datavorous/entrance-exam-dataset huggingface dataset:
In the following sequence of reactions the P is:

Options:
A
B
C
D

The correct answer is:

Solution:
According to the question, In the first step of the reaction, when cyclo-pentyl-chloride reacts with magnesium in the presence of dry ether than cyclo-pentyl-magnesium-chloride (grignard reagent) is formed.

In the second step of the reaction, when grignard reagent is react with a compound, which has active hydrogen than grignard reagent abstract hydrogen from active hydrogen compound and converted in to alkane.

So cyclo-pentane is formed as the product(second step of the reaction) of this reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2021 (27 Aug Shift 1)
In the following sequence of reactions, the product D is

CHâ¡CHâHBr AâHBr Bâalc KOH CâNaNH2 D

Options:
A Ethanol
B Ethyne
C Ethanal
D Ethene

The correct answer is:
Ethyne

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following sequence of reactions:
Toluene→KMnO4A→SOCl2B→BaSO4H2/PdC
The Product C is:

Options:
A C6H5CHO
B C6H5COOH
C C6H5CH3
D C6H5CH2OH

The correct answer is:
C6H5CHO

Solution:
Treatment of an alkylbenzene with potassium permanganate results in oxidation to give benzoic acid. The position directly adjacent to an aromatic group is called the “benzylic” position. The reaction only works if there is a hydrogen attached to the carbon. Hence, toluene, on oxidation with potassium permanganate, gives benzoic acid (A).

Carboxylic acids reacts with thionyl chloride to form acid chlorides. Hence, benzoic acid, on reaction with thionyl chloride (SOCl2), gives benzoyl chloride (B).

Next reaction is the Rosenmund reaction. It is a reaction where acid chlorides are converted into aldehydes by using hydrogen gas over palladium poisoned by barium sulfate. Hence, benzoyl chloride upon reduction with H2/Pd or BaSO4, forms benzaldehyde (C).

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following sequence of reactions:
Toluene→KMnO4A→SOCl2B→BaSO4H2/PdC
The Product C is:

Options:
A C6H5CHO
B C6H5COOH
C C6H5CH3
D C6H5CH2OH

The correct answer is:
C6H5CHO

Solution:
Treatment of an alkylbenzene with potassium permanganate results in oxidation to give benzoic acid. The position directly adjacent to an aromatic group is called the “benzylic” position. The reaction only works if there is a hydrogen attached to the carbon. Hence, toluene, on oxidation with potassium permanganate, gives benzoic acid (A).

Carboxylic acids reacts with thionyl chloride to form acid chlorides. Hence, benzoic acid, on reaction with thionyl chloride (SOCl2), gives benzoyl chloride (B).

Next reaction is the Rosenmund reaction. It is a reaction where acid chlorides are converted into aldehydes by using hydrogen gas over palladium poisoned by barium sulfate. Hence, benzoyl chloride upon reduction with H2/Pd or BaSO4, forms benzaldehyde (C).

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2015 (04 Apr)
In the following series of chemical reactions, identify $Z$
$\mathrm{C}_3 \mathrm{H}_7 \mathrm{OH} \xrightarrow[\text { 160-180 } 0^{\circ} \mathrm{C}]{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} X \xrightarrow{\mathrm{Br}_2} Y \xrightarrow[\text { Alc. } \mathrm{KOH}]{\text { Excessof }} Z$

Options:
A

B

C

D $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following series of reactions:

i A + 2 HCl + 5 H 2 O â¶ 2 NaCl + X

ii Xâ¶370 KÎHBO2â¶Red heatÎZ

iii 4 BF 3 + 3 LiAlH 4 â¶ Y + 3 LiF + 3 AlF 3

The compounds A, X, Z and Y may be selected from following options

Options:
A A X Z Y Na 2 B 4 O 7 H 3 BO 3 B 2 O 3 B 2 H 6
B A X Z Y H 3 BO 3 Na 2 B 4 O 7 B 2 H 6 B 2 O 3
C A X Z Y B 2 H 6 B 2 O 3 H 3 BO 3 Na 2 B 4 O 7
D A X Z Y B 2 O 3 H 2 B O 3 Na 2 B 4 O 7 B 2 H 6

The correct answer is:
A X Z Y Na 2 B 4 O 7 H 3 BO 3 B 2 O 3 B 2 H 6

Solution:
Na 2 B 4 O 7 + 2 HCl + 5 H 2 O â¶ 2 NaCl + 4 H 3 BO 3 Borax (A) Orthoboric aicd (X)
H 3 BO 3 â¶ Î , 3 7 0 K HBO 2 + H 2 O X Metaboric acid
4HBO2âÎ, Red heat2B2O3Boron trioxide z+H2O
4BF3â+â3LiAlH4â2B2H6(Diborane)â+â3LiFâ+â3AlF3

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the following sets of reactants which two sets best exhibit the amphoteric character of Al2O3 . xH2O ?
Set I Al 2 O 3 · xH 2 O s + OH aq - Set II Al 2 O 3 x · H 2 O s + H 2 O l Set III Al 2 O 3 x · H 2 O s + H aq + Set IV Al 2 O 3 · xH 2 O s + NH 3 aq

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

The correct answer is:
1 and 3

Solution:
For a substance to be called amphoteric, it should be able to react as a base as well as an acid. Hence, when it is reacted with a base, it should act as an acid, and when it is reacted with an acid, it should behave as a base.
The reactions for amphoteric nature of Al2O3. x H2O(s) are
Al 2 O 3 . x H 2 O s → OH aq - Al OH 4 - aq
Aluminate anion
Al 2 O 3 . x H 2 O s → H aq + Al H 2 O 6 aq 3 + .
[Al(H2O)6]3+ is octahedral complex ion
[Al(OH)4]-1 tetrahedral at pH = 13

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In the following sets of reactants which two sets best exhibit the amphoteric character of Al2O3 . xH2O ?
Set I Al 2 O 3 · xH 2 O s + OH aq - Set II Al 2 O 3 x · H 2 O s + H 2 O l Set III Al 2 O 3 x · H 2 O s + H aq + Set IV Al 2 O 3 · xH 2 O s + NH 3 aq

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

The correct answer is:
1 and 3

Solution:
For a substance to be called amphoteric, it should be able to react as a base as well as an acid. Hence, when it is reacted with a base, it should act as an acid, and when it is reacted with an acid, it should behave as a base.
The reactions for amphoteric nature of Al2O3. x H2O(s) are
Al 2 O 3 . x H 2 O s → OH aq - Al OH 4 - aq
Aluminate anion
Al 2 O 3 . x H 2 O s → H aq + Al H 2 O 6 aq 3 + .
[Al(H2O)6]3+ is octahedral complex ion
[Al(OH)4]-1 tetrahedral at pH = 13

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2014 (09 Apr Online)
In the following skew conformation of ethane, H'-C-C-H″ dihedral angle is:

Options:
A 58o
B 120o
C 149o
D 151o

The correct answer is:
149o

Solution:
The structure given in question is in staggered form. Hence, the two planes are 120° apart.
Given difference in angle of H-C-H bond is 29.
∴ The dihedral angle will be 120°+29°=149°

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
In the following statements -
(i) Relation is a special case of a function.
(ii) Function is a special case of a relation.
(iii) Both relation and function are same.

Options:
A (iii) is True, (i) and (ii) are false
B (i) is True, (ii) and (iii) are false
C (ii) is True, (i) and (iii) are false
D All (i), (ii) and (iii) are true

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

Solution:
Follow the line from each number in the domain to the range. This is a relation but not a function.

This is a relation & also a function.

⇒(ii) is true and (i) and (iii) are false.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2020 (23 Sep Shift 1)
In the following structure, the double bonds are marked as I, II, III and IV.
Geometrical isomerism is not possible at the site(s):

Options:
A I and III
B III
C I
D III and IV

The correct answer is:
I

Solution:
The different group should be attached to each sp2 hybridised C-atom.
It has 2 CH3 group at I-C, so it will not show geometrical isomerism.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2017 (09 Apr Online)
In the following the correct bond order sequence is:

Options:
A O22->O2+>O2->O2
B O2+>O2->O22->O2
C O2+>O2>O2->O22-
D O2>O2->O22->O2+

The correct answer is:
O2+>O2>O2->O22-

Solution:
O2(16 electrons)

σ1s2,σ1s*2,σ2s2,σ2s*2,σ2p22

π2px2=π2py2,π2px*1=π2py*1,σ2pz*

Bond order of O2⇒2
Bond order of O2-⇒1.5
Bond order of O22-⇒1
Bond order of O2+⇒2.5

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (25 Jul Shift 2)
In the following Wheatstone bridge $P / Q=R / S$. If key $K$ is closed, then the galvanometer will show deflection

Options:
A In left side
B In right side
C No deflection
D In either side

The correct answer is:
In either side

Solution:
Pressing the key does not disturb current in all resistances as the bridge is balanced.
Therefore, deflection in the galvanometer in whatever direction it was, will stay.

Topic: Physics, Current Electricity, JEE Main
In the following, which one of the diodes is reverse biased?

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2006
In the follwing displacement $(x)$ vs time $(t)$ graph, at which among the points $P, Q$, and $R$ is the object's speed increasing?

Options:
A R only
B P only
C $\mathrm{Q}$ and $\mathrm{R}$ only
D $P, Q, R$

The correct answer is:
R only

Solution:
Sol. $\quad$ |slope| is increasing at point $\mathrm{R}$

Topic: Physics, Motion In One Dimension, JEE Main
In the formation of primary rainbow, the sunlight rays emerge at minimum deviation from rain-drop after

Options:
A One internal reflection and one refraction
B One internal reflection and two refractions
C Two internal reflections and one refraction
D Two internal reflections and two refractions

The correct answer is:
One internal reflection and two refractions

Solution:
Both the primary and secondary rainbows are phenomena that formed by the reflection and refraction of sunlight in tiny water droplets. Different colors of sunlight have different refractive index so they will be reflected in slightly different directions inside a water droplet and dispersed to form rainbow. The sunlight rays after being refracted twice and reflected once by the water droplet form a primary rainbow and emerge at minimum deviation.

Topic: Physics, Ray Optics, JEE Main
In the formula, a=3bc2, a and c have dimensions of electric capacitance and magnetic field intensity respectively. The dimensions of b in S.I. system are

Options:
A [M-3L-2T4Q4]
B [M-3T4Q4]
C [M-3T3Q]
D [M-3L2T4Q-4]

The correct answer is:
[M-3L-2T4Q4]

Solution:
Given, a=3bc2
â b=a3c2
Writing dimensions for a and c, we have
b=Q/VB2=Q[ML2T-2Q-1]-1MT-1Q-12
=[M-3L-2T4Q4]

Topic: Physics, Units and Dimensions, JEE Main
In the formula X=5YZ2 , X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?

Options:
A M-1 L-2 T4 A2
B M-2 L0 T-4 A-2
C M-2 L-2 T6 A3
D M-3 L-2 T8 A4

The correct answer is:
M-3 L-2 T8 A4

Solution:
Y=XZ2
=M-1L-2T4A2ML0T-2A-12
=M-3L-2T8A4

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2019 (10 Apr Shift 2)
In the four numbers first three are in G.P. and last three are in A.P. whose common difference is 6 . If the first and last numbers are same, then first will be

Options:
A $2$
B $4$
C $6$
D $8$

The correct answer is:
$8$

Solution:
Let the numbers be $\frac{a}{r}, a, a r, 2 a r-a$\ldots(i)
Where first three numbers are in G.P. and last three are in A.P. Given that the common difference of A.P. is 6 , so
$a r-a=6$\ldots(ii)
Also given, $\frac{a}{r}=2 a r-a \Rightarrow \frac{a}{r}=2(a r-a)+a$
$\Rightarrow \quad \frac{a}{r}=2(6)+a$
$\Rightarrow\left(\frac{a}{r}\right)-a=12 \Rightarrow$ $\Rightarrow a(1-r)=12 r \Rightarrow r=-\frac{1}{2}$
From (i) we get, $a\left[\left(-\frac{1}{2}\right)-1\right]=6$ or $a=-4$
Required numbers from (i) are $8,-4,2,8$.

Topic: Mathematics, Sequences and Series, JEE Main
In the Freundlich adsorption isotherm, the value of 1n is between 0 and -----.

Options:

The correct answer is:
1

Solution:
X M =KP 1 n
⇒0≤ 1 n <1

Topic: Chemistry, Surface Chemistry, JEE Main
In the fully developed angiospermic female gametophyte, what is the ratio of haploid, diploid and triploid cells?

Options:
A 5:1:1
B 3:1:3
C 6:1 :0
D 6:0:1

The correct answer is:
6:1 :0

Solution:
The nucleus of the female gametocyte undergoes three mitotic division progressing from 2 nucleate to 4 nucleate to ultimately 8 nucleate stage. Six of the eight nuclei gets surrounded by a cell wall and the remaining two polar nuclei form a large central cell. Three cells move towards the chalazal end and the other three move towards the micropylar end. All the cells in the female gametocytes are haploid but the two polar nuclei fuse to form a diploid cell. Hence the ratio of haploid: diploid: triploid is 6: 1 :0

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In the fusion reaction, ${ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+\mathrm{Q}, \mathrm{Q}$ is energy released. If ' $\mathrm{c}$ ' is the speed of light and ' $\mathrm{m}$ ' is the mass of each deuterium nucleus then the mass of helium nucleus formed is

Options:
A $2 \mathrm{~m}+\frac{\mathrm{Q}}{\mathrm{c}^2}$
B $\frac{\mathrm{Q}}{\mathrm{mc}^2}$
C $\mathrm{~m}+\frac{\mathrm{Q}}{\mathrm{c}^2}$
D $2 \mathrm{~m}-\frac{\mathrm{Q}}{\mathrm{c}^2}$

The correct answer is:
$2 \mathrm{~m}-\frac{\mathrm{Q}}{\mathrm{c}^2}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Nuclear Physics, JEE Main
In the Gabriel's phthalimide synthesis, phthalimide, is treated first with

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} / \mathrm{KOH}$
B Ethanolic Na
C Ethanol and $\mathrm{H}_2 \mathrm{SO}_4$
D Ether and $\mathrm{LiAlH}_4$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} / \mathrm{KOH}$

Solution:
The correct option is $\mathrm{1}_2 \mathrm{H}_5$ I/Alc. $\mathrm{KOH}$
Gabriel phthalimide is firstly treated with alcoholic $\mathrm{KOH}$ to from potassium phthalimide which on further reaction with ethyl iodide gives N-Ethyl phthalimide.
The product $\mathrm{N}$-Ethyl phthalimide is hydrolysed with aq. $\mathrm{NaOH}$ to form primary amine, i.e. ethylamine. This reaction is known as Gabriel phthalimide synthesis. Thus option (1) is correct.
Step 1: When potassium hydroxide is introduced to the phthalimide, an acidbase reaction occur. The hydroxide ion deprotonates the imide. The resulting proton is more acidic than any simple amine, generating a strong nucleophile the potassium phthalimide ion.
The nucleophilic potassium imide ion attacks the electrophilic carbon of the alkyl halide. The nitrogen atom subsequently replaces the halogen (Fluorine, Chlorine, Bromine or lodine) in the alkyl halide and bonds with the carbon itself. This results in the formation of an N-Alkyl Phthalimide.
The product $\mathrm{N}$-Alkyl phthalimide is hydrolysed with aq. $\mathrm{NaOH}$ to form primary amine, i.e. Alkylamine.
product in Gabriel's phthalimide synthesis

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the Gattermann reaction, a diazonium group is replaced by \( \mathrm{X} \) using \( \mathrm{Y} ; \mathrm{X} \) and \( \mathrm{Y} \) are

Options:
A \(\mathrm{X} \text{ and } \mathrm{Y} =\mathrm{Cl}^{-} \quad =\mathrm{Cu} / \mathrm{HCl} \)
B \(\mathrm{X} \text{ and } \mathrm{Y} =\mathrm{Cl}^{+} \quad \mathrm{Y}=\mathrm{CuCl}_{2} / \mathrm{HCl} \)
C \(\mathrm{X} \text{ and } \mathrm{Y} =\mathrm{Cl}^{-} \quad \mathrm{Y}=\mathrm{CuCl}_{2} / \mathrm{HCl} \)
D \(\mathrm{X} \text{ and } \mathrm{Y} =\mathrm{Cl}_{2} \mathrm{Y}=\mathrm{Cu}_{2} \mathrm{O} / \mathrm{HCl} \)

The correct answer is:
\(\mathrm{X} \text{ and } \mathrm{Y} =\mathrm{Cl}^{-} \quad =\mathrm{Cu} / \mathrm{HCl} \)

Solution:
When the benzene diazonium salt reacts with the halogen in the presence of copper powder, then it is known as the Gatterman reaction.

ArN2+Cl- + HCl →Cu ArCl + N2 + CuCl

So, X-Cl- & Y=Cu/HCl.

Topic: Chemistry, Amines, JEE Main
In the given arrangement, the rod is free to rotate about the hinge and it is in constant contact with the equilateral triangular block of base length 2a3. If the block is moving horizontally with a speed v=20 ms-1, then find the magnitude of the angular velocity of the rod (in rad.s-1) at the instant when Î¸=30o. [Given that a=1ââm]

Options:

The correct answer is:
5

Solution:
h x =tanθ
x=hcotθ
dx dt =−h  cosec 2 θ dθ dt
ω = v h sin 2 θ= 20 1 × 1 4 =5 rad/s

Topic: Physics, Rotational Motion, JEE Main
In the given arrangement, the slits S1 and S2 are fixed at small separation d while the slits S3 and S4 have variable separation z. The screen is fixed and the slits are symmetric with respect to the centre of the screen. A parallel beam of light is incident as shown in figure.

The minimum value of separation z so that the intensity at O is zero is

Options:
A λDd
B 2λDd
C λD2d
D λD3d

The correct answer is:
λDd

Solution:
For zero intensity at O,z2=Î»D2d
â z=Î»Dd

Topic: Physics, Wave Optics, JEE Main
In the given below alkali metal ions which has lowest ionic mobility in aqueous solutions?

Options:
A Rb+
B Cs+
C Li+
D Na+

The correct answer is:
Li+

Solution:
As we know that hydration power decreases on moving down the group hence among alkali metals Li has excessive hydration & hence it has low mobility in aqeous solution.

Topic: Chemistry, s Block Elements, JEE Main
In the given bonds which one is most ionic

Options:
A $\mathrm{Cs}-\mathrm{Cl}$
B $\mathrm{Al}-\mathrm{Cl}$
C $\mathrm{C}-\mathrm{Cl}$
D $\mathrm{H}-\mathrm{Cl}$

The correct answer is:
$\mathrm{Cs}-\mathrm{Cl}$

Solution:
Cs is more electropositive.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In the given chemical reaction, colors of theFe2+ and Fe3+ ions, are respectively :

5Fe2++MnO4-+8H+→Mn2++4H2O+5Fe3+

Options:
A Yellow, Green
B Yellow, Orange
C Green, Orange
D Green, Yellow

The correct answer is:
Green, Yellow

Solution:
Configuration	Example	Colour

3d6	Fe+2	Green

3d5	Fe+3	Yellow

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (01 Sep Shift 2)
In the given circuit, a charge of $+80 \mu \mathrm{C}$ is given to upper plate of a $4 \mu \mathrm{F}$ capacitor. At steady state, the charge on the upper plate of the $3 \mu \mathrm{F}$ capacitor is

Options:
A $60 \mu \mathrm{C}$
B $48 \mu \mathrm{C}$
C $80 \mu \mathrm{C}$
D $0 \mu \mathrm{C}$

The correct answer is:
$48 \mu \mathrm{C}$

Solution:
The net charge on upper plates both capacitor $(2 \mu \mathrm{F}$ and $3 \mu \mathrm{F}$ ) will be $+80 \mu \mathrm{C}$, so charge on upper plate of $3 \mu \mathrm{C}$ capacitor will be
$$
\begin{aligned}
& q=\left(\frac{3}{3+2}\right)(80 \mu \mathrm{C}) \\
& q=\frac{3}{5} \times 80=48 \mu \mathrm{C}
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be x2. The value of x is _____ .

Options:

The correct answer is:
3

Solution:
Left part of the resistors are connected in parallel, hence R1=am3.

Similarly, for the right part of the resistors we can write, R2=a2m.

Therefore, equivalent resistance will be,

R=am3+a2m

Differentiating both sides w.r.t m to observe critical points, we get dRdm=a3+a-12m2=0

⇒a3=a2m2 ⇒m=32

Thus, x=3.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jun Shift 2)
In the given circuit all resistance is of the value of R ohm each. The equivalent resistance between A and B is:

Options:
A 5R2
B 3R
C 5R3
D 2R

The correct answer is:
2R

Solution:
Resistance of ideal wire would be zero, and hence potential will be the same of all C point as shown in the figure. So effective path of current is shown in figure,

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (15 Apr)
In the given circuit all resistances are of value $R$ ohm each. The equivalent resistance between $A$ and $B$ is:

Options:
A
$2 R$

B
$\frac{5 R}{2}$

C
$\frac{5 R}{3}$

D
$3 R$

The correct answer is:
$2 R$

Solution:
$\mathrm{R}_{\text {senies }}=\mathrm{R}_1+\mathrm{R}_2+\ldots . \ldots+\mathrm{R}_{\mathrm{n}}$
$\frac{1}{\mathrm{R}_{\text {parallel }}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\ldots \ldots+\frac{1}{\mathrm{R}_{\mathrm{n}}}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In the given circuit all resistors are identical. Their resistance is equal to 20 Ω. Find the equivalent resistance between A and B (in Ω).

Options:

The correct answer is:
09.00

Solution:
Equivalent circuit is shown below:-

Therefore, Re⁡q⁡=r5+r4=9r20=9 Ω

Topic: Physics, Current Electricity, JEE Main
In the given circuit, an ideal voltmeter connected across the 10 Ω resistance reads 2 V . The internal resistance r , of each cell is:

Options:
A 0 Ω
B 1.5 Ω
C 0.5 Ω
D 1 Ω

The correct answer is:
0.5 Ω

Solution:
Current through battery =210+215=6+430=13 A
Also, net resistance Req=15×1015+10+2+2r
=8+2r
⇒13=1.5+1.58+2r
8+2r=9
r=12=0.5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Apr Shift 1)
In the given circuit.C1=2μF,C2=0.2μF,C3=2μF,C4=4μF,C5=
2μF,C6=2μF. The charge stored on capacitor C4 is ______ μC.

Options:

The correct answer is:
4

Solution:
From the diagram of the circuit, C3, C4, C5 are in series, so the total capacitance is

1C'=12+14+12⇒C'=45 μF

C' & C2 are parallel, the total capacitance is

C''=45+0.2 μF=1 μF

C1, C'', C6 are in series. So the equivalent capacitance is,

1Ceq=12+1+12⇒Ceq=0.5 μF

Using Q=CeqV.

The charge through the battery, Q =10×12=5 μC

From the diagram above,

Q'=5μC×0.80.8+0.2=4 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (11 Apr Shift 2)
In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
∵1 μF & 2 μF are in parallel.
∴ The equivalent capacitance of the series combination is, Ceq is=3CC+3. So, the total charge supplied by the battery is,
⇒Q=CeqE=3CEC+3.∴ The potential difference across the parallel combination of 1 μF and 2μF is, ∆V=Q3=CEC+3. So charge on 2 μF capacitor is, Q2=C2∆V=2CEC+3⇒Q22E=CC+3⇒Q22E=C+3-3C+3
⇒Q22E=1-3C+3⇒ So the graph is a hyperbola. With downward curve line. i.e.

Topic: Physics, Capacitance, JEE Main
In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
∵1 μF & 2 μF are in parallel.
∴ The equivalent capacitance of the series combination is, Ceq is=3CC+3. So, the total charge supplied by the battery is,
⇒Q=CeqE=3CEC+3.∴ The potential difference across the parallel combination of 1 μF and 2μF is, ∆V=Q3=CEC+3. So charge on 2 μF capacitor is, Q2=C2∆V=2CEC+3⇒Q22E=CC+3⇒Q22E=C+3-3C+3
⇒Q22E=1-3C+3⇒ So the graph is a hyperbola. With downward curve line. i.e.

Topic: Physics, Capacitance, JEE Main, JEE Main 2015 (04 Apr)
In the given circuit diagram, a connecting wire of negligible resistance is joining the points B and D, the current in the connecting wire is

Options:
A zero
B 2 A
C 0.4 A
D 4 A

The correct answer is:
2 A

Solution:
Reff=45+65=2Ω

i=202=10A

I=4i5-3i5=+i5=2A

Topic: Physics, Current Electricity, JEE Main
In the given circuit diagram, a wire is joining points B and D. The current in this wire is:

Options:
A 0.4A
B 2A
C 4A
D zero

The correct answer is:
2A

Solution:
Reff=45+65=2Ω
i=202
=10A
I=4i5-3i5=+i5=2A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (09 Jan Shift 1)
In the given circuit diagram the current through the 1 Î© resistor is I. Find the value of 2I (in A)?

Options:

The correct answer is:
5

Solution:
V-202+V2+V-5=0
V-20+V+2 V-5=0â4V-20-10=0

V=304=152 V
I=15-102=52 Aâ2I=5 A

Topic: Physics, Current Electricity, JEE Main