datavorous/entrance-exam-dataset huggingface dataset:
The ammeter reading in the given circuit is

Options:
A \( \frac{15}{32} \mathrm{~A} \)
B \( \frac{14}{33} \mathrm{~A} \)
C \( \frac{17}{33} \mathrm{~A} \)
D \( \frac{15}{31} \mathrm{~A} \)

The correct answer is:
\( \frac{15}{32} \mathrm{~A} \)

Solution:
For parallel connection of cell,
Equivalent emf of cell is, Eeq=3×1+6×21+2=5V,
Equivalent internal resistance is, req=2×12+1=23 Ω
Simplified circuit is shown in below figure,

From Kirchoff's law,
Current through 10 Ω resistor is, I=510+23=1532 A

Topic: Physics, Current Electricity, JEE Main
The ammeter reading in the given circuit is

Options:
A 1433 A
B 1532Â A
C 1733Â A
D 1531A

The correct answer is:
1532 A

Solution:
Eeq=3×1+6×21+2=5V, req=2×12+1=23Ω

I=510+2/3=1532 A

Topic: Physics, Current Electricity, JEE Main
The ammonia evolved from the treatment of $0.30 \mathrm{~g}$ of an organic compound for the estimation of nitrogen was passed in $100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ sulphuric acid. The excess of acid required $20 \mathrm{~mL}$ of $0.5 \mathrm{M}$ sodium hydroxide solution hydroxide solutio for complete neutralization. The organic compound is

Options:
A
acetamide

B
thiourea

C
urea

D
benzamide

The correct answer is:
urea

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2004
The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The percentage nitrogen in the organic compound is

Options:

The correct answer is:
46.60

Solution:
Equivalents of NH3 evolved
= 1 0 0 Ã 0 Â· 1 Ã 2 1 0 0 0 - 2 0 Ã 0 Â· 5 1 0 0 0 = 1 1 0 0
Percent of nitrogen in the unknown organic compound
= 1 1 0 0 Ã 1 4 0.3 Ã 1 0 0 = 4 6 . 6

Topic: Chemistry, Practical Chemistry, JEE Main
The ammonia NH3 released on quantitative reaction of 0.6 g urea NH2CONH2 with sodium hydroxide NaOH can be neutralized by

Options:
A 200 ml of 0.4 NHCl
B 200 ml of 0.2 NHCl
C 100 ml of 0.2 NHCl
D 100 ml of 0.1 NHCl

The correct answer is:
100 ml of 0.2 NHCl

Solution:
NH2CONH2 + 2NaOH →2NH3 +Na2CO3

By, stoichiometry, 1 mol of urea gives 2 mol NH3

Moles of urea ==0.6 gm60 gm/mol=0.01 mol

So, 0.01 mole urea produce 0.02 mole NH3

To neutralize ammonia HCl is required

NH3 +HCl→ NH4Cl

By equation, it is clear that 1 mol of ammonia is neutralized by 1mol of HCl.

So, 0.02 mol of ammonia neutralized by 0.02 mol of HCl

∴Mole of HCl=0.02 mole

Equivalent of HCl=normality×volume=0.2 N×100 mL=0.02 equivalent

As n-factor of HCl is 1, so number of equivalents=number of moles=0.02 moles

Hence, option 3 is correct.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The amonia evolved from $2 \mathrm{~g}$ of a compound in Kjeldahl's estimation of nitrogen neutralizes $10 \mathrm{~mL}$ of $2 \mathrm{M}$ $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution. The weight percentage of nitrogen in the compound is

Options:
A 28
B 14
C 56
D 7

The correct answer is:
28

Solution:
$\begin{array}{c}
2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \\
10 \mathrm{ml} \\
2 \mathrm{M}
\end{array}$
millimole of $\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{\mathrm{mmol} \text { of } \mathrm{NH}_{3}}{2}=20$ $\mathrm{mmol} \mathrm{NH}_{3}=\mathrm{mmol}$ of $N=40$
$W_{N}=\frac{40 \times 14}{1000}=\frac{560}{1000}=0.56 \mathrm{~g}$
$\%$ of $N=\frac{0.56}{2} \times 100=28$

Topic: Chemistry, General Organic Chemistry, JEE Main
The amorphous form of silica is:

Options:
A quartz
B kieselguhr
C tridymite
D cristobalite

The correct answer is:
kieselguhr

Solution:
Kieselguhr, an amorphous form of silica is used in filtration plants.
Quartz is one of the most common minerals in the earth’s crust having a specific crystalline form.
Cristobalite is a high temperature polymorph of silica, having same chemical formula as quartz but distinct crystal structure.
Tridymite is also a high temperature polymorph of silica and occurs as hexagonal crystals.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (09 Apr Shift 2)
The amount (in grams) of sucrose (mol wt.=342 g) that should be dissolved in 100 g water in order to produce a solution with a 105.0 oC difference between the freezing point and boiling point is
(Given that K f =1.86â K kg mol â1 and K b =0.51â K kg mol â1 for water)

Options:
A 34.2 g
B 72.2 g
C 342 g
D 460 g

The correct answer is:
72.2 g

Solution:
Boiling point Tb=100+âTb=100+kbm
Freezing point Tf=0-âTf=-kfm
Tb-Tf=100+kbm-(-kfm)
105=100+0.51 m+1.86 m
2.37m=5 or m=52.37=2.11
â´ Weight of sucrose to be dissolved in 100 g water
=2.11Ã3421000Ã100=72.2 g

Topic: Chemistry, Solutions, JEE Main
The amount (in grams) of sucrose (mol.wt.=342âg) that should be dissolved in 100âg water in order to produce a solution with a 105.0 o C difference between the boiling point and freezing point is
(Given that kÆ=1.86 Kkgmolâ1 and kb=0.52 Kkgmolâ1 for water)
Report your answer by rounding it up to the nearest whole number.

Options:

The correct answer is:
72

Solution:
Let the molality of the soluiton is m.
ΔT f = K f ×m and ΔT b = K f b×m
ΔT f + ΔT b =( K f + K b )×m=(1.86+0.51)×m
=2.37×m
Hence, m= 5 2.37 =2.11
and 2.11×100 1000 =0.211 moles of sucrose must be dissolved in 100 gram of water.
Required mass of sucrose = 0.211 mol×342 g/mole=72.2 g

Topic: Chemistry, Solutions, JEE Main
The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S in the presence of conc. HCl (assuming 100% conversion) is

Options:
A 0.25 mol
B 0.50 mol
C 0.333 mol
D 0.125 mol

The correct answer is:
0.125 mol

Solution:
2H3AsO4+5H2S→Conc.HClAs2S5+8H2O
2 moles of Arsenic Acid → 1 mole of Arsenic Pentasulphide
1 moles of Arsenic Acid → 1/2 mole of Arsenic Pentasulphide
Molar mass of H3AsO4=141; Molar mass of As2S5=308
∴ Number of moles of H3AsO4=35.5141=0.25
∴ Number of moles of As2S5=0.252=0.125 mol

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2016 (09 Apr Online)
The amount of $\mathrm{BaSO}_4$ formed upon mixing 100 $\mathrm{mL}$ of $20.8 \% \mathrm{BaCl}_2$ solution with $50 \mathrm{~mL}$ of $9.8 \%$ $\mathrm{H}_2 \mathrm{SO}_4$ solution will be: $(\mathrm{Ba}=137, \mathrm{Cl}=35.5, \mathrm{~S}=32, \mathrm{H}=1$ and $\mathrm{O}=16$ )

Options:
A
$23.3 \mathrm{~g}$

B
$11.65 \mathrm{~g}$

C
$30.6 \mathrm{~g}$

D
$33.2 \mathrm{~g}$

The correct answer is:
$23.3 \mathrm{~g}$

Solution:
$\mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{BaSO}_4+2 \mathrm{HCl}$
$\begin{array}{llll}208 \mathrm{~g} & 98 \mathrm{~g} & 233 \mathrm{~g} & 73 \mathrm{~g} \\ 20.8 \mathrm{~g} & 9.8 \mathrm{~g} & 23.3 \mathrm{~g} & 7.3 \mathrm{~g}\end{array}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2014 (12 Apr Online)
The amount of charge in F (Faraday) required to obtain one mole of iron from Fe3O4 is ______. (Round off the answer to the nearest integer)

Options:

The correct answer is:
3

Solution:
Fe3O4 is a mixed oxide(Fe+2O . Fe+32O3)

The net change in oxidation can be calculated as follows,

Fe3O4⟶+8e-3Fe

For 3 moles of formation of iron, 8 moles of electrons are involved.

One mole of electrons charge is called one Faraday.

Charge for 1 mole Fe=83F

=2.67 F

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The amount of electricity in Coulomb required for the oxidation of 1 mol of H2O to O2 is ______ ×105C.

Options:

The correct answer is:
2

Solution:
H2O→12O2+2H++2e-

Electricity required to oxidise 1mol H2O to O2=2F

WE=Q96500

mole ×n-factor =Q96500

1×2=Q96500

Q=2×96500C

=1.93×105C

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The amount of electricity required to produce \( 1 \) mole of \( \mathrm{Cu} \) from \( \mathrm{CuSO}_{4} \) solution will be

Options:
A \( 1 \mathrm{~F} \)
B \( 2.33 \mathrm{~F} \)
C \( 2 \mathrm{~F} \)
D \( 1.33 \mathrm{~F} \)

The correct answer is:
\( 2 \mathrm{~F} \)

Solution:
CuSO4 ⇌ Cu2++ SO42-
At cathode: Cu2+aq+ 2e- ⟶Cu(s)
At anode: 2H2Ol⟶ O2g+ 4H+aq+ 4e-
2e- gained by Cu2+ at cathode to reduce to Cus. Thus, for the production of one mole of copper from copper sulphate, 2 F of electricity is required.

Topic: Chemistry, Electrochemistry, JEE Main
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly: (surface tension of soap solution = 0.03 N m-1)

Options:
A 50.1×10-4 J
B 30.16×10-4 J
C 5.06×10-4 J
D 3.01×10-4 J

The correct answer is:
3.01×10-4 J

Solution:
Soap bubble has two free surface.

Therefore, amount of energy required =2S×A=20.03×4π2×10-22=3.01×10-4 J

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The amount of $\mathrm{H}_{2} \mathrm{O}_{2}$ present in 1 litre of $1.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{O}_{2}$ solution, is:

Options:
A $25.5 \mathrm{~g}$
B $3.0 \mathrm{~g}$
C $8.0 \mathrm{~g}$
D $2.5 \mathrm{~g}$

The correct answer is:
$25.5 \mathrm{~g}$

Solution:
Molecular weight of $\mathrm{H}_{2} \mathrm{O}_{2}=34$ Equivalent weight of $\mathrm{H}_{2} \mathrm{O}_{2}=17$

$\therefore 1 \mathrm{~L}$ of $1 \mathrm{~N} \mathrm{H}_{2} \mathrm{O}_{2} \mathrm{has}=17 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$

$\therefore 1$ Lof $1.5 \mathrm{NH}_{2} \mathrm{O}_{2}$ has $=1.5 \times 17=25.5 \mathrm{~g} \mathrm{of} \mathrm{H}_{2} \mathrm{O}_{2}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The amount of heat energy required to freeze 4.5 g of water at 6 Â°C to ice at 0 Â°C is [S=4190 J kg-1 K-1, L=3.33Ã105 J kg-1]

Options:
A 1612 J
B 1512 J
C 1132 J
D 1499 J

The correct answer is:
1612 J

Solution:
Total heat loss Q=ms∆T+mL
⇒ Q=(4.5×10-3) 41906-0+(4.5×10-3)(3.33×105)
=113+1499=1612 J

Topic: Physics, Thermodynamics, JEE Main
The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0 °C to 50 °C when no work is done is (R is the universal gas constant)

Options:
A 250R
B 750R
C 175R
D 500R

The correct answer is:
500R

Solution:
ΔQ=ΔU+ΔW

Here ΔW=0

ΔQ=ΔU=nCVΔT

ΔQ=4×5R2(50)=500R

Hence option (4).

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The amount of heat produced in an electric circuit depends upon the current $(I)$, resistance $(R)$ and time $(t)$. If the error made in the measurements of the above quantities are $2 \%, 1 \%$ and $1 \%$ respectively then the maximum possible error in the total heat produced will be

Options:
A
$1 \%$

B
$2 \%$

C
$6 \%$

D
$3 \%$

The correct answer is:
$6 \%$

Solution:
Given : $\frac{\Delta I}{I} \times 100=2 \%$;
$\frac{\Delta R}{R} \times 100=1 \%$ and $\frac{\Delta t}{t} \times 100=1 \%$
Heat produced, $H=I^2 R t$
Maximum possible error in heat produced,
$$
\begin{aligned}
& =\frac{\Delta H}{H} \times 100 \\
& =2\left(\frac{\Delta I}{I} \times 100\right)+\frac{\Delta R}{R} \times 100+\frac{\Delta t}{t} \times 100 \\
& =2 \times 2 \%+1 \%+1 \%=6 \%
\end{aligned}
$$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2012 (19 May Online)
The amount of oxygen in 3.6 moles of water is:

Options:
A 57.6 g
B 115.2 g
C 18.4 g
D 28.8 g

The correct answer is:
57.6 g

Solution:
∴ 1 mole of H2O containS=16 g O
3.6 mole of H2O contains=16×3.6 g O=57.6 g O

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2014 (09 Apr Online)
The amount of work done in an adiabatic expansion from temperature $T$ to $T_1$ is

Options:
A $R\left(T-T_1\right)$
B $\frac{R}{\gamma-1}\left(T-T_1\right)$
C $R T$
D $R\left(T-T_1\right)(\gamma-1)$

The correct answer is:
$\frac{R}{\gamma-1}\left(T-T_1\right)$

Solution:
$W_{a d i}=\frac{R}{\gamma-1}\left(T_i-T_f\right)=\frac{R}{\gamma-1}\left(T-T_1\right)$

Topic: Physics, Thermodynamics, JEE Main
The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $\underset{\sim}{D}$ is ( $T=$ surface tension of the solution)

Options:
A $4 \pi\left(D^2-d^2\right) T$
B $8 \pi\left(D^2-d^2\right) T$
C $\pi\left(D^2-d^2\right) T$
D $2 \pi\left(D^2-d^2\right) T$

The correct answer is:
$2 \pi\left(D^2-d^2\right) T$

Solution:
$W=T \times 8 \pi\left(r_2^2-r_1^2\right)=T \times 8 \pi\left(\frac{D^2}{4}-\frac{d^2}{4}\right)$
$=2 \pi\left(D^2-d^2\right) T$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The amphoteric hydroxide is:

Options:
A $\mathrm{Be}(\mathrm{OH})_{2}$
B $\mathrm{Ca}(\mathrm{OH})_{2}$
C $\mathrm{Mg}(\mathrm{OH})_{2}$
D $\operatorname{Sr}(\mathrm{OH})$

The correct answer is:
$\mathrm{Be}(\mathrm{OH})_{2}$

Solution:
$\mathrm{Be}(\mathrm{OH})$, is amphoteric in nature.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V, what increase, in plate potential will keep the plate current unchanged?

Options:
A 5 V

B 10 V

C 0.2 V

D 50 V

The correct answer is:
10 V

Solution:
μ=ΔVpΔVg
⇒ ΔVp= -μ×ΔVs
=-50-20=10 V

Topic: Physics, Semiconductors, JEE Main
The amplitude and the time period in a S.H.M. is $0.5 \mathrm{~cm}$ and $0.4 \mathrm{sec}$ respectively. If the initial phase is $\pi / 2$ radian, then the equation of S.H.M. will be

Options:
A $y=0.5 \sin 5 \pi t$
B $y=0.5 \sin 4 \pi t$
C $y=0.5 \sin 2.5 \pi t$
D $y=0.5 \cos 5 \pi t$

The correct answer is:
$y=0.5 \cos 5 \pi t$

Solution:
$\begin{gathered}y=a \sin (\omega t+\varphi) \\ =a \sin (2 \pi T t+\varphi) \\ y=0.5 \sin (2 \pi 0.4 t+\pi 2) \\ y=0.5 \sin (5 \pi t+\pi 2) \\ =0.5 \cos 5 \pi t\end{gathered}$

Topic: Physics, Oscillations, JEE Main
The amplitude of 15sin(1000πt) is modulated by 10sin(4πt) signal. The amplitude modulated signal contains frequency (ies) of

(A) 500 Hz

(B) 2 Hz

(C) 250 Hz

(D) 498 Hz

(E) 502 Hz

Choose the correct answer from the options given below:

Options:
A A only
B A, D and E only
C B only
D A and B only

The correct answer is:
A, D and E only

Solution:
The frequency components of amplitude modulated signal are fc-fm, fc, fc+fm

Here, carrier wave frequency is

fc=100π2π=500 Hz

Modulating wave frequency is

fm=4π2π=2 Hz

So, the frequencies in amplitude modulated signal are =498 Hz, 500 Hz, 502 Hz

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to α times its original magnitude, where α equals :

Options:
A 0.729
B 0.6
C 0.7
D 0.81

The correct answer is:
0.729

Solution:
A=A0 e-bt2m
After 5 second
0.9 A 0 = A 0 e - b 5 2 m ...... (i)
After 10 more second
A = A 0 e - b 1 5 2 m ...... (ii)
From (i) and (ii)
A = 0.729 A 0

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (07 Apr)
The amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass =500 g, Decay constant =20 g s-1 then how much time is required for the amplitude of the system to drop to half of its initial value? ln2=0.693

Options:
A 34.65 s
B 17.32 s
C 0.034 s
D 15.1 s

The correct answer is:
34.65 s

Solution:
A=A0e-γt=A0e-bt2m

A02=A0e-bt2m

bt2m=ln2

t=2mbln2=2×500×0.69320

t=34.65 second.

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is: ______ cm.

Options:

The correct answer is:
2

Solution:
Let x be the displacement at which its kinetic energy will be 25% more than the potential energy.

Given here, A=3 cm and KE=PE+PE4

So, KE=54PE

⇒12mω2A2-x2=54×12mω2x2

⇒A2-x2=54x2

⇒9x24=A2

⇒x=23A

∴ x=23×3 cm

⇒x=2 cm

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The amplitude of a simple pendulum is 10cm. When the pendulum is at a displacement of 4cm from the mean position, then what is the ratio of kinetic and potential energies at that point?

Options:

The correct answer is:
5.25

Solution:
K.EP.E=12mw2A2-x212mw2 x2
K.EP.E=A2-x2x2
=102-4242
=8416
K.EP.E=214=5.25

Topic: Physics, Oscillations, JEE Main
The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbondioxide will be close to (ln 5 = 1.601, ln 2 = 0.693).

Options:
A 231 s
B 142 s
C 208 s
D 161 s

The correct answer is:
161 s

Solution:
A= A 0 e âbt 2m
b is related to viscosity
8=10 e â b a 40 2m
5=10 e â b C o 2 t 2m
ln 5 4 ln2 = 40 t Ã1.3
t= 40Ã( ln2 )Ã1.3 ln5â2ln2
= 40Ã0.693Ã1.3 1.601â2Ã0.693
â161sec

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (09 Apr Online)
The amplitude of a wave disturbance propagating in positive direction of x-axis is given by y=11+x2 at t=0 and by y=11+x-12 at t=2 s, where x and y are in meters. The shape of the wave disturbance does not change during propagation. The velocity of the wave is

Options:
A 0.5 ms-1
B 2.0 ms-1
C 1.0 ms-1
D 4.0 ms-1

The correct answer is:
0.5 ms-1

Solution:
In a wave equation, x and t must be related in the form (x-vt). Therefore, we rewrite the given equation as
y=11+x-vt2
For t=0, it becomes y=11+x2
And for t=2, it becomes
y=11+x-2v2=11+x-12
â´2v=1 or v=0.5 ms-1

Topic: Physics, Waves and Sound, JEE Main
The amplitude of a wave, represented by displacement equation
$y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

Options:
A $\frac{a+b}{a b}$
B $\frac{\sqrt{a}+\sqrt{b}}{a b}$
C $\frac{\sqrt{a} \pm \sqrt{b}}{a b}$
D $\sqrt{\frac{a+b}{a b}}$

The correct answer is:
$\sqrt{\frac{a+b}{a b}}$

Solution:
The displacement equation presented by
$\begin{aligned} & y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t \\ & =\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin \left(\omega t+\frac{\pi}{2}\right)\end{aligned}$
Here phase difference $=\frac{\pi}{2}$
The amplitude of wave,
$A=\sqrt{\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2}=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{a b}}$

Topic: Physics, Waves and Sound, JEE Main
The amplitude of $e^{e^{-i \theta}}$ is equal to

Options:
A $\sin \theta$
B $-\sin \theta$
C $e^{\cos \theta}$
D $e^{\sin \theta}$

The correct answer is:
$-\sin \theta$

Solution:
$\begin{aligned} & \text { Let } z=e^{e^{-i \theta}}=e^{\cos \theta-i \sin \theta}=e^{\cos \theta} e^{-i \sin \theta} \\ & z=e^{\cos \theta}[\cos (\sin \theta)-i \sin (\sin \theta)] \\ & z=e^{\cos \theta} \cos (\sin \theta)-i e^{\cos \theta} \sin (\sin \theta) \\ & \operatorname{amp}(z)=\tan ^{-1}\left[-\frac{e^{\cos \theta} \sin (\sin \theta)}{e^{\cos \theta} \cos (\sin \theta)}\right] \\ & =\tan ^{-1}[\tan (-\sin \theta)]=-\sin \theta\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is 6.0 × 10–7 T. The maximum value of electric field in the electromagnetic wave is

Options:
A 2×1015 V m–1
B 180 V m–1
C 6.0 × 10–7 V m–1
D 5×1014 V m–1

The correct answer is:
180 V m–1

Solution:
The relation between the electric field and magnetic field is given by

E0 = cB0 .

The given data is

B0=6×10-7 T

c=3×108 m s-1

Thus, the maximum value of the electric field is

E0= 3×108×6×10–7=180 V m-1.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The amplitude of upper and lower side bands of AM wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5 V amplitude are a10 V and b10 V respectively. Then the value of ab is_____.

Options:

The correct answer is:
1

Solution:
In radio communications, a sideband is a band of frequencies higher than or lower than the carrier frequency, that are the result of the modulation process. The signal components above the carrier frequency constitute the upper sideband (USB), and those below the carrier frequency constitute the lower sideband (LSB).

a10=b10=μAC2
⇒ab=1

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The angle between curves $y^2=4 x$ and $x^2+y^2=5$ at $(1,2)$ is

Options:
A $\tan ^{-1}(3)$
B $\tan ^{-1}(2)$
C $\frac{\pi}{2}$
D $\frac{\pi}{4}$

The correct answer is:
$\tan ^{-1}(3)$

Solution:
For curve $y^2=4 x \Rightarrow \frac{d y}{d x}=\frac{4}{2 y}$
$\left(\frac{d y}{d x}\right)_{(1,2)}=1 \quad$ and for curve $x^2+y^2=5 \Rightarrow \frac{d y}{d x}=\frac{-x}{y}$
$\therefore\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{-1}{2}$
$\therefore$ Angle between the curves is $\theta=\tan ^{-1}\left|\frac{\frac{-1}{2}-1}{1+\left(\frac{-1}{2}\right)}\right|=\tan ^{-1}(3)$

Topic: Mathematics, Application of Derivatives, JEE Main
The angle between lines x-22=y-3-2=z-51 and x-21=y-32=z-52 is ________

Options:
A 30o
B 60o
C 45o
D 90o

The correct answer is:
90o

Solution:
Given: Lines are x-22=y-3-2=z-52
and   x-21=y-32=z-52
Direction ratio of lines are 2,-2,1 and 1,2,2
Then, angle between lines are
cos⁡θ=2.1+-22+1.24+4+11+4+4=0
θ=90o

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the asymptotes of the hyperbola $x^2-3 y^2=3$ is

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
Given equation of hyperbola is
$\begin{aligned}
x^2-3 y^2 & =3 \\
\Rightarrow \quad \frac{x^2}{3}-\frac{y^2}{1} & =1
\end{aligned}$
Here, $a^2=3$ and $b^2=1, a>b$.
Now, the equation of asymptote of this hyperbola is,
$\begin{aligned} y & = \pm \frac{b}{a} x \\ \Rightarrow \quad y & = \pm \frac{1}{\sqrt{3}} x\end{aligned}$
Let slope of asymptote (i) is, $m_1=\frac{1}{\sqrt{3}}$ and slope of asymptote (ii) is, $m_2=\frac{-1}{\sqrt{3}}$
Let $\theta$ be the angle between both asymptote, then
$\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{1 / \sqrt{3}+1 / \sqrt{3}}{1-1 / 3}\right|$
$\begin{array}{cc}\Rightarrow & \tan \theta=\left|\frac{2 / \sqrt{3}}{2 / 3}\right|=\sqrt{3} \\ \Rightarrow & \tan \theta=\tan 60^{\circ} \\ \Rightarrow & \theta=\pi / 3\end{array}$

Topic: Mathematics, Hyperbola, JEE Main
The angle between the chords of the circle x2+y2=100, which passes through the point 7,1 and also divides the circumference of the circle into two arcs whose lengths are in the ratio 2:1, is equal to

Options:
A π6
B π3
C π2
D 2π3

The correct answer is:
π2

Solution:
Let the chord is AB which subtends an angle θ at the centre 0,0
⇒θ+2θ=360o
⇒θ=120o=∠AOB
Let the distance of O from AB=h
Then, cos⁡60o=h10=12⇒h=5
Let the equation of the chord is y-1x-7=m
⇒mx-y+1-7m=0 whose distance from 0,0 is equal to 5
⇒0-0+1-7m1+m2=5
⇒1-14m+49m2=25+25m2
⇒24m2-14m-24=0⇒m1m2=-1
⇒ Chords are perpendicular

Topic: Mathematics, Circle, JEE Main
The angle between the curves $y=\sin x$ and $y=\cos x$ is

Options:
A $\tan ^{-1}(2 \sqrt{2})$
B $\tan ^{-1}(3 \sqrt{2})$
C $\tan ^{-1}(3 \sqrt{3})$
D $\tan ^{-1}(5 \sqrt{2})$

The correct answer is:
$\tan ^{-1}(2 \sqrt{2})$

Solution:
$\begin{aligned} & \text { If } \sin x=\cos x \Rightarrow x=\pi / 4 \\ & \qquad y=\sin x \Rightarrow\left(\frac{d y}{d x}\right)_{x=\pi / 4}=\frac{1}{\sqrt{2}} \\ & \text { If } y=\cos x \Rightarrow\left(\frac{d y}{d x}\right)_{x=\pi / 4}=\frac{-1}{\sqrt{2}} \\ & \tan \theta=\frac{m_1-m_2}{1+m_1 m_2}=2 \sqrt{2} \Rightarrow \theta=\tan ^{-1}(2 \sqrt{2})\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
The angle between the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2}$ and the plane $x+y+4=0$, is

Options:
A $0^{\circ}$
B $30^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
$\sin \theta=\frac{2+1}{\sqrt{2} \times \sqrt{9}} \Rightarrow \theta=45^{\circ}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main