datavorous/entrance-exam-dataset huggingface dataset:
A $5 \mathrm{~m}$ long aluminium wire $\left(Y=7 \times 10^{10} \mathrm{Nm}^{-2}\right.$ ) of diameter $3 \mathrm{~mm}$ supports a $40 \mathrm{~kg}$ mass. In order to have the same elongation in the copper wire $\left(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\right)$ of the same
length under the same weight, the diameter should now be (in $\mathrm{mm}$ )

Options:
A $1.75$
B $1.5$
C $2.5$
D $5.0$

The correct answer is:
$2.5$

Solution:
$l=\frac{F L}{\pi r^{2} Y}$
$\Rightarrow r^{2} \propto \frac{1}{Y}(F, L$ and $l$ are constant) $\frac{r_{2}}{r_{1}}=\left[\frac{Y_{1}}{Y_{2}}\right]^{1 / 2}=\left[\frac{7 \times 10^{10}}{12 \times 10^{10}}\right]^{1 / 2}$
$\Rightarrow \quad r_{2}=1.5 \times\left(\frac{7}{12}\right)^{1 / 2}=1.145 \mathrm{~mm}$
$\therefore$ diameter $=2.29 \mathrm{~mm}$.

Topic: Physics, Mechanical Properties of Solids, JEE Main
A \( 5 \mathrm{cc} \) solution of \( \mathrm{H}_{2} \mathrm{O}_{2} \) liberates \( 0.508 \mathrm{~g} \) of iodine from an acidified \( \mathrm{KI} \) solution. Calculate the strength of \( \mathrm{H}_{2} \mathrm{O}_{2} \) solution in terms of volume strength at STP.

Options:

The correct answer is:
4.48

Solution:
Volume of H2O2=5 cc

Weight of I2 liberated = 0.508 g

Reaction of H2O2 with KI,

H2O2+2I-→I2+H2O.

Here, valency factor of iodine 2I-→I2=Change in oxidation number per mole of iodine = 2.

Equivalent weight of I2=Molecular weight of I2valency factor=2542=127 g.

Milli equivalents= Given weight of I2Equivalent weight of I2=0.508127

=Normality×Volume.

Now,

Milli equivalents of H2O2=Milli equivalents of I2

N1V1(H2O2)=N2V2(I2)

Substituting the values, we get:

N1×51000=0.508127⇒N1=0.508×10005×127=0.8 N.

Now,

for H2O2,

5.6× Normality =Volume strength

⇒5.6×0.8=4.48 V.

Topic: Chemistry, Redox Reactions, JEE Main
A 5 metrelong wire is fixed to the ceiling. A weight of $10 \mathrm{~kg}$ is hung at the lower end and is 1 metre above the floor. The wire was elongated by $1 \mathrm{~mm}$. The energy stored in the wire due to stretching is

Options:
A Zero
B 0.05 joule
C 100 joule
D 500 joule

The correct answer is:
0.05 joule

Solution:
$W=\frac{1}{2} \times F \times l=\frac{1}{2} m g l$
$=\frac{1}{2} \times 10 \times 10 \times 1 \times 10^{-1}=0.05 \mathrm{~J}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A 5 V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.

Options:
A ~0.86 A
B ~0.5 A
C ~1.5 A
D ~0.43 A

The correct answer is:
~0.43 A

Solution:
Here only D1 will work and we know for silicon diode, potential drop on D1 will be 0.7 V

I=5-0.710=0.43 A

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is

Options:
A
$0.27 A P_2$ to $P_1$

B
$0.03 A P_1$ to $P_2$

C
$0.03 A P_2$ to $P_1$

D
$0.27 \mathrm{~A} \mathrm{P} P_1$ to $P_2$

The correct answer is:
$0.03 A P_2$ to $P_1$

Solution:
$$
\begin{aligned}
& V_{P_2}-V_{P_1}=\frac{\frac{5}{2}+\frac{0}{10}-\frac{2}{1}}{\frac{1}{2}+\frac{1}{10}+\frac{1}{1}} \\
& I=\frac{V_{P_2}-V_{P_1}}{10}=0.03 \text { from } P_2 \rightarrow P_1
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2008
A 5-ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is :

Options:
A $0.04 \mathrm{ohm}$
B $0.2 \mathrm{ohm}$
C $5 \mathrm{ohm}$
D $0.4 \mathrm{ohm}$

The correct answer is:
$0.04 \mathrm{ohm}$

Solution:
$$
\begin{aligned}
P & =I^2 R \\
\Rightarrow \quad & I=25 \times R \\
\Rightarrow \quad & R=\frac{1}{25}=0.04 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
A 6.50 molal solution of KOHaq. has a density of 1.89 g cm-3. The molarity of the solution is mol dm-3. (Round off to the Nearest Integer).

[Atomic masses: K:39.0 u; O:16.0 u; H:1.0 u]

Options:

The correct answer is:
9

Solution:
6.5 molal KOH=1000 gm solvent has 6.5 moles KOH

so, wt. of solute =6.5×56

=364 gm

wt. of solution =1000+364=1364

Volume of solution =13641.89 mℓ

Molarity =mole of soluteVsolution in Litre

=9.00

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A ${ }^{6 \mu F}$ capacitor is charged from 10 volts to 20 volts. Increase in energy will be

Options:
A $18 \times 10^{-1} \mathrm{~J}$
B $9 \times 10^{-1} \mathrm{~J}$
C $4.5 \times 10^{-4} \mathrm{~J}$
D $9 \times 10^{-6} \mathrm{~J}$

The correct answer is:
$9 \times 10^{-1} \mathrm{~J}$

Solution:
$\begin{aligned} & \Delta E=E_{\text {Final }}-E_{\text {ineial }}=\frac{1}{2} C\left(V_{\text {final }}^2-V_{\text {ntial }}^2\right) \\ & =\frac{1}{2} \times 6 \times\left(20^2-10^2\right) \times 10^{-6} \\ & =3 \times(400-100) \times 10^{-6}=3 \times 300 \times 10^{-6}=9 \times 10^{-4} \mathrm{~J}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
A $6 \%$ solution of urea is isotonic with

Options:
A $1 \mathrm{M}$ solution of glucose
B $0.05 \mathrm{M}$ solution of glucose
C $6 \%$ solution of glucose
D $25 \%$ solution of glucose

The correct answer is:
$1 \mathrm{M}$ solution of glucose

Solution:
Molarity of urea $=\frac{\frac{6}{60}}{\frac{100}{1000}}=1 \mathrm{M}$
Hence, $1 \mathrm{M}$ solution of glucose is isotonic with $6 \%$ urea solution.

Topic: Chemistry, Structure of Atom, JEE Main
A 6 V battery of internal resistance 1 Î© is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of E.M.F 4 V and internal resistance 1 Î© is joined to the point A as shown. The distance of point P from A is Î± Ã10 cm. Find the value of Î±, for which there is no current through the galvanometer. (resistance of AB wire is 5 Î©)

Options:

The correct answer is:
8

Solution:
xAD=45×100=80 cm=8×10 cm

Topic: Physics, Current Electricity, JEE Main
A 6 volt battery is connected to the terminal of a three metre long wire of uniform thickness and resistance of 100 $\mathrm{ohm}$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be:

Options:
A 2 volt
B 3 volt
C 1 volt
D 1.5 volt

The correct answer is:
1 volt

Solution:
$$
\begin{gathered}
\therefore \text { Voltage on } 50 \mathrm{~cm}=\frac{6}{300} \times 50 \\
=1 \text { volt }
\end{gathered}
$$

Topic: Physics, Current Electricity, JEE Main
A 8V Zener diode along with a series resistance R is connected across a 20 V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be _____ Ω.

Options:

The correct answer is:
480

Solution:
Let I1 be the current through Zener diode and I2 through load resistor.

âââââââ

As maximum current flows through Zener diode, hence I2=0.

∴Maximum current, I=I1+I2=25 mA+0=25 mA

Here, Vo is voltage drop across Zener =8 V.
Since I is maximum, hence minimum value of R

R=Vi-VoI=20-825×10-3=480 Ω

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A 9 V battery with an internal resistance of 0.5 Ω is connected across an infinite network, as shown in the figure. All ammeters A1, A2, A3 and voltmeter V are ideal.
Choose the correct statement.

Options:
A Reading of A1 is 18 A.
B Reading of V is 9 V.
C Reading of V is 7 V.
D Reading of A1 is 2 A.

The correct answer is:
Reading of A1 is 2 A.

Solution:
In the given infinite network problems, let, X be an equivalent resistance. If we connect one section of an infinite network with the replacing network using an equivalent resistance, then in this new equivalent circuit the equivalent resistance will still remain the same.

Let, Req=x
x=4x4+x+2
x=8+6x4+x
4x+x2=8+6x
x2-2x-8=0.

On solving,

x = 4 , x= -2
Here, the negative value of resistance is not possible.

So, resistance is x=4 Ω

Reading of A1 is I=VX+r ...(1),

where, V=9 Volt, r = 0·5 Ω , R=4 Ω.

Then, from the above equation,

I =94+0.5=2 A.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (08 Apr Online)
A 10 Ω, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 μs. The average e.m.f. induced in the coil is _____ V.

Options:

The correct answer is:
400

Solution:
The initial current in the coil will be, i=VR=2010=2 A.

Average e.m.f. developed in the coil will be,

ε=-∆ϕ∆t=0-Lit=20×10-3×2100×10-6=400 V.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A 10 μC charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:

Options:
A 7μC,3μC
B 8μC,2μC
C 5μC,5μC
D 9μC,1μC

The correct answer is:
5μC,5μC

Solution:
Let the charges be x μC, q-x μC, where q=10 μC

The force between them is

F=Kx(q-x)r2.

For the force to be maximum,

dFdx=0⇒dFdx=K(q-2x)r2=0⇒x=q2=5 μC

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A $10 \mathrm{~cm}$ long wire is placed horizontally on the surface of water and is gently pulled up with a force of $2 \times 10^{-2} \mathrm{~N}$ to keep the wire in equilibrium. The surface tension, in $\mathrm{Nm}^{-1}$, of water is

Options:
A 0.1
B 0.2
C 0.001
D 0.00

The correct answer is:
0.1

Solution:
$T=\frac{F}{2 l}=\frac{2 \times 10^{-2}}{2 \times 10 \times 10^{-2}}=0.1 \mathrm{~N} / \mathrm{m}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A 10Î¼F capacitor and a 20Î¼F capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected such that those same polarities are connected to each other and no external voltage is applied. The potential difference across capacitors is

Options:
A 4009V
B 8003V
C 400 V
D 200 V

The correct answer is:
4009V

Solution:
Cs=10×2010+20=20030=203 μF
Q =CsV
Q =203 μF×200 V
Q =40003 μC
Now, V = 4000 μC3×30 μF=400090 V=4009 V

Topic: Physics, Electrostatics, JEE Main
A 10 Î¼F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. What is the capacitance value (in Î¼F) of the second capacitor?

Options:

The correct answer is:
15

Solution:
q1=10×50=500 μC, C1=10 μF, C2=?, q2=0
As, V=q1+q2C1+C2
C1+C2=q1+q2V=500+020=25 μF
C2=25-C1=25-10=15 μF

Topic: Physics, Electrostatics, JEE Main
A 10 Î¼F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is

Options:
A 15 Î¼F
B 30 μF
C 20 μF
D 10 μF

The correct answer is:
15 Î¼F

Solution:
q1=10×50=500 μC, C1=10 μF, C2=?, q2=0
As, V=q1+q2C1+C2
C1+C2=q1+q2V=500+020=25 μF
C2=25-C1=25-10=15 μF

Topic: Physics, Electrostatics, JEE Main
A 10 μF capacitor is fully charged to a potential difference of 50 V After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

Options:
A 15 μF
B 30 μF
C 20 μF
D 10 μF

The correct answer is:
15 μF

Solution:
V=C1V1+C2V2C1+C2
20=10×50+020+C
C=15μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A 10 g mixture of hydrogen and helium is contained in a vessel of capacity 0.0125 m3 at 6 bar and 27 °C. The mass of helium in the mixture is ..... g.

(Given : R=8.3JK-1 mol-1 (Atomic masses of H and He are 1u and 4u, respectively)

Options:

The correct answer is:
8

Solution:
PV=nmixRT

nmix=6×12.50.083×300≈3

Out of total 3 moles Let mole of He=x

Mole of H2=3-x

Then total mass of the mixture will be = 4x+23-x=10

Number of moles of He gas x=2 mol

Mass of He= number of moles × molar mass = 8 g

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A $10 \mathrm{~g}$ of a gas at atmospheric pressure is cooled from $273^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ keeping the volume constant, its pressure would become

Options:
A $1 / 2$ atm
B $1 / 273$ atm
C 2 atm
D $273 \mathrm{atm}$

The correct answer is:
$1 / 2$ atm

Solution:
$T_1=273^{\circ} \mathrm{C}=273+273^{\circ} \mathrm{K}=546^{\circ} \mathrm{K}$
$T_2=0^{\circ} \mathrm{C}=273+0^{\circ} \mathrm{C}=273^{\circ} \mathrm{K}$
$P_1=1 . P_2=$ ?
According to Gay-Lussac's law
$\frac{P_1}{T_1}=\frac{P_2}{T_2} \quad \therefore \quad P_2=\frac{P_1 T_2}{T_1}=\frac{1 \times 273^{\circ} \mathrm{K}}{546^{\circ} \mathrm{K}} \quad$ atm; $\quad \frac{1}{2} \mathrm{~atm}$.

Topic: Chemistry, States of Matter, JEE Main
A $1.0 \mathrm{~kg}$ ball drops vertically onto the floor with a speed of $25 \mathrm{~m} / \mathrm{s}$. It rebounds with an initial speed of $10 \mathrm{~m} / \mathrm{s}$. The impulse action on the ball during contact will be

Options:
A 15 N-s
B 25 N-s
C 35 N-s
D 45 N-s

The correct answer is:
35 N-s

Solution:
$\begin{aligned} & \mathrm{m}=1.0 \mathrm{~kg}, \mathrm{v}_1=25 \mathrm{~m} / \mathrm{s}, \mathrm{v}_2=10 \mathrm{~m} / \mathrm{s} \\ & \text { Impulse }=\text { change in momentum } \\ & \mathrm{J}=\Delta \mathrm{P} \\ & \mathrm{J}=\mathrm{m}\left(\mathrm{v}_1-\mathrm{v}_2\right) \\ & \mathrm{J}=1.0 \times(25-(-10)) \\ & =1 \times(25)-1 \times(-10)=35 \mathrm{~N}-\mathrm{s} \\ & \therefore \mathrm{J}=35 \mathrm{~N}-\mathrm{s}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A $10 \mathrm{~kW}$ transmitter emits radio waves of wavelength $500 \mathrm{~m}$. The number of photons emitted per second by the transmitter is of the order of

Options:
A
$10^{37}$

B
$10^{31}$

C
$10^{25}$

D
$10^{43}$

The correct answer is:
$10^{31}$

Solution:
Power $=\frac{n h c}{\lambda}$
(where, $n=$ no. of photons per second)
$$
\Rightarrow \quad n=\frac{10 \times 10^3 \times 500}{6.6 \times 10^{-34} \times 3 \times 10^8} \simeq 10^{31}
$$

Topic: Physics, Communication System, JEE Main, JEE Main 2012 (12 May Online)
A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m s-1, at right angles to the horizontal component of the earth's magnetic field of 0.3×10-4 Wb m-2. The value of the induced emf in the wire is:

Options:
A 0.3×10-3 V
B 1.1×10-3 V
C 2.5×10-3 V
D 1.5×10-3 V

The correct answer is:
1.1×10-3 V

Solution:
The wire makes an angle 45° with the magnetic field lines. Therefore, the induced emf will be,

ε=Bvlsin450

=0.3×10-4×5×10×12

=1.1×10-3 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A 0.10 M solution of a monoprotic acid (d = 1.01 g/cm3) is 5% ionized. What is the freezing point of the solution ? The molecular weight of the acid is 300 g/mol and Kf(H2O) = 1.86o C/m.

Options:
A -0.189oC
B -0.194oC
C -0.199oC
D None of these

The correct answer is:
-0.199oC

Solution:
Mass of 1 litre of solution = 1010 g
Mass of solvent = 1 0 1 0 - 3 0 0 Ã 0.1 â 9 8 0 g
Molality(m) = 0.1 0.98 â 0.102 ;
Î T f = K f . m . i = 1 + Î± K f Â· m
Î T f = 1.05 Ã 1.86 Ã 0.102 = 0.199 o C ;
T f = 0 - 0.199 = - 0.199 o C

Topic: Chemistry, Solutions, JEE Main
A 1.0 M solution of Cd2+ is added to excess iron and the system is allowed to reach equilibrium. What is the concentration in mol of Cd2+?

Cd 2+ aq +Fe s âCd s + Fe 2+ aq ; E o =0.037âV

Given: log 18 = 1.25
Report your answer upto two decimal places.

Options:

The correct answer is:
0.05

Solution:
Cd 2 + aq + Fe s â Cd s + Fe 2 + aq At eqm. 1 - x - - x
At equilibrium, E o = 0.0591 2 log Fe 2 + Cd 2 +
0.037 = 0.0591 2 log x 1 - x
x = Fe 2 + â 0.947 M
â´ Cd 2 + = 0.053 M

Topic: Chemistry, Electrochemistry, JEE Main
A $10 \mathrm{mg}$ effervescent tablet containing sodium bicarbonate and oxalic acid releases $0.25 \mathrm{~mL}$ of $\mathrm{CO}_{2}$ at $\mathrm{T}=298.15 \mathrm{~K}$ and $\mathrm{P}=1$ bar. If molar volume of $\mathrm{CO}_{2}$ is $25.0 \mathrm{~L}$ under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of $\left.\mathrm{NaHCO}_{3}=84 \mathrm{~g} \mathrm{~mol}^{-1}\right]$

Options:
A 0.84
B 33.6
C 16.8
D 8.4

The correct answer is:
8.4

Solution:
$$

2 \mathrm{NaHCO}_{3}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}

$$

Moles of $\mathrm{CO}_{2}$ evolved $=\frac{0.25}{25 \times 10^{3}}=10^{-5}$

$$

\begin{aligned}

\therefore \quad & \text { Moles of } \mathrm{NaHCO}_{3}=10^{-5} \\

\therefore \quad \text { Mass of } \mathrm{NaHCO}_{3} &=84 \times 10^{-5} \mathrm{~g} \\

&=0.84 \times 10^{-3} \mathrm{~g}=0.84 \mathrm{mg}

\end{aligned}

$$

$\therefore \quad \%$ by weight $=\times 100=8.4 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A 10 micro-farad capacitor is charged to $500 \mathrm{~V}$ and then its plates are joined together through a resistance of $10 \mathrm{ohm}$. The heat produced in the resistance is

Options:
A $500 \mathrm{~J}$
B $250 \mathrm{~J}$
C $125 \mathrm{~J}$
D $1.25 \mathrm{~J}$

The correct answer is:
$1.25 \mathrm{~J}$

Solution:
$U=\frac{1}{2} C V^2=\frac{1}{2} \times 10 \times 10^{-6} \times(500)^2=1.25 \mathrm{~J}$

Topic: Physics, Capacitance, JEE Main
A $10 \mathrm{ohm}$ resistance, $5 \mathrm{mH}$ coil and $10 \mu \mathrm{F}$ capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency

Options:
A Is halved
B Is doubled
C Remains unchanged
D In quadrupled

The correct answer is:
Remains unchanged

Solution:
At resonance, \(\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
Hence no effect of \(R\). The resonance frequency remains same when \(R\) is changed.

Topic: Physics, Alternating Current, JEE Main
A $10 \Omega$ resistance, $5 \mathrm{mH}$ coil and $10 \mu \mathrm{F}$ capacitor are joined in series. When alternating current source of suitable frequency is joined to this combination, the circuit resonates. If the resistance is halved, the resonant frequency is

Options:
A quadrupled.
B halved.
C doubled.
D remains unchanged.

The correct answer is:
remains unchanged.

Solution:
Concept based.
At resonance, in LCR circuit inductive and capacitive reactance are equal:
$\begin{aligned}
& X_L=X_C \\
& \Rightarrow 2 \pi \mathrm{fL}=\frac{1}{2 \pi \mathrm{fC}} \\
& \Rightarrow \mathrm{f}^2=\left(\frac{1}{4 \pi^2 \mathrm{LC}}\right)
\end{aligned}$
The resonant frequency is independent of the resistance.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (08 Aug Shift 2)
A 10 Ω resistance is connected across 220 V-50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is:

Options:
A 2.5 ms
B 1.5 ms
C 3.0 ms
D 4.5 ms

The correct answer is:
2.5 ms

Solution:
⇒i=i0sinωt

When i=i0

i0=i0sinωt1⇒ωt1=π2 …(1)

When i=i02

i02=i0sinωt2⇒ωt2=π4 …(2)

Time taken by current from maximum value to rms value, ⇒t1-t2=π2ω-π4ω=π4ω=π4×2πf

=18×50

=1400sec

=2.5 ms

Topic: Physics, Alternating Current, JEE Main
A 10 Ω resistance is connected across 220 V-50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is:

Options:
A 2.5 ms
B 1.5 ms
C 3.0 ms
D 4.5 ms

The correct answer is:
2.5 ms

Solution:
⇒i=i0sinωt

When i=i0

i0=i0sinωt1⇒ωt1=π2 …(1)

When i=i02

i02=i0sinωt2⇒ωt2=π4 …(2)

Time taken by current from maximum value to rms value, ⇒t1-t2=π2ω-π4ω=π4ω=π4×2πf

=18×50

=1400sec

=2.5 ms

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A 10 V battery with internal resistance 1 Ω and a 15 V battery with internal resistance 0.6 Ω are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to:

Options:
A 11.9 V
B 13.1 V
C 12.5Â V
D 24.5Â V

The correct answer is:
13.1 V

Solution:
The equivalent emf of the battery parallel combination is given as
Equation = E1r1 + E2r21r1 + 1r2
= 101 + 150.61 1 + 1 0.6= 10+ 15061+ 106
=1058
=13.1 V
∴ The reading measured by voltmeter =13.1 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2015 (10 Apr Online)
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

Options:
A 1
B 4
C 2
D 3

The correct answer is:
3

Solution:
According to Bohr's postulates, if an electron absorbs a photon with energy equal to the difference between the energies of an excited state and the ground state, it will jump to a higher energy orbital.

Considering that the energy of the colliding electron is 10.2 eV or 12.09 eV, depending on how much energy is lost during the collision. n=3 is the maximum excited state. Hence, there are spectral lines in a total of

C23=3!2!1!=3

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A 12 kg bomb at rest explodes into two pieces of 4 kg and 8 kg. If the momentum of 4 kg piece is 20 Ns, the kinetic energy of the 8 kg piece is

Options:
A 25 J
B 20 J
C 50 J
D 40 J

The correct answer is:
25 J

Solution:
The explosion is due to internal forces so the momentum of the system is conserved.
Applying conservation of momentum,
0=m1v+m2v′ 0=20+8v′
v′=-208=-52 m s-1
After the explosion,
Then, KE=12mv′2
=12×8×522=12×8×254=25 J

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A 1.2 of solution of $\mathrm{NaCl}$ is isotonic with 7.2 of solution of glucose. Calculate the van't Hoff's factor of $\mathrm{NaCl}$ solution

Options:
A 2.36
B 1.50
C 1.95
D 1.00

The correct answer is:
1.95

Solution:
Vant hoff factor of $\mathrm{NaCl}$ about 1.95 because it will be ionise into two ions.
$\mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}$

Topic: Chemistry, Solutions, JEE Main
A $12 \mathrm{ohm}$ resistor and a 0.21 henry inductor are connected in series to an ac source operating at 20 volts, 50 cycle/second. The phase angle between the current and the source voltage is

Options:
A $30^{\circ}$
B $40^{\circ}$
C $80^{\circ}$
D $90^{\circ}$

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

Solution:
$\tan \phi=\frac{\omega L}{R}=\frac{2 \pi \times 50 \times 0.21}{12}=5.5 \Rightarrow \phi=80^{\circ}$

Topic: Physics, Alternating Current, JEE Main
A 12 V, 60 W lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?

Options:
A 0.37 A
B 0.27 A
C 2.7 A
D 3.7 A

The correct answer is:
0.27 A

Solution:
For an ideal transformer,

Powerinput=Poweroutput⇒VPIP=Poweroutput⇒IP=60220≈0.27 A

Topic: Physics, Alternating Current, NEET, NEET 2023 (All India)
A 12 V battery connected to a coil of resistance 6 Ω through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is :

Options:
A 10 mH
B 8 mH
C 5 mH
D 12 mH

The correct answer is:
10 mH

Solution:
The current i through the circuit is given by

i=VR= 12 V6 Ω= 2 A

The formula to calculate the induced emf in the coil can be written as

ε=Ldidt ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the inductance of the coil.

20 V=L×2 A1×10-3 s⇒L= 20 V×1×10-3 s2 A= 10 mH

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature, 27oC. The amount of heat transferred to the gas, so that R.M.S. velocity of molecules is doubled, is about.
R=8.3 J K mole-1

Options:
A 14 kJ
B 10 kJ
C 6 kJ
D 0.9 kJ

The correct answer is:
10 kJ

Solution:
The formula of root-mean-square speed, vrms=3RTM. For doubling the vrms, temperature, should be made 4-time is, Tf=273+27×4 K=1200 K, the formula of, ∆Q=∆U+WW=0,
⇒∆Q=∆U=nCV ∆T=1528×52R ∆T
⇒∆Q=1528×52×8.3×1200-300 J≈10 KJ.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (09 Jan Shift 2)