datavorous/entrance-exam-dataset huggingface dataset:
In a vessel, the gas is at pressure $P$, if the mass of all the molecules is halved of their speed is double, then the resultant pressure will be :

Options:
A $2 P$
B $P$
C $P / 2$
D $4 P$

The correct answer is:
$2 P$

Solution:
The equation of pressure is
$$
P=\frac{1}{3} \frac{m \mathrm{~N}}{v^2} v_{r m s}^2
$$
where $m=$ mass of each molecules
$$
\begin{aligned}
& N=\text { number of molecules } \\
& V=\text { volume of the gas }
\end{aligned}
$$
According to the question:
mass of all molecules is halved an velocity is double then
$$
\begin{aligned}
\mathrm{P}^{\prime}=\frac{1}{3}\left(\frac{m}{2}\right) \times \frac{\mathrm{N}}{\mathrm{V}} \times( & \left(2 \mathrm{~V}_{\mathrm{ras}}\right)^2 \\
& =\frac{2}{3} \frac{m \mathrm{~N}}{\mathrm{~V}} \mathrm{~V}_{\mathrm{ras}}^2 \\
& =2 \mathrm{P}
\end{aligned}
$$
compare with eq (i)
$$
\begin{aligned}
& \mathrm{P}^{\prime}=2 \mathrm{P} \\
& \mathrm{P}^{\prime}=2 \mathrm{P}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2013 (Karnataka)
In a Wheatstone's bridge all the four arms have equal resistance $R$. If the resistance of the galvanometer arm is also $R$, the equivalent resistance of the combination as seen by the battery is:

Options:
A $\frac{R}{4}$
B $\frac{R}{2}$
C $R$
D $2 R$

The correct answer is:
$R$

Solution:
Resistance seen by the battery $=$ Equivalent resistance between $A$ and $B$ is $R$.

Topic: Physics, Current Electricity, NEET, NEET 2003
In a wire stretched by hanging a weight from its end, the elastic potential energy per unit volume in terms of longitudinal strain σ and modulus of elasticity Y is

Options:
A Yσ22
B Yσ2
C 2Yσ22
D Y2σ2

The correct answer is:
Yσ22

Solution:
Energy density=12×stress×strain,
Y=stressσ or stress=Yσ,
∴ Energy density=12 Yσ×σ=Yσ22

Topic: Physics, Mechanical Properties of Solids, NEET
In a YDSE, if the slits are of unequal widths,

Options:
A fringes will not be formed
B the positions of minimum intensity will not be completely dark
C bright fringe will not be formed at the centre of the screen
D distance between two consecutive bright fringes will not be equal to the distance between two consecutive dark fringes.

The correct answer is:
the positions of minimum intensity will not be completely dark

Solution:
If width of the slits are different, the ray undergoing interference will be of different intensity. So, when they will be out of phase, they will not completely cancel each other, resulting in minimum non zero intensity.

Topic: Physics, Wave Optics, NEET
In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

Options:
A 12
B 18
C 24
D 30

The correct answer is:
18

Solution:
In Young’s double slit experiment, a fringe width = λD d
Given, that x 2 − x 1 =12 λ 1 D d ...(i)(where   λ 1 =600  nm)
Also, x 2 - x 1 = k λ 2 Dâ¡ dâ¡ (Where k is the number of fringes)
Dividing (i) and (ii)
1= 12 λ 1 k λ 2 ∴   k= 12×600 400 =18
∴ The number of fringes observed now = 18.

Topic: Physics, Wave Optics, NEET
In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:

Options:
A 8
B 9
C 12
D 6

The correct answer is:
12

Solution:
Wavelength of light used is inversely proportional to number of fringes observed.

Hence, we can write:

n1λ1=n2λ2⇒n2=n1λ1λ2=8×600×10-9400×10-9=12

Topic: Physics, Wave Optics, NEET, NEET 2022 (Phase 1)
In a Young's double slit experiment, I0 is the maximum intensity and Î² is the fringe width. Intensity at point P which is distance x from central maxima is

Options:
A I0cosπxβ
B 4I0cos2πxβ
C I0cos2πxβ
D I04cos2πxβ

The correct answer is:
I0cos2πxβ

Solution:
Path difference at point P=xdD

∴ phase deff at. P=2πλxdD=2πxβ

∴ Intensity at P=I+I+2I cos2πxβ

=2I1+cos2πxβ ∵I0=4I

=4I cos2πxβ=I0 cos2πxβ

Topic: Physics, Wave Optics, NEET
In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.

Options:
A $5 \frac{\lambda}{2}$
B $10 \frac{\lambda}{2}$
C $9 \frac{\lambda}{2}$
D $11 \frac{\lambda}{2}$

The correct answer is:
$9 \frac{\lambda}{2}$

Solution:
In a YDSE, the path difference for $\mathrm{n}$ th minima is given by
$$
\Delta y=(2 n-1) \frac{\lambda}{2}
$$
For 5 th minima, $\mathrm{n}=5$
$$
\therefore \quad \Delta y=[2(5)-1] \frac{\lambda}{2}=\frac{9 \lambda}{2}
$$

Topic: Physics, Wave Optics, NEET, NEET 2019 (Odisha)
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

Options:
A 15.6 mm
B 1.56 mm
C 7.8 mm
D 9.75 mm

The correct answer is:
7.8 mm

Solution:
y1=n1Î»1Dd for bright fringes
y2=n2Î»2Dd for bright fringes
To coincide
n1Î»1=n2Î»2
n1Ã650=n2Ã520
â´ n1n2=45
For minimum value of n1 and n2
n1=4 n2=5
y1min=4Ã650ÃnmÃ150m0.5Ã10-3m
=7800Ã10-6m
=7.8 mm

Topic: Physics, Wave Optics, NEET
In a Young's double slit experiment, the fringe width is found to be 2 mm , when light of wavelength 6000 â« is used. Find the final fringe width if the whole apparatus is immersed in water having refractive index as 1.33 .

Options:
A 0.5 mm
B 1Â mm
C 1.5 mm
D 2Â mm

The correct answer is:
1.5 mm

Solution:
Fringe width Î²=2âmm=2Ã10â3 m
Î»=6000 â«
Î¼=1.33
Î²â²=?
Î²â²=Î²Î¼
=21.33

= 1.5 mm

Topic: Physics, Wave Optics, NEET
In a Young's double slit experiment with light of wavelength λ, the separation of slits is d and distance of screen is D such that D≫d≫λ . If the Fringe width is β , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

Options:
A β4
B β3
C β6
D β2

The correct answer is:
β4

Solution:
As Imax=4I
Then Inet=12 Imax=2I
â2I=2I1+cosÏ
â1+cosÏ=1
âcosÏ=0
âÏ=Ï2,3Ï2,5Ï2, â¦
Ï=2ÏÎ»Îx
âÏ2=2ÏÎ»Îx
âÎx=Î»4,3Î»4,5Î»4, â¦
Also path difference Îx=dsinâ¡Î¸ =Î»4
At desired location
âd ynD=Î»4
â yn=Î»D4d=Î²4 âµÎ²=Î»Dd
â Separation between central maxima and desired y=Î²4

Topic: Physics, Wave Optics, NEET
In a zero order reaction for every $10^{\circ}$ rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$, the rate of the reaction will become

Options:
A 256 times
B 512 times
C 64 times
D 128 times

The correct answer is:
512 times

Solution:
For $10^{\circ}$ rise in temperature, $n=1$ so rate $=2^n=2^1=2$
When temperature is increased from $10^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$, change in temperature $=100-10=90^{\circ} \mathrm{C}$, i.e., $n=9$
So, rate $=2^9=512$ times
Alternate method With every $10^{\circ}$ rise in temperature, rate becomes double, so $\frac{r^{\prime}}{r}=2^{\left(\frac{100-10}{10}\right)}=2^9=512$ times.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2012 (Screening)
In a zero order reaction, what is the mathematical relationship between the rate and time?

Options:
A Rate ∝ t
B Rate ∝ 1/t
C Rate ∝ t2
D Rate ∝ ln(t)

The correct answer is:
Rate ∝ 1/t

Solution:
In a zero order reaction, the rate is inversely proportional to time, so Rate ∝ 1/t.

Topic: Chemistry, Chemical Kinetics, NEET
In accordance with the study of industrial melanism in England, which of these is NOT correct?

Options:
A Before industrialization, light-winged moths were more on trees than dark-winged moths.
B Dark winged or melanic moths were a mutant variety.
C After industrialization, melanic moths reduced number.
D Lichens could grow over tree trunk in a non-polluted atmosphere.

The correct answer is:
After industrialization, melanic moths reduced number.

Solution:
Industrial melanism is a classic example of natural selection. It was seen in England that moth (Biston betularia) were available in two varieties. Black and White in colour. In earlier times the white moth were able to survive as these used to camouflage on the barks while black moth were spotted easily. In 1950 industrilisation began in England. This resulted in release of soot and this was deposited on the barks of trees. Thus the barks of trees became black and thus black or dark moths were favoured and survived white white moths were easily spotted by predators. These black moths are also known as melanic moths.

Topic: Biology, Evolution, NEET
In acidic medium, H2O2 changes Cr2O7-2 to CrO5 which has two -O-O- bonds Oxidation state of Cr in CrO5 is:

Options:
A +5
B +3
C +6
D -10

The correct answer is:
+6

Solution:
Chromium forms six bonds with the oxygen atoms and oxygen is more electronegative than the chromium. Due to this chromium shows oxidation state in CrO5 is +6.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2014
In acidic medium, H2O2 changes Cr2O7-2 to CrO5 which has two (–O–O) bonds. Oxidation state of Cr in CrO5 is

Options:
A + 3
B + 5
C + 6
D – 10

The correct answer is:
+ 6

Solution:
Let oxidation number of Cr as x.
x+(2)+4(-1)=0
x=+6

Topic: Chemistry, Redox Reactions, NEET
In addition (chain growth) polymerization, what type of compounds are used as monomers?

Options:
A Saturated compounds
B Unsaturated compounds
C Ionic compounds
D Aromatic compounds

The correct answer is:
Unsaturated compounds

Solution:
Unsaturated compounds, such as alkenes and alkadienes, are commonly used as monomers in addition (chain growth) polymerization.

Topic: Chemistry, Polymers, NEET
In addition to animals, where else is glycogen found?

Options:
A Plants
B Bacteria
C Fungi
D Viruses

The correct answer is:
Fungi

Solution:
Glycogen is also found in yeast and fungi.

Topic: Chemistry, Biomolecules, NEET
In addition to determining thermodynamic properties, for what other purpose are electrochemical cells extensively used?

Options:
A Generating electrical energy.
B Separating elements from their compounds.
C Determining the color of solutions.
D Analyzing the radioactivity of solutions.

The correct answer is:
Generating electrical energy.

Solution:
Electrochemical cells are extensively used for generating electrical energy in addition to determining thermodynamic properties.

Topic: Chemistry, Electrochemistry, NEET
In addition to flavouring agents, for what other purpose are some families of aldehydes, ketones, and carboxylic acids manufactured, according to the paragraph?

Options:
A Medicinal drugs
B Fertilisers
C Solvents and materials preparation
D Cleaning agents

The correct answer is:
Solvents and materials preparation

Solution:
The paragraph mentions that some families of aldehydes, ketones, and carboxylic acids are manufactured for use as solvents and for preparing materials like adhesives, paints, resins, perfumes, plastics, and fabrics. Therefore, they have applications beyond flavouring agents.

Topic: Chemistry, Aldehydes, Ketones and Carboxylic Acids, NEET
In addition to its use in explosives and fertilizers, what is another application of nitric acid mentioned in the paragraph?

Options:
A Production of rocket fuels
B Etching of metals
C Neutralization of acids
D Preparation of organic acids

The correct answer is:
Etching of metals

Solution:
The paragraph states that nitric acid is used in the etching of metals, which is mentioned as one of its major applications.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
In addition to using water filters, what is another preventive measure to reduce lead exposure in older homes with lead pipes?

Options:
A Replacing all plumbing with plastic pipes
B Regularly flushing the water system
C Increasing the water pressure
D Adding more lead to the pipes

The correct answer is:
Regularly flushing the water system

Solution:
Flushing the water system in older homes with lead pipes can help reduce lead exposure by removing standing water that may have absorbed lead from the pipes.

Topic: Chemistry, Environmental Chemistry, NEET
In aerobic respiration, how many times more ATP is produced than anaerobic respiration?

Options:
A 10
B 19
C 3
D 30

The correct answer is:
19

Solution:
In aerobic respiration during complete oxidation of one glucose molecule, 38 ATP are produced. However, in anaerobic respiration, only 2 ATP are produced. Thus, in aerobic respiration, 19 times more ATP is produced than in anaerobic respiration.

Topic: Biology, Respiration in Plants, NEET
In aerobic respiration, there is complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, releasing _____________ molecules of CO2

Options:
A Two
B Three
C Four
D Six

The correct answer is:
Three

Solution:
Pyruvate is a three-carbon compound thus its complete oxidation release three molecules of carbon dioxide, of which 1 molecule is liberated in link reaction, and 2 molecules are liberated during Kreb's cycle.

Topic: Biology, Respiration in Plants, NEET
In agarose gel electrophoresis, DNA molecules are separated on the basis of their:

Options:
A Charge the only
B Size only
C Charge to size ratio
D All of the above

The correct answer is:
All of the above

Solution:
In agarose gel electrophoresis, DNA molecules are separated on the basis of their charge, size and charge to size ratio.

Topic: Biology, Biotechnology - Principles and Processes, NEET
In agarose gel electrophoresis,

Options:
A DNA migrates towards the negative electrode
B Supercoiled plasmids migrate slower than their nicked counterpart
C Larger molecules migrate faster than smaller molecules
D Ethidium bromide can be used to visualize the DNA

The correct answer is:
Ethidium bromide can be used to visualize the DNA

Solution:
Ethidium bromide is an intercalating agent commonly used as a fluorescent tag (nucleic acid stain) in molecular biology laboratories for techniques such as agarose gel electrophoresis. It is commonly abbreviated as EtBr, which is also an abbreviation for bromoethane. Ethidium bromide is a DNA interchelator, inserting itself into the spaces between the base pairs of the double helix

Topic: Biology, Biotechnology - Principles and Processes, NEET
In alcohol fermentation:

Options:
A triose phosphate is the electron donor while acetaldehyde is the electron acceptor
B triose phosphate is the electron donor while pyruvic acid is the electron acceptor
C there is no electron donor
D oxygen is electron acceptor

The correct answer is:
triose phosphate is the electron donor while acetaldehyde is the electron acceptor

Solution:
During fermentation, glucose undergoes glycolysis and yields pyruvic acid, which then transformed into acetyl CoA. Then glyceraldehyde phosphate is produced, which then transforms to acetaldehyde and this acetaldehyde transforms into ethanol. In alcohol fermentation triose phosphate is the electron donor, at the same time as acetaldehyde is the electron acceptor.
Related Theory
Alcoholic fermentation is also known as ethanol fermentation where biological elements such as glucose, fructose, and sucrose are converted into cellular energy and thereby produce ethanol and carbon dioxide as their by-product.

Topic: Biology, Respiration in Plants, NEET, NEET 2003
In alimentary canal, pyloric sphincter prevents the entry of food from -

Options:
A oesophagus to stomach
B stomach to intestine
C ileum to colon
D bile duct to the intestine

The correct answer is:
stomach to intestine

Solution:
Three major parts of the human stomach are -
A. The cardiac portion - Oesophagus opens into this region.
B. Fundic portion
C. Pyloric portion - It opens into the first part of small intestine.
The pyloric sphincter regulates the opening of the stomach into the duodenum.

Topic: Biology, Digestion and Absorption, NEET
In all the following, the gynoecium is unilocular with multiple ovules except

Options:
A Trifolium
B Lupin
C Sesbania
D Petunia

The correct answer is:
Petunia

Solution:
In Fabaceae, the gynoecium has superiorovary, and is monocarpellary, unilocular with many ovules, style single.
Examples: Pulses (gram, arhar, sem, moong, soyabean; edible oil (soyabean, groundnut); dye (Indigofera); fibres (sunhemp); fodder (Sesbania, Trifolium), ornamentals (lupin, sweet pea); medicine (muliathi).
In Solanaceae, the gynoecium is bicarpellary obligately placed, syncarpous; ovary superior, bilocular, placenta swollen with many ovules.
Examples: Tomato, brinjal, potato, spice (chilli); medicine (belladonna, ashwagandha); fumigatory (tobacco); ornamentals (petunia).

Topic: Biology, Morphology of Flowering Plants, NEET
In alveoli of the lungs, the air at the site of gas exchange, is separated from the blood by

Options:
A Alveolar epithelium only
B Alveolar epithelium and capillary endothelium
C alveolar epithelium capillary endothelium and tunica adventitia
D Alveolar epithelium capillary endothelium a thin layer of tunica media and tunica adventiti

The correct answer is:
Alveolar epithelium and capillary endothelium

Solution:
Alveoli are the site of the respiratory exchange of gases. Oxygen from the alveolar air diffuses through the alveolar epithelium and the capillary endothelium into the capillary blood and carbon dioxide diffuses in the opposite direction.

Topic: Biology, Breathing and Exchange of Gases, NEET
In Amoeba, encystation is

Options:
A withdrawal of pseudopodia and secretes a one-layered hard covering or cyst around itself.
B withdrawal of pseudopodia and secretes a two-layered hard covering or cyst around itself.
C withdrawal of pseudopodia and secretes a three-layered hard covering or cyst around itself.
D withdrawal of pseudopodia and secretes a four-layered hard covering or cyst around itself.

The correct answer is:
withdrawal of pseudopodia and secretes a three-layered hard covering or cyst around itself.

Solution:
Under unfavourable condition the Amoeba withdraws its pseudopodia and secretes a three-layered hard covering or cyst around itself. This phenomenon is termed as encystation.

Topic: Biology, Reproduction in Organisms, NEET
In Amoeba the

(i)

vacuole is important for osmoregulation and excretion. In many cells, as in protists,

(ii)

vacuoles are formed by engulfing the particles.

Options:
A (i) food; (ii) contractile
B (i) osmoregulatory; (ii) contractile
C (i) food; (ii) osmoregulatory
D (i) contractile; (ii) food

The correct answer is:
(i) contractile; (ii) food

Solution:
In Amoeba the contractile vacuole is important for osmoregulation and excretion. In many cells, as in protists, food vacuoles are formed by engulfing the food particles.

Topic: Biology, Cell - The Unit of Life, NEET
In amorphous solids, are their properties such as mechanical strength, refractive index, and electrical conductivity the same in all directions?

Options:
A Yes, they are the same in all directions.
B No, they vary in different directions.
C They are not mentioned in the paragraph.
D Only mechanical strength is the same in all directions.

The correct answer is:
Yes, they are the same in all directions.

Solution:
In amorphous solids, properties like mechanical strength, refractive index, and electrical conductivity are the same in all directions because of their isotropic nature, as stated in the paragraph.

Topic: Chemistry, Solid State, NEET
In an AC circuit, a resistance R is connected in series with an inductance L. If the phase angle between the voltage and the current is π4, then the value of the inductive reactance is

Options:
A R4
B R2
C R
D R8

The correct answer is:
R

Solution:
If the phase angle is ϕ, then

tanϕ=XLR⇒tanπ4=XLR⇒XL=R

Topic: Physics, Alternating Current, NEET
In an $A C$ circuit the emf (e) and the current (i) at any instant are given respectively by $e=E_0 \sin \omega t$ $i=I_0 \sin (\omega t-\phi)$
The average power in the circuit over one cycle of AC is

Options:
A $\frac{E_0 I_0}{2}$
B $\frac{E_0 I_0}{2} \sin \phi$
C $\frac{E_0 I_0}{2} \cos \phi$
D $E_0 I_0$

The correct answer is:
$\frac{E_0 I_0}{2} \cos \phi$

Solution:
Key Idea: The power is defined as the rate at which work is being done in the circuit.
Power $=$ rate of work done in one complete cycle.
$\begin{aligned}
& P_{a \nu}=\frac{W}{T} \\
& P_{a \nu}=\frac{\left(E_0 I_0 \cos \phi\right) T / 2}{T} \\
& P_{a \nu}=\frac{E_0 I_0 \cos \phi}{2}
\end{aligned}$
where $\cos \phi$ is called the power factor of an AC circuit.

Topic: Physics, Alternating Current, NEET, NEET 2008 (Screening)
In an $\mathrm{AC}$ circuit an alternating voltage $e=200 \sqrt{2} \sin 100 t$ volt is connected to a capacitor of capacity $1 \mu \mathrm{F}$. The rms value of the current in the circuit is

Options:
A $100 \mathrm{~mA}$
B $200 \mathrm{~mA}$
C $20 \mathrm{~mA}$
D $10 \mathrm{~mA}$

The correct answer is:
$20 \mathrm{~mA}$

Solution:
Given, $e=200 \sqrt{2} \sin 100 t$
and
$$
\begin{aligned}
C & =1 \mu \mathrm{F} \\
E_{\mathrm{rms}} & =200 \mathrm{~V} \\
X_C & =\frac{1}{\omega C}=\frac{1}{1 \times 10^{-6} \times 100}=10^4 \Omega \\
i_{\mathrm{rms}} & =\frac{E_{\mathrm{rms}}}{X_C} \\
i_{\mathrm{rms}} & =\frac{200}{10^4}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2011 (Screening)
In an AC circuit, the current is 3 A and voltage 210 V and power is 63 W. The power factor is

Options:
A 0.11
B 0.09
C 0.08
D 0.10

The correct answer is:
0.10

Solution:
Power P=I.V.cosθ
∴ cosθ=PIV=633×210=0.1

Topic: Physics, Alternating Current, NEET
In an AC circuit, the instantaneous values of emf and current are E=200sin300t V and i=2sinâ¡300t+Ï3 A respectively. The average power consumed (in W) is

Options:
A 200
B 100
C 50
D 400

The correct answer is:
100

Solution:
The average power consumed in an AC circuit is given by
P=V0I02cos⁡ϕ
Where ϕ is phase angle and V0 and I0 are the peak values of voltage and current respectively.
Given, V0=200 V, I0=2 A, ϕ=π3.
P=200×22cos⁡π3
=200×22×12=100 W

Topic: Physics, Alternating Current, NEET
In an AC circuit, the instantaneous values of emf and current are e=200sinâ¡314t volt and I=sinâ¡314t+Ï3 A. The average power consumed in watt is

Options:
A 200
B 100
C 50
D 25

The correct answer is:
50

Solution:
Vrms=2002, Irms=12
â´ P=VrmsIrmscosâ¡Ï=200212cosâ¡Ï3=50 W

Topic: Physics, Electromagnetic Induction, NEET
In an AC circuit, V and I are given by V=150sinâ¡150t V and I=150sinâ¡150t+Ï3 A. The power dissipated in the circuit is

Options:
A zero
B 5625 W
C 150 W
D 106 W

The correct answer is:
5625 W

Solution:
Power P=12 V0I0cos⁡ϕ
=0.5×150×150×cos⁡60°=225004
=5625 W

Topic: Physics, Alternating Current, NEET
In an acidified solution of K2Cr2O7, H2O2 is added. Then

Options:
A solution turns green due to formation of Cr2O3
B solution turns yellow due to formation of K2CrO4
C a blue coloured compound CrOO22 is formed
D solution gives green precipitate of Cr(OH)3

The correct answer is:
a blue coloured compound CrOO22 is formed

Solution:
Hydrogen peroxide acts as an oxidising as well as a reducing agent in both acidic and alkaline media.

When H2O2 is added to an acidified potassium dichromate solution, the blue coloured compound CrO5 or CrO(O2)2 is formed. The chemical reaction is as follows

Cr2O72-+2H++4H2O2→2CrOO22+5 H2O

Topic: Chemistry, s Block Elements, NEET
In an adiabatic expansion of air, the volume increases by 5%. The corresponding percentage change in pressure will be 1.0575=1.07

Options:
A 7%
B 6%
C 4%
D 3%

The correct answer is:
7%

Solution:
P1V1γ=P2V2γγ=75, V2=1.05 V1
P1V1γ=P21.05 V1γ
P2=P11.05γ
So % change P2-P1P1×100
=P11.05γP1×100=1-1.05γ1.05γ×100=7%

Topic: Physics, Thermodynamics, NEET
In an adiabatic expansion of air, the volume increases by 5%. What is the percentage change in pressure? 1.0575=1.07

Options:
A 7%
B 5%
C 4%
D 3%

The correct answer is:
7%

Solution:
P1V1γ=P2V2γγ=75,V2=1.05V1
P1V1γ=P2(1.05V⁡)γ
P2=P1(1.05)γ
so% change P2-P1P1×100
=P1(1.05)γ-P1P1×100=1-(1.05)γ(1.05)γ×100=7%

Topic: Physics, Thermodynamics, NEET