datavorous/entrance-exam-dataset huggingface dataset:
Removal of gonads cannot be considered as a contraceptive option. Why?

Options:

None

Solution:
Removal of gonads or its parts is a surgical method of sterilization. Vasectomy and tubectomy are carried out in males and females respectively. It will lead to infertility \& both male and female will be dependent on hormones in their remaining life to regulate functioning of reproductive organs. These are very effective but their irreversibility is very poor, so they are not good options for contraception.

Topic: Biology, Reproductive Health
Removal of introns and joining the exons in a defined order in a transcription unit is called :

Options:
A Capping
B Splicing
C Tailing
D Transformation

The correct answer is:
Splicing

Solution:
No Solution Available

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2009 (Mains)
Removal of proximal convoluted tubule from the nephron will result in:

Options:
A More diluted urine
B More concentrated urine
C No change in quality and quantity of urine
D No urine formation

The correct answer is:
More diluted urine

Solution:
The removal of proximal convoluted tubule from the nephron results in a lack of reabsorption of high threshold substance from renal tubules and obligatory reabsorption of water is also affected leading to more diluted urine. Since proximal convoluted tubules are mainly associated with reabsorption of much water by osmosis, reabsorption of glucose and amino acids by secondary active transport, and other salts and ions as Na+, K+, vitamins etc., by primary active transport.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2015 (Phase 1)
Renin is secreted by

Options:
A juxtaglomerular cells of juxtamedullary nephrons in response to high blood pressure
B juxtaglomerular cells of juxtamedullary nephrons in response to low blood pressure
C juxtaglomerular cells of cortical nephrons in response to high blood pressure
D juxtaglomerular cells of cortical nephrons in response to increased osmomolarity.

The correct answer is:
juxtaglomerular cells of juxtamedullary nephrons in response to low blood pressure

Solution:
Nephron is the structural and functional unit of kidney. Each nephron consists of glomerulus and the renal tubule. Based on the length of Loop of Henle two types of Nephrons are seen. Cortical and Juxta-medullary nephrons. Cortical nephrons have short loop of henle as it extends in the cortex only and majority of nephrons are of this type. This type of nephron is mainly responsible for excretion function. Juxta medullary nephrons have long loop and extends in the medulla (Juxta means near). This is mainly responsible for concentration of urine so that water loss is prevented. This is done with the help of JGA i.e. Juxta Glomerular Apparatus. This includes Juxta Glomerular cells and Macula densa. The Juxta cells secrete Renin in response to low blood pressure and try to increase the blood pressure by RAAS (Renin Angiotension Aldosterone System).

Topic: Biology, Excretory Products and their Elimination, NEET
Rennet enzyme was purified by

Options:
A A. Flemming
B S.A. Waksman
C Watson and Crick
D Christian Hansen

The correct answer is:
Christian Hansen

Solution:
Rennet is a complex set of enzymes produced by ruminant mammals. Chymosin, pepsin and lipase are the main component of this enzyme.Rennet enzyme was purified for the first time by Christian Hensen in 1874 from calf stomach and is especially used during cheese formation.

Topic: Biology, Biotechnology and Its Applications, NEET
Replacement of $\mathrm{Cl}$ of chlorobenzene to give phenol requires drastic conditions but 4-dinitrochlorobenzene is readily replaced because

Options:
A $\mathrm{NO}_2$ make ring electron rich at ortho and para
B $\mathrm{NO}_2$ withdraws $e^{-}$from meta position
C denotes $\mathrm{e}^{-}$at meta position
D $\mathrm{NO}_2$ withdraws $e^{-}$from ortho/para positions

The correct answer is:
$\mathrm{NO}_2$ withdraws $e^{-}$from ortho/para positions

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Replacement of \( \mathrm{Cl} \) of Chlorobenzene to give phenol requires drastic conditions, but \( \mathrm{Cl} \) of \( 2,4 \) -
dinitro chlorobenzene is readily replaced. This is because,

Options:
A \( -\mathrm{NO}_{2} \) group makes the ring electron rich at o- and p- positions.
B \(-\mathrm{NO}_{2} \) group withdraws electrons from \( \mathrm{m} \)-position.
C \( -\mathrm{NO}_{2} \) donate electrons at \( \mathrm{m} \) - position.
D \( -\mathrm{NO}_{2} \) withdraws electrons from o- and p-positions.

The correct answer is:
\( -\mathrm{NO}_{2} \) withdraws electrons from o- and p-positions.

Solution:
The reaction involved is
The presence of \( -\mathrm{NO}_{2} \) group at o- and p- positions withdraws electron density from the benzene ring, therefore,
facilitates the attack of the nucleophile on haloarenes. Thus, the carbanion formed, is stabilized through resonance as
shown below:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Replum is present in the ovary of flower of

Options:
A Pea
B Lemon
C Mustard
D Sunflower

The correct answer is:
Mustard

Solution:
Mustard (Brassicaceae) possess false septum in ovary having parietal placentation.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Mains)
Represented below is the inheritance pattern of a certain type of traits in humans. Which one of the following conditions could be an example of this pattern?

Options:
A Phenylketonuria
B Sickle cell anaemia
C Haemophilia
D Thalassemia

The correct answer is:
Haemophilia

Solution:
Colourblindness and haemophilia are two main sex-linked or X-linked recessive diseases. The gene of haemophilia goes to son from mother and to daughter from father. In this disease, the blood is fail to clot when exposed to air result in continuous bleeding and leads to death.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2012 (Mains)
Reproduction in bacteria is shown in following diagram. Select the correct match.

Options:
A

Mode of Reproduction	Conditions
Budding	Favourable

B

Mode of Reproduction	Conditions

Binary fission	Favourable

C

Mode of Reproduction	Conditions
Spore formation	Unfavourable

D

Mode of Reproduction	Conditions
Budding	Unfavourable

The correct answer is:

Mode of Reproduction	Conditions

Binary fission	Favourable

Solution:
Reproduction: Mainly by fission (during favorable conditions). These also reproduce by spore formation (during unfavorable conditions) or by primitive type of DNA transfer from one bacteria to another.
Heterotrophs are most abundant in nature. These also play important role in curdling of milk, production of antibiotics, nitrogen fixation in legumes, etc.

Topic: Biology, Biological Classification, NEET
Reproductive organs in humans can be categorized into two categories primary and secondary. Which of the following is not a secondary sex organ?

a Prostate gland b Mammary gland c Seminal vesicle d Ovary

Options:
A Only b and d
B Both d and c
C Only d
D a, b and c

The correct answer is:
Only d

Solution:
Reproductive organs in humans are divided into two categories: primary and secondary. Primary organs are those that form the sex cells i.e. sperms and ova. These are formed by testes in males and ovaries in females. These are collectively known as gonads. Hence gonads are considered primary sex organs. Apart from these structures, the remaining structures concerned with reproduction are secondary organs. These include the epididymis, vasa deferens, penis, urethra, ejaculatory duct, accessory glands like seminal vesicles, prostate glands and Cowper's gland in males. In females the secondary sex organs include oviduct, uterus, vagina, vulva, vestibular glands and mammary glands.

Topic: Biology, Human Reproduction, NEET
Resistance of 0.2 M solution of an electrolyte is 5 0 Ω . The specific conductance of the solution is 1.4 S m -1 . The resistance of 0.5 M solution of the same electrolyte is 28 0 Ω . The molar conductivity of 0.5 M solution of the electrolyte in S m 2 mol -1 is :

Options:
A 5 × 1 0 - 4
B 5 × 1 0 - 3
C 5 × 1 0 3
D 5 × 1 0 2

The correct answer is:
5 × 1 0 - 4

Solution:
K = 1 R l a
K 0.2 M = 1 5 0 l a ...... (I)
K 0.5 M = 1 28 0 l a ...... (II)
K 0.5 M K 0.2 M = 1 2 8 0 × 5 0
K 0.5 M = 1.4 2 8 0 × 5 0 = 1 4
Λ m = K C = 1 4 × 1 0.5 × 1 0 3 = 5 × 1 0 - 4 Ω -1 m 2 mol -1
C = concentration in mol/m 3 = 0.5 × 1 0 3 moles/m 3

Topic: Chemistry, Electrochemistry, JEE Main
Resistance of 0.2 M solution of an electrolyte is 5 0 Ω . The specific conductance of the solution is 1.4 S m -1 . The resistance of 0.5 M solution of the same electrolyte is 28 0 Ω . The molar conductivity of 0.5 M solution of the electrolyte in S m 2 mol -1 is :

Options:
A 5 × 1 0 - 4
B 5 × 1 0 - 3
C 5 × 1 0 3
D 5 × 1 0 2

The correct answer is:
5 × 1 0 - 4

Solution:
K = 1 R l a
K 0.2 M = 1 5 0 l a ...... (I)
K 0.5 M = 1 28 0 l a ...... (II)
K 0.5 M K 0.2 M = 1 2 8 0 × 5 0
K 0.5 M = 1.4 2 8 0 × 5 0 = 1 4
Λ m = K C = 1 4 × 1 0.5 × 1 0 3 = 5 × 1 0 - 4 Ω -1 m 2 mol -1
C = concentration in mol/m 3 = 0.5 × 1 0 3 moles/m 3

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2014 (06 Apr)
Resistance of a 0.1M KCl solution in a conductance cell is 300 ohm and specific conductance of 0.1M KCl is 1.33 Ã10-2ohm-1cm-1 . The resistance of 0.1M NaCl solution in the same cell is 400ohm. The equivalent conductance of the 0.1M NaCl solution (in ohm-1cm2/gmeq.) is

Options:

The correct answer is:
100

Solution:
κ=1RlA
i.e. lA=R×κ=300×1.33×10-2≃4.0cm-1
κNaCl=1RlA=1400×4.0
Λm(NaCl⁡)=κ×1000M=4.0400×10000.1=100

Topic: Chemistry, Electrochemistry, JEE Main
Resistance of a conductivity cell (cell constant 129 m-1) filled with 74.5ppm solution of KCl is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. ∧1∧2=x×10-3. The value of x is____Given, molar mass of KCl is 74.5 g mol-1)

Options:

The correct answer is:
1000

Solution:
Given lA=129 m-1

KCl solution 1:

74.5ppm, R1=100 Ω

KCl solution 2:

149ppm, R2=50 Ω

Here, ppm1ppm2=M1M2

and κ = 1RG* where G* = cell constant

Since value of cell constant is same in both the cases.

k1k2 = R2R1

∧1∧2=k1×1000M1k2×1000M2

=K1 K2×M2M1 = R2 R1×M2M1

=50100×14974.5 = 1

=∧1∧2=1,000×10-3

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $0.1 \mathrm{M}$ is $100 \Omega$.
The conductivity of this solution is $1.29 \mathrm{~S} \mathrm{~m}^{-1}$. Resistance of the same cell when filled with $0.2 \mathrm{M}$ of the same solution is $520 \Omega$. The molar conductivity of $0.02 \mathrm{M}$ solution of the electrolyte will be

Options:
A
$124 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$

B
$1240 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$

C
$1.24 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$

D
$12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$

The correct answer is:
$12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$

Solution:
There is one mistake in Question paper.
Assuming concentration of solution is $0.2 \mathrm{M}$ instead of $0.02 \mathrm{M}$. Since resistance of $0.2 \mathrm{M}$ is $520 \Omega$.
$$
\begin{aligned}
\mathrm{R} & =100 \Omega \\
\mathrm{K} & =\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{a}}\right) \\
1.29 & =\frac{1}{100}\left(\frac{\ell}{\mathrm{a}}\right) \\
\left(\frac{\ell}{\mathrm{a}}\right) & =129 \mathrm{~m}^{-1} \\
\mathrm{R} & =520 \Omega, \mathrm{C}=0.2 \mathrm{M} \\
\mathrm{K} & =\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{a}}\right)=\frac{1}{520}(129) \Omega^{-1} \mathrm{~m}^{-1} \\
\mu & =\mathrm{K} \times \mathrm{V}_{\text {in } \mathrm{cm}} \\
& =\frac{1}{520} \times 129 \times \frac{1000}{0.2} \times 10^{-6} \mathrm{~m}^3 \\
& =\frac{129}{520} \times \frac{1000}{0.2} \times 10^{-6} \\
& =1.24 \times 10^{-3} \\
& =12.4 \times 10^{-4}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2006
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3 \%$ each, then the error in the value of resistance of the wire is

Options:
A $3 \%$
B $6 \%$
C zero
D $1 \%$

The correct answer is:
$6 \%$

Solution:
$R=V / I$
$(\Delta R / R) \times 100=(\Delta V / V) \times 100+(\Delta I / I) \times 100=3+3=6 \%$
Answer: (2) $6 \%$

Topic: Physics, Mathematics in Physics, JEE Main
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3 \%$ each, then error in the value of resistance of the wire is

Options:
A
$6 \%$

B
zero

C
$1 \%$

D
$3 \%$

The correct answer is:
$6 \%$

Solution:
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{i}}$
$\Rightarrow \quad\left|\frac{\Delta \mathrm{R}}{\mathrm{R}}\right|=\left|\frac{\Delta \mathrm{V}}{\mathrm{V}}\right|+\left|\frac{\Delta \mathrm{i}}{\mathrm{i}}\right|$
$\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3$
$\Rightarrow \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=0.03$
Similarly, $\frac{\Delta i}{i}=0.03$
Hence $\frac{\Delta \mathrm{R}}{\mathrm{R}}=0.06$
So percentage error is $\frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=6 \%$

Topic: Physics, Mathematics in Physics, JEE Main
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3 \%$ each, then error in the value of resistance of the wire is

Options:
A
$6 \%$

B
zero

C
$1 \%$

D
$3 \%$

The correct answer is:
$6 \%$

Solution:
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{i}}$
$\Rightarrow \quad\left|\frac{\Delta \mathrm{R}}{\mathrm{R}}\right|=\left|\frac{\Delta \mathrm{V}}{\mathrm{V}}\right|+\left|\frac{\Delta \mathrm{i}}{\mathrm{i}}\right|$
$\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3$
$\Rightarrow \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=0.03$
Similarly, $\frac{\Delta i}{i}=0.03$
Hence $\frac{\Delta \mathrm{R}}{\mathrm{R}}=0.06$
So percentage error is $\frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=6 \%$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2012 (Offline)
Resistance of a wire at $0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$ and $t^{\circ} \mathrm{C}$ is found to be $10 \Omega, 10.2 \Omega$ and $10.95 \Omega$ respectively. The temperature $t$ in Kelvin scale is _________

Options:

The correct answer is:
748

Solution:
$\begin{aligned} & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}_0}=\alpha \Delta \mathrm{T}\end{aligned}$
Case-I
$\begin{aligned} & 0{ }^{\circ} \mathrm{C} \rightarrow 100^{\circ} \mathrm{C} \\ & \frac{10.2-10}{10}=\alpha(100-0)\end{aligned}$ ...(1)
Case-II
$0^{\circ} \mathrm{C} \rightarrow \mathrm{t}^{\circ} \mathrm{C}$
$\frac{10.95-10}{10}=\alpha(\mathrm{t}-0)$ ...(2)
$\begin{aligned} & \Rightarrow \frac{\mathrm{t}}{100}=\frac{0.95}{0.2}=475^{\circ} \mathrm{C} \\ & \mathrm{t}=475+273=748 \mathrm{~K}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Resistance of an Ideal ammeter is:

Options:
A Infinite
B $1 \Omega$
C Zero
D $10 \Omega$

The correct answer is:
Zero

Solution:
Ammeter
- An Ammeter is a device which is connected in series of the circuit to find the current.
- Ampere is the unit of current.
- Ammeter is connected in series. So, if it is having a high resistance, it will affect the reading as the current will be reduced in the circuit.
- So, an ammeter should have very small (ideally zero) resistance so that perfect reading can be obtained.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Resistance of the wire is measured as 2 Ω and 3 Ω at 10°C and 30°C respectively. Temperature coefficient of resistance of the material of the wire is

Options:
A 0.033C∘-1
B -0.033C∘-1
C 0.011C∘-1
D 0.055°C-1

The correct answer is:
0.033C∘-1

Solution:
Resistance of a material is related to temperature by expression R=R01+αΔT

Here, at 10°C, we have 2=R01+10α ...i

And at 30°C, we have 3=R01+30α ...ii

From equation i, R0=21+10α

Putting this value in equation ii,

3=21+10α1+30α⇒3+30α=2+60α

⇒1=30α

Hence, temperature coefficient of the material of wire is α=130=0.033°C-1

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Resistances of 100 Î© and 200 Î© are connected in series with 220 V mains. When a voltmeter of 1000 Î© resistance is connected in parallel to 100 Î©, then the reading of voltmeter is

Options:
A 68.5 volts
B 6.87 volts
C 587.5 volts
D 58.75 volts

The correct answer is:
68.5 volts

Solution:
Req=320011 Ω
V=220320011×100011=220032=68.75 V

Topic: Physics, Current Electricity, NEET
Resonance in carbonate ion CO32- is

Which of the following is true?

Options:
A It is possible to identify each structure individually by some physical or chemical method.
B All these structures are in dynamic equilibrium with each other.
C Each structure exists for equal amount of time.
D CO32- has a single structure i.e., resonance hybrid of the above three structures.

The correct answer is:
CO32- has a single structure i.e., resonance hybrid of the above three structures.

Solution:
Resonating structure are hypothetical and resonance hybrid is real structure which is weighted average of all the resonating structures.

So, CO32- has a single structure i.e. resonance hybrid of the three structures.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Resonance structure of molecule does not have

Options:
A Identical arrangement of atoms
B Nearly the same energy content
C The same number of paired electrons
D Identical bonding

The correct answer is:
Identical bonding

Solution:
Resonance structure of molecule does not have identical bonding.

Topic: Chemistry, General Organic Chemistry, JEE Main
Respiratory system plays an important role of enabling gas exchange and supply of oxygen to the animals. Although, the basic mechanism of the system remains same, animals display variation in their respiratory system. In the following columns match the animals with their organ for respiration.

Animal	Respiratory Organ

1	Kangaroo	P	Moist cuticle

2	Pomfret	Q	Lungs

3	Cockroach	R	Gills

4	Earthworm	S	Tracheal tubes

Options:
A 1-Q, 2-P, 3-S, 4-R
B 1-Q, 2-R, 3-S, 4-P
C 1-Q, 2-R, 3-P, 4-S
D 1-Q, 2-P, 3-R, 4-S

The correct answer is:
1-Q, 2-R, 3-S, 4-P

Solution:
The respiratory system is the organ system helping animals in breathing and gas exchange to utilize oxygen for respiration process. Animals show variation in their way of obtaining the oxygen depending on their complexity habitant among other factors.
Kangaroo is a mammal and terrestrial animal that obtains oxygen from the air via the lungs.
Pomfret is a type of fish that uses gills for breathing in water. The cockroach is simpler invertebrate compared to kangaroo and pomfret using tube-like structure called trachea.
Lastly, the earthworm is an annelid that uses its moist cuticle region for exchanging gases enabling breathing.

Topic: Biology, Breathing and Exchange of Gases, NEET
Resting membrane potential is maintained by

Options:
A
hormones

B
neurotransmitters

C
ion pumps

D
None of these

The correct answer is:
ion pumps

Solution:
The ionic gradients across the resting membranes are maintained by the active transport of ions by the sodium- potassium pumps which transport $3 \mathrm{Na}^{+}$outward for every $2 \mathrm{~K}^{+}$into the cell.

Topic: Biology, Neural Control and Coordination
Resting membrane potential is maintained by

Options:
A Hormones
B Neurotransmitters
C Ion pump
D None of the above

The correct answer is:
Ion pump

Solution:
Resting membrane potential is maintained by the Ion pump.

Topic: Biology, Neural Control and Coordination, NEET
Resting membrane potential is maintained by

Options:
A hormones
B neurotransmitters
C ion pumps
D none of the above

The correct answer is:
ion pumps

Solution:
During the resting-state of the neuron:
- The high concentration of K+ and negatively charged proteins and low concentration of Na+ in axoplasm.
- In contrast, the fluid outside the axon contains a low concentration of K+, a high concentration of Na+, and thus forms a concentration gradient.
- These ionic gradients across the resting membrane are maintained by the Na-K pump which actively transports 3Na+ outwards for 2K+ into the cell.
- The membrane is polarised.
- Potential difference is termed as resting potential.

Topic: Biology, Neural Control and Coordination, NEET
Reverse bias applied to a junction diode

Options:
A lowers of the potential barrier
B raises the potential barrier
C increases the majority carrier current
D increases the minority carrier current

The correct answer is:
raises the potential barrier

Solution:
Reverse bias increases the potential barrier

Topic: Physics, Semiconductors, JEE Main
Reversible expansion of an ideal gas under isothermal and adiabatic conditions are as shown in the figure.
$A B \rightarrow$ Isothermal expansion
$A C \rightarrow$ Adiabatic expansion
Which of the following option is not correct?

Options:
A $\Delta S_{\text {isothermal }}>\Delta S_{\text {adiabatic }}$
B $T_A=T_B$
C $W_{\text {isothermal }}>W_{\text {adiabatic }}$
D $T_c>T_A$

The correct answer is:
$T_c>T_A$

Solution:
From first law of thermodynamics,
$$
\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}
$$
In adiabatic expansion, $q=0$
$$
\therefore \quad \Delta \mathrm{U}=\mathrm{W}
$$
During expansion of a gas $w$ is negative i.e $w < 0$ or $\Delta \mathrm{U} < 0$.
We know that, $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$
$$
\begin{array}{lclc}
\therefore & \mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T} < 0 & \text { or } & \Delta \mathrm{T} < 0 \\
\therefore & \mathrm{T}_{\mathrm{C}}-\mathrm{T}_{\mathrm{A}} < 0 & \text { or } & \mathrm{T}_{\mathrm{C}} < \mathrm{T}_{\mathrm{A}}
\end{array}
$$
Thus, option (4) is incorrect while the remaining options are correct.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2019 (Odisha)
RH2 (ion exchange resin) can replace Ca2+ ions in hard water as:

RH2 + Ca2+⟶ RCa + 2H+

If 1 L of hard water after passing through RH2 has pH = 3, then hardness in parts per million of Ca2+ is :

Options:
A 10 ppm
B 40 ppm
C 100 ppm
D 20 ppm

The correct answer is:
20 ppm

Solution:
[ H+ ] =10 - 3 M [ âµ pH = 3 = - log [ H+ ]
=10 - 3 moles per litre
[ Ca2 + ] = 5 Ã 10 - 4 moles per litre
= 5 Ã 10 - 1 moles of Ca2 + per 103 L
= 5 Ã 10 - 1Ã 40 g Ca2 + per 106 g of H2O
Ca2 + = 20 ppm = (H).

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
RMgX+CO2→etherdryY⟶H3O+RCOOH

What is Y in the above reaction?

Options:
A R3CO-Mg+X
B RCOO-X+
C RCOO2Mg
D RCOO-Mg+X

The correct answer is:
RCOO-Mg+X

Solution:
Grignard reagent on reaction with carbondioxide in presence of dry ether forms an addition product which on hydrolysis produces carboxylic acid.

Hence, Y in the above reaction is RCOO-Mg+X

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
RNA and DNA are chiral molecules, their chirality is due to

Options:
A chiral bases
B chiral phosphate ester units
C D-sugar component
D L-sugar component

The correct answer is:
D-sugar component

Solution:
The constituents of nucleic acids are nitrogenous bases, sugar and phosphoric acid. The sugar present in DNA is D(-)-2-deoxyribose and the sugar present in RNA is D(-) ribose. Due to these D(-)-sugar components, DNA and RNA molecules are chiral molecules.

Topic: Chemistry, Biomolecules, JEE Main
RNA is different from DNA because RNA contains

Options:
A
ribose sugar and thymine

B
ribose sugar and uracil

C
deoxyribose sugar and thymine

D
deoxyribose sugar and uracil

The correct answer is:
ribose sugar and uracil

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, JEE Main