datavorous/entrance-exam-dataset huggingface dataset:
Amongst $\mathrm{LiCl}, \mathrm{RbCl}, \mathrm{BeCl}_2$ and $\mathrm{MgCl}_2$ the compounds with the greatest and the least ionic character, respectively, are

Options:
A $\mathrm{LiCl}$ and $\mathrm{RbCl}$
B $\mathrm{RbCl}$ and $\mathrm{BeCl}_2$
C $\mathrm{RbCl}$ and $\mathrm{MgCl}_2$
D $\mathrm{MgCl}_2$ and $\mathrm{BeCl}_2$

The correct answer is:
$\mathrm{RbCl}$ and $\mathrm{BeCl}_2$

Solution:
According to Eaian's rule largest cation and smallest anion form ionic bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with the greatest and the least ionic character, respectively are :

Options:
A MgCl2 and BeCl2
B RbCl and MgCl2
C LiCl and RbCl
D RbCl and BeCl2

The correct answer is:
RbCl and BeCl2

Solution:
(i) % Ionic character = 16(XA – XB) + 3.5 (XA – XB)2
X Rb = 0 · 8 , X Cl = 3 · 0 , XBe=2
For Rb and Cl the electronegativity difference is highest so having more ionic character. For Be and Cl, electronegativity difference is the lowest so having less ionic character

Covalent character in compound depends upon the charge and size of ions. As charge in cation increases, covalent character increases and as size of cation increases covalent character decreases. Charge factor dominate over size factor. Since Be2+ has higher charge and smallest size therefore, BeCl2 has maximum covalent character and least ionic character

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2014 (19 Apr Online)
Amongst $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \quad\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2} ]\right.$, $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right.]$ and $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ the paramagnetic species are

Options:
A $\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
B $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right],\left[\mathrm{NiCl}_{4}\right]^{2-}$
C $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} \cdot\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{NiCl}_{4}\right]^{2-}$
D $\left[\mathrm{Nl}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right\}\left[\mathrm{Ni}(\mathrm{CO})_{4}\right\}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$

The correct answer is:
$\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$

Solution:
Species having unpaired electrons are paramagnetic
$$
\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{d}^{8}
$$

(i) In $[\left.\mathrm{NiCl}_{4}\right]^{2-}$ and $[\left.\operatorname{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ ligands $\mathrm{Cl}^{-}$ and
$\mathrm{H}_{2} \mathrm{O}$ are weak ligands, therefore no pairing will be possible. Thus, there are two unpained electrons.
(II) In [Ni $\left.\left(P h_{3}\right)_{2} C l_{2}\right\}$. although $P P h_{3}$, has d-acceptance nature but presence of Cl, makes electrons unpaired.

Topic: Chemistry, Coordination Compounds, JEE Main
Amongst SF4,XeF4,CF4 and H2O, the number of species with two lone pairs of electrons is

Options:

The correct answer is:
2

Solution:
Electrons which are not involved in the bond formation are called the lone pair of electrons.

As we can see from the structures XeF4, H2O are having two lone pairs on the central atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Amongst the compound given, the one that would form a brilliant colored dye on treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of β - naphthol is

Options:
A
B
C
D

The correct answer is:

Solution:
This reaction occurs as follows

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Amongst the element with following electronic configurations, which one of them may have the highest ionization energy ?

Options:
A $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^1$
B $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3$
C $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^2$
D $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^3$

The correct answer is:
$[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3$

Solution:
$[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3$ has highest ionization energy (half-filled)

Topic: Chemistry, Structure of Atom, NEET, NEET 2009 (Mains)
Amongst the following alcohols which would react fastest with conc. $\mathrm{HCl}$ and $\mathrm{ZnCl}_2$ ?

Options:
A
pentanol

B
2-methyl butanol

C
2-pentanol

D
2-methyl butan-2-ol

The correct answer is:
2-methyl butan-2-ol

Solution:
$\mathrm{ZnCl}_2+$ conc. $\mathrm{HCl}$ is Lucas reagent. Lucas reagent reacts fastest with tertiary alcohol. $3^{\circ}$ alcohol $+$ Lucas reagent $=$ Immediate turbidity

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (22 Apr Online)
Amongst the following compound, the optically active alkane having lowest molecular mass is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2004
Amongst the following compounds, which one is an antacid?

Options:
A Ranitidine
B Meprobamate
C Terfenadine
D Brompheniramine

The correct answer is:
Ranitidine

Solution:
Antacids are actually alkaline ions that directly neutralize the gastric acids of the stomach. Substances such as sodium hydrogen carbonate or a mixture of aluminium and magnesium hydroxide are used as antacids. Ranitidine is an antacid.

Tranquilizers are a class of chemical compounds used for the treatment of stress, and mild or even severe mental diseases.Meprobamate is relatively mild tranquilizers suitable for relieving tension.

Synthetic drugs, brompheniramine (Dimetapp) and terfenadine (Seldane), act as antihistamines. They interfere with the natural action of histamine by competing with histamine for binding sites of receptor where histamine exerts its effect.

So, the correct answer is Ranitidine is an antacid.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Amongst the following, identify the species with an atom in +6 oxidation state.

Options:
A CrO 2 Cl 2
B Cr CN 6 3 -
C Cr 2 O 3
D MnO 4 -

The correct answer is:
CrO 2 Cl 2

Solution:
Cr + 6 O - 2 2 Cl - 1 2
⇒x+2(−2)+2(−1)=0x=6
Cr + 3 CN 6 - 3
⇒x+6(−1)=−3x=3
Cr + 3 2 O - 2 3
⇒2x+3(−2)=0x=3
Mn + 7 O 4 - 2 - 1
⇒x+4(−2)=−1x=+7

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (19 Apr Online)
Amongst the following statements regarding adsorption, those that are valid are:
(a) ΔH becomes less negative as adsorption proceeds.
(b) On a given adsorbent, ammonia is adsorbed more than nitrogen gas.
(c) On adsorption, the residual force acting along the surface of the adsorbent increases
(d) With increase in temperature, the equilibrium concentration of adsorbate increases.

Options:
A (d) and (a)
B (b) and (c)
C (a) and (b)
D (c) and (d)

The correct answer is:
(a) and (b)

Solution:
(a) When gas is adsorbed on metal surface.
ΔH become less negative with progress of reaction because number of active sites decreases.
(b) Gas with greater value of critical temperature TC absorbed more. As TCNH3>TCN2 So NH3 absorbed more than N2.

Statement (c) is wrong because residual forces decrease as adsorption proceeds further.

Statement (d) is also wrong because temperature opposes extent of adsorption.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Amongst the following statements, select the set having statements which was proposed by Dalton.
(1) All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
(2) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P
(3) Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
(4) Matter consists of indivisible atoms

Options:
A (1), (2), (3)
B (1), (3), (4)
C (1), (2), (4)
D (1), (2), (3), (4)

The correct answer is:
(1), (3), (4)

Solution:
Statements (1), (3) and (4) are of Dalton's theory.
Statement (2) - When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P represents Avogadro law.

Topic: Chemistry, Structure of Atom, JEE Main
Amongst the following statements, that which was not proposed by Dalton was :

Options:
A chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
B all the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
C when gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P.
D matter consists of indivisible atoms.

The correct answer is:
when gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P.

Solution:
When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Amongst the following, the form of water with the lowest ionic conductance at 298K is:

Options:
A distilled water
B saline water used for intravenous injection
C water from a well
D sea water

The correct answer is:
distilled water

Solution:
Distilled water have the lowest ionic conductance due to presence of less number of ions in this purified water

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Amongst the following, the linear species is:

Options:
A NO2
B Cl2O
C O3
D N3-

The correct answer is:
N3-

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Amongst the following, the major product of the given chemical reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction proceeds via cyclic bromonium ion with anti addition mechanism.

And, the major product will be formed according to the following reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Amongst the following the most basic compound is

Options:
A
benzylamine

B
aniline

C
acetanilide

D
p-nitroaniline

The correct answer is:
benzylamine

Solution:
-NH2 group is not linked with benzene ring.

Topic: Chemistry, Amines, JEE Main
Amongst the following the most basic compound is

Options:
A
benzylamine

B
aniline

C
acetanilide

D
p-nitroaniline

The correct answer is:
benzylamine

Solution:
-NH2 group is not linked with benzene ring.

Topic: Chemistry, Amines, JEE Main, JEE Main 2005
Amongst the following, the number of species having the linear shape is

XeF2,I3+,C3O2,I3-,CO2,SO2,BeCl2 and BCl2⊖

Options:

The correct answer is:
5

Solution:
Using VSEPR theory, following are the shapes:

Thus, XeF2, C3O2, I3-, CO2 & BeCl2 have linear structure.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Amongst the following, the total number of compounds soluble in aqueous NaOH is

Options:

The correct answer is:
4

Solution:
All carboxylic acids and phenols are soluble in aqueous NaOH. Four compounds are soluble in aqueous NaOH.

Topic: Chemistry, General Organic Chemistry, JEE Main
Amongst the following, the total number of species which are diamagnetic is _______.

K 4 Fe CN 6 , K 3 Cr CN 6 , K 3 Co CN 6 ,K 2 Ni CN 4 , Co NH 3 6 3+ , K 2 TiF 6 , Pt NH 3 4 2+

Options:

The correct answer is:
6

Solution:
Fe CN 6 4â , Co CN 6 3â
Ni CN 4 2â , Co NH 3 6 3+ , TiF 6 2â
Pt NH 3 4 2+ all are diamagnetic

Topic: Chemistry, Coordination Compounds, JEE Main
Amorphous substances show
(A) Short and long range order
(B) Short range order
(C) Long range order
(D) Have no sharp M.P.

Options:
A $A$ and $C$ are correct
B $B$ and $C$ are correct
C C and $D$ are correct
D $B$ and $D$ are correct

The correct answer is:
$B$ and $D$ are correct

Solution:
Amorphous solids have short range order but no sharr in melting point.

Topic: Chemistry, Solid State, JEE Main
Amount of electricity that can deposit $10 \mathrm{k} \mathrm{gm}$ of silver from $\mathrm{AgNO}_3$ solution is

Options:
A 1 ampere
B 1 coulomb
C 1 faraday
D None of the above

The correct answer is:
1 faraday

Solution:
$\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow$ Ag; $E_{A g}=\frac{\text { Atomic Mass }}{1}=108$
Number of faraday $=\frac{W_{A g}}{E_{A g}}=\frac{108}{108}=1$

Topic: Chemistry, Electrochemistry, JEE Main
Amount of oxalic acid present in a solution can be determined by its titration with $\mathrm{KMnO}_4$ solution in the presence of $\mathrm{H}_2 \mathrm{SO}_4$. The titration gives unsatisfactory result when carried out in the presence of $\mathrm{HCl}$, because $\mathrm{HCl}$

Options:
A
gets oxidised by oxalic acid to chlorine

B
furnishes $\mathrm{H}^{+}$ions in addition to those from oxalic acid

C
reduces permanganate to $\mathrm{Mn}^{2+}$

D
oxidises oxalic acid to carbon dioxide and water

The correct answer is:
reduces permanganate to $\mathrm{Mn}^{2+}$

Solution:
$\mathrm{HCl}$ being stronger reducing agent reduces $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$ and result of the titration becomes unsatisfactory.

Topic: Chemistry, d and f Block Elements, JEE Main
Amount of oxalic acid present in a solution can be determined by its titration with $\mathrm{KMnO}_4$ solution in the presence of $\mathrm{H}_2 \mathrm{SO}_4$. The titration gives unsatisfactory result when carried out in the presence of $\mathrm{HCl}$, because $\mathrm{HCl}$

Options:
A
gets oxidised by oxalic acid to chlorine

B
furnishes $\mathrm{H}^{+}$ions in addition to those from oxalic acid

C
reduces permanganate to $\mathrm{Mn}^{2+}$

D
oxidises oxalic acid to carbon dioxide and water

The correct answer is:
reduces permanganate to $\mathrm{Mn}^{2+}$

Solution:
$\mathrm{HCl}$ being stronger reducing agent reduces $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$ and result of the titration becomes unsatisfactory.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2008
Amount of solar energy received on the earth's surface per unit area per unit time is defined a solar constant. Dimension of solar constant is:

Options:
A ML2T-2
B ML0T-3
C M2L0T-1
D MLT-2

The correct answer is:
ML0T-3

Solution:
Solar constant =EnergyTime Area

=M1L2T-2TL2=M1L0T-3

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Amplitude modulated wave is represented by VAM=101+0.4cos2π×104tcos2π×107t. The total bandwidth of the amplitude modulated wave is

Options:
A 10kHz
B 20MHz
C 20kHz
D 10MHz

The correct answer is:
20kHz

Solution:
Expression for modulation is given by,

st=Ac1+μcos2πfmtcos2πfct, where fm and fc are the frequency of the modulating signal and the carrier signal respectively.

Comparing it with the given expression in question, we get,

fm=104 Hz

Now, Bandwidth =2fm=20 kHz

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Amplitude modulation is the process of

Options:
A superimposing a high frequency signal on a low frequency carrier signal
B superimposing a low frequency signal on a high frequency carrier signal
C single sideband communication
D amplitude shift and phase shift

The correct answer is:
superimposing a low frequency signal on a high frequency carrier signal

Solution:
Modulation is the phenomenon of superimposing a low audio frequency information signals (called the modulation signals) on a high-frequency wave (called, the carrier wave). When we modulate the amplitude of the carrier wave to transmit the information, it is known as amplitude modulation.

Topic: Physics, Communication System, JEE Main
Amylopectin is a polymer of

Options:
A
$\alpha-D-$ glucose

B
amino acid

C
$\beta-D-$ glucose

D
amylase.

The correct answer is:
$\alpha-D-$ glucose

Solution:
Amylopectin is a polymer of $\alpha-D$-glucose.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2012 (12 May Online)
Amylopectin is composed of:

Options:
A β-D- glucose , C1-C4 and C1-C6 linkages
B α-D- glucose, C1-C4 and C2-C6 linkages
C α-D- glucose, C1-C4 and C1-C6 linkages
D β-D- glucose, C1-C4 and C2-C6 linkages

The correct answer is:
α-D- glucose, C1-C4 and C1-C6 linkages

Solution:
Amylopectin is insoluble in water and constitutes about 80-85% starch.

Amylopectin is a water-soluble polysaccharide and highly branched polymer of a-glucose units found in plants. It is one of the two components of starch, the other being amylose. Amylopectin bears a straight/linear chain along with a number of side chains which may be branched further.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Amylopectin is composed of:

Options:
A β-D- glucose , C1-C4 and C1-C6 linkages
B α-D- glucose, C1-C4 and C2-C6 linkages
C α-D- glucose, C1-C4 and C1-C6 linkages
D β-D- glucose, C1-C4 and C2-C6 linkages

The correct answer is:
α-D- glucose, C1-C4 and C1-C6 linkages

Solution:
Amylopectin is insoluble in water and constitutes about 80-85% starch.

Amylopectin is a water-soluble polysaccharide and highly branched polymer of a-glucose units found in plants. It is one of the two components of starch, the other being amylose. Amylopectin bears a straight/linear chain along with a number of side chains which may be branched further.

Topic: Chemistry, Biomolecules, JEE Main
Amylopectin is composed of

Options:
A Î± -D-glucose with C1âC4 and C1âC6 linkage
B Î± -D-glucose with C1âC4 and C2âC6 linkage
C Î² -D-glucose with C1âC4 and C2âC6 linkage
D Î² -D-glucose with C1âC4 and C1âC6 linkage

The correct answer is:
Î± -D-glucose with C1âC4 and C1âC6 linkage

Solution:
Amylopectin is a homopolymer of α D-glucose having C1—C4 β -glycocidic linkage (for linear chain formation) and C1—C6 linkage (for branch chain formation).

Topic: Chemistry, Biomolecules, JEE Main
An 8 kg metal block of dimensions 16 cmÃ8 cmÃ6 cm is lying on a table with its face of largest area touching the table. If g=10 ms-2, then the minimum amount of work done in making it stand with its length vertical is

Options:
A 8 J
B 6.4 J
C 4 J
D 12.8 J

The correct answer is:
4 J

Solution:
When face of the largest area is touching the table, height of centre of gravity above the table hâ¡ 1 = 6 2 = 3 cm
With its length vertical, height of centre of gravity would become hâ¡ 2 = 1 6 2 = 8 cm
Minimum work required W = (P.E.)2 â (P.E.)1
= mg(h2 â h1)
Wâ¡ = 8 Ã 1 0 8 - 3 1 0 0 = 4 J

Topic: Physics, Work Power Energy, JEE Main
An 80 Î¼C charge is given to the 4 Î¼F capacitor in the circuit shown in the figure so that the upper plate A is positively charged. An unknown resistance R is connected in the left limb. As soon as the switch S in the central limb is closed, a current of 2 A flows through the 2 Î© resistor in the central limb. The capacitive time constant for the circuit is

Options:
A 56 μ s
B 8 μ s
C 200 μ s
D 40 Î¼ s

The correct answer is:
40 Î¼ s

Solution:
Potential difference across the central limb = 16 V = Potential difference across 16 Ω = Potential difference across the left limb.
⇒ current through 16 Ω = 1 A
⇒ current through the left limb = 1 A and
R=11 Ω
∴τc=162+2ohms×4×10-6 F=40 μs

Topic: Physics, Current Electricity, JEE Main
An 80 kg person is parachuting and is experiencing a downward acceleration of 2.8 m s-2. The mass of the parachute is 5 kg. If the upward force on the open parachute is k×102 N, then what is the value of k? (Take g=9.8 m s-2)

Options:

The correct answer is:
5.95

Solution:
Mass of the person M=80 kg
Mass of the parachute m=5 kg
∴ Total mass of the system =M+m=85kg
Downward acceleration a=2.8 ms-2
Let upward force =F
Applying Newton's II law of motion to this system
F=m+Mg-m+Ma
or F=m+Mg-a
F=859.8-2.8N
=85×7 N
F=595 N=5.95×102 N

k=5.95

Topic: Physics, Laws of Motion, JEE Main
An 1800 W toaster, a 1.3 kW electric fan and a 100 W lamp are plugged in the same 120 V circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. the sum of current drawn by the three devices) through the circuit?

Options:
A 40 A
B 120 A
C 18 A
D 27 A

The correct answer is:
27 A

Solution:
Power consumed
P=P1+P2+P3 .....(i)
And P=VI
From Equation. (i), we get
P1+P2+P3=VI
â 1800+1300+100 W=120I
â 3200=120I
I=26.66 A
â27 A

Topic: Physics, Current Electricity, JEE Main
An $A B_2$ type structure is found in

Options:
A $\mathrm{NaCl}$
B $\mathrm{Al}_2 \mathrm{O}_3$
C $\mathrm{CaF}_2$
D $\mathrm{N}_2 \mathrm{O}$

The correct answer is:
$\mathrm{CaF}_2$

Solution:
$A B_2$ type of structure is present in $\mathrm{CaF}_2$
$\therefore A B_2 \rightleftharpoons A^{2+}+2 B^{-} ; \quad \mathrm{CaF}_2 \rightleftharpoons \mathrm{Ca}^{2+}+2 F^{-}$

Topic: Chemistry, Solid State, JEE Main
An AC current is given by I=I1sinωt+I2cosωt.
A hot wire ammeter will give a reading:

Options:
A I12-I222
B I12+I222
C I1+I22
D I1+I222

The correct answer is:
I12+I222

Solution:
I=I1sinωt+I2cosωt

∴I0=I12+I22

∴Irms=I02=I12+I222

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (17 Mar Shift 1)
An AC source rated 220 V,50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :

Options:
A 2.5 ms
B 25 ms
C 2.5 s
D 0.25 ms

The correct answer is:
2.5 ms

Solution:
i=i0cos(ωt)
i=i0 at t=0
i=i02 at ωt=π4
t=π4ω=π4(2πf)=18f
t=1400=2.5 ms

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (18 Mar Shift 1)
An AC circuit drawing power from a source of angular frequency 50 rad s-1 has a power factor of 0.6. In this condition, a resistance of 100 Î© is present and the current is lagging behind the voltage. If a capacitor is connected in series, then the required capacitance that will result in a power factor of unity is

Options:
A 30 Î¼F
B 150 μF
C 50 Î¼F
D 200 Î¼F

The correct answer is:
150 μF

Solution:
As the current lags the voltage, the circuit must contain a combination of resistance and inductance.

Power factor cos⁡ϕ=0.6

⇒RR2+XL2=0.6
⇒R2+XL2=R0.62

⇒XL2=R20.62-R2
⇒XL2=R2×0.640.36
∴XL=0.8 R0.6=4R3

Now we want the power factor cos⁡ϕ=1

This is the condition of resonance in which

XL=XC

∴XC=4R3

⇒1ωC=4R3

⇒C=34×50×100=150 μF

Topic: Physics, Alternating Current, JEE Main
An AC circuit has an inductor and a resistor of resistance R in series, such that XL=3R. Now, a capacitor is added in series such that XC=2R. the ratio of the new power factor with the old power factor of the circuit is 5:x. The value of x is

Options:

The correct answer is:
1

Solution:
First Recall the formula of power factor, cosϕ=RZ ⇒cosϕinitial =RZ=RXL2+R2=R9R2+R2=110.

⇒cosϕfinal =RZ=RXL-XC2+R2=RR2+R2=12

⇒cosϕfinal cosϕinitial =5x⇒x=1

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (27 Aug Shift 2)
An AC e.m.f E=4 cosâ¡1000t V is applied to an LR circuit containing inductance 3 mH and resistance 4 Î©. The amplitude of the current is

Options:
A 47 A
B 1.0Â A
C 47 A
D 0.8 A

The correct answer is:
0.8 A

Solution:
i0=V0Z ,
Z=R2+ωL2
=42+1000×3×10-32=5 Ω
i0=45
i0=0.8 A

Topic: Physics, Alternating Current, JEE Main