datavorous/entrance-exam-dataset huggingface dataset:
Movement and accumulation of ions across a membrane against their concentration gradient can be explained by:

Options:
A Active transport
B Osmosis
C Facilitated diffusion
D Passive transport

The correct answer is:
Active transport

Solution:
One of the main way of transport of solute or molecules in plant cells is called active transport. Active transport uses energy to transport and pump molecules against a concentration gradient. This is carried out by specific membrane-bound proteins or pumps. These protein pumps use energy (derived from ATP) to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration (‘uphill’ transport).

Topic: Biology, Transport in Plants, NEET, NEET 2023 (All India)
Movement of substances in xylem is unidirectional while in phloem it is bidirectional. Explain

Options:

None

Solution:
Xylem is involved in the one way transport of water and minerals from soil to root' $\rightarrow$ stem $\rightarrow$ leaves. Several forces like imbibition, root pressure and finally transpiration pull. Act in this mechanism, It is a undirectional process as there is continuous loss of water at the body surface of plants. The main function of Phloem is to transport food from source to sink where source is the part of plant responsible for food synthesis and sink are the organs requiring food for their growth and development. These source and sink parts of a plants may vary in different phases of growth, thus the food needs to travel in both upwards and downward direction. So, phloem shows bidirectional movement of substances.

Topic: Biology, Transport in Plants
Mowing grass lawn facilitates better maintenance because:

Options:
A Wounding stimulates regeneration
B Removal of apical dominance and stimulation of intercalary meristem
C Removal of apical dominance
D Removal of apical dominance and promotion of lateral meristem

The correct answer is:
Removal of apical dominance and stimulation of intercalary meristem

Solution:
Apical dominance is a common observation in vascular plants. If apical meristem is intact & growing, the lateral tips growth is suppressed. According to Thirnann & Skoog the Auxin, synthesized in apical meristem, in translocated downwards & inhibits the growth of lateral tips. That is why mowing -grass lawns facilitates better maintenance.

Topic: Biology, Plant - Growth and Development, NEET
Mr. Vipin, a famous liar, is known to speak the truth 5 out of 6 times. His blind folded friend Shubham throws a pair of dice and asked Vipin the result, who says the sum of numbers on the pair of dice is 9. The probability that the sum of numbers on the pair of dice is actually 9 is k, then the value of 52k is equal to

Options:

The correct answer is:
20

Solution:
Let, A be an event that Vipin speaks the truth
âPA=56
and B be an event such that the sum on the pair of dice is equal to 9
âPB=436=19
PBA=PBâ©APA=PBâPABPBPAB+PBCPAcBc
=19Ã5619Ã56+89Ã16=513

Topic: Mathematics, Probability, JEE Main
Mucus secreted by the goblet cells, combined with bicarbonates maintains the alkaline medium of the intestine. The source of bicarbonates is:

Options:
A Intestinal mucosa
B Gall bladdar
C Pancreas
D Brunner's glands

The correct answer is:
Pancreas

Solution:
The intestinal mucosal epithelium has goblet cells which secrete mucus. The secretions of the brush border cells of the mucosa along with the secretions of the goblet cells constitute the intestinal juice or succus entericus. This juice contains a variety of enzymes like disaccharidases (e.g., maltase), dipeptidases, lipases, nucleosidases, etc. The mucus along with the bicarbonates from the pancreas protects the intestinal mucosa from acid as well as provide an alkaline medium (pH 7.8) for enzymatic activities.

Topic: Biology, Chemical Coordination and Integration, NEET
Multicellular organisms have division of labour. Explain.

Options:

None

Solution:
In unicellular organisms, there is no division of labour. The single cell of the organisms is capable of performing all the vital activities of life i.e., respiration, movement, digestion and reproduction, etc. Respiration, nutrition and excretion in most of these unicellular organisms takes place through general body surface. No special organs for these are present in them because they are too small to need them. Most of these unicellular organisms reproduce by simple binary division, to maintain their continuity. However in some, sexual reproduction has also been observed.

Topic: Biology, Cell - The Unit of Life
Muon $\left(\mu^{-1}\right)$ is negatively charged $(|\mathrm{q}|=|\mathrm{e}|)$ with a mass $\mathrm{m}_\mu=200 \mathrm{~m}_{\mathrm{e}}$, where $\mathrm{m}_{\mathrm{e}}$ is the mass of the electron and e is the electronic charge. If $\mu^{-1}$ is bound to a proton to form a hydrogen like atom, identify the correct statements
(A) Radius of the muonic orbit is 200 times smaller than that of the electron
(B) the speed of the $\mu^{-1}$ in the $n$th orbit is $\frac{1}{200}$ times that of the election in the nth orbit
(C) The lonization energy of muonic atom is 200 times more than that of an hydrogen atom
(D) The momentum of the muon in the nth orbit is 200 times more than that of the electron

Options:
A
(A), (B), (D)

B
(B), (D)

C
(C),(D)

D
(A), (C), (D)

The correct answer is:
(A), (C), (D)

Solution:
$$
\begin{aligned}
&\text { (A) Radius of muon }=\frac{\text { Radius of hydrogen }}{200} \\
&\text { Radius of } \mathrm{H} \text { atom }=r=\frac{\in_0 n^2 h^2}{\pi m e^2} \\
&\text { Radius of muon }=r_\mu=\frac{\in_0 n^2 h^2}{\pi \times 200 m e^2} \\
&r_\mu=\frac{r}{200}
\end{aligned}
$$
(B) Velocity relation given is wrong
(C) Ionization energy in $e^{-} \mathrm{H}$ atom
$$
E=\frac{+m e^4}{8 \epsilon_0^2 n^2 h^2}
$$
$$
E_\mu=\frac{200 m e^4}{8 \epsilon_0^2 n^2 h^2}=200 E
$$
(D) Momentum of $\mathrm{H}$-atom
$$
m v r=\frac{n h}{2 \pi}
$$
Hence (A), (C), (D) are correct.

Topic: Physics, Atomic Physics, JEE Main
Muon $\left(\mu^{-1}\right)$ is negatively charged $(|\mathrm{q}|=|\mathrm{e}|)$ with a mass $\mathrm{m}_\mu=200 \mathrm{~m}_{\mathrm{e}}$, where $\mathrm{m}_{\mathrm{e}}$ is the mass of the electron and e is the electronic charge. If $\mu^{-1}$ is bound to a proton to form a hydrogen like atom, identify the correct statements
(A) Radius of the muonic orbit is 200 times smaller than that of the electron
(B) the speed of the $\mu^{-1}$ in the $n$th orbit is $\frac{1}{200}$ times that of the election in the nth orbit
(C) The lonization energy of muonic atom is 200 times more than that of an hydrogen atom
(D) The momentum of the muon in the nth orbit is 200 times more than that of the electron

Options:
A
(A), (B), (D)

B
(B), (D)

C
(C),(D)

D
(A), (C), (D)

The correct answer is:
(A), (C), (D)

Solution:
$$
\begin{aligned}
&\text { (A) Radius of muon }=\frac{\text { Radius of hydrogen }}{200} \\
&\text { Radius of } \mathrm{H} \text { atom }=r=\frac{\in_0 n^2 h^2}{\pi m e^2} \\
&\text { Radius of muon }=r_\mu=\frac{\in_0 n^2 h^2}{\pi \times 200 m e^2} \\
&r_\mu=\frac{r}{200}
\end{aligned}
$$
(B) Velocity relation given is wrong
(C) Ionization energy in $e^{-} \mathrm{H}$ atom
$$
E=\frac{+m e^4}{8 \epsilon_0^2 n^2 h^2}
$$
$$
E_\mu=\frac{200 m e^4}{8 \epsilon_0^2 n^2 h^2}=200 E
$$
(D) Momentum of $\mathrm{H}$-atom
$$
m v r=\frac{n h}{2 \pi}
$$
Hence (A), (C), (D) are correct.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Mutual inductance in the figure shown is -

Options:
A Zero
B μ0b2πlogeab
C μ0b2πyloge1+ba
D Î¼0b2Ïlogeâ¡1+ab

The correct answer is:
μ0b2πloge⁡1+ab

Solution:
ϕ=MI=∫B.dA
MI=∫aa+bμ0I2πx×bdx
M=μ0b2π∫aa+bdxx
M=μ0b2πloge⁡a+ba
M=μ0b2πloge⁡1+ba

Topic: Physics, Electromagnetic Induction, JEE Main
Mycorrhizal association plays an important role in which of the following bio-geochemical cycle?

Options:
A Carbon cycle
B Sulphur cycle
C Nitrogen cycle
D Phosphorous cycle

The correct answer is:
Phosphorous cycle

Solution:
Mycorrhizae is a mutual relationship between fungi and plant. Both benefit from each other. In it, the fungus colonises the root system of the host plant either intra- or extra- cellularly. The fungal group is known to absorb phosphorus from soil and pass it to the host plant. Such associations also provide other benefits like resistance to root-borne pathogens, tolerance to salinity and drought.
The movement and exchange of organic and inorganic matter between living and non-living component in a cyclical manner refers to nutrient cycle. These are also called as biogeochemical cycles as there is involvement of biotic and abiotic factors and chemical reactions. There are many important cycles which help in maintaining the biosphere like carbon cycle, nitrogen cycle etc.

Topic: Biology, Microbes in Human Welfare, NEET
N0/4 atoms of X(g) are converted into X+(g) by energy E1. N0 /4 atoms of X(g) are converted into Xâ(g) by energy E2. Hence, ionisation potential and electron affinity of X(g) are :

Options:
A 4 E 1 N 0 , 4 E 1 - E 2 N 0
B 4 E 1 N 0 , 4 E 2 N 0
C E 1 - E 2 N 0 , 4 E 2 N 0
D None is correct

The correct answer is:
4 E 1 N 0 , 4 E 2 N 0

Solution:
X g â¶ X + g + e -
If I ionisationh energy then
â´ N 0 4 I = E 1
â´ I = 4 E 1 N 0
If E is electron-affinity then
X g + e - â¶ X - g
N 0 4 E = E 2
â´ E = 4 E 2 N 0

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
n1 capacitors, each of capacitance C1 are connected in parallel and are charged by a source of potential difference V and n2 capacitors, each of capacitance C2 are connected in series and are charged by a source of potential difference 4V. If both the combinations have the same total energy, then C1=mC2n1n2. What is the value of m?

Options:

The correct answer is:
16

Solution:
A parallel combination of n1,capacitors each of capacitance C1is, Cp=n1C1

Total energy stored in parallel combination,

Up=12Cp(V)2=12n1C1V2 ….(i)

A series combination of n2 capacitors each of capacitance C2are connected to 4V source.

Cs=C2n2

Total energy stored in series combination,

Us=12Cs(4V)2=12C2n216V2 ...ii

According to give12C2p16V2=12n1C1V2

n probelm,

Us=Up

From equations, i and ii

∴ C1=16C2n1n2

∴ m=16

Topic: Physics, Electrostatics, JEE Main
$(n-1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$-polygon. The vacant vertex has a position vector $a$ with respect to the centre of the polygon. Find the position vector of centre of mass.

Options:

None

Solution:
Let us consider $\boldsymbol{b}$ be the position vector of the centre of mass of a regular $n$-polygon.
$(n-1)$ equal point masses are placed at $(n-1)$ vertices of the regular $n$-polygon, then for its centre of mass $r_{c m}$ when $m$ is placed at $n$th vertex.
$$
r_{C M}=\frac{(n-1) m b+m a}{(n-1) m+m}
$$
If mass $m$ is placed at $n$th remaining vertex or mass lies at centre then
$$
\begin{aligned}
&r_{C M}=0 \\
&\frac{(n-1) m b+m a}{(n-1) m+m}=0 \\
&(n-1) m b+m a=0 \\
&\vec{b}=-\frac{\vec{a}}{(n-1)}
\end{aligned}
$$
(-) sign shows that $C M$ lies other side from $n$th vertex geometrical centre of $n$-polygon i.e., $\vec{b}$ is opposite to the vector $\vec{a}$ from centre to $n$th vertex.

Topic: Physics, System of Particles and Rotational Motion
N2+3H2â¶2NH3

1 mole N2 and 4 moles H2 are taken in 15L flask at 27oC. After complete conversion of N2 into NH3, 5L of H2O is added. Pressure set up in the flask is (assume water dissolves NH3)

Options:
A 3Ã0.0821Ã30015 atm
B 2Ã0.0821Ã30010 atm
C 1Ã0.0821Ã30015 atm
D 1Ã0.0821Ã30010 atm

The correct answer is:
1Ã0.0821Ã30010 atm

Solution:
N 2 + 3 H 2 â¶ 2 NH 3 Initial mole 1 4 0 Reacted mole 1 3 Left mole 0 4-3=1 2
H2 left = 1 mol
NH3 formed = 2 mol
But NH3 g dissolves in water forming NH4OH . Residual gas is only H2 which exerts pressure.
Volume =15-5=10 L (5L occupied by H2O added)
â´p=nVRT=1Ã0.0821Ã30010 atm

Topic: Chemistry, States of Matter, JEE Main
N2 and O2 are converted to mono cations N2+ and O2+ respectively. The wrong statement with respect to given changes in the following is

Options:
A the N – N bond weakens
B the O – O bond order increases
C the O-O bond length decreases
D N2+ becomes diamagnetic

The correct answer is:
N2+ becomes diamagnetic

Solution:
B.O or N2=3 and N2+=2.5

As here bond order decreases so N-N bond get weak N2+ becomes paramagnetic
In case of O2 and O2+
B.O or O2=2 and O2+=2.5
as bond order increases so O-O bond length decreases.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
N2 and O2 are converted to monopositive cations N2+ and O2+ respectively. Which set is having correct statements?
1. In O2+ , paramagnetism decreases
2. N2+ becomes diamagnetic
3. In O2+ , the bond order increases
4. In N2+ the N-N bond is weakened.

Options:
A 3, 4
B 1, 4
C 1, 3, 4
D 1, 2, 3, 4

The correct answer is:
1, 3, 4

Solution:
M.O. configuration of N2 is
Ï1s2Ï*1s2Ï2s2Ï*2s2Ï2px2Ï2py2Ï2pz1
B.O. of N2=10-42=3
M.O. configuration of N2+ is
Ï1s2Ï*1s2Ï2s2Ï*2s2Ï2px2Ï2py2Ï2pz1
B.O. of N2+=9-42=2.5
Hence, N2+ becomes paramagnetic so statement 2 is incorrect
M.O. of configuration of O2 is
Ï1s2Ï*1s22Ï2s2Ï*2s2Ï2pz2Ï2px2Ï2py2Ï*2px1Ï*2py1
So B.O. of O2=10â62=2
M.O. configuration of O2+ is
Ï1s2Ï*1s2Ï2s2Ï*2s2Ï2pz2Ï2px2Ï2py2Ï*2px1
B.O. of O2+=10-52=2.5
B.O.∝Bond energy ∝ 1 Bond length

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
N2g+3H2g⇌2NH3g20 g 5 g

Consider the above reaction. If $20$g of dinitrogen reacts with $5$g of dihydrogen, then the limiting reagent of the reaction and number of moles of NH3 formed respectively are

Options:
A H2,1.42 moles
B H2,0.71 moles
C N2,1.42 moles
D N2,0.71 moles

The correct answer is:
N2,1.42 moles

Solution:
N2g+3H2g⇌2NH3g20 g 5 g Given mass2028 52 Number of moles

To find the limiting reagent we can divide given number of moles with Stoichiometric coefficients, whichever will be the lower will be limiting reagent.

N2→20281=2028 H2→523=56

∴ N2 is the Limiting Reagent.

According to the reaction one mole of nitrogen gas gives two moles of ammonia.

∴ nNH3=2×nN2=2×2028

Number of moles of ammonia formed will be =1.42

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (29 Jul Shift 1)
$\mathrm{N}_2$ is found in a litre flask under $100 \mathrm{kPa}$ pressure and $\mathrm{O}_2$ is found in another 3 litre flask under $320 \mathrm{kPa}$ pressure. If the two flasks are connected, the resultant pressures is

Options:
A $310 \mathrm{kPa}$
B $210 \mathrm{kPa}$
C $420 \mathrm{kPa}$
D $265 \mathrm{kPa}$

The correct answer is:
$265 \mathrm{kPa}$

Solution:
Total volume of two flasks $=1+3=4$
If $P_1$ the pressure of gas $N_2$ in the mixture of $N_2$ and $O_2$ then
$P=100 \mathrm{kPQ}$,
$P_1=$ $V=1$ litre, $V_1=4$ litre
applying Boyle's law $P V=P_1 V_1$
$100 \times 1=P_1 \times 4$
$P_1=25$
If $\mathrm{P}_2$ is the pressure of $\mathrm{O}_2$ gos in the mixture of $\mathrm{O}_2$ and $N_2$ then, $320 \times 3=P_2 \times 4 ; P_2=240$
Hence, Total pressure $P=P_1+P_2=25+240$
$=265 \mathrm{kPa}$

Topic: Chemistry, States of Matter, JEE Main
$\mathrm{N}_2 \mathrm{O}_5$ decomposes to $\mathrm{NO}_2$ and $\mathrm{O}_2$ and follows first order kinetics. After $50$ minutes, the pressure inside the vessel increases from $50 \mathrm{~mm} \mathrm{~Hg}$ to $87.5 \mathrm{~mm} \mathrm{~Hg}$. The pressure of the gaseous mixture after $100$ minutes at constant temperature will be ______.

Options:
A
$136.25 \mathrm{~mm} \mathrm{~Hg}$

B
$106.25 \mathrm{~mm} \mathrm{~Hg}$

C
$175.0 \mathrm{~mm~Hg}$

D
$116.25 \mathrm{~mm} \mathrm{~Hg}$

The correct answer is:
$106.25 \mathrm{~mm} \mathrm{~Hg}$

Solution:
$\mathrm{N}_2 \mathrm{O}_5 \rightarrow 2 \mathrm{NO}_2+\frac{1}{2} \mathrm{O}_2$
$\begin{array}{llll}\text { At } \mathrm{t}=0 & 50 & 0 & 0 \\ \text { At } \mathrm{t}=50 \mathrm{~min} & 50-\mathrm{p}_1 & 2 \mathrm{p}_1 & \frac{\mathrm{p}_1}{2}\end{array}$
Total pressure at $50$ minutes
$\begin{aligned}
&=50-\mathrm{p}_1+2 \mathrm{p}_1+\frac{\mathrm{p}_1}{2}=87.5 \\
&50+\frac{3 \mathrm{p}_1}{2}=87.5 \\
&\frac{3 \mathrm{p}_1}{2}=37.5 \\
&\therefore \quad \mathrm{p}_1=\frac{37.5 \times 2}{3}=25
\end{aligned}$
At $\mathrm{t}=100 \mathrm{~min} \quad 50-\mathrm{p}_2 \quad 2 \mathrm{p}_2 \quad \frac{\mathrm{p}_2}{2}$
$50$ minutes is half life period
For $100$ minutes i.e. for $2$ half lives $50-p_2=12.5$
$\therefore \mathrm{p}_2=37.5 \mathrm{~mm} \text { of } \mathrm{Hg}$
Total pressure at $100$ minutes
$=50-\mathrm{p}_2+2 \mathrm{p}_2+\frac{\mathrm{p}_2}{2}$
$=50+\frac{3 \mathrm{p}_2}{2}=50+\frac{3}{2} \times 37.5$
$=50+56.25$
$=106.25 \mathrm{~mm}$ of $\mathrm{Hg}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
N2O5 decomposes to NO2 and O2 follows the first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 minutes at constant temperature will be:

Options:
A 136.5 mm Hg
B 106.25 mm Hg
C 175.0 mm Hg
D 116.25 mm Hg

The correct answer is:
106.25 mm Hg

Solution:
N2O5→2NO2+12O2t=0 50 0 0t=50mins  50-p1     2p1     p12t=100mins   50-p2    2p2      p22
50-p1+2p1+p12=87.5
50+3p12=87.5
3p12=37.5
p1=37.5×23=25
50 minutes is half-life period.
For 100 minutes ( half-life 2)
50-p2=12.5
p2=37.5 mm Hg
Total pressure at 100 minutes.
=50-p2+2 p2+p22
50+3p22=50+32×37.5
=50+56.25
=106.25 mm of Hg.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2018 (15 Apr)
N2O5( g)→2NO2( g)+12O2( g)
In the above first order reaction the initial concentration of N2O5 is 2.40×10-2 mol L-1 at 318 K. The concentration of N2O5 after 1 hour was 1.60×10-2 mol L-1. The rate constant of the reaction at 318 K is ×10-3 min-1 (Nearest integer):
[ Given :log3=0.477,log5=0.699]

Options:

The correct answer is:
7

Solution:
K=2.303tlogN2O50 N2O51
=2.30360log2.41.6=6.76×10-3 min-1≈7×10-3 min-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
NaBH4+I2→X+Y+Z
BF 3  +  LiAlH 4 → 450 K X + P
BF3+LiAlH4→X+Q+R

X, Y, Z, P, Q and R in the reactions are

Options:
A

X	Y	Z	P	Q	R

Na4B4O7	NaI	HI	HF	AlF 3	LiF

B

X	Y	Z	P	Q	R

B2H6	NaI	H2	NaF	LiF	AlF 3

C

X	Y	Z	P	Q	R

B2H6	BH3	NaI	B3N3H6	Al2F6	AlF3

D

X	Y	Z	P	Q	R

BH3	B2H6	H2	B3N3H6	LiF	AlF3

The correct answer is:

X	Y	Z	P	Q	R

B2H6	NaI	H2	NaF	LiF	AlF 3

Solution:
2NaBH4+I2→B2H6(X)+2NaI(Y)+H2(Z)
2BF 3  + 6NaH → 450 K B 2 H 6 X +  6NaF P
4BF3+3LiAlH4→2B2H6(X)+3LiF(Q)+3AlF3(R)

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
NaNO3→ Δ Agas+B

A+HCl→ CuCl2 Cgas+H2O

Colour of C and A are respectively.

Options:
A Colourless, reddish brown
B Greenish yellow, Colourless
C Yellow, reddish brown
D Colourless, Greenish yellow

The correct answer is:
Greenish yellow, Colourless

Solution:
NaNO3→ Δ NaNO2+O2colorless
O2+HCl→ CuCl2 Cl2green is yellow+H2O

Topic: Chemistry, s Block Elements, JEE Main
NaOCl is used as a bleaching agent and sterilising agent. It can be synthesised by the action of

Options:
A NaCl with H2O
B NH4Cl with NaOH
C Cl2 with cold and dilute NaOH
D Cl2 with hot and concentrated NaOH

The correct answer is:
Cl2 with cold and dilute NaOH

Solution:
NaOCl is used as a bleaching agent and sterilising agent. It is formed by the action of Cl2 with cold and dilute NaOH.
2NaOH+Cl2 â¶NaCl+NaOCl+H2O
Cold and dil.
6NaOH+3Cl2 â¶5NaCl+NaClO3+3 H2O
Hot and conc.

Topic: Chemistry, s Block Elements, JEE Main
n-digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed is :

Options:
A 9
B 7
C 8
D 6

The correct answer is:
7

Solution:
n- digit number are formed using only three digits 2, 5 and 7 with repetition =3n.

Since, 36<900 and 37>900

So, n=7.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2018 (15 Apr)
$n$ - digit numbers are formed using only three digits 2,5 and 7 . The smallest value of $n$ for which 900 such distinct numbers can be formed, is

Options:
A
6

B
8

C
9

D
7

The correct answer is:
7

Solution:
Required $n$ digit numbers is $3^n$ as each place can be filled by $2,5,7$.
So smallest value of $n$ such that $3^n>900$. Therefore $n=7$.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
N divisions on the main scale of a vernier calipers coincide with N+1 divisions of the vernier scale. If each division of main scale is a units, then the least count of the calipers is

Options:
A a N
B a
C aN+1
D NaN+1

The correct answer is:
aN+1

Solution:
It is given that N+1 Divisions on vernier scaleV.S.D.=N Divisions on main scale M.S.D.
⇒ 1 V.S.D.=NN+1M.S.D.

Least count of vernier callipers L.C.=1 M.S.D.-1 V.S.D.

⇒ L.C.=a-aNN+1=aN+1 unit

Topic: Physics, Experimental Physics, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
$N$ divisions on the main scale of a vernier calliper coincide with $(N+1)$ divisions of the vernier scale. If each division of main scale is ' $a$ ' units, then the least count of the instrument is

Options:
A
$a$

B
$\frac{a}{N}$

C
$\frac{N}{N+1} \times a$

D
$\frac{a}{N+1}$

The correct answer is:
$\frac{a}{N+1}$

Solution:
No of divisions on main scale $=N$
No of divisions on vernier scale $=N+1$ size of main scale division $=a$
Let size of vernier scale division be $b$ then we have
$$
a N=b(N+1) \Rightarrow \mathrm{b}=\frac{a N}{N+1}
$$
Least count is $\mathrm{a}-\mathrm{b}=\mathrm{a}-\frac{a N}{N+1}$ $=a\left[\frac{N+1-N}{N+1}\right]=\frac{a}{N+1}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2012 (19 May Online)
$n g$ of substance $X$ reacts with $m g$ of substance $Y$ to form $p g$ of substance $R$ and $q g$ of substance $S$. This reaction can be represented as, $X+Y=R+S$. The relation which can be established in the amounts of the reactants and the products will be

Options:
A $n-m=p-q$
B $n+m=p+q$
C $n=m$
D $p=q$

The correct answer is:
$n+m=p+q$

Solution:
$\begin{aligned} & n+m=p+q \\ & X_{n g}+Y_{m g} \rightleftharpoons R_{p g}+S_{\ell g}\end{aligned}$
$n+m=p+q$ by law of conservation of mass

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
NH4HS(s)⇌NH3(g)+H2S(g)
The equilibrium pressure at 25°C is 0.660 atm. What is Kp for the reaction?

Options:
A 0.109 (atm)2
B 0.218 (atm)2
C 1.89 (atm)2
D 2.18 (atm)2

The correct answer is:
0.109 (atm)2

Solution:
NH 4 HS s ⇌ NH 3 g P + H 2 S g P
Equilibrium pressure is 2P here
2P=0.660
P=0.330
Kp=PNH3.PH2S
=0.33Ã0.33=0.109(atm)2

Topic: Chemistry, Chemical Equilibrium, JEE Main
[NiCl4]2-,[PtCl4]2- and [PdCl4]2- are respectively :-

Options:
A high spin, low spin, high spin
B low spin, low spin, low spin
C high spin, low spin, low spin
D low spin, high spin, high spin

The correct answer is:
high spin, low spin, low spin

Solution:
[NiCl4]2-→ Ni is 3d series element
→Cl- weak ligand
→ High spin
[PdCl4]2-→ Pd is 4d series element
→ Low spin
[PtCl4]2-→ Pt is 5d series element
→ Cl − act as strong field ligand for Pt and Pd as they are 5d, 4d elements respectively with high effective nuclear charge.

Topic: Chemistry, Coordination Compounds, JEE Main
n identical cells are joined in series with two cells A and B with reversed polarities. EMF of each cell is E and internal resistance is r. The potential difference across the cell A or B is (n>4)

Options:
A 2 E n
B 2 E 1 - 1 n
C 4 E n
D 2 E 1 - 2 n

The correct answer is:
2 E 1 - 2 n

Solution:
Total number of cells = n
Total resistance = nr
n-2 cells provide a forward emf, 2 cells provide a reverse emf so:
Total emf =n-2E-2E=ne-4E=En-4
Current, I =total emftotal resistance
I=En-4nr
PD across A or B (voltage across internal resisitance)
V = Ir + E
= E n - 4 / nr r + E
= E n - 4 /n + E
= E n - 4 ) / n + 1
= E n - 4 + n / n
= E 2 n - 4 / n
= 2 E n - 2 / n
= 2 E 1 - 2 / n

Topic: Physics, Current Electricity, JEE Main
n identical cells each of e.m.f. $E$ and internal resistance $r$ are connected in series. An external resistance $R$ is connected in series to this combination. The current through $R$ is

Options:
A $\frac{n E}{R+n r}$
B $\frac{n E}{n R+r}$
C $\frac{E}{R+n r}$
D $\frac{n E}{R+r}$

The correct answer is:
$\frac{n E}{R+n r}$

Solution:
Current through $R$ is maximum when total internal resistance of the circuit is equal to external resistance.

Topic: Physics, Current Electricity, JEE Main