datavorous/entrance-exam-dataset huggingface dataset:
Five identical cells each of internal resistance 1 Ω and emf 5 V are connected in series and in parallel with an external resistance R. For what value of R, current in series and parallel combination will remain the same?

Options:
A 1 Ω
B 5 Ω
C 25 Ω
D 10 Ω

The correct answer is:
1 Ω

Solution:
i1=5εR+5r ...1

I2=εR+r5 ...2

From equation 1 and 2

i1=i2

5εR+5r=εR+r5⇒R=r=1Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Five identical conducting plates, each of face area $\mathrm{A}$ are placed parallel to each other with a separation of ' $d$ ' between two adjacent plates as shown in the figure. The plates $\mathrm{M}$ and $\mathrm{N}$ are given charges $Q_1$ and $Q_2$ respectively and the remaining plates are neutral. If the outermost plates are grounded, the potential difference between the plates $\mathrm{M}$ and $\mathrm{N}$ is ____

Options:
A $\frac{2 \mathrm{~d}\left(\mathrm{Q}_1-\mathrm{Q}_2\right)}{\epsilon_0 \mathrm{~A}}$
B $\frac{\mathrm{d}\left(\mathrm{Q}_1-\mathrm{Q}_2\right)}{2 \epsilon_0 \mathrm{~A}}$
C $\frac{4 \mathrm{~d}\left(\mathrm{Q}_1-\mathrm{Q}_2\right)}{\epsilon_0 \mathrm{~A}}$
D $\frac{\mathrm{d}\left(\mathrm{Q}_1-\mathrm{Q}_2\right)}{4 \epsilon_0 \mathrm{~A}}$

The correct answer is:
$\frac{\mathrm{d}\left(\mathrm{Q}_1-\mathrm{Q}_2\right)}{2 \epsilon_0 \mathrm{~A}}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 1)
Five identical plates of equal area A are placed parallel to each other and at equal distance d from each other as shown in the figure. The effective capacity of the system between the terminals A and B is

Options:
A 35ε0Ad
B 54ε0Ad
C 53Îµ0Ad
D 45ε0Ad

The correct answer is:
53ε0Ad

Solution:
Plates 1-2, 2-3, 3-4 and 4-5 form a total of four capacitors.
The capacity of each capacitor is, C0 = Îµ0Ad
From figure, it is clear that, Ceq = 53C0 = 53Îµ0Ad

Topic: Physics, Electrostatics, JEE Main
Five identical resistors each of resistance R=1500 Î© are connected to a 300 V battery as shown in the circuit. The reading of the ideal ammeter A is

Options:
A 15A
B 35A
C 25A
D 45A

The correct answer is:
35A

Solution:
The circuit can be reduced as follows
Reading of ammeter =I
=300R/3=900R
As. R=1500 Î©, I=35 A

Topic: Physics, Current Electricity, JEE Main
Five kingdom system of classification suggested by R.H. Whittaker is not based on:

Options:
A Presence or absence of a well defined nucleus
B Mode of reproduction
C Mode of nutrition
D Complexity of body organization

The correct answer is:
Presence or absence of a well defined nucleus

Solution:
R. H. Whittaker's five-kingdom classification is mainly based on Cell structure, Thallus organization, Mode of Nutrition, Mode of reproduction and Phylogenetic relationship.

Topic: Biology, Biological Classification, NEET, NEET 2014
Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :

Options:
A CvT2-T1
B -RTV2-V1
C -RT ln V2/V1
D Zero

The correct answer is:
Zero

Solution:
W=-Pext ΔV

In expansion against vacuum Pext=0

So work done is zero.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (04 Sep Shift 2)
Five moles of an ideal gas has pressure $p_0$ volume $V_0$ and temperature $T_0$. The gas is expanded to volume $3 V_0$ along a path, so that the pressure $p$ is changed as function of volume $V$ as $p=p_0\left(V / V_0\right)$. The pressure is then reduced to $p_0$ maintaining the volume constant. The gas undergoes an isobaric compression till the volume and temperature become $V_0$ and $T_0$, respectively. The total work done by the gas during the entire process is

Options:
A $p_0 V_0 / 3$
B $3 p_0 V_0$
C $5 p_0 V_0 / 3$
D $2 p_0 V_0$

The correct answer is:
$2 p_0 V_0$

Solution:
The given situation is shown in the following diagram
Work done in expansion (process I)
$=\int_{V_i}^{V_f} p d V=\int_{V_0}^{3 V_0} \frac{p_0 V}{V_0} \cdot d V$
$=\frac{p_0}{V_0}\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0}=\frac{p_0}{2 V_0}\left(9 V_0^2-V_0^2\right)=4 p_0 V_0$
In process $\mathrm{II}$, there is no work done as $V=$ constant.
In process III, work done is given as
$W=p_0 \int_{V_i}^{V_f} d V=p_0 \int_{3 V_0}^{V_0} d V$
$=p_0\left(V_0-3 V_0\right)=-2 p_0 V_0$
So, total work done in entire process is
$W_{\text {Total }}=4 p_0 V_0+\left(-2 p_0 V_0\right)=2 p_0 V_0$

Topic: Physics, Thermodynamics, JEE Main
Five moles of an ideal gas has pressure $p_0$ volume $V_0$ and temperature $T_0$. The gas is expanded to volume $3 V_0$ along a path, so that the pressure $p$ is changed as function of volume $V$ as $p=p_0\left(V / V_0\right)$. The pressure is then reduced to $p_0$ maintaining the volume constant. The gas undergoes an isobaric compression till the volume and temperature become $V_0$ and $T_0$, respectively. The total work done by the gas during the entire process is

Options:
A $p_0 V_0 / 3$
B $3 p_0 V_0$
C $5 p_0 V_0 / 3$
D $2 p_0 V_0$

The correct answer is:
$2 p_0 V_0$

Solution:
The given situation is shown in the following diagram
Work done in expansion (process I)
$=\int_{V_i}^{V_f} p d V=\int_{V_0}^{3 V_0} \frac{p_0 V}{V_0} \cdot d V$
$=\frac{p_0}{V_0}\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0}=\frac{p_0}{2 V_0}\left(9 V_0^2-V_0^2\right)=4 p_0 V_0$
In process $\mathrm{II}$, there is no work done as $V=$ constant.
In process III, work done is given as
$W=p_0 \int_{V_i}^{V_f} d V=p_0 \int_{3 V_0}^{V_0} d V$
$=p_0\left(V_0-3 V_0\right)=-2 p_0 V_0$
So, total work done in entire process is
$W_{\text {Total }}=4 p_0 V_0+\left(-2 p_0 V_0\right)=2 p_0 V_0$

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 2)
Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is -12, then the greatest number amongst them is

Options:
A 27
B 7
C 212
D 16

The correct answer is:
16

Solution:
Let term be a-2d,a-d,a,a+d,a+2d
Sum =25⇒5a=25⇒a=5
Product =2520
5-2d5-d55+d5+2d=2520
⇒25-4d225-d2=504
⇒625-100d2-25d2+4d4=504
⇒4d4-125d2+625-504=0
⇒4d4-125d2+121=0
⇒4d4-121d2-4d2+121=0
⇒d2-14d2-121=0
⇒d=±1,d=∓112
d=±1, does not give -12 as a term
∴d=112
∴ Largest term =5+2d=5+11=16

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Five numbers are in AP with common difference $\neq 0 .$ If the 1 st, 3 rd and 4 th terms are in GP, then

Options:
A the 5th term is always $0 .$
B the 1 st term is always $0 .$
C the middle term is always $0 .$
D the middle term is always -2.

The correct answer is:
the 5th term is always $0 .$

Solution:
Let the tine number is in AP are
$$
(a-2 d),(a-d), a,(a+d),(a+2 d)
$$
where, $d \neq 0$
Given, 1 st, $3 r d$ and 4 th terms are in GP.
$$
\begin{array}{lr}
\Rightarrow \quad a^{2}=(a-2 d)(a+d) \\
\Rightarrow \quad a^{2}=a^{2}-2 a d+a d-2 d^{2} \\
\Rightarrow \quad 2 d^{2}+a d=0 \Rightarrow d(2 d+a)=0 \\
\because \quad d \neq 0
\end{array}
$$
$\therefore \quad a+2 d=0 \Rightarrow a=-2 d$
Hence, terms are
$$
-4 d,-3 d,-2 d,-d, 0
$$
$\therefore$ The fifth term is always $0 .$

Topic: Mathematics, Sequences and Series, JEE Main
Five numbers are selected from 1, 2, 3, 4, 5, 6, 7, 8 and 9. The probability that their product is divisible by 5 or 7 is

Options:
A 14
B 34
C 56
D 16

The correct answer is:
56

Solution:
A= product of 5 numbers is divisible by 5
B= product of 5 numbers is divisible by 7
Total cases =C95
PA∪B=PAC∩BCC=1-PAC∩BC=1-C57C59
=1-21126=1-16
=56

Topic: Mathematics, Probability, JEE Main
Five persons $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ are seated in a circular arrangement. If each of them is given a cap of one of the three colours red, blue and green, then the number of ways of distributing the caps such that the persons seated in adjacent seats get different coloured caps, is

Options:
A 30
B 15
C 60
D 40

The correct answer is:
30

Solution:
There are 5 caps and 3 colours.
$\therefore \quad$ At least one colour will get repeated. As adjacent caps should be of different colours, no colour can repeat thrice.
$\therefore \quad$ Exactly two colours will repeat twice.
$\therefore \quad$ Colour of the caps are selected in 3 ways as follows:
Red-Red-Green-Green-Blue, Red-Red-Green-Blue-Blue, Red-Green-Green-Blue-Blue.
Now, while distributing the caps from above combinations, we choose any one of the 5 persons and give single colour cap. And remaining four caps can be distributed in alternate colour sequence, clock-wise or anticlock-wise.
This can be done in $5 \times 2$ ways.
$\therefore \quad$ Required number of ways $=3 \times 5 \times 2=30$.

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2023 (13 May Shift 1)
Five rods of same dimensions are arranged as shown in figure. They have thermal conductivities K1, K2, K3, K4 and K5. When points A and B are maintained at different temperature, no heat would flow through central rod, if

Options:
A K1K4=K2 K3
B K1=K4Â andÂ Â K2=K3
C K1K4=K2K3Â
D K1K2=K3K4.

The correct answer is:
K1K4=K2 K3

Solution:
The equivalent electrical circuit, figure in these cases is of Wheatstone bridge. No current would flow through central rod CD when the bridge is balanced. The condition for balanced Wheatstone bridge is PQ=RS (in terms of resistances)
K2K1=K4K3
Or K1K4=K2K3

Topic: Physics, Thermal Properties of Matter, NEET
Five students are selected from n students such that the ratio of number of ways in which 2 particular students are selected to the number of ways 2 particular students not selected is $2: 3$. Then the value of $\mathrm{n}$ is

Options:
A 5
B 6
C 11
D not possible

The correct answer is:
11

Solution:
Five students are selected from $\mathrm{n}$ students. Number of ways in which 2 particular students are sélected $=\mathrm{n}^{-2} \mathrm{C}_3$
Number of ways in which 2 particular students are not selected $={ }^{\mathrm{n}-2} \mathrm{C}_5$
$\therefore \quad$ According to the given condition,
$\frac{{ }^{n-2} C_3}{{ }^{n-2} C_5}=\frac{2}{3}$
$\begin{aligned} & \Rightarrow \frac{(\mathrm{n}-2) !}{3 !(\mathrm{n}-5) !} \times \frac{5 !(\mathrm{n}-7) !}{(\mathrm{n}-2) !}=\frac{2}{3} \\ & \Rightarrow(\mathrm{n}-5)(\mathrm{n}-6)=30 \\ & \Rightarrow \mathrm{n}=11\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
Five students are to be arranged on a platform such that the boy $B_1$ occupies the second position and such that the girl $\mathrm{G}_1$ is always adjacent to the girl $G_2$. Then, the number of such possible arrangements is

Options:
A $4$
B $7$
C $8$
D $6$

The correct answer is:
$8$

Solution:
There are 5 positions. Given that $\mathrm{B}_1$ occupies $2^{\text {nd }}$ position
$\therefore \quad \mathrm{B}_1$ can be arranged in 1 way. As $\mathrm{G}_1$ and $\mathrm{G}_2$ are always together, none of them can take $1^{\text {st }}$ position.
$\therefore \quad \mathrm{G}_1, \mathrm{G}_2$ and one of the remaining students can be arranged on $3^{\text {rd }}, 4^{\text {th }}$ and $5^{\text {th }}$ position when $\mathrm{G}_1$ and $\mathrm{G}_2$ are always together in $2 ! \times 2$ ! Ways.
And remaining 2 students can be arranged in 2 ! Ways.
$\therefore \quad$ The required number of arrangements $=2 ! \times 2 ! \times 2 !=8$

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2023 (11 May Shift 1)
Flocculation value of $\mathrm{BaCl}_2$ is much less than that of $\mathrm{KCl}$ for sol $\mathrm{A}$ and flocculation value of $\mathrm{Na}_2 \mathrm{SO}_4$ is much less than that of $\mathrm{NaBr}$ for sol $\mathrm{B}$. The correct statement among the following is :

Options:
A
Both the sols A and B are negatively charged.

B
Sol A is positively charged and Sol B is negatively charged.

C
Both the sols A and B are positively charged.

D
Sol A is negatively charged and sol B is positively charged.

The correct answer is:
Sol A is positively charged and Sol B is negatively charged.

Solution:
In first case the given compounds have same anion but different cations having different charge hence they will precipitate negatively charged sol i.e. 'A'.
In second case the given compounds have similar cation but different anion with different charge. Hence they will precipitate positively charged sol. i.e. ' $\mathrm{B}$ '.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2013 (22 Apr Online)
Fluoride pollution mainly affects:

Options:
A brain
B heart
C teeth
D kidney

The correct answer is:
teeth

Solution:
Fluoride is a mineral found in bones, teeth, the thyroid gland, and the skin. It is important for the optimal formation of dental enamel and bone mineralization. However, in abundance, it has negative effects on the joints, bones and causes fluorosis in the teeth.
Related Theory
- Teeth are affected by fluoride immediately. Since it results in tooth decay and decolouration quite quickly. it gets distributed into the contaminated water: great caution must be taken. Knock knee syndrome is caused by Fluoride pollution.

Topic: Biology, Environmental Issues, NEET, NEET 2003
Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction?

Options:
A NaF/Cu
B Cu2O/H2O
C Only Heat
D NaNO2/Cu

The correct answer is:
Only Heat

Solution:
Only Heat

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2016 (10 Apr Online)
Fluorine has low electron affinity than chlorine because of

Options:
A Bigger radius of fluorine, less density
B Smaller radius of fluorine, high density
C Smaller radius of chlorine, high density
D Smaller radius of chlorine, less density

The correct answer is:
Smaller radius of fluorine, high density

Solution:
Electron affinity is defined as the amount of energy released on the addition of an electron to an atom.

Fluorine has a lower electron affinity than chlorine because of its smaller size with 2p-orbitals, where the interelectronic repulsion is more.

The orbital configuration of F7 can be shwon as under :

The orbital configuration of Cl17 can be shown as under :

In fluorine, the valence electrons are present in the 2p-orbitals, which are smaller in size than the 3p-orbitals of chlroine. Due to a small orbital size, the electrons in fluorine are crowded, and experience greater inter-electronic repulsions. This high electron density makes fluorine reluctant to accept an electron, and therefore, it requires a supply of energy for the addition of an electron to it. This decreases the amount of energy released by it, and thus, fuorine has a lower negative electron affinity than chlorine.

So, the correct reason for fluorine having a lower electron affinity than chlorine is, its small atomic radius, and high electron density.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Fluorine is a better oxidising agent than $\mathrm{Br}_2$. It is due to

Options:
A Small size of fluorine
B More electron repulsion in fluorine
C More electronegativity of fluorine
D Non-metallic nature of fluorine

The correct answer is:
More electronegativity of fluorine

Solution:
\(F_2\) is a better oxidising agent than \(B r_2\). It is due to more electronegativity of fluorine.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Fluorobenzene $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{~F}\right)$ can be synthesized in the laboratory

Options:
A
by heating phenol with $\mathrm{HF}$ and $\mathrm{KF}$

B
from aniline by diazotisation followed by heating the diazonium salt with $\mathrm{HBF}_4$

C
by direct fluorination of benzene with $\mathrm{F}_2$ gas

D
by reacting bromobenzene with NaF solution

The correct answer is:
from aniline by diazotisation followed by heating the diazonium salt with $\mathrm{HBF}_4$

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2006
Focal length of a convex lens will be maximum for

Options:
A Blue light
B Yellow light
C Green light
D Red light

The correct answer is:
Red light

Solution:
$f \propto \frac{1}{(\mu-1)}$ and $\mu \propto \frac{1}{\lambda}$. Hence $f \propto \lambda$ and $\lambda_r\gt\lambda_v$

Topic: Physics, Ray Optics, JEE Main
Focal length of a plane mirror is

Options:
A Zero
B Infinite
C Very less
D Indefinite

The correct answer is:
Infinite

Solution:
A plane mirror is a flat reflecting surface. Thus, in the case of a plane mirror, when the parallel rays of light strike the mirror they get reflected back parallel to each other. So, they never meet or we can say they meet at infinity. So, the focal length of the plane mirror is Infinity.

Topic: Physics, Ray Optics, JEE Main
Foetal ejection reflex in human female is induced by

Options:
A Differentiation of mammary glands
B Pressure exerted by amniotic fluid
C Release of oxytocin from pituitary
D Fully developed foetus and placenta

The correct answer is:
Fully developed foetus and placenta

Solution:
No Solution Available

Topic: Biology, Human Reproduction, NEET, NEET 2009 (Mains)
Fog is a colloidal solution of

Options:
A
solid particles dispersed in gas

B
solid particles dispersed in a liquid

C
liquid particles dispersed in gas

D
gaseous particles dispersed in a liquid

The correct answer is:
liquid particles dispersed in gas

Solution:
Fog has liquid as dispersed phase and gas as dispersion medium.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2012 (19 May Online)
Following are expressions for four plane simple harmonic waves
(i) $\quad y_1=A \cos 2 \pi\left(n_1 t+\frac{x}{\lambda_1}\right)$
(ii) $y_2=A \cos 2 \pi\left(n_1 t+\frac{x}{\lambda_1}+\pi\right)$
(iii) $y_3=A \cos 2 \pi\left(n_2 t+\frac{x}{\lambda_2}\right)$
(iv) $y_4=A \cos 2 \pi\left(n_2 t-\frac{x}{\lambda_2}\right)$
The pairs of waves which will produce destructive interference and stationary waves respectively in a medium, are

Options:
A
(iii, iv), (i, ii)

B
(i, iii), (ii, iv)

C
(i, iv), (ii, iii)

D
(i, ii), (iii, iv)

The correct answer is:
(i, ii), (iii, iv)

Solution:
In case of destructive interference Phase difference $\phi=180^{\circ}$ or $\pi$ So wave pair (i) and (ii) will produce destructive interference.
Stationary or standing waves will produce by equations (iii) \& (iv) as two waves travelling along the same line but in opposite direction.

Topic: Physics, Waves and Sound, JEE Main
Following are expressions for four plane simple harmonic waves
(i) $\quad y_1=A \cos 2 \pi\left(n_1 t+\frac{x}{\lambda_1}\right)$
(ii) $y_2=A \cos 2 \pi\left(n_1 t+\frac{x}{\lambda_1}+\pi\right)$
(iii) $y_3=A \cos 2 \pi\left(n_2 t+\frac{x}{\lambda_2}\right)$
(iv) $y_4=A \cos 2 \pi\left(n_2 t-\frac{x}{\lambda_2}\right)$
The pairs of waves which will produce destructive interference and stationary waves respectively in a medium, are

Options:
A
(iii, iv), (i, ii)

B
(i, iii), (ii, iv)

C
(i, iv), (ii, iii)

D
(i, ii), (iii, iv)

The correct answer is:
(i, ii), (iii, iv)

Solution:
In case of destructive interference Phase difference $\phi=180^{\circ}$ or $\pi$ So wave pair (i) and (ii) will produce destructive interference.
Stationary or standing waves will produce by equations (iii) \& (iv) as two waves travelling along the same line but in opposite direction.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2012 (07 May Online)
Following are the functions of a particular hormone:

a) controller of circadian rhythm in humans
b) helps in pigmentation, metabolism, menstrual cycle
c) controls the defence mechanism of the body.

Name the hormone.

Options:
A Epinephrine
B Gastrin
C Melatonin
D Insulin.

The correct answer is:
Melatonin

Solution:
The pineal gland is present in the forebrain. It secretes a hormone known as Melatonin. This hormone is responsible for sleep- wake cycle. Other functions are maintaining the temperature of body, pigmentation, metabolism, menstrual cycle as well as controlling the defense mechanism of the body. Melatonin increases in darkness that causes initiation of sleep.

Topic: Biology, Chemical Coordination and Integration, NEET
Following are the roles of predation, except

Options:
A it helps in maintaining species diversity in a community.
B it is a potent force of organic evolution.
C predators act as 'conduits' of energy transfer across trophic levels.
D the population of prey is kept under control by the predators.

The correct answer is:
it is a potent force of organic evolution.

Solution:
Predation is nature's way of transferring energy across trophic levels in the food chain. Predators also help in maintaining species diversity in a community by reducing the intensity of competition among prey species. It also keeps the prey population under control. Predation has not been associated with organic evolution.

Topic: Biology, Organisms and Populations, NEET
Following compounds are given
A. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$
B. $\mathrm{CH}_3 \mathrm{COCH}_3$
C.
$$
\text { D. } \mathrm{CH}_3 \mathrm{OH}
$$
Which of the above compound(s), on being warmed with iodine solution and $\mathrm{NaOH}$, will give iodoform?

Options:
A A, C and D
B Only B
C A, B and C
D A and B

The correct answer is:
A, B and C

Solution:
$$
\text { Compounds having either }
$$
$$
\text { group or }
$$
$\mathrm{CH}_3 \mathrm{CHOH}-$ group, give iodoform when warmed with $\mathrm{I}_2$ and $\mathrm{NaOH}$. Thus,
give iodoform when warmed with $\mathrm{I}_2$ and $\mathrm{NaOH}$. (Remember, NaOI oxidises $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ to $\mathrm{CH}_3 \mathrm{CHO}$, thus it gives positive iodoform test.)

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2010 (Mains)
Following compounds are given
(i) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
(ii) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$
(iii)
(iv) $\mathrm{CH}_{3} \mathrm{OH}$
Which of the above compound(s) on being warmed with iodine solution and $\mathrm{NaOH}$, will give iodoform?

Options:
A (i), (iii) and (iv)
B Only(ii)
C (i), (ii) and (iii)
D (i) and (ii)

The correct answer is:
(i), (ii) and (iii)

Solution:
Compounds having either
group or $\mathrm{CH}_{3} \mathrm{CHOH}-$ group, give iodoform when warmed with $\mathrm{I}_{2}$ and $\mathrm{NaOH}$. Thus, compounds
give iodoform when heated with $\mathrm{I}_{2}$ and $\mathrm{NaOH}$. (Note: $\mathrm{NaOI}$ oxidises $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ to $\mathrm{CH}_{3} \mathrm{CHO}$, and gives positive iodoform test.)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Following data is available about \( 3 \) nuclei \( \mathrm{P}, \mathrm{Q} \& \mathrm{R} \) decrease order of stability is given by
\(\begin{array}{|l||l||l|l||} \hline \text{Element} & \mathrm{P} & \mathrm{Q} & \mathrm{R} \\ \hline \text{Atomic mass number A} & 10 & 5 & 6 \\ \hline \text{Binding energy (MeV)} & 100 & 60 & 66 \\ \hline \end{array}\)

Options:
A \( \mathrm{Q}>\mathrm{R}>\mathrm{P} \)
B \( P>Q>R \)
C \( P>R>Q \)
D \( \mathrm{Q}>\mathrm{P}>\mathrm{R} \)

The correct answer is:
\( \mathrm{Q}>\mathrm{R}>\mathrm{P} \)

Solution:
B.EAP=10010=10
B.EAQ=605=12Q>R>P
B.EAR=666=11

Topic: Physics, Nuclear Physics, JEE Main
Following diagram performs the logic function of:

Options:
A AND gate
B NAND gate
C OR gate
D XOR gate

The correct answer is:
AND gate

Solution:
$$
\begin{aligned}
& x=\overline{\mathrm{A} \cdot \mathrm{B}}, y=\bar{x}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}} \\
& \Rightarrow \text { AND-gate. }
\end{aligned}
$$

Topic: Physics, Semiconductors, JEE Main
Following figure shows dependence of molar conductance of two electrolytes on concentration. Λm0 is the limiting molar conductivity.

The number of

Incorrect

statement(s) from the following is _____

(A) Λm0 for electrolyte A is obtained by extrapolation
(B) For electrolyte B, Λm Vs c graph is a straight line with intercept equal to Λm0
(C) At infinite dilution, the value of degree of dissociation approach zero for electrolyte B.
(D) Λm for any electrolyte A or B can be calculated using λ° for individual ions.

Options:

The correct answer is:
2

Solution:
Following are the correct versions of the two incorrect statements:
(A) Λm0 for electrolyte A is cannot be obtained by extrapolation as the curve for A is diverging away from the y-axis

(C) At infinite dilution, value of degree of dissociation approaches one for electrolyte B

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (29 Jan Shift 1)