datavorous/entrance-exam-dataset huggingface dataset:
$\mathrm{Cu}_{\mathrm{aq}}^{+}$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction :

$2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

choose correct $\mathrm{E}^{\circ}$ for above reaction if

$\mathrm{E}^{\circ} \mathrm{Cu}^{2+}=0.34 \mathrm{~V}$ and $\mathrm{E}^{\circ} \mathrm{Cu}^{2+}=0.15 \mathrm{~V}$

Options:
A $-0.38 \mathrm{~V}$
B $+0.49 \mathrm{~V}$
C $+0.38 \mathrm{~V}$
D $-0.19 \mathrm{~V}$

The correct answer is:
$+0.38 \mathrm{~V}$

Solution:
$2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Cu}$

$2 \mathrm{e}^{-}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Cu} ; \mathrm{E}_{1}^{\circ}=0.34 \mathrm{~V} ; \ldots$ (i)

$\mathrm{e}^{-}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Cu}^{+} ; \mathrm{E}_{2}^{\circ}=0.15 \mathrm{~V} ; \ldots$ (ii)

$\mathrm{Cu}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Cu} ; \mathrm{E}_{3}^{\circ}=?$ ...(iii)

Now, $\Delta \mathrm{G}_{1}^{\circ}=-\mathrm{nFE}_{1}^{\circ}=-2 \times 0.34 \mathrm{~F}$

$\Delta \mathrm{G}_{2}^{\circ}=-1 \times 0.15 \mathrm{~F}, \Delta \mathrm{G}_{3}^{\circ}=-1 \times \mathrm{E}_{3}^{\circ} \mathrm{F}$

Again, $\Delta \mathrm{G}_{1}^{\circ}=\Delta \mathrm{G}_{2}^{\circ}+\Delta \mathrm{G}_{3}^{\circ}$

$\Rightarrow-0.68 \mathrm{~F}=-0.15 \mathrm{~F}-\mathrm{E}_{3} \mathrm{~F}$

$0 \Rightarrow \mathrm{E}_{3}^{\circ}=0.68-0.15=0.53 \mathrm{~V}$

$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}\left(\mathrm{Cu}^{+} / \mathrm{Cu}\right)$

$=0.53-0.15=0.38 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main
Cus+Sn2+0.001M→Cu2+0.01M+Sns

The Gibbs free energy change for the above reaction at 298 K is x×10-1 kJ mol-1. The value of x is____[nearest integer]

[Given : ECu2+/Cu⊖=0.34 V;ESn2+/Sn⊖=-0.14 V;F=96500Cmol-1]

Options:

The correct answer is:
983

Solution:
Ecell°=ESn2+Sn°-ECu2+Cu°=-0.14-0.34=-0.48

Ecell=Ecell°-0.0592logCu2+Sn2+

=-0.48-0.0295log10-210-3

=-0.48-0.0295

=-0.5095

ΔG=-nFEcell

=-2×96500-0.5095

=98333.5=98.33kJ=983.3×10-1kJ

x=983.3Nearest integer=983

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Curie temperature is the temperature above which

Options:
A Ferromagnetic material becomes diamagnetic material
B Ferromagnetic material becomes paramagnetic material
C Paramagnetic material becomes diamagnetic material
D Paramagnetic material becomes ferromagnetic material

The correct answer is:
Ferromagnetic material becomes paramagnetic material

Solution:
Above curie temperature domains break down, hence ferromagnetic substances become paramagnetic.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2008 (Mains)
Curie temperature is the temperature above which

Options:
A Ferromagnetic material becomes diamagnetic material
B Ferromagnetic material becomes paramagnetic material
C Paramagnetic material becomes diamagnetic material
D Paramagnetic material becomes ferromagnetic material

The correct answer is:
Ferromagnetic material becomes paramagnetic material

Solution:
Above curie temperature domains break down, hence ferromagnetic substances become paramagnetic.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Curie temperature is the temperature above which

Options:
A
a ferromagnetic material becomes paramagnetic

B
a paramagnetic material becomes diamagnetic

C
a ferromagnetic material becomes diamagnetic

D
a paramagnetic material becomes ferromagnetic

The correct answer is:
a ferromagnetic material becomes paramagnetic

Solution:
No Solution Available

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2003
Curie temperatures is the temperature above which

Options:
A ferromagnetic material becomes paramagnetic material
B paramagnetic material becomes diamagnetic material
C paramagnetic materíal becomes ferromagnetic material
D ferromagnetic material becomes diamagnetic material

The correct answer is:
ferromagnetic material becomes paramagnetic material

Solution:
Ferromagnetism decreases with rise in temperature. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the Curie temperature of the substance.
Note: The Curie temperature of iron is $770^{\circ} \mathrm{C}$ and that of nickle in $358^{\circ} \mathrm{C}$.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Current I through a given $\mathrm{p}-\mathrm{n}$ junction when a voltage $\mathrm{V}$ is applied across it is given to be $\mathrm{I}=\mathrm{I}_0\left(e^{\frac{V}{2 V_T}}-1\right)$ where $I_0$ and $V_T$ are constants. If $r_d(I)$ is the dynamic resistance of the junction, then $r_d\left(1000 I_0\right)=\alpha r_d\left(10 I_0\right)$, where $\alpha$ is approximately equal to

Options:
A 10
B $1 / 10$
C $1 / 100$
D $1 / 1000$

The correct answer is:
$1 / 100$

Solution:
We have
Dynamic resistance $r_d(I)=\frac{\text { Change in voltage }}{\text { Change in current }}$
Give $r_d\left(1000 I_0\right)=\alpha r_d\left(100 I_0\right)$
$\frac{\Delta \mathrm{V}}{1000 \mathrm{I}_0\left(\mathrm{e}^{\frac{\mathrm{V}}{2 \mathrm{~V}_{\mathrm{T}}}}-1\right)}=\frac{\alpha \Delta \mathrm{V}}{10 \mathrm{I}_0\left(\mathrm{e}^{\frac{\mathrm{V}}{2 \mathrm{~V}_{\mathrm{T}}}}-1\right)}$
$\Rightarrow \quad \alpha=\frac{1}{100}$

Topic: Physics, Semiconductors, JEE Main
Current in the circuit is wattless, if

Options:
A Inductance in the circuit is zero
B Resistance in the circuit is zero
C Current is alternating
D Resistance and inductance both are zero

The correct answer is:
Resistance in the circuit is zero

Solution:
Current in the circuit is wattless,
Because power $=i^2 R$, if $R=0$, then $P=0$.

Topic: Physics, Alternating Current, JEE Main
Current is flowing in conducting wire shaped as shown in the figure. The radius of the curved part is R and the length of straight portion is very large. The value of the magnetic field at the centre O is B=Î¼0IÏRkÏ+14, where the value of k is

Options:
A 32
B 38
C 92
D 98

The correct answer is:
38

Solution:
BA=0
BB=Î¼04Ï2Ï-Ï/2IR
BC=Î¼0I4ÏR
â´ B0=BA+BB+BC=Î¼oI4ÏR3Ï2+1
B=Î¼oIÏR38Ï+14
B=Î¼oIÏRÎºÏ+14
When Îº=38

Topic: Physics, Magnetic Effects of Current, JEE Main
current is passed so that the amount of heat liberated is $H_1$ and $H_2$ respectively. The ratio $\frac{H_1}{H_2}$ has the value

Options:
A $\frac{R_2}{R_1}$
B $\frac{R_1}{R_2}$
C $\frac{R_1^2}{R_2^2}$
D $\frac{R_2^2}{R_1^2}$

The correct answer is:
$\frac{R_2}{R_1}$

Solution:
Heat produced $H=\frac{V^2 t}{4.2 R}=H \propto \frac{1}{R}$ Hence $\frac{H_1}{H_2}=\frac{R_2}{R_1}$

Topic: Physics, Current Electricity, JEE Main
Current measured by the ammeter A in the reported circuit when no current flows through 10Ω resistance, will be _____ A.

Options:

The correct answer is:
10

Solution:
This is a Wheatstone circuit. For the current to be zero in 10 Ω the Wheatstone circuit should be balanced. Therefore,

R3=46⇒R=2 Ω

Now, the equivalent resistance across the battery will be,

Req=2+4×3+62+4+3+6=3.6 Ω

Therefore, the current measured by the ammeter is i=VR=363.6=10 A.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Current sensitivity of a moving coil galvanometer is $5 \mathrm{div} / \mathrm{mA}$ and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

Options:
A $40 \Omega$
B $25 \Omega$
C $250 \Omega$
D $500 \Omega$

The correct answer is:
$250 \Omega$

Solution:
Given that,
Current sensitivity $\left(S_j\right)=5 \mathrm{div} /(\mathrm{mA})=5 \times 10^3 \mathrm{div} / \mathrm{A}$
Voltage sensitivity $\left(S_V\right)=20 \operatorname{div} / \mathrm{V}$
Since Voltage sensitivity $\left(S_V\right)=S_i / R$
Resistance $(R)=S_{/} / S_V=\left(5 \times 10^3 \mathrm{div} / \mathrm{A}\right) /(20 \mathrm{div} / \mathrm{N})=250 \Omega$
So option 3 is correct.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Current through \(\mathrm{ABC}\) and \(\mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime}\) is \(\mathrm{I}\). What is the magnetic field at \(\mathrm{P}\) ? \(\mathrm{BP}=\mathrm{PB}^{\prime}=\mathrm{r}\left(\right.\) Here \(\mathrm{C}^{\prime} \mathrm{B}^{\prime} \mathrm{PBC}\) are collinear)

Options:
A \(\mathrm{B}=\frac{1}{4 \pi} \frac{2 \mathrm{I}}{\mathrm{r}}\)
B \(\mathrm{B}=\frac{\mu_0}{4 \pi}\left(\frac{2 \mathrm{I}}{\mathrm{r}}\right)\)
C \(\mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi}\left(\frac{\mathrm{I}}{\mathrm{r}}\right)\)
D Zero

The correct answer is:
\(\mathrm{B}=\frac{\mu_0}{4 \pi}\left(\frac{2 \mathrm{I}}{\mathrm{r}}\right)\)

Solution:
Hints:
\(\mathrm{B}=2\left[\frac{\mu_0}{4 \pi} \frac{I}{\mathrm{r}}\right]\)

Topic: Physics, Magnetic Effects of Current, JEE Main
Currents of a 10 ampere and 2 ampere are passed through two parallel thin wires $A$ and $B$ respectively in opposite directions. Wire $A$ is infinitely long and the length of the wire $B$ is $2 \mathrm{~m}$. The force acting on the conductor $B$, which is situated at $10 \mathrm{~cm}$ distance from $A$ will be

Options:
A
$8 \times 10^{-5} \mathrm{~N}$

B
$5 \times 10^{-5} \mathrm{~N}$

C
$8 \pi \times 10^{-7} \mathrm{~N}$

D
$4 \pi \times 10^{-7} \mathrm{~N}$

The correct answer is:
$8 \times 10^{-5} \mathrm{~N}$

Solution:
Force acting on conductor $B$ due to conductor $A$ is given by relation $F=\frac{\mu_0 I_1 I_2 l}{2 \pi r}$
$l$-length of conductor $B$
$r$-distance between two conductors
$$
\therefore F=\frac{4 \pi \times 10^{-7} \times 10 \times 2 \times 2}{2 \times \pi \times 0.1}=8 \times 10^{-5} \mathrm{~N}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (26 May Online)
Curve $16 x^2+8 x y+y^2-74 x-78 y+212=0$ represents

Options:
A Parabola
B Hyperbola
C Ellipse
D None of these

The correct answer is:
Parabola

Solution:
$\Delta \neq 0, h^2=a b$ i.e., parabola

Topic: Mathematics, Parabola, JEE Main
Curve $16 x^2+8 x y+y^2-74 x-78 y+212=0$ represents

Options:
A Parabola
B Hyperbola
C Ellipse
D None of these

The correct answer is:
Parabola

Solution:
$\Delta \neq 0, h^2=a b$ i.e., parabola.

Topic: Mathematics, Ellipse, JEE Main
Curve $x y=c^2$ is said to be

Options:
A Parabola
B Rectangular hyperbola
C Hyperbola
D Ellipse

The correct answer is:
Rectangular hyperbola

Solution:
$x y=c^2 \cdot$ Rectangular hyperbola $a^2=b^2$

Topic: Mathematics, Hyperbola, JEE Main
Curved surfaces of a plano-convex lens of refractive index μ1 and a plano-concave lens of refractive index μ2 have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses

Options:
A 1μ2-μ1
B 1μ1-μ2
C μ2-μ1
D μ1-μ2

The correct answer is:
μ1-μ2

Solution:
1f1=μ1-11R

1f2=μ2-11-R⇒1feq=1f1+1f2=μ1-1R-μ2-1R

1feq=μ1-μ2R⇒Rfeq=μ1-μ2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Cuscuta is an example of

Options:
A ectoparasitism
B brood parasitism
C predation
D endoparasitism

The correct answer is:
ectoparasitism

Solution:
Cuscuta (Amarbel) is a total stem parasite that grows on a number of plants like Citrus, Duranta, Zizyphus, etc. It sends a number of haustoria into the host. Xylem and phloem of haustorium make connection with sap and food channel of host.
Predation is an interaction in which member of one species capture, kill and eat up member of other species. Example insectivorous plants. Endoparasite lives inside the host body.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2012 (Mains)
CuSO 4 (aq) → H 2 S↑ M↓ → Excess of KCN N + O
Then final product N and O are respectively.

Options:
A [Cu(CN)4]3-,(CN)2
B CuCN,(CN)2
C [Cu(CN)4]2-,(CN)2
D Cu(CN)2,K2S

The correct answer is:
[Cu(CN)4]3-,(CN)2

Solution:
CuSO 4 (aq) → H 2 S 2CuS Black ppt. (M) → KCN Excess [Cu(CN) 4 ] 3− N  +  (CN) 2 O ↑

Topic: Chemistry, d and f Block Elements, JEE Main
Cyanide process is used for the extraction of

Options:
A
barium

B
aluminium

C
boron

D
silver

The correct answer is:
silver

Solution:
Forms a soluble complex which is precipitated with zinc

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2002
Cyanohydrin of which of the following compound can forms lactic acid

Options:
A CH3CH2CHO
B CH3CHO
C HCHO
D CH3COCH3

The correct answer is:
CH3CHO

Solution:
We know that carbonyl compound react with cyanide ion to form cyanohydrin. This on further hydrolysis gives carboxylic acid.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Cyclic photophosphorylation results in the formation of :

Options:
A ATP
B $\mathrm{NADPH}$
C ATP and NADPH
D ATP, NADPH and $\mathrm{O}_2$

The correct answer is:
ATP

Solution:
No Solution Available

Topic: Biology, Respiration in Plants, NEET, NEET 2009 (Mains)
Cyclo [18] carbon is an allotrope of carbon with molecular formula $\mathrm{C}_{18}$. It is a ring of 18 carbon atoms, connected by single and triple bonds. The total number of triple bonds present in this cyclocarbon are

Options:
A 9
B 10
C 12
D 6

The correct answer is:
9

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Cyclohexane, a hydrocarbon floats on water because

Options:
A It is immiscible with water
B Its density is low as compared to water
C It is non-polar substance
D It is immiscible and lighter than water

The correct answer is:
It is immiscible and lighter than water

Solution:
Cyclohexane, is immiscible and lighter than water. Hence, floats on the surface of water.

Topic: Chemistry, Hydrocarbons, JEE Main
Cyclohexene is _________ type of an organic compound.

Options:
A Benzenoid aromatic
B Benzenoid non-aromatic
C Acyclic
D Alicyclic

The correct answer is:
Alicyclic

Solution:
Benzenoid aromatic compounds: The benzenoid aromatic compounds are referred to as those compounds which contain only benzene rings in the whole structure. The benzene ring may be one or more than one. Examples of Benzenoid aromatic compounds are benzene-benzene contains a single ring, naphthalene- naphthalene contains two benzene rings.

Non-benzenoid aromatic compounds: The non-benzenoid aromatic compounds are referred to as those compounds which do not contain any benzene ring in their structure. Example Furan- Furan is a five membered ring which contains four carbon atoms and one oxygen atom.
For a compound to be aromatic it should follow Huckel’s rule. The Huckel rule says that for a ring to be aromatic in nature it should follow the given rules shown below.
(1) The ring should be planar
(2) The ring should be complete delocalized by pi-electrons
(3) The compound should possess

4n+2π electrons in the ring.

Acyclic compounds are just opposite to cyclic compounds because their molecules don't form any ring. It is called open-chain compounds because they have a linear structure.

Alicyclic compounds are organic compounds that are both aliphatic and cyclic. These are the saturated or unsaturated,hydrocarbons containing non-aromatic rings of carbon atoms. Alicyclic compounds may have one or more aliphatic side chains attached.

is Alicyclic

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Cyclohexene Product
The correct name of the product obtained is:

Options:
A cis-1,2-dibromocyclohexane
B cis-1,4 -dibromocyclohexane
C trans-1,2-dibromocyclohexane
D trans-1,4-dibromocyclohexane

The correct answer is:
trans-1,2-dibromocyclohexane

Solution:
Cyclohexene reacts with the bromine and forms the brominium cation with three membered ring. After that the another bromine react from the another plane and formed the trans-1,2-dibromocyclohexane as a product.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Cyclohexylamine when treated with nitrous acid yields (P). On treating (P) with PCC results in (Q). When (Q) is heated with dil. NaOH we get (R) The final product (R) is:

Options:
A
B
C
D

The correct answer is:

Solution:
Cyclohexylamine when treated with nitrous acid yields cyclohexanol.On treating cyclohexanol with PCC which is a mild oxidising agent and oxidises the alcohols into aldehyde (in case of primary alcohol) or ketones (in case of secondary alcohol) in the presence of dichloromethane results cyclohexanone. When cyclohexanone is heated with dilute NaOH dehydration of aldol takes place by the removal of water molecules we get the final product.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Cyclopentadienyl anion is

Options:
A benzenoid and aromatic
B non-benzenoid and aromatic
C non-benzenoid and non-aromatic
D non-benzenoid and anti- aromatic

The correct answer is:
non-benzenoid and aromatic

Solution:
Cyclopentadienyl anion. It is cyclic, planar with $(4 n+2) \pi e^{-}$. Therefore aromatic but structure is not similar to benzene.

Topic: Chemistry, Coordination Compounds, JEE Main
Cylindrical rod of copper of length $2 \mathrm{~m}$ and cross-sectional area $2 \mathrm{~cm}^2$ is insulated at its curved surface. The one end of rod is maintained in steam chamber and other is maintained in ice at $0^{\circ} \mathrm{C}$. The thermal conductivity of copper is $386 \mathrm{Js}^{-1} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ ). Find the temperature at a point which is at a distance of $120 \mathrm{~cm}$ from the colder end.

Options:
A $80^{\circ} \mathrm{C}$
B $50^{\circ} \mathrm{C}$
C $60^{\circ} \mathrm{C}$
D $70^{\circ} \mathrm{C}$

The correct answer is:
$60^{\circ} \mathrm{C}$

Solution:
For cylindrical wire,
$$
\begin{aligned}
l & =2 \mathrm{~m}, A=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2 \\
\theta_1 & =100^{\circ} \mathrm{C}, \theta_2=0^{\circ} \mathrm{C} \\
\theta & =?
\end{aligned}
$$
At steady state, temperature gradient of rod remains constant.
$$
\begin{array}{ll}
\text { i.e., } & \frac{\theta_1-\theta}{x_1}=\frac{\theta-\theta_2}{x_2} \\
\Rightarrow & \frac{100^{\circ}-\theta}{80 \times 10^{-2}}=\frac{\theta-0^{\circ}}{120 \times 10^{-2}} \\
\Rightarrow & 100^{\circ}-\theta=\frac{80}{120} \times \theta \\
\Rightarrow & 100^{\circ}-\theta=\frac{2}{3} \theta \Rightarrow \frac{5 \theta}{3}=100^{\circ} \\
\Rightarrow & \theta=\frac{100^{\circ} \times 3}{5}=60^{\circ} \mathrm{C}
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
Cystic fibrosis is a/an

Options:
A sex-linked recessive disorder
B autosomal dominant disorder
C autosomal recessive disorder
D sex-linked dominant disorder

The correct answer is:
autosomal recessive disorder

Solution:
Mendelian disorders are mainly determined by alteration or mutation in the single gene. These disorders are transmitted to the offspring on the same lines as we have studied in the principle of inheritance. The pattern of inheritance of such Mendelian disorders can be traced in a family by the pedigree analysis. Most common and prevalent Mendelian disorders are Haemophilia, Cystic fibrosis, Sickle-cell anaemia, Colour blindness, Phenylketonuria, Thalassemia, etc.
Cystic fibrosis, sickle-cell anaemia and phenylketonuria are examples of autosomal recessive disorders, while haemophilia and colour blindness are sex-linked disorders (both of them are X-linked recessive disorders). Examples of autosomal dominant includes myotonic dystrophy.

Topic: Biology, Principles of Inheritance and Variation, NEET
Cytoskeleton is made up of :

Options:
A Proteinaceous filaments
B Calcium carbonate granules
C Callose deposits
D Cellulose microfibrils

The correct answer is:
Proteinaceous filaments

Solution:
No Solution Available

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2009 (Mains)
d1 and d2 are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an α particle is approaching a gold nucleus. For d1=xd2, the value of x will be _____ .

Options:

The correct answer is:
3

Solution:
The relation between impact parameter and scattering angle is b=kze2Ecotθ2

⇒b∝cotθ2

⇒d1∝cot60°2

⇒d2∝cot90°2

d1d2=cot60°2cot90°2=3⇒d1=3d2⇒x=3

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
D + - 2-Chloro-2-phenylethane gets racemises slowly when small amount of SbCl5, is added to a solution in toluene, it due to the formation of

Options:
A Carbanion
B Carbocation
C Free radical
D Carbene

The correct answer is:
Carbocation

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
dGdpT=

Options:
A V
B S
C - S
D - V

The correct answer is:
V

Solution:
We know that,
Gibbs free energy, G=H-TS
and H=E+pV
So G=E+pV-TS
On differentiating, we get
dG=dE+pdV+Vdp-TdS-SdT ...(i)
For an infinitesimal stage of a reversible process,
dq=dE+dW
and dS=dqT ∵W=pdV
TdS=dE+pdV ...(ii)
From equation (i) and (ii), we get
dG=dE+pdV+Vdp-dE-pdV-SdT
=Vdp-SdT
At constant temperature, dT=0
So dG=Vdp
dGdpT=V

Topic: Chemistry, Thermodynamics (C), JEE Main
$\alpha-D-(+)$-glucose and $\beta-D-(+)-g l u c o s e ~ a r e$

Options:
A
conformers

B
epimers

C
anomers

D
enantiomers

The correct answer is:
anomers

Solution:
$\alpha-D(+)$ glucose and $\beta-D(+)$ glucose are anomers.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2008
D−glucose → Tollen′s reagent A ;
D−glucose → Br 2 -water B
A and B are:

Options:
A gluconic acid, gluconic acid
B glucaric acid, gluconic acid
C salt of gluconic acid, gluconic acid
D salt of gluconic acid, glucaric acid

The correct answer is:
salt of gluconic acid, gluconic acid

Solution:
Tollen's reagent is a mild oxidising agent and easily oxidises the aldehyde group, but can not oxidise the ketone or alcohol group and form salt of gluconic acid.

Br2/H2O is a mild oxidising agent and easily oxidises the aldehyde group, but cannot oxidise the ketone or alcohol group.

Topic: Chemistry, Biomolecules, JEE Main
D is

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction sequence is as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
$\frac{d^3 y}{d x^3}+2\left[1+\frac{d^2 y}{d x^2}\right]=1$ has degree and order as

Options:
A degree $=1$, order $=3$
B degree $=2$, order $=3$
C degree $=3$, order $=3$
D degree $=1$, order $=2$

The correct answer is:
degree $=1$, order $=3$

Solution:
Clearly degree $=1$, order $=3$.

Topic: Mathematics, Differential Equations, JEE Main
$\frac{d}{d x}\left[\cos \left(1-x^2\right)^2\right]=$

Options:
A $-2 x\left(1-x^2\right) \sin \left(1-x^2\right)^2$
B $-4 x\left(1-x^2\right) \sin \left(1-x^2\right)^2$
C $4 x\left(1-x^2\right) \sin \left(1-x^2\right)^2$
D $-2\left(1-x^2\right) \sin \left(1-x^2\right)^2$

The correct answer is:
$4 x\left(1-x^2\right) \sin \left(1-x^2\right)^2$

Solution:
$\begin{aligned} \frac{d}{d x}\left[\cos \left(1-x^2\right)^2\right]=- & \sin \left(1-x^2\right)^2 \frac{d}{d x}\left(1-x^2\right)^2 \\ = & 4 x\left(1-x^2\right) \sin \left(1-x^2\right)^2\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left\{\cos \left(\sin x^2\right)\right\}=$

Options:
A $\sin \left(\sin x^2\right) \cdot \cos x^2 \cdot 2 x$
B $-\sin \left(\sin x^2\right) \cdot \cos x^2 \cdot 2 x$
C $-\sin \left(\sin x^2\right) \cdot \cos ^2 x \cdot 2 x$
D None of these

The correct answer is:
$-\sin \left(\sin x^2\right) \cdot \cos x^2 \cdot 2 x$

Solution:
$\frac{d}{d x}\left\{\cos \left(\sin x^2\right)\right\}=-\sin \left(\sin x^2\right) \cos x^2 \cdot 2 x$.

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left\{e^{-a x^2} \log (\sin x)\right\}=$

Options:
A $e^{-a x^2}(\cot x+2 a x \log \sin x)$
B $e^{-a x^2}(\cot x+a x \log \sin x)$
C $e^{-a x^2}(\cot x-2 a x \log \sin x)$
D None of these

The correct answer is:
$e^{-a x^2}(\cot x-2 a x \log \sin x)$

Solution:
$\begin{aligned} & \frac{d}{d x}\left\{e^{-e x^2} \log (\sin x)\right\} \\ & =e^{-a x^2}(-2 a x) \cdot \log (\sin x)+e^{-a x^2} \cot x \\ & =e^{-e x^2}[\cot x-2 a x \log (\sin x)] .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} e^{x \sin x}=$

Options:
A $e^{x \sin x}(x \cos x+\sin x)$
B $\mathrm{e}^{x \sin x}(\cos x+x \sin x)$
C $e^{x \sin x}(\cos x+\sin x)$
D None of these

The correct answer is:
$e^{x \sin x}(x \cos x+\sin x)$

Solution:
$\begin{aligned} & \text { Let } y=e^{x \sin x} \Rightarrow \log y=x \sin x \\ & \therefore \frac{1}{y} \frac{d y}{d x}=\sin x+x \cos x \text { or } \frac{d y}{d x}=e^{x \sin x}(\sin x+x \cos x) \text {. }\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main