datavorous/entrance-exam-dataset huggingface dataset:
In the given figure, each diode has a forward bias resistance of 30 Ω and infinite resistance in reverse bias. The current I1 will be :

Options:
A 2.0 A
B 3.75 A
C 2.73 A
D 2.35 A

The correct answer is:
2.0 A

Solution:
We know that in reverse bias ideal diode does not pass current it only works in forward bias so, the current in the third branch is zero, it means the net resistance of the upper two branch is, 1Req=1160+1160⇒Req=80 Ω

⇒R'eq=80+20=100 Ω

⇒i=200100=2 A

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (01 Sep Shift 2)
In the given figure, if i1=3sinÏt and i2=4cosÏt, then i3 is

Options:
A 5 sin Ït+53Â°
B 5 sin ωt+37°
C 5 sin ωt+45°
D 5 cos ωt+53°

The correct answer is:
5 sin Ït+53Â°

Solution:
i3=i1+i2

âi3 = 3 sinÏt + 4 cosÏt

âi3 = 5 35 sinÏt + 45 cosÏt

âi3 = 5 cos53Â° sinÏt + sin53Â° cosÏt

âi3=5 sin Ït+53Â°

Topic: Physics, Current Electricity, JEE Main
In the given figure, if the velocity of the block C at a particular instant is -20 m s-1j^, then the velocity of the rod (A) at that instant will be : (string is attached to wedge)

Options:
A 15 m s-1j^
B -15 m s-1j^
C 20 m s-1j^
D -20 m s-1j^

The correct answer is:
15 m s-1j^

Solution:
In the given figure, if the velocity
l1'+l2'=0
20-u=0
u=20 m s-1
By wedge constraint
usin37∘=v cos37∘
v=u×34=15 m s-1
v→=15j^

Topic: Physics, Laws of Motion, JEE Main
In the given figure magnetic field at the center of ring O is 82 T. Now it is turned through 90° about xx' axis, so that two semicircular parts are mutually perpendicular. The new value of magnetic field (in Tesla) at centre is n. Find n.

Options:

The correct answer is:
16

Solution:
Given, μ0i2R=82 T

In 2nd case,
B→=μ0i4R2+μ0i4R2=μ0i22R=4 T

Topic: Physics, Magnetic Effects of Current, JEE Main
In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be _____ cm.

Options:

The correct answer is:
40

Solution:
Even if the radius of wire is doubled, the balancing point would not change asxl-x=R1R2 , which is independent of area in case of uniform wire. Thus, the balancing length remains the same i.e., 40 cm.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jul Shift 2)
In the given figure $\mathrm{R}_1=10 \Omega, \mathrm{R}_2=8 \Omega, \mathrm{R}_3=4 \Omega$ and $\mathrm{R}_4=8 \Omega$. Battery is ideal with emf $12 \mathrm{~V}$. Equivalent resistant of the circuit and current supplied by battery are respectively :

Options:
A $10.5 \Omega$ and $1.14 \mathrm{~A}$
B $12 \Omega$ and $1 \mathrm{~A}$
C $10.5 \Omega$ and $1 \mathrm{~A}$
D $12 \Omega$ and $11.4 \mathrm{~A}$

The correct answer is:
$12 \Omega$ and $1 \mathrm{~A}$

Solution:
Here $\mathrm{R}_2, \mathrm{R}_3, \mathrm{R}_4$ are in parallel
$\begin{aligned}
& \frac{1}{\mathrm{R}_{234}}=\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}+\frac{1}{\mathrm{R}_4} \\
& \mathrm{R}_{234}=2 \Omega
\end{aligned}$
$R_{234}$ is in series with $R_1$ so
$\begin{aligned}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{234}+\mathrm{R}_1=2+10=12 \Omega \\
& \mathrm{i}=\frac{12}{12}=1 \mathrm{Amp}
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (05 Apr Shift 1)
In the given figure switches S1 and S2 are in open condition. The resistance across ab when the switches S1 and S2 are closed is ________Ω.

Options:

The correct answer is:
10

Solution:
when switch S1 and S2 are closed

12×612+6+2+6×126+12

7218+2+7218=4+2+4=10 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (20 Jul Shift 2)
In the given figure the area of the frictionless piston is 1 m2 and it is being held in equilibrium by an external force. Now the external force is slowly removed and heat is supplied to the gas due to which the piston displaces by L2 where L=1 m. The natural length of the ideal springs is L=1 m. Spring constant is given as K=100 N m-1. The pressure of the gas in the final situation is

Options:
A 50 N m-2
B 100 N m-2
C 200 N m-2
D 400 N m-2

The correct answer is:
100 N m-2

Solution:
In the case of equilibrium, net force on the piston becomes zero. So,
PA-Kx=Kx⇒P=2KxA
=2×1001×12
=100 N/m2

Topic: Physics, Thermodynamics, JEE Main
In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6 Ω. Calculate the power dissipated in the whole circuit:

Options:
A 1.32 W
B 4.4W
C 0.65 W
D 2.2 W

The correct answer is:
2.2 W

Solution:
The above circuit in the question is arranged like that, so all the resistance are connected in parallel and the net resistance of the parallel resistor becomes,

1RP=1R1+1R2+1R3+1R4=14+16+112+18⇒RP=4830 Ω

Now the net resistance will be, Rnet=4830+0.6 Ω, so the net power, P=V2R=(2.2)22.2=2.2 W

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (26 Aug Shift 1)
In the given figure, the graph of \( y=p(x)=x^{4}+a x^{3}+b x^{2}+c x+d \) is given.
The product of all the imaginary roots of \( p(x)=0 \) is

Options:
A \( 1 \)
B \( 2 \)
C \( \frac{1}{3} \)
D \( \frac{1}{4} \)

The correct answer is:
\( 1 \)

Solution:
Given,

px=x4+ax3+bx2+cx+d

It is passing through 1,0

Hence, p1=0

⇒a+b+c+d=0

It is passing through 0, 2. Hence, we get,

⇒p0=2

⇒d=2

The product of all the roots of the bi-quadratic equation is given as,

a constant term/coefficient of x4=2

From the graph given, we can see that the real roots of px are given as,

x=1, 2 (as it intersects the real axis at 1, 0 and 2, 0)

Thus, the product of the real roots is 2.

Hence, the product of the imaginary roots can be given as,

Product of all the roots/Product of real roots.

=22

=1

Topic: Mathematics, Quadratic Equation, JEE Main
In the given figure the magnetic flux through the loop increases according to the relation ϕB(t)=10t2+20t, where ϕB is in milliwebers and t is in seconds. The magnitude of current through R=2 Ω resistor at t=5 s is _____ mA.

Options:

The correct answer is:
60

Solution:
|ϵ|=dϕdt=20t+20 mV

|i|=|ϵ|R=10t+10 mA

At t=5

|i|=60 mA

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (27 Jul Shift 2)
In the given figure the total charge stored in the combination of capacitors is 100 µC. The value of ‘x’ is ______________.

Options:

The correct answer is:
5

Solution:
All the capacitors are in parallel combination and therefore potential drop across each capacitor would be the same.

Charge on the capacitor can be written as, Q=CV.

The charge on C1 is Q1=2×10=20 μC

The charge on C2 is Q2=x×10=10x μC

The charge on C3 is Q3=3×10=30 μC

By the given data.

C1+C2+C3=100 μC

⇒102+x+3=100

⇒x=5

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (15 Apr Shift 1)
In the given figure the work done by the battery after the switch S is closed is

Options:
A 1 0 0 μ J
B - 1 0 0 μ J
C 8 0 μ J
D - 8 0 μ J

The correct answer is:
8 0 μ J

Solution:
Before the switch S is closed
Q i = C eq V
= 5 × 4 = 2 0 μ C
When the switch S is closed
Q f = CV
= 1 0 × 4 = 4 0 μ C
W cell = Δ QE = Q f - Q i × E
= 2 0 × 4 = 8 0 μ J

Topic: Physics, Electrostatics, JEE Main
In the given figure, there is a circuit of potentiometer of length AB=10 m. The resistance per unit length is 0.1 Ω per cm. Across AB, a battery of emf E and internal resistance r is connected. The maximum value of emf measured by this potentiometer is:

Options:
A 5 V
B 2.25 V
C 6 V
D 2.75 V

The correct answer is:
5 V

Solution:
Maximum voltage that can be measured by this potentiometer will be equal to potential drop across AB,

RAB=10×0.1×100=100 ohm

∴ VAB=620+100×100=6×100120=5 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (25 Jul Shift 1)
In the given figure, three capacitors each of capacitance \( 6 \mathrm{pF} \) are connected in series. The total capacitance of the combination becomes

Options:
A \( 2 \times 10^{-12} \mathrm{~F} \)
B \( 3 \times 10^{-12} F \)
C \( 6 \times 10^{-12} F \)
D \( 9 \times 10^{-12} F \)

The correct answer is:
\( 2 \times 10^{-12} \mathrm{~F} \)

Solution:
The reciprocal of equivalent capacitance of any number of capacitors joined in series is equal to sum of the reciprocals of individual capacitances.
For n capacitors connected in series, total capacitance would be
1CS=∑i=1i=n1Ci
Or, 1CS=1C1+1C2+1C3+…
∴1Cs=16+16+16=36=12
∴CS=2 pF=2×10-12 F

Topic: Physics, Capacitance, JEE Main
In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times that of Q. In the case of the same rotational kinetic energy, the ratio of rotational inertias I1I2 will be x: 1 . The value of x will be ______.

Options:

The correct answer is:
9

Solution:
12I1ω12=12I2ω22

I1v3R2=I2vR2

I1I2=91

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (27 Jul Shift 2)
In the given figure, what is the angle of prism

Options:
A $A$
B $B$
C $C$
D $D$

The correct answer is:
$C$

Solution:
Correct option is 3. $C$
Angle of prism is the angle between incident and emergent surfaces.

Topic: Physics, Ray Optics, JEE Main
In the given figure when galvanometer shows no deflection current flowing through \( 5 \Omega \) resistance will be

Options:
A \( 0.5 A \)
B \( 0.6 A \)
C \( 1.5 A \)
D \( 2.0 \mathrm{~A} \)

The correct answer is:
\( 0.6 A \)

Solution:
Let i be the current through the arm ADC. Then current through the arm ABC=2.1-i. As there is no deflection in the galvanometer, the two arms are parallel to each other, therefore their potential drop across the arm will be same. Hence,
20+5i=8+2(2.1-i)

⇒25i=21-10i⇒35i=21
⇒i=2135=35=0.6A

Topic: Physics, Current Electricity, JEE Main
In the given figure, when key $K$ is opened, the reading of the ammeter $A$ will be

Options:
A $50 \mathrm{~A}$
B $2 \mathrm{~A}$
C $0.5 \mathrm{~A}$
D $\frac{10}{9} A$

The correct answer is:
$2 \mathrm{~A}$

Solution:
The circuit will be as shown
$i=\frac{10}{5}=2 A$

Topic: Physics, Current Electricity, JEE Main
In the given network of ideal cells and resistors, R1=1.0 Î©, R2=2.0 Î©, E1=2 V and E2=E3=4 V. The potential difference between the point a and b is

Options:
A 83 V
B 103 V
C 94 V
D 134 V

The correct answer is:
103 V

Solution:
The equivalent emf of the combination can be calculated by applying the parallel combination of batteries

Vab=εeq=ε1r1+ε2r2+ε3r31r1+1r2+1r3

⇒Vab=εeq=22+42+4212+12+12

⇒Vab=εeq=103 V

Topic: Physics, Current Electricity, JEE Main
In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells ε1ε2 is:

Options:
A 53
B 85
C 43
D 32

The correct answer is:
53

Solution:
E1=kℓ1 …(1)

E1+E2=kℓ2 …(2)

E1E1+E2=ℓ1ℓ2=250400=58

8E1=5E1+5E2

3E1=5E2

E1E2=53

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (25 Jul Shift 2)
In the given reaction 3-Bromo-2, 2-dimethyl butane ⟶C2H5OH'A'(Major Product) Product A is:

Options:
A 2-Ethoxy-3, 3-dimethyl butane
B 1-Ethoxy-3, 3 -dimethyl butane
C 2-Ethoxy-2, 3-dimethyl butane
D 2-Hydroxy-3, 3-dimethyl butane

The correct answer is:
2-Ethoxy-2, 3-dimethyl butane

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (20 Jul Shift 1)
In the given reaction

as main product, 'X' will be

Options:
A
B
C
D (CH3)2COOCH3

The correct answer is:

Solution:
The reaction is Baeyar-villiger oxidation. In this reaction the migrating power of 2° -alkyl is more than the 1° -alkyl group.

Topic: Chemistry, Aldehydes and Ketones, NEET
In the given reaction

C6H5-O-CH2-CH3âExcess HI/Î X+Y

X and Y will respectively be

Options:
A C6H5I and CH3-CH2-I
B C6H5-OH and CH3-CH2-I
C C6H5I and CH3-CH2OH
D C6H5OH and CH2=CH2

The correct answer is:
C6H5-OH and CH3-CH2-I

Solution:
This reaction occurs by SN2 mechanism as follows to give phenol and ethyl iodide

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the given reaction

CH2=CH-CHO→[X]CH2=CH-CH2OH

[X] will be

Options:
A H2,Ni
B H2,Pt
C NaBH4
D H2 /Wilkinson catalyst

The correct answer is:
NaBH4

Solution:
CH2=CH-CHO→NaBH4CH2=CH-CH2OH
NaBH4 is selective reagent for the reduction of only carbonyl group.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the given reaction

CH3-CH2COOHââi CH3Li excessii H3OââX

[X] will be

Options:
A CH3-CH2COOLi
B CH3-CH2-C||O-CH3
C CH3-CH2-CH2OH
D CH3-CH2-CH2-CH3

The correct answer is:
CH3-CH2-C||O-CH3

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the given reaction
+  CH 3 −CHO → (i) NaOH (ii) Δ [X]
[X] will be

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the given reaction
CH3âCOOHâii NaCN(iii) H2O/Hâi RedâP+Br2X
'X' will be

Options:
A
B COOH-CH2-CH2-COOH
C
D

The correct answer is:

Solution:
Here the final product is malonic acid

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the given reaction cycle

X, Y and Z respectively are

Options:
A X-CaCO3, Y-NaCl, Z-KCl
B X-CaCO3, Y-NaCl, Z-HCl
C X-CaO, Y-NaCl+CO2, Z-NaCl
D X-CaO, Y-NaCl+CO2, Z-KCl

The correct answer is:
X-CaCO3, Y-NaCl, Z-HCl

Solution:
When calcium chloride reacts with sodium carbonate, a double replacement reaction takes place. The products of this reaction are calcium carbonate and sodium chloride.CaCl2+Na2CO3→CaCO3X+2NaClY

To Obtain Calcium chloride from Calcium carbonate we need to react calcium carbonate with Hydro chloric acid.

CaCO3X+2HClZ→CaCl2+CO2g+H2Ol

∴ (X) CaCO3

(Y) NaCl

(Z) HCl

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (12 Apr Shift 1)
In the given reaction
→ (i) HCHO/ H ⊕   (ii) HOH  [X] will be

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction is Princ reaction which takes place as follows
(i)
(ii)

Topic: Chemistry, Hydrocarbons, JEE Main
In the given reaction

Major product [X] will be

Options:
A 2-chloro-3-methylbutane
B 1-chloro-3-methylbutane
C 2-chloro-2-methylbutane
D 2-chloropentane

The correct answer is:
2-chloro-2-methylbutane

Solution:
Here the major product X is 2-chloro-2-methylbutane

Topic: Chemistry, Hydrocarbons, JEE Main
In the given reaction
R− CH 2 −COOH → Red  P+Br 2 A → NH 3 Alanine
The name of I st reaction and group R is -

Options:
A Hell Vohlard Zelinsky reaction, - C6H5
B Hell Vohlard Zelinsky reaction, - CH3
C Schimdt Reaction,
D Claisen Schimdt reaction, - CH3

The correct answer is:
Hell Vohlard Zelinsky reaction, - CH3

Solution:
Alanine is
and product in given reaction is
A→R−CH−COOH           ​    |              Br
So R=-CH3
I st reaction is Hell Vohlard Zelinsky reaction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the given reaction sequence

C6H5CHO+CH3-NO2→(i)NaOH(ii)ΔX

[X] will be

Options:
A C6H5-CHOH-CH2-CHO
B C6H5CH2OH
C C6H5-COOH
D C6H5-CH=CH-NO2

The correct answer is:
C6H5-CH=CH-NO2

Solution:
Here the reaction is as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the given reaction sequence, the major product 'C' is :

C8H10⟶HNO3H2SO4A⟶Br2ΔB⟶alcoholicKOHC

Options:
A

B

C

D

The correct answer is:

Solution:
In first step nitration of benzene ring will take place at para position, since ethyl group is ortho-para directing in nature.

Then halogenation reaction will take place at benzyl position followed by dehydrohalogenation reaction with alcoholic KOH.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (24 Jun Shift 1)
In the given reaction sequence

The product C is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
In the given reaction,

,

the major product 'X' will be

Options:
A
B
C
D

The correct answer is:

Solution:
It is a case of intramolecular aldol condensation and the reaction is as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the given reaction,

The number of π electrons present in the product 'P' is ____________.

Options:

The correct answer is:
4

Solution:
In an aldol condensation, an enolate ion reacts with a carbonyl compound in the presence of acid/base catalyst to form a β-hydroxy aldehyde or β-hydroxy ketone, followed by dehydration to give a conjugated enone. It is a useful carbon-carbon bond-forming reaction.

As there are 2 pi bonds in the major product(P), the number of pi electrons will be equal to 4.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (29 Jun Shift 2)
In the given reaction

the number of sp2 hybridised carbon (s) in compound 'X' is _______.

Options:

The correct answer is:
8

Solution:
In the first step oxidation of cyclohexanol to cyclohexanone by acidified potassium dichromate takes place.

In the second step cyclohexanone reacts with grignard reagent followed by hydrolysis to form a tertiary alcohol.

In last step dehydration of tertiary alcohol leads to the formation of

In the above molecule the number of sp2 hybridized carbons are 8.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (29 Jun Shift 2)
In the given reaction,
the product 'A' is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2012 (12 May Online)
In the given reaction

[X] will be

Options:
A
B
C
D

The correct answer is:

Solution:
It involves alkylation through carbanion mechanism

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In the given reaction

[X] will be

Options:
A HCHO
B
C
D HCN

The correct answer is:

Solution:
The reaction occurs as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the given reactions identify A and B.

Options:
A A : 2–Pentyne B : trans – 2 – butene
B A : n – Pentane B : trans – 2 – butene
C A : 2 – Pentyne B : Cis – 2 – butene
D A : n – Pentane B : Cis – 2 – butene

The correct answer is:
A : 2–Pentyne B : trans – 2 – butene

Solution:
In palladium-catalyzed hydrogenations of alkynes, the Lindlar catalyst facilitates the addition of hydrogen to the alkyne to afford the corresponding alkene. When this catalyst is used, the addition of the hydrogen substituent always proceeds via syn addition (addition of the substituent on the same side of the bond).

But-2-yne when treated with sodium in liquid ammonia gives trans-2-butene. The reaction of alkyne with sodium in liquid ammonia produces sodamide in the process converting alkynes to alkenes. Mechanism of Birch Reduction: -Sodium in liquid ammonia produces solvated electrons.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (01 Feb Shift 2)
In the given reactions identify the reagent A and reagent B

Options:
A A-CrO3 B-CrO3
B A- CrO3 B-CrO2Cl2
C A-CrO2Cl2 B-CrO2Cl2
D A-CrO2Cl2 B-CrO3

The correct answer is:
A- CrO3 B-CrO2Cl2

Solution:
Toluene or substituted toluene is converted to benzylidene diacetate on treating with chromic oxide in acetic anhydride. The benzylidene diacetate can be hydrolysed to corresponding benzaldehyde with aqueous acid.

Chromyl chloride oxidises
methyl group to a chromium complex, which on hydrolysis
gives corresponding benzaldehyde. The reaction is known as Etard reaction.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (30 Jan Shift 1)
In the given sequence of reactions

C will be

Options:
A
B
C
D

The correct answer is:

Solution:
Here C is 2-bromo-4methyl amino benzene

Topic: Chemistry, Amines, JEE Main
In the given structure of the human brain below:

Cerebral cortex, thalamus, limbic system, corpora quadrigemina, hypothalamus, brain stem, cerebral aqueduct, cerebellar hemisphere.

How many structures are associated with the forebrain of human?

Options:
A Three
B Four
C Five
D Six

The correct answer is:
Four

Solution:
Forebrain contains cerebrum/cerebral cortex, thalamus, hypothalamus, limbic system, cerebral hemisphere, corpus callosum, etc.
The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passes through the midbrain. The dorsal portion of the midbrain consists mainly of four round swellings (lobes) called corpora quadrigemina.
The hindbrain comprises pons, cerebellum and medulla (also called the medulla oblongata).

Topic: Biology, Neural Control and Coordination, NEET
In the given TLC, the distance of spot A & B are \(5 \mathrm{~cm} ~\&~ 7 \mathrm{~cm}\), from the bottom of TLC plate, respectively.
\(\mathrm{R}_{\mathrm{f}}\) value of \(\mathrm{B}\) is \(x \times 10^{-1}\) times more than \(\mathrm{A}\). The value of \(x\) is ______.

Options:

The correct answer is:
15

Solution:
$\mathrm{R}_{\mathrm{f}}=\frac{\text { Distance moved by substance from base line }}{\text { Distance moved by solvent from base line }}$
$\begin{aligned} & \left(R_f\right)_A=\frac{4}{8} \quad\left(R_f\right)_B=\frac{6}{8} \\ & \frac{\left(R_f\right)_B}{\left(R_f\right)_A}=\frac{6}{8} \times \frac{8}{4} \\ & \left(R_f\right)_B=1.5\left(R_f\right)_A \\ & x=15\end{aligned}$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (09 Apr Shift 2)
In the given transformation, which of the following is the most appropriate reagent?

Options:
A
$\mathrm{NH}_2 \mathrm{NH}_2, \stackrel{(-)}{\mathrm{OH}}$

B
$\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$

C
$\mathrm{Na}$, Liq. $\mathrm{NH}_3$

D
$\mathrm{NaBH}_4$

The correct answer is:
$\mathrm{NH}_2 \mathrm{NH}_2, \stackrel{(-)}{\mathrm{OH}}$

Solution:
$\mathrm{ZnHg} / \mathrm{HCl}$ can't be used due to the presence of acid sensitive group i.e. $\mathrm{OH}$
and $\mathrm{Na} / \mathrm{Liq} . \mathrm{NH}_3$ and $\mathrm{NaBH}_4$ convert $-\mathrm{CO}-$ into $-\mathrm{CH}(\mathrm{OH})-$

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2012 (Offline)
In the given transformation, which of the following is the most appropriate reagent?

Options:
A
$\mathrm{NH}_2 \mathrm{NH}_2, \stackrel{(-)}{\mathrm{OH}}$

B
$\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$

C
$\mathrm{Na}$, Liq. $\mathrm{NH}_3$

D
$\mathrm{NaBH}_4$

The correct answer is:
$\mathrm{NH}_2 \mathrm{NH}_2, \stackrel{(-)}{\mathrm{OH}}$

Solution:
$\mathrm{ZnHg} / \mathrm{HCl}$ can't be used due to the presence of acid sensitive group i.e. $\mathrm{OH}$
and $\mathrm{Na} / \mathrm{Liq} . \mathrm{NH}_3$ and $\mathrm{NaBH}_4$ convert $-\mathrm{CO}-$ into $-\mathrm{CH}(\mathrm{OH})-$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the given transformation, which of the following is the most appropriate reagent?

Options:
A Zn-Hg/HCl
B Na, liq. NH3
C NaBH4
D NH2 - NH2, OH-

The correct answer is:
NH2 - NH2, OH-

Solution:
-OH group and alkene are acid-sensitive groups so Clemmensen reduction cannot be used and NaBH 4 reduces only.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the graph between v and Z for the Mosley' equation, v=aZ-b , the intercept OX is 1 on v axis.

What is the frequency v when the atomic number Z is 52?

Options:
A 7.14 s-1
B 7 s-1
C 2401 s-1
D 2601 s-1

The correct answer is:
2601 s-1

Solution:
v=aZ-ab
ab=1
a=tanâ¡45Â°=1
v=52-1=51
v=512=2601 s-1

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In the graph shown, the potential energy of the earth-satellite system is shown by a solid line as a function of distance \( r \) (the separation between the earth's centre and satellite). The total energy of the two objects which may or may not be bounded to the earth is shown in the figure by dotted lines. Mark the correct statement(s).

Options:
A e object having the total energy \( E_{1} \) is bounded one.
B e object having the total energy \( E_{2} \) is bounded one.
C th the objects are bounded.
D th the objects are unbounded.

The correct answer is:
e object having the total energy \( E_{1} \) is bounded one.

Solution:
TE=PE+KE

For the system to be bounded one, the total energy of the system must be negative.

Therefore, the object having the negative total energy is bounded and if the total energy is positive is then object is unbounded.

Topic: Physics, Gravitation, JEE Main
In the Grignard reaction, which metal forms an organometallic bond?

Options:
A Sodium
B Titanium
C Magnesium
D Palladium

The correct answer is:
Magnesium

Solution:
Grignard reagent is a sigma bonded organometallic compound in which $\mathrm{Mg}$ is bonded with one alkyl and one halogen group. This can be prepared as
$\mathrm{R}-\mathrm{X}+\mathrm{Mg} \stackrel{\text { ether }}{\longrightarrow} \mathrm{R}-\mathrm{Mg}-\mathrm{X}$

Topic: Chemistry, Hydrocarbons, JEE Main