datavorous/entrance-exam-dataset huggingface dataset:
In rabbit, end of a long bone is connected in another by

Options:
A tendon
B ligaments
C muscle
D cartilage

The correct answer is:
ligaments

Solution:
Ligaments consist of mainly collagen fibres and some elastic fibres. It connects end of a long bone to another.Tendon is a very dense, strong fibrous connective tissue made of collagen fibres. Tendon connects a skeletal muscle to a bone.Cartilage is a solid but semi-rigid and flexible connective tissue. Cartilage like the fibrous tissues, is a vascular and nutrients must diffuse into it from nearby tissues.Muscles are composed of muscle tissue, that contracts to affect a particular movement.

Topic: Biology, Locomotion and Movement, JIPMER, JIPMER 2009
In radioactive decay process, the negatively charged emitted $\beta$ -particles are

Options:
A the electrons present inside the nucleus
B the electrons produced as a result of the decay of neutrons inside the nucleus
C the electrons produced as a result of collisions between atoms
D the electrons orbiting around the nucleus

The correct answer is:
the electrons produced as a result of the decay of neutrons inside the nucleus

Solution:
Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$ -decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state. In negative $\beta$ -decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted \beta-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

Topic: Physics, Nuclear Physics, JEE Main
In radioactive reaction

ZAXâ Z+1A X1â Z+2A X2â ZA-4 X3â Z+1A-4 X4

Successive emission of particles is

Options:
A β-, β-, α
B β-, β-, β+, α
C β-, β-, α, α
D β-, β-, α, β-

The correct answer is:
β-, β-, α, β-

Solution:

Topic: Physics, Nuclear Physics, JEE Main
In Ramsay and Rayleigh's isolation of noble gases from air, the nitrogen of the air is finally converted into

Options:
A $\mathrm{NaNO}_{2}$ only
B $\mathrm{NO}$ and $\mathrm{NO}_{2}$
C $\mathrm{NaNO}_{3}$ only
D $\mathrm{NaNO}_{2}$ and $\mathrm{NaNO}_{3}$

The correct answer is:
$\mathrm{NaNO}_{2}$ and $\mathrm{NaNO}_{3}$

Solution:
In Ramsay and Rayleigh's isolation of noble gases from air, the nitrogen of the air finally converted into $\mathrm{NaNO}_{2}$ and $\mathrm{NaNO}_{3}$.
Electric
$\stackrel{\text { discharge }}{\longrightarrow} 2 \mathrm{NO}$
$2 \mathrm{NO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g)$
$2 \mathrm{NO}_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{NaNO}_{2}+\mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In reaction $\mathrm{A}+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D}$, initial concentration of $\mathrm{B}$ was $1.5$ times of $\mathrm{[A]}$, but at equilibrium the concentrations of $\mathrm{A}$ and $\mathrm{B}$ became equal. The equilibrium constant for the reaction is :

Options:
A
$8$

B
$4$

C
$12$

D
$6$

The correct answer is:
$4$

Solution:
Hence $\mathrm{K}_{\mathrm{C}}=\frac{(2 x)^2 \times x}{(a-x)(1.5 a-2 x)^2}$
Given, at equilibrium
$\begin{array}{ll}\therefore & (a-x)=(1.5 a-2 x) \\ \therefore & a=2 x\end{array}$
On solving $K_C=4$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2013 (09 Apr Online)
In reaction $\mathrm{BF}_3+3 \mathrm{LiBH}_4 \rightarrow 3 \mathrm{LiF}+X ; X$ is

Options:
A $\mathrm{B}_4 \mathrm{H}_{10}$
B $\mathrm{B}_2 \mathrm{H}_6$
C $\mathrm{BH}_3$
D $\mathrm{B}_3 \mathrm{H}_8$

The correct answer is:
$\mathrm{B}_2 \mathrm{H}_6$

Solution:
$2 \mathrm{BF}_3+3 \mathrm{LiBH}_4 \rightarrow 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{LiF}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In reaction N2O4gâ2NO2g , The observed molecular weight 80 gmol-1 at 350 K. The percentage dissociation of N2O4g at 350 K is

Options:
A 10%
B 15%
C 20%
D 18%

The correct answer is:
15%

Solution:
Or N2O4→2NO2101-α2α
Mass remains same
Average molar man =MN2O41+α
So, =28+641+α=921+α
1+α=9280=2320
α=2320-1=320
%α=320×100=15%

Topic: Chemistry, Chemical Equilibrium, NEET
In recent years, DNA sequences (nucleotide sequence) of $m t$ DNA and Y chromosomes were considered for the study of human evolution, because:

Options:
A they are small and therefore, easy to study
B they are uniparental in origin and do not take part an recombination
C their structure is known in greater detail
D they can be studied from the samples of fossil remains

The correct answer is:
they are uniparental in origin and do not take part an recombination

Solution:
Sequence of both $m t$ DNA and $\mathrm{Y}$ chromosomes are considered for the study of human evolution because they are uniparental in origin. mtDNA is inherited along with the maternal cytoplasm and $\mathrm{Y}$ chromosome is inherited from father. So they do not take part in recombination. In addition, mtDNA has a higher mutation rate then nuclear DNA so that it is more useful for short term evolutionary studies.
Related Theory
The Ychromosomeisinheritedfromthefather; therefore, paternal social class was included in the Y chromosome analysis. Mitochondrial DNA is inherited from the mother; thus, maternal social class was included in the analysis along with paternal social class for mitochondrial DNA. More than $27 \%$ of the variance in Y-haplogroups can be attributed to differences between populations, whereas mtDNA showed much lower heterogeneity between populations (less than 5\%), suggesting a strong influence of patrilocal social structure.

Topic: Biology, Evolution, NEET, NEET 2003
In recent years, DNA sequences of mitochondrial' DNA and Y-chromosomes were considered for the study of human evolution because

Options:
A They can be studied from the sample of fossil remains
B They are small and therefore, easy to study
C They are uniparental in origin and do not take part in recombination
D Their structure is known in greater detail

The correct answer is:
They are uniparental in origin and do not take part in recombination

Solution:
Today, besides the autosomal chromosomes, Y-chromosomes and mitochondrial DNA are being studied for the study of human evolution, as they are uniparental in origin and do not take part in recombination.

Topic: Biology, Evolution, NEET
In resonance hybrid of ozone molecule, $0-0$ bond length is

Options:
A $128 \mathrm{pm}$
B $134 \cdot 5 \mathrm{pm}$
C $121 \mathrm{pm}$
D $148 \mathrm{pm}$

The correct answer is:
$121 \mathrm{pm}$

Solution:
In resonance hybrid of ozone molecule, $\mathrm{O}-\mathrm{O}$ bond length is $128 \mathrm{pm}$.
Explanation:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In respect of the equation $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{o}} / \mathrm{RT}}$ in chemical kinetics, which one of the following statements is correct?

Options:
A
$\mathrm{A}$ is adsorption factor

B
$\mathrm{E}_{\mathrm{a}}$ is energy of activation

C
$\mathrm{R}$ is Rydberg's constant

D
$\mathrm{k}$ is equilibrium constant

The correct answer is:
$\mathrm{E}_{\mathrm{a}}$ is energy of activation

Solution:
No Solution Available

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2003
In reverse bias $p n$ -junction diode depletion layer width

Options:
A decreases
B increases
C remain same
D Can't predicted

The correct answer is:
increases

Solution:
In reverse bias $p n$ -junction diode depletion layer width increases.

Topic: Physics, Semiconductors, MHT CET, MHT CET 2011
In root nodules of leguminous plants, the function of leg-hemoglobin during biological nitrogen fixation is

Options:
A Convert N2 to NH3
B Convert NH3 Nitrite
C Transport oxygen for nitrogenase activity
D Protect nitrogenase from oxygen

The correct answer is:
Protect nitrogenase from oxygen

Solution:
Leghemoglobin is an oxygen-carrying protein, somewhat similar to the hemoglobin of blood, that helps maintain the low oxygen tension necessary for effective nitrogen fixation. This is an essential component for nitrogen fixation by legumes. Leghemoglobin is produced as a result of symbiotic association between bacteroid and plant. The major role of leghemoglobin involves protection of nitrogenase enzyme from denaturation, if exposed to atmospheric concentration of oxygen, but at the same time supply of ample amount of oxygen to bacteria for respiration.

Topic: Biology, Mineral Nutrition, NEET
In Rutherford experiment, for head-on collision of a-particles with a gold nucleus, the impact
parameter is

Options:
A of the order of \( 10^{-10} \mathrm{~m} \)
B zero
C of the order of \( 10^{-6} \mathrm{~m} \)
D of the order of \( 10^{-14} \mathrm{~m} \)

The correct answer is:
zero

Solution:
(B)
For head on collision the impact parameter is zero as it retraces the path.

Topic: Physics, Atomic Physics, JEE Main
In $S=a+b t+c t^2$. $S$ is measured in metres and $t$ in seconds. The unit of $c$ is

Options:
A None
B $m$
C $ms^{-1}$
D $ms^{-2}$

The correct answer is:
$ms^{-2}$

Solution:
$c t^2$ must have dimensions of $L$
$\Rightarrow$ c must have dimensions of $L / T^2$ i.e. $L T^{-2}$.

Topic: Physics, Units and Dimensions, JEE Main
In SI units, the dimensions of ϵ0μ0 is:

Options:
A AT2M-1L-1
B A2T3M-1L-2
C A-1TML3
D AT-3ML3/2

The correct answer is:
A2T3M-1L-2

Solution:
As we know, c=1μ0ε0
∴ ε0μ0=ε02μ0ε0=ε0c
∴ε0μ0= ε0c=M-1L-3T4A2[LT-1]
=[M-1L-2T3A2]

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2019 (08 Apr Shift 1)
In SN2 reactions, the correct order of reactivity for the following compounds :
CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is :

Options:
A CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl
B CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl
C CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl
D (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl

The correct answer is:
CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

Solution:
SN2 involves the formation of a pentavalent intermediate, and the reactivity depends on steric hindrance. Lesser the steric hindrance, more is the reactivity of S N 2 .

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (06 Apr)
In satellite communication, the uplink frequency band used is:

Options:
A 420-890 MHz
B 5.925-6.425 GHz
C 76-88 MHz
D 3.7-4.2 GHz

The correct answer is:
5.925-6.425 GHz

Solution:
In satellite telecommunication, a downlink is the link from a satellite down to one or more ground stations or receivers, and an uplink is the link from a ground station up to a satellite. Uplink frequency band for satellite communication is 5.925-6.425 GHz.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (11 Apr Shift 2)
In sea anemone, the symmetry is

Options:
A radial
B bilateral
C spherical
D absent

The correct answer is:
radial

Solution:
In radial symmetry, body can be divided into similar halves by more than two planes passing through the main axis.
Cnidarians or coelenterates (like Hydra, sea anemone, Aurelia, Obelia) are radially symmetrical animals with cell-tissue grade of body organization.

Topic: Biology, Photosynthesis in Higher Plants, JIPMER, JIPMER 2008
In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be :

Options:
A reduced by 14L
B increased by 2L
C reduced by 34L
D increased to 4L

The correct answer is:
reduced by 34L

Solution:
As resonance frequency remains unchanged, therefore ω'=ω

⇒1L'C'=1LC

⇒ L'C'=LC

⇒L'4C=LC

⇒L'=L4

∵ Inductance must be decreased by 3L4.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (01 Feb Shift 1)
In series LR circuit, \( R=18 \Omega \) and impedance is \( 33 \Omega \). An r.m.s. voltage \( 220 \mathrm{~V} \) is applied across the circuit. The true power consumed in a.c. circuit is:

Options:
A \( 220 \mathrm{~W} \)
B \( 400 \mathrm{~W} \)
C \( 600 \mathrm{~W} \)
D \( 800 \mathrm{~W} \)

The correct answer is:
\( 800 \mathrm{~W} \)

Solution:
Reactance of an inductor is given by, XL=ωL, where L represents inductance.
Net impedance of LR circuit is given by, ZL=XL2+R2=33 Ω(given)
Power factor of this circuit is given by, cosϕ=RZL
⇒cosϕ=1833
Now, average power in AC circuit is given by, P=erms2ZLcosϕ
⇒P=220233×1833=800 W

Topic: Physics, Alternating Current, JEE Main
In silicon dioxide

Options:
A Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded
to two silicon atoms
B Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bounded to two silicons atoms.
C Silicon atom is bonded to two oxygen atoms
D There are double bonds between silicon and oxygen atoms

The correct answer is:
Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded
to two silicon atoms

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In silicon dioxide

Options:
A
Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms

B
Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms

C
Silicon atoms is bonded to two oxygen atoms

D
There are double bonds between silicon and oxygen atoms

The correct answer is:
Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2005
In simple harmonic motion, the total mechanical energy of given system is $E$. If mass of oscillating particle $P$ is doubled then the new energy of the system for same amplitude is:

Options:
A $E$
B $E / \sqrt{2}$
C $2 E$
D $E \sqrt{2}$

The correct answer is:
$E$

Solution:
$\text {T.E. }=\frac{1}{2} \mathrm{kA}^2$
since A is same T.E. will be same.

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (04 Apr Shift 2)
In single slit diffraction experiment, when the distance of separation between the
slit and screen is doubled, the angular separation between fringes

Options:
A increases.
B decreases.
C remains same.
D first increases and then decreases.

The correct answer is:
remains same.

Solution:
(C)

Topic: Physics, Wave Optics, MHT CET, MHT CET 2020 (19 Oct Shift 1)
In single slit diffraction pattern

Options:
A central fringe has negligible width than others
B all fringes are of same width
C central fringes do not exist
D None of the above

The correct answer is:
None of the above

Solution:
In single slit diffraction, the central fringe has maximum intensity and has the width double than other fringes.

Topic: Physics, Wave Optics, JEE Main
In situ conservation is carried out through:

Options:
A National parks
B Wild life sanctuaries
C Biosphere reserves
D All the above

The correct answer is:
All the above

Solution:
In situ conservation is maintenance of biological diversity in natural habitats like forest and nature preserves by setting national parks, wildlife sanctuaries, Biosphere reserves.

Topic: Biology, Biodiversity and its Conservation, NEET
In solid argon, the atoms are held together by

Options:
A Ionic bonds
B Hydrogen bonds
C Vander Waals forces
D Hydrophobic forces

The correct answer is:
Vander Waals forces

Solution:
The melting and boiling points of argon is low hence, in solid argon atoms are held together by weak Vander Waal's forces.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In solid state which among the following exists as covalent crystals?

Options:
A Sulphur
B Phosphorus
C Iodine
D Silicon

The correct answer is:
Silicon

Solution:
Silicon is covalent network solid.
Each Si atom is in sp3 hybridisation.
Sulphur (S8) â molecular solid
phosphorus (P4) â molecular solid
Iodine (I2) â molecular solid

Topic: Chemistry, Solid State, JEE Main
In some members of which of the following pairs of families, pollen grains retain their viability for months after release?

Options:
A Poaceae; Rosaceae
B Poaceae; Leguminosae
C Poaceae; Solanaceae
D Rosaceae; Leguminosae

The correct answer is:
Rosaceae; Leguminosae

Solution:
The period for which pollen grains remain viable is highly variable and to some extent depends on the prevailing temperature and humidity. In some cereals such as rice and wheat, pollen grains lose viability within 30 minutes of their release, and in some members of Rosaceae, Leguminosae and Solanaceae, they maintain viability for months.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2021
In some region, the gravitational field is zero. The gravitational potential in this region

Options:
A Must be variable
B Must be constant
C Cannot be zero
D Must be zero

The correct answer is:
Must be constant

Solution:
$I=\frac{-d V}{d r}$. If $I=0$ then $V=$ constant

Topic: Physics, Gravitation, JEE Main
In some solutions, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as:

Options:
A
Ideal solutions

B
Colloidal solutions

C
True solutions

D
Buffer solutions

The correct answer is:
Buffer solutions

Solution:
Solutions which resist the change in the value of $\mathrm{pH}$ when small amount of acid or base is added to them are known as buffers.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2014 (11 Apr Online)
In some solutions, the concentration of H 3 O + remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as :

Options:
A Ideal solutions
B Colloidal solutions
C True solutions
D Buffer solutions

The correct answer is:
Buffer solutions

Solution:
In acidic Buffer solutions when
pH=pKa+log Salt Acid
pH is not disturbed i.e H 3 O + , remains constant on addition of small quantity of acid or Base.

Topic: Chemistry, Ionic Equilibrium, NEET
In Sommerfeld's modification of Bohr's theory, the trajectory of an electron in a hydrogen atom is

Options:
A a perfect ellipse
B a closed ellipse - like curve, narrower at the perihelion position and flatter at the aphelion position
C a closed loop on spherical surface
D a rosette

The correct answer is:
a closed loop on spherical surface

Solution:
No Solution Available

Topic: Chemistry, Structure of Atom, JEE Main
In standard Young's double-slit experiment, visible light λ∈350nm, 750nm is used in both the slits. Distance between the slits is 2 mm, and the distance of the screen from the slits is 1 m. A thin glass slab μ = 1.5 of thickness 6 μm is placed in front of slit S2. Which among the following wavelengths is missing at point P, which is directly in front of slit S2?(Neglect dispersion of light)

Options:
A 400 nm
B 500 nm
C 600 nm
D 700 nm

The correct answer is:
400 nm

Solution:
Net path difference = Î¼ - 1 t - yd D
=32-16 Î¼m-d/2dD=1 Î¼m
For Destructive interference
Îx=1 Î¼m=2n+1Î»2âÎ»=22n+1 Î¼m
Î» = 2000 nm 2 + 1 = 2000 3 ( n=1 )
Î»=2000 nm4+1=20005nm=400 nm ( n=2 )

Topic: Physics, Wave Optics, JEE Main
In stationary wave

Options:
A Strain is maximum at nodes
B Strain is maximum at antinodes
C Strain is minimum at nodes
D Amplitude is zero at all the points

The correct answer is:
Strain is minimum at nodes

Solution:
At nodes pressure change (strain) is maximum

Topic: Physics, Waves and Sound, JEE Main
In stationary waves, distance between a node and its nearest antinode is $20 \mathrm{~cm}$. The phase difference between two particles having a separation of $60 \mathrm{~cm}$ will be

Options:
A Zero
B $\pi / 2$
C $\pi$
D $3 \pi / 2$

The correct answer is:
$3 \pi / 2$

Solution:
$\begin{aligned} & \frac{\lambda}{4}=20 \Rightarrow \lambda=80 \mathrm{~cm}, \text { also } \Delta \phi=\frac{\lambda}{2 \pi} \cdot \Delta x \\ & \Rightarrow \Delta \phi=\frac{60}{80} \times 2 \pi=\frac{3 \pi}{2}\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
In steady state heat conduction, the equations that determine the heat current jr [heat flowing per unit time per unit area] and temperature Tr in space are exactly the same as those governing the electric field Er and electrostatic potential Vr with the equivalence given in the table below.

Heat flow
Electrostatics

Tr
Vr

jr
Er

We exploit this equivalence to predict the rate Q of total heat flowing by conduction from the surfaces of spheres of varying radii, all maintained at the same temperature. If Q∝Rn, where R is the radius, then the value of n is

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

The correct answer is:
2

Solution:
Temperature difference causes heat to flow.

So, potential difference is equivalent to temperature difference.

As, dVdR=-E=-kQR2

We can write for heat flux,

dTdR=-kQR2

Here, Q= heat transferred.

⇒Q∝R2

when dTδR= constant.

∴n=2

Topic: Physics, Thermal Properties of Matter, KVPY, KVPY 2018 (SB/SX)
In stratosphere most of the ozone formation is assisted by :

Options:
A γ-rays
B cosmic rays
C visible radiations
D ultraviolet radiation

The correct answer is:
ultraviolet radiation

Solution:
Ozone in the stratosphere is a product of UV radiations acting on dioxygen O2 molecules.

O2( g)⟶UVO(g)+O(g)

O(g)+O2( g)⇌UVO3( g)

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (27 Aug Shift 2)
In sulphur estimation. 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is _______(Nearest Integer)

(Given: Atomic mass Ba: 137u:S:32u,O:16u )

Options:

The correct answer is:
42

Solution:
Molecular mass of Barium Sulphate(BaSO4) = 233 g
%sulphur=32233× weight of BaSO4 formed weight of organic compound ×100

=32233×1.44390.471×100

=42.10

Nearest integer: 42

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 1)
In ten observations, the mean of all \( 10 \) numbers is \( 15 \), the mean of the first six observations is \( 16 \) and the mean of the last five observations is \( 12 \). The sixth number is

Options:
A \( 6 \)
B \( 9 \)
C \( 12 \)
D \( 3 \)

The correct answer is:
\( 6 \)

Solution:
Let the mean of the last four observations be A2. Then, by the formula for combined mean, we get,
15=6Ã16+4ÃA26+4
or 150=96+4A2
â´A2=544
Let the sixth number is x, then taking the sixth number as a collection, the combined mean of this collection and the collection of the last four is 12.
â´ By the definition of combined mean
12=1Ãx+4Ã5441+4
â´60=x+54â´x=6
Hence, the sixth number =6

Topic: Mathematics, Statistics, JEE Main
In ten observations, the mean of all 10 numbers is 15, the mean of the first six observations is 16 and the mean of the last five observations is 12. The sixth number is

Options:
A 6
B 9
C 12
D 3

The correct answer is:
6

Solution:
Let the mean of the last four observations be A2. Then, by the formula for combined mean, we get,
15=6Ã16+4ÃA26+4
or 150=96+4A2
â´A2=544
Let the sixth number is x, then taking the sixth number as a collection, the combined mean of this collection and the collection of the last four is 12.
â´ By the definition of combined mean
12=1Ãx+4Ã5441+4
â´60=x+54â´x=6
Hence, the sixth number =6

Topic: Mathematics, Statistics, JEE Main
In terms of energy bands, semiconductors can be defined as those materials that have:

Options:
A Almost an empty conduction band and almost filled valence band with a very narrow energy gap between the conduction and valence band.
B Almost an empty conduction band and almost filled valence band with a very high energy gap between the conduction and valence band
C Almost and empty conduction band and low filled valence band with a very narrow energy gap between the conduction and valence band
D Almost full conduction band and almost filled valence band with a very narrow energy gap between the conduction and valence band

The correct answer is:
Almost an empty conduction band and almost filled valence band with a very narrow energy gap between the conduction and valence band.

Solution:
CONCEPT:
Forbidden energy gap ( $\Delta \mathrm{Eg}$ ): Energy gap between the conduction band and valence band is known as the forbidden energy gap i.e.,
$\Delta E_g=(C . B)_{\min }-(V . B)_{\max }$
- No free electron is present in the forbidden energy gap.
- Width of the forbidden energy gap depends upon the nature of the substance.
- As temperature increases, the forbidden energy gap decreases very slightly.

Topic: Physics, Semiconductors, JEE Main