datavorous/entrance-exam-dataset huggingface dataset:
\(n\) identical resistance are taken in which \(\frac{n}{2}\) resistors are joined in series in the left gap and the remaining \(\frac{n}{2}\) resistances are joined in parallel in the right gap of a metre bridge. Balancing length in \(\mathrm{cm}\) is

Options:
A \(100 \cdot \frac{n^2}{n^2+4}\)
B \(100 \cdot \frac{n^2}{n^2+1}\)
C \(400 \cdot \frac{1}{n^2+4}\)
D \(400 \cdot \frac{1}{n^2+1}\)

The correct answer is:
\(100 \cdot \frac{n^2}{n^2+4}\)

Solution:
Meter bridge is shown in the figure below,
When \(\frac{n}{2}\) resistances are joined in series in left gap each of resistance \(R_1\), then equivalent resistance in left gap.
\(R=\frac{R_1}{2}+\frac{R_1}{2}+\frac{R_1}{2}+\ldots \frac{n}{2} \cdot \text { times }=\frac{R_1 n}{2}\)
When \(\frac{n}{2}\) resistors are joined in parallel in right gap, then the equivalent resistance in right gap.
\(\begin{aligned}
\frac{1}{S} & =\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\ldots. \frac{n}{2} \text { times }=\frac{n}{2 R_1} \\
\Rightarrow \quad S & =\frac{2 R_1}{n}
\end{aligned}\)
If \(l\) be the balancing length in the meter bridge wire, then
\(\begin{aligned}
& \frac{R}{S}=\frac{l}{100-l} \Rightarrow \frac{\frac{R_1 n}{2}}{\frac{2 R_1}{n}}=\frac{l}{100-l} \\
& \frac{n^2}{4}=\frac{l}{100-l} \Rightarrow l=\frac{100 n^2}{n^2+4}
\end{aligned}\)

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
$n$ identical waves each of intensity $I_0$ interfere with each other. The ratio of maximum intensities if the interference is (i) coherent and (ii) incoherent is :

Options:
A
$n^2$

B
$\frac{1}{n}$

C
$\frac{1}{n^2}$

D
$n$

The correct answer is:
$n$

Solution:
$\frac{\text { (Maximum intensity) coherent interference }}{\text { (Maximum intensity) coherent interference }}$ $=\frac{\mathrm{n}^2 \mathrm{I}_0}{\mathrm{nI}_0}=\mathrm{n}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (23 Apr Online)
n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.

A→B: Isothermal expansion at temperature T so that the volume is doubled from V1 to V2=2V1 and pressure changes from P1 to P2
B→C: Isobaric compression at pressure P2 to initial volume V1.

C→A: Isochoric change leading to change of pressure from P2 to P1

Total work done in the complete cycle ABCA is:

Options:
A nRTln2-12
B nRTln2
C nRTln2+12
D 0

The correct answer is:
nRTln2-12

Solution:
Wisothermal=nRTlnV2V1\

Wisobaric=PΔV=nRΔT

Wisochoric=0

W1=nRTln2 V V=nRTln2

W2=nRT2-T=-nRT2

W3=0

⇒Wnet=W1+W2+W3

Wnet=nRTln2-12

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (24 Feb Shift 1)
$\mathrm{N}$ molecules each of mass $\mathrm{m}$ of gas $\mathrm{A}$ and $2 \mathrm{~N}$ molecules each of mass $2 \mathrm{~m}$ of gas $\mathrm{B}$ are contained in a vessel which is maintained at temperature $T$. The mean square velocity of the molecules of gas B is denoted by $V_2^2$ and the mean square of the $\mathrm{X}$-component velocity of the molecules of gas $A$ is denoted by $V_1^2$, then $\frac{V_1}{V_2}$ is

Options:
A $2$
B $1$
C $\frac{2}{3}$
D $\sqrt{\frac{2}{3}}$

The correct answer is:
$\sqrt{\frac{2}{3}}$

Solution:
The mean square velocity of gas molecule is given by
$\begin{aligned} & v^2=\frac{3 K T}{m} \\ & \text { for gas A, } \\ & v_A^2=\frac{3 K T}{m} \\ & v^2=v_x^2+v_y^2+v_z^2 \quad\left(\because v_x^2=v_y^2=v_z^2\right) \\ & v^2=3 v_x^2 \Rightarrow v_x^2=\frac{v^2}{3} \\ & v_1^2=v_x^2=\frac{3 K T}{3 m}=\frac{\dot{K} T}{m}\end{aligned}$
$\mathrm{v}_1=\sqrt{\frac{\mathrm{KT}}{\mathrm{m}}}$...(1)
For gas B,
$$
\mathrm{v}_{\mathrm{B}}^2=\frac{3 \mathrm{KT}}{2 \mathrm{~m}}
$$
$\mathrm{v}_2^2=\frac{3 \mathrm{KT}}{2 \mathrm{~m}} ; \mathrm{v}_2=\sqrt{\frac{3 \mathrm{KT}}{2 \mathrm{~m}}}$...(2)
Divide equation (1) by ( 2 ), we have
$$
\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\sqrt{\frac{\mathrm{KT}}{\mathrm{m}}}}{\sqrt{\frac{3 \mathrm{KT}}{2 \mathrm{~m}}}}
$$

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2023 (14 May Shift 1)
N moles of a polyatomic gas(f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is:

Options:
A 6
B 3
C 4
D 2

The correct answer is:
4

Solution:
The equivalent degree of freedom feq of the mixture is

feq=n1f1+n2f2n1+n2

For diatomic gas feq=5.

⇒5=(n)(6)+(2)(3)n+2

⇒5n+10=6n+6

⇒n=4

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (29 Jan Shift 2)
N moles of an ideal diatomic gas is enclosed in a cylinder at temperature T. suppose on supplying heat to the gas, its temperature remains constant but n moles get dissociated into atoms. Heat supplied to the gas is

Options:
A Zero
B 12nRT
C 32nRT
D 32RT

The correct answer is:
12nRT

Solution:
Since the gas is enclosed in a vessel, therefore, during heating process, volume of the gas remains constant. Hence, no work is done by the gas. It means heat supplied to the gas is used to increase its internal energy only.
Initial internal energy of the gas is U 1 = N 5 2 R T
Since n moles get dissociated into atoms, therefore, after heating, vessel contains N - n moles of diatomic gas and 2n moles of a mono-atomic gas. Hence the internal energy for the gas, after heating, will be equal to U 2 = N - n 5 2 R T + 2 n 3 2 R T = 5 2 NRT + 1 2 nRT
Hence, the heat supplied = increase in internal energy
= U 2 - U 1 = 1 2 nRT

Topic: Physics, Thermodynamics, JEE Main
$n$ moles of an ideal gas undergo a process $A \rightarrow B$ as shown in the figure. Maximum temperature of the gas during the process is

Options:
A
$\frac{9 P_0 V_0}{n R}$

B
$\frac{3 P_0 V_0}{2 n R}$

C
$\frac{9 P_0 V_0}{2 n R}$

D
$\frac{9 P_0 V_0}{4 n R}$

The correct answer is:
$\frac{9 P_0 V_0}{4 n R}$

Solution:
For given graph, equation of $P$ - $V$ line is
$$
P-2 P_0=\frac{2 P_0-P_0}{V_0-2 V_0}\left(V-V_0\right)
$$
So, $P=3 P_0-\frac{P_0}{V_0} V$ as $P V=n R T$
$$
\Rightarrow\left(3 P_0-\frac{P_0}{V_0} V\right) V=n R T ;
$$
For maximum temperature $\frac{d T}{d V}=0$
$$
\Rightarrow \frac{d T}{d V}=3 P_0-\frac{2 P_0}{V_0} V=0 \Rightarrow V=\frac{3}{2} V_0
$$
Also $P=\frac{3}{2} P_0$
So, $T=\frac{P V}{n R}=\frac{1}{n R} \times \frac{3}{2} P_0 \times \frac{3}{2} V_0=\frac{9 P_0 V_0}{4 n R}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (12 May Online)
n moles of an ideal gas undergoes a process AâB as shown in the diagram. The maximum temperature of the gas during the process is

Options:
A 3P0V02nR
B 9P0V04nR
C 9P0V02nR
D 9P0V0nR

The correct answer is:
9P0V04nR

Solution:
Since the P-V graph of the process is a straight line and two points (V0, 2P0) and (2V0, P0) are known, its equation will be
P-P0=2P0-P0V0-2V0V-2V0=P0V02V0-V
â´ P=3P0-P0VV0
According to the equation for an ideal gas,
T=pVnR
=3P0-P0VV0VnR
=3P0V0V-P0V2nRV0 ...(i)
For T to maximum, dTdV=0
3P0V0-2P0V=0
or V=3V02 ...(ii)
Putting this value in Eq. (i), we get
Tmax=3P0V03V02-P094V02nRV0=9P0V04nR

Topic: Physics, Thermodynamics, JEE Main
n moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:

Options:
A 9Â P0V02nR
B 9Â P0V0nR
C 9 P0V04nR
D 3Â P0V02nR

The correct answer is:
9 P0V04nR

Solution:
Temperature T will be maximum where product of PV is max equation of line
P=−P0V0V+3P0
Multiplying both sides by V,
PV=−P0V0V2+3P0V
For maxima, d(PV) dV =0
⇒               −2 P 0 V 0 V+3 P 0 =0     ⇒  V= 3 V 0 2
and P=3P02
⇒   T= PV nR = 9 P 0 V 0 4 nR
Alternate method :
Since initial and final temperature are equal.
Hence maximum temperature is at middle of line.
PV=nR T max
⇒Tmax=3P023V02nR=9P0V04nR

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2016 (03 Apr)
n moles of an ideal gas with constant volume heat capacity Cv undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is:

Options:
A 4nRCv+nR
B 4nRCv-nR
C nRCv+nR
D nRCv-nR

The correct answer is:
nRCv+nR

Solution:
From First Law of thermodynamics,
ΔQ=ΔU+Δw
ΔwΔQ=1-ΔUΔQ=1-ncvΔTncpΔT
=1-cvR+cv
=RR+cv
As, ncv=Cv
ΔwΔQ=nRCv+nR

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (10 Apr Shift 1)
→Δ(n) moles of CO2 evolved during given reaction?
What is the value of 'n' here

Options:

The correct answer is:
5.00

Solution:
β -keto acids and Gem-dicarboxylic acids give CO2 on heating.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
n moles of diatomic gas in a cylinder is at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. The change in the total kinetic energy of the gas is

Options:
A 0
B 52nRT
C 32nRT
D 12nRT

The correct answer is:
12nRT

Solution:
Initial kinetic energy of the system,
Ui=N52RT
Final kinetic energy of the system,
Uf=2n32 RT+N-n52RT
=12nRT+52NRT
ÎU=Q=12nRT

Topic: Physics, Thermodynamics, NEET
N moles of diatomic gas in a cylinder is at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. The change in the total kinetic energy of the gas is

Options:
A 0
B 52nRT
C 32nRT
D 12nRT

The correct answer is:
12nRT

Solution:
Initial kinetic energy of the system,
Ui=N52RT
Final kinetic energy of the system,
Uf=2n32 RT+N-n52RT
⇒Uf=12nRT+52NRT
ΔU=Q=12nRT.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2017 (09 Apr Online)
N moles of diatomic gas in a cylinder is at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. The change in the total kinetic energy of the gas is

Options:
A 0
B 52nRT
C 32nRT
D 12nRT

The correct answer is:
12nRT

Solution:
Initial kinetic energy of the system,
Ui=N52RT
Final kinetic energy of the system,
Uf=2n32 RT+N-n52RT
=12nRT+52NRT
ΔU=Q=12nRT

Topic: Physics, Thermodynamics, NEET
NO2 and N2O4 are two forms of nitrogen dioxide. One exists in gaseous state while other in liquid state. The nature of NO2 and N2O4 forms are

Options:
A both are paramagnetic
B both are diamagnetic
C NO2 is paramagnetic while N2O4 is diamagnetic
D NO2 is diamagnetic while N2O4 is paramagnetic

The correct answer is:
NO2 is paramagnetic while N2O4 is diamagnetic

Solution:
NO2 is an odd electron molecule so NO2 is paramagnetic
The dimer, N2O4 , has no unpaired electron so it is diamagnetic.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5g→4NO2g+O2g.
The initial concentration of N2O5 is 3.00 mol L-1 and it is 2.75 mol L-1 after 30 minutes. The rate of formation of NO2 is:

Options:
A 1.667×10-2  mol L-1  min-1
B 8.333×10-3 mol L-1min-1
C 4.167×10-3 mol L-1min-1
D 2.083×10-3 mol L-1 min-1

The correct answer is:
1.667×10-2  mol L-1  min-1

Solution:
Rate of reaction in terms of rate of disappearance of N2O5(s)
-dN2O5dt=12dNO2dt=2dO2dt=rate of Reaction
-(-3+1.75)30=rate of reaction
--0.2530=8.3×10-3mol L-1min-1
=253000=8.3×10-3=12× rate of formation of O2
8.3×10-3=12dNO2dt
1.66×10-2m/min =d[NO2]dt

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (12 Apr Shift 2)
n-Propanol on treatment with concentrated HBr gives P. The product P on reaction with KCN gave the product $\mathrm{Q}$. The product Q on heating with aqueous acidic solution, furnished the product R. The product 'R' is

Options:
A Propanoic acid
B Propanamide
C Butanoic acid
D Butanamide

The correct answer is:
Butanoic acid

Solution:
n-propanol on treatment with concentrated HBr produces n-propyl bromide. n-propyl bromide on reaction with KCN to give butanenitrile.butanenitrile on hydrolysis give butanoic acid.

CH3CH2CH2OH→HBrCH3CH2CH2BrCH3CH2CH2Br→KCNCH3CH2CH2CNCH3CH2CH2CN→H3O+CH3CH2CH2COOH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
' $n$ ' small drops of same size fall through air with constant velocity $5 \mathrm{~cm} / \mathrm{s}$. They coalesce to form a big drop. The terminal velocity of the big drop is

Options:
A $7 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}$
B $5 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}$
C $3 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}$
D $9 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}$

The correct answer is:
$5 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}$

Solution:
$$
\begin{aligned}
& \text { Volume }=\frac{4}{3} \pi r^3 n=\frac{4}{3} \pi R^3 \\
& \therefore \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r} \therefore \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{\frac{1}{3}}
\end{aligned}
$$
Terminal velocity, $\mathrm{v} \propto \mathrm{r}^2$
$$
\begin{aligned}
& \frac{\mathrm{v}_2}{\mathrm{v}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^2=\mathrm{n}^{2 / 3} \\
& \therefore \mathrm{v}_2=\mathrm{n}^{2 / 3} \mathrm{v}_1=5 \mathrm{n}^{2 / 3} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Na and Li are placed in dry air, we get:

Options:
A NaOH,Na2O,Li2O
B Na2O,Li2O
C Na2O,Li2O,Li3N,NH3
D Na2O, Li3N, Li2O

The correct answer is:
Na2O, Li3N, Li2O

Solution:
Dry air contains N2 and O2. Na does not form nitride.

Topic: Chemistry, s Block Elements, JEE Main
Na and Mg crystallize in BCC and FCC type crystals respectively, then the number of atoms of Na and Mg present in the unit cell of their respective crystal is

Options:
A
4 and 2

B
9 and 14

C
14 and 9

D
2 and 4

The correct answer is:
2 and 4

Solution:
BCC - points are at corners and one in the centre of the unit cell
Number of atoms per unit cell $=8 \times \frac{1}{8}+1=2$
FCC - points are at the corners and also centre of the six faces of each cell
Number of atoms per unit cell $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2002
$\mathrm{Na}^{+}$ion is isoelectronic with

Options:
A $\mathrm{Li}$
B $M g^{+2}$
C $\mathrm{Ca}^{+2}$
D $\mathrm{Ba}^{+2}$

The correct answer is:
$M g^{+2}$

Solution:
Electrons in $\mathrm{Na}^{+}=11-1=10$
Electrons in $\mathrm{Mg}^{2+}=12-2=10$

Topic: Chemistry, Structure of Atom, JEE Main
$\mathrm{Na}^{+} / \mathrm{K}^{+}$pump is associated with

Options:
A passive transport
B active transport
C osmosis
D imbibition.

The correct answer is:
active transport

Solution:
Sodium-potassium pump $\left(\mathrm{Na}^{+}-\mathrm{K}^{+}\right)$is a membrane transport protein that exchanges sodium ions $\left(\mathrm{Na}^{+}\right)$for potassium ions $\left(\mathrm{K}^{+}\right)$. It transports $3 \mathrm{Na}^{+}$outwards for $2 \mathrm{~K}^{+}$into the cell, thus maintaining the differential concentrations of each ion across the plasma membrane. The process requires energy in the form of ATP, being a form of active transport.

Topic: Biology, Transport in Plants, AIIMS, AIIMS 2012
NaClO3 is used in spacecraft to produce O2. The daily consumption of pure O2 by a person is 492âL at 1âatm, 300âK. How much amount of NaClO3 (in grams) is required to produce O2 for the daily consumption of a person at 1âatm, 300âK?

NaClO3s+FesâO2g+NaCls+FeOs R=0.082 L atm molâ1Kâ1

Options:
A 21.3
B 1115
C 2130
D 4260

The correct answer is:
2130

Solution:
Moles of NaClO3= moles of O2
moles of O2=PVRT=1Ã4920.082Ã300=20 moles
As molecular mass of NaClO3=106.5
NaClO3=20Ã106.5=2130 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
NaCl reacts with conc. H2SO4 and K2Cr2O7 to give reddish fumes (B), which react with NaOH to give yellow solution (C). (B) and (C) respectively are ;

Options:
A CrO2Cl2, Na2CrO4
B Na2CrO4, CrO2Cl2
C CrO2Cl2, KHSO4
D CrO2Cl2, Na2Cr2O7

The correct answer is:
CrO2Cl2, Na2CrO4

Solution:
Chromyl chloride (B) is produced when a chloride ion reacts with conc.H2SO4 and K2Cr2O7.
Passing the chromyl chloride gas through NaOH turns the solution yellow. This is due to the formation of Na2CrO4 (C).

The reaction is as follows,

NaCl + conc. H2SO4 + K2Cr2O7→CrO2Cl2(B)(Reddish brown) + KHSO4 + NaHSO4 +H2O

CrO2Cl2 + NaOH → Na2CrO4(C)Yellow Colour + NaCl +H2O

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (27 Jan Shift 1)
NaClO3 is used, even in spacecrafts, to produce O2. The daily consumption of pure O2 by a person in 492 L at 1 atm, 300K. How much amount of NaClO3, in grams, is required to produce O2 for the daily consumption of a person at 1 atm, 300K ? _____

NaClO3s+Fes→O2g+NaCls+FeOs
R=0.082 L atm mol-1 K-1

Options:

The correct answer is:
2130

Solution:
Mol of NaClO3 = Mol of O2

Mol of O2=PVRT=1×4920.082×300=20 mol

Mass of NaClO3 =20×106.5=2130 g

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2020 (08 Jan Shift 2)
NaH is an example of:

Options:
A $Electron-rich hydride$
B $Metallic hydride$
C $Saline hydride$
D $Molecular hydride$

The correct answer is:
$Saline hydride$

Solution:
NaH is an ionic hydride which is also known as saline hydride.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Name a peptide hormone which acts mainly on hepatocytes and adipocytes and enhances cellular glucose uptake and utilisation.

Options:
A Insulin
B Glucagon
C Secretin
D Gastrin

The correct answer is:
Insulin

Solution:
Insulin is a peptide hormone which acts mainly on hepatocytes and adipocytes and enhances cellular glucose uptake and utilisation decreasing the blood glucose level.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2016 (Phase 2)
Name a peptide hormone which acts mainly on hepatocytes and adipocytes and enhances cellular glucose uptake and utilisation.

Options:
A Insulin
B Glucagon
C Secretin
D Gastrin

The correct answer is:
Insulin

Solution:
Insulin is a peptide hormone which acts mainly on hepatocytes and adipocytes and enhances cellular glucose uptake and utilisation decreasing the blood glucose level.

Topic: Biology, Chemical Coordination and Integration, NEET
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Options:

None

Solution:
A mixture of camphor and \(\mathrm{CaSO}_4\) can be separated by the following methods:
(i) Camphor is a sublimable solid while \(\mathrm{CaSO}_4\) is not, therefore, sublimation of mixture gives camphor on the sides of funnel while \(\mathrm{CaSO}_4\) is left in the china dish.
(ii) Camphor is soluble in organic solvents like \(\mathrm{CHCl}_3, \mathrm{CCl}_4\) etc while \(\mathrm{CaSO}_4\) is not. Therefore, when the mixture is shaken with the solvent, camphor goes into solution while \(\mathrm{CaSO}_4\) remains as residue. It is filtered and evaporation of solvent gives camphor.

Topic: Chemistry, General Organic Chemistry
Name important defence mechanisms in plants against herbivory.

Options:

None

Solution:
There are various defence measures for animals against predators. But plants, as they cannot move away, have certain defence mechanisms against herbivory. Their main defences are chemical toxins, such as strychnine, a poison produced by tropical vine, morphine by opium poppy, nicotine produced by tobacco plant. Apart from these chemicals, the common defence measure is presence of spines (modified leaves) on the leaves, stems of the plant, modifications of leaves into thorns, development of sharp silicated edges in leaves which prevent them against damage caused by herbivores.

Topic: Biology, Organisms and Populations
Name of the compound given below is:

Options:
A 4-ethyl-3-methyloctane
B 3-methyl-4-ethyloctane
C 2,3-diethylheptane
D 5-ethyl-6-methyloctane

The correct answer is:
4-ethyl-3-methyloctane

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Name of the structure of silicates in which three oxygen atoms of $\left[\mathrm{SiO}_4\right]^{4-}$ are shared is

Options:
A Pyrosilicate
B Sheet silicate
C Linear chain silicate
D Three dimensional silicate

The correct answer is:
Three dimensional silicate

Solution:
Three dimensional silicate structures are formed when three oxygen atoms of each $\left[\mathrm{SiO}_4\right]^{4-}$ tetrahedral are shared.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Name the EM wave used for studying crystal structure of solids

Options:
A Microwave
B Visible radiation
C Ultraviolet
D X – rays

The correct answer is:
X – rays

Solution:
X â Rays radiation is a form of electromagnetic radiation. X-rays have a wavelength in the range of 0.01 nm to 10 nm, corresponding to frequencies in the range 30 PHz to 30 EHz. Therefore, they are used in studying crystals structure of solids.

Topic: Physics, Electromagnetic Waves, JEE Main
Name the end product in the following series of reaction.

CH 3 COOH → Δ NH 3 A → Δ P 4 O 10 B

Options:
A CH4
B CH3OH
C CH3CN
D CH3COONH4

The correct answer is:
CH3CN

Solution:
P4O10 is a powerful dehydrating agent.
CH3COOH→ΔNH3CH3CONH2→ΔP4O10CH3CN

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Name the end product in the following series of reactions
C6H5COOH→NH3P→ΔQ→P2O5R→LiAlH4S

Options:
A Aniline
B Benzylamine
C Acetonitrile
D Acetamide

The correct answer is:
Benzylamine

Solution:
Here the final product is Benzylamine

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Name the end product in the following series of reactions
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{NH}_3} A \xrightarrow[\mathrm{P}_2 \mathrm{O}_5]{\Delta} B$

Options:
A $\mathrm{CH}_4$
B $\mathrm{CH}_3 \mathrm{OH}$
C Acetonitrile
D Ammonium acetate

The correct answer is:
$\mathrm{CH}_4$

Solution:
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3 \mathrm{CONH}_2 \xrightarrow[\mathrm{A}_2 \mathrm{O}_5]{\Delta} \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{N}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Name the functions of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae

Options:

None

Solution:
(a) Corpus luteum : The corpus luteum secretes large amounts of progesterone which is essential for maintenance of the endometrium.
(b) Endometrium is necessary for implantation of the fertilized ovum and other events of pregnancy.
(c) The acrosome is filled with enzymes that help during fertilization of the ovum.
(d) Sperm tail : Tail facilitates sperm motility which is essential for fertilization.
(e) Fimbriae : Fimbriae help in collection of the ovum after ovulation.

Topic: Biology, Human Reproduction
Name the law that stated that the sum of allelic frequencies in a population remains constant. What are the five factors that influence these values?

Options:

None

Solution:
Hardy-Weinberg Principle states that the sum of allelic frequencies in a population is stable and is constant from generation to generation or total genes and their alleles in a population i.e., the gene pool remains constant. This is called genetic equilibrium.
The five factors that influence these values of allelic frequencies are
(i) Gene Migration or Gene Flow Gene frequencies change in the original as well as in the new population. When migration of a section of population to another place occurs, new genes/ alleles are added to the new population and existing genes are lost from the old population. Gene flow would occur if this gene migration, happens multiple times.
(ii) Mutation is the sudden heritable change occuring suddenly in a gene. It alters the genetic frequency or genetic make up of an individual.
(iii) Genetic Drift refers to the elimination of the genes of certain traits when fragment population migrates or dies due to natural calamity. It is an evolutionary force operating in small populations whereby gene frequency changes by chance leading to loss of some genes or gain of others irrespective of their selective advantages or disadvantages.
(iv) Genetic Recombination this phenomenon occurs during gamete formation when chromosomes pass from parents to offsprings displaying a new combination of characteristics.
(v) Natural Selection It is a phenomenon by which some members of population, processing traits that enable them to grow and reproduce at higher rates are favoured. Thus, they leave more surviving offspring in the next generation than others.

Topic: Biology, Evolution