datavorous/entrance-exam-dataset huggingface dataset:
The vector sum of two forces is perpendicular to their vector differences. In that case, the force:

Options:
A are equal to each other
B are equal to each other in magnitude
C are not equal to each other in magnitude
D cannot be predicted

The correct answer is:
are equal to each other in magnitude

Solution:
$(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0$
$\begin{aligned}
& \Rightarrow A^2-\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{A}+B^2=0 \\
& \Rightarrow \mathrm{A}=\mathrm{B} \quad(\because \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A})
\end{aligned}$

Topic: Physics, Mathematics in Physics, NEET, NEET 2003
The vector used in the first successful recombinant DNA was from:

Options:
A a bacteria that causes pneumonia.
B a bacteria against which the first antibiotic was discovered.
C a bacteria that can be detected by WIDAL test.
D None of these

The correct answer is:
a bacteria that can be detected by WIDAL test.

Solution:
The construction of the first recombinant DNA emerged from the possibility of linking a gene encoding antibiotic resistance with a native plasmid (autonomously replicating circular extra-chromosomal DNA) of Salmonella typhimurium. Stanley Cohen and Herbert Boyer accomplished this in 1972 by isolating the antibiotic resistance gene by cutting out a piece of DNA from a plasmid which was responsible for conferring antibiotic resistance.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is:

Options:
A $45^{\circ}$
B $90^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
According to the question:
$\begin{aligned}
|\vec{A}+\vec{B}| & =|\vec{A}-\vec{B}| \\
\Rightarrow |\vec{A}+\vec{B}|^2 & =|\vec{A}-\vec{B}|^2
\end{aligned}$
$\begin{aligned}
& A^2+B^2+2 A B=A^2+B^2-2 A B \\
& 4 \vec{A} \cdot \vec{B}=0 \\
& \vec{A} \cdot \vec{B}=0 \\
& \Rightarrow A B \cos \theta=0 \\
& \Rightarrow \theta=90^{\circ} \\
& {[\text {As } A=B \neqeq 0] }
\end{aligned}$

Topic: Physics, Mathematics in Physics, NEET, NEET 2006
The velocities are in ground frame and the cylinder is performing pure rolling on the plank, velocity of point 'A' would be

Options:
A 2VC
B 2VC-VP
C 2VC+VP
D 2VC-VP

The correct answer is:
2VC-VP

Solution:
In case of pure rolling relative velocity of bottom most point is zero
Vc-rω=Vp ...(i)
VA=VC+rω=2Vc-Vp

Topic: Physics, Rotational Motion, NEET
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, what is the wavelength associated with this velocity?

Options:
A 3.652 × 10–38m
B 0.112 × 10–38m
C 2.468 × 10–38m
D 1.516 × 10–38m

The correct answer is:
1.516 × 10–38m

Solution:
Correct Option is : (D)
1.516 × 10–38m

Topic: Chemistry, Structure of Atom, NEET
The velocity displacement graph of a particle moving along a straight line is -

The most suitable acceleration-displacement graph will be

Options:
A
B

C

D

The correct answer is:

Solution:
The equation for the given v - x graph is
v=â v 0 x 0 x+ v 0 ...(i)
Differentiating the above equation w.r.t. x we get
dv dx =â v 0 x 0
Multiplying both sides of the above equation by v, we get
v dv dx =â v 0 x 0 Ãv=â v 0 x 0 [ â v 0 x 0 x+ v 0 ] From (i),
â´a= v 0 2 x 0 2 xâ v 0 2 x 0 ...(ii) [ âµa=v dv dx ]
On comparing equation (ii) with equation of a straight line y=mx+c where m is the slope of the line and c is its intercept on y -axis we get, m= v 0 2 x 0 2 [therefore, m will always be positive]
â The slope m or tanÎ¸=0 or â² Î¸ â² is an acute angle.
Also, the comparison gives -
c= â v 0 2 x 0 or the intercept will be negative.
These conditions are met only in this graph.

Topic: Physics, Motion In One Dimension, NEET
The velocity of a body of mass 2 kg as a function of t is given by v=2t i + t2j. Find the momentum acting on it, at time t= 2s.

Options:
A 5i+5j
B 5i-5j
C 8i+8j
D 8i-8j

The correct answer is:
8i+8j

Solution:
Correct Option is : (C)
8i+8j

Topic: Chemistry, Laws of Motion, NEET
The velocity of a particle is zero at time t=2 s, then

Options:
A displacement must be zero in the interval t=0 to t=2 s.
B acceleration may be zero at t=2 s.
C velocity must be zero for t>2 s.
D acceleration must be zero at t=2 s.

The correct answer is:
acceleration may be zero at t=2 s.

Solution:
Let us consider an example in which particle is projected vertically upwards from ground at t=0 and it reaches the highest point at t=2 s.
Then, at that instant (t=2 s) velocity v=0, but acceleration =-g.
Here, displacement is non zero for the duration t=0 to t=2 s. For t>2 s the ball again acquires velocity. In this example option (1), (3) and (4) are incorrect.
Let us consider another example in which a particle is moving on a horizontal plane and it comes to rest permanently at t=1 s, then this is one of the special case in which acceleration of the particle is zero at t=2 s.

Topic: Physics, Motion In One Dimension, NEET
The velocity of a small ball of mass M and density ρ, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is ρ2, then the viscous force acting on the ball will be:

Options:
A 3Mg2
B Mg
C 12Mg
D 2Mg

The correct answer is:
12Mg

Solution:
When velocity becomes constant then,

Fv+Fb=Mg

Fv=Mg-Fb

=Vρbg-Vρgg

=Vgρ-ρ2

=Vgρ2

=(Vρ)g2=Mg2

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2021
The velocity of electromagnetic radiation in a medium of permittivity $\varepsilon_0$ and permeability $\mu_0$ is given by

Options:
A $\sqrt{\frac{\varepsilon_0}{\mu_0}}$
B $\sqrt{\mu_0 \varepsilon_0}$
C $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
D $\sqrt{\frac{\mu_0}{\varepsilon_0}}$

The correct answer is:
$\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

Solution:
Velocity of electromagnetic radiation is the velocity of light (c), ie,
$$
c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
$$
where $\mu_0$ is the permeability and $\varepsilon_0$ is the permittivity of free space.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2008 (Screening)
The velocity of kerosene oil in a horizontal pipe is 5 m s-1. If g=10 m s-1 then the velocity head of oil will be

Options:
A 1.25 m
B 12.5 m
C 0.125 m
D 125 m

The correct answer is:
1.25 m

Solution:
Velocity head h = v 2 2 g = 5 2 2 × 1 0 = 1 . 2 5 m

Topic: Physics, Mechanical Properties of Fluids, NEET
The velocity of kerosene oil in a horizontal pipe is 5 m s-1. If g=10 m s-2, then the velocity head of oil will be

Options:
A 1.25 m
B 12.5Â m
C 0.125Â m
D 125Â m

The correct answer is:
1.25 m

Solution:
In Bernoulli's theorem,
Velocity head=v22g
Velocity head=(5)22×10
Velocity head=1.25 m

Topic: Physics, Mechanical Properties of Fluids, NEET
The velocity of sound waves in air is 330 m s-1. For a particular sound in air, a path difference of 40 cm is equivalent to a phase difference of 1.6Ï. The frequency of this wave is

Options:
A 165Â Hz
B 150Â Hz
C 660 Hz
D 330Â Hz

The correct answer is:
660 Hz

Solution:
Phase difference =2ÏÎ»Ã path difference
â 1.6Ï=2ÏÎ»Ã40
â Î»=50cmâ¡=0.5 m
â v=nÎ»
â 330=0.5Ãn
â n=660 Hz

Topic: Physics, Waves and Sound, NEET
The velocity of the electromagnetic wave is parallel to:

Options:
A $\vec{B} \times \vec{E}$
B $\vec{E} \times \vec{B}$
C $\vec{E}$
D $\vec{B}$

The correct answer is:
$\vec{E} \times \vec{B}$

Solution:

Topic: Physics, Electromagnetic Waves, NEET, NEET 2002
The velocity $v$ of a particle at time $t$ is given by $v=a t+\frac{b}{t+c}$, where $a, b$ and $c$ are constants. The dimensions of $a, b$ and $c$ are:

Options:
A L, LT and $\mathrm{LT}^2$
B $\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$
C $\mathrm{L}^2, \mathrm{~T}$ and $\mathrm{LT}^2$
D $\mathrm{LT}^2$, LT and L.

The correct answer is:
$\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$

Solution:
Using principle of homogeniety.
Dimension of $v=$ Dimension of $a t=$
Dimenstion of $\frac{b}{t+c}$
$\begin{array}{l}
\therefore a t =\left[\mathrm{LT}^{-1}\right] \\
\Rightarrow |a| =\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-2}\right] \\
\text {And } \frac{[b]}{[t]} =\left[\mathrm{LT}^{-1}\right] \\
{[b]} =\left[\mathrm{LT}^{-1}\right][\mathrm{T}]=[\mathrm{L}] \\
\text {And } c =\mathrm{T} \\
\Rightarrow c =[\mathrm{T}]
\end{array}$

Topic: Physics, Units and Dimensions, NEET, NEET 2006
The velocity $(v)$ - time $(t)$ plot of the motion of a body is shown below:
The acceleration (a) - time $(t)$ graph that best suits this motion is :

Options:
A

B

C

D

The correct answer is:

Solution:
Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially. After that, the slope of v-t curve is constant and positive.
After some time, velocity becomes constant and acceleration is zero.
After that, the slope of v-t curve is constant and negative.

Topic: Physics, Motion In One Dimension, NEET, NEET 2024
The velocity-time graph of a particle moving along a straight line is as shown in the diagram. The rate of acceleration and retardation is constant and is equal to 5âm s-2. If the average velocity during the motion is 20âm s-1, then the maximum velocity of the particle is

Options:
A 20 m s-1
B 25âm s-1
C 30 m s-1
D 40 m s-1

The correct answer is:
25âm s-1

Solution:
Average speed =20âm s-1
Total distance Total time = 2 0 â Area of graph 2 0 + t = 2 0
â 1 2 2 0 + t + 2 0 - t Ã 5 t = 2 0 2 0 + t â t = 5 s

Maximum velocity =5t=5Ã5=25âm s-1

Topic: Physics, Motion In One Dimension, NEET
The ventral diaphragm of cockroach is present in between

Options:
A Thorax and abdomen
B Pericardial and perivisceral sinuses
C Perivisceral and perineural sinuses
D Pericardial and perineural sinuses

The correct answer is:
Perivisceral and perineural sinuses

Solution:
The body of the cockroach is divided into three sinuses. Pericardial on the dorsal side, perineural on the ventral side and perivisceral in the center. Two diaphragms separate these three sinuses. The ventral diaphragm is between perivisceral and perineural sinuses and the dorsal diaphragm is between pericardial and perivisceral sinuses. There is ostia present, which acts as communication between these sinuses.

Topic: Biology, Structural Organisation in Animals, NEET
The ventricles are relax and the ventricular pressure falls causing the closure of semilunar valves, the above events are related to

Options:
A ventricular systole.
B ventricular failure.
C ventricular diastole.
D More than one option is correct

The correct answer is:
ventricular diastole.

Solution:
As the ventricular muscle relaxes, pressure on the blood within the ventricles drops even further. This causes closing of semilunar valves producing second heart sound called 'dub' and the phase is called ventricular diastole. Since at the same time atria is also under diastole state, it is called joint diastole stage. Eventually, it drops below the pressure in the atria. When this occurs, blood flows from the atria into the ventricles, pushing open the tricuspid and mitral valves.

Topic: Biology, Body Fluids and Circulation, NEET
The ventricles are relaxed and the ventricular pressure falls causing the closure of semilunar valves. These events are related to

Options:
A atrial depolarization
B ventricular systole
C ventricular diastole
D joint diastole

The correct answer is:
ventricular diastole

Solution:
As ventricular muscle relaxes, pressure on the blood within the ventricles drops even further. Eventually, it drops below the pressure in the atria. When this occurs, blood flows from the atria into the ventricles, pushing open the tricuspid and mitral valves.

Topic: Biology, Body Fluids and Circulation, NEET
The ventricular muscles relax and the ventricular pressure falls causing the closure of semilunar valves. These events are related to

Options:
A atrial depolarization
B ventricular systole
C ventricular diastole
D joint diastole

The correct answer is:
ventricular diastole

Solution:
As ventricular muscles relax, pressure on the blood within the ventricles drops even further. Eventually, it drops below the pressure in the atria. When this occurs, blood flows from the atria into the ventricles, pushing open the tricuspid and mitral valves.

Topic: Biology, Body Fluids and Circulation, NEET
The vertebrate Myxine has been classified under which division of phylum Chordata?

Options:
A Cyclostomata
B Gnathostomata
C Agnatha
D Chondrichthyes

The correct answer is:
Agnatha

Solution:
Myxine (hagfish) is a jawless vertebrate, which is classified under class Cyclostomata, under the division (superclass) of Agnatha. All living members of the class Cyclostomata are ectoparasites on some fishes. Other examples include Petromyzon.

Topic: Biology, Animal Kingdom, NEET
The vertebrate shown in this figure is

Options:
A Ornithorhynchus
B Macropus
C Pteropus
D Balaenoptera

The correct answer is:
Pteropus

Solution:
The Indian flying fox (Pteropus medius), also known as the greater Indian fruit bat, is one of the largest bats in the world. Bats are mammals of the order Chiroptera with forelimbs adapted as wings, they are the only mammals that is capable for flight with spread-out digits covered by a thin membrane called patagium.

Topic: Biology, Animal Kingdom, NEET
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:

Options:
A 10
B 10-6
C 106
D 100

The correct answer is:
10-6

Solution:
Forward bias resistance
= ÎV ÎI = 0.1 10Ã 10 â3 =10âÎ©
Reverse bias resistance = 10 10 â6 = 10 7 âÎ©
Ratio of resistances
= Forwardâ biasâ resistance Reverseâ biasâ resistance = 10 â6

Topic: Physics, Semiconductors, NEET
The V-I characteristic of a silicon diode is shown in the Fig. Calculate the resistance of the diode at ID= 15 mA

Options:
A 5 ohms
B 10 ohms
C 15 ohms
D 20 ohms

The correct answer is:
10 ohms

Solution:
Correct Option is : (B)
10 ohms

Topic: Chemistry, Semiconductors, NEET
The V-I characteristic of a silicon diode is shown in the Fig. Calculate the resistance of the diode at VD= –10 V.

Options:
A 2.0 × 107ohms
B 2.0 × 108ohms
C 1.0 × 107ohms
D 1.0 × 108ohms

The correct answer is:
1.0 × 107ohms

Solution:
Correct Option is : (C)
1.0 × 107ohms

Topic: Chemistry, Semiconductors, NEET
The virologist _______ showed that the viruses could be crystallized.

Options:
A W. M. Stanley
B T.O. Diener
C M.W. Beijerinek
D D.J. Ivanowsky

The correct answer is:
W. M. Stanley

Solution:
Wendell Meredith Stanley, an American virologist studied the tobacco mosaic virus, which affects the tobacco leaves. He experimented by using the infected tobacco leaves and was successful in extracting the virus in form of crystals.
T.O. Diener -Viriods are nothing but free DNA.
M.W. Beijerinek -Dutch microbiologist, 'Contagiumvivumfluidum'
D.J. Ivanowsky - Coined the term 'virus'

Topic: Biology, Biological Classification, NEET
The virus that created a scare in the country and drastically affected egg and chicken consumption is

Options:
A HPV
B Hepatitis virus
C Bird flu virus
D Herpes virus

The correct answer is:
Bird flu virus

Solution:
The bird flu virus is called H5N1 or avian influenza virus. Bird flu is a highly pathogenic avian influenza type. This influenza is most often spread by contact with the infected birds. The virus is found in secretions from the nostrils, mouth, and eyes of infected birds as well as their droppings.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
The viscous drag acting on a metal sphere of diameter $1 \mathrm{~mm}$, falling through a fluid of viscosity $0.8 \mathrm{~Pa} s$ with a velocity of $2 \mathrm{~m} \mathrm{~s}^{-1}$ is equal to

Options:
A $1.5 \times 10^{-3} \mathrm{~N}$
B $20 \times 10^{-3} \mathrm{~N}$
C $15 \times 10^{-3} \mathrm{~N}$
D $30 \times 10^{-3} \mathrm{~N}$

The correct answer is:
$15 \times 10^{-3} \mathrm{~N}$

Solution:
Viscous drag acting on sphere is given by
$\begin{aligned} F_V & =6 \pi \eta r V \\ & =6 \times \frac{22}{7} \times 0.8 \times \frac{10^{-3}}{2} \times 2 \\ & =15.08 \times 10^{-3} \mathrm{~N} \\ & \simeq 15 \times 10^{-3} \mathrm{~N}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2023 (Manipur)
The vitamin whose content increases following the conversion of milk into curd by lactic acid bacteria is

Options:
A Vitamin C
B Vitamin D
C Vitamin B12
D Vitamin E

The correct answer is:
Vitamin B12

Solution:
The vitamin B12 content increases following the conversion of milk into curd by lactic acid bacteria (LAB).

Topic: Biology, Microbes in Human Welfare, NEET
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

Options:
A 453.8m and this is a characteristic microwave wavelength
B 219.3m and this is a characteristic microwave wavelength
C 453.8m and this is a characteristic radiowave wavelength
D 219.3m and this is a characteristic radiowave wavelength

The correct answer is:
219.3m and this is a characteristic radiowave wavelength

Solution:
Correct Option is : (D)
219.3m and this is a characteristic radiowave wavelength

Topic: Chemistry, Structure of Atom, NEET
The void space in a primitive unit cell is:

Options:
A 48% void space
B 24% void space
C 96% void space
D 50% void space

The correct answer is:
48% void space

Solution:
a = 2r
Volume of the cube =a3=2r3=8r3
Packing fraction=Volume of one atomVolume of the cube
=43 Ïr38r3=Ï6=0.52
Void fraction = 1 â 0.52 = 0.48
Void space =48 %

Topic: Chemistry, Solid State, NEET
The voltage and current in a conductor are measured as (50 Â± 2) V and (5 Â± 0.2) A. The percentage error in the calculation of resistance is

Options:
A 4%
B 5%
C 7%
D 8%

The correct answer is:
8%

Solution:
R=VI
ÎRR=ÎVV+ ÎII
â´ % error in R=250Ã100+0.25Ã100=8%

Topic: Physics, Units and Dimensions, NEET
The voltage applied to the Coolidge X-ray tube is increased by 25%. As a result, the short wave limit of continuous X-ray spectrum shifts by Δλ. The initial voltage applied to the tube is

Options:
A hc4eΔλ
B hc5eΔλ
C 4hc5eΔλ
D 5hc4eΔλ

The correct answer is:
hc5eΔλ

Solution:
12mv2max=eV=hcλ
⇒ λ=hceV⇒λ1=hceV1, λ2=hceV2⇒ λ1-λ2=hceV1-hceV2
⇒ Δλ=hce1V1-45V1=hceV1×15
V=hc5⋅eΔλ

Topic: Physics, Atomic Physics, NEET
The voltage drop between the points A and B in the given figure is

Options:
A 24 V
B 14 V
C 32 V
D 48 V

The correct answer is:
48 V

Solution:
The current through each circuit element is 2 A, so

VA-2×6-12-2×9+4-2×5=VB⇒VA-VB=48 V

Topic: Physics, Current Electricity, NEET
The voltage E and the current I in an instrument are represented by the equations:

E=2cos⁡ωt V

I=2sin⁡ωt A

The average power dissipated in the instrument will be

Options:
A zero
B 1.0 W
C 4 W
D 2.0 W

The correct answer is:
zero

Solution:
Current is
I=2sin⁡ωt A
Voltage is
E=2cos⁡ωt=2sin⁡ωt+π2 V
The phase difference is of π2, so power factor is zero. Therefore average power dissipated in the instrument is zero.

Topic: Physics, Alternating Current, NEET
The voltage gain of an amplifier stage is 100. The gain expressed in dB is :

Options:
A 100
B 20
C 40
D 10

The correct answer is:
20

Solution:
Y=10log100
=20log10
Y=20dB

Topic: Physics, Semiconductors, NEET
The voltage gain of an amplifier with $9 \%$ negative feedback is 10. The voltage gain without feedback will be

Options:
A 90
B 10
C 1.25
D 100

The correct answer is:
100

Solution:
The process of injecting a fraction of output energy of some device back to the input is known as feedback. When the feedback energy (voltage or current) is out of phase with the input signal and thus opposes it, it is called negative feedback.
Voltage gain with feedback is
$A_{v f}=\frac{A_\nu}{1+\beta A_\nu}$
where $A_\psi$ is voltage gain without feedback and $\beta$ is negative feedback.
Given $A_{v f}=10, \beta=9 \%=\frac{9}{100}$
$\begin{array}{ll}
\therefore & 10=\frac{A_\nu}{1+\frac{9}{100} A_\nu} \\
\text {or } & 10+\frac{9}{10} A_\nu=A_\nu \\
\text {or } & 0.1 A_\nu=10 \text { or } A_\nu=100
\end{array}$

Topic: Physics, Semiconductors, NEET, NEET 2008 (Screening)
The volume occupied by $1.8 \mathrm{~g}$ of water vapour at $374^{\circ} \mathrm{C}$ and 1 bar pressure will be [Use $\mathrm{R}=0.083$ bar $\mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ]

Options:
A $96.66 \mathrm{~L}$
B $55.87 \mathrm{~L}$
C $3.10 \mathrm{~L}$
D $5.37 \mathrm{~L}$

The correct answer is:
$5.37 \mathrm{~L}$

Solution:
According to ideal gas equation,
$$
\begin{aligned}
& \mathrm{pV}=\mathrm{nRT} \\
& \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{p}}=\frac{\mathrm{w}}{\text { M.wt }} \frac{\mathrm{RT}}{\mathrm{p}} \ldots(\mathrm{i})\left[\because \mathrm{n}=\frac{\mathrm{w}}{\mathrm{M} \cdot \mathrm{wt}}\right]
\end{aligned}
$$
Given, $\mathrm{w}=1.8 \mathrm{~g}, \mathrm{~T}=374^{\circ} \mathrm{C}$
$$
=(374+273) \mathrm{K}=647 \mathrm{~K}
$$
$$
\mathrm{p}=1 \text { bar, } \mathrm{R}=0.083 \mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1}
$$
On substituting the given values in Eq. (i), we get
$$
\begin{aligned}
\mathrm{V} & =\frac{1.8 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}} \times \frac{0.083 \mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1} \times 647 \mathrm{~K}}{1 \mathrm{bar}} \\
& =5.37 \mathrm{~L}
\end{aligned}
$$

Topic: Chemistry, States of Matter, NEET, NEET 2019 (Odisha)
The volume occupied by an atom is greater than the volume of the nucleus by a factor of about:

Options:
A $10^1$
B $10^5$
C $10^{10}$
D $10^{15}$

The correct answer is:
$10^{15}$

Solution:
$\frac{\text { Volume of atom }}{\text { Volume of nucleus }} =\frac{\frac{4}{3} \pi\left(10^{-10}\right)^3}{\frac{4}{3} \pi\left(10^{-15}\right)^3}=10^{15}$

Topic: Physics, Nuclear Physics, NEET, NEET 2003
The volume occupied by formed elements in the blood is

Options:
A 30%
B 45%
C 55%
D 75%

The correct answer is:
45%

Solution:
Blood is composed of -
A. Plasma
- Straw coloured viscous fluid. - Forming nearly 55% of the blood.
- Plasma = 90-92 % Water + 6-8% Proteins + Minerals + Simple molecules (glucose, amino acids, lipids, etc).
B. Formed Elements
- Forming nearly 45% of the blood.

Topic: Biology, Body Fluids and Circulation, NEET
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is

Options:
A 5.6×103 m3
B 5.6×10-3 m3
C 5.6 m3
D 5.6×106 m3

The correct answer is:
5.6 m3

Solution:
Molecular mass of water is 18 g mol-1

Therefore, number of moles in 4.5 kg of water will be,

n=450018=250

Volume occupied by one mole of gas at STP is 22.4 Litre

Therefore, volume occupied by 250 mol of water will be

V=22.4×10-3×250

⇒ V=5.6 m3

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2022 (Phase 1)
The volume of 0.1 N dibasic acid sufficient to neutralize 1g of a base that furnishes 0.04 mole of OH- in aqueous solution is :

Options:
A 400 mL
B 600 mL
C 200 mL
D 800 mL

The correct answer is:
200 mL

Solution:
N1V1=N2V2
0.1Ã V 1 = 0.04 2
V1=0.2 L
=0.2Ã1000=200 ml

Topic: Chemistry, Redox Reactions, NEET
The volume of a gas at 20Â°C is 200 ml . If the temperature is reduced to -20Â°C at constant pressure, its volume will be

Options:
A 172.6 ml
B 17.26Â ml
C 192.7Â ml
D 19.27Â ml

The correct answer is:
172.6 ml

Solution:
VâT
â V1V2=T1T2
â 200V2=273+20273-20=293253
â V2=200Ã253293=172.6 ml

Topic: Physics, Thermodynamics, NEET
The volume of air in the dead space fo human lungs is normally

Options:
A 230 mL
B 210 mL
C 190 mL
D 150 mL

The correct answer is:
150 mL

Solution:
150 mL of air is always gets stuck in the lungs which are called anatomically dead space. The dead space represents the volume of ventilated air that does not participate in gas exchange during the respiratory process.

Topic: Biology, Breathing and Exchange of Gases, NEET