datavorous/entrance-exam-dataset huggingface dataset:
The minimum value of the expression $7-20 x+11 x^2$ is

Options:
A $\frac{177}{11}$
B $-\frac{177}{11}$
C $-\frac{23}{11}$
D $\frac{23}{11}$

The correct answer is:
$-\frac{23}{11}$

Solution:
Given $f(x)=7-20 x+11 x^2$
$f^{\prime}(x)=-20+22 x$
Put $f^{\prime}(x)=0$ i.e., $-20+22 x=0$
$\Rightarrow x=10 / 11 \text { and } f^{\prime \prime}(x)=22\gt0$
Hence at $x=10 / 11, f(x)$ will have minimum value,
$\therefore f\left(\frac{10}{11}\right)=7-\frac{200}{11}+\frac{100 \times 11}{121}=7-\frac{200}{11}+\frac{100}{11}=-\frac{23}{11}$

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum value of the expression E=z2+z-32+z-6i2 (where z is a complex number) is equal to

Options:

The correct answer is:
30

Solution:
Let z=x+iy
E=z z Â¯ + zâ3 z Â¯ â3 + zâ6i z Â¯ +6i
=3z z Â¯ â3 z+ z Â¯ +9+6 zâ z Â¯ i+36
z+ z Â¯ =2x;zâ z Â¯ =2iy
=3x2+y2-6x-12y+45
=3x2+y2-2x-4y+15
E=3x-12+y-22+10
Hence, Emin=30 when x=1 and y=2 i.e., z=1+2i

Topic: Mathematics, Complex Number, JEE Main
The minimum value of the function fx=tanâ¡x3+2tanâ¡x, âxâ0,Ï2 is

Options:
A 0
B 12
C 13
D 16

The correct answer is:
0

Solution:
If x∈0,π2, tanx∈0,∞

Let, tanx=t
i.e. y=t3+2t, t∈0,∞
⇒3y+2ty=t
⇒t=3y1-2y≥0
⇒y∈0,12
Hence, the minimum value of the function=0

Topic: Mathematics, Functions, JEE Main
The minimum value of the function $f(x)=x \log x$ is

Options:
A -e
B e
C $\frac{1}{\mathrm{e}}$
D $-\frac{1}{\mathrm{e}}$

The correct answer is:
$-\frac{1}{\mathrm{e}}$

Solution:
$f(x)=x \log x$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x}=1+\log \mathrm{x}$
When $1+\log \mathrm{x}=0 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \Rightarrow\left[\mathrm{f}^{\prime \prime}(\mathrm{x})\right]_{\mathrm{x}=\frac{1}{\mathrm{e}}}=\mathrm{e}>0$
Thus $\mathrm{f}(\mathrm{x})$ is minimum at $\mathrm{x}=\frac{1}{\mathrm{e}}$ and
$f(x)=\left(\frac{1}{e}\right) \log \left(\frac{1}{e}\right)=\frac{-1}{e}$

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum value of the sum of numbers \( a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, 1, a^{8} \) and \( a^{10} \) with \( \mathrm{a}>0 \) is.

Options:

The correct answer is:
8

Solution:
Using AM ≥ GM
a-5+a-4+a-3+a-3+a-3+1+a8+a108≥a-5·a-4·a-3·a-3·a-3·1·a8·a1018≥a18a1818
⇒ a - 5 + a - 4 + 3 a - 3 + 1 + a 8 + a 1 0 ≥ 8.1
∴ Minimum value is 8

Topic: Mathematics, Sequences and Series, JEE Main
The minimum value of the sum of the squares of the roots of x2+3-ax=2a-1 is

Options:
A 6
B 4
C 5
D 8

The correct answer is:
6

Solution:
Let α,β be the roots of the equation

x2+3-ax+1-2a=0

Then, sum of roots α+β=a-3

And product of roots αβ=1-2a

We know that α2+β2=α+β2-2αβ

∴α2+β2=a-32-21-2a

=a2-2a+7

=a-12+6

∴ Minimum value of α2+β2=6 at a=1.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The minimum value of x which satisfies the inequality sin-1x2≥cos-1x2 is

Options:
A 12
B 12
C 32
D 13

The correct answer is:
12

Solution:
sin-1x2-cos-1x2≥0
sin-1x+cos-1xsin-1x-cos-1x≥0
⇒sin-1x≥cos-1x
Clearly from graph
x∈12,1

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The minimum value of x which satisfies the inequality sin-1xâ¥cos-1x is Î», then the value of 2Î» is (use 2=1.41)

Options:

The correct answer is:
1.41

Solution:
Using the graphs of sin-1x & cos-1x, we can see that
sin-1x≥cos-1x for x∈12,1
So, the minimum value of x is λ=12

Hence, 2λ=2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The minimum value of $(x-\alpha)(x-\beta)$ is

Options:
A $0$
B $\alpha \beta$
C $\frac{1}{4}(\alpha-\beta)^2$
D $-\frac{1}{4}(\alpha-\beta)^2$

The correct answer is:
$-\frac{1}{4}(\alpha-\beta)^2$

Solution:
Let $y=(x-\alpha)(x-\beta)$
$\begin{aligned} y & =x^2-(\alpha+\beta) x+\alpha \beta \\ y^{\prime} & =2 x-(\alpha+\beta)\end{aligned}$
For maximum or minimum, put $y^{\prime}=0$
$\begin{array}{rrrl}\Rightarrow & 2 x-(\alpha+\beta) & =0 \\ \Rightarrow & x & =\frac{\alpha+\beta}{2} \\ \Rightarrow & y^{\prime \prime} & =2=+ \text { ve }\end{array}$
The minimum value at $x=\frac{\alpha+\beta}{2}$ is
$\begin{aligned} y & =\left(\frac{\alpha+\beta}{2}-\alpha\right)\left(\frac{\alpha+\beta}{2}-\beta\right) \\ & =\frac{1}{4}(\beta-\alpha)(\alpha-\beta)=-\frac{1}{4}(\alpha-\beta)^2\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
The minimum velocity (in $\mathrm{ms}^{-1}$ ) with which a car driver must traverse a flat curve of radius $150 \mathrm{~m}$ and coefficient of friction $0.6$ to avoid skidding is

Options:
A
60

B
30

C
15

D
25

The correct answer is:
30

Solution:
The condition to avoid skidding, $\mathrm{v}=\sqrt{\mu \mathrm{rg}}=\sqrt{0.6 \times 150 \times 10}=30 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2002
The minimum velocity v with which charge q should be projected so that it manages to reach the centre of the ring starting from the position shown in figure is

Options:
A v=kQqmR2-2
B v=kQq2mR2-2
C v=kQqmR
D v=kQq2mR1-2

The correct answer is:
v=kQqmR2-2

Solution:
Ui+Ki=Uf+Kf
kQq2R+12mv2=kQqR+O
12mv2=kQqR1-12
v=2kQqmR1-12
v=kQqmR2-2

Topic: Physics, Electrostatics, JEE Main
The minimum voltage required to electrolyse alumina in the Hall-Heroult process is

Given, Δ G f ∘ Al 2 O 3 = - 1520 kJ mol -1

Δ G f ∘ CO 2 = - 394 kJ mol -1

Options:
A 1.575 V
B 1.60 V
C 1.312 V
D - 2.62 V

The correct answer is:
1.60 V

Solution:
2 Al 2 O 3 s + 3 C → 1 2 e - 4 Al s + 3 CO 2 g
Δ G â = 3 Δ G f â CO 2 - 2 Δ G â Al 2 O 3
= - 3 × 3 9 4 - 2 - 1 5 2 0
= - 1 1 8 2 + 3 0 4 0 = + 1858 kJ
Δ G â = - nFE â
1 8 5 8 × 1 0 0 0 = - 1 2 × 9 6 5 0 0 × E â
∴ E â = - 1.60 V
Thus, voltage requires = 1.60 V

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ( Ksp of PbCl2=3.2×10-8 ;atomic mass of Pb=207 u ) is:

Options:
A 1.798 L
B 0.36 L
C 17.98 L
D 0.18 L

The correct answer is:
0.18 L

Solution:
(Ksp)PbCl2=32×10-9
PbCl2=Pb2+ s +2Cl- 2s
Ksp= Pb2+ 2Cl-2
Ksp=4s3 = 32×10-9
s3 = 8×10-9
s=2×10-3M
wM.w×1VL=2×10-3
0.1278×1VL=2×10-3
VL=0.1×1000278×2=0.18 L

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2018 (15 Apr)
The minimum volume of water required to dissolve $0.1 \mathrm{~g}$ lead (II) chloride to get a saturated solution $\left(K_{\mathrm{SP}}\right.$ of $\mathrm{PbCl}_2=3.2 \times 10^{-8}$; atomic mass of $\mathrm{Pb}=207 \mathrm{~u})$ is :

Options:
A
$1.798 \mathrm{~L}$

B
$0.36 \mathrm{~L}$

C
$17.95 \mathrm{~L}$

D
$0.18 \mathrm{~L}$

The correct answer is:
$0.18 \mathrm{~L}$

Solution:
$\left(\mathrm{K}_{\mathrm{sp}}\right)_{\mathrm{PbCl}_2}=3.2 \times 10^{-8}=32 \times 10^{-9}$
$\mathrm{PbCl}_2 \rightleftharpoons\underset{\mathrm{s}}{\mathrm{Pb}^{2+}}+\underset{2 \mathrm{s}}{2 \mathrm{Cl}^{-}}$
$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2$
$\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{s}^3=32 \times 10^{-9}$
$\mathrm{~s}^3=8 \times 10^{-9}$
$\mathrm{~s}=2 \times 10^{-3} \mathrm{M}$
$\frac{\mathrm{w}}{\mathrm{M} . \mathrm{W}.} \times \frac{1}{\mathrm{~V}_{\mathrm{L}}}=2 \times 10^{-3}$
$\frac{0.1}{278} \times \frac{1}{\mathrm{~V}_{\mathrm{L}}}=2 \times 10^{-3}$
$\mathrm{~V}_{\mathrm{L}}=\frac{0.1 \times 1000}{278 \times 2}=0.18 \mathrm{~L}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The minors of -4 and 9 and the co-factors of -4 and 9 in determinant $\left|\begin{array}{ccc}-1 & -2 & 3 \\ -4 & -5 & -6 \\ -7 & 8 & 9\end{array}\right|$ are respectively

Options:
A 42,$3 ;-42,3$
B $-42,-3 ; 42,-3$
C 42,$3 ;-42,-3$
D 42,$3 ; 42,3$

The correct answer is:
$-42,-3 ; 42,-3$

Solution:
Minor of $-4=\left|\begin{array}{cc}-2 & 3 \\ 8 & 9\end{array}\right|=-42,\left|\begin{array}{ll}-1 & -2 \\ -4 & -5\end{array}\right|=-3$
and cofactor of $-4=(-1)^{2+1}(-42)=42$,
cofactor of $9=(-1)^{3+3}(-3)=-3$.

Topic: Mathematics, Determinants, JEE Main
The mirror image of the point 1,2,3, in a plane is -73,-43,-13. Which of the following points lies on this plane?

Options:
A 1, 1, 1
B 1,-1, 1
C -1,-1, 1
D -1,-1,-1

The correct answer is:
1,-1, 1

Solution:
Assume, Q be the midpoint of 1,2,3 & -73,-43,-13

So, coordinate of

Q=1-732,2-432,3-132=-23,13,43

The direction ratio of normal to the plane is

1+73, 2+43, 3+13=103,103,103 ⇒1,1,1

So, equation of plane is x+y+z=1.

So, 1,-1,1 lies on this plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The mismatched combinations are

A. Chlorophyll–Co

B. Water hardness–EDTA

C. Photography-Ag(CN)2-

D. Wilkinson catalyst-Ph3P3RhCl

E. Chelating ligand–D-Penicillamine

Choose the correct answer from the options given below.

Options:
A

A, C and E only

B

A and C only

C

A and E only

D

D and E only

The correct answer is:

A and C only

Solution:
The chlorophyll molecule has a central magnesium atom engulfed by a nitrogen-containing structure called a porphyrin ring; attached to the ring is a long carbon-hydrogen side chain.

Hardness of water is determined by titrating with a standard solution of ethylene diamine tetra acetic acid (EDTA) which is a complexing agent. Since EDTA is insoluble in water, the disodium salt of EDTA is taken for this experiment. EDTA can form four or six coordination bonds with a metal ion.

Silver bromide is used in photography as a component of an emulsion that helps develop a photographic image. Silver bromide is sensitive to light, and when suspended in gelatin, silver bromide's grains create a photographic emulsion.

Wilkinson's catalyst is the common name for chlorido­tris(triphenylphosphine)­rhodium(I), a coordination complex of rhodium with the formula Ph3P3RhCl

D-Penicillamine is chelating ligand.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The missing value in the following figure is

Options:

The correct answer is:
4

Solution:
x=(2-1)1!=1

w=(12-8)4!=424

z=(7-4)3!=36

hence y=(5-3)2!=22

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The mixture of chloroxylenol and terpineol is an example of :

Options:
A Antiseptic
B Pesticide
C Disinfectant
D narcotic analgesic

The correct answer is:
Antiseptic

Solution:
Commonly used antiseptic, dettol is a mixture of chloroxylenol and terpineol.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The mode of propagation used by short wave broadcast services having a frequency range from a few MHz up to 30 MHz is

Options:
A ground wave
B space wave
C sky wave
D all of these

The correct answer is:
sky wave

Solution:
Sky wave propagation is used by short wave broadcast services having a frequency range from few MHz up to 30 to 40 MHz

Long-distance communication can be achieved by ionosphere reflection of radio waves back towards the earth as shown in Fig. This mode of propagation is called skywave propagation.

Topic: Physics, Communication System, JEE Main
The mode of the distribution

Options:
A $5$
B $6$
C $8$
D $10$

The correct answer is:
$6$

Solution:
Since frequency is maximum for $6 . \therefore$ Mode $=6$.

Topic: Mathematics, Statistics, JEE Main
The mode of the following items is $0,1,6,7,2,3,7,6,6,2,6,0,5,6,0$

Options:
A $0$
B $5$
C $6$
D $2$

The correct answer is:
$6$

Solution:
Since 6 occurs most times (i.e., 5 times) than any other 15 observations.
$\therefore \text { Mode }=6 \text {. }$

Topic: Mathematics, Statistics, JEE Main
The modulation frequency of an AM radio station is 250kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot?

Options:
A 2750 kHz
B 2000 kHz
C 2250 kHz
D 2900kHz

The correct answer is:
2000 kHz

Solution:
Carrier wave frequency =250×10=2500 kHz
Frequency used 2500±250k Ω=2250 kHz to 2750 kHz
The allotted frequency showed be such that f±250 kHz should not overlap with previous one. Hence possible frequency =2000 kHz

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The modulation index for an A.M. wave having maximum and minimum peak to peak voltages of 14 mV and 6 mV respectively is :

Options:
A 1.4
B 0.4
C 0.2
D 0.6

The correct answer is:
0.4

Solution:
As we know, modulation index is given by,

μ=Amax-AminAmax+Amin

=14-614+6=0.4

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The modulus and amplitude of $\frac{1+2 i}{1-(1-i)^2}$ are

Options:
A $\sqrt{2}$ and $\frac{\pi}{6}$
B $1$ and $0$
C $1$ and $\frac{\pi}{3}$
D 1 and $\frac{\pi}{4}$

The correct answer is:
$1$ and $0$

Solution:
$\frac{1+2 i}{1-(1-i)^2}=\frac{1+2 i}{1-(1-1-2 i)}=\frac{1+2 i}{1+2 i}=1+0 i$
Modulus $=1$
Amplitude $\theta=\tan ^{-1} \frac{0}{1}=0$

Topic: Mathematics, Complex Number, JEE Main
The modulus of $\frac{(1+i \sqrt{3})(2+2 i)}{(\sqrt{3}-i)}$ is

Options:
A
2
B
4
C $3 \sqrt{2}$
D $2 \sqrt{2}$

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

Solution:
$\begin{aligned} & \frac{(1+i \sqrt{3})(2+2 i)}{\sqrt{3}-i}=\frac{2+2 \sqrt{3} i+2 i-2 \sqrt{3}}{\sqrt{3}-i} \\ & =\frac{(2-2 \sqrt{3})+(2 \sqrt{3}+2) i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} \\ & =\frac{2 \sqrt{3}-6+2 i-2 \sqrt{3} i+6 i+2 \sqrt{3} i-2 \sqrt{3}-2}{3+1} \\ & =\frac{8 i-8}{4}=-2+2 i \\ & \therefore \text { Modulus }=\sqrt{(-2)^2+(2)^2}=2 \sqrt{2}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The molal elevation constant is the ratio of the elevation in B.P. to

Options:
A Molarity
B Molality
C Mole fraction of solute
D Mole fraction of solvent

The correct answer is:
Molality

Solution:
$\Delta T_b=K_b \times m$ or $K_b=\Delta T_b / m$

Topic: Chemistry, Solutions, JEE Main
The molal elevation constant of water $=0.52^{\circ} \mathrm{C}$. The boiling point of 1.0 molal aqueous $\mathrm{KCl}$ solution (assuming complete dissociation of $\mathrm{KCl}$ ), therefore, should be

Options:
A $100.52^{\circ} \mathrm{C}$
B $101.04^{\circ} \mathrm{C}$
C $99.48^{\circ} \mathrm{C}$
D $98.96^{\circ} \mathrm{C}$

The correct answer is:
$101.04^{\circ} \mathrm{C}$

Solution:
$\begin{aligned} & \Delta T_b=i m k_b=0.52 \times 1 \times 2=1.04 \\ & T_b=100+1.04=101.04^{\circ} \mathrm{C} .\end{aligned}$

Topic: Chemistry, Solutions, JEE Main
The molality of a 10%(v/V) solution of di-bromine solution in CCl4 (carbon tetrachloride) is x '. x= _____×10-2M. (Nearest integer)

[Given : molar mass of Br2=160 g mol-1

atomic mass of C=12 g mol-1

atomic mass of Cl=35.5 g mol-1

density of dibromine =3.2 g cm-3

density of CCl4=1.6 g cm-3

Options:

The correct answer is:
139

Solution:
10% v/V means 10 ml solute in 90 ml solvent.

mass of soluteBr2 =10×3.2=32g

mass of solventCCl4 =90×1.6 g

molality(m) =mass of solutemolar mass of solute×1000mass of solvent

m=32×1000160×90×1.6=1.388

m=138.8×10-2=139

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The molar conductances of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_3 \mathrm{COONa}$ at infinite dilution are $126.45,426.16$ and $91 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The molar conductance of $\mathrm{CH}_3 \mathrm{COOH}$ at infinite dilution is

Options:
A $201.28 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
B $390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
C $698.28 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
D $540.48 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

The correct answer is:
$390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution:
$\wedge_m^o\left(\mathrm{CH}_3 \mathrm{COOH}\right)=$
$\wedge^0\left(\mathrm{CH}_3 \mathrm{COONa}\right)+\wedge^{\circ}(\mathrm{HCl})-\wedge^{\circ}(\mathrm{NaCl})$
$=91+426.16-126.45=390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Topic: Chemistry, Electrochemistry, JEE Main
The molar conductivities $\wedge_{\mathrm{NaOAC}}^{\circ}$ and $\wedge_{\mathrm{HCl}}^{\circ}$ at infinite dilution in water at $25^{\circ} \mathrm{C}$ are $91.0$ and $426.2 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol}$ respectively. To calculate $\wedge_{\mathrm{HOAc}}^O$, the additional value required is

Options:
A
$\wedge_{\mathrm{H}_2 \mathrm{O}}^{\circ}$

B
$\wedge_{\mathrm{KCl}}^{\circ}$

C
$\wedge_{\mathrm{NaOH}}^{\circ}$

D
$\wedge_{\mathrm{NaCl}}^{\circ}$

The correct answer is:
$\wedge_{\mathrm{NaCl}}^{\circ}$

Solution:
$$
\begin{aligned}
& \lambda_{\mathrm{CH}_3 \mathrm{COONa}}^0=\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0 \\
& \lambda_{\mathrm{HCl}}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{C}}^0 \cdots \cdots \cdots \cdots \cdot \\
& \lambda_{\mathrm{NaCl}}^0=\lambda_{\mathrm{Na}}^0+\lambda_{\mathrm{Cl}}^0 \cdots \cdots \cdots \cdots \\
& \lambda_{\mathrm{CH}_3 \mathrm{COOH}}^0=(1)+(2)-(3)
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2006
The molar conductivities of $\mathrm{HCl}, \mathrm{NaCl}, \mathrm{CH}_{3} \mathrm{COOH}$, and $\mathrm{CH}_{3} \mathrm{COONa}$ at infinite dilution follow the order

Options:
A $\mathrm{HCl}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{NaCl}>\mathrm{CH}_{3} \mathrm{COONa}$
B $\mathrm{CH}_{3} \mathrm{COONa}>\mathrm{HCl}>\mathrm{NaCl}>\mathrm{CH}_{3} \mathrm{COOH}$
C $\mathrm{HCl}>\mathrm{NaCl}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{COONa}$
D $\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{COONa}>\mathrm{HCl}>\mathrm{NaCl}$

The correct answer is:
$\mathrm{HCl}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{NaCl}>\mathrm{CH}_{3} \mathrm{COONa}$

Solution:
$\lambda_{\infty} \quad \mathrm{CH}_{3} \mathrm{COO}^{-}=40.9 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{Na}^{+}=50.10 \times 10^{-4} \quad \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{Cl}^{-} \quad=\quad 76.35 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{H}^{+}=349 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{x} \mathrm{HCl}=\lambda_{u^{t}}+\lambda_{\mathrm{c}^{-}} / \lambda_{\mathrm{Cl}, \mathrm{coon}}=\lambda_{\mathrm{al}, \mathrm{coo}^{-}}+\lambda_{\mathrm{u}^{+}}$
$\lambda_{\mathrm{Ncl}}=\lambda_{\mathrm{N}}+\lambda_{\mathrm{c} \mathrm{r}^{-}} / \lambda_{\mathrm{CH}_{3} \mathrm{com} \mathrm{N}}=\lambda_{\mathrm{CH}_{3} \mathrm{coo}^{-}}+\lambda_{\mathrm{N}^{+}}$
$\lambda_{\text {HC }}=425.35 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\mathrm{cH}_{3} \mathrm{coon}}=389.9 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\text {Nacl }}=126.45 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\mathrm{Cl}, \mathrm{coON}}=91 \mathrm{sm}^{2} / \mathrm{mol}$

Topic: Chemistry, Electrochemistry, JEE Main
The molar conductivity at infinite dilution of barium chloride, sulphuric arid and hydrochloric acid are 280, 860 426 Scm2mol-1 respectively. The molar conductivity at infinite dilution of barium sulphate is ___ Scm2 mol-1 ( Round off to the Nearest Integer).

Options:

The correct answer is:
288

Solution:
From Kohlrausch's law

Λm∞BaSO4=λm∞Ba2++λm∞SO42-

Λm∞BaSO4=Λm∞BaCl2+Λm∞H2SO4-2Λm∞(HCl)

=280+860-2(426)

=288 Scm2 mol-1

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The molar conductivity for electrolytes \(A\) and \(B\) are plotted against \(C^{1 / 2}\) as shown below. Electrolytes \(A\) and \(B\) respectively are :

Options:
A A: strong electrolyte ; B: weak electrolyte
B A: weak electrolyte ; B: weak electrolyte
C A: weak electrolyte ; B: strong electrolyte
D A: strong electrolyte ; B: strong electrolyte

The correct answer is:
A: weak electrolyte ; B: strong electrolyte

Solution:
A $\rightarrow$ Weak electrolyte
B $\rightarrow$ Strong electrolyte

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The molar conductivity is maximum for the solution of concentration

Options:
A $0.001 \mathrm{M}$
B $0.005 \mathrm{M}$
C $0.002 M$
D $0.004 M$

The correct answer is:
$0.001 \mathrm{M}$

Solution:
Since molar conductance $\propto \frac{1}{\text { Molarity }}$

Topic: Chemistry, Electrochemistry, JEE Main
The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is Λm1 and that of 20 moles another identical cell heaving 80 mL NaCl solution is Λm2, The conductivities exhibited by these two cells are same.

The relationship between Λm2 and Λm1 is

Options:
A Λm2=2Λm1
B Λm2=Λm1/2
C Λm2=Λm1
D Λm2=4Λm1

The correct answer is:
Λm2=2Λm1

Solution:
Λm=κ×1000M

⇒Λm∝1M

Λm1Λm2=M2M1=20801020=14×21=12

⇒Λm2=2Λm1

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The molar conductivity of aqueous solution of a weak acid HA is 200  Scm 2 mol −1 and its pH is 4. What is the value of pK a of HA at 25 o C ?
(Given: λ m o NaA  = 100  Scm 2 mo1 −1 ,  λ m o HCl  = 425  Scm 2 mol −1 ,  λ m o (NaCl) = (125  Scm 2 mo1 −1 )λ m o NaA  = 100  Scm 2 mo1 −1 ,  λ m o HCl  = 425  Scm 2 mol −1 ,  λ m o (NaCl) = 125  Scm 2 mo1 −1 ))

Options:

The correct answer is:
4

Solution:
â§ M o (HA)= â§ M â (HCl)+ â§ M â (NaA)â â§ M â (NaCl)
=425+100â125=400â Scm 2 mol â1
pH=4,â [H + ]= 10 â4 =Î±C
Î±= â§m â§ m â = 200 400 =0.5
K a = (Cα)α (1−α) = 10 −4 (0.5) (1−α) = 10 −4 ,  pK a =4

Topic: Chemistry, Electrochemistry, JEE Main
The molar heat capacity C of an ideal gas in a process is given by aT, where a is constant. If the ratio γ=CpCv is known for the gas, then the work done by one mole of gas during heating from T0 to η T0 through the same process will be

Options:
A 1alnη
B alnη-RTOη-1γ-1
C alnη-γ-1RT0
D Î·-1Î³-1RT0

The correct answer is:
alnη-RTOη-1γ-1

Solution:
Q=∫nCdT=∫ToηToaT=alnη
∆U=Rη-1T0γ-1
W=Q-∆U=alnη-RTOη-1γ-1

Topic: Physics, Thermodynamics, JEE Main
The molar heat capacity $\left(\mathrm{C}_{\mathrm{p}}\right)$ of $\mathrm{CD}_2 \mathrm{O}$ is 10 cals at $1000 \mathrm{~K}$. The change in entropy associated with cooling of $32 \mathrm{~g}$ of $\mathrm{CD}_2 \mathrm{O}$ vapour from $1000 \mathrm{~K}$ to $100 \mathrm{~K}$ at constant pressure will be:
$(\mathrm{D}=$ deuterium, atomic mass $=2 \mathrm{u})$

Options:
A
$23.03 \mathrm{cal} \mathrm{deg}^{-1}$

B
$-23.03 \mathrm{cal} \mathrm{deg}^{-1}$

C
$2.303 \mathrm{cal} \mathrm{deg}^{-1}$

D
$-2.303 \mathrm{cal} \mathrm{deg}^{-1}$

The correct answer is:
$-23.03 \mathrm{cal} \mathrm{deg}^{-1}$

Solution:
Given, $C_p=10$ cals at $1000 \mathrm{~K}$
$$
\begin{aligned}
&T_1=1000 \mathrm{~K} . T_2=100 \mathrm{~K} \\
&m=32 \mathrm{~g}
\end{aligned}
$$
$$
\Delta S=?
$$
at constant pressure
$$
\begin{aligned}
&\triangle S=C_p \ln \frac{T_2}{T_1} \\
&=2.303 \times C_p \log \frac{T_2}{T_1} \\
&=2.303 \times 10 \log \frac{100}{1000} \\
&=-23.03 \mathrm{cal} \mathrm{deg}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2014 (11 Apr Online)
The molar heat capacity for an ideal gas at constant pressure is 20.785 J K-1 mol-1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is____

(Given: R=8.314 J K-1 mol-1)

Options:

The correct answer is:
2

Solution:
Molar specific heat capacity at constant pressure is defined as the amount of heat energy required to raise the temperature of one mole of a substance through 1K or 1°C at constant pressure. It is denoted by Cp,m.

Molar specific heat capacity at constant volume is defined as the amount of heat energy required to raise the temperature of one mole of a substance through 1K or 1°C at constant volume. It is denoted by Cv,m.

Cv,m=1n∆U∆T

Cp,m=Cv,m+R

⇒Cv,m=20.785-8.314=12.471 Jk-1ml-1

ΔU=nCv,mΔT

⇒n=500012.471×200=2512.471≈2

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (27 Jul Shift 1)
The molar heat capacity of a monoatomic ideal gas undergoing the process PV1/2=constant is

Options:
A 152R
B 232R
C 72R
D Zero

The correct answer is:
72R

Solution:
For a polytropic process, specific heat of an ideal gas is given by,
c=RÎ³-1+R1-n n=12 , for this polytropic process
c=R53-1+2R
=3R2+2R=7R2

Topic: Physics, Thermodynamics, JEE Main