datavorous/entrance-exam-dataset huggingface dataset:
When 92U235 is bombarded with one neutron, fission occurs and the products are three neutrons, 36Kr94 , and

Options:
A 56Ba141
B 54Xe139
C 56Ba139
D 58I142

The correct answer is:
56Ba139

Solution:
When a slow neutron strikes a U235 nucleus it is absorbed by the nucleus and the following reaction occurs.
92U235 + 0n1→ 36Kr94 + 56Ba139 + 30n1+ energy
Hence, 56Ba139 is another product.

Topic: Physics, Nuclear Physics, JEE Main
When $100 \mathrm{~g}$ of a liquid $\mathrm{A}$ at $100^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of a liquid $B$ at temperature $75^{\circ} \mathrm{C}$, the temperature of the mixture becomes $90^{\circ} \mathrm{C}$. The temperature of the mixture, if $100 \mathrm{~g}$ of liquid $A$ at $100^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of liquid $\mathrm{B}$ at $50^{\circ} \mathrm{C}$, will be:

Options:
A $85^{\circ} \mathrm{C}$
B $60^{\circ} \mathrm{C}$
C $80^{\circ} \mathrm{C}$
D $70^{\circ} \mathrm{C}$

The correct answer is:
$80^{\circ} \mathrm{C}$

Solution:
Heat loss $=$ Heat gain $=\mathrm{mS} \Delta \theta$

$\mathrm{So}, \mathrm{m}_{\mathrm{A}} \mathrm{S}_{\mathrm{A}} \Delta \theta_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{S}_{\mathrm{B}} \Delta \theta_{\mathrm{B}}$

$\begin{array}{l}

\Rightarrow 100 \times \mathrm{S}_{\mathrm{A}} \times(100-90)=50 \times \mathrm{S}_{\mathrm{B}} \times(90-75) \\

2 \mathrm{~S}_{\mathrm{A}}=1.5 \mathrm{~S}_{\mathrm{B}} \Rightarrow \mathrm{S}_{\mathrm{A}}=\frac{3}{4} \mathrm{~S}_{\mathrm{B}}

\end{array}$

Now, $100 \times \mathrm{S}_{\mathrm{A}} \times(100-\theta)=50 \times \mathrm{S}_{\mathrm{B}} \times(\theta-50)$

$2 \times\left(\frac{3}{4}\right) \times(100-\theta)=(\theta-50)$

$\begin{array}{l}

300-3 \theta=2 \theta-100 \\

400=5 \theta \Rightarrow \theta=80^{\circ} \mathrm{C}

\end{array}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
When $100 \mathrm{~g}$ of a liquid $\mathrm{A}$ at $100^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of a liquid $B$ at temperature $75^{\circ} \mathrm{C}$, the temperature of the mixture becomes $90^{\circ} \mathrm{C}$. The temperature of the mixture, if $100 \mathrm{~g}$ of liquid $A$ at $100^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of liquid $\mathrm{B}$ at $50^{\circ} \mathrm{C}$, will be:

Options:
A $85^{\circ} \mathrm{C}$
B $60^{\circ} \mathrm{C}$
C $80^{\circ} \mathrm{C}$
D $70^{\circ} \mathrm{C}$

The correct answer is:
$80^{\circ} \mathrm{C}$

Solution:
Heat loss $=$ Heat gain $=\mathrm{mS} \Delta \theta$

$\mathrm{So}, \mathrm{m}_{\mathrm{A}} \mathrm{S}_{\mathrm{A}} \Delta \theta_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{S}_{\mathrm{B}} \Delta \theta_{\mathrm{B}}$

$\begin{array}{l}

\Rightarrow 100 \times \mathrm{S}_{\mathrm{A}} \times(100-90)=50 \times \mathrm{S}_{\mathrm{B}} \times(90-75) \\

2 \mathrm{~S}_{\mathrm{A}}=1.5 \mathrm{~S}_{\mathrm{B}} \Rightarrow \mathrm{S}_{\mathrm{A}}=\frac{3}{4} \mathrm{~S}_{\mathrm{B}}

\end{array}$

Now, $100 \times \mathrm{S}_{\mathrm{A}} \times(100-\theta)=50 \times \mathrm{S}_{\mathrm{B}} \times(\theta-50)$

$2 \times\left(\frac{3}{4}\right) \times(100-\theta)=(\theta-50)$

$\begin{array}{l}

300-3 \theta=2 \theta-100 \\

400=5 \theta \Rightarrow \theta=80^{\circ} \mathrm{C}

\end{array}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (11 Jan Shift 2)
When 100 ml of 0.2 M H2SO4 is added to 100 ml of 0.2 M NaOH. What is the nature of the resulting solution obtained?

Options:
A Neutral
B Slightly basic
C Basic
D Acidic

The correct answer is:
Acidic

Solution:
M.eq.âofâ0.2âMâ H 2 SO 4 â=â 2âÃâ0.2âM 1000 âÃâ100â=â0.04
M.eq. of â2âMâNaOHâ=â 0.2 1000 Ãâ100â=â0.02
milliâmolesâofâ H + âleftâ=â0.04â0.02â=â0.02
Totalâvolumeâ [H + ]â=â200â=â .02 200 â=â0.0001â=â 10 â4 âM
pH=4 .

Topic: Chemistry, Ionic Equilibrium, JEE Main
When 200 mL of 0.2 M acetic acid is shaken with 0.6 g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1 M. The mass of acetic acid adsorbed per gram of carbon is g.

Options:

The correct answer is:
2

Solution:
Mole of acetic acid absorbed (on 0.6 gram charcoal) =(0.2 X200)10-3 – (0.1 X 200)10-3=20 X 10-3

Weight of CH3COOH absorbed (on 0.6 gram charcoal)=(20 X10-3 ) 60 = 1200 X10-3 gram = 1.2 gram

Mass of CH3COOH absorbed per gram=1.20.6=2grams

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
When 500 calories heat is given to the gas X in an isobaric process, its work done comes out as 142.8 calories. The gas X is

Options:
A O 2
B N H 3
C He
D SO2

The correct answer is:
O 2

Solution:
∵ΔT= W nR
∴ Q= nC p (ΔT)= nC p W nR = C p W R
C p  =  QR W  =  500 × 2 142.8  = 7
C p  = 7, indicates that the gas is diatomic. Thus, it should be O 2

Topic: Chemistry, Thermodynamics (C), JEE Main
When 800 mL of 0.5M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is x×10-2M. (Molar mass of nitric acid is 63 g mol-1)

Options:

The correct answer is:
54

Solution:
The number of moles of nitric acid, nHNO3=0.5×0.8

=0.4 mole

nHNO3remains=0.4-11.563

=0.4-0.1825

=0.2175

Molarity =0.2175400×1000

=0.21750.4

=0.5437 mole/lit.

≃0.54 mole/lit.

=54×10-2 mol/lit.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (26 Jul Shift 1)
When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is:

Options:
A 109.0 g
B 98.1 g
C 9.81 g
D 10.9 g

The correct answer is:
9.81 g

Solution:
4e-+4H++C6H5NO2 → C6H4OHNH2+H2O
Molecular Weight of p-aminophenol = 109 g
W (Deposited)   = ZIt
=EF ×I ×t E=M4
W=109 ×9.65 ×60 ×604 ×96500=9.81 g

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2018 (16 Apr Online)
When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of $A g=107.87$ taking as 108$)$ the amount of silver deposited is

Options:
A $10.8 \mathrm{mg}$
B $5.4 \mathrm{mg}$
C $16.2 \mathrm{mg}$
D $21.2 \mathrm{mg}$

The correct answer is:
$10.8 \mathrm{mg}$

Solution:
$W_{A g}=\frac{E_{A g} \times Q}{96500}=\frac{108 \times 9.65}{96500} \quad=1.08 \times 10^{-2} g m=10.8 \mathrm{mg}$

Topic: Chemistry, Electrochemistry, JEE Main
When \( 2^{301} \) is divided by \( 5 \), then find least positive remainder.

Options:

The correct answer is:
2.00

Solution:
2301=2.2300=2.4150=25-1150

Here all terms, except last term are divisible by 5

∴ Remainder =2(last term)=2-1150=2

Topic: Mathematics, Binomial Theorem, JEE Main
When A2 and B2 are allowed to react, the equilibrium constant of the reaction at 27°C is found (KC=4) .
A2(g)+B2(g)⇌2AB(g)
What will be the equilibrium concentration of AB?

Options:
A 1.33M
B 2.66 M
C 0.66 M
D 0.33 M

The correct answer is:
0.66 M

Solution:
KC=[AB]2[A2][B2]
4=[2x/4]22-x44-x4
4=4x2x2-6x+8
x2-6x+8=x2
x=86=43
AB=2x4=24×43=0.66 .

Topic: Chemistry, Chemical Equilibrium, JEE Main
When a $4 \mathrm{~kg}$ mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by $2 \mathrm{cms}$, The work required to be done by an external agent in stretching this spring by $5 \mathrm{cms}$ will be $\left(g=9.8\right.$ metres/sexc $\left.\mathrm{c}^2\right)$

Options:
A 4.900 joule
B 2.450 joule
C 0.495 joule
D 0.245 joule

The correct answer is:
2.450 joule

Solution:
$K=\frac{F}{x}=\frac{40}{2 \times 10^{-2}}=0.2 \mathrm{Nm}$
Work done $=\frac{1}{2} \mathrm{Kx}^2=\frac{1}{2} \times(0.2) \times(0.05)^2=2.5 \mathrm{~J}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
When a 60 W electric heater is immersed in a gas for 100 s in a constant volume container with adiabatic walls, the temperature of the gas rises by 5°C. The heat capacity of the given gas is JK-1 (Nearest integer)

Options:

The correct answer is:
1200

Solution:
Heat capacity = Heat absorbed change in temperature

1 Watt = 1 Joule per second

Hence, heat absorbed =60×100J

Change in temperature =5°C

Heat capacity=60×100 J5

=1200JK-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (08 Apr Shift 1)
When a $100 \mathrm{Kg}$ mass is suspended from wire of length 1 meter and and area $1 \mathrm{~cm}^2$, it length increases.
Find the elastic potential energy stored in the wire

Options:
A $5 \times 10^{-2} \mathrm{~J}$
B $4\times 10^{-2} \mathrm{~J}$
C $6\times 10^{-2}\mathrm{~J}$
D $9\times 10^{-2} \mathrm{~J}$

The correct answers are:
$5 \times 10^{-2} \mathrm{~J}$

Solution:
$\checkmark$ Answer
Given $\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^2$ and Young Modulus of the material of the wire $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
Stress $=$ Load/Area
$=\frac{m g}{A}=\frac{100 \times 10}{10^4}=10^7 \mathrm{~N} / \mathrm{m}^2$
Now
Young Modulus=Stress/Strain
So strain $=$ Stress/Young Modulus
$=5 \times 10^{-5}$
Now we know that
Strain $=$ Increase in length/Initial length
$\Rightarrow$ Increase in Length=strain * Initial length
$=5 \times 10^{-5} \mathrm{~m}$
Elastic Potential energy
$\begin{aligned} & =\frac{1}{2} \text { stress } \times \text { strain } \times \text { Volume } \\ & =5 \times 10^{-2} \mathrm{~J}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
When a ball is thrown with a velocity of 50 m s-1 at an angle 30° with the horizontal, it remains in the air for seconds. (take g=10 m s-2

Options:
A 5
B 2.5
C 1.25
D 0.625

The correct answer is:
5

Solution:
The vertical component of velocity is, 50sin30° therefore, u=50sin30° and when the ball will be thrown vertically the net displacement along the vertical direction of body will be zero, since body will again come to ground after time t.

Now using the equation of uniform motion, s=ut+12at2

⇒0=50sin30°t+12-gt2⇒5t2-50sin30°t=0 ⋯⋯g=10 m s-2⇒t5t-25=0

Solving the equation for t,

t=0, and t=5 s

Thus, the ball will remain in air for 5 s

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
When a battery of emf 8 V with internal resistance 0.5 Î© is charged by a 120 V DC supply using a series resistance of 15.5 Î©, then the terminal voltage of the battery is

Options:
A 11.5 V

B 20 V
C 21.5 V
D 12.3 V

The correct answer is:
11.5 V

Solution:
Here, emf of the battery =8 V
The internal resistance of the battery, r=0.5 Ω
The voltage of D. C supply = 120 V
External resistance, R=15.5 Ω
When the battery of emf 8 V is charged from a DC supply of 120 V, the effective emf in the circular is
ε=120-8=112 V
The total resistance of the circuit =R+r
=15.5+0.5 Ω=16 Ω
∴ Current in the circuit during charging
I=εR+r=11216=7 A
∴ The terminal voltage of the battery,
V=emf of the battery+voltage drop across the battery
=8+Ir=8+70.5=11.5 V

Topic: Physics, Current Electricity, JEE Main
When a battery of emf 8 V with internal resistance 0.5 Î© is charged by a 120 V DC supply using a series resistance of 15.5 Î©, then what is the terminal voltage (in volt) of the battery?

Options:

The correct answer is:
11.5

Solution:
Here, emf of the battery =8 V
The internal resistance of the battery, r=0.5 Î©
The voltage of D. C supply = 120 V
External resistance, R=15.5 Î©
When the battery of emf 8 V is charged from a DC supply of 120 V, the effective emf in the circular is
Îµ=120-8=112 V
The total resistance of the circuit =R+r
=15.5+0.5 Î©=16 Î©
â´ Current in the circuit during charging
I=ÎµR+r=11216=7 A
â´ The terminal voltage of the battery,
V=emf of the battery+voltage drop across the battery
=8+Ir=8+70.5=11.5 V

Topic: Physics, Current Electricity, JEE Main
When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called :

Options:
A Scattering
B Chromatic aberration
C Spherical aberration
D Polarisation

The correct answer is:
Chromatic aberration

Solution:
Due to difference in refractive index of a material for different wavelengths, the focal length will be different for different colours. This will cause a separation of parallel beam of white light. This phenomenon is known as chromatic aberration.

Polarisation is a property applying to transverse waves that specify the geometrical orientation of the oscillations.

When energy waves (such as light, sound, and various electromagnetic waves) are caused to depart from a straight path due to imperfections in the medium it is called Scattering.

Spherical aberration is present when the outer parts of a lens do not bring light rays into the same focus as the central part.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (24 Jan Shift 2)
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index

Options:
A equal to that of glass
B less than one
C greater than that of glass
D less than that of glass

The correct answer is:
equal to that of glass

Solution:
If biconvex lens behave like plane sheet, ray will pass undeviated through it only when medium has same refractive index as that of biconvex lens.

Topic: Physics, Ray Optics, JEE Main
When a block moves down a smooth inclined plane of inclination Î¸, its velocity on reaching the bottom is v. When it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v/n, where n is a number greater than 1. The coefficient of friction between the block and the rough surface is

Options:
A Î¼=1-1n2tanâ¡Î¸
B μ=1-1n2cot⁡θ
C μ=1-1n212tanθ
D μ=1-1n212cotθ

The correct answer is:
μ=1-1n2tan⁡θ

Solution:
For smooth incline, v2=2gsinθ×s ……(1)
For rough incline, vn2=2gsinθ-μcosθ×s ……(2)

On solving (1) and (2), we obtain,

μ=1-1n2tanθ

Topic: Physics, Laws of Motion, JEE Main
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes L+l. The elastic potential energy stored in the extended wire is

Options:
A Mgl
B MgL
C 12Mgl
D 12 MgL

The correct answer is:
12Mgl

Solution:
12Kl2= energy
Kl=Mg
K=Mgl
∴ Energy =12 Kl2=12Mgl l2
=Mgl2

Topic: Physics, Mechanical Properties of Solids, JEE Main
When a body is taken from the equator to the poles, its weight

Options:
A Remains same
B increases
C decreases
D increase at N-pole & decreases at S-pole

The correct answer is:
increases

Solution:
Because acceleration due to gravity increase so weight also increased.

Topic: Physics, Gravitation, NEET
When a body moves with a constant speed along a circle

Options:
A no acceleration is produced on the body
B no work is done on it
C its velocity remains constant
D no force acts on the body

The correct answer is:
no work is done on it

Solution:
When a body moves with constant speed v along a circle, it has a centripetal acceleration but no linear acceleration.
The centripetal force F acting on the body is at right angle to the displacement s.
Since,
$$
\begin{aligned}
W & =F \cdot s=F s \cos \theta \\
& =0 \quad\left(\because \cos 90^{\circ}=0\right)
\end{aligned}
$$
Hence, no work is done on it.

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
When a body of mass $8 \mathrm{~kg}$ is attached to a spring balance, the reading of the balance is $20 \mathrm{~cm}$. Instead of $8 \mathrm{~kg}$, if another body of mass $M$ is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is $\frac{\pi}{5} \mathrm{~s}$, then the value of $\mathrm{M}$ is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $4 \mathrm{~kg}$
B $6 \mathrm{~kg}$
C $8 \mathrm{~kg}$
D $9 \mathrm{~kg}$

The correct answer is:
$4 \mathrm{~kg}$

Solution:
Mass of body, $\mathrm{m}=8 \mathrm{~kg}$
Period of oscillation, $\mathrm{T}=\frac{\pi}{5} \mathrm{~s}$
Reading of the balance, $x=20 \mathrm{~cm}=0.2 \mathrm{~m}$
For mass $8 \mathrm{~kg}$
$\mathrm{mg}=\mathrm{kx} \Rightarrow \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}$
$=\frac{8 \times 10}{0.2}=400 \mathrm{~N} / \mathrm{m}$
For mass $\mathrm{M}$
$\begin{aligned} & \omega^2=\frac{k}{M} \\ & M=\frac{k T^2}{4 \pi^2} \quad\left(\because \omega=\frac{2 \pi}{T}\right) \\ & =\frac{400 \times \pi^2}{4 \pi^2}=4 \mathrm{~kg}\end{aligned}$

Topic: Physics, Oscillations, JEE Main
When a body slides down from rest along a smooth inclined plane making an angle of 30° with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time αT, where α is a constant greater than 1. The co-efficient of friction between the body and the rough plane is 1xα2-1α2 where x=__________.

Options:

The correct answer is:
3

Solution:
On smooth incline,

a=gsin30°

by S=ut+12at2

S=12g2T2=g4T2 ...(i)

On rough incline

a=gsin30°-μgcos30°

by S=ut+12at2

S=14 g(1-3μ)(αT)2 ...(ii)

By (i) and (ii)

14gT2=14g(1-3μ)α2T2

⇒1-3μ=1α2 ⇒μ=α2-1α213

⇒x=3.00

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (01 Sep Shift 2)
When a brown compound of \( M n(A) \) is treated with \( H C l \), it gives a gas (B). The gas (B) taken in
excess reacts with \( \mathrm{NH}_{3} \) to give an explosive compound (C).
The compounds A, B and C are;

Options:
A \( \mathrm{A}=\mathrm{MnO}_{2}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{3} \)
B \( A=M n O, B=C l_{2}, C=N H_{4} C l \)
C \( \mathrm{A}=\mathrm{Mn}_{3} \mathrm{O}_{4}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{3} \)
D \( \mathrm{A}=\mathrm{MnO}_{3}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{2} \)

The correct answer is:
\( \mathrm{A}=\mathrm{MnO}_{2}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{3} \)

Solution:
When a brown compound of \( M n(A) \) is treated with \( H C l \), it gives a gas (B) as shown below.
\( \mathrm{MnO}_{2}(\mathrm{~A})+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}(\mathrm{~B}) \)
The gas (B) taken in excess reacts with \( \mathrm{NH}_{3} \) to give an explosive compound (C) as shown below.
\( 3 \mathrm{Cl}_{2}(B)+\mathrm{NH}_{3} \rightarrow \mathrm{NCl}_{3}(C)+3 \mathrm{HCl} \)
Therefore, the compounds \( A, B \) and \( C \) are
\( \mathrm{MnO}_{2}, \mathrm{Cl}_{2} \), and \( \mathrm{NCl}_{3} \).

Topic: Chemistry, d and f Block Elements, JEE Main
When a buffer solution, sodium acetate and acetic acid is diluted with water

Options:
A Acetate ion concentration increase
B \( H^{+} \)ion concentration increases
C \( O H^{-} \)ion conc. Increases
D \( H^{+} \)ion concentration remains unaltered

The correct answer is:
\( H^{+} \)ion concentration remains unaltered

Solution:
pH or H+ of a buffer does not change with dilution as per the equation:

pH=pka +log[Salt][Acid]
Therefore, when sodium acetate and acetic acid is diluted with water H+ ion concentration remains unaltered.

Topic: Chemistry, Ionic Equilibrium, JEE Main
When a bus suddenly takes a turn, the passengers are thrown outwards because of

Options:
A Inertia of motion
B Acceleration of motion
C Speed of motion
D Both (b) and (c)

The correct answer is:
Inertia of motion

Solution:
According to law of inertia of motion a body will continue moving with constant velocity until unless any force is acting on it. When a bus is moving straight the passengers inside are also moving straight but when suddenly bus turn left passengers still continue to move straight because no force is acting on passengers so they will move with same magnitude and direction i.e. outwards when bus turns .

Topic: Physics, Laws of Motion, JEE Main
When a capacitor is connected in series LR circuit, the alternating current flowing in the circuit.

Options:
A remains constant.
B increases.
C increases.
D decreases.

The correct answer is:
increases.

Solution:
Impedance of series LR circuit
$Z=\sqrt{R^2+X_L^2}$
When a capacitor is also connected in circuit, Then impedance
$Z_L=\sqrt{R^2+\left(X_L-X_C\right)^2}$
Clearly impedance of circuit decreases $\left(Z_2$I=\frac{V}{Z} \propto \frac{1}{Z}$ in the circuit increases.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (10 Aug Shift 1)
When a capacitor is connected in series LR circuit, the alternating current flowing in the circuit

Options:
A increases.
B decreases.
C remains constant.
D is zero.

The correct answer is:
increases.

Solution:
$z=\sqrt{{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$
With capacitor the value of $z$ decreases. So, current increases.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2020 (20 Oct Shift 2)
When a capacitor is connected to a battery

Options:
A an alternating current flow in the circuit
B no current flow in the circuit
C a current flow for sometime and finally it decreases to zero
D current keeps on increasing and reaches maximum after sometime

The correct answer is:
a current flow for sometime and finally it decreases to zero

Solution:
When a capacitor is connected to a batter (DC source), then it is impossible to flow a steady current but a transient current flows through the capacitors plates, i.e. current flow for sometime and finally it decreases to zero.

Topic: Physics, Capacitance, JEE Main
When a capillary is dipped in water, water rises to a heights $h$. If the length of the capillary is made less than $h$, then

Options:
A The water will come out
B The water will not come out
C The water will not rise
D The water will rise but less than height of capillary

The correct answer is:
The water will not come out

Solution:
Correct option is 2. the water will not come out When a capillary is dipped in water, the water rises to a height h. If the length of the capillary is made less than h, then
The water will not come out but rise to the top and forms a dome over the top of the tube.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
When a car is approaching the observer, the frequency of horn is 100 Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be x3Hz where x=_____ .

Options:

The correct answer is:
200

Solution:
Apparent frequency when source of moving towards observer, f1=100=f0cc-vs ...1

Here, f0=frequency of observer, c=speed of sound and vs= speed of source.

Apparent frequency when source of moving away from observer,

f2=50=f0cc+vs ...2

From above two equations, we have

f1f2=2=c+vsc-vs

⇒2c-2vs=c+vs

⇒3vs=c

Putting the value in equation 1, we have

100=f0c2c3=32f0

⇒f0=2003 Hz

Hence, x=200.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (26 Jul Shift 1)
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops coming at an angle 60∘ from the horizontal. On further increasing the speed of the car to 1+βv, this angle changes to 45∘. The value of β is close to :

Options:
A 0.50
B 0.41
C 0.37
D 0.73

The correct answer is:
0.73

Solution:
tan=60°=vrgv ........1

tan45°=vrg(β+1)v ..........2

3v=(β+1)v

β=3-1=0.732

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (06 Sep Shift 2)
When a card is drawn from a well shuffled pack of cards. what is the probability of getting a Queen?

Options:
A $2 / 13$
B $1 / 13$
C $1 / 26$
D $1 / 52$

The correct answer is:
$1 / 13$

Solution:
Since, $n(S)=52$ and $n(E)=4$
$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{4}{52}=\frac{1}{13}$

Topic: Mathematics, Probability, JEE Main
When a cardiac muscle cell receives a signal to contract, its neighbours are also stimulated to contract. This is due to

Options:
A desmosomes
B intercalated disc
C tight junctions
D sarcoplasm

The correct answer is:
intercalated disc

Solution:
Unique junctions called intercalated discs (gap junctions) link the cells together and define their borders. Intercalated discs are the major portals for cardiac cell-to-cell communication, which is required for coordinated muscle contraction and maintenance of circulation.

Topic: Biology, Structural Organisation in Animals, NEET