datavorous/entrance-exam-dataset huggingface dataset:
Consider the above reaction. The number of π electrons present in the product 'P' is

Options:

The correct answer is:
2

Solution:
SN2 reaction takes place and the number of pi-electrons in the product is 2.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Consider the above reaction. The percentage yield of amide product is (Round off to the Nearest Integer). (Given : Atomic mass : C:12.0u,H:1.0u N:14.0u,O:16.0u,Cl:35.5u)

Options:

The correct answer is:
77

Solution:
1 mole 1 mole 1 mole

=140.5gm =169gm =273gm

∴0.140gm 169140.5×0.140

L.R. =0.168gm<0.388gm

excess

∴ Theoretical amount of given product formed

=273140.5×0.140=0.272gm

But its actual amount formed is 0.210gm. Hence, the percentage yield of product.

=0.2100.272×100=77.20≈77

OR

Mole of Ph-CoCl=0.140140=10-3 mol

by mol-mol analysis =10-3 mol.

Theoritical mass of product =10-3×273= 273×10-3 g

Observed mass of product =210×10-3 g

% yield of product =210×10-3273×10-3×100=76.9%=77

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Consider the above reaction where 6.1 g of benzoic acid is used to get 7.8 g of m-bromo benzoic acid. The percentage yield of the product is ___ .

(Round off to the Nearest integer)

[Given : Atomic masses : C=12.0 u, H:1.0 u, O:16.0 u, Br=80.0 u]

Options:

The correct answer is:
78

Solution:
Moles of Benzoic acid =6.1122

= moles of m-bromobenzoic acid

So, weight of m-bromobenzoic acid

=6.1122×201 gm

=10.05 gm

% yield =Actual weightTheoretical weight×100

=7.810.05×100

=77.61%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider the acidity of the carboxylic acids:

Options:
A
$\mathrm{PhCOOH}$

B
$\mathrm{O}-\mathrm{NO}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{COOH}$

C
$\mathrm{p}-\mathrm{NO}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{COOH}$

D
$\mathrm{m}-\mathrm{NO}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{COOH}$

The correct answer is:
$\mathrm{O}-\mathrm{NO}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{COOH}$

Solution:
No Solution Available

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2004
Consider the atomization of Br2(g) according to the reaction :
Br 2 g â 2 Br g {Given antiloge(-7.02)=9x10-4}
If the heat of atomization of bromine gas is 82 KJ/mol ; standard molar entropies of Br(g) and Br2(g) are 175 and 245.4 JK-1 respectively, calculate degree of dissociation when the total pressure is 40 atm at 500 K.
(assume Î±<<1 in your calculation)

Options:
A 2.37 Ã 1 0 - 3
B 3.20 × 1 0 - 4
C 4.30 × 1 0 - 5
D 3.60 × 1 0 - 2

The correct answer is:
2.37 × 1 0 - 3

Solution:
For Br 2 g â 2 Br g
Î S reac â = 2 Ã 1 7 5 - 2 4 4 Â· 4 = 105.6 J K - 1
Î H reac â = 8 2 KJ m - 1
Î G reac â = Î H reac â - T Î S â = 8 2 0 0 0 - 5 0 0 Ã 105.6
= 2 9 Â· 2 Ã 1 0 3 J = - RT ln K P
ln K P = - 2 9 7 0 0 8 Â· 3 1 4 Ã 5 0 0 = - 7.02
â´â K P =9Ã 10 â4
K P = 4 Î± 2 P 1â Î± 2
as Î±<<1âââ´â1â Î± 2 =1
â´â K P =4Ã Î± 2 P
âµâPâ=â40
â´â9Ã 10 â4 =4â Î± 2 â40
â´â Î± 2 = 9 160 Ã 10 â4
= 9 16 Ã 10 â5
=â2.37Ã 10 â3

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50 % more in solid 2 than in 1 . What is the approximate packing efficiency in solid 2 ?

Options:
A 90 %
B 45 %
C 75 %
D 65 %

The correct answer is:
90 %

Solution:
If radius of A is r radius of B will be 2r . Atoms in the body diagonal touch.

Therefore:
r+2r+2r+r=3 a

6 r=3 a or a=6r3

Packing Efficiency
=1×43πr3+1×43π(2r)3a3=43πr3×9(6r/3)3
=4×9×π×r3×333×216×r3=90.5 %

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Consider the cell at 25°C

ZnZn2+aq,1 M‖Fe3+(aq),Fe2+aqPts

The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x×10-2. The value of x is ______. (Nearest integer)

Given: E∘Fe3+|Fe2+=0.77V,  E∘Zn2+|Zn=-0.76V

Options:

The correct answer is:
24

Solution:
Ecell0=0.77-0.76

=1.53 V

1.50=1.53-0.062logFe2+Fe3+2

logFe2+Fe3+=0.030.06=12

Fe2+Fe3+=101/2=10

Fe3+Fe2+=110

Fe3+Fe2++Fe3+=11+10=14.16

=0.2402

=24×10-2

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Consider the cell

PtsH2g,1atmH+aq,1M|Fe3+aq,Fe2+aqPts

When the potential of the cell is 0.712 V at 298 K, the ratio Fe2+/Fe3+ is
(Nearest integer)
Given: Fe3++e-=Fe2+,E°Fe3+,Fe2+Pt=0.771 2.303RTF=0.06 V

Options:

The correct answer is:
10

Solution:
Given cell reaction:
PtsH2g,1atmH+aq,1M||Fe3+aq,Fe2+aqPts

at anode(oxidation) H2⟶2H++2e-

At cathode(reduction) Feaq3++e-⟶Feaq2+

E°cell=Ecathodeo - Eanodeo=EoH2|H+ +EoFe3+∣Fe2+=0.771 V

Thus, using Nernst equation,
⇒E=E°-0·061logFe2+Fe3+

⇒0.712=0+0.771-0.061logFe2+Fe3+

⇒logFe2+Fe3+=0.0590.06≈1

Fe2+Fe3+=10

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Consider the cell Pt(s)H2( s)(latm)H+aq,H+=1||Fe3+(aq),Fe2+(aq)∣Pt(s)

Given: EFe3+/Fe2+°=0.771 V and EH+/12H2°=0 V, T=298 K

If the potential of the cell is 0.712 V the ratio of concentration of Fe2+ to Fe3+ is
(Nearest integer)

Options:

The correct answer is:
10

Solution:
Cell reaction which occurs:
12H2( g)+Fe3+ (aq.) ⟶H+(aq)+Fe2+ (aq.)
Using Nernst' equation:

E=Eo-0.0591logFe2+Fe3+

⇒0.712=(0.771-0)-0.0591logFe2+Fe3+

⇒logFe2+Fe3+=(0.771-0712)0.059=1

⇒Fe2+Fe3+=10

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to

Options:
A $q_2$
B Only the positive charges
C All the charges
D $+q_1$ and $-q_1$

The correct answer is:
All the charges

Solution:
The electric field is due to all charges present whether inside or outside the given surface.

Topic: Physics, Electrostatics, JEE Main
Consider the circles $x^{2}+y^{2}+2 a x+c=0$ and $x^{2}+y^{2}+2 b y+c=0$
The two circles touch each other if

Options:
A $c=\sqrt{a^{2}+b^{2}} \quad$
B $\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
C $\quad c=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
D $c=\frac{1}{a^{2}+b^{2}}$

The correct answer is:
$\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

Solution:
Two circles touch each other, iff distance between two centres $=$ Sum of radius of two circles
$\sqrt{a^{2}+b^{2}}=\sqrt{a^{2}-c}+\sqrt{b^{2}-c}$
On squaring both sides, we get
$a^{2}+b^{2}=a^{2}-c+b^{2}-c+2 \sqrt{\left(a^{2}-c\right)\left(b^{2}-c\right)}$
$\Rightarrow \mathrm{c}=\sqrt{\left(\mathrm{a}^{2}-\mathrm{c}\right)\left(\mathrm{b}^{2}-\mathrm{c}\right)}$
Again, squaring both sides, we get
$$
\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~b}^{2}-\mathrm{a}^{2} \mathrm{c}-\mathrm{b}^{2} \mathrm{c}+\mathrm{c}^{2}
$$
$\Rightarrow a^{2} b^{2}=\left(a^{2}+b^{2}\right) c \Rightarrow \frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

Topic: Mathematics, Circle, JEE Main
Consider the circuit shown below. The current in the 4 Ω resistor is

Options:
A 0.25 A
B 0.50Â A
C 0.75Â A
D 1.00Â A

The correct answer is:
0.25 A

Solution:
Given circuit below
Resistance between B and C on the right-hand side of the circuit
=10Ã1010+10=10020=5 Î©
because 1 Î©+4 Î©+5 Î©=10 Î© are in parallel with 10 Î© .
Resistance between A and D on the right-hand side of the circuit.
=10Ã1010+10=10020=5 Î©
because 5 Î© (equivalent resistance of 1 Î©, 4 Î©, 5 Î© and 10 Î© ) is in series with 3 Î© and 2 Î© .
Hence, resistance between A and D will be 5 Î©
Equivalent resistance of the circuit,
R=5+2+5=12 Î©
Current drawn from the battery,
I=12 V12 Î©=1 A
At the junction A , the current of 1 A is divided equally between the 10 Î© resistance and the remaining circuit of the 10 Î© resistance.
At the junction B , the current of 0.5 A is divided equally between the 10 Î© resistor and remaining circuit of resistance 10 Î© .
â´ Current through the 4 Î© resistor =0.25 A .

Topic: Physics, Current Electricity, NEET
Consider the circuit shown below where all resistors are of $1 \mathrm{k} \Omega$
If a current of magnitude $1 \mathrm{~mA}$ flows through the resistor marked $\mathrm{X}$, what is the potential difference measured between point $\mathrm{P}$ and $\mathrm{Q}$ ?

Options:
A $21 \mathrm{~V}$
B $68 \mathrm{~V}$
C $55 \mathrm{~V}$
D $34 \mathrm{~V}$

The correct answer is:
$34 \mathrm{~V}$

Solution:
Using KCL
At point $A$
Current is $3 \mathrm{~mA}$
At point $C$
Current is $8 \mathrm{~mA}$
At point $E$
Current is $21 \mathrm{~mA}$
At point $G$
Current through $\mathrm{GH}$ is 34 ma
$\begin{array}{l}
\therefore \mathrm{V}_{\mathrm{PQ}}=\mathrm{V}_{\mathrm{GH}}=\mathrm{i} \mathrm{R}_{\mathrm{GH}} \\
=34 \mathrm{~V}
\end{array}$

Topic: Physics, Current Electricity, JEE Main
Consider the circuit shown in the figure. The current $I_3$ is equal to

Options:
A $5 \mathrm{amp}$
B $3 \mathrm{amp}$
C $-3 \mathrm{amp}$
D $-5 / 6 a m p$

The correct answer is:
$-5 / 6 a m p$

Solution:
Suppose current through different paths of the circuit is as follows.
After applying KVL for loop (1) and loop (2)
We get $28 i_1=-6-8 \Rightarrow i_1=-\frac{1}{2} \mathrm{~A}$ and $54 i_2=-6-12 \Rightarrow i_2=-\frac{1}{3} A$
Hence $i_3=i_1+i_2=-\frac{5}{6} \mathrm{~A}$

Topic: Physics, Current Electricity, JEE Main
Consider the circuit shown in the figure.

Which of the following statements are correct ?

Options:
A Current through the battery is 5 A

B Current through branch AD is 2 A
C Current through branch BC is 3 A
D The potential difference between points D and C is 4 V

The correct answer is:
Current through the battery is 5 A

Solution:
By using the concept of Wheatstone bridge
VD â VC = 0
â´ Req=12Ã612+6=4 Î©
â´ Ibattery=204=5 A

Topic: Physics, Current Electricity, JEE Main
Consider the combination of two capacitors C1 and C2, with C2>C1, when connected in parallel, the equivalent capacitance is 10 times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, C2C1.

Options:
A 4+15
B 2+15
C 9
D 154

The correct answer is:
4+15

Solution:
When connected in parallel

Ceq=C1+C2

When in series

Ceq'=C1C2C1+C2

C1+C2=10C1C2C1+C2

C1+C22=10C1C2

C12+C22-8C1C2=0

dividing by C12

C2C12-8C2C1+1=0

Let C2C1=x

x2-8x+1=0

x=8±64-42=4+15

This question was Bonus in JEE Main 2021 exam because of incorrect data in original question.

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ______ ×10-1 g. (in nearest integer)

Options:

The correct answer is:
464

Solution:
C4H10+132O2⟶4CO2+5H2O

Moles of H2O=7218=4

Moles of C4H10 used =15×4

Weight of C4H10 used =45×58

=46.4 gm

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Consider the complexes.
(i) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{ClBr}\right]$
(ii) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
(iii) $\left[\mathrm{Pd}(\mathrm{en}) \mathrm{Cl}_2\right]$
(iv) [Pd(en) $\mathrm{ClBr}]$
(v) $\left[\mathrm{Pd}(\mathrm{en})_2\right] \mathrm{Cl}_2 \quad$ (en = ethylenediamine)
The total number of geometrical isomers of (A) is same as the total number of geometrical isomers of

Options:
A (ii)
B (iii)
C (iv)
D (v)

The correct answer is:
(ii)

Solution:
(i) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{ClBr}\right]$
(ii) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
$M a_2 b c$
$M a_2 b_2$
G.I $=2$
G.I $=2$
cis and trans
cis and trans
(iii) $\left[\mathrm{Pd}(\mathrm{en}) \mathrm{Cl}_2\right]$
(iv) $[\mathrm{Pd}(\mathrm{en}) \mathrm{ClBr}]$
$\mathrm{Maab}_2$
$M a b c$
$\mathrm{G} . \mathrm{I}=3$
G.I $=3$
(v) $\mathrm{Pd}(\mathrm{en})_2 \mathrm{Cl}_2$
$\mathrm{Ma}_2 \mathrm{~b}_2$
$\mathrm{G} . \mathrm{I}=4$
So, only (i) and (ii) have two, i.e. same number of geometrical isomers.

Topic: Chemistry, Coordination Compounds, JEE Main
Consider the conic $\mathrm{ex}^{2}+\pi y^{2}-2 \mathrm{e}^{2} \mathrm{x}-2 \pi^{2} \mathrm{y}+\mathrm{e}^{3}+\pi^{3}=\pi \mathrm{e}$. Suppose $\mathrm{P}$ is any point on the conic and $\mathrm{S}_{1}, \mathrm{~S}_{2}$ are the foci of the conic, then the maximum value of $\left(\mathrm{PS}_{1}+\mathrm{PS}_{2}\right)$ is $-$

Options:
A $\pi \mathrm{e}$
B $\sqrt{\pi \mathrm{e}}$
C $2 \sqrt{\pi}$
D $2 \sqrt{\mathrm{e}}$

The correct answer is:
$2 \sqrt{\pi}$

Solution:
$e x^{2}+\pi y^{2}-2 e^{2} x-2 \pi^{2} y+e^{3}+\pi^{3}=\pi e$ $e\left(x^{2}-2 e x+e^{2}\right)+\pi\left(y^{2}-2 \pi y+\pi^{2}\right)=\pi e$ $\frac{(x-e)^{2}}{\pi}+\frac{(y-\pi)^{2}}{e}=1$
$a^{2}=\pi \Rightarrow a=\Gamma \pi \quad \quad \pi>e$
$P S_{1}+P S_{2}=2 a$ Major axis is $\|$ to axis
$P S_{1}+P S_{2}=2 \cdot F \bar{\pi}$

Topic: Mathematics, Ellipse, JEE Main
Consider the coordination compound, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$. In the formation of this complex, the species which acts as the Lewis acid is:

Options:
A
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

B
$\mathrm{Cl}^{-}$

C
$\mathrm{Co}^{3+}$

D
$\mathrm{NH}_3$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (11 Apr Online)
Consider the cubic equation $x^{3}+a x^{2}+b x+c=0$, where $a, b, c$ are real numbers. Which of the following statements is correct?

Options:
A If $a^{2}-2 b < 0$, then the equation has one real and two imaginary roots
B If $a^{2}-2 b \geq 0$, then the equation has all real roots
C If $a^{2}-2 b>0$, then the equation has all real and distinct roots
D If $4 a^{3}-27 b^{2}>0$, then the equation has real and distinct roots

The correct answer is:
If $a^{2}-2 b < 0$, then the equation has one real and two imaginary roots

Solution:
$\begin{array}{l}
f(x)=x^{3}+a x^{2}+b x+c \\
f^{\prime}(x)=3 x^{2}+2 a x+b \\
D=4 a^{2}-4.3 \quad b=4\left(a^{2}-3 b\right)
\end{array}$
If $a^{2} < 2 b \Rightarrow a^{2} < 3 b \Rightarrow D < 0 \Rightarrow f^{\prime}(x)=0$ has non real roots
Hence $\mathrm{f}(\mathrm{x})=0$ has 1 real and two imaginary roots

Topic: Mathematics, Quadratic Equation, JEE Main
Consider the cubic fx=x3-3x+a, where aâ0, 2. Then, the equation fx=0 has

Options:
A 3 real solutions
B 2 real solutions
C 1 real solution
D no real solutions

The correct answer is:
3 real solutions

Solution:
f â² x =3 x 2 â3
=3x-1x+1
Also, f-1=a+2>0
f1=a-2<0
Hence, fx=0 has 3 real solutions.

Topic: Mathematics, Application of Derivatives, JEE Main
Consider the data on \( \mathrm{x} \) taking the values \( 0,2,4,8, \ldots ., 2^{\mathrm{n}} \) with frequencies \( { }^{\mathrm{n}} \mathrm{C}_{0},{ }^{\mathrm{n}} \mathrm{C}_{1},{ }^{\mathrm{n}} \mathrm{C}_{2}, \ldots,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \) respectively. If the mean of this data is \( \frac{728}{2^{\mathrm{n}}} \), then \( \mathrm{n} \) is equal to ....... .

Options:

The correct answer is:
6

Solution:
xi (observation)	0	2	22................	2n

fi (frequency)	C0n	C1n	C2n.............	Cnn

x=ΣfixiΣfi

0×C0n+2×C1n+22×C2n.....2n×CnnC0n+C1n+C2n.....+Cnn=3n-12n=7282n

⇒ 3n=36

⇒ n=6

Topic: Mathematics, Binomial Theorem, JEE Main
Consider the definite integrals A=â«0Ïxsinâ¡xcos2â¡xdx and B=â«0Ï2sinâ¡xcos2â¡xdx. Then,

Options:
A A=2B
B A=πB
C A=π2B
D B=2A

The correct answer is:
A=πB

Solution:
Applying a+b-x property in A, and adding, we get,

2A=â«0Ïx+Ï-xsinâ¡xcos2â¡xdx
â2A=Ïâ«0Ïsinâ¡xcos2â¡xdx
â´2A=Ïâ2â«0Ï2sinxcos2xdx
âA=ÏB

Topic: Mathematics, Definite Integration, JEE Main
Consider the differential equation :
$$
\frac{d y}{d x}=\frac{y^3}{2\left(x y^2-x^2\right)}
$$
Statement-1: The substitution $z=y^2$ transforms the above equation into a first order homogenous differential equation.
Statement-2: The solution of this differential equation is $y^2 e^{-y^2} / x=C$.

Options:
A
Both statements are false.

B
Statement-1 is true and statement- 2 is false.

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

D
Both statements are true.

The correct answer is:
Both statements are true.

Solution:
Given differential equation is
$$
\frac{d y}{d x}=\frac{y^3}{2\left(x y^2-x^2\right)}
$$
By substituting $z=y^2$, we get diff. eqn. as
$$
\frac{d z}{d x}=\frac{2 z^2}{2\left(x z-x^2\right)}=\frac{z^2}{x z-x^2}
$$
Now, $\frac{d x}{d z}=\frac{x}{z}-\frac{x^2}{z^2}=\frac{x}{z}\left[1-\frac{x}{z}\right] \approx \mathrm{F}\left(\frac{x}{z}\right)$
Hence, statement-1 is true.
Now, $y^2 e^{-y^2 / x}=\mathrm{C}$ satisfies the given diff. equation
$\therefore$ It is the solution of given diff. equation. Thus, statement $-2$ is also true.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2013 (22 Apr Online)
Consider the differential equation, y2dx+x-1ydy=0. If value of y is 1 when x=1, then the value of x for which y=2, is

Options:
A 32-1e
B 32-e
C 12+1e
D 52+1e

The correct answer is:
32-1e

Solution:
Given differential equation is
y2dx=1y-xdy

⇒y2dxdy+x=1y
⇒dxdy+1y2x=1y3
It is a linear differential equation whose integrating factor I.F. =e∫dyy2=e-1y
Solution of a given differential equation can be written as
xe-1y=∫e-1y1y3dy=I
Let -1y=t⇒dyy2=dt⇒I=∫-tet dt
=et1-t+C, (Integrating by parts)
⇒ Solution of differential equation is xe-1y=e-1y1+1y+C
Since for x=1, we have y=1,
⇒1.e-1=e-11+1+C⇒C=-1e
⇒ Solution with given condition is xe-1y=e-1y1+1y-e-1
or x=1+1y-e1y-1
So, xy=2=1+12-e12-1=32-1e

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter 0.1 μm. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that

Options:
A its size increases, and intensity increases
B its size decreases, and intensity decreases
C its size decreases, but intensity increases
D its size increases, but intensity decreases

The correct answer is:
its size decreases, but intensity increases

Solution:
sinθ=mλa

when a increases, θ decreases,

width decreases

width decreases so intensity will increases

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Consider the dissociation of the weak acid \(\mathrm{HX}\) as given below
\(\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}\)
\(\left[\mathrm{K}_{\mathrm{a}}\right.\) : dissociation constant \(]\)
The osmotic pressure of \(0.03 \mathrm{M}\) aqueous solution of \(\mathrm{HX}\) at \(300 \mathrm{~K}\) is _______ \(\times 10^{-2}\) bar (nearest integer).
[Given : \(\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)]

Options:

The correct answer is:
76

Solution:
$\begin{aligned}
& \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5} \\
& 0.03 \mathrm{M} \\
& 0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x} \\
& \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}} \\
& 0.03-\mathrm{x} \approx 0.03\left(\mathrm{~K}_{\mathrm{a}} \text { is very small }\right) \\
& \frac{\mathrm{x}^2}{0.03}=1.2 \times 10^{-5} \\
& \mathrm{x}=6 \times 10^{-4}
\end{aligned}$
Final solution : $0.03-x+x+x$
$\begin{aligned}
& =0.03+x=0.03+6 \times 10^{-4} \\
& \Pi=\left(0.03+\left(6 \times 10^{-4}\right)\right) \times 0.083 \times 300 \\
& =76.19 \times 10^{-2} \approx 76 \times 10^{-2}
\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Consider the elements Mg, Al,S,P and Si, the correct increasing order of their first ionisation enthalpy is:

Options:
A Al<Mg<Si<S<P
B Al<Mg<S<Si<P
C Mg<Al<Si<P<S
D Mg<Al<Si<S<P

The correct answer is:
Al<Mg<Si<S<P

Solution:
In general from left to right in a period, ionisation enthalpy increases due to effective nuclear charge increases. but due to extra stability of half filled and full filled electronic configuration, required ionisation enthalpy is more from neighbouring elements.
i.e. first ionisation enthalpy order is

Al<Mg<Si<S<P

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Consider the equation log2sinx1+cosx=2, xâ-Ï2,3Ï2. If the sum of the roots is pÏq, where G.C.D p,q=1, then the value of p2+q2 is

Options:

The correct answer is:
10

Solution:
log2sinx1+cosx=2 â¦.(i)
2sinxâ 1, 2sinx>0, 1+cosx>0
âsinxâ 12, sinx>0 and xâ 0 or a multiple of Ï
âxâ0,Ï-Ï4,3Ï4 (feasible region)

Hence, from (i), we get,

2sinx2=1+cosx
â2sin2x=1+cosx
â2cos2x+cosx-1=0
â2cosx-1cosx+1=0
âcosx=12 â¦cosx+1>0
âx=Ï3
âp=1,q=3
âp2+q2=10

Topic: Mathematics, Trigonometric Equations, JEE Main
Consider the equation x2+2x-n=0, where n∈N and n∈5, 100. The number of different values of n so that the given equation has integral roots, is

Options:

The correct answer is:
8

Solution:
x2+2x-n=0; n∈5, 100
D will be a perfect square,

D=4+4n=41+n
⇒1+n is a perfect square
⇒1+n=9, 16, 25, 36, 49, 64, 81, 100
⇒n=8, 15, 24, 35, 48, 63, 80, 99
Therefore, 8 values are possible.

Topic: Mathematics, Quadratic Equation, JEE Main
Consider the equation $x^n-5 x^{n-1}+6 x^{n-2}=0$, where $n$ is a positive integer. What is the sum of all possible real roots of this equation?

Options:
A $5$
B $6$
C $0$
D $1$

The correct answer is:
$5$

Solution:
To find the sum of all possible real roots of the equation, we can factor out $x^{n-2}$ from the equation:$x^{n-2}\left(x^2-5 x+6\right)=0$
Now, we solve $x^2-5 x+6=0$ for real roots. Factoring this quadratic equation, we get:$(x-2)(x-3)=0$
So, the roots of $x^2-5 x+6=0$ are $x=2$ and $x=3$.
Since the original equation is $x^n-5 x^{n-1}+6 x^{n-2}=0$, the sum of all possible real roots is the sum of the roots of $x^2-5 x+6=0$, which is $2+3=5$.
Therefore, the correct answer is 1) 5 .

Topic: Mathematics, Quadratic Equation, JEE Main
Consider the equation \( x^{2}+2 x-n=0 \), where \( n \in N \) and \( n \in[5,100] . \) The total number of different values of \( n \) so that the given equation has integral roots is

Options:
A \( 8 \)
B \( 3 \)
C \( 6 \)
D \( 4 \)

The correct answer is:
\( 8 \)

Solution:
We have,

x2+2x-n=0

We know that for a Quadratic equation ax2+bx+c=0, a≠0 the roots are x=-b±b2-4ac2a.

⇒x=-1± n+1

Thus, n+1 should be a perfect square for integral roots.

Now,

n∈5,100⇒n+1∈6,101

Perfect square values of n+1 are 9,16,25,36,49,64,81,100

Hence, number of values is 8.

Topic: Mathematics, Quadratic Equation, JEE Main
Consider the equations given below and find the sum of x, y and z .

(i) Mg+xHNO3very dilute 5%âMgNO32+H2

(ii) Cu+yHNO3(dil.)â3CuNO32+2NO +4H2O

(iii) I2+zHNO3â2HIO3 +10NO2+4H2O

Options:

The correct answer is:
20

Solution:
(i) Mg+2HNO3very dilute 5%âMgNO32+H2P
(ii) Cu+8HNO3(dil.)â3CuNO32+2NOQ+4H2O
(iii) I2+10HNO3â2HIO3R+10NO2+4H2O
Here x = 2, y = 8 and z = 10
Hence, x + y + z = 2 + 8 + 10 = 20

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Consider the family of lines 5x+3y-2+λ3x-y-4=0 and x-y+1+μ2x-y-2=0. The equation of a straight line that belongs to both the families is

Options:
A 5x-2y-7=0
B 3x+y-2=0
C 5x+2y-3=0
D 2x+y-1=0

The correct answer is:
5x-2y-7=0

Solution:
All lines of the family 5x+3y-2=0 and 3x-y-4=0 are concurrent.

The point of concurrency is 1,-1
Similarly, the point of concurrency of lines of the other family is 3,4
â The line of both families is the line passing through 1,-1 & 3,4
â Equation of the required line is 5x-2y-7=0

Topic: Mathematics, Straight Lines, JEE Main
Consider the figure here. A particle of mass m is constrained to move inside a smooth vertical groove of radius R and is connected to a light spring of spring constant K in equilibrium. O is centre of the groove. x and y are horizontal and vertical axes respectively. Different physical parameters are related as 2KR=7mg. Angular frequency of the oscillations, if the particle is slightly displaced from the shown equilibrium position is

Options:
A gR
B g2R
C 2gR
D 2gR

The correct answer is:
2gR

Solution:
Restoring torque on small displacement dθ
τ=KR2dθ+mg2Rdθ
⇒α=τl=KR2+mg2RdθmR2
=Km+g2Rdθ=ω2dθ
⇒ω=Km+g2R=2gR

Topic: Physics, Oscillations, JEE Main