datavorous/entrance-exam-dataset huggingface dataset:
The equilibrium constants \(K_{P 1}\) and \(K_{P 2}\) for the reaction \(X \rightleftharpoons 2 Y\) and \(Z \rightleftharpoons P+Q\), respectively are in the ratio of \(1:\) 9. If the degree of dissociation of \(X\) and \(Z\) be equal then the ratio of total pressure at these equilibria is:

Options:
A 1:36
B 1:1
C 1:3
D 1:9

The correct answer is:
1:36

Solution:
X( g )â2Y( g )
1 0
At equilibrium 1âx 2x
Total no. of moles = 1+x
K p 1 = 2x 1+x .p 1 2 1âx 1+x .p 1 = 4x 2 1âx p 1 1+x = 4x 2 p 1 1â x 2
Z( g )âP( g )+Q( g )
1 0 0
At equilibrium 1âx x x
Similarly, K p 2 = x 2 1âx p 2 1+x = x 2 p 2 1â x 2
â´ K p1 K p2 = 1 9
So, 4 ×  p 1 p 2  =  1 9
⇒ p 1 p 2  =  1 36

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium Cu 0 + Cu II ⇌ 2Cu I

In aqueous medium at 25oC shifts towards the left in the presence of

(1) Cl- (2) CN- (3) SCN- (4) NO3-

Options:
A 1,2
B 2,3
C 1,2,3
D 2,3,4

The correct answer is:
1,2,3

Solution:
Cu2+ ions will react with CN- and SCN- forming [Cu(CN)4]3- and [Cu(SCN)4]3- leading the reaction in the backward direction.
Cu2++2CN-→CuCN2
2Cu(CN)2→2CuCN+CN2
CuCN+3CN-→[CuCN)43-
Cu2++4SCN-→[CuSCN)43-
Cu2+ also combines with CuCl2 which reacts with Cu to produce CuCl pushing the reaction in the backward direction.
CuCl2+Cu→2CuCl↓

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightleftharpoons 2 \mathrm{CrO}_4^{2-}\) is shifted to the right in :

Options:
A an acidic medium
B a basic medium
C a neutral medium
D a weakly acidic medium

The correct answer is:
a basic medium

Solution:
$\mathrm{Cr}_2 \mathrm{O}_7^{2-} \stackrel{\mathrm{OH}^{-}}{\rightleftharpoons} 2 \mathrm{HrO}_4^{2-}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2024 (08 Apr Shift 2)
The equilibrium, NH4HS(s)⇌NH3(g)+H2S(g) , is followed to set-up at 127°C in a closed vessel. The total pressure at equilibrium was 20 atm. The KC for the reaction is

Options:
A 0.092 M2
B 0.085 M2
C 3.045 M2
D None of these

The correct answer is:
0.092 M2

Solution:
pNH3=pH2S=202=10 atm
KP=pNH3ÃpH2S
=100 atm2
KP=KC(RT)În
100=KC(0.0821Ã400)2
KC=0.092 M2 .

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium SO2Cl2g ⇌SO2g +Cl2g is attained at 250C in a closed container and an inert gas, helium is introduced. Which of the following statement is correct?

Options:
A More chlorine is formed
B Concentration of SO2 is reduced
C More SO2Cl2 is formed
D Concentration of SO2, Cl2, SO2 and Cl2 do not change

The correct answer is:
Concentration of SO2, Cl2, SO2 and Cl2 do not change

Solution:
According to lechatelier principle there will be no Change in concentration of the species at equilibrium if any inert gas is introduced at constant volume, only pressure is increased when some inert gas is introduced

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium which remains uneffected by change in pressure of the reactants is

Options:
A $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$
B $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
C $2 O_{3}(g) \rightleftharpoons 3 O_{2}(g)$
D $2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_{4}(g)$

The correct answer is:
$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$

Solution:
The following equilibria remains unaffected by pressure change :
$\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equivalent capacitance between $A$ and $B$ in the circuit given below is:

Options:
A
$4.9 \mu \mathrm{F}$

B
$3.6 \mu \mathrm{F}$

C
$5.4 \mu \mathrm{F}$

D
$2.4 \mu \mathrm{F}$

The correct answer is:
$2.4 \mu \mathrm{F}$

Solution:
The simplified circuit of the circuit given in question as follows:
The equivalent capacitance between C \& D capacitors of $2 \mu \mathrm{F}, 5 \mu \mathrm{F}$ and $5 \mu \mathrm{F}$ are in parallel.
$\therefore \quad \mathrm{C}_{\mathrm{CD}}=2+5+5=12 \mu \mathrm{F}(\because$ In parallel grouping $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2+\ldots .+\mathrm{C}_{\mathrm{n}}$ )
Similarly equivalent capacitance between $\mathrm{E}$ $\& \mathrm{BC}_{\mathrm{EB}}$ $=4+2=6 \mu \mathrm{F}$
Now equivalent capacitance between A \& B $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}$
$\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{12}{5}=2.4 \mu \mathrm{F}(\because$ In series grouping, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\ldots \ldots \ldots+\frac{1}{\mathrm{C}_{\mathrm{n}}}$ )

Topic: Physics, Capacitance, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The equivalent capacitance between A and B in the circuit is given below.

Options:
A 3.6 μF
B 2.4 μF
C 4.9 μF
D 5.4 μF

The correct answer is:
2.4 μF

Solution:
Simplified circuit,
⇒1Ceq = 16+112+16=512 ⇒Ceq= 125=2.4 μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2018 (15 Apr)
The equivalent capacitance between A and B, where two concentric spherical shells having radius a and b are connected as shown in the figure.

Options:
A 4ÏÎµ0abb-a+4ÏÎµ0b
B 4πε0b
C 4πε0abb-a
D none

The correct answer is:
4πε0abb-a+4πε0b

Solution:
Between the inner shell and outer shell, the capacitance of the system is

CAB=4πε0abb-a

Between the outer shell and infinity

CB∞=4πε0b

⇒Ctotal=4πε0abb-a+4πε0b

Topic: Physics, Electrostatics, JEE Main
The equivalent capacitance of the combination shown in the figure is:

Options:
A 3C
B C2
C 2C
D 3C/2

The correct answer is:
2C

Solution:
Redrawing the circuit between A & B,

Capacitor 3 can be eliminated

Capacitor 1 & 2 are in parallel.

CAB=C1+C2

=C+C

=2C

Topic: Physics, Capacitance, JEE Main
The equivalent capacitance of the combination shown is

Options:
A 2 C
B 53 C
C C2
D 4 C

The correct answer is:
2 C

Solution:
The potential difference across the branches EF & GH is zero. So, the equivalent capacitance will be only due to the combination of two capacitors in parallel.

Ceq=C+C

⇒Ceq=2 C

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The equivalent conductance of 1 M benzoic acid is 12.8ohm-1cm2 and if the conductance of benzoate ion and H+ ion at infinite dilution are 42 and 288.42âohm-1cm2 respectively. Calculate percentage of degree of dissociation.
Report your answer by rounding upto nearest whole number.

Options:

The correct answer is:
4

Solution:
Λm (C6H5COOH)o=Λ (C6H5COO-)o+Λ(H+)o
=42+288.42=330.42
α=ΛmcΛmo=12.8330.42=3.9%

Topic: Chemistry, Electrochemistry, JEE Main
The equivalent conductance of $\frac{M}{32}$ solution of a weak monobasic acid is 8.0 mho $\mathrm{cm}^2$ and at infinite dilution is 400 mho $\mathrm{cm}^2$, The dissociation constant of this acid is -

Options:
A $1.25 \times 10^{-4}$
B $1.25 \times 10^{-5}$
C $1.25 \times 10^{-6}$
D $6.25 \times 10^{-4}$

The correct answer is:
$1.25 \times 10^{-5}$

Solution:
$$
\begin{aligned}
\alpha= & \frac{\mathrm{A}_{\mathrm{C}}}{\mathrm{A}_{\infty}}=\frac{8}{400} \\
\mathrm{~K}_{\mathrm{a}} & =\mathrm{C}^2 \\
& =\frac{1}{32} \times \frac{8}{400} \times \frac{8}{400} \\
& =1.25 \times 10^{-5}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2009 (Mains)
The equivalent conductance of monobasic acid at infinite dilution is 348 ohm-1cm2eq-1 . If the resistivity of the solution containing 15 g acid (mol. wt 49) in litre is 18.5 ohm cm. What is the % degree of dissociation of acid? Report your answer up to two decimal places without rounding up.

Options:

The correct answer is:
50.74

Solution:
λ N = k×1000 N
= 1000×49×1000 18.5×15×1000
=176.6
α=λvλ∞
=176.6348=0.5074

Topic: Chemistry, Electrochemistry, JEE Main
The equivalent conductance of NaCl at concentration C and at infinite dilution are λ C and λ ∞ , respectively. The correct relationship between λ C and λ ∞ is given as :
(where the constant B is positive)

Options:
A λ C = λ ∞ + B C
B λ C = λ ∞ - B C
C λ C = λ ∞ - B C
D λ C = λ ∞ + B C

The correct answer is:
λ C = λ ∞ - B C

Solution:
By Debye Huckel Equation
Λ m c = Λ m â - A + B Λ m â C
Dividing by Z Factor of Electrolyte
Λ m c z = Λ m â z - A + B Λ m â z C
λ c = λ ∞ - B C

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2014 (06 Apr)
The equivalent conductance of NaCl at concentration C and at infinite dilution are λ C and λ ∞ , respectively. The correct relationship between λ C and λ ∞ is given as :
(where the constant B is positive)

Options:
A λ C = λ ∞ + B C
B λ C = λ ∞ - B C
C Î» C = Î» â - B C
D λ C = λ ∞ + B C

The correct answer is:
λ C = λ ∞ - B C

Solution:
By Debye Huckel Equation
⇒ λ c  =  λ ∞ − B C

Topic: Chemistry, Electrochemistry, JEE Main
The equivalent conductances of two strong electrolytes at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ (where ions move freely through a solution) at $25^{\circ} \mathrm{C}$ are given below:
$\wedge^{\circ} \mathrm{CH}_3 \mathrm{COONa}=91.0 \mathrm{~S} \mathrm{~cm}^2$ /equiv
$\wedge_{\mathrm{HCl}}^{\circ}=426.2 \mathrm{~S} \mathrm{~cm^2}$ / equiv
What additional information/quantity one needs to calculate $\wedge^{\circ}$ of an aqueous solution of acetic acid?

Options:
A
$\wedge^{\circ}$ of $\mathrm{NaCl}$

B
$\wedge^{\circ}$ of $\mathrm{CH}_3 \mathrm{COOK}$

C
The limiting equivalent conductance of $\mathrm{H}^{+}\left(\wedge_{\mathrm{H}^{+}}^{\circ}\right)$

D
$\wedge^{\circ}$ of chloroacetic acid $\left(\mathrm{C} / \mathrm{CH}_2 \mathrm{COOH}\right)$

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

Solution:
From Kohlrausch's law
$\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\circ}=\Lambda_{\mathrm{CH}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}$
Hence, (A) is the correct answer.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2007
The equivalent conductivity of 0.1 M weak acid is 100 times lesser than that at infinite dilution. The degree of dissociation of weak electrolyte at 0.1 M is -

Options:
A 100
B 10
C 0.01
D 0.001

The correct answer is:
0.01

Solution:
The equivalent conductivity of 0.1 M weak acid is 100 times lesser than that at infinite dilution.
â§v = â§0100
â´ Degree of dissociation Î± = â§vâ§0= â§0100â§0= 0.01

Topic: Chemistry, Electrochemistry, JEE Main
The equivalent resistance between A and B of the network shown in figure:

Options:
A 112R3
B 14 R
C 21 R
D 83 R

The correct answer is:
83 R

Solution:
The given circuit can be redrawn as shown below:

Since the circuit is in balanced wheatstone configuration, R2R=3R6R.

The 9 Ω resistor can be omitted. Now, the circuit has two branches having two resistors connected in series.

Therefore, the resistance of upper branch is R+3R=4R and lower branch is 2R+6R=8R.

The equivalent resistance will therefore be:

Req=4R8R4R+8R=32R212R=83R

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The equivalent resistance between A and B as shown in figure is:

Options:
A 10 kΩ
B 5 kΩ
C 20 kΩ
D 30 kΩ

The correct answer is:
5 kΩ

Solution:
The equivalent circuit of the given configuration can be drawn as follows:

With reference to the above diagram, the equivalent resistance of the given configuration can be calculated as follows:
1R=120 Ω+120 Ω+110 Ω=15 Ω⇒R= 5Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The equivalent resistance between A and B is _______.

.

Options:
A 23Ω
B 12Ω
C 32Ω
D 13Ω

The correct answer is:
23Ω

Solution:
Point D and G are at same potential. Same is the case with point E and F.

Now, resistance between DC and GC are in parallel combination. Similarly, resistance between CE and CF are in parallel combination. The modified circuit will be as shown below considering Req=R1R2R1+R2 for parallel combination.

Therefore, we can write

1Req=12+112+14+16+12

=6+1+3+2+612=32

⇒Req=23Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The equivalent resistance between the points $a$ and $b$ of the electrical network shown in the figure is

Options:
A $6 r$
B $4 r$
C $2 r$
D $r$

The correct answer is:
$r$

Solution:
The equlvalent circuit for the given electrical network is

This is a balanced Wheatstone bridge, so the arm containing $2 r$ not in use. So, we have

So, the equivalent resistance between a and bis
$$
\frac{1}{r^{\prime}}=\frac{1}{2 r}+\frac{1}{2 r} \Rightarrow r^{\prime}=\frac{2 r}{2}=r
$$

Topic: Physics, Current Electricity, JEE Main
The equivalent resistance of resistors connected in series is always

Options:
A Equal to the mean of component resistors
B Less than the lowest of component resistors
C In between the lowest and the highest of component resistors
D Equal to sum of component resistors

The correct answer is:
Equal to sum of component resistors

Solution:
Equivalent resistance of series combination $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3+\ldots \ldots$ Thus equivalent resistance of the resistors connected in series is always equal to the sum of component resistors.

Topic: Physics, Current Electricity, JEE Main
The equivalent resistance of series combination of two resistors is s. When they are connected in parallel, the equivalent resistance is p. If s=np, then the minimum value for n is ______.

(Round off to the Nearest Integer)

Options:

The correct answer is:
4

Solution:
R1+R2=s …1

R1R2R1+R2=p …2

R1R2=sp

R1R2=np2

R1+R2=nR1R2R1+R2

R1+R22R1R2=n

for minimum value of n

R1=R2=R

∴n=(2R)2R2=4

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The equivalent resistance of the circuit shown below between points a and b is :

Options:
A 16 Ω
B 3.2 Ω
C 24 Ω
D 20 Ω

The correct answer is:
3.2 Ω

Solution:
The resistance in ab is parallel to all other resistances. Since all the resistances other than Rb is 4 Ω the potential is zeroVx=Vy as the upper branch will become a case of balanced wheatstone bridge between Vx&Vy. So, the simplified circuit looks like the one below.

Hence, the equivalent resistance is

1Req=18+18+116=516⇒Req=165 Ω=3.2 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The equivalent weight of a certain trivalent element is $20$. Molecular weight of its oxide is

Options:
A $152$
B $56$
C $168$
D $68$

The correct answer is:
$168$

Solution:
Equivalent weight $=\frac{\text { molecular weight }}{\text { valence factor }}$
$\therefore$ Molecular weight $=20 \times 3=60$
It's oxide is $\mathrm{M}_{2} \mathrm{O}_{3}$ and molecular weight
$$
=2 \times 60+16 \times 3=120+48=168
$$

Topic: Chemistry, Redox Reactions, JEE Main
The equivalent weight of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in acidic medium is expressed in terms of its molecular weight (M) as

Options:
A $M / 3$
B $M / 4$
C $M / 6$
D $M / 7$

The correct answer is:
$M / 6$

Solution:
In acidic medium,
$$
\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \longrightarrow \mathrm{Cr}^{3+}
$$
Equivalent weight of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=\frac{\text { Molecular weight }}{\text { Change in oxidation state of } \mathrm{c}}$
$$
=\frac{M}{2(6-3)}=\frac{M}{6}
$$

Topic: Chemistry, d and f Block Elements, JEE Main
The equivalent weight of \(\mathrm{KIO}_3\) in the given reaction is (\(\mathrm{M}=\) molecular mass) :
\(2 \mathrm{Cr}(\mathrm{OH})_3+4 \mathrm{OH}^{-}+\mathrm{KIO}_3 \rightarrow 2 \mathrm{CrO}_4{ }^{2-}+5 \mathrm{H}_2 \mathrm{O}+\mathrm{KI}\)

Options:
A \(\mathrm{M}\)
B \(\mathrm{M} / 2\)
C \(\mathrm{M} / 6\)
D \(\mathrm{M} / 8\)

The correct answer is:
\(\mathrm{M} / 6\)

Solution:
Hint : \(\mathrm{Cr}(\mathrm{OH})_3+4 \overline{\mathrm{O}} \mathrm{H}+\mathrm{K~} \mathrm{I}^{+5} \mathrm{O}_3 \longrightarrow 2 \mathrm{CrO}_4^{-2}+5 \mathrm{H}_2 \mathrm{O}+\mathrm{K~} \mathrm{I}^{-1}\)
Change in oxidation state of iodine \(=6\)
\(\therefore\) Equivalent weight of \(\mathrm{KIO}_3=\mathrm{M} / 6\)

Topic: Chemistry, Redox Reactions, JEE Main
The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of CoCl3.4NH3 is _______ . (Round off to the Nearest Integer).

Options:

The correct answer is:
2

Solution:
1. trans- CoCl3.4NH3
or

trans- CoNH34Cl2Cℓ

As we know that ethylene diamine is a bidentate ligand and ammonia is a mono dentate ligand.

It means overall two ethylene diamine is required to replace the all neutral ligands (four ammonia) from the coordination sphere of this complex.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The error in the measurement of the length and the breadth of a rectangular table is $1 \%$. If the length and breadth of the table are $1 \mathrm{~m}$ and $50 \mathrm{~cm}$ respectively, then the area of the table including error is

Options:
A $(0.5 \pm 0.1) \mathrm{m}^2$
B $(0.5 \pm 0.01) \mathrm{m}^2$
C $(5000 \pm 10) \mathrm{cm}^2$
D $(5000 \pm 1) \mathrm{cm}^2$

The correct answer is:
$(0.5 \pm 0.01) \mathrm{m}^2$

Solution:
According to the question, Length of the rectangle, $l=1 \mathrm{~m}$ and breadth, $b=50 \mathrm{~cm}=0.5 \mathrm{~m}$
$\therefore$ Area of rectangle, $A=l \times b$
$\begin{aligned}
& \therefore & \ln A & =\ln l+\ln b \\
& \Rightarrow & \frac{\delta A}{A} & =\frac{\delta l}{l}+\frac{\delta b}{b}
\end{aligned}$
Here, $\frac{\delta l}{l}=\frac{\delta b}{b}=1 \%=0.01$
$\begin{aligned}
\therefore \quad \delta A & =A\left(\frac{\delta l}{l}+\frac{\delta b}{b}\right) \quad[\because A=l \times b=1 \times 0.5] \\
& =1 \times 0.5(0.01+0.01)=0.01
\end{aligned}$
Area of the table including error is $A^{\prime}=A \pm \delta A$
$\Rightarrow \quad A^{\prime}=(0.5 \pm 0.01) \mathrm{m}^2$

Topic: Physics, Mathematics in Physics, JEE Main
The escape velocities of two planets A and B are in the ratio 1 : 2. If the ratio of their radii respectively is 1 : 3, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be:

Options:
A 43
B 32
C 23
D 34

The correct answer is:
34

Solution:
The formula to calculate the mass M of any planet of radius R and density ρ, considering it to be a sphere, is given by

M= 43πR3ρ.....................(1)

The escape velocity vA for planet A can be expressed as

VA=2GMARA=2GρA43πRA3RA=CρA.RA............................(2)

where, G is Gravitational constant and C= 8πG3 is a constant.

Similarly, the escape velocity vB of planet B can be expressed as

vB= CρBRB......................(3)

Divide equation (2) by equation (3) and solve with the known parameters to obtain the ratio of the densities of the planets.

VAVB=CρARACρBRBρAρB= VARBVBRAρAρB= VARBVBRA2= 12×312= 94............................(4)

The acceleration due to gravity gA for planet A is given by

gA=GMARA2=G43πRA3×ρARA2= C22ρARA..........................(5)

Similarly, the acceleration due to gravity gB can be expressed as

gB= C22ρBRB...........................(6)

Divide equation (5) by equation (6) and solve with the help of known parameters to calculate the required ratio of the accelerations due to gravity in both the planets.

gAgB=C22ρARAC22ρBRB=ρAρBRARB= 94×13=34

=14×R2R1=34

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The escape velocity for a body projected vertically upwards from the surface of earth is $11 \mathrm{~km} / \mathrm{s}$. If the body is projected at an angle of $45^{\circ}$ with the vertical, the escape velocity will be

Options:
A
$11 \sqrt{2} \mathrm{~km} / \mathrm{s}$

B
$22 \mathrm{~km} / \mathrm{s}$

C
$11 \mathrm{~km} / \mathrm{s}$

D
$\frac{11}{\sqrt{2}} \mathrm{~km} / \mathrm{s}$

The correct answer is:
$11 \mathrm{~km} / \mathrm{s}$

Solution:
Escape velocity of a body is independent of the angle of projection. Hence, changing the angle of projection is not going to effect the magnitude of escape velocity

Topic: Physics, Gravitation, JEE Main, JEE Main 2003
The escape velocity for a planet from the surface is ve. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be

Options:
A ve

B 1.5ve

C 1.5ve
D 2ve

The correct answer is:
1.5ve

Solution:
Taking the potential at a large distance from the planet as zero, the potential at the centre of the planet =-3GM2R.
12mv2=m0--3GM2R
or v2=3GM2R=3Rg

⇒ v2=322RG=32ve2 ∵ ve=2RG
⇒ v=1.5ve

Topic: Physics, Gravitation, JEE Main
The escape velocity from the earth is about 11 km s-1. The escape velocity from a planet having twice the radius and the same mean density as the earth is

Options:
A 22 km s-1
B 11 km s-1
C 5.5 km s-1
D 15.5 km s-1

The correct answer is:
22 km s-1

Solution:
v=2GMR=2G43πR2d
v ∝ Rd
v=2v0

Topic: Physics, Gravitation, JEE Main
The escape velocity of a body depends upon mass as

Options:
A
$m^0$

B
$m^1$

C
$m^2$

D
$m^3$

The correct answer is:
$m^0$

Solution:
$\mathrm{v}_{\mathrm{esc}}=\sqrt{2 \mathrm{gR}}$, where $\mathrm{R}$ is radius of the planet
Hence escape velocity is independent of m

Topic: Physics, Gravitation, JEE Main, JEE Main 2002
The escape velocity of a sphere of mass $m$ from earth having mass $M$ and radius $R$ is given by

Options:
A $\sqrt{\frac{2 G M}{R}}$
B $2 \sqrt{\frac{G M}{R}}$
C $\sqrt{\frac{2 G M m}{R}}$
D $\sqrt{\frac{G M}{R}}$

The correct answer is:
$\sqrt{\frac{2 G M}{R}}$

Solution:
We have a sphere of mass $m$ going from earth's surface to infinity. At infinity, its kinetic energy and potential energy will both be equal to zero.
By law of conservation of energy,
$\begin{aligned}& (K E+P E)_{\text {surface }}=(K E+P E)_{\text {space }} \\& \Rightarrow \frac{1}{2} m v_e^2-\frac{G M m}{R_e}=0 \\& \Rightarrow \frac{1}{2} m v_e^2=\frac{G M m}{R_e} \\& \Rightarrow v_e^2=\frac{2 G M}{R_e} \\& \Rightarrow v_e=\sqrt{\frac{2 G M}{R}} \text { is our required answer. }\end{aligned}$

Topic: Physics, Gravitation, JEE Main
The essential photosynthetic pigment has a _______ at carbon atom 3 of the pyrrole ring-II of chlorophyll a.

Options:
A Methyl group
B Ethyl group
C Carboxyl group
D Magnesium

The correct answer is:
Methyl group

Solution:
There is two main type of chlorophyll- Chlorophyll a and Chlorophyll b. Chlorophyll consists of methyl group whereas chlorophyll b has an aldehyde group. The chlorophyll molecule is chemically composed of two parts- tetrapyrrole heads the porphyrin ring and attached to it is a long hydrocarbon tail called phytol. Magnesium is present in the center of the ring whereas the methyl group is attached to its side chain which is replaced by aldehyde group in chlorophyll b

Topic: Biology, Photosynthesis in Higher Plants, NEET
The ester among the following is

Options:
A Calcium lactate
B Ammonium acetate
C Sodium acetate
D None of these

The correct answer is:
None of these

Solution:
Esters are $R C O O R_m^{\prime}$ where $O R^{\prime}=$ alkoxx group and $R=H$ or alkyl or aryl group.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The ether that undergoes electrophilic substitution reactions is

Options:
A $\mathrm{CH}_3 \mathrm{OC}_2 \mathrm{H}_5$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OCH}_3$
C $\mathrm{CH}_3 \mathrm{OCH}_3$
D $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OCH}_3$

Solution:
Only alkyl aryl ethers e.g., $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OCH}_3$ undergoes electrophilic substitution reactions.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The ether when treated with HI produces
(1) (2)
(3) (4)

Options:
A 1, 3
B 1, 2
C 1, 4
D all are formed

The correct answer is:
1, 4

Solution:
This reaction occurs by SN1 mechanism as stable benzyl carbocation is possible it means I- goes with benzyl carbocation to give benzyl iodide the other product is phenol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The Eu2+ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu=63 ]

Options:
A 4f66s2
B 4f7
C 4f76s2
D 4f6

The correct answer is:
4f7

Solution:
EuZ=63=[Xe]4f76s2
Eu+2Z=63=Xe4f7

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (31 Aug Shift 2)