datavorous/entrance-exam-dataset huggingface dataset:
The equation of motion of a projectile are given by x=36 t m and 2y=96t-9.8 t2 m. The angle of projection is

Options:
A sin −1 ( 4 5 )
B sin −1 ( 3 5 )
C sin −1 ( 4 3 )
D sin −1 ( 3 4 )

The correct answer is:
sin −1 ( 4 5 )

Solution:
x=36t
â´ vx=dxdt=36 mâ s-1

y=48tâ4.9 t 2
â´ vy=48â9.8t
so at t=0 v x =36 and vy=48 m âs-1 So, angle of projection Î¸= tan â1 ( v y v x )= tan â1 ( 4 3 ) Or Î¸= sin â1 (4â/â5)

Topic: Physics, Motion In Two Dimensions, NEET
The equation of motion of a projectile is y=12x-34x2. The horizontal component of velocity is 3 m s-1. the range of the projectile is

Options:
A 18 m

B 16 m

C 12 m

D 21.6 m

The correct answer is:
16 m

Solution:
y = 12 x - 3 4 x 2

The range is the point where y = 0
thus 0 = 12 x - 3 4 x 2
⇒ x = 0
or x=16 m

Topic: Physics, Motion In Two Dimensions, NEET
The equation of state corresponding to 8 g of O2 is

Options:
A PV=8RT
B PV=RT4
C PV=RT
D PV=RT2

The correct answer is:
PV=RT4

Solution:
8g of oxygen is equivalent to 14 mole.
∴PV=μRT=RT4

Topic: Physics, Thermodynamics, NEET
The equation of state corresponding to 8g of O2 is

Options:
A pV=8RT
B pV=RT
C pV=0.25 RT
D pV=0.5 RT

The correct answer is:
pV=0.25 RT

Solution:
No of moles = 8 g 3 2 g = 0 . 2 5 moles
â´ pV = nRT â 0 . 2 5 RT

Topic: Chemistry, States of Matter, NEET
The equation of state for $5 \mathrm{~g}$ of oxygen at a pressure $P$ and temperature $T$, when occupying a volume $V$, will be:
(where $R$ is the gas constant)

Options:
A $P V=(5 / 32) R T$
B $P V=5 R T$
C $P V=(5 / 2) R T$
D $P V=(5 / 16) R T$

The correct answer is:
$P V=(5 / 32) R T$

Solution:
We know that $P V=\underline{n} R T$
$$
\begin{gathered}
\therefore n=\frac{m}{\text { moleculer mass }}=\frac{5}{32} \\
\Rightarrow \quad P V=\left(\frac{5}{32}\right) R T
\end{gathered}
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2004
The equation of the progressive wave is Y=3sin⁡πt3-x5+π4 where X and Y are in meter and time in second. Which of the following is correct?

Options:
A Velocity V=1.5 m/s
B Amplitude A=3Â cm
C Frequency F=0.2Â Hz
D Wavelength Î»=10Â m

The correct answer is:
Wavelength λ=10 m

Solution:
y=3sinâ¡2Ït6-x10+Ï4
Compare y=Asinâ¡2ÏtT-xÎ»+Ï4
x and y are meter
â´ Î»=10 m

Topic: Physics, Waves and Sound, NEET
The equation of Verhulst-Pearl logistic growth is $\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{rN}\left[\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right]$. From this equation, K indicates:

Options:
A Biotic potential
B Carrying capacity
C Population density
D Intrinsic rate of natural increase

The correct answer is:
Carrying capacity

Solution:
In the equation $\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{rN}\left(\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right), \mathrm{K}$ represents carrying capacity.

Topic: Biology, Organisms and Populations, NEET, NEET 2024
The equilibrium concentrations of the species in the reaction A+B⇋C+D are 2,3,10 and 6 mol L-1 respectively at 300K. ∆G∘ for the reaction is

R =2 cal/mol K

Options:
A -13.73 cal
B -1372.60 cal
C -137.26 cal
D -1381.80 cal

The correct answer is:
-1381.80 cal

Solution:
A+B⇔ C+D at equilibrium 2 3 10 6

Keq = [C] [D] [A] [B] Keq =10×62×3=10∆Gº = –RT ln K

Where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant of the reaction.

ΔG =-2.303×2×300×log1010 =-2.303×2×300×1 =-1381.8 cal

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2023 (All India)
The equilibrium constant at 25°C for the reaction, 2NO(g)+Br2(g)⇌2NOBr(g) is equal to 160 atm-1. If the partial pressures of NO, Br2 and NOBr in a flask at 25° C are 0.01, 0.1 and 0.04 atm respectively. It can be said that

Options:
A there is equilibrium in the flask
B there reaction will proceed in the forward direction
C the reaction will proceed in the backward direction
D the partial pressure of NOBr finally will be 0.05 atm

The correct answer is:
there is equilibrium in the flask

Solution:
Qp=pNOBr(g)2pNO(g)2.pBr2(g)
=(0.04atm)2(0.1atm)2×(0.1atm)
=16×10-4atm21×10-4atm×1×10-1atm
=160atm-1
Since Qp=Kp , therefore there is equilibrium in the flask.

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant expression for a gas reaction is,
\(K_c=\frac{\left[\mathrm{NH}_3\right]^2[\mathrm{NO}]^4}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]^5\left[\mathrm{H}_2 \mathrm{O}\right]^6}\)
Write the balanced chemical equation corresponding to this expression.

Options:
A \(2 \mathrm{NH}_3(g)+4 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)+3 \mathrm{H}_2 \mathrm{O}(g)\)
B \(2 \mathrm{NH}_3(g)+4 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)
C \(2 \mathrm{NH}_3(g)+4 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)
D \(2 \mathrm{NH}_3(g)+4 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)+3 \mathrm{H}_2 \mathrm{O}(g)\)

The correct answer is:
\(2 \mathrm{NH}_3(g)+4 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)+3 \mathrm{H}_2 \mathrm{O}(g)\)

Solution:
Correct Option is : (D)
\(2 \mathrm{NH}_3(g)+4 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)+3 \mathrm{H}_2 \mathrm{O}(g)\)

Topic: Chemistry, Ionic Equilibrium, NEET
The equilibrium constant for a reaction A+B⇋C+D is 1×10-2 at 298 K and is 2 at 273 K. The chemical process resulting in the formation of C and D is:

Options:
A Exothermic
B Endothermic
C Unpredictable
D There is no relationship between ΔH and K

The correct answer is:
Exothermic

Solution:
log K 2 K 1 = ΔH 2.303R 1 T 1 − 1 T 2
T 1 > T 2 â&â K 2 > K 1
ΔH =−Ve

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for a reaction is 10. What will be the value of ∆G⊖? (R = 8.314 JK–1mol–1, T = 300 K)

Options:
A –5.744 kJ mol–1
B –5.995 kJ mol–1
C –6.452 kJ mol–1
D –6.687 kJ mol–1

The correct answer is:
–5.744 kJ mol–1

Solution:
Correct Option is : (A)
–5.744 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
The equilibrium constant for ionization of aliphatic amine RNH2 (g) in water is
RNH2(g)+H2O(l)⇌RNH3+(aq)+OH-(aq)
is 8×10-6 at 25°C . Partial pressure of amine at equilibrium is 0.5 bar. The pOH of the aqueous solution of amine will be

Options:
A 2.7
B 4.7
C 3.7
D 5.7

The correct answer is:
2.7

Solution:
RNH2(g)0.5+H2O(l)⇌RNH3+(aq)x+OH-(aq)x for it equilibrium constant is given as
K=[RNH3+][OH-][RNH2]
Here K = 8×10-6
8×10-6=x×x0.5
x=2×10-3
[OH]-=2×10-3M
pOH=-log2×10-3
=2.7 .

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for the decomposition of water,
H2O(g)⇌H2(g)+12O2(g) is given by

Options:
A K=α3p1/2(1-α)(2-α)1/2
B K=α3/2p1/2(1-α)(2+α)1/2
C K=α3p1/22
D K=α3p3/2(1-α)(2+α)1/2

The correct answer is:
K=α3/2p1/2(1-α)(2+α)1/2

Solution:
teqH2O(g)1-α⇌H2(g)α+12O2(g)α2
Δng=32-1=12
KP=nH2×nO21/2nH2O×PΣnΔng
=α×α21/2(1-α)×P1+α/21/2
=α3/2P1/2(1-α)(2+α)1/2

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for the disproportionation of HgCl2 into HgCl+ and HgCl3- is
Given HgCl + + Cl − ⇌ HgCl 2 ; K 1 =3× 10 6
HgCl2+Cl-⇌HgCl3-;K2=9.0

Options:
A 27×106
B 3.3×10-7
C 3.3×10-6
D 3×10-6

The correct answer is:
3×10-6

Solution:
K1=[HgCl2][HgCl+][Cl-];K2=[HgCl3-][HgCl2][Cl-]
So K3=[HgCl+][HgCl3-][HgCl2]2=K2K1=93×106=3×10-6

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for the disproportionation of HgCl2 into HgCl+ and HgCl3- of
Given HgCl++Cl-⇌HgCl2;K1=3×106
HgCl2+Cl-⇌HgCl3-;K2=9.0

Options:
A 27×106
B 3.3×10-6
C 3.2×10-7
D 3×10-6

The correct answer is:
3×10-6

Solution:
K1=[HgCl2][HgCl+][Cl-]
K2=[HgCl3-][HgCl2][Cl-]
So K3=[HgCl+][HgCl3-][HgCl2]2
=K2K1=93×106=3×10-6.

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for the following reaction is

Options:
A 1011
B 10-11
C 1029
D 10-29

The correct answer is:
1011

Solution:
Subtracting (ii) from (i) gives the desired equation,
Hence equilibrium constant = 1 0 - 9 1 0 - 2 0 = 1 0 1 1 .

Topic: Chemistry, General Organic Chemistry, NEET
The equilibrium constant for the given reaction is approximately 10â3

HPO 4 2â aq + HCO 3 â aq â H 2 PO 4 â aq + CO 3 2â aq

Which is strongest conjugate base in the given reaction?

Options:
A HPO42-aq
B HCO3-aq
C H2PO4-aq
D CO32-aq

The correct answer is:
CO32-aq

Solution:
CO 3 2− + H 2 PO 4 − ⇌ HPO 4 2− + HCO 3 2− Kc= 10 3
Since the value of K c for this reaction is greater than 1, it means this reaction is favourable in forward direction. Therefore, CO32- is the stronger conjugate base than HPO42-

Topic: Chemistry, Ionic Equilibrium, NEET
The equilibrium constant for the reaction,

Ag 2 O(s)⇌2Ag(s) +  1 2 O 2 (g) is given by

Options:
A K=2Ag12O2Ag2O
B K=[Ag]2[O2]12[Ag2O]
C K=AgO2Ag2O
D none of the above

The correct answer is:
none of the above

Solution:
K=[Ag]2[O2]12 Ag2Os
By convention, [Solid] = 1
i.e., Ags=1 and Ag2Os=1
So K=[O2]12 .

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for the reaction given is 3.6×10-7
OCl − aq + H 2 O l ⇌ HOCl aq + OH − aq .
What is Ka for HOCl?

Options:
A 2.77×10-8
B 3.6×10-7
C 6×10-4
D 2.8×10-6

The correct answer is:
2.77×10-8

Solution:
Keq=KwKa
3.6×10-7=KwKa
Ka=10-143.6×10-7=2.77×10-8

Topic: Chemistry, Ionic Equilibrium, NEET
The equilibrium constant for the reaction,

N2g+O2(g)â2NO(g)

at temperature T is 4Ã10-4. The value of Kc for the reaction

NO(g)â12N2g+12O2(g)

at the same temperature is

Options:
A 2.5×102
B 50
C 4×10-4
D 0.02

The correct answer is:
50

Solution:
N2g+O2(g)⇌2NO(g)
Kc=NO2N2O2=4×10-4
NO⇌12N2g+12O2(g)
Kc′=N21/2O21/2[NO]
=1Kc=14×10-4=50

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant for the reaction
N2(g)+O2(g)⇌2NO(g) is 4×10-4 at 2000 K.
In presence of a catalyst, the equilibrium is attained ten times faster. Therefore, the equilibrium constant, in presence of the catalyst, at 2000 K is

Options:
A 40×10-4
B 4×10-4
C 4×10-3
D Difficult to compute without more data

The correct answer is:
4×10-4

Solution:
The presence of catalyst does not affect the magnitude of equilibrium constant. In presence of a catalyst the equilibrium is attained sooner.

Topic: Chemistry, Chemical Kinetics, NEET
The equilibrium constant for the reaction
N 2 g + O 2 g ⇌2NO g
At temperature T is 4×10-4 . the value of KC fro the reaction
NO g ⇌ 1 2 N 2 g + 1 2 O 2 g at the same temperature is

Options:
A 4×10-6
B 2.5×102
C 0.02
D 50

The correct answer is:
50

Solution:
N 2 g + O 2 g ⇌2NO g
KC=[NO]2[N2][O2]=4×10-4
NO⇌ 1 2 N 2 g + 1 2 O 2 g
KC=[N2]1/2[O2]1/2[NO]
1KC=1×1044=50

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant K, for the reaction
2 HI g â H 2 g + I 2 g at room temperature is 2.85 and that at 698 K is 1.4 Ã 10-2. This implies that the forward reaction is

Options:
A Exothermic
B Endothermic
C Exergonic
D Unpredictable

The correct answer is:
Exothermic

Solution:
On increasing temperature, equilibrium constant is decreasing. It means that backward reaction is favoured. According to Le-Chatelier's principle, increase in temperature favours endothermic reaction. Thus backward reaction is endothermic, consequently forward reaction should be exothermic.

Topic: Chemistry, Ionic Equilibrium, NEET
The equilibrium constant of the following are reactions
N2+3H2⇌2NH3 K1
N2+O2⇌2NO K2
H2+12O2→H2O K3
The equilibrium constant (K) of the reaction
NH3+52O2↔K2NO+3H2O , will be

Options:
A K2K33/K1
B K2K3/K1
C K23K3/K1
D K1K33/K2

The correct answer is:
K2K33/K1

Solution:
N2+3H2⇌2NH3 K1} Reverse it.
N2+O2⇌2NO K2
H2+12O2→H2O K3}×3
2NH3+52O2⇌2NO+3H2O KC
KC=K2K33K1

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant of the reaction A2 g + B2 gâ2AB g at 373 K is 50. If 1âL of flask containing 1 mole of A 2 â(g) is connected to 2âL flask containing 2 moles B2 g at 100 o C , the amount of AB produced at equilibrium at 100 o C would be

Options:
A 0.93Â mol
B 1.87 mol
C 2.80Â mol
D 3.74Â mol

The correct answer is:
1.87 mol

Solution:
A2 g + B2 g â 2AB g
Initially: 1 3 M ââ 2 3 M
At equilibrium: 1 3 âx Mâ 2 3 âx M â2x M
Keq=50 =4 x213-x23-x
or 50 =36 x21-3x2-3x
or 50 9x2 â 9x + 2=36 x2
or 450 x2 â 450x + 100=36x2
or 414x2 â 450x + 100=0
or x= +450â 450 2 â4Ã414Ã100 2Ã414
or x=+450-202500-1656002Ã414
or x=450-369002Ã414=450â192.12Ã414
= 0.31 M
â´ moles of AB produced = 0.31 Ã 6=1.86

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant of the reaction: $\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow \mathrm{Cu}_{(a q)}^{2+}+2 \mathrm{Ag}_{(s)}$ $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ at $298 \mathrm{~K}$ is:

Options:
A $2.4 \times 10^{16}$
B $4.0 \times 10^{10}$
C $4.0 \times 10^{15}$
D $2.4 \times 10^{10}$

The correct answer is:
$4.0 \times 10^{15}$

Solution:
As $\mathrm{E}_{\mathrm{cell}}^0=\frac{0.0591}{n} \log \mathrm{K}_c$
$\begin{aligned}
& \therefore 0.46=\frac{0.0591}{2} \log \mathrm{K}_c \\
& \therefore \log \mathrm{K}_c=\frac{2 \times 0.46}{0.0591}=15.57 \\
& \text {or } \mathrm{K}_c=\text { Antilog } 15.57=3.7 \times 10^{15}
\end{aligned}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2007
The equilibrium constants for the reaction, Br2⇌2Br , at 500 K and 700 K are 1×10-10 and 1×10-5 respectively. The reaction is

Options:
A endothermic
B exothermic
C fast
D slow

The correct answer is:
endothermic

Solution:
As the equilibrium constants for the reaction, Br2⇌2Br , at 500 K and 700 K are 1×10-10 and 1×10-5 respectively it means the value of equilibrium constant is increasing with increase temperature so it is an endothermic reaction.

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constants for the reactions are
H 3 PO 4   ⇌ K 1   H +  +  H 2 PO 4 − ;  K 1
H 2 PO 4 − ⇌ K 2 H +  +  HPO 4 −2 ;  K 2
HPO 4 2−   ⇌ K 2   H +  +  PO 4 3− ;  K 3
The equilibrium constant for
H3PO4â3H++PO43- will be

Options:
A K1/K2/K3
B K1×K2×K3
C K2/K1K3
D K1+K2+K3

The correct answer is:
K1×K2×K3

Solution:
K1=[H+][H2PO4-][H3PO4]
K2=[H+][HPO42-][H2PO4-]
K3=[H+][PO43-][HPO42-]
Multiplying these three
K 1  ×  K 2  ×  K 3  =  H + 3 PO 4 3− H 3 PO 4 .

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constants of the following are:
N2+3H2⇌2 NH3K1N2+O2⇌2 NOK2H2+12O2→H2O K3
The equilibrium constant (K) of the reaction:
2N⁡H⁡3+52O⁡2⇌K⁡2N⁡O⁡+3H⁡2O⁡, will be:

Options:
A K1 K33K2
B K2 K33K1
C K2 K3K1
D K23 K3K1

The correct answer is:
K2 K33K1

Solution:
(I) N2+3H2⇌2NH3; K1=NH32N2 H23
(II) N2+O2⇌2NO; K2=NO2N2 O2
(III) H2+12O2→H2O; K3=H2OH2 O212
II+3×III-I will give
2N⁡H⁡3+52O⁡2⇌K⁡2N⁡O⁡+3H⁡2O⁡;
∴ K=K2×K33K1

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2017
The equilibrium reaction PCl5(s)⇌PCl3(s)+Cl2(g) is carried in a closed vessel. What is the effect on equilibrium concentration of Cl2(g) by adding PCl5 at constant temperature?

Options:
A Unaffected
B Increases
C Decreases
D Cannot be described without the value of Kp

The correct answer is:
Increases

Solution:
On adding more PCl5 , equilibrium shifts forward, hence concentration of Cl2(g) increases.

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium state of a thermodynamic system is described by
A. Pressure
B. Total heat
C. Temperature
D. Volume
E. Work done

Options:
A A, B and E only
B B, C and D only
C A, B and C only
D A, C and D only

The correct answer is:
A, C and D only

Solution:
Equilibrium state of a thermodynamic system is defined by state variables $P, V$ and $T$. Work done and total heat are path variables.

Topic: Physics, Thermodynamics, NEET, NEET 2024 (Re-NEET)
The equilibrium that exists in aqueous solution, CH 3 COOH ⇌ CH 3 COO − +  H + if dil HCl is added at constant temperature, then

Options:
A The equilibrium constant will increase
B The equilibrium constant will decrease
C Concentration of CH3COO- will decrease
D Concentration of CH3COO- will increase

The correct answer is:
Concentration of CH3COO- will decrease

Solution:
Due to common ion effect of H+ , the ionisation of CH3COOH is suppress hence the concentration of CH3COO- will decrease.

Topic: Chemistry, Ionic Equilibrium, NEET
The equivalence of a thin cotyledon in monocots is

Options:
A Scutellum
B Prophyll
C Coleoptile
D Coleorhiza

The correct answer is:
Scutellum

Solution:
Single cotyledon of the embryo in cereal grain is represented by scutellum. Coleoptile represents the covering of the stem. Coleorhiza represents the covering of root. The scutellum can also refer to the equivalence of a thin cotyledon in monocots. It is very thin with high surface area and serves to absorb nutrients from the endosperm during germination.

Topic: Biology, Morphology of Flowering Plants, NEET
The equivalent capacitance between the points A and B is

Options:
A 2C0
B C0
C 4C0
D 6C0

The correct answer is:
2C0

Solution:
Ceq=2C03+4C03=2C0

Topic: Physics, Electrostatics, NEET
The equivalent capacitance of the arrangement shown in figure is

Options:
A $25 \mu \mathrm{F}$
B $20 \mu \mathrm{F}$
C $30 \mu \mathrm{F}$
D $15 \mu \mathrm{F}$

The correct answer is:
$20 \mu \mathrm{F}$

Solution:
We know that the, circuit can be modified as
$\therefore C_{\text {net }}=15+5=20 \mu \mathrm{F}$

Topic: Physics, Capacitance, NEET, NEET 2023 (Manipur)
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, NEET, NEET 2021
The equivalent capacitance of the system shown in the following circuit is:

Options:
A 9 μF
B 2 μF
C 3 μF
D 6 μF

The correct answer is:
2 μF

Solution:
For parallel grouping(the loop part), equivalent capacitance can be written as

Cp=3+3 μF=6 μF.

Now, this part is in series with the left capacitor. So we can write

Ceq=3×63+6 μF=2 μF

Topic: Physics, Capacitance, NEET, NEET 2023 (All India)
The equivalent conductance of 1M benzoic acid is 12.8ohm-1cm2 and if the conductance of benzoate ion and H+ ion at infinite dilution are 42 and 288.42ohm-1cm2 respectively. Then its degree of dissociation is

Options:
A 0.039%
B 3.9%
C 0.35%
D 39%

The correct answer is:
3.9%

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, NEET
The equivalent conductance of $\frac{\mathrm{M}}{32}$ solution of a weak monobasic acid is $8.0 \mathrm{~mho~} \mathrm{cm}^2$ and at infinite dilution is $400 \mathrm{~mho~} \mathrm{cm}^2$. The dissociation constant of this acid is

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

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

Solution:
Degree of dissociation, $\alpha=\frac{\Lambda^c}{\Lambda^{\infty}}$ where, $\Lambda^c$ and $\Lambda^{\infty}$ are equivalent conductances at a given concentration and at infinite dilution respectively.
$$
\Rightarrow \quad \alpha=\frac{8.0}{400}=2 \times 10^{-2}
$$
From Ostwald's dilution law (for weak monobasic acid)
$$
\begin{aligned}
\mathrm{K}_{\mathrm{a}} & =\frac{\mathrm{C} \alpha^2}{(1-\alpha)} \\
\text { or } \quad & \mathrm{C} \alpha^2 \\
& =\frac{1}{32}\left(2 \times 10^{-2}\right)^2 \\
\text { or } & =1.25 \times 10^{-5}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2009 (Screening)
The equivalent conductance of a 0.2 N solution of an electrolyte was found to be 200  Ω −1   cm 2   eq −1 . The cell constant of the cell is 2  cm −1 . The resistance of the solution is

Options:
A 50Ω
B 400Ω
C 100Ω
D None of these

The correct answer is:
50Ω

Solution:
Λ v =200  Ω −1 cm 2 eq −1
C N =0.2 N,  cm −1
K=ÎvÃCN1000=200Ã0.21000Ã10=125Î©-1cm-1
R=1K.la=25Ã2=50Î©.

Topic: Chemistry, Electrochemistry, NEET
The equivalent conductance of a substance is 1/3rd of the molar conductance. The time required to electrolyse 3 moles the substance using a current of 9 ampere is

Options:
A 4808 min
B 1608 min
C 2408 min
D 2008 min

The correct answer is:
1608 min

Solution:
Since the equilibrium mass is 13 of molecular mass, therefore 1 mole of substance will be liberated by using 3 moles of electrons M3++3e-âM(s)
So 9Ã96500=9Ãt
or t = 96500 sec =  96500 60 ≈1608 min

Topic: Chemistry, Electrochemistry, NEET
The equivalent conductances of two strong electrolytes at infinite dilution in H 2 O (where ions move freely through a solution) at 25 â are given below :

Î CH 3 COONa o = 91.0 S cm 2 / equivalent

Î HCl o = 426.2 S cm 2 / equivalent .

What additional information/quantity one needs to calculate Î o of an aqueous solution of acetic acid?

Options:
A Λ o of chloroacetic acid ClCH 2 COOH
B Λ o of NaCl
C Λ o of CH 3 COOK
D The limiting equivalent conductance of H + λ H + o

The correct answer is:
Λ o of NaCl

Solution:
According to Kohlrausch's law, the molar conductivity at infinite dilution Λ o for weak electrolyte, CH 3 COOH is
Λ CH 3 COOH o = Λ CH 3 COONa o + Λ HCl o - Λ NaCl o
So, for calculating the value of Λ CH 3 COOH o , value of Λ NaCl o should also be known.

Topic: Chemistry, Electrochemistry, NEET
The equivalent conductivity of 0.1 M weak acid is 100 times less 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:
Λv=Λ°100
So α=ΛvΛ°=Λ°100Λ°=0.01

Topic: Chemistry, Electrochemistry, NEET
The equivalent conductivity of 0.1 M weak acid is 100 times less 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:
Λv=Λ°100
So α=ΛvΛ°=Λ°100Λ°
=0.01

Topic: Chemistry, Electrochemistry, NEET
The equivalent of a structural gene is:

Options:
A Muton
B Cistron
C Operon
D Recon

The correct answer is:
Cistron

Solution:
Cistron is a segment of DNA coding for a polypeptide. Eukaryotic structural gene is monocistronic whereas prokaryotic structural gene is polycistronic.

Replicon: Unit of DNA that undergoes replication. Muton: Unit of DNA which undergoes mutation. Recon: Unit of DNA which undergoes recombination. Operon is a co-ordinate group of genes such as structural gene, operator gene, promoter gene, regulator gene which function together and regulate a metabolic pathway as a unit, e.g., lac operon, trp operon, ara operon, his operon, val operon etc.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2016 (Phase 2)
The equivalent resistance between A and B for the mesh shown in the figure is

Options:
A 7.2 Ω
B 16 Ω
C 30 Ω
D 4.8 Ω

The correct answer is:
16 Ω

Solution:
When the resistance is connected in series then the formula of equivalent resistance, R=R1+R2+R3+....+Rn, but when they are connected in parallel then net resistance is, 1R=1R1+1R2+1R3+...+1Rn, so in the above diagram the equivalent resistance will be,

,

in the above branch, resistance are connected in parallel, so new equivalent resistance will be,

So, the final resistance will be, Rnet=4+4+8 Ω=16 Ω.

Topic: Physics, Current Electricity, NEET, NEET 2020 (Phase 2)
The equivalent resistance between the points A and B will be (each resistance is 15 Ω)

Options:
A 30 Î©
B 8 Î©
C 10Â Î©
D 40Â Î©

The correct answer is:
8 Î©

Solution:
The circuit can be shown as given below

The equivalent resistance between A and C.
RDC=15×15+1515+15+15
=15×3015+30
=15×3045=10 Ω
Now, between A and B, the resistance of the upper part ADCB,
R1=15+10+15=40 Ω
Between A and B, the resistance of the middle part AOB
R2=15+15=30 Ω
Therefore, equivalent resistance between A and B
1R=1R1+1R2+1R3
=140+130+115
=3+4+8120
=15120
⇒ R=12015=8 Ω

Topic: Physics, Current Electricity, NEET
The equivalent resistance of the infinite network given below is:

Options:
A $2 \Omega$
B $(1+\sqrt{2}) \Omega$
C $(1+\sqrt{3}) \Omega$
D $(1+\sqrt{5}) \Omega$

The correct answer is:
$(1+\sqrt{3}) \Omega$

Solution:
Let net resistance of the given infinite network be ' \(R\) '
Now, the circuit can be modified as

Now, \(R_{\text {net }}=R=1+1+\frac{R}{R+1}\)
\(\begin{aligned}
& \therefore R=2+\frac{R}{R+1} \\
& R^2+R=2 R+2+R \\
& R^2-2 R-2=0 \\
& R=\left(\frac{2 \pm \sqrt{4+8}}{2}\right) \\
& R=(1+\sqrt{3}) \Omega
\end{aligned}\)

Topic: Physics, Current Electricity, NEET, NEET 2022 (Phase 2)