datavorous/entrance-exam-dataset huggingface dataset:
The dipole moment of LiH is 1.964×10-29 Coulomb meter and the interatomic distance between Li and H is 1.596 Å. The percentage ionic character of LiH is

Options:
A 82.5%
B 63.2%
C 76.8%
D 90.5%

The correct answer is:
76.8%

Solution:
Î´=1.964Ã10-291.596Ã10-10=1.2306Ã10-19C
%ionic character=1.2306Ã10-19Ã1001.602Ã10-19=76.8%

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The dipole moments of the given species are such that

Options:
A BF3>NF3>NH3
B NF3>BF3>NH3
C NH3>NF3>BF3
D NH3>BF3>NF3

The correct answer is:
NH3>NF3>BF3

Solution:
The dipole moments of the given species are as follows
NH3>NF3>BF3
BF3 is a planar molecule, so it has zero dipole moment. NH3 has more dipole moment than NF3 . The resultant dipole moment of three N-H bonds and dipole moment due to lone pair of electrons are acting in the same direction, but in case of NF3 the resultant dipole moment of the three N-F bonds is acting opposite to the direction of dipole moment due to lone pair of electrons.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The direction of propagation of electromagnetic wave is along

Options:
A Electric field vector, E
B Magnetic field vector, B
C E→×B→
D B→×E→

The correct answer is:
E→×B→

Solution:
The direction of propagation of the electromagnetic wave is always perpendicular to the plane in which E→ and B→ lies. So the direction is along the direction of E→×B→

Topic: Physics, Electromagnetic Waves, NEET
The direction of the gravitational intensity at an arbitrary point P is indicated by the arrow

Options:
A d,
B e,
C f,
D g.

The correct answer is:
e,

Solution:
Correct Option is : (B)
e,

Topic: Chemistry, Gravitation, NEET
The discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1$ has $2 \mathrm{~kg}$ mass and $0.2 \mathrm{~m}$ radius and initial angular velocity of $50 \mathrm{rad} \mathrm{s}^{-1}$. Disc $\mathrm{D}_2$ has $4 \mathrm{~kg}$ mass, $0.1 \mathrm{~m}$ radius and initial angular velocity of $200 \mathrm{rad} \mathrm{s}^{-1}$. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad $s^{-1}$ ) of the system is:

Options:
A 60
B 100
C 120
D 40

The correct answer is:
100

Solution:
M.J. of $\mathrm{D}_1$ is $\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}=\frac{2(0.2)^2}{2}$ $=0.04 \mathrm{~kg} \mathrm{~m}^2$ initial angular velocity of $D_1$ is $\omega_1=50 \mathrm{rad} \mathrm{S}^{-1}$ M.J of $\mathrm{D}_2$ is $\mathrm{I}_2=\frac{4 \times(0.1)^2}{2}=0.02 \mathrm{~kg} m_2$
initial angular velocity of $\mathrm{D}_2$ is $\omega_2=$ 200 and $\mathrm{S}^{-1}$. According to the law of conservation of angular moment
$$
\begin{gathered}
L_1=L_2 \\
\therefore \mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2=\left(\mathrm{I}_1+\mathrm{I}_2\right) \\
\Rightarrow \omega=\frac{l_1 w_1+l_2 w_2}{\mathrm{I}_1+1} \\
\therefore \omega=\frac{(0.04)^2 \times 50+(0.02)^2 \times 200}{0.04+0.02} \\
=\frac{2+4}{0.06}=100 \mathrm{rads}^{-1}
\end{gathered}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2013 (Karnataka)
The disease caused due to cigarette smoking and blockage of blood supply to heart muscle respectively are:

Options:
A Angina and coronary artery disease
B Heart failure and goitre
C Diabetes and angina
D Emphysema and atherosclerosis

The correct answer is:
Emphysema and atherosclerosis

Solution:
Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased. One of the major causes of this is cigarette smoking.
Coronary Artery Disease (CAD): Coronary Artery Disease, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused by deposits of calcium, fat, cholesterol and fibrous tissues, which makes the lumen of arteries narrower.

Topic: Biology, Breathing and Exchange of Gases, NEET
The disease chikungunya is transmitted by:

Options:
A house fly
B Aedes mosquito
C cockroach
D female Anopheles

The correct answer is:
Aedes mosquito

Solution:
Aedes mosquito transmits chikungunya and dengue.

Topic: Biology, Human Health and Diseases, NEET
The disease sickle-cell anaemia is caused by the substitution of ___(i)____ by ____(ii)____ at the ____(iii)____ position of ____(iv)_____ globin chain of the haemoglobin molecule.

Which of the following CORRECTLY fills the blanks in the above statement?

Options:
A (i) valine, (ii) glutamic acid, (iii) sixth, (iv) beta
B (i) glutamic acid, (ii) valine, (iii) sixth, (iv) beta
C (i) glutamic acid, (ii) valine, (iii) fifth, (iv) beta
D (i) valine, (ii) glutamic acid, (iii) fifth, (iv) beta

The correct answer is:
(i) glutamic acid, (ii) valine, (iii) sixth, (iv) beta

Solution:
Sickle cell anemia is a type of genetic blood disorder which is inherited from the parents. The person suffering from it develops an abnormal haemoglobin that leads to elongated and curved or sickle shaped RBC. Such condition occurs when the individual inherits two abnormal copies of gene for beta globin chain of haemoglobin from each parent. Normally, the red pigment molecule is made of four polypeptide chains two α-chains and two β -chains. Sickle cell anaemia results in glutamic acid (Glu) is replaced by valine (Val) at sixth position in each of two β globin chain of haemoglobin. Such mutation is called as substitution mutation.

Topic: Biology, Principles of Inheritance and Variation, NEET
The disease that is caused by fungal pathogen reported by E. Kurosawa in rice plant is:

Options:
A Foolish seeding
B Rust diseases
C Smut disease
D Both (A) and (B)

The correct answer is:
Foolish seeding

Solution:
The ‘bakanae’ (foolish seedling) disease of rice seedlings, was caused by a fungal pathogen Gibberella fujikuroi. E. Kurosawa (1926) reported the appearance of symptoms of the disease in rice seedlings when they were treated with sterile filtrates of the fungus. The active substances were later identified as gibberellic acid.

Topic: Biology, Plant - Growth and Development, NEET
The disorder caused due to acute chest pain due to failure of O₂ supply to heart muscles is known as

Options:
A hypertension
B angina pectoris
C Heart attack
D Cardiac arrest

The correct answer is:
angina pectoris

Solution:
The disorder caused due to acute chest pain due to failure of O₂ supply to heart muscles is known as angina pectoris.

Topic: Biology, Body Fluids and Circulation, NEET
The disorder in which walls of the alveoli gets damaged, resulting in a decrease in the surface area, is called:

Options:
A Asthma
B Allergy
C Emphysema
D Pneumonia

The correct answer is:
Emphysema

Solution:
Emphysema is a respiratory disease. In this condition, millions of the lungs' tiny air sacs (alveoli) got rupture. As these thin, fragile air sacs become damaged or destroyed. Smoking is responsible for the vast majority of cases of emphysema.

Topic: Biology, Breathing and Exchange of Gases, NEET
The dispersive powers of flint glass and crown glass are 0.053 and 0.034 respectively and their mean refractive indices are 1.68 and 1.53 for white light. Calculate the angle of the flint glass prism required to form an achromatic combination with a crown glass prism of refracting angle 4°.

Options:
A 2°
B 4°
C 5°
D 6°

The correct answer is:
5°

Solution:
Given-Glass type	Flint	Crown
Dispersive power	0.053	0.034
Refractive Index of Mean Colour	1.68	1.53
Refracting Angle	?	4°

Condition for Achromatism =ω1ω1=f1f2
⇒ ω1ω2=μ1-1A1μ2-1A2
⇒ 0.0530.034=1.68-1⋅A1.53-1⋅4
⇒ 0.0530.034=0.68.A0.53×4
A=4.85≈5

Topic: Physics, Ray Optics, NEET
The displacement of a particle along the $x$ axis is given by $\mathrm{x}=\mathrm{a} \sin ^2 \omega \mathrm{t}$. The motion of the particle corresponds to

Options:
A simple harmonic motion of frequency $\omega / \pi$
B simple harmonic motion of frequency $3 \omega / 2 \pi$
C non simple harmonic motion
D simple harmonic motion of frequency $\omega / 2 \pi$

The correct answer is:
non simple harmonic motion

Solution:
For a particle executing SHM acceleration (a) $\propto-\omega^2$ displacement $(x)$
Given
$\mathrm{x}=\mathrm{a} \sin ^2 \omega \mathrm{t}$
Differentiating the above equation w.r.t, we get $\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{a} \omega(\sin \omega \mathrm{t})(\cos \omega \mathrm{t})$
Again differentiating, we get
$$
\begin{aligned}
& \frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=\mathrm{a}=2 \mathrm{a} \omega^2\left[\cos ^2 \omega \mathrm{t}-\sin ^2 \omega \mathrm{t}\right] \\
& =2 \mathrm{a} \omega^2 \cos 2 \omega \mathrm{t}
\end{aligned}
$$
The given equation does not satisfy the condition for SHM [Eq. (i)] . Therefore, motion is not simple harmonic.

Topic: Physics, Oscillations, NEET, NEET 2010 (Screening)
The displacement of a particle executing simple harmonic motion is given by,
y=A0+Asinωt+Bcosωt.
Then, the amplitude of its oscillation is given by

Options:
A A0+A2+B2
B A2+B2
C A02+A+B2
D A+B

The correct answer is:
A2+B2

Solution:
y=A0+Asinωt+Bcosωt
=A0+A2+B2·AA2+B2sinωt+BA2+B2cosωt
=A0+A2+B2sinωt+φ
Amplitude =A2+B2

Topic: Physics, Oscillations, NEET, NEET 2019 (All India)
The displacement of a particle from its mean position (in m) is given by, y=0.2sinâ¡10Ït+1.5Ï cosâ¡(10Ït+1.5Ï).The motion of particle is

Options:
A periodic but not S.H.M
B non-periodic
C simple harmonic motion with a period 0.1 s
D simple harmonic motion with a period 0.2 s

The correct answer is:
simple harmonic motion with a period 0.1 s

Solution:
y=0.2sinâ¡10Ït+1.5Ïcosâ¡(10Ït+1.5 Ï)
=0.1sinâ¡2(10Ït+1.5Ï)
âµsinâ¡2A=2 sinA cosA
=0.1sinâ¡20Ït+3.0Ï
â´ Time period T=2ÏÏ=2Ï20Ï=110=0.1 s

Topic: Physics, Oscillations, NEET
The displacement of a particle (in meter) from its mean position is given by the equation y=0.2cos2⁡πt2-sin2⁡πt2, The motion of the above particle is

Options:
A not simple harmonic
B simple harmonic with amplitude 0.2 m
C simple harmonic with the period double that of a second's pendulum
D simple harmonic with amplitude 0.4Â m

The correct answer is:
simple harmonic with amplitude 0.2 m

Solution:
The equation of motion is

y= 0.2cos2â¡Ït2-sin2â¡Ït2 , or
y = 0.2cosÏt

Above equation is an equation of SHM, Hence the motion of the particle is SHM.

Where,
Ï = Ï = 2ÏT
âT=2 second
Amplitude =0.2 m

Topic: Physics, Oscillations, NEET
The displacement of a particle performing simple harmonic motion is given by, x=8 sinâ¡Ït+6cosâ¡Ït, where distance is in cm and time is in second. The amplitude of motion is

Options:
A 10 cm
B 2 cm
C 14 cm
D 3.5 cm

The correct answer is:
10 cm

Solution:
A harmonic oscillation of constant amplitude and of single frequency is called simple harmonic oscillation.
Here, x=8sin⁡ωt+6cos⁡ωt
So, a1=8 cm and a2=6 cm
∴ Amplitude of motion
A=a12+a22
= 82+62
= 64+36= 100=10 cm

Topic: Physics, Oscillations, NEET
The displacement of a particle performing simple harmonic motion is given by, x=8 sinâ¡Ït+6cosâ¡Ït, where distance is in cm and time is in second. The amplitude of motion is

Options:
A 10 cm
B 2Â cm
C 14Â cm
D 3.5Â cm

The correct answer is:
10 cm

Solution:
A harmonic oscillation of constant amplitude and of single frequency is called simple harmonic oscillation.
Here, x=8sinâ¡Ït+6cosâ¡Ït
So, a1=8 cm and a2=6 cm
â´ Amplitude of motion
A=a12+a22
= 82+62
= 64+36= 100=10 cm

Topic: Physics, Oscillations, NEET
The displacement of a particle varies with time (t)âas:âs=a t 2 âb t 3. The acceleration of the particle at any given time (t) will be equal to

Options:
A 2at-3bt2

B 2a-6bt
C 2a-6b

D None of these

The correct answer is:
2a-6bt

Solution:
sâ=âa t 2 âb t 3 (given).

Differentiating, âsâ wrt âtâ we get velocity.

â´ââââvâ=â ds dt â=â2atâ3b t 2

andââaâ=â dv dt =2aâ6bt

Topic: Physics, Motion In One Dimension, NEET
The displacement of a travelling wave $y=C \sin \frac{2 \pi}{\lambda}$ (at $-x$ ) where $t$ is time, $x$ is distance and $\lambda$ is the wavelength, all in S.I. units. Then the frequency of the wave is

Options:
A $\frac{2 \pi \lambda}{a}$
B $\frac{2 \pi a}{\lambda}$
C $\frac{\lambda}{a}$
D $\frac{a}{\lambda}$

The correct answer is:
$\frac{a}{\lambda}$

Solution:
$y=c \sin \frac{2 \pi}{\lambda}($ at $-x)$
$y=c \sin \left(\frac{2 \pi}{\lambda}\right.$ at $\left.-\frac{2 \pi}{\lambda} x\right)$
Comparing with $y=A \sin (\omega \hbar-k x)$
$\omega=2 \pi f=\frac{2 \pi a}{\lambda}$
$f=\frac{a}{\lambda}$

Topic: Physics, Waves and Sound, NEET, NEET 2024 (Re-NEET)
The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10-2cos⁡πt metre. The time at which the maximum speed first occurs is

Options:
A 0.5 s
B 0.75Â s
C 0.125Â s
D 0.25Â s

The correct answer is:
0.5 s

Solution:
x=(2×10-2)cos⁡πt
Here, a=2×10-2 m=2 cm
At t=0, x=2 cm , ie, the object is at the positive extreme, so to acquire maximum speed (i.e., to reach mean position) it takes 14th of the time period.
∴ Required time = T4
where ω=2πT=π
⇒ T=2 s
So, the required time = T4=24=0.5 s

Topic: Physics, Oscillations, NEET
The displacement of an object attached to a spring and executing simple harmonic motion is given by X = 2 Ã 10 - 2 ( cosâÏt ) . The time at which the maximum speed first occurs is

Options:
A 0.25 s
B 0.75 s
C 0.125 s
D 0.5 s

The correct answer is:
0.5 s

Solution:
Given : displacement x = 2 Ã 10 - 2 cos Ït
Velocity Î½ = d x dt = - 2 Ã 10 - 2 Ï sin Ï t
For the first time when Î½ = Î½ max , sin Ï t = 1
or sin Ï t = sin Ï 2 or Ï t = Ï 2
or t = 1 2 s =0.5âs.

Topic: Physics, Oscillations, NEET
The displacement of two identical particles executing SHM are represented by equations

x1=4sinâ¡10t+Ï6 and x2= 5 cosÏt

For what value of Ï energy of both the particles is same?

Options:
A 16 unit
B 6 unit
C 4 unit
D 8 unit

The correct answer is:
8 unit

Solution:
E1= E2
∴ 12m1ω12A12=12m2ω22A22
but m1=m2
∴ ω12×16=ω22×25
∴ 100×16=ω2×25
ω=8 units

Topic: Physics, Oscillations, NEET
The displacement of two identical particles executing SHM are represented by equations

x1=4sinâ¡10t+Ï6 and x2=5cosâ¡Ït

For what value of Ï, the energy of both the particles is the same?

Options:
A 16 unit
B 6 unit
C 4 unit
D 8 unit

The correct answer is:
8 unit

Solution:
E=12mA2ω2 i.e., E∝Aω2
As E1=E2
or A1ω12=A2ω22
or A1ω1=A2ω2
or 4×10=5×ω
or ω=8 unit

Topic: Physics, Oscillations, NEET
The displacement s - velocity (v) graph of a particle if it starts moving from rest with a uniform acceleration which is parallel to its instantaneous direction of motion is

Options:
A
B
C
D

The correct answer is:

Solution:
Since initial velocity is zero,

v2=2ass=v22a

It is similar to graph of parabola given by y=x2

Topic: Physics, Motion In One Dimension, NEET
The displacement time graph of a moving particle is shown below

The instantaneous velocity of the particle is negative at the point

Options:
A C
B E
C D
D F

The correct answer is:
E

Solution:
At E, the slope of the curve is negative, i.e. for a displacement-time graph the slope represents the velocity and at E, the velocity is −ve as the slope is −ve .

Topic: Physics, Motion In One Dimension, NEET
The displacement ' $x$ ' (in meter) of a particle of mass ' $m$ ' (in kg) moving in one dimension under the action of a force, is related to time ' $t$ ' (in sec) by $t=\sqrt{x}+3$. The displacement of the particle when its velocity is zero will be:

Options:
A $4 \mathrm{~m}$
B $0 \mathrm{~m}$ (zero)
C $6 \mathrm{~m}$
D $2 \mathrm{~m}$

The correct answer is:
$0 \mathrm{~m}$ (zero)

Solution:
According to the question
$t=\sqrt{x}+3$
$\Rightarrow \sqrt{x}=t-3$
Squaring both sides we get
$x=(t-3)^2$
Now velocity $V=\frac{d x}{d t}=\frac{d}{d t}(t-3)^2$ $=2(t-3)$ velocity becomes zero means
$\begin{aligned}
& 2(t-3)=0 \Rightarrow t=3 \mathrm{sec} . \\
& \text {At } \begin{array}{l}
t=3 \mathrm{~s} \\
\\
x=(3-3)^2=0 m
\end{array}
\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2013 (Karnataka)
The displacement x of a body of mass 1 kg on a horizontal smooth surface as a function of time t is given by x=t44. The work done in the first second is

Options:
A 14 J
B 12 J
C 34 J
D 54 J

The correct answer is:
12 J

Solution:
v=dxdt=14Ã4t3=t3=1 m s-1
Work done = Change in K.E.
= 1 2 Ã 1 Ã 1 - 1 2 Ã 1 Ã 0
=12 J

Topic: Physics, Work Power Energy, NEET
The displacement $x$ of a particle varies with time $t$ as $x=a e^{-a t}+B^{\beta x}$, where $a, b, \alpha$ and $\beta$ are positive constants. The velocity of the particle will:

Options:
A be independent of $\beta$
B drop to zero when $\alpha=\beta$
C go on decreasing with time
D go on increasing with time

The correct answer is:
go on increasing with time

Solution:
From question
$$
\begin{aligned}
& x=a e^{-\alpha \mathrm{t}}+b e^{\beta t} \\
& V=\frac{d x}{d t}=-a \alpha e^{-x t}+b \beta_e^{\mathrm{B}+} \\
& \Rightarrow \mathrm{V}=-a \alpha e^{-t}+b \beta \varepsilon^{\beta t}
\end{aligned}
$$
$\therefore$ Velocity will decrease for certain value of $t$.

Topic: Physics, Motion In One Dimension, NEET, NEET 2005
The displacement x of a particle varies with time t as x=ae-αt+beβt, where a, b, α and β are positive constants. The velocity of the particle will

Options:
A be independent of β
B drop to zero, when α=β
C decrease with time
D increase with time

The correct answer is:
increase with time

Solution:
x=ae-Î±t+beÎ²t
dxdt=-aÎ±e-Î±t+bÎ²eÎ²t
âv=bÎ²eÎ²t-aÎ±e-Î±t
As t increases, e-Î±t decreases and eÎ²t increases.
So v will increase with time.

Topic: Physics, Motion In One Dimension, NEET
The displacement-time graphs for two sound waves A and B are shown in the figure, then the ratio of their intensities IAIB is equal to

Options:
A 1 :â4
B 1 :â16
C 1 :â2
D 1 :â1

The correct answer is:
1 :â1

Solution:
The intensity can be given by

I=2π2ρ2A2v

Hence,
IAIB=fA2AA2fB2AB2
IAIB=122212
IAIB=1

Topic: Physics, Waves and Sound, NEET
The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis as shown in the figure. The ratio of their respective velocity is:

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

The correct answer is:
1:3

Solution:
Slope of displacement-time graph is velocity.
v1v2=tanθ1tanθ2=tan30°tan45°=13
⇒v1:v2=1:3

Topic: Physics, Motion In One Dimension, NEET, NEET 2022 (Phase 1)
The displacement-time x-t graph of a body is given as

The corresponding velocity-time v-t graph will be

Options:
A
B
C
D

The correct answer is:

Solution:
As we known that, the velocity of a body is given by the slope of displacement - time graph. So initially slope of the graph is positive and after sometimes, it becomes zero and then will be negative.

Topic: Physics, Motion In One Dimension, NEET
The dissociation constant of a substituted benzoic acid is 1.0Ã10-4 at 25oC. The pH of 0.01 M solution of its sodium salt is

Options:
A 7
B 6
C 8
D 4

The correct answer is:
8

Solution:
For a salt of a weak acid and strong base
pH=7+12pKa+12logC
=7+ 1 2 −log 1× 10 −4 + 1 2 log 0.01
=8.

Topic: Chemistry, Ionic Equilibrium, NEET
The dissociation constant of a weak acid is $1 \times 10^{-4}$. In order to prepare a buffer solution with a pH $=5$, the [Salt]/ [Acid] ratio should be:

Options:
A $4: 5$
B $10: 1$
C $5: 4$
D $1: 10$

The correct answer is:
$10: 1$

Solution:
$\mathrm{pH}=\mathrm{pK}_a+\log \frac{[\text { Salt }]}{\text { Acid }}$
$$
\begin{aligned}
& 5=\log \left[1 \times 10^{-4}\right]+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 5=4+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& \frac{[\text { Salt }]}{[\text { Acid }]}=10: 1
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2013 (Karnataka)
The dissociation constant of a weak monoprotic acid, which is 0.01% ionised in 1.00M solution, is

Options:
A 1×10-8
B 1×10-4
C 1×10-6
D 10-5

The correct answer is:
1×10-8

Solution:
K =  α 2 C 1−α ; 1−α≈1
â´Kâ=â Î± 2 Câ=â 0.01 100 2 Ãâ1
=1Ã10-8 .

Topic: Chemistry, Ionic Equilibrium, NEET
The dissociation constants for acetic acid and $\mathrm{HCN}$ at $25^{\circ} \mathrm{C}$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$, respectively. The equilibrium constant for the equilibrium,
$\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \quad \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-}$
would be

Options:
A $3.0 \times 10^5$
B $3.0 \times 10^{-5}$
C $3.0 \times 10^{-4}$
D $3.0 \times 10^4$

The correct answer is:
$3.0 \times 10^4$

Solution:
Given, $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
$$
\begin{array}{r}
\mathrm{K}_{\mathrm{a}}=1.5 \times 10^{-5} \ldots \text { (i) } \\
\mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-} ; \quad \mathrm{K}_{\mathrm{a}}=4.5 \times 10^{-10} \ldots \text { (ii) } \\
\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} \\
\mathrm{K}=\text { ? }
\end{array}
$$
On substracting Eq. (ii) from Eq. (i), we get
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CN}^{-} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} ; \\
& \mathrm{K}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{a}_1}}=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=\frac{10^5}{3}=3.33 \times 10^4
\end{aligned}
$$
While adding two equations, dissociation constants are multiplied and when subtracting the equations, dissociation constants are divided.
Alternative
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} ; \mathrm{K}_{\mathrm{a}}=1.5 \times 10^{-5} \\
& \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=1.5 \times 10^{-5} \\
& \mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-} ; \mathrm{K}_{\mathrm{a}}=4.5 \times 10^{-10} \\
& \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}=4.5 \times 10^{-10} \\
& \mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} \\
& \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HCN}]\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CN}^{-}\right]\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\
&
\end{aligned}
$$
From Eqs. (i), (ii) and (iii),
$$
\mathrm{K}_c=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=3.33 \times 10^4
$$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2009 (Screening)
The dissociation energies of H2 and O2 are 104 and 118 kcal mol-1 respectively. The heat of reaction
12H2(g)+12O2(g)→O-H(g) is 10 kcal. The bond energy of O-H bond is

Options:
A 101 kcal/mol
B 111 kcal/mol
C 10.1 kcal/mol
D 11.1 kcal/mol

The correct answer is:
101 kcal/mol

Solution:
ΔHrxn=ΣBE (reactants) -ΣBE (Products)
ΔHrxn=12.BE(H-H)+12BE(O=O)-1BE(O-H)
So 10 kcal/mol
=12×104 kcal/mol +12×118 kcal/mol -BE(O-H)
or BE(O-H)=101 kcal/mol

Topic: Chemistry, Thermodynamics (C), NEET
The dissociation equilibrium of a gas $A B_2$ can be represented as
$2 A B_2(g) \rightleftharpoons 2 A B(g)+B_2(g)$
The degree of dissociation is ' $x$ ' and is small compared to 1. The expression relating the degree of dissociation $(x)$ with equilibrium constant $K_P$ and total pressure $p$ is

Options:
A $\left(2 K_p / p\right)$
B $\left(2 K_p / p\right)^{1 / 3}$
C $\left(2 K_p / p\right)^{1 / 2}$
D $\left(K_p / p\right)$

The correct answer is:
$\left(2 K_p / p\right)^{1 / 3}$

Solution:
\(\begin{array}{cccr}
2 A B_{2(g)} & \rightleftharpoons & 2 A B_{(g)} & + & B_{2(g)} & \\
2 & & 0 & & 0 & \text { (initially) } \\
2(1-x) & & 2 x & & x & \text { (at equilibrium) }
\end{array}\)
\(\begin{aligned} \text {Amount of moles at equilibrium } & =2(1-x)+2 x+x \\ & =2+x\end{aligned}\)
\(\begin{aligned} & K_p= \frac{\left[p_{A B}\right]^2\left[p_{B_2}\right]}{\left[p_{A B_2}\right]^2} \\ & K_p= \frac{\left(\frac{2 x}{2+x} \times P\right)^2 \times\left(\frac{x}{2+x} \times P\right)}{\left(\frac{2(1-x)}{2+x} \times P\right)^2}=\frac{\frac{4 x^3}{2+x} \times P}{4(1-x)^2} \\ & K_p= \frac{4 x^3 \times P}{2} \times \frac{1}{4} \quad(\because 1-x \approx 1 \& 2+x \approx 2) \\ & x=\left(\frac{8 K_p}{4 P}\right)^{1 / 3} \quad \Rightarrow x=\left(\frac{2 K_p}{P}\right)^{1 / 3}\end{aligned}\)

Topic: Chemistry, Biomolecules, NEET, NEET 2008 (Screening)