Statement-1 and Statement-2 both are true.
Statement-2 is the correct explanation of Statement-1.

C
Statement-1 is true and Statement- 2 is false.

D
Statement-1 and Statement-2 both are true.
Statement-2 is not correct explanation of Statement-1.

The correct answer is:
Statement-1 is false and Statement-2 is true.

Solution:
In ideal gases the molecules are considered as point particles and for point particles, there is no internal excitation, no vibration and no rotation. For an ideal gas the internal energy can only be translational kinetic energy and for real gas both kinetic as well as potential energy.

Topic: Physics, Kinetic Theory of Gases, JEE Main
There are two identical chambers, completely thermally insulated from surroundings. Both chambers have a partition wall dividing the chambers in two compartments. Compartment $1$ is filled with an ideal gas and Compartment $3$ is filled with a real gas. Compartments $2$ and $4$ are vacuum. A small hole (orifice) is made in the partition walls and the gases are allowed to expand in vacuum.
$\mathrm{Statement-1:}$ No change in the temperature of the gas takes place when ideal gas expands in vacuum. However, the temperature of real gas goes down (cooling) when it expands in vacuum.
$\mathrm{Statement-2:}$ The internal energy of an ideal gas is only kinetic. The internal energy of a real gas is kinetic as well as potential.

Options:
A
Statement-1 is false and Statement-2 is true.

B
Statement-1 and Statement-2 both are true.
Statement-2 is the correct explanation of Statement-1.

C
Statement-1 is true and Statement- 2 is false.

D
Statement-1 and Statement-2 both are true.
Statement-2 is not correct explanation of Statement-1.

The correct answer is:
Statement-1 is false and Statement-2 is true.

Solution:
In ideal gases the molecules are considered as point particles and for point particles, there is no internal excitation, no vibration and no rotation. For an ideal gas the internal energy can only be translational kinetic energy and for real gas both kinetic as well as potential energy.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2013 (09 Apr Online)
There are two identical containers $C_1$ and $C_2$ containing to identical gases. Gas in $C_1$ is reduced to half of its original volume adiabatically, while the gas in container $C_2$ is also reduced to half of its initial volume isothermally. Find the ratio of final pressure in these containers. ( $\gamma$ be the adiabatic constant).

Options:
A $2: 1$
B $1: 2$
C $2^y: 1$
D $2^{\gamma-1}: 1$

The correct answer is:
$2^{\gamma-1}: 1$

Solution:
Since, container are identical, so initial volume and pressure will be same for identical gases.
For container $C_1$,
$$
p_1 V_1^\gamma=p_2 V_2^2
$$
(adiabitic)
$$
p_2=\left(\frac{V_1}{V_2}\right)^\gamma p_1=\left(\frac{V_1}{V_1 / 2}\right)^\gamma p_1=2^\gamma p_1
$$
For container $C_2$,
$$
\begin{aligned}
p_1 V_1 & =p_2^{\prime} V_2 \text { (isothermal) } \\
p_2^{\prime} & =\left(\frac{V_1}{V_2}\right) p_1=\left(\frac{V_1}{V_1 / 2}\right) p_1=2 p_1
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\frac{p_2}{p_2{ }^{\prime}}=\frac{2^\gamma p_1}{2 p_1}=2^{\gamma-1}: 1
$$

Topic: Physics, Thermodynamics, JEE Main
There are two identical small holes of area of cross-section on the opposite sides of a tank containing a liquid of density, The difference in height between the holes is $h$.
Tank is resting on a smooth horizontal surface. Horizontal force which will has to be

Options:
A pgh
B $2 \mathrm{gh} / \mathrm{pa}$
C 2 pagh
D $\mathrm{pgh} / \mathrm{a}$

The correct answer is:
2 pagh

Solution:
Net force $\left(\right.$ reaction) $=F=F_B-F_A=\frac{d p_B}{d t}-\frac{d p_A}{d t}$
$=a v_B \rho \times v_B-a v_A \rho \times v_A$
$F=a \rho\left(v_B^2-v_A^2\right)$
$\therefore$ According to Bernoulli's theorem
$p_A+\frac{1}{2} \rho v_A^2+\rho g h=p_B+\frac{1}{2} \rho v_B^2+0$
$\Rightarrow \frac{1}{2} \rho\left(v_B^2-v_A^2\right)=\rho g h \Rightarrow v_B^2-v_A^2=2 g h$
From equation (i), $F=2 a p g h$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
There are two identical small holes of the area of cross-section a on the opposite sides of a tank containing a liquid of density ρ. The difference in height between the holes is h. The tank is resting on a smooth horizontal surface. The horizontal force which will have to be applied on the tank to keep it in equilibrium is -

Options:
A ghρa
B 2ghρa
C 2ρagh
D ρgha

The correct answer is:
2ρagh

Solution:
Thrust force
F=F1- F2=Ïav1 2 â Ïav2 2
=Ïa 2gh1-Ïa 2gh2
=2Ïag h1 â h2
=2 Ïagh

Topic: Physics, Mechanical Properties of Fluids, JEE Main
There are two infinitely long straight current-carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is 1:1. The magnetic field at point P is_________.

Options:
A μ0I4πxyx2+y2-(x+y)
B μ0Ixy4πx2+y2-(x+y)
C μ0I4πxyx2+y2+(x+y)
D μ0Ixy4πx2+y2+(x+y)

The correct answer is:
μ0I4πxyx2+y2+(x+y)

Solution:
B=μ0I4πy1+sinθ1+μ0I4πx1+sinθ2
B=μ0I4πy1+xx2+y2+μ0I4πx1+yx2+y2

B=μ0I4πx+μ0I4πy+μ0I4πxyx2+y2+yxx2+y2⇒B=μ0I4π1x+1y+μ0I4πx2+y2xy=μ0I4πx+y+x2+y2xy

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (01 Sep Shift 2)
There are two long co-axial solenoids of same length $l$. The inner and outer coils have radii $r_{1}$ and $r_{2}$ and number of turns per unit length $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$, respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is :

Options:
A $\frac{n_{1}}{n_{2}}$
B $\frac{n_{2}}{n_{1}} \cdot \frac{r_{1}}{r_{2}}$
C $\frac{n_{2}}{n_{1}} \cdot \frac{r_{2}^{2}}{r_{1}^{2}}$
D $\frac{n_{2}}{n_{1}}$

The correct answer is:
$\frac{n_{2}}{n_{1}}$

Solution:
The rate of mutual inductance is given by

$$

\mathrm{M}=\mu_{0} \mathrm{n}, \mathrm{n}_{2} \pi \mathrm{r}_{1}^{2} \quad \ldots(\mathrm{i})

$$

The rate of self inductance is given by

$\mathrm{L}=\mu_{0} \mathrm{n}_{1}^{2} \pi \mathrm{r}_{1}^{2} \quad \ldots(\mathrm{ii})$

Dividing (i) by (ii) $\Rightarrow \frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (11 Jan Shift 1)