datavorous/entrance-exam-dataset huggingface dataset:
Eight chairs are numbered 1 to 8 . Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.

Options:

None

Solution:
Since, there are two women so they have to choose chairs from 1-4 chairs; so number of arrangements $={ }^4 P_2$ ways. Now, 3 men have to choose chairs from remainig 6 chairs so number of arrangements $={ }^6 P_3$ ways.
$\therefore$ Total number of possible arrangement
$$
\begin{aligned}
&={ }^4 P_2 \times{ }^6 P_3=\frac{4 !}{4-2 !} \times \frac{6 !}{6-3 !} \\
&=\frac{4 !}{2 !} \times \frac{6 !}{3 !}=\frac{4 \times 3 \times 2 !}{2 !} \times \frac{6 \times 5 \times 4 \times 3 !}{3 !} \\
&=4 \times 3 \times 6 \times 5 \times 4=1440
\end{aligned}
$$

Topic: Mathematics, Permutation Combination
Eight copper wire of length l and diameter d are joined in parallel to form a single composite conductor of resistance R. If a single copper wire of length 2l have the same resistance R then its diameter will be _____ d.

Options:

The correct answer is:
4

Solution:
Each wire has resistance, r=ρ4lπd2

Eight wire in parallel, then equivalent resistance is

R=r8=ρl2πd2

Single copper wire of length 2l has resistance R=ρ2l×4πd12.

Comparing both equations, we get,

ρ8lπd12=ρl2πd2

⇒d1=4d

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Eight drops of a density \( \rho \) and each of radius \( a \) are falling through air with a constant velocity \( 375 \mathrm{~cm} \mathrm{~s}^{-1} \). When the eight drops coalesce to from a single drop the terminal velocity of the new drop will be

Options:
A \( 2.4 \times 10^{-2} \mathrm{~ms}^{-1} \)
B \( 0.75 \times 10^{-2} \mathrm{~ms}^{-1} \)
C \( 25 \times 10^{-2} \mathrm{~ms}^{-1} \)
D \( 15 \times 10^{-2} \mathrm{~ms}^{-1} \)

The correct answer is:
\( 15 \times 10^{-2} \mathrm{~ms}^{-1} \)

Solution:
Terminal velocity of single drop,
v=3.75cms-1=3.75×10-2 ms-1
Terminal velocity of the big drop
V=n23v
=4×3.75×10-2=15×10-2ms-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Eight drops of mercury, each of same radius and same charge combine to form a bigger drop. The ratio of capacitance of the bigger drop to that of each smaller drop is

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

The correct answer is:
$2: 1$

Solution:
Number of drops, $\mathrm{n}=8$
Let, $\mathrm{r}$ is radius of smaller drop and $\mathrm{R}$ is the radius of Bigger drop.
Volume of smaller drops \& bigger drop remains same
$$
8 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi r^3 \Rightarrow R=2 r
$$
Capacitance of spherical conductor, $C=4 \pi \epsilon_0 R$ Ratio of capacitance of bigger drop to smaller drop is given as
$$
\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=\frac{4 \pi \epsilon_0 \mathrm{R}}{4 \pi \epsilon_0 \mathrm{r}}=\frac{\mathrm{R}}{\mathrm{r}}=\frac{2 \mathrm{r}}{4}=2: 1
$$
$\frac{C}{C^{\prime}}=\frac{2}{1}$

Topic: Physics, Capacitance, JEE Main
Eight drops of mercury, each of same radius and same charge combine to form a bigger drop. The ratio of capacitance of the bigger drop to that of each smaller drop is

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

The correct answer is:
$2: 1$

Solution:
Number of drops, $\mathrm{n}=8$
Let, $\mathrm{r}$ is radius of smaller drop and $\mathrm{R}$ is the radius of Bigger drop.
Volume of smaller drops \& bigger drop remains same
$$
8 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi r^3 \Rightarrow R=2 r
$$
Capacitance of spherical conductor, $C=4 \pi \epsilon_0 R$ Ratio of capacitance of bigger drop to smaller drop is given as
$$
\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=\frac{4 \pi \epsilon_0 \mathrm{R}}{4 \pi \epsilon_0 \mathrm{r}}=\frac{\mathrm{R}}{\mathrm{r}}=\frac{2 \mathrm{r}}{4}=2: 1
$$
$\frac{C}{C^{\prime}}=\frac{2}{1}$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
Eight equal drops of water are falling through air with a steady speed of 10 cm s-1. If the drops coalesce, the new velocity is:-

Options:
A 16 cm s-1
B 40 cm s-1
C 5 cm s-1
D 10 cm s-1

The correct answer is:
40 cm s-1

Solution:
The terminal velocity can be given by

vT=2(ρ-σ)r2g9η.

Clearly, vT∝r2

Let the radius of the small drops be r1 and the radius of the big drop be r2.

So,

84π3r13=4π3r23⇒r1=r22

Hence, taking the ratio
vT1vT2=r1r22
⇒10vT2=14
⇒vT2=40 cm s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways, in which they can be transported, is

Options:
A 1120
B 3360
C 1680
D 560

The correct answer is:
1680

Solution:
Let us find the different ways so that 8 persons can travel in 3 cars.

C1C2C3332233323

Hence we have 3 ways.

Now the number of ways to distribute 8 persons such that they can travel in 3 cars with any car carrying maximum of 3 persons is 8!3!3!2!×3

=1680

Therefore, the required answer is 1680

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Eight point charges of charge q each are placed on the eight corners of a cube of side a. A solid neutral metallic sphere of radius a3 is placed with its centre at the centre of the cube. As a result, charges are induced on the sphere, which form a certain pattern on its surface. The electric potential of the sphere is

Options:
A –12 3 Kq a
B 18 4 Kq a
C - 8 2 Kq a
D 16 3 Kq a

The correct answer is:
16 3 Kq a

Solution:
Net induced charge on sphere is zero so its potential at its own centre will be zero.
V0 = potential due to charges placed at eight corners
V0=8Kq32a⇒V0=16Kq3 a

Topic: Physics, Electrostatics, JEE Main
Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V, when the radiation of 250 nm is used. The work function of the metal is

Options:
A 4 eV
B 5.5 eV
C 4.5 eV
D 5 eV

The correct answer is:
4.5 eV

Solution:
λ=250 nm=2500 A
E=hcλ=124002500=4.96 eV
KE=e×stopping potential=0.5 eV
E=W0+KE
4.96=Wo + 0.5
W0=4.46≈4.5 eV

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2018 (15 Apr)
Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5 \mathrm{~V}$ when the radiation of $250 \mathrm{~nm}$ is used. The work function of the metal is :

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

B
$5.5 \mathrm{~eV}$

C
$4.5 \mathrm{~eV}$

D
$5 \mathrm{~eV}$

The correct answer is:
$4.5 \mathrm{~eV}$

Solution:
$\lambda=250 \mathrm{~nm}=2500 Å$
$\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{12400}{2500}=4.96 \mathrm{~eV}$
$\mathrm{KE}=$ stopping potential $=0.5 \mathrm{~eV}$
$\mathrm{E}=\mathrm{W}_0+\mathrm{K} . \mathrm{E}$.
$4.96=\mathrm{W}_0+0.5$
$\mathrm{~W}_0=4.46 \approx 4.5 \mathrm{~eV}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Elbow joint is an example of :

Options:
A Ball and socket joint
B Pivot joint
C Hinge joint
D Gliding joint

The correct answer is:
Hinge joint

Solution:
No Solution Available

Topic: Biology, Locomotion and Movement, NEET, NEET 2009 (Mains)
Electric current (I) is measured by galvanometer, the current being proportional to the tangent of the angle $(\theta)$ of deflection. If the deflection is read as $45^{\circ}$ and an error of $1 \%$ is made in reading it, the percentage error in the current is

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

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

Solution:
$\begin{aligned} & \quad I \propto \tan \theta \\ & d I=\sec ^2 \theta d \theta \\ & \frac{\Delta I}{I} \times 100=\sec ^2 \theta \frac{\Delta \theta}{\theta} \times 100 \\ & \text { when } \theta=\frac{\pi}{4} \\ & \frac{\Delta \theta}{\theta} \times 100=\frac{\pi}{4} \times \frac{1}{100} \times 100=\frac{\pi}{4} \\ & \Rightarrow \frac{\Delta I}{I} \times 100=\left[\sec ^2\left(\frac{\pi}{4}\right)\right] \times \frac{\pi}{4}=2 \times \frac{\pi}{4}=\frac{\pi}{2}\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
Electric displacement is given by D=ÎµE, here, Îµ=electric permittivity, E=electric field strength
The dimensions of electric displacement are

Options:
A [ML-2TA]
B [L-2T-1A]
C [L-2TA]
D None of these

The correct answer is:
[L-2TA]

Solution:
Electric displacement, D=εE
Unit of D=C2Nm2NC
∴ D=Cm2=[AT][L2]=[L-2TA]

Topic: Physics, Units and Dimensions, JEE Main
Electric field in a certain region is given by E→=Ax2i^+By3j^. The SI unit of A and B are :

Options:
A N m3 C-1; N m2 C-1
B N m2 C-1; N m3 C-1
C N m3 C; N m2 C
D N m2 C; N m3 C

The correct answer is:
N m2 C-1; N m3 C-1

Solution:
E→=Ax2i^+By3j^

Here, the units of Ax2 and By3 is equal to unit of electric field.

The unit of electric field is N C-1.

So, the SI unit of A is Ax2=N C-1⇒A=N m2 C-1

And the SI unit of B is By3=N C-1⇒B=N m3 C-1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Electric field in a certain region is given by $\vec{E}=\left(\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}\right)$. The SI unit of $A$ and $B$ are:

Options:
A $\mathrm{Nm}^3 \mathrm{C}^{-1} ; \mathrm{Nm}^2 \mathrm{C}^{-1}$
B $\mathrm{Nm}^2 \mathrm{C}^{-1} ; \mathrm{Nm}^3 \mathrm{C}^{-1}$
C $\mathrm{Nm}^3 \mathrm{C} ; \mathrm{Nm}^2 \mathrm{C}$
D $\mathrm{Nm}^2 \mathrm{C} ; \mathrm{Nm}^3 \mathrm{C}$

The correct answer is:
$\mathrm{Nm}^2 \mathrm{C}^{-1} ; \mathrm{Nm}^3 \mathrm{C}^{-1}$

Solution:
Electric field in a certain region is given by,
$\vec{E}=\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}$
$\left[\frac{\mathrm{A}}{\mathrm{x}^2}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{A}]=\mathrm{Nm}^2 \mathrm{C}^{-1}$
$\left[\frac{\mathrm{B}}{\mathrm{y}^3}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{B}]=\mathrm{Nm}^3 \mathrm{C}^{-1}$

Topic: Physics, Units and Dimensions, JEE Main
Electric field lines
from positive charge and
in negative charge

Options:
A start; start
B start; end
C end; end
D end; start

The correct answer is:
start; end

Solution:
Electric field lines starts from positive charge and end at negative charge.
Generally it is considered that an electric field is described by the closeness of electric field lines.
Since the electrostatic field is always directed away from positive charge and end towards negative charges, so the field lines must go away from positive charges to negative ones. Electric field lines never form closed loop.

Topic: Physics, Electrostatics, JEE Main
Electric field of a plane electromagnetic wave propagating through a non-magnetic medium is given by E=20cos2×1010t-200x V m-1. The dielectric constant of the medium is equal to:
(Take μr=1 )

Options:
A 2
B 13
C 9
D 3

The correct answer is:
9

Solution:
Equate the wave equation with the equation of EM wave, E=E0cosωt-kx V m-1 here the given equation of EM wave, E=20cos2×1010t-200x V m-1⇒v=ωk=2×1010200=108 m s-1
The refractive index,⇒n=cv=3×108108⇒3=μrεr μr=1given)
εr=9

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Electric lines of force about negative point charge are

Options:
A Circular, anticlockwise
B Circular, clockwise
C Radial, inward
D Radial, outward

The correct answer is:
Radial, inward

Solution:
Electric lines force due to negative charge are radially inward

Topic: Physics, Electrostatics, JEE Main
Electric potential at a point P due to a point charge of 5×10-9 C is 50 V. The distance of P from the point charge is:
(Assume, 14πε0=9×109 N m2 C-2)

Options:
A 9 cm
B 3 cm
C 0.9 cm
D 90 cm

The correct answer is:
90 cm

Solution:
The electric potential for a point charge is given by

V=Q4πε0r

The data given is

V=50 VQ=5×10-9 C

So, the distance of the point is

r=14πε0QV
⇒r=9×109×5×10-950
⇒r=0.9 m=90 cm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Electric power is transmitted over long distances through conducting wires at high voltage because

Options:
A High voltage travels faster
B Power loss is large
C Power loss is less
D Generator produced electrical energy at a very high voltage

The correct answer is:
Power loss is less

Solution:
Power loss in transmission $P_L=\frac{P^2 R}{V^2} \Rightarrow P_L \propto \frac{1}{V^2}$

Topic: Physics, Current Electricity, JEE Main
Electrode potential data are given below :
$\begin{aligned} & \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{-1}(\mathrm{aq}) ; E^{\circ}=+0.77 \mathrm{~V} \\ & \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) ; E^{\circ}=-1.66 \mathrm{~V} \\ & \mathrm{Br}_2(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) ; E^{\circ}=+1.08 \mathrm{~V}\end{aligned}$
Based on the data given above, reducing power of $\mathrm{Fe}^{2+}, \mathrm{Al}$ and $\mathrm{Br}^{-}$will increase in the order

Options:
A $\mathrm{Br}^{-} \lt \mathrm{Fe}^{2+} \lt \mathrm{Al}$
B $\mathrm{Fe}^{2+} \lt \mathrm{Al} \lt \mathrm{Br}^{-}$
C $\mathrm{Al} \lt \mathrm{Br}^{-} \lt \mathrm{Fe}^{2+}$
D $\mathrm{Al} \lt \mathrm{Fe}^{2+} \lt \mathrm{Br}^{-}$

The correct answer is:
$\mathrm{Br}^{-} \lt \mathrm{Fe}^{2+} \lt \mathrm{Al}$

Solution:
Reducing power, i.e. the tendency to lose electrons increases as the reduction potential decreases.

Topic: Chemistry, Electrochemistry, JEE Main
Electrode potential data are given below:
$\mathrm{Fe}^{+3}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{+2}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{~V}$
$\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{(\mathrm{s})} ; \mathrm{E}^{\circ}=-1.66 \mathrm{~V}$
$\mathrm{Br}_2(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+1.08 \mathrm{~V}$
Based on the data, the reducing power of $\mathrm{Fe}^{2+}$, $\mathrm{Al}$ and $\mathrm{Br}^{-}$will increase in the order

Options:
A $\mathrm{Br}^{-} < \mathrm{Fe}^{2+} < \mathrm{Al}$
B $\mathrm{Fe}^{2+} < \mathrm{Al} < \mathrm{Br}^{-}$
C $\mathrm{Al} < \mathrm{Br}^{-} < \mathrm{Fe}^{2+}$
D $\mathrm{Al} < \mathrm{Fe}^{2+} < \mathrm{Br}^{-}$

The correct answer is:
$\mathrm{Br}^{-} < \mathrm{Fe}^{2+} < \mathrm{Al}$

Solution:
$\begin{array}{rrrr} & \text { Fe } & \text { A1 } & \text { B } \\ E_{\text {Red }}^{\circ} & 0.77 & -1.66 & 1.08 \\ E_{\text {Oxi }}^{\circ} & -0.77 & 1.66 & -1.08\end{array}$
Hence, reducing power $\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Br}$

Topic: Chemistry, Electrochemistry, JEE Main
Electrode potential data given below
Cl2+2H2O→2ClO-+4H++2e-;E°=-1.61volt
ClO-+2H2O→ClO3-+4H++4e-;E°=-0.50volt
Based on these data which is the spontaneous reaction.

Options:
A Cl2+ClO-+ClO3-
B ClO-→Cl2+ClO3-
C ClO3-→Cl2+ClO-
D ClO-+Cl2→ClO3-

The correct answer is:
ClO-→Cl2+ClO3-

Solution:
Given electron potential data is-
Cl2+2H2O→2ClO–+4H++2e–;E°=-1.61 Volt
ClO–+2H2O→ClO3–+4H++4e–;E°=-0.50 Volt
For reaction to be spontaneous E0 reaction should +ve , Thus substracting equation 1 from equation 2.

Topic: Chemistry, Electrochemistry, JEE Main
Electrode potentials $\left(\mathrm{E}^{\circ}\right)$ are given below :
$\begin{aligned}
& \mathrm{Cu}^{+} / \mathrm{Cu}=+0.52 \mathrm{~V} ,\\
& \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V}, \\
& \frac{1}{2} \mathrm{I}_2(\mathrm{~s}) / \mathrm{I}^{-}=+0.54 \mathrm{~V}, \\
& \mathrm{Ag}^{+} / \mathrm{Ag}=+0.88 \mathrm{~V}.
\end{aligned}$
Based on the above potentials, strongest oxidizing agent will be :

Options:
A
$\mathrm{Cu}^{+}$

B
$\mathrm{Fe}^{3+}$

C
$\mathrm{Ag}^{+}$

D
$\mathrm{I}_2$

The correct answer is:
$\mathrm{Ag}^{+}$

Solution:
Higher the value of reduction potential stronger will be the oxidising hence based on the given values $\mathrm{Ag}^{+}$will be strongest oxidizing agent.

Topic: Chemistry, Electrochemistry, JEE Main
Electrode potentials $\left(\mathrm{E}^{\circ}\right)$ are given below :
$\begin{aligned}
& \mathrm{Cu}^{+} / \mathrm{Cu}=+0.52 \mathrm{~V} ,\\
& \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V}, \\
& \frac{1}{2} \mathrm{I}_2(\mathrm{~s}) / \mathrm{I}^{-}=+0.54 \mathrm{~V}, \\
& \mathrm{Ag}^{+} / \mathrm{Ag}=+0.88 \mathrm{~V}.
\end{aligned}$
Based on the above potentials, strongest oxidizing agent will be :

Options:
A
$\mathrm{Cu}^{+}$

B
$\mathrm{Fe}^{3+}$

C
$\mathrm{Ag}^{+}$

D
$\mathrm{I}_2$

The correct answer is:
$\mathrm{Ag}^{+}$

Solution:
Higher the value of reduction potential stronger will be the oxidising hence based on the given values $\mathrm{Ag}^{+}$will be strongest oxidizing agent.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2013 (09 Apr Online)
Electrolysis of a solution of HSO 4 - ions produces S 2 O 8 2â . Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mol of S 2 O 8 2â per hour ?

Options:
A 71.50 A
B 35.70 A
C 142.96 A
D 285.93 A

The correct answer is:
71.50 A

Solution:
2 HSO 4 â â S 2 O 8 2â +2 H + +2 e â â
So, required rate = 1 mol/hr of S 2 O 8 2− = 2 moles of HSO 4 â
= 2 Ã 9 6 5 0 0 C 3 6 0 0 s = 2 Ã 9 6 5 3 6 A â 53.6 A
So required current = 4 3 Ã 53.6 A = 71.47 A

Topic: Chemistry, Electrochemistry, JEE Main
Electrolysis of aqueous $\mathrm{HCl}$ solution produces

Options:
A $\mathrm{H}_2$ gas at the anode
B $\mathrm{H}_2$ gas at the cathode
C $\mathrm{Cl}_2$ gas at the cathode
D $\mathrm{Cl}_2$ and $\mathrm{O}_2$ gases both at the anode

The correct answer is:
$\mathrm{H}_2$ gas at the cathode

Solution:
$2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2(g)}$ at cathode.

Topic: Chemistry, Electrochemistry, JEE Main
Electrolysis of concentrated aqueous solution of NaCl results in

Options:
A increase in pH of the solution
B decrease in pH of the solution
C O2 liberation at the cathode
D H2 liberation at the anode

The correct answer is:
increase in pH of the solution

Solution:
The cathode half-reaction in the process is 2H2O+2e-⟶H2+2OH-

which increases OH- and, therefore pH and the anode half-reaction.

2Cl-⟶Cl2+2e- does not produce or
consume H+ or OH-.

Overall, the reaction can be written as 2NaCl+2H2O⟶2NaOH+H2+Cl2

Topic: Chemistry, Electrochemistry, JEE Main
Electrolysis of dilute aq. $\mathrm{NaCl}$ solution was carried out by passing $10 \mathrm{~mA}$ current. The time required to liberate $0.01$ mole of $\mathrm{H}_2$ gas at the cathode is (1 Faraday $=96500 \mathrm{C} \mathrm{mol}^{-1}$ )

Options:
A
$9.65 \times 10^4 \mathrm{~s}$

B
$19.3 \times 10^4 \mathrm{~s}$

C
$28.95 \times 10^4 \mathrm{~s}$

D
$38.6 \times 10^4 \mathrm{~s}$

The correct answer is:
$19.3 \times 10^4 \mathrm{~s}$

Solution:
$2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
For $0.01$ mole of $\mathrm{H}_2, 0.02$ mole of electrons are consumed Charge required $=0.02 \times 96500 \mathrm{C}=i \times t$
$$
\text { Time required }=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^4 \mathrm{~s}
$$

Topic: Chemistry, Electrochemistry, JEE Main
Electrolysis of fused sodium hydride liberate hydrogen at the

Options:
A Anode
B Cathode
C Cathode and anode both
D None of these

The correct answer is:
Anode

Solution:
$\mathrm{NaH} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{H}^{-}$
At anode :
$\begin{aligned} & H^{-} \rightarrow H+e^{-} \\ & H+H \rightarrow H_2\end{aligned}$

Topic: Chemistry, Hydrogen, JEE Main
Electrolysis of hot aqueous solution of NaCl gives NaClO4 , i.e., sodium perchlorate,
NaCl+4H2OâNaClO4+4H2â
How many Faraday are required to obtain 1000 g of sodium perchlorate? Report your answer by rounding it up to nearest whole number.

Options:

The correct answer is:
66

Solution:
Equivalent mass of NaClO4=Molecular massChange in oxidation state of chloride
=122.5+7-(-1)=15.13
Number of Faraday = Number of equivalent of NaClO4 formed
=100015.13=66

Topic: Chemistry, Electrochemistry, JEE Main
Electrolysis of molten sodium chloride leads to the formation of

Options:
A $\mathrm{Na}$ and $\mathrm{H}_2$
B $\mathrm{Na}$ and $\mathrm{O}_2$
C $\mathrm{H}_2$ and $\mathrm{O}_2$
D $\mathrm{Na}$ and $\mathrm{Cl}_2$

The correct answer is:
$\mathrm{Na}$ and $\mathrm{Cl}_2$

Solution:
Electrolysis of molten sodium chloride leads to the formation of $\mathrm{Na}$ metal and $\mathrm{Cl}_2$ gas.
\(\underset{\text{Molten}}{2 \mathrm{NaCl}} \xrightarrow{\text{Electrolysis}} \underset{\text{Cathode}}{2 \mathrm{Na}}+\underset{\text{Anode}}{\mathrm{Cl}_2}\)

Topic: Chemistry, s Block Elements, JEE Main
Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to

Options:
A $m \propto I^2$
B $m \propto Q$
C $m \propto Q^2$
D None of these

The correct answer is:
$m \propto Q$

Solution:
It is Faraday's law.

Topic: Chemistry, Electrochemistry, JEE Main
Electrolytes when dissolved in water dissociates into ions because

Options:
A They are unstable
B The water dissolves it
C The force of repulsion increases
D The forces of electrostatic attraction are broken down by water

The correct answer is:
The forces of electrostatic attraction are broken down by water

Solution:
- Electrolytes are ionic compounds that can conduct electricity when dissolved in water. This is because they dissociate into their constituent ions in water.
- Water is a polar molecule, meaning it has a positive and a negative end due to the uneven sharing of electrons between oxygen and hydrogen atoms. The positive end of the water molecule is attracted to the negative ions (anions) of the electrolyte, and the negative end of the water molecule is attracted to the positive ions (cations) of the electrolyte.
- These attractions between the water molecules and the ions of the electrolyte disrupt the electrostatic forces of attraction between the ions themselves. This weakening of the forces holding the ions together allows them to separate and move freely in solution, which is what makes the solution conductive.
So, the key point is that the polarity of water molecules disrupts the electrostatic attraction between ions in an electrolyte, causing them to dissociate.

Topic: Chemistry, Electrochemistry, JEE Main
Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conduction

Options:
A The resistance increases with increasing temperature
B The resistance decreases with increasing temperature
C The flow of current does not generate heat
D The resistance is independent of the length of the conductor

The correct answer is:
The resistance decreases with increasing temperature

Solution:
Electrolytic conduction resistance decreases with increasing temperature.

Topic: Chemistry, Electrochemistry, JEE Main
Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out

Options:
A in the presence of NaCl
B in presence of fluorite
C in presence of cryolite which forms a melt with lower melting temperature
D in the presence of cryolite which forms a melt with higher melting temperature

The correct answer is:
in presence of cryolite which forms a melt with lower melting temperature

Solution:
Cryolite is used to lower the meeting point of alumina (Al2O3) as well as to make it good conductor of electric current.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Electrolytic reduction of nitrobenzene in weakly acidic medium gives:

Options:
A N-phenylhydroxylamine
B nitrosobenzene
C aniline
D p-hydroxyaniline

The correct answer is:
aniline

Solution:

Topic: Chemistry, Amines, JEE Main
Electromagnetic radiation of frequency $n$, wavelength $\lambda$, travelling with velocity $v$ in air, enters a glass slab of refractive index $\mu$. The frequency, wavelength and velocity of light in the glass slab will be respectively

Options:
A $\frac{n}{\mu}, \frac{\lambda}{\mu}, \frac{v}{\mu}$
B $\mathrm{n}, \frac{\lambda}{\mu}, \frac{v}{\mu}$
C $n, \lambda, \frac{v}{\mu}$
D $\frac{\mathrm{n}}{\mu^{\prime}}, \frac{\lambda}{\mu,} v$

The correct answer is:
$\mathrm{n}, \frac{\lambda}{\mu}, \frac{v}{\mu}$

Solution:
The correct option is $\mathrm{2~} n, \frac{\lambda}{\mu^{\prime}} \frac{v}{\mu}$
In going from one medium to another, frequency remains the same but wavelength and velocity decreases with the increase in refractive index, i.e., wavelength becomes $\lambda^{\prime}=\frac{\lambda}{\mu}$ and velocity becomes $v=\frac{v}{\mu}$

Topic: Physics, Ray Optics, JEE Main