NEET-UG 2020

Least count of screw gauge

= $\frac{Pitch}{Noofdivisionsoncircularscale}$

0.01 mm = $\frac{Pitch}{50}$

Pitch = 0.5 mm

$9.99-0.0099$ = 9.9801 m
Here, 9.98 is having the least number of decimal places of two, so answer should also have only two decimal places. So answer is 9.98 m.

$Stress=\frac{Force}{Area}$

$=\left[\frac{ML{T}^{-2}}{L^2}\right]=\left[M{L}^{-1}{T}^{-2}\right]$

$1eV=1.6\times 10^{-19}J$

$1J=\frac{1}{1.6\times {10}^{-19}}eV$

$10^{-20}J=\frac{{10}^{-20}}{1.6\times {10}^{-19}}eV$

= 0.06 eV

From the Kinematic equation
v² = u² + 2gh
v = 80 m/s
u = 20 m/s
h = $\frac{v^2-u^2}{2g}$
$=\frac{6400-400}{20}$
= 300 m

Acceleration of the masses is given to be :
$a=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_{}\right)}$
where, $m_1>m_2$
$a=\frac{\left(6-4\right)g}{6+4}$
$a=\frac{g}{5}$

Let's assume both the particles are lying on X-axis.
Let the 5 kg particle is at the origin.
Then the X-coordinate of the centre of mass of the system is,

$x_{cm}=\frac{m_{{}_1}{x}_1+m_2{x}_2}{m_1+m_2}$

= $\frac{5\times 0+100\times 10}{5+10}$

= 66.66 cm $\simeq$ 67 cm

Torque is the moment of Force applied.
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
$\overrightarrow{\tau }=2\hat{k}\times 3\hat{j}$
$\overrightarrow{\tau }=-6\hat{i}$ Nm

We know W = mg

At a height h, above the surface of earth, weight is given as,

$mgh=\frac{mg}{\left[{\left(1+\frac{h}{R}\right)}^2\right]}$

$=\frac{72}{{\left(1+\frac{R/2}{R}\right)}^2}$

$=\frac{72}{{\left(\frac{3}{2}\right)}^2}$

$=\frac{4}{9}\times 72=32N$

The weight of water in the capillary is balanced by the surface tension force.

mg = 2$\pi$r.T.cos$\theta$

Since constant, m is directly proportional to r.

m $\propto$ r

$\frac{m_2}{m_1}=\frac{r_2}{r_1}$

$\frac{m_2}{5}=\frac{2r}{r}$

$\therefore$ $m_2$ = 10 grams.

$Stress=\frac{Mg}{A}$

$Strain=\frac{\Delta L}{L}=\frac{L_{-1}-L}{L}$

Young's modulus = $\frac{Stress}{Strain}$

= $\frac{MgL}{A\left(L_1-L\right)}$

The degree of freedom for monoatomic gas is 3. So, average thermal energy per molecule,
K.E._avg = $\frac{3}{2}{k}_{B}T$

As per the formula,
$\lambda =\frac{1}{\sqrt{2}n\pi d^2}$

From the ideal gas equation, PV = nRT also
Volume (V) = $\frac{mass(M)}{density(\rho )}$
So, PM = $\rho$RT
P = 249 $\times$ 10³ N/m²
M = 2 $\times$ 10^-3 kg
T = 333 K
$\rho$ = $\frac{\left(249\times {10}^3\right)\left(2\times {10}^{-3}\right)}{8.3\times 300}$
= 0.2 kg/m³

$\Delta Q=ms\Delta T$

$\Delta Q=\frac{4}{3}\rho s\Delta T$

$\frac{\Delta Q_1}{\Delta Q_2}=\frac{r_1^3}{r_2^3}=1.5^3=\frac{27}{8}$

The system is insulated thermally. So no heat exchange takes place. Therefore the process is adiabatic.

Displacement, y = A.sin $\omega$ t
Velocity, $\upsilon$ = $\frac{dy}{dt}A\omega .\cos \omega t$
Acceleration, a = $-A{\omega }^2.\sin \omega t$
$=A{\omega }^2.\sin \left(\omega t+\pi \right)$
So the phase difference between y and a is $\pi$.

Difference of $f_A>f_B$

$f_A-f_B$ = 6 Hz

$f_A$ = 530 Hz

$f_B$ = 524 Hz (original)

Electric potential due to electric dipole (V)

$=\frac{1}{4\pi {\epsilon }_0}\frac{p.\cos \theta }{r^2}$

$V=\frac{9\times {10}^9\times 16\times {10}^9\times \cos 60}{0.36}$

V = 200 V

Potential V is constant. Electric field E is the differentiation of potential V with respect to distance. Differentiation of a constant is zero. So E = 0.

The electric field for the conducting sphere, away from the surface is given by,

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

$=\frac{9\times {10}^9\times 3.2\times {10}^{-7}}{225\times {10}^{-4}}$

$=0.128\times 10^6$

$=1.28\times 10^5$ N/C

Initially, $\frac{P}{10}=\frac{l_1}{l_2}=\frac{3}{2}$
$\Rightarrow$ $P=\frac{30}{2}=15\Omega$
Now Resistance, $R=\frac{\rho l}{A}$
$\frac{R_1}{R_2}=\frac{l_1}{l_2}$
$\Rightarrow$ $\frac{15}{1}=\frac{1.5}{l_2}$
$l_2=0.1m$
$=1.0\times 10^{-1}m$

As per colour coding Short explanation must be given to understand the trick
Yellow = 4; Violet = 7 ; Brown = 1
Tolerance for Gold is 5%

First colour gives first digit, second colour gives the second digit and third colour gives the multiplier and fourth colour gives tolerance.

$\therefore$ Resistance R = $47\times 10^1\pm 5\%=470\pm 5\%$

The relation between the mobility and drift velocity is,

Mobility, $\mu =\frac{{\upsilon }_d}{E}$

$=\frac{7.5\times {10}^{-4}}{3\times {10}^{-10}}$

$=2.5\times 10^6{m}^2{V}^{-1}{s}^{-1}$

For very low temperatures the graph will not be straight line for metals.

The temperature coefficient of resistance is negative for semiconductor and insulators, and it is positive for metals.

The capacitance increases with dielectric constant K

$C=K{C}_0$

(or)$K=\frac{C}{C_0}=\frac{30}{6}=5$

$\epsilon =K{\epsilon }_0=$$5\times 8.85\times 10^{-12}=$

$0.44\times 10^{-10}$

Magnetic field at centre of solenoid = ${\mu }_{0}nI$
$n=\frac{N}{L}=\frac{100}{50\times {10}^{-2}}$
= 200 turns/m
I = 25A
Now, substituting the values,
B = 4$\pi$ $\times$ 10^-7 $\times$ 200 $\times$ 2.5
= 6.28 $\times$ 10^-4 T

$X_m=599$

${\mu }_r=1+X_m=600$

$\mu ={\mu }_r{\mu }_0=600\times 4\pi \times 10^{-7}$

$=2.4\pi \times 10^{-4}Tm{A}^{-1}$

i_rms = C$\omega$ $\epsilon$_rms

C = 40 $\times$ 10^-6 F

$\omega$ = 2$\pi f$ = 100$\pi$

$\epsilon$_rms = 200 V

i_rms = 200 $\times$ 40 $\times$ 10^-6 $\times$ 2$\pi$ $\times$ 50
= 2.5 A

The power factor for the LCR circuit will be :
When L is removed,

$\tan \varphi =\frac{|X_C|}{R}$

$\Rightarrow \tan \frac{\pi }{3}=\frac{X_C}{R}$ ...(i)

When C is removed,

$\tan \varphi =\frac{|X_L|}{R}$

$\Rightarrow \tan \frac{\pi }{3}=\frac{X_L}{R}$ ...(ii)

Since, X_L = X_C, the circuit is in resonance,

Z = R

Power factor = cos $\varphi$ = $\frac{R}{Z}$ = 1

The total energy of an electromagnetic wave is equally shared by E and B. So the ratio is 1 : 1.

Since the light emerges normally from the other surface, the angle of, emergence

e = 0

For the triangular prism, we know

r₁ + r₂ = A

But, e = r₂ = 0

So, r₁ = A

For surface 1,

Snell's Law is $\sin i=\mu .\sin r_1$

$\sin i=\mu .\sin A$

For small angles $\theta$ = sin$\theta$

So, i = $\mu$.A

Refractive index $\mu$ is equal to tangent of Brewster's angle i_b
$\mu$ = tan i_b
1 < $\mu$ < $\infty$
1 < tan i_b < $\infty$
tan^-1(1) < i_b < tan^-1($\infty$)
45${}^{\circ }$ < i_b < 90${}^{\circ }$

Limit of Resolution for a Telescope is

$\theta =1.22\frac{\lambda }{d}$

Given, $\lambda$ = 600 $\times$ 10^-9; d = 2 m

By substituting the values, we get

$\theta$ = 3.66 $\times$ 10^-7 rad

Given that the slit distance is made half and the screen distance made double than the original value, then,
Fringe width, $\beta =\frac{\lambda D}{d}$
Now, $d^{\prime }=\frac{d}{2}$ and $D^{\prime }=2D$
So, ${\beta }^{\prime }=\frac{\lambda (2D)}{d/2}=\frac{4\lambda D}{d}$
$\beta$ = 4$\beta$

${}_{92}^{235}U$ + ${}_0^{1}n$ = ${}_{36}^{89}U$ + ${}_0^{1}n$ + ${}_Z^{A}X$

Balancing atomic number on both side, we get

92 + 0 = 36 + Z

So, Z = 56

Balancing mass number on both side, we get

235 + 1 = 89 + 3 + A

So, A = 144

It gives ${}_{56}^{144}Ba$

Singly ionized neon has electron count more than one. Bohr's model is valid for atoms of single electron. So option (c) is not valid.

$$E=m{c}^2=0.5\times 10^{-3}\times {\left(3\times {10}^8\right)}^2=4.5\times 10^{13}J$$

$v=\frac{3}{2}{v}_0$
and $v^{\prime }=\frac{v}{2}=\frac{3}{4}{v}_0$
Here we see $v^{\prime }<v^{\prime }<v_0$
Below threshold frequency, no photo-electric emission takes place.

Energy = intensity $\times$ time $\times$ area
= 20 $\times$ 60 $\times$ 20
E = 24 $\times$ 10³ J

The de-Broglie wavelength and the accelerated potential is related as follows :

$\lambda =\frac{12.27}{\sqrt{V}}$$A^{\circ }$

$\sqrt{V}=\frac{12.27\times {10}^{-10}}{1.227\times {10}^{-11}}=10^2$

$V=10^4$ volts

For transistor action, the input is always forward biased and output is always reverse biased. For this Base has to be very thin and lightly doped.

The width of depletion layer will increase with reverse bias.

$Y=\overline{\overline{A}+\overline{B}}=\overline{\overline{A.B}}$
$=A.B\Rightarrow AND$ Gate

Number of atoms
= $\frac{w}{molarmass}\times N_A\times Atomicity$

Number of Mg atoms = $\frac{1}{24}\times N_A$

Number of O atoms = $\frac{1}{32}\times 2\times N_A$

Number of Li atoms = $\frac{1}{7}\times N_A$

Number of Ag atoms = $\frac{1}{108}\times N_A$

No. of Protons = 71 = No. of Electrons
No. of Neutrons = Mass No. $-$ No. of Protons
= 175 $-$ 71 = 104

Change in oxidation state of carbon is from $-$ 4 to + 4

${}^n{N}_2=\frac{7}{28}=0.25$
${}^n{A}_r=\frac{8}{40}=0.2$
Now, applying Dalton's law of partial pressure,
${}^P{N}_2=\frac{{}^n{N}_2}{{}^n{N}_2{+}^n{A}_r}\times P_{Total}$
$=\frac{0.25}{0.45}\times 27$ bar
= 15 bar

Total [OH^- ] = 2s + 0.1 $\simeq$ 0.1

Ionic product = [Ni²⁺][OH^-]²

$\Rightarrow$ 2 $\times$ 10^-15 = s(0.1)²

$\Rightarrow$ s = 2 $\times$ 10^-13

$\Delta T_f=k_{f}m$
= 5.12 $\times$ 0.078
= 0.399 K = 0.4 K

Pure ethanol molecules are hydrogen bonded. On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them. This weakness the intermolecular attractive interactions and the solution shows positive deviation from Raoult's law.

$\Delta G$ = $\Delta G$ ${}^{\circ }$ + RT ln Q

At equilibrium $\Delta G$ = 0, Q = K_eq

So, ${\Delta }_r{G}^{\circ }$ = $-$ RT in K_eq

${\Delta }_r{G}^{\circ }$ = $-$ 8.314 J mol^-1 K^-1 $\times$ 300K $\times$ ln (2 $\times$ 10¹³)

Given reaction,
2Cl(g)$\to$Cl₂(g)
This is an endothermic reaction because it requires energy to break bonds.
So, the reverse reaction is exothermic
${\Delta }_{r}H<0$
Also, two gaseous atoms combine together to form 1 gaseous molecule.
So, the randomness,
${\Delta }_{r}S<0$

Free expansion, so p_ex = 0
So, w = $-$ p_ex$\Delta$ V = 0
Since, adiabatic process, so q = 0
Since both q and w are equal to zero. Then according to first law of thermodynamics $\Delta$ E = q + w
$\Delta$ U = 0
Hence, $\Delta$ T = 0

During the electrolysis of dil. sulphuric acid using Pt electrodes following reaction will take place.
At cathode :
$4H_{(aq)}^{+}+4e^{-}\to 2H_{2(g)}$
At anode :
$2H_2{O}_{(I)}\to O_{2(g)}+4H_{(aq)}^{+}+4e^{-}$
At the anode oxygen gas will be released.

1 equivalent of any substance is deposited by 1F of charge.
We have 20g calcium.
No. of equivalent = $\frac{givenmass}{equivalentmass}$
Ca²⁺ + 2e^$$-$$ $\to$ Ca(s)
v.f. = 2
According to Faraday's 1st law
Charge passed in Faradey = g. equivalent of product
= $\frac{20}{40}\times 2$ = 1 F
So, 1 F of charge is required.

The correct answer to this question is collision frequency. Here's an explanation for each option to understand why:

Option A - Heat of Reaction: The heat of reaction, also known as the enthalpy change $(\Delta H)$, is determined by the difference in the energy levels of the reactants and products. It is a characteristic feature of a particular chemical reaction and is not typically affected by the concentration of reactants except in cases of extremely high or non-ideal concentrations where volume changes might somewhat influence pressure and therefore the heat of reaction in gases. Under normal conditions, however, changing the concentration of reactants does not alter the heat of reaction.

Option B - Threshold Energy: Threshold energy is the minimum energy that reactant molecules must possess in order to undergo a successful collision, leading to a chemical reaction. This value is inherent to the specific reaction and is related to the strength of the bonds in the reactants as well as the energy required to form the activated complex during the reaction. The threshold energy does not change simply because the concentration of reactants is altered.

Option C - Collision Frequency: Collision frequency corresponds to how often reacting particles collide in a given time interval. When the concentration of reactants is increased, there are more reactant particles per unit volume. This statistically leads to an increased number of collisions per unit time, thereby raising the collision frequency. This is in accordance with the collision theory of chemical kinetics.

Option D - Activation Energy: Activation energy is the minimum energy barrier that must be overcome for reactants to be converted into products. It is a property of a particular reaction and is related to the nature of the reactants and the reaction pathway. Altering the concentration of reactants does not change the activation energy for that reaction.

Thus, the only factor among the provided options that changes with an increase in the concentration of reactants is indeed the collision frequency.

$k=\left(\frac{2.303}{t}\right)$ log$\left(\frac{A_0}{A}\right)$ for first order reaction

$4.606\times 10^{-3}=\left(\frac{2.303}{t}\right)$ log$\left(\frac{2}{0.2}\right)$

$\Rightarrow$ t = 500 s

For BCC, lattice $\sqrt{3}a=4r$
$r=\frac{\sqrt{3}a}{4}$
So, $r=\frac{\sqrt{3}\times 288}{4}$

In colloidal solution, the potential difference between the fixed layer and the diffused layer of opposite charge is known as Zeta potential. The presence of equal and similar charges on colloidal particles are largely responsible in providing stability to the colloidal solution.
Zeta potential $\propto$ Stability of colloidal particle

Unununnium
Atomic number = 111
IUPAC official name : Roentgenium (Rg) not Darmstadtium.

CO : Neutral oxide
BaO : Basic oxide
Al₂O₃ : Amphoteric oxide
Cl₂O₇ : Acidic oxide

Due to their symmetrical structure, BF₃, BeF₂, CO₂ and 1, 4-dichloro benzene molecules have a zero dipole moment.

For He₂ molecule
Total number of electron = 4
Electronic configuration is $\sigma$1s², $\sigma$*1s²
Bond order = $\frac{[N_b-N_a]}{2}$
= $\frac{[2-2]}{2}$
= 0
So bond order = 0
So bond order = 0 (Remove this line, not needed.)
Since, bond order is zero, so He₂ molecule does not exist.

The iron obtained from blast furnace contains about 4% carbon and many impurities like S, P, Si, Mn in due to it's malleable nature they can smaller amount. This is known as pig iron and cast into variety of shapes.

Potassium (K) activities many enzymes participate in the oxidation of glucose to produce ATP and helps in the transmission of nerve signal along with Na.

Since CaCl₂ and MgCl₂ are more soluble than NaCl, on passing HCl(g) through A solution containing CaCl₂, MgCl₂ and NaCl, then NaCl crystallizes out (common ion effect).

The carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin.

$\bullet$ Mixture of CO and H₂ gases is known as water gas or synthesis gas.
$\bullet$ Temporary hardness of water is due to bicarbonates of calcium and magnesium.
$\bullet$ Diborane (B₂H₆) is an electron deficient hydride.
$\bullet$ H₂O₂ is non-planar molecule having open book like structure.

Dry ice is used as refrigerant. C₆₀ contains 20 six membered ring, 12 five membered rings. In the presence of a zeolite catalyst (HZSM-5), ethanol, methanol and larger alcohols can be converted into gasoline at 300-400${}^{\circ }$ C. CO, is a tasteless, colorless, and odorless gas.

Oxidation state of Cr in CrO₄^2- and Cr₂O₇^2- is + 6.

Electronic configuration of $Cr:[Ar]3d^{5}4s^1$

Electronic configuration of $C{r}^{2+}$:[Ar]3d⁴

No. of unpaired electrons = 4

Spin only magnetic moment = $\sqrt{n(n+2)}$

n = number of unpaired e^$$-$$

Spin only magnetic moment = $\sqrt{4(4+2)}$

= $\sqrt{24}$ BM

= 4.9 BM

Spectrochemical series (as given in NCERT) :
$L^{-}<B{r}^{-}<SC{N}^{-}<C{l}^{-}<S^{-2}<F^{-}<O{H}^{-}<C_2{O_4}^{2-}<H_{2}O<NC{S}^{-}<EDT{A}^{4-}<N{H}_3<en<C{N}^{-}<CO$

Paper chromatography is a type of partition chromatography in which a special quality paper known as chromatography paper is used.

More the number of $\alpha$$-$ H atoms, more will be the hyper conjugation effect hence more will be the stability of carbocation.

Wurtz reaction is used to prepare symmetrical alkanes like

R₁$-$ X + 2Na + X$-$ R₁$\frac{\text{buildrel}Dryether}{⟶}$

R₁$-$R₁ + R₁$-$ R₂ + R₂$-$ R₂ + 2NaX

Using Wurtz reaction, n-Heptane can not be produced in good production because it is unsymmetrical alkane.

Since $\beta$-hydrogen is abstracted it is $\beta$-elimination.
Since more substituted alkene is formed (stable and major product), it follows Zaitsev's rule or Saytzev's rule.
Since H and Br are removed, it is dehydrohalogenation.

In the presence of dil. OH^- , benzaldehyde and acetophenone will react to undergo cross-aldol condensation.

Aliphatic and aromatic primary amines gives carbylamine reaction (isocyanide test).

Naturally occurring polymer, natural rubber is cis-1, 4-polyisoprene.

Lysine is a basic amino acid it contains more number of $-$ NH₂ groups as compared to $-$ COOH groups due to their symmetrical.

Sucrose is a disaccharide composed of two monosaccharide units, glucose and fructose, joined together via a glycosidic bond. Upon hydrolysis, this bond is broken, and the two constituent monosaccharides are released. Hydrolysis of sucrose results in the formation of one molecule of glucose and one molecule of fructose. However, it is important to specify the configurations of these monosaccharides as they exist in specific forms.

Sucrose $\frac{\text{buildrel}Hydrolysis}{⟶}$ $\alpha$-D-Glucose + $\beta$-D-Fructose

Glucose in sucrose is in the $\alpha$-D form while fructose is in the $\beta$-D configuration. Thus, the correct answer reflecting the products of sucrose hydrolysis is:

Option B: $\alpha$-D-Glucose + $\beta$-D-Fructose

This reflects the correct stereochemistry and identity of the monosaccharides produced from the hydrolysis of sucrose.

Cetyltrimethyl ammonium bromide
(C₁₉H₄₂N ⁺Br^$$-$$) cationic detergent.

Inclusion bodies are nuclear or cytoplasmic aggregates which are stainable substances, usually proteins, and formed due to viral multiplication or genetic disorders in human beings these bodies are either intracellular or extracellular abnormalities and they are specific to certain diseases. These are not involved in ingestion of food particles.

Golgi bodies are site of formation of glycoproteins and glycolipids in eukaryotic cells.

Inulin is a mixture of linear fructose polymers with different chain length and a glucose molecule at each C2 end. Adjacent fructose units are linked through glycosidic bond. The inulin that is used for medicine is most commonly obtained by soaking chicory roots in hot water.

Insulin is a protein composed of 51 amino acids and acts as a hormone which is secreted by beta-cells of pancreas. Adjacent amino acids are attached through peptide bond.

Collagen is the most abundant protein in animal world. Collagen is the major insoluble fibrous protein found in the extracellular matrix and in connective tissue. RuBisCO is the most abundant protein in the whole of the Biosphere.

Malonate is the competitive inhibitor of catalytic activity of succinic dehydrogenase, Collagen is proteinaceous in nature and possesses peptide bonds, Chitin is a homopolymer present in the cell wall of fungi and exoskeleton of arthropods, Abrin and Ricin are toxins, secondary metabolites.

When I^A and I^B are present together, they express same type of sugar is wrong statement with reference to the gene ‘I’ that controls ABO blood group because I^A and I^B are completely dominant over I^O , but when I^A and I^B are present together, they both express their own types of sugar and thus behaving as codominant alleles.

Glutamic Acid - is an acidic amino acid.
Lysine - is a basic amino acid.
Valine - is an aliphatic amino acid.
Tyrosine - is an aromatic amino acid.

During the G₁ phase of the cell cycle, a cell checks for the availability of growth factors and nutrients. If the necessary resources are not available, the cell can exit the cell cycle and enter the G0 stage, which is a non-dividing, metabolically inactive phase. Cells in the G₀ stage may stay in this phase temporarily or permanently, depending on the cell type and environmental conditions.

Some cells in the G₀ stage, such as neurons and muscle cells, remain in this phase permanently and do not re-enter the cell cycle, whereas other cells, such as liver cells and lymphocytes, can re-enter the cell cycle in response to specific signals or stimuli.