datavorous/entrance-exam-dataset huggingface dataset:
A $70 \mathrm{~kg}$ man leaps vertically into the air from a crouching position. To take the leap the man pushes the ground with a constant force $F$ to raise himself. The center of gravity rises by $0.5 \mathrm{~m}$ before he leaps. After the leap the c.g. rises by another $1 \mathrm{~m}$. The maximum power delivered by the muscles is : (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

Options:
A
$6.26 \times 10^3$ Watts at the start

B
$6.26 \times 10^3$ Watts at take off

C
$6.26 \times 10^4$ Watts at the start

D
$6.26 \times 10^4$ Watts at take off

The correct answer is:
$6.26 \times 10^3$ Watts at take off

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2013 (23 Apr Online)
A $70 \mathrm{~kg}$ man stands in contact against the inner wall of a hollow cylindrical drum of radius $3 \mathrm{~m}$ rotating about its vertical axis with $200 \mathrm{riv} / \mathrm{min}$. The coefficient of friction between the wall and his clothing is $0.15$. What is the
minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Options:

None

Solution:
Given, $m=70 \mathrm{~kg}, r=3 \mathrm{~m}, n=200 \mathrm{rpm}=200 / 60 \mathrm{rps}$, and $\mu=0.15$
Horizontal forceprovided bythe wall on the man $\mathrm{N}=$ centripetal force $=m r \omega^2$. Frictional force, acting vertically upwards, opposes the weight of the man.
If floor is removed, man would stuck the wall, when $\mathrm{mg}=f$ $ < \mu \mathrm{N}$ i.e. $g < \mu r \omega^2$.
$\therefore$ Minimum angular speed of rotation of the cylinder is $\omega=\sqrt{ }(g / \mu r)=\sqrt{[10 /(0.15 \times 3)}]=4.7 \mathrm{rad} / \mathrm{s}$.

Topic: Physics, Laws of Motion
A 72 Ω galvanometer is shunted by a resistance of 8 Ω. The percentage of the total current which passes through the galvanometer is

Options:
A 0.1%
B 10%
C 25%
D 0.25%

The correct answer is:
10%

Solution:
Let G be resistance of galvanometer and I' be the current passing through the galvanometer produces full scale deflection. If I is the maximum current, and since, G and S are in parallel so, the potential difference between them will be same.

Then,

I'G=I2S⇒72I'=8I2⇒I2=9I'⇒I=I'+I2=10I'

So the percentage of total current passes through galvanometer is

I'I×100=I'10I'×100=10%

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A $90 \mathrm{~kg}$ body placed at $2 \mathrm{R}$ distance from surface of earth experiences gravitational pull of : $\text { ( } \mathrm{R}=\text { Radius of earth, } \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} \text { ) }$

Options:
A $100 \mathrm{~N}$
B $300 \mathrm{~N}$
C $225 \mathrm{~N}$
D $120 \mathrm{~N}$

The correct answer is:
$100 \mathrm{~N}$

Solution:
Here $g_s=$ gravitational acceleration at surface
$\begin{aligned}
& \text {Value of } g=g_5\left(1+\frac{h}{R}\right)^{-2} \\
& =g_5(1+2)^{-2}=\frac{g_5}{9}
\end{aligned}$
$\text {Force }=\mathrm{mg}=90 \times \frac{\mathrm{g}_5}{9}=100 \mathrm{~N}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is

Options:
A 15.2%
B 32.1%
C 21.8%
D 11.07%

The correct answer is:
32.1%

Solution:
Moles =â Weightâinâgm Molecularâweight
Moles of CaOâ=â 1.62 56
Moles of CaCl2=1.6256
Mass of CaCl 2 â=â 1.62 56 âÃâ111â=â3.21âgm
Mass % =  Weight in gm Total weight given  × 100
% of CaCl2=3.2110Ã100=32.1%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A $100 \mathrm{~L}$ cylinder containing $\mathrm{H}_2$ exerted a pressure of $4 \mathrm{~atm}$ at $300 \mathrm{~K}$. It was accidentally opened and some $\mathrm{H}_2$ was escaped. When it was closed, it exerted a pressure of $3 \mathrm{~atm}$ at $300 \mathrm{~K}$. The number of moles of $\mathrm{H}_2$ remaining in the cylinder is equal to
(Assume $\mathrm{H}_2$ as an ideal gas; $\mathrm{R}=$ gas constant)

Options:
A $\frac{1}{2 \mathrm{R}}$
B $\mathrm{R}$
C $\frac{1}{\mathrm{R}}$
D $2 \mathrm{R}$

The correct answer is:
$\frac{1}{\mathrm{R}}$

Solution:
$\mathrm{n}_1=\frac{\mathrm{P}_1 \mathrm{~V}}{\mathrm{RT}}=\frac{4 \mathrm{~V}}{\mathrm{RT}}$ and $\mathrm{n}_2=\frac{\mathrm{P}_2 \mathrm{~V}}{\mathrm{RT}}=\frac{3 \mathrm{~V}}{\mathrm{RT}}$ $\Rightarrow$ Number of moles remaining $=$ $\frac{4 \mathrm{~V}}{\mathrm{RT}}-\frac{3 \mathrm{~V}}{\mathrm{RT}}=\frac{1}{\mathrm{R}}$

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2023 (15 May Shift 1)
A 100 ml solution of 0.1 N-HCl was titrated with 0.2 N-NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25 N-KOH solution. The volume of KOH required for completing the titration is

Options:
A 16 ml
B 32Â ml
C 35Â ml
D 70Â ml

The correct answer is:
16 ml

Solution:
In the neutralization of acid and base NÃV of both must be equivalent
NÃV of HCl = 0.1Ã100 = 10
NÃV of NaOH= 0.2Ã30=6
N1V1ml=NÃVNaOHml+NÃVmlKOH
0.1Ã100 = 0.2Ã30+0.25ÃV
10 = 6+0.25 V
V=40025 V = 16 ml

Topic: Chemistry, Redox Reactions, JEE Main
A 100 m long wire having cross-sectional area 6.25×10-4 m2 and Young's modulus is 1010 N m-2 is subjected to a load of 250 N, then the elongation in the wire will be :

Options:
A 6.25×10-3 m
B 4×10-4 m
C 6.25×10-6 m
D 4×10-3 m

The correct answer is:
4×10-3 m

Solution:
As we know, elongation in wire is given by,

∆l=FlAY

⇒∆l=250×1006.25×10-4×1010

⇒∆l=4×10-3 m

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A $100 \mathrm{~m}$ long wire having cross-sectional area $6.25 \times 10^{-4} \mathrm{~m}^2$ and Young's modulus is $10^{10}$ $\mathrm{Nm}^{-2}$ is subjected to a load of $250 \mathrm{~N}$, then the elongation in the wire will be :

Options:
A $6.25 \times 10^{-3} \mathrm{~m}$
B $4 \times 10^{-4} \mathrm{~m}$
C $6.25 \times 10^{-6} \mathrm{~m}$
D $4 \times 10^{-3} \mathrm{~m}$

The correct answer is:
$4 \times 10^{-3} \mathrm{~m}$

Solution:
$\Delta \ell=\frac{F \ell}{Y A}=\frac{250 \times 100}{10^{10} \times 6.25 \times 10^{-4}}=40 \times 10^{-4} \mathrm{~m}$
$=4 \times 10^{-3} \mathrm{~m}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A 100 mark examination was administered to a class of 50 students. Despite only integer marks being given, the average score of the class was 47.5 . Then, the maximum number of students who could get marks more than the class average is

Options:
A 25
B 35
C 45
D 49

The correct answer is:
49

Solution:
Total number of students

=50

Average marks of student =47.5

∴ Total marks of students

=50×47.5=2375

Now, the student get integer marks

Hence, the maximum number of students we will divide total mark by 48 .

∴ 237b48=49

Topic: Mathematics, Statistics, JEE Main
A $100$ mL, $0.1=\mathrm{M}$ solution of ammonium acetate is diluted by adding $100 \mathrm{mL}$ of water. The pH of the resulting solution will be $\left(p K_{a}\right.$ of acetic acid is nearly equal to $\mathrm{pK}_{b}$ of $\mathrm{NH}_{4} \mathrm{OH}$ )

Options:
A 4.9
B 5
C 7
D 10

The correct answer is:
7

Solution:
Ammonium acetate is a salt of weak acid and weak base. When solutions of such salts diluted resulting pH of solution is calculated from following relationship
$$
\mathrm{pH}=7+\frac{1}{2} \mathrm{p} \mathrm{K}_{a}-\frac{1}{2} \mathrm{p} K_{\mathrm{b}}
$$
$\because p K_{a}=p K_{b}$
$\therefore \quad \mathrm{pH}$ of diluted solution $=7$

Topic: Chemistry, Ionic Equilibrium, JEE Main
A 100 mL solution of CH3CH2MgBr on treatment with methanol produces 2.24 mL of a gas at STP. The weight of gas produced is____mg (nearest integer)

Options:

The correct answer is:
3

Solution:
In the above reaction the Ethyl group of grignard reagent acts as a strong base.

As per the stoichiometric equation the number of moles of methanol is equal to the number of moles of ethane.

Volume of methanol=Volume of ethane=2.24ml

Hence, number of moles of ethane

n=2.24×10-322.4=10-4

Weight of the gas= Number of moles× Molecular mass

W=n×M

=10-4×30=3mg

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A 100 mL solution was made by adding 1.43 g of Na2CO3.xH2O. The normality of the solution is 0.1 N. The value of x is _______

(The atomic mass of Na is 23g/mol)

Options:

The correct answer is:
10

Solution:
Equivalent of solute =0.1×0.1

Mole of solute Na2CO3,xH2O=[0.1×0.1]12

Mass of Na2CO3.xH2O=[0.1×0.1]12×[106+18x]=1.43

⇒ [106+18x=286

18 x=180

x=10

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A 100Ω resistance, a 0.1μF capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.

Options:
A 0.70H
B 70.3mH
C 7.03×10-5H
D 70.3 H

The correct answer is:
70.3 H

Solution:
C=0.1μF=10-7 F

Resonant frequency =60 Hz

ωo=1LC

2πf0=1LC⇒L=14π2f02C

by putting values L≃70.3 Hz.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A 100 V, AC source of frequency 500 Hz is connected to an LCR circuit with L=8.1 mH,
C=12.5 Î¼F, R=10 Î© all connected in series as shown in the figure. What is the quality factor of the circuit?

Options:
A 2.02
B 2.54
C 50.54
D 200.54

The correct answer is:
2.54

Solution:
Q=1RLC
⇒ Q=1108.1×10-312.5×10-6
⇒ Q=11081×103125
⇒ Q=81×1025×5
⇒ Q=9×25=2.54

Topic: Physics, Alternating Current, JEE Main
A 100V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index?

Options:
A 0.3
B 0.6
C 0.4
D 0.5

The correct answer is:
0.6

Solution:
AC=100
AC+Am=160
AC-Am=40
AC=100,Am=60
μ=AmAc=0.6

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A $100 \mathrm{~V}$ voltmeter of internal resistance $20 \mathrm{k} \Omega$ in series with a high resistance $R$ is connected to a $110 \mathrm{~V}$ line. The voltmeter reads $5 \mathrm{~V}$, the value of $R$ is

Options:
A $210 \mathrm{k} \Omega$
B $315 \mathrm{k} \Omega$
C $420 \mathrm{k} \Omega$
D $440 \mathrm{k} \Omega$

The correct answer is:
$420 \mathrm{k} \Omega$

Solution:
\(\begin{aligned}
& \mathrm{I}=\frac{110}{\mathrm{R}+20 \mathrm{k} \Omega} \\
& \text {Voltmeter reads } 5 \mathrm{~V}=\frac{(20 \mathrm{k} \Omega)(110)}{(\mathrm{R}+20 \mathrm{k} \Omega)} \\
& \mathrm{R}+20=4 \times 110=440 \\
& \mathrm{R}=440-20=420 \mathrm{k} \Omega \\
& R=420 \mathrm{k} \Omega
\end{aligned}\)
(C) is correct option.

Topic: Physics, Current Electricity, JEE Main
A $10.0 \mathrm{~W}$ electrical heater is used to heat a container filled with $0.5 \mathrm{~kg}$ of water. It is found that the temperature of the water and the container rise by $3 \mathrm{K}$ in $15 \mathrm{~minutes}$. The container is then emptied, dried, and filled with $2 \mathrm{~kg}$ of an oil. It is now observed that the same heater raises the temperature of the container-oil system by $2 \mathrm{K}$ in $20 \mathrm{~minutes}$. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is

Options:
A $2.5 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
B $5.1 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
C $3.0 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
D $1.5 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$

The correct answer is:
$2.5 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$

Solution:
$\mathrm{Pt}=\mathrm{m}_{\mathrm{w}} \mathrm{S}_{\mathrm{w}} \Delta \mathrm{T}+\mathrm{m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}} \Delta \mathrm{T}$ $10 \times 15 \times 60=0.5 \times 4200 \times 3+\mathrm{m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}} \times 3$ $9000=6300+\mathrm{m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}} 3$
$\mathrm{~m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}}=900 \mathrm{~J} / \mathrm{k}$.
Now, for oil
$10 \times 20 \times 60=2 \times \mathrm{S}_{0} \times 2+900 \times 2$
$12000-1800=4 \mathrm{~S}_{0}$
$\mathrm{~S}_{0}=\frac{10200}{4}=2.51 \times 10^{3} \mathrm{~J} / \mathrm{kg}-\mathrm{k}$

Topic: Physics, Current Electricity, JEE Main
A 100âW sodium lamp radiates energy uniformly in all directions. The wavelength of the sodium light is 589 nm. What is the energy per photon in (eV) associated with the sodium light? [ h = 6 .63 Ã 10 -34 J s ]

Options:
A 2.11
B 1.22
C 0.5
D 3.21

The correct answer is:
2.11

Solution:
Given, power of the lamp, P = 100 W
The wavelength of the sodium light, λ = 5 8 9 nm = 589 × 10 −9 m
Planck constant h = 6.63 × 10-34 J s
The energy of each photon
E=hcλ=6.63×10-34×3×108589×10-9
∵ c=3×108 m s-1
= 3 . 3 8 × 1 0 - 1 9 J
= 3 . 3 8 × 1 0 - 1 9 1 . 6 × 1 0 - 1 9 eV
= 2.11 eV

Topic: Physics, Dual Nature of Matter, JEE Main
A 110 V,50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be _____A.

Options:

The correct answer is:
0

Solution:
Let current in the inductance is iL, through the capacitor is iC and through resistance is iR at any instance as shown in the figure.

According to the Kirchoff's junction law, iR=iL+iC.

At resonance XC=XL hence iL=iC. Also, current through the capacitor and inductor have phase difference of π rad. Therefore, they will form destructive interference and hence the current through the resistance will be 0.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A $120 \mathrm{~m}$ long train is moving in a direction with speed $20 \mathrm{~m} / \mathrm{s}$. A train $B$ moving with $30 \mathrm{~m} / \mathrm{s}$ in the opposite direction and $130 \mathrm{~m}$ long crosses the first train in a time

Options:
A 6s
B 36s
C 38s
D None of these

The correct answer is:
None of these

Solution:
Total distance $=130+120=250 \mathrm{~m}$
Relative velocity $=30-(-20)=50 \mathrm{~m} / \mathrm{s}$
Hence $t=250 / 50=5 \mathrm{~s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A $12.5$ ev electron beam is used to bombard gaseous hydrogen at ground state. The energy level upto which the hydrogen atoms would be excited is

Options:
A 2
B 3
C 4
D 1

The correct answer is:
3

Solution:
$12.5 < 12.75$

$\therefore$ 3rd state

Topic: Physics, Atomic Physics, JEE Main
A $12.5 \mathrm{eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:

Options:
A
2 lines in the Lyman series and 1 line in the Balmar series

B
3 lines in the Lyman series

C
1 line in the Lyman series and 2 lines in the Balmar series

D
3 lines in the Balmer series

The correct answer is:
2 lines in the Lyman series and 1 line in the Balmar series

Solution:
$$
\begin{aligned}
&\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{12.5 \times 1.6 \times 10^{-19}} \\
&=993 \mathrm{~A}^{\circ} \\
&\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\
&
\end{aligned}
$$
(where Rydberg constant , $\mathrm{R}=1.097 \times 10^7$ )
or, $\frac{1}{993 \times 10^{-10}}=1.097 \times 10^7\left(\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
Solving we get $\mathrm{n}_2=3$
Spectral lines
Total number of spectral lines $=3$
Two lines in Lyman series for $\mathrm{n}_1=1, \mathrm{n}_2=2$ and $\mathrm{n}_1=1, \mathrm{n}_2=3$ and one in Balmer series for $\mathrm{n}_1=2, \mathrm{n}_2=3$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2013 (25 Apr Online)
A $150 \mathrm{~m}$ long train is moving with a uniform velocity of $45 \mathrm{~km} / \mathrm{h}$. The time taken by the train to cross a bridge of length 850 meters is

Options:
A 56 sec
B 68 sec
C 80 sec
D 92 sec

The correct answer is:
80 sec

Solution:
Total distance to be covered for crossing the bridge $=$ length of train + length of bridge
$\begin{aligned} & =150 \mathrm{~m}+850 \mathrm{~m}=1000 \mathrm{~m} \\ & \text { Time }=\frac{\text { Distance }}{\text { Velocity }}=\frac{1000}{45 \times \frac{5}{18}}=80 \mathrm{sec}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A $1.85 \mathrm{~g}$ sample of an arsenic-containing pesticide was chemically converted to $\mathrm{AsO}_{4}{ }^{3-}$ (atomic mass of $\mathrm{As}=$ 74.9) and titrated with $\mathrm{Pb}^{2+}$ to form $\mathrm{Pb}_{3}\left(\mathrm{AsO}_{4}\right)_{2}$ If $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{Pb}^{2+}$ is required to reach the equivalence point, the mass percentages of arsenic in the pesticide sample is closest to

Options:
A 8.1
B $2.3$
C $5.4$
D $3.6$

The correct answer is:
$5.4$

Solution:
$3 \mathrm{pb}^{2+}+2 \mathrm{ASO}_{4}^{3-} \longrightarrow \mathrm{pb}_{3}\left(\mathrm{AsO}_{4}\right)_{2}$
$\mathrm{n}=\mathrm{M} \times \mathrm{V} \quad \mathrm{n}=\frac{2}{3} \times 2 \times 10^{-3}$
$=0.1 \times \frac{20}{1000}=0.00133$
$=2 \times 10^{-3}$
$\eta_{\mathrm{AS}}=\eta_{\mathrm{ASO}_{4}^{3}}=0.00133$
$\mathrm{~W}_{\mathrm{AS}}=0.00133 \times 74.9=0.0996$
$\%$ of $\mathrm{AS}=\frac{0.0996}{1.85} \times 100=5.4 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A 200 Ω resistance has a certain colour code as per standard colour coding. If one replaces the red colour by green in the code, the new resistance will be

Options:
A 500 Ω
B 400 Ω
C 300 Ω
D 100 Ω

The correct answer is:
500 Ω

Solution:
When red is replaced by green, then the first significant figure in the value of resistance changes from 2 to 5. So the new resistance is 500 Ω

Topic: Physics, Current Electricity, JEE Main
A 200 Ω resistor has certain colour code. If one replaced the red colour by green in the code, the new resistance will be:

Options:
A 300 Ω
B 100 Ω
C 400 Ω
D 500 Ω

The correct answer is:
500 Ω

Solution:
Colour band A and B indicate the first two significant figures of resistance in "ohm" , while the C band gives the decimal multiplier i.e., the number of zeros that follows the two significant figures A and B.

The last band D indicates the tolerance in percentage about the indicate value or in another ward it represents the percentage accuracy of the indicated value.

As we know red corresponds to digit 2 and green corresponds to digit 5 , new resistance will be 500 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A 210 meter long train is moving due North at a of $25 \mathrm{~m} / \mathrm{s}$. A small bird is flying due South a little above the train with speed $5 \mathrm{~m} / \mathrm{s}$. The time taken by the bird to cross the train is

Options:
A 6s
B 7s
C 9s
D 10s

The correct answer is:
7s

Solution:
Relative velocity of bird w.r.t train $=25+5=30 \mathrm{~m} / \mathrm{s}$ time taken by the bird to cross the train $t=\frac{210}{30}=7 \mathrm{sec}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A 220 V,50 Hz AC source is connected to a 25 V,5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be

Options:

The correct answer is:
975

Solution:
Resistance of the bulb can be calculated as,

P=V2RB⇒RB=V2P=2525=125 Ω

The current through the bulb for peak brightness should be,

i=25125=15 A

Now, irms=15=220RB+R

RB+R=1100

R=1100-125=975

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A $220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac source is connected to an inductance of $0.2 \mathrm{H}$ and a resistance of $20 \mathrm{ohm}$ in series. What is the current in the circuit

Options:
A $10 \mathrm{~A}$
B $5 \mathrm{~A}$
C $33.3 \mathrm{~A}$
D $3.33 \mathrm{~A}$

The correct answer is:
$3.33 \mathrm{~A}$

Solution:
$i=\frac{220}{\sqrt{(20)^2+(2 \times \pi \times 50 \times 0.2)^2}}=\frac{220}{66}=3.33 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main
A 220 volt, 1000 watt bulb is connected across a 110 volt mains supply. The power consumed will be

Options:
A
$750 \mathrm{watt}$

B
$500 \mathrm{watt}$

C
$250 \mathrm{watt}$

D
$1000 \mathrm{watt}$

The correct answer is:
$250 \mathrm{watt}$

Solution:
$\mathrm{p}=\frac{\mathrm{V}^2}{\mathrm{R}}$ or $\mathrm{R}=\frac{(220)^2}{1000}$
Power consumed $=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{110 \times 110}{220 \times 220} \times 1000=250$ watt

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
A 250 mL sample of a 0.2 M Cr3+ is electrolysed with a current of 96.5 amp; if the remaining concentration of Cr3+ ion is 0.1 M, the duration of electrolysis is
given : atomic mass of Cr = 52

Options:
A 75 sec
B 150 sec
C 225 sec
D 25 sec

The correct answer is:
75 sec

Solution:
Number of moles of Cr3+ before electrolysis =MV (mol)1000
=0.2Ã2501000=0.05
Number of moles of Cr3+ after electrolysed =MV1000
=0.1Ã2501000=0.025
Number of moles of Cr3+ electrolysed =0.05-0.025=0.025
Mass of chromium electrolysis =0.025Ã52=1.3 g
Equivalent mass of chromium =Atomic massValency of ions
=523=17.33
W=ItE96500âWÃ96500IÃE=1.3Ã9650096.5Ã17.33
t=75.14 sec
â75 sec.

Topic: Chemistry, Electrochemistry, JEE Main
A 300mL bottle of soft drink has 0.2MCO2 dissolved in it. Assuming CO2 behaves as an ideal gas, the volume of the dissolved CO2 at STP is _____ mL. (Nearest integer)

Given: At STP, molar volume of an ideal gas is 22.7L mol-1

Options:

The correct answer is:
1362

Solution:
Moles dissolved in soft drinks of CO2=molarity × volume = 0.2M×300×10-3L

=0.06 Mole

Volume of 0.06 mole of CO2 at S.T.P

=0.06×22.7

=1.362 L

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A $500 \mathrm{~g}$ ball is attached to one end of an aluminum wire of area of cross-section $0.5 \mathrm{~mm}^2$ and an unstretched length of $1.4 \mathrm{~m}$. The other end of the wire is fixed to the top of a vertical pole. The ball rotates about the pole in a horizontal plane such that the angle between the wire and the horizontal is $30^{\circ}$. The increase in the length of the wire is ____ $\mathrm{mm}$. (Young's modulus of aluminum $=0.7 \times 10^{11} \mathrm{Nm}^{-2}$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$)

Options:
A 0.1
B 0.2
C 0.3
D 0.4

The correct answer is:
0.4

Solution:
No solution. Refer to answer key.

Topic: Physics, Mechanical Properties of Solids, JEE Main
A 500 g ball is released from a height of 4 m. Each time it makes contact with the ground, it loses 25% of its kinetic energy. Find the kinetic energy it possess just after the 3rd hit

Options:
A 15Â J
B 11.25Â J
C 8.44 J
D none of these

The correct answer is:
8.44 J

Solution:
After first hit K.E1=e2 K.E
3/4 K.E=e2 K.E â e2=3/4
K.E. after third hit= ( e 2 ) 3
KE= 27 64 mgh = 276412.10.4
= 27516=13516= 8.44 J

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A 500 kg car takes a round turn of radius 50 m with a velocity of 36 km h-1. The centripetal force is

Options:
A 250 N

B 750 N
C 1000 N
D 1200 N

The correct answer is:
1000 N

Solution:
Centripetal force in a circular motion is given as
F=mv2r=50010250=1000 N

Topic: Physics, Laws of Motion, JEE Main
A $500 \mathrm{~kg}$ car takes a round turn of radius $50 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} / \mathrm{hr}$. The centripetal force is

Options:
A 250 N
B 750 N
C 1000 N
D 1200 N

The correct answer is:
1000 N

Solution:
$v=36 \frac{\mathrm{km}}{\mathrm{h}}=10 \frac{\mathrm{m}}{\mathrm{s}} 4 \quad F=\frac{\mathrm{mv}^2}{r}=\frac{500 \times 100}{50}=1000 \mathrm{~N}$.

Topic: Physics, Motion In Two Dimensions, JEE Main