datavorous/entrance-exam-dataset huggingface dataset:
The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A body starting from the rest at the top comes back to rest at the bottom if the coefficient of friction for the lower half is given by

Options:
A μ = sinθ
B μ = cotθ

C μ = 2cosθ

D μ = 2tanθ

The correct answer is:
μ = 2tanθ

Solution:
For the upper half v2=u2+2al/2=2(gsinθ)l/2=glsinθ For lower half ⇒ 0=u2+2g(sinθ−μcosθ)l2 ⇒ −glsinθ=gl(sinθ−μcosθ) ⇒ μcosθ=2sinθ⇒μ=2tanθ

Topic: Physics, Laws of Motion, NEET
The upright pyramid of number is absent in

Options:
A pond
B forest
C lake
D grassland

The correct answer is:
forest

Solution:
Pyramid of number represents the number of individuals per unit area at various trophic levels. It is always upright in grassland, pond and lake ecosystems. But in forest or single tree ecosystem, it is spindle-shaped and if parasitic food chain is considered, then it will be an inverted pyramid.

Topic: Biology, Ecosystem, NEET, NEET 2012 (Screening)
The U-shaped and C-shaped structures are

Options:
A Hyoid bone and duodenum
B Sternum bone and jejenum
C Hyoid bone and ileum
D Hyoid bone and colon

The correct answer is:
Hyoid bone and duodenum

Solution:
The tongue is attached to the mandible, the styloid processes of the temporal bones, and the hyoid bone. The hyoid is unique in that it only distantly/indirectly articulates with other bones. The tongue is positioned over the floor of the oral cavity.
The duodenum of the large intestine is a C-shaped structure and continues with jejunum a coiled structure. The colon is a 5-6âft long, inverted, U-shaped part of the large intestine present in the lower gastrointestinal tract.

Topic: Biology, Digestion and Absorption, NEET
The uterus open into vagina through a narrow

Options:
A clitoris
B hymen
C cervix
D pelvis

The correct answer is:
cervix

Solution:
The female reproductive system consists of a pair of ovaries, a pair of oviducts, uterus, vagina, and vulva. The uterus is also known as the womb or hysteria. The parts of the uterus are fundus, body, and cervix. Cervix is between the body of uterus and vagina and this shows two openings internal and external os. Thus uterus opens into the vagina through cervix.

Topic: Biology, Human Reproduction, NEET
The uterus opens into ....A.... through a narrow ......B...... The cavity of the .......B....... is called .......C....... canal which along with .......A....... forms the .......D....... canal.

Options:
A

A	B	C	D

Vagina	Cervix	Birth	Cervical

B

A	B	C	D

Vagina	Cervix	Cervical	Birth

C

A	B	C	D

Cervix	Vagina	Birth	Cervical

D

A	B	C	D

Cervix	Vagina	Cervical	Birth

The correct answer is:

A	B	C	D

Vagina	Cervix	Cervical	Birth

Solution:
The uterus is single and it is also called womb. The shape of the uterus is like an inverted pear. It is supported by ligaments attached to the pelvic wall. The uterus opens into vagina through a narrow cervix. The cavity of the cervix is called cervical canal which alongwith vagina forms the birth canal.

Topic: Biology, Human Reproduction, NEET
The UV-visible absorption bands in the spectra of lanthanoid ions are ' $X$ ', probably because of the excitation of electrons involving ' $Y$ '. The ' $X$ ' and ' $Y$ ', respectively, are :

Options:
A Broad and f orbitals
B Narrow and f orbitals
C Broad and d and f orbitals
D Narrow and d and f orbitals

The correct answer is:
Narrow and f orbitals

Solution:
In lanthanoids, absorption bands are narrow because of excitation within f-level.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2024 (Re-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
=ΔVΔI=0.110×10−3=10 Ω
Reverse bias resistance =1010−6=107 Ω
Ratio of resistances
= Forward  bias  resistance Reverse  bias  resistance = 10 −6

Topic: Physics, Semiconductors, NEET
The V-I graph for a wire of copper of length L and cross-section area A is shown in the figure below. The slope of the graph will be

Options:
A Less if the experiment is repeated at a higher temperature
B More if a wire of silver having the same dimension is used
C Doubled if the length of the wire is doubled
D Doubled if length of the wire is halved

The correct answer is:
Doubled if the length of the wire is doubled

Solution:
On doubling the length of wire its resistance is doubled and slope of V-I graph is doubled

Topic: Physics, Current Electricity, NEET
The vacant space in bcc lattice unit cell is:

Options:
A 26%
B 48%
C 23%
D 32%

The correct answer is:
32%

Solution:
Step 1: Packing efficiency of BCC unit cell
- Packing efficiency of BCC unit cell is given by,
\(\begin{aligned}
& \text { Packing efficiency }=\frac{\text { Volume occupied by two spheres } \times 100}{\text { Volume of unit cell }} \\
& =\frac{2 \times \frac{4}{3} \pi r^3 \times 100}{a^3} \\
& =68 \%
\end{aligned}\)
- The \(\%\) volume occupied by atoms in a unit cell is \(68 \%\)
Step 2: % Vacant space in BCC unit cell
- % vacant space is given by,
\(\begin{aligned}
& \% \text { Vacant space }=100-\% \text { of the volume occupied by atoms } \\
& =100-68 \\
& =32 \%
\end{aligned}\)
- % Vacant space is \(32 \%\)

Topic: Chemistry, Solid State, NEET, NEET 2015 (Phase 2)
The vacuole in the plant cell contains

Options:
A stroma
B matrix
C tonoplast
D cell sap

The correct answer is:
cell sap

Solution:
Plant cell is a eukaryotic type cell with well defined nucleus. It has a cell wall and chloroplast (for phosynthesis). The plant cell contains a large organelle, vacuole. It contains water, cell sap, excretory products and other substances. The vacuole is enclosed by a single membrane called tonoplast. The membrane facilitates movement of a number of ions and other materials against concentration gradient.

Topic: Biology, Cell - The Unit of Life, NEET
The vacuoles are surrounded by a thin membrane called

Options:
A Plasmodesmata
B Hydathodes
C Tonoplasm
D Tonoplast

The correct answer is:
Tonoplast

Solution:
Large sap vacuoles are present in plants and they are surrounded by thin membrane called tonoplast, and the fluid present inside is called tonoplasm. Tonoplast facilitates the movement of ions against the concentration gradient from cytoplasm to cell sap.

Topic: Biology, Cell - The Unit of Life, NEET
The valence shell electronic configuration of an element is ns 2 np 5. The element will belong to the group of -

Options:
A Alkali metals
B Inert metals
C Noble gases
D Halogens

The correct answer is:
Halogens

Solution:
Halogen Familyâns2np5
All halogens contain seven electrons in their outermost shell. All other shells are completely filled.
These elements require only one electron to complete their octet.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The valency of an element whose atomic weight is 12 and sp. heat 0.25 is

Options:
A 2
B 3
C 4
D None of these

The correct answer is:
2

Solution:
Approx. atomic weight =6.4specific heat=6.40.25=25.6
Valency =approx.atomic weight eq. weight
=25.612=2 .

Topic: Chemistry, Redox Reactions, NEET
The value of acceleration due to gravity at a depth of 1600 km is equal to

Options:
A 4.9 m s-2
B 9.8 m s-2
C 7.35 m s-2
D 19.6 m s-2

The correct answer is:
7.35 m s-2

Solution:
Given,
Depth d=1600 km
We know that,
gd=g1-dR
Here, g=9.8 m s-2
R=6400 km
gd=g1-16006400
=9.81-14
=9.84-14=9.8×34
gd=7.35 m s-2

Topic: Physics, Gravitation, NEET
The value of BOD of highly polluted water is

Options:
A more than 17 ppm
B more than 5 ppm
C 1 ppm
D 5 ppm

The correct answer is:
more than 17 ppm

Solution:
Highly polluted water can have a BOD more than 17.

Topic: Chemistry, Environmental Chemistry, NEET
The value of $\gamma\left(=\frac{C_p}{C_V}\right)$ for hydrogen, helium and another ideal diatomic gas $\mathrm{X}$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to

Options:
A $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
B $\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
C $\frac{5}{3}, \frac{7}{5} \frac{7}{5}$
D $\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$

The correct answer is:
$\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$

Solution:
The poisson's ratio,
$$
\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}
$$
where, $C_p=$ molar heat capacity constant pressure and $\mathrm{C}_{\mathrm{V}}=$ molar heat capacity at constant volume Also, $C_p=C_V+R$ (from Mayer's relation) $C_V=\frac{f}{2} \mathrm{R} \quad$ (where, $\mathrm{f}=$ degree of freedom)
$$
\Rightarrow \quad \mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}
$$
So, Eq. (i) becomes,
$$
\Rightarrow \quad \gamma=1+\frac{2}{f}
$$
For hydrogen gas, which is diatomic, the degree of freedom is 5 ( 3 translational, 2 rotational).
$$
\therefore \quad \gamma=1+\frac{2}{5}=\frac{7}{5}
$$
For helium gas, which is monoatomic, the degree of freedom is 3 ( 3 translational only).
$$
\therefore \quad \gamma=1+\frac{2}{3}=\frac{5}{3}
$$
The diatomic gas $\mathrm{X}$ also have vibrational motion, so degree of freedom is 7 ( 3 translational, 2 rotational and 2 vibrational).
$$
\therefore \quad \gamma=1+\frac{2}{7}=\frac{9}{7}
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2019 (Odisha)
The value of coefficient of volume expansion of glycerin is 5×10-4 K-1. The fractional change in the density of glycerin for a rise of 40 oC in its temperature is:

Options:
A 0.020
B 0.025
C 0.010
D 0.015

The correct answer is:
0.020

Solution:
ρ=ρ01-γ∆t
∆ρρ0=γ∆T=5×10-440=0.02

Topic: Physics, Mechanical Properties of Solids, NEET, NEET 2015 (Phase 2)
The value of \(\Delta G^{\circ}\) for the phosphorylation of glucose in glycolysis is \(13.8 \mathrm{~kJ} / \mathrm{mol}\). Find the value of Kc at \(298 \mathrm{~K}\).

Options:
A \(7.64 \times 10^4\)
B \(1.26 \times 10^{-5}\)
C \(2.18 \times 10^3\)
D \(3.81 \times 10^{-3}\)

The correct answer is:
\(3.81 \times 10^{-3}\)

Solution:
Correct Option is : (D)
\(3.81 \times 10^{-3}\)

Topic: Chemistry, Ionic Equilibrium, NEET
The value of electric potential at a distance of 9 cm from the point charge $4 \times 10^{-7} \mathrm{C}$ is $\left[\right.$ Given $\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\right]$ :

Options:
A $4 \times 10^2 V$
B 44.4 V
C $4.4 \times 10^5 \mathrm{~V}$
D $4 \times 10^4 \mathrm{~V}$

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

Solution:
$V=\frac{k q}{r}$
$=9 \times 10^9 \times \frac{4 \times 10^{-7}}{9 \times 10^{-2}}$
$=4 \times 10^4 \mathrm{~V}$

Topic: Physics, Electrostatics, NEET, NEET 2024 (Re-NEET)
The value of equilibrium constant of the reaction $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2$ is 8.0
The equilibrium constant of the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ will be

Options:
A $\frac{1}{16}$
B $\frac{1}{64}$
C 16
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{64}$

Solution:
$\begin{gathered}
\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \\
\mathrm{K}_1=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]} \\
\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\
K_2=\frac{\left[\mathrm{HI}^2\right.}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
\end{gathered}$
From Eqs (i) and (ii)
$\begin{array}{l}
& K_1^2=\frac{1}{K_2} \\
\because & K_1=8.0 \\
\therefore & K_2=\frac{1}{K_1^2}=\frac{1}{8^2}=\frac{1}{64}
\end{array}$

Topic: Chemistry, Coordination Compounds, NEET, NEET 2008 (Screening)
The value of ∆fH⊖for NH3is – 91.8 kJ mol–1. Calculate enthalpy change for the following reaction : 2NH3(g) → N2(g) + 3H2(g)

Options:
A ∆rH⊖= -83.6 kJ mol–1
B ∆rH⊖= -91.8 kJ mol–1
C ∆rH⊖= +83.6 kJ mol–1
D ∆rH⊖= +91.8 kJ mol–1

The correct answer is:
∆rH⊖= +91.8 kJ mol–1

Solution:
Correct Option is : (D)
∆rH⊖= +91.8 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
The value of g on the surface of earth is smallest at the equator because

Options:
A The centripetal force is maximum at equator
B The centripetal force is least at equator
C The centrifugal force is maximum at equator
D The angular speed of earth is least at equator

The correct answer is:
The centrifugal force is maximum at equator

Solution:
Due to rotation of earth the effective acceleration due to gravity g′=g-Rω2cos2λ.
For a given point on the surface of earth g decreases as ω increases. The angular speed of earth is maximum at equator hence, the value of g on the surface of the earth is smallest.

Topic: Physics, Gravitation, NEET
The value of $\Delta H$ for the raction $X_2(g)+4 Y_2(g)^{39} 2 X Y_4(g)$ is less than zero. Formation of $X Y_4(g)$ will be favoured at

Options:
A low pressure and low temperature
B high temperature and low pressure
C high pressure and low temperature
D high temperature and high pressure

The correct answer is:
high pressure and low temperature

Solution:
$$
\begin{aligned}
& X_2(g)+4 Y_2 \rightleftharpoons 2 X Y_4(g) \\
& \Delta H < 0 \text { and } \Delta n < 0
\end{aligned}
$$
$\therefore$ The forward reaction is favoured at high pressure and low temperature. (According to co-chatelie's principle)

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2011 (Screening)
The value of ΔH for the reaction

X2(g) + 4Y2(g) ⇌ 2XY4(g) is less than zero.

​Formation of XY4(g) will be favoured at

Options:
A high temperature and high pressure
B low pressure and low temperature
C high temperature and low pressure
D high pressure and low temperature

The correct answer is:
high pressure and low temperature

Solution:
X2(g) + 4Y2(g) ⇌ 2XY4(g)
Δng = –ve and ΔH = –ve (exothermic)
According to Le-Chatelier's principle:
The reaction is favoured in forward direction at low temperature and high pressure.

Topic: Chemistry, Chemical Equilibrium, NEET
The value of Henry's law constant for few gases at 298 K is given below. Arrange these gases in the increasing order of their solubility.
N2 : 76.48 kbar, O2 : 34.86 kbar, H2 : 69.16 kbar

Options:
A O2<N2<H2
B O2<H2<N2
C H2<N2<O2
D N2<H2<O2

The correct answer is:
N2<H2<O2

Solution:
As higher the value of KH , lower is the solubility of gas in the liquid so the correct order of solubility of these gases is as follows
KHN276.48<H269.16<O234.86

Topic: Chemistry, Solutions, NEET
The value of Henry's law constant for some gases at 293 K is given below. Arrange the gases in the increasing order of their solubility.
He: 144.97 kbar, H2 : 69.16 kbar,
N2 : 76.48 kbar, O2 : 34.86 kbar

Options:
A He<N2<H2<O2
B O2<H2<N2<He
C H2<N2<O2<He
D He<O2<N2<H2

The correct answer is:
He<N2<H2<O2

Solution:
Higher the value of KH , lower is the solubility of gas in the liquid.
The correct order of solubility of these gases is as follows
He144.97<N276.48<H269.16<O234.86
KH=144.97,76.48,69.16,34.86 kbar

Topic: Chemistry, Solutions, NEET
The value of inductance L for which the current is maximum in a series LCR A.C. circuit with C = 1 0 Î¼ F and Ï = 1 0 0 0 s - 1 is

Options:
A 1 mH

B Cannot be calculated unless R is known
C 10 mH

D 100 mH

The correct answer is:
100 mH

Solution:
In series LCR, the current is maximum at resonance.
â´ Resonant frequency, Ï=1LC
â´ Ï2=1LC or, L=1Ï2C
Given, Ï = 1 0 0 0 s - 1 and C = 1 0 Î¼ F
â´ L=11000Ã1000Ã10Ã10-6=0.1 H=100 mH

Topic: Physics, Alternating Current, NEET
The value of inductance L for which the current is maximum in series LCR circuit with C=10 Î¼F and Ï=1000 rad s-1.

Options:
A 10 mH
B 50 mH
C 200 mH
D 100 mH

The correct answer is:
100 mH

Solution:
In resonance condition, maximum current flows in the circuit.
Current in the LCR circuit is given by
i=VR2+XL-XC2
For current to be maximum denominator should be minimum.
XL-XC2=0
⇒ XL=XC
⇒ ωL=1ωC
∴ L=1ω2C
=110002×10×10-6
=110 H
∴ L=0.1 H=100 mH

Topic: Physics, Alternating Current, NEET
The value of $\mathrm{K}_{s p}$ of $\mathrm{CaCO}_3$ and $\mathrm{CaC}_2 \mathrm{O}_4$ are $4.7 \times 10^{-9}$ and $1.3 \times 10^{-9}$ respectively at $25^{\circ} \mathrm{C}$. If the mixture of these two is washed with water, what is the concentration of $\mathrm{Ca}^{2+}$ ions in water?

Options:
A $5.831 \times 10^{-5} \mathrm{M}$
B $6.856 \times 10^{-5} \mathrm{M}$
C $3.606 \times 10^{-5} \mathrm{M}$
D $7.746 \times 10^{-5} \mathrm{M}$

The correct answer is:
$7.746 \times 10^{-5} \mathrm{M}$

Solution:
$\begin{aligned}
\mathrm{CaCO}_3 & \rightarrow \mathrm{Ca}^{2+}+\mathrm{CO}_3{ }^{2-} \\
\mathrm{CaC}_2 \mathrm{O}_4 & \rightarrow \mathrm{Ca}^{2+}+\mathrm{C}_2 \mathrm{O}_4^{2-} \\
y & \\
{\left[\mathrm{Ca}^{2+}\right] } & =x+y \\
x(x+y) & =4.7 \times 10^{-9} \\
y(x+y) & =1.3 \times 10^{-9}
\end{aligned}$
$\begin{aligned}
& \frac{x}{y}=3.6 \\
& \therefore \quad x=3.6 y \\
& y(3.6 y+y)=1.3 \times 10^{-9} \\
& y=1.6 \times 10^{-5} \\
& {\left[\mathrm{Ca}^{2+}\right] }=(x+y)=\left(1.6 \times 10^{-5}\right) \\
&+\left(5.76 \times 10^{-5}\right) \\
&=7.46 \times 10^{-5} \mathrm{M}
\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2013 (Karnataka)
The value of Kc for the reaction 2HI (g)
H2(g) + I2(g) is 1 × 10–4At a given time, the composition of the reaction mixture is [HI] = 2 × 10–5mol, [H2] = 1 × 10–5mol and [I2] = 1 × 10–5mol. In which direction will the reaction proceed?

Options:
A Forward
B Reverse
C At equilibrium
D Cannot be determined

The correct answer is:
Reverse

Solution:
Correct Option is : (B)
Reverse

Topic: Chemistry, Ionic Equilibrium, NEET
The value of log10K for a reaction A⇌B is
(Given: ΔrHo298K=-54.07 kJmol-1,ΔrSo298K=10J K-1mol-1
and R=8.314 JK-1mol-1;2.303×8.314×298=5705 )

Options:
A 90
B 100
C 5
D 10

The correct answer is:
10

Solution:
ΔGo=ΔHo-TΔSo=-54.07×1000–298×10=-57050 Jmol-1
ΔG°=-2.303RTlog10K
-57050=-5705log10K
log10K=10

Topic: Chemistry, Chemical Equilibrium, NEET
The value of n in, MnO 4 −  +  8H +  +  ne − → Mn 2−  +  4H 2 O is

Options:
A 5
B 4
C 3
D 2

The correct answer is:
5

Solution:
MnO4-+8H+ââMn2+â+2+4H2O
To balance charge on both sides add 5e! to L.H.S.
MnO 4 −  +  8H ⊕  +  5e − → Mn 2+ +  4H 2 O

Topic: Chemistry, d and f Block Elements, NEET
The value of β of a transistor is 19. The value of α will be

Options:
A 0.93
B 0.98
C 0.99
D 0.95

The correct answer is:
0.95

Solution:
IE=IB+IC
â IEIC=IBIC+1
â 1Î±=1Î²+1=119+1
â 1Î±=2019
â Î±=1920=0.95

Topic: Physics, Semiconductors, NEET
The value of Planck's constant is $6.63 \times 10^{-34} \mathrm{Js}$. The speed of light is $3 \times 10^{17} \mathrm{nms}^{-1}$. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of $6 \times 10^{15} \mathrm{~s}^{-1} ?$

Options:
A 10
B 25
C 50
D 75

The correct answer is:
50

Solution:
Given, Planck's constant,
$$
h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}
$$
Speed of light, $c=3 \times 10^{17} \mathrm{~nm} \mathrm{~s}^{-1}$
Frequency of quanta
$$
v=6 \times 10^{15} \mathrm{~s}^{-1}
$$
Wavelength, $\lambda=$ ?
We know that, $v=\frac{c}{\lambda}$
or
$$
\begin{aligned}
\lambda & =\frac{c}{v} \\
& =\frac{3 \times 10^{17}}{6 \times 10^{15}} \\
& =0.5 \times 10^2 \mathrm{~nm} \\
& =50 \mathrm{~nm}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2013 (All India)
The value of Planck's constant is 6.63 $\times 10^{-34} \mathrm{Js}$. The velocity of light is $3.0 \times$ $10^8 \mathrm{~ms}^{-1}$. Which value is closest to the wavelength in nanometers of quantum of light with frequency of $8 \times 10^{15} \mathrm{~s}^{-1}$ ?

Options:
A $2 \times 10^{-25}$
B $5 \times 10^{-18}$
C $4 \times 10^1$
D $3 \times 10^7$

The correct answer is:
$4 \times 10^1$

Solution:
$\begin{aligned}
& c=v \times \lambda ; \lambda=\frac{c}{v} \\
& \Rightarrow \lambda=\frac{3 \times 10^8}{8 \times 10^{15}}=0.375 \times 10^{-7} \mathrm{~m} \\
& =0.375 \times 10^{-7} \times 10^9 \mathrm{~nm} \\
& =0.375 \times 10^2 \mathrm{~nm} \\
& =37.5 \mathrm{~nm} \\
& \approx 4 \times 10 \mathrm{~nm}
\end{aligned}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2003
The value of Planck's constant is:

Options:
A $6.63 \times 10^{-34} \mathrm{Js}^{-1}$
B $6.63 \times 10^{-34} \mathrm{~m}^2 \mathrm{~s}^{-1}$
C $6.63 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2$
D $6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}^{-1}$

The correct answer is:
$6.63 \times 10^{-34} \mathrm{~m}^2 \mathrm{~s}^{-1}$

Solution:
The value of Planck's constant
$h=6.63 \times 10^{-34} \mathrm{kgm}^2 \mathrm{~s}^{-1}$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2002
The value of sink temperature, when the efficiency of the engine is 100% , is

Options:
A Zero
B 300 K

C 273 K

D 400 K

The correct answer is:
Zero

Solution:
Efficiency Î·=T1-T2T1
where T2=sink temperature,
T1=source temperature.
For 100% efficiency, Î·=1
â´ T2T1=0
â T2=0 K

Topic: Physics, Thermodynamics, NEET
The value of the current I1, I2 and I3 flowing through the circuit given below is

Options:
A I1=-3 A, I2=2 A, I3=-1 A
B I1=2 A, I2=-3 A, I3=-1 A
C I1=3 A, I2=-1 A, I3=-2 A
D I1=1 A, I2=-3 A, I3=-2 A

The correct answer is:
I1=2 A, I2=-3 A, I3=-1 A

Solution:
Applying Kirchoff's law
10-6IB+14-4IA=0 ...(loop 1)
24=6IB+4IA
12=3IB+2IA .....(i)
IA=IB+IC .....(junction)
10-6IB+2IC=0 .....(loop 2)
10=6IB-2IC
10=6IB-2IA-IB .....(ii)
On solving Equations (i) and (ii), we get
IB=2 A, IA=3 A, IC=1 A
Now as per given circuit,
I1=IB=2 A, I2=-IA=-3 A
I3=-IC= -1 A

Topic: Physics, Current Electricity, NEET
The value of the regression coefficient 'Z' is dependent on:

Options:
A Species richness
B Area or region
C Flora present in the habitat
D Both A and B

The correct answer is:
Both A and B

Solution:
Biodiversity includes all the flora and fauna on Earth. All the varieties of organisms come under biodiversity. German naturalist Alexander von Humboldt made the observation regarding the relationship between species richness and the area of the habitat. As per his observation, species richness increased with increasing area of habitat upto a limit. When plotted in a graph, this relationship forms a rectangular hyperbola On a log scale, the relationship is a straight line having the equation
Log S=Log C+Z Log A . S= species richness, A= area of region, C=Y- intercept and Z= regression co-efficient (Slope). Ecologists' study has shown that value of Z lies in the rang of 0.1-0.2 irrespective of the organism group or region unless the area is very large like entire continents. Thus, it is independent of species type or area.

Topic: Biology, Biodiversity and its Conservation, NEET
The value of the resistor, RS, needed in the DC voltage regulator circuit shown here, equals:

Options:
A Vi+VLn+1IL
B Vi-VLn IL
C Vi+VLn IL
D Vi-VLn+1IL

The correct answer is:
Vi-VLn+1IL

Solution:
Across zener diode VL=constant
âµVi=VL+n+1IL RS
⇒Vi-VL=n+1IL RS
⇒RS=Vi-VLn+1IL

Topic: Physics, Semiconductors, NEET
The value of the spin only magnetic moment for one of the following configurations is 2.84 BM. The correct one is

Options:
A d5 (in strong field ligand)
B d3 (in weak as well as strong field)
C d4 (in weak field ligand)
D d4 (in strong ligand field)

The correct answer is:
d4 (in strong ligand field)

Solution:
The given value of μ (spin only)
μ=(n+2)n
2.84=n(n+2) BM
So, n = 2.
Among the given configurations, d4 system in strong ligand field will have 2-unpaired e- in t2g set of orbitals as shown below

Topic: Chemistry, Coordination Compounds, NEET
The value of transistor parameters, Î² and Î±, respectively, for the given figure is

Options:
A 49, 0.98
B 32, 0.64
C 29, 0.97
D 45, 0.90

The correct answer is:
29, 0.97

Solution:
ICIE=α
β=ICIB

Topic: Physics, Semiconductors, NEET