datavorous/entrance-exam-dataset huggingface dataset:
What will be the expression of Kp for the given reaction if the total pressure inside the vessel is P and degree of dissociation of the reactant is a? The reaction

N 2 O 4 ⇄  2 NO 2

Options:
A 4a2P1+a2
B 4a2P1-a2
C a2P1-a2
D a21-a2

The correct answer is:
4a2P1-a2

Solution:
Moles at t=0 Moles at equilibrium N2O411-aâ2NO202a
Total moles at equilibrium =1-a+2a=1+a
P N 2 O 4 =( 1âa 1+a )P;â P NO 2 =( 2a 1+a )P
Kp=PNO2PN2O4=2a1+a2P21-a1+aP=4a2P1-a2

Topic: Chemistry, Chemical Equilibrium, JEE Main
What will be the height of image when an object of $2 \mathrm{~mm}$ is placed on the axis of a convex mirror at a distance $20 \mathrm{~cm}$ of radius of curvature $40 \mathrm{~cm}$

Options:
A $20 \mathrm{~mm}$
B $10 \mathrm{~mm}$
C $6 \mathrm{~mm}$
D $1 \mathrm{~mm}$

The correct answer is:
$1 \mathrm{~mm}$

Solution:
$\frac{I}{O}=\frac{f}{f-u} \Rightarrow \frac{I}{2}=\frac{20}{20+20}=\frac{1}{2} \Rightarrow I=1 \mathrm{~mm}$

Topic: Physics, Ray Optics, JEE Main
What will be the input of $A$ and $B$ for the Boolean expression $\overline{(A+B)} \cdot \overline{(A \cdot B)}=1$

Options:
A 0,0
B 0,1
C 1,0
D 1,1

The correct answer is:
0,0

Solution:
The given Boolean expression can be written as
$\begin{aligned}
& Y=(\overline{A+B}) \cdot(\overline{A \cdot B})=(\bar{A} \cdot \bar{B}) \cdot(\bar{A}+\bar{B})=(\bar{A} \cdot \bar{A}) \cdot \bar{B}+\bar{A}(\bar{B} \cdot \bar{B}) \\
& =\bar{A} \cdot \bar{B}+\bar{A} \bar{B}=\bar{A} \bar{B}
\end{aligned}$
$\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ \hline 1 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{array}$

Topic: Physics, Semiconductors, JEE Main
What will be the magnitude of the electric field at point O as shown in the figure? Each side of the figure is l and perpendicular to the other?

Options:
A 14πε0ql2
B 14πε0q2l2(22-1)
C q4πε0(2l)2
D 14πε02q2l2(2)

The correct answer is:
14πε0q2l2(22-1)

Solution:
E1=kql2=E2

E3=kq(2ℓl=kq2l2

E=2kql2-kq2l2=kq2l2(22-1)

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (27 Jul Shift 2)
What will be the major product when m-cresol is reacted with propargyl bromide HC≡C-CH2Br in presence of K2CO3 in acetone?

Options:
A
B
C
D

The correct answer is:

Solution:
The m-cresol is the m-methyl phenol. Phenol is acidic in nature. It gives the acid-base reaction with base. So, when it reacts with the base, it will form the phenolate ion. After that phenolate ion act as a nucleophile. This nucleophile gives the nucleophilic substitution reaction with the alkyl halide. It will follow the SN2 reaction mechanism.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2019 (12 Apr Shift 2)
What will be the maximum value of Σn+l+m for all unpaired electrons of chlorine in its second excited state?

Options:
A 25
B 20
C 28
D 27

The correct answer is:
25

Solution:
Cl17 (second excited state) =3p33d2

Topic: Chemistry, Structure of Atom, JEE Main
What will be the minimum pressure required to compress \(500 \mathrm{dm}^3\) of air at 1 bar to \(200 \mathrm{dm}^3\) at \(30^{\circ} \mathrm{C}\) ?

Options:

None

Solution:
\(\begin{aligned} P_1 &=1 \text { bar, } & V_1=500 \mathrm{dm}^3 \\ P_2 &=?, & V_2=200 \mathrm{dm}^3 \end{aligned}\)
As temperature remains constant at \(30^{\circ} \mathrm{C}\),
\(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2\)
1 bar \(\times 500 \mathrm{dm}^3=\mathrm{P}_2 \times 200 \mathrm{dm}^3\)
or \(\mathrm{P}_2=\frac{500}{200} \mathrm{bar}=2.5 \mathrm{bar}\)

Topic: Chemistry, States of Matter
What will be the most suitable combination of three resistors A=2 Ω, B=4 Ω, C=6 Ω so that 223Ω is equivalent resistance of combination?

Options:
A Parallel combination of A and C connected in series with B
B Parallel combination of A and B connected in series with C
C Series combination of A and C connected in parallel with B.
D Series combination of B and C connected in parallel with A.

The correct answer is:
Parallel combination of A and B connected in series with C

Solution:
If the 2 Ω resistance and the 4 Ω resistance are connected in parallel to each other, then R'=4×24+2=43 Ω. If the resistance of 6 Ω is now connected in series to the previous combination.

Req=2×42+4+6=223 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (24 Jun Shift 2)
What will be the nature of existence of an amino acid (containing one amino and one carboxylic acid group) in solution of pH < pKa1 ?

Options:
A It exists an anion
B It exists as cation
C It exists as zwitter ion
D It exists as neutral species with no charge

The correct answer is:
It exists as cation

Solution:
At low pH (acidic solution) COO- ion acts as the base and accepts proton to form the cation.
H 3 N + â C | R | H â C || O â O â Zwitterâion â In acidic medium H 3 N + â C | R | H â C || O âOH Cation

Topic: Chemistry, Biomolecules, JEE Main
What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L min-1 to 0.48 L min-1? The radius of the tap and viscosity of water are 0.5 cm and 10-3 Pa s, respectively. (Density of water : 103 kg m-3)

Options:
A Unsteady to steady flow
B Remains steady flow
C Remains turbulent flow
D Steady flow to unsteady flow

The correct answer is:
Steady flow to unsteady flow

Solution:
The nature of flow is determined by Reynolds
Number.

Re=ρvDη

ρ→ density of fluid ; η→ coefficient of velocityv→ velocity of flow D→ Diameterofpipe

From NCERT

If Re<1000 → flow is steady

1000<Rp<2000→ flow becomes unsteady

Re>2000→ flow is turbulent

Reinitial =103×0.18×10-3π×0.5×10-22×60×1×10-210-3

=382.16

Re final=103×0.48×10-3π×0.5×10-22×60×1×10-210-3

=1019.09

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (16 Mar Shift 2)
What will be the normality of the salt solution obtained by neutralising $x$ mL. $y$ (N) HCl with $y \quad$ mL $x$ (N) $\mathrm{NaOH}$, and finally adding $(x+y) \mathrm{mL}$ distilled water

Options:
A $\frac{2(x+y)}{x y} N$
B $\frac{x y}{2(x+y)} N$
C $\left(\frac{2 x y}{x+y}\right) N$
D $\left(\frac{x+y}{x y}\right) N$

The correct answer is:
$\frac{x y}{2(x+y)} N$

Solution:
The reaction for given reaction is $\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
Number of moles	xy	xy	0	0
Added	$\quad N V \quad N V$
Number of moles left 0	0	xy	xy
$\therefore N($ solution $)=\frac{\text { Number of milliequivalent }}{\text { vol. of solution in mililetre }}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
What will be the order of reaction and rate constant for a chemical change having log  t 50% vs. log concentration (a) curves as?

Options:
A 0, 1
B 1, 1
C 2, 2
D 1, 0

The correct answer is:
0, 1

Solution:
For a reaction t12∝a1-n
=Ka1-n
or logt1/2=1-nloga+logK
y=mx+c
∴1-n=tan450=1
∴n=0

Topic: Chemistry, Chemical Kinetics, JEE Main
What will be the order of reaction for a chemical change having the graph between \( \log \mathrm{t}_{1 / 2} \mathrm{vs} \log \) a as shown below? ( where, \( \mathrm{a}= \) initial concentration of reactant; \( \mathrm{t}_{1 / 2}= \) half \( - \) life )

Options:
A zero order
B first order
C second order
D none of these

The correct answer is:
zero order

Solution:
The general expression for the half-life of nth order reaction is as follows:

t1/2=k'·a1-n

Here, k' is not rate constant

Taking log on both sides,

log t1/2=log k'+(1-n) log a

The given graph is a straight line with slope 45°, thus the equation of the line will be:

log t1/2=log a + C

On comparing both the equations, 1-n=1⇒n=0

Thus, the given graph represents a zero-order reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
What will be the pH of a solution by mtextxing 40 mL of 0.1M HCl with 10 mL of 0.45M NaOH?

Options:
A 12
B 10
C 8
D 6

The correct answer is:
12

Solution:
40mL of 0.1 M HCl=0.11000Ã40
=0.004 mole
10 mL of 0.45 M NaOH=0.451000Ã10=0.0045 mole
â´NaOH left unused =0.0005 mole
Volume of solution =50 mL
â´Molarity of NaOH=0.000550Ã1000
=0.01M
Or OH-=10-2 or H+=10-12 pH=12

Topic: Chemistry, Ionic Equilibrium, JEE Main
What will be the pH of a solution formed by mixing 10 mL 0.1 M NaH2PO4 and 15 mL 0.1 M Na2HPO4 ?

[Given: for H3PO4 pKa1=2.12,pKa2=7.2]

Options:
A 7.0
B 6.9
C 7.4
D 7.5

The correct answer is:
7.4

Solution:
pH=pKa+log⁡conjugate baseacid
Number of moles of NaH2PO4=101000L×0.1 mol L-1
=0.001 moles
Number of moles of Na2HPO4=151000×L×0.1 mol L-1
=0.0015 moles
pH=7.2+log⁡0.00150.001
=7.37≈7.4

Topic: Chemistry, Ionic Equilibrium, JEE Main
What will be the pH of a solution formed by mixing 40 cm3 of 0.1 M HCl with 10 cm3 of 0.45 M NaOH?

Options:
A 10
B 8
C 5
D 12

The correct answer is:
12

Solution:
Number of moles of HCl =MV1000=0.1×401000=0.004
Number of moles of NaOH =MV1000=0.45×101000=0.0045
Remaining moles of NaOH after neutralization = 0.0045 – 0.004 = 0.0005
Molarity of OH- =0.000550×1000=0.01M
So pOH =2
So pH = 14 – 2 = 12

Topic: Chemistry, Ionic Equilibrium, JEE Main
What will be the potential difference between two points \( \mathrm{A} \) and \( \mathrm{B} \) in front of an infinitely long line of charge of linear charge density \( \lambda \), as shown in figure ?

Options:
A \( \frac{\lambda}{2 \pi \varepsilon_{0}} \ln 2 \)
B \( -\frac{\lambda}{2 \pi \varepsilon_{0}} \ln 2 \)
C \( \frac{\lambda}{4 \pi \varepsilon_{0}} \ln 2 \)
D \( -\frac{\lambda}{4 \pi \varepsilon_{0}} \ln 2 \)

The correct answer is:
\( \frac{\lambda}{2 \pi \varepsilon_{0}} \ln 2 \)

Solution:
VA- VB = -∫BAE→.dr→
VA- VB = -∫2rr2λ4π ε0rdr
VA- VB =2λ4π ε0r ∫r2r1rdr
=2λ4π ε0 ln 2rr
=2λ4π ε0ln 2

Topic: Physics, Electrostatics, JEE Main
What will be the pressure of the gaseous mixture when \(0.5 \mathrm{~L}\) of \(\mathrm{H}_2\) at \(0.8\) bar and \(2.0 \mathrm{~L}\) of dioxygen at \(0.7\) bar are introduced in a \(1 \mathrm{~L}\) vessel at \(27^{\circ} \mathrm{C}\) ?

Options:

None

Solution:
Partial pressure of \(\mathrm{H}_2\) in \(1 \mathrm{~L}\) vessel
\(\begin{aligned}
&\mathrm{P}_1=0.8 \mathrm{bar}, \mathrm{V}_1=0.5 \mathrm{~L} \\
&\mathrm{P}_2=? \mathrm{~V}_2=1.0 \mathrm{~L}
\end{aligned}\)
Temperature remaining constant,
Partial pressure of \(\mathrm{O}_2\) in \(1 \mathrm{~L}\) vessel
\(\mathrm{P}_1^{\prime}=0.7 \mathrm{bar}, \mathrm{V}_1^{\prime}=2.0 \mathrm{~L}, \mathrm{P}_2^{\prime}=? \mathrm{~V}_2^{\prime}=1.0 \mathrm{~L}\)
Temperature remaining constant
or, \(\mathrm{P}_2^1=1.4\) bar or, \(\mathrm{P}_{\mathrm{O}_2}=1.4\) bar
Total pressure \(=\mathrm{P}_{\mathrm{H}_2}+\mathrm{P}_{\mathrm{O}_2}=0.4 \mathrm{bar}+1.4 \mathrm{bar}=1.8 \mathrm{bar}\)

Topic: Chemistry, States of Matter
What will be the primary valency of cobalt in the complex \( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \), according to Werner's theory?

Options:
A \( 5 \)
B \( 1 \)
C \( 3 \)
D \( 2 \)

The correct answer is:
\( 3 \)

Solution:
Negative ions satisfy the primary valencies in the given complex. Primary valency represent the oxidation state of central metal ion and in this oxidation state is +3, so primary valency is 3. Out of three chloride ion, two chloride ions are of dual nature i.e. shows both primary and secondary whereas one chloride ion shows only primary.

Topic: Chemistry, Coordination Compounds, JEE Main
What will be the product for the given reaction?
$\mathrm{CH}_3 \mathrm{CONH}_2 \xrightarrow{\mathrm{NaOBr}}$ ?

Options:
A $\mathrm{CH}_3 \mathrm{Br}$
B $\mathrm{CH}_4$
C $\mathrm{CH}_3 \mathrm{COBr}$
D $\mathrm{CH}_3 \mathrm{NH}_2$

The correct answer is:
$\mathrm{CH}_3 \mathrm{NH}_2$

Solution:
Explanation: Amide on reacting with $\mathrm{NaOBr}$ will give an amine as shown below.
$\mathrm{CH}_3 \mathrm{CONH}_2 \xrightarrow{\mathrm{NaOBr}} \mathrm{CH}_3 \mathrm{NH}_2$

Topic: Chemistry, Amines, JEE Main
What will be the projection of vector A→=i^+j^+k^ on vector B→=i^+j^ ?

Options:
A 2(i^+j^+k^)
B 2i^+j^+k^
C 2(i^+j^)
D (i^+j^)

The correct answer is:
(i^+j^)

Solution:
AcosθB^=AA→·B→ABB^=A→·B→BB^
=22i^+j^2=i^+j^

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
What will be the radius of the octahedral void?

Options:
A \( 0.414 \) of radius of atom
B \( 0.732 \) times of radius of atoms
C \( 0.324 \) of radius of atom
D \( 0.224 \) times of radius of atom

The correct answer is:
\( 0.414 \) of radius of atom

Solution:
In th figure a sphere fitted in an octahedral void is shown by shaded part

Since ∆ABC is right angle triangle.

AC2=AB2+CB2(2R)2=(R+r)2+(R+r)24R2=2(R+r)22R2=(R+r)22R2=(R+r)22R=(R+r)r=2R-Rr =0.414 R

Here R is the radius of atom.

Topic: Chemistry, Solid State, JEE Main
What will be the ratio of acceleration due to gravity on the surface of the earth to the acceleration due to gravity at a height 3R above the earth's surface? (R = radius of earth)

Options:

The correct answer is:
16

Solution:
The value of acceleration due to gravity changes with height (ie, altitude). If g′ is the acceleration due to gravity at a point, at height h above the surface of earth, then
g′=GMR+h2
but, g=GMR2
∴ g′g=GMR+h2×R2GM=R2R+h2
Here, g′=GMR+h2=GMR+3R2
=GM4R2=GM16R2=ge16

⇒ geg'=16

Topic: Physics, Gravitation, JEE Main
What will be the relation between the T1 of gas 1 with M1=56 and T2 of gas 2 with M2=44 if the average speed of gas 1 is equal to most probable speed of gas 2?

Options:
A T1=T22
B T1=T2
C T1=(T2)12
D T1=1T2

The correct answer is:
T1=T2

Solution:
cav=8RTπM;cmp=2RTM
Given that, 8RT1πM1=2RT2M2
8T1πM1=2T2M2
4T1π×56=T244
T1T2=56×π44×4=1
T1=T2

Topic: Chemistry, States of Matter, JEE Main
What will be the resistance of the shunt when $5 \%$ of the main current is passed through a galvanometer of resistance G?

Options:
A $\frac{G}{20}$
B $\frac{G}{21}$
C $\frac{G}{5}$
D $\frac{G}{19}$

The correct answer is:
$\frac{G}{19}$

Solution:
Let the total passing be $\mathrm{I}$.
Thus current passing through galvanometer $\mathrm{I}_{\mathrm{G}}=0.05 \mathrm{I}$
Also, current passing through shunt resistance $\mathrm{I}_{\mathrm{S}}=\mathrm{I}-0.05 \mathrm{I}=0.95 \mathrm{I}$
Using $I_{S} S=I_{G} G$
$\therefore \frac{S}{G}=\frac{0.05}{0.95}=\frac{1}{10}$
$\Longrightarrow S=\frac{G}{19}$

Topic: Physics, Current Electricity, MHT CET, MHT CET 2020 (20 Oct Shift 2)
What will be the stress at -20 Â°C , if a steel rod with a cross-sectional area of 150 mm2 is stretched between two fixed points? The tensile load at 20 Â°C is 5000 N. (Assume Î±=11.7 Ã10-6 Â°C-1 and Y=200 Ã1011N m-2 )

Options:
A 12.7 ×106 N m-2
B 84.2 Ã106 N m-2
C 127 Ã106 N m-2
D 0.842 Ã106 N m-2

The correct answer is:
127 Ã106 N m-2

Solution:
Let L = free length at 0 °C
L0= final stretched length in each case.
L1 and L2 are free length at +20 °C and - 20 °C respectively.
We know s2=s1+2αLΔT
Here, s1 and s2 are load deformation.
F2LAY=F1LAY+2αLΔT
σ2=F2A=F1A+2αΔTY=5000150 ×10-6+2 11.7 ×10-6 20 2 ×1011
=127 ×106 N m-2

Topic: Physics, Thermal Properties of Matter, JEE Main
What will be the sum of $\mathrm{pH}$ and $\mathrm{pOH}$ in an aqueous solution?

Options:
A 7
B $p k_w$
C Zero
D 1

The correct answer is:
$p k_w$

Solution:
$p H+p O H=p K_w$

Topic: Chemistry, Ionic Equilibrium, JEE Main
What will be the sum of principal and azimuthal quantum number for last electron of Argon?

Options:

The correct answer is:
4

Solution:
Electronic configuration of
Ar : 1s2,2s22p6,3s23p6
Therefore, quantum numbers for last electron are n=3ââ&âl=1, so n+l=4

Topic: Chemistry, Structure of Atom, JEE Main
What will be the time period of the pendulum of length l suspended from the ceiling of a cart which is sliding without friction on an inclined plane of inclination Î¸ ?

Options:
A 2 π l⁡ g cos θ
B 2 Ï 3 lâ¡ g cos Î¸
C 4 π 2 l⁡ g cos θ
D 3 Ï 4 lâ¡ 2g cos Î¸

The correct answer is:
2 π l⁡ g cos θ

Solution:
Here, point of suspension has an acceleration. a → = g sin θ (down the plane). Further, g → can be resolved into two components g sin θ (along the plane) and g cos θ (perpendicular to plane).
g → eff = g → - a →
= g cosθ (perpendicular to plane)
T = 2 π l⁡ | g → eff |
= 2 π l⁡ g cosθ

Topic: Physics, Oscillations, JEE Main
What will be the value of x, y and z in the following equation?
H2C2O4+xH2O2→yCO2+zH2O

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

The correct answer is:
1, 2, 2

Solution:
(i) The half reaction for oxidation is,
H2C2O4âCO2
Balancing carbon atoms on both sides, H2C2O4â2CO2
Balancing hydrogen atoms on both sides, H2C2O4â2CO2+2H+
Balancing the charge on both sides,
H2C2O4â2CO2+2H++2e- ...(i)
(ii) The half reaction for reduction isâ H 2 O 2 → H 2 O
Balancing oxygen atoms on both sides, H2O2â2H2O
Balancing hydrogen atoms, H2O2+2H+â2H2O
Balancing the charge,
H2O2+2H++2e-â2H2O ...(ii)
Now, adding both equations, H2C2O4+H2O2â2CO2+2H2O
This is balanced equation so x, y, z are 1,2,2 respectively.

Topic: Chemistry, Redox Reactions, JEE Main
What will be the volume of O2 Liberated at NTP by passing 5 A current For 193 Sec. through acidified water.

Options:
A 56 mL
B 112 mL
C 158 mL
D 965 mL

The correct answer is:
56 mL

Solution:
Wâ=â iâÃâtâÃâE F
= 5 Ã 1 9 3 Ã 8 9 6 5 0 0
âµ At NTP, volume of 32 g O2 = 22400 mL
â´ Volume of 0.08 g O2 = 2 2 4 0 0 Ã 0 . 0 8 3 2
= 56 mL

Topic: Chemistry, Electrochemistry, JEE Main
What will happen if decomposers are completely removed from the ecosystem?

Options:
A There will be no energy flow.
B There will be an increase in the number of detritivores
C Herbivores will not receive solar energy
D The movement of minerals in an ecosystem will be blocked.

The correct answer is:
The movement of minerals in an ecosystem will be blocked.

Solution:
Thee organic content of the dead and decaying organisms are broken down to inorganic substances by the process called decomposition. It is brought about by decomposers. Decomposition is mainly an oxygen dependent process and its rate is dependent on chemical composition and climatic factors. Decomposers play a vital role in energy flow through the ecosystem and also in making available the essential nutrients to the producers. If decomposers are removed from the ecosystem, then minerals required by the producers will not be available and disrupt the food chain.

Topic: Biology, Ecosystem, NEET
What will happen if fresh water lake gets contaminated by addition of polluted water with high BOD?

Options:
A Number of submerged aquatic plants in the lake will increase
B Number of aquatic animals in the lake will increase
C Amount of dissolved oxygen in the lake will decrease
D The lake will remain unaffected

The correct answer is:
Amount of dissolved oxygen in the lake will decrease

Solution:
Higher BOD of water indicates higher amount of biodegradable organic materials present in it. Since BOD is inversely proportional to the dissolved oxygen (DO), therefore, the dissolved oxygen in the lake will decrease.

Topic: Biology, Environmental Issues, NEET, NEET 2023 (Manipur)
What will happen if $\mathrm{LiAlH}_4$ is added to an ester

Options:
A Two units of alcohol are obtained
B One unit of alcohol and one unit of acid is obtained
C Two units of acids are obtained
D None of these

The correct answer is:
Two units of alcohol are obtained

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
What will happen if the secretion of parietal cells of gastric glands is blocked with an inhibitor?

Options:
A Enterokinase will not be released from the duodenal mucosa and so trypsinogen is not converted to trypsin
B Gastric juice will be deficient in chymosin
C Gastric juice will be deficient in pepsinogen
D In the absence of $\mathrm{HCl}$ secretion, inactive pepsinogen is not converted into the active enzyme pepsin.

The correct answer is:
In the absence of $\mathrm{HCl}$ secretion, inactive pepsinogen is not converted into the active enzyme pepsin.

Solution:
Parietal cells/oxyntic cells, their function is to secrete $\mathrm{HCl}$ and Castle's intrinsic factor. Pepsinogen, is converted into active pepsin by $\mathrm{HCl}$, i.e. by change in $\mathrm{pH}$.

Topic: Biology, Digestion and Absorption, NEET, NEET 2008 (Mains)
What will happen if the secretion of parietal cells of gastric glands is blocked with an inhibitor?

Options:
A Gastric juice will be deficient in chymosin
B Gastric juice will be deficient in pepsinogen
C In the absence of $\mathrm{HCl}$ secretion, inactive pepsinogen is not converted into the active enzyme pepsin
D Enterokinase will not be released from the duodenal mucosa and so trypsinogen is not converted to trypsin

The correct answer is:
In the absence of $\mathrm{HCl}$ secretion, inactive pepsinogen is not converted into the active enzyme pepsin

Solution:
Gastric glands are numerous microscopic tubular glands formed by the epithelium of stomach. The parietal cells (oxyntic cells) are large and most numerous on the side walls of gastric glands. They secrete hydrochloric acid and castle intrinsic factor. The peptic cells (zymogen) cells of gastric glands secrete gastric digestive enzymes as proenzymes-pepsinogen and prorennin and small amount of gastric amylase and gastric lipase. The hydrochloric acid maintains a strongly acidic $\mathrm{pH}$ of about 1.5 to 2.5 in the stomach. HCl converts pepsinogen and prorennin to pepsin and rennin respectively. The secretion of intestinal glands is called succus entericus. It contains many enzymes, maltase, lactase, sucrase, enterokinase, nucleotidase etc. The enterokinase converts proenzyme trypsinogen into active trypsin.

Topic: Biology, Digestion and Absorption, NEET, NEET 2008 (Screening)
What will happen if the stretch receptors of the urinary bladder wall are totally removed ?

Options:
A There will be no micturition
B Urine will continue to collect normally in bladder
C Micturition will continue
D Urine will not collect in the bladder

The correct answer is:
There will be no micturition

Solution:
No Solution Available

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2009 (Mains)
What will happen to the capacitance of a capacitor that has a dielectric between its plates when the capacitor, initially connected to a DC source, has its battery disconnected and then the dielectric is removed?

Options:
A Increase
B Decrease
C Remain constant
D Insufficient information to determine

The correct answer is:
Decrease

Solution:
Correct Option is : (B)
Decrease

Topic: Chemistry, Capacitance, NEET
What will happen to the concentrations of reactants of a reactionwhen Qc is less than Kc?

Options:
A They will remain constant.
B They will increase.
C They will decrease.
D They will become zero.

The correct answer is:
They will decrease.

Solution:
When Qc is not equal to Kc, the concentrations of reactants will decrease until Qc equals Kc.

Topic: Chemistry, Chemical Equilibrium, NEET
What will happen to the current in a solenoid connected to a battery if an iron core is inserted into it?

Options:
A Increase
B Decrease
C Remain unchanged
D Insufficient information to determine

The correct answer is:
Decrease

Solution:
Correct Option is : (B)
Decrease

Topic: Chemistry, Electromagnetic Induction, NEET
What will happen when a block of copper metal is dropped into a beaker containing a solution of 1 M ZnSO4?

Options:
A The copper metal will dissolve with evolution of oxygen gas.
B The copper metal will dissolve with evolution of hydrogen gas.
C No reaction will occur.
D The copper metal will dissolve and zinc metal will be deposited.

The correct answer is:
No reaction will occur.

Solution:
Zn is above Cu in electrochemical series. So, Zn will reduce Cu2+ to Cu.
Cu + Zn2+ → No reaction
Displacement reaction is feasible in opposite direction only.
Cu2+ + Zn → Zn2+ + Cu ↓

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2016 (09 Apr Online)
What will happen when a block of copper metal is dropped into a beaker containing a solution of 1 M ZnSO4?

Options:
A The copper metal will dissolve with evolution of oxygen gas.
B The copper metal will dissolve with evolution of hydrogen gas.
C No reaction will occur.
D The copper metal will dissolve and zinc metal will be deposited.

The correct answer is:
No reaction will occur.

Solution:
Zn is above Cu in electrochemical series. So, Zn will reduce Cu2+ to Cu.
Cu + Zn2+ → No reaction
Displacement reaction is feasible in opposite direction only.
Cu2+ + Zn → Zn2+ + Cu ↓

Topic: Chemistry, Electrochemistry, JEE Main