datavorous/entrance-exam-dataset huggingface dataset:
An ether (A), $\mathrm{C}_5 \mathrm{H}_{12} \mathrm{O}$, when heated with excess of hot concentrated $\mathrm{HI}$ produced two alkyl halides which when treated with $\mathrm{NaOH}$ yielded compounds (B) and (C). Oxidation of (B) and (C) gave a propanone and an ethanoic acid respectively. The IUPAC name of the ether (A) is :

Options:
A
$2-$ethoxypropane

B
ethoxypropane

C
methoxybutane

D
$2-$methoxybutane

The correct answer is:
$2-$ethoxypropane

Solution:
hence the IUPAC name of compound is

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (09 Apr Online)
An ether is more volatile than an alcohol having the same molecular formula. This is due to

Options:
A
alcohols having resonance structures

B
inter-molecular hydrogen bonding in ethers

C
inter-molecular hydrogen bonding in alcohols

D
dipolar character of ethers

The correct answer is:
inter-molecular hydrogen bonding in alcohols

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2003
An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with a liquid of density 0.95 g mL-1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in gmol-1 is : (Given : R=0.082 L atmK-1mol-1)

Options:
A 35
B 50
C 75
D 125

The correct answer is:
125

Solution:
Weight of gas =40.5-40=0.5 g

Mass of liquid =135.5-40=95 g.

Volume of glass vessel =95.0 g0.95 g/ml

=100ml.

∴ PV=WMRT

0.82 atm100×10-3=0.5M0.082250

M=125 g/mol

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2022 (26 Jun Shift 1)
An exam consists of 3 problems selected randomly from a collection of 10 problems. For a student to pass, he needs to solve correctly at least two of three problems. If the student knows to solve exactly 5 problems, then the probability that the students pass the exam is

Options:
A 12
B 13
C 34
D 56

The correct answer is:
12

Solution:
Let Ai be an event that student solve i problems correctly for i=0,1,2,3
PA2âªA3=PA2+PA3
âµ Students knows to solve exactly 5 problems correctly
P A 2 = 5 C 2 ×   5 C 1 10 C 3 ,P A 3 = 5 C 3 10 C 3
PA2âªA3=60120=12

Topic: Mathematics, Probability, JEE Main
An example of a disproportionation reaction is

Options:
A 2KMnO4→K2MnO4+MnO2+O2
B 2MnO4-+10I-+16 H+→2 Mn2++5I2+8H2O
C 2Â CuBrâCuBr2+Cu
D Â 2Â NaBr+Cl2â2NaCl+Br2

The correct answer is:
2 CuBr→CuBr2+Cu

Solution:
Disproportionation reaction: In this type of reaction, same elemental species gets oxidised as well as reduced.

Na+1Br-1+Cl20→2Na+1Cl-1+Br20-not disproportionate,

2KMn+7O4→K2Mn+6O4+Mn+4O2+O2-not disproportionate,

2Mn+7O4-+10J-+16H+→2Mn+2+SI2+8H2O-not disproportionate

-It is disproportionate.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2019 (12 Apr Shift 1)
An example of a non-stoichiometric compound is

Options:
A $\mathrm{Al}_2 \mathrm{O}_3$
B $\mathrm{Fe}_3 \mathrm{O}_4$
C $\mathrm{NiO}_2$
D $\mathrm{PbO}$

The correct answer is:
$\mathrm{Fe}_3 \mathrm{O}_4$

Solution:
$\mathrm{Fe}_3 \mathrm{O}_4$ is a non-stoichiometric compound because in it the ratio of the cations to the anions becomes different from that indicated by the chemical formula.

Topic: Chemistry, Solid State, JEE Main
An example of a perfect diamagnet is a superconductor. This implies that when a superconductor is put in a magnetic field of intensity B, the magnetic field Bs inside the superconductor will be such that

Options:
A Bs=B.
B Bs=0.
C Bs<B but Bs≠0.
D Bs=–B

The correct answer is:
Bs=0.

Solution:
In diamagnetic material, the magnetic field strength decreases. So, in a perfect diamagnetic, it has to be zero.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2014 (19 Apr Online)
An example of a seed with endosperm, perisperm and caruncle is -

Options:
A Castor
B Cotton
C Coffee
D Lily

The correct answer is:
Castor

Solution:
No Solution Available

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2009 (Mains)
An example of axile placentation is -

Options:
A Marigold
B Argemone
C Dianthus
D Lemon

The correct answer is:
Lemon

Solution:
No Solution Available

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2009 (Mains)
An example of endomycorrhiza is

Options:
A Nostoc
B Glomus
C Agaricus
D Rhizobium

The correct answer is:
Glomus

Solution:
Nostoc is a blue-green algae capable of fixing nitrogen.
Glomus is associated with endomycorrhiza.
Agaricus or mushroom is an edible fungus of class-Basidiomycetes.
Rhizobium is a symbiotic nitrogen-fixing bacterium found in the root nodules of leguminous plants.

Topic: Biology, Mineral Nutrition, NEET, NEET 2010 (Mains)
An example of halide ore is

Options:
A Galena
B Bauxite
C Cinnabar
D Cryolite

The correct answer is:
Cryolite

Solution:
Cryolite $\left(\mathrm{Na}_3 \mathrm{AlF} \mathrm{F}_6\right) \rightarrow$ Halide ore
$\left.\begin{array}{rr}\text { Galena } & (\mathrm{PbS}) \\ \text { Cinnaber } & (\mathrm{HgS})\end{array}\right\}$ Sulphide ore
Bauxite $\mathrm{Al}_2 \mathrm{O}_3 \cdot 2 \mathrm{H}_2 \mathrm{O} \rightarrow$ Oxide ore

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
An example of halide ore is

Options:
A Galena
B Bauxite
C Cinnabar
D Cryolite

The correct answer is:
Cryolite

Solution:
Cryolite Na3AlF6âHalide ore
Galena PbSCinnabar HgSSulphide ore
Bauxite Al2O3.2H2OâOxide ore

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
An example of solid sol is:

Options:
A Paint
B Gem stones
C Butter
D Hair cream

The correct answer is:
Gem stones

Solution:
Gem stones are solid sol.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
An excess of AgNO3 solution is added to 100mL of a 0.2M dichloridotetraaquachromium (III) chloride. The number of millimoles of AgCl precipitated would be -----.

Options:

The correct answer is:
20

Solution:
100mL of 0.2M solution means 0.02 moles of complex ∵1001000×0.2=0.02
CrH2O4Cl2Cl has one ionizable Cl-.
∴0.02 moles of the complex precipitates the same amount of AgCl .
∴ Millimoles of AgCl precipitated =20

Topic: Chemistry, Coordination Compounds, JEE Main
An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is
(for a photon of wavelength Î», energy E=1240 eVÎ»(in nm))

Options:
A n=6
B n=5
C n=7
D n=4

The correct answer is:
n=5

Solution:
E1=1240λ1=1240108.5≈11.43 eV
E2=1240λ2=124030.4≈40.79 eV
ETotal=E1+E1=52.22 eV
52.22=13.6221-1n2
52.22=54.41-1n2
0.96=1-1n2
1n2=0.04
n2=10.04=1004=25
n=5

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
An excited hydrogen atom returns to the ground state. The wavelength of emitted photo is λ . The principal quantum number of the excited state will be

Options:
A λRλR-11/2
B λR+1λR1/2
C [λR(λR+1)]1/2
D 1λR(λR+1)1/2

The correct answer is:
λRλR-11/2

Solution:
1λ=R1n12-1n22
=R112-1n2
n=λRλR-11/2

Topic: Chemistry, Structure of Atom, JEE Main
An exothermic reaction, A→B , has an activation energy of 15 kcal mol-1 and the energy of reaction is 5 kcal mol-1 . The activation energy in kcal mol-1 for the reaction, B→A is

Options:

The correct answer is:
20

Solution:
(E a ) r  =  (E a ) f  + ΔE = 15 kcal  mol -1 +5kcalmol −1
=20 kcal  mol −1 .

Topic: Chemistry, Chemical Kinetics, JEE Main
An exothermic reaction X→Y has an activation energy 30 kJ mol-1. If energy change ΔE during the reaction is -20 kJ, then the activation energy for the reverse reaction in kJ is

(Integer answer)

Options:

The correct answer is:
50

Solution:
X⟶Y

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (26 Feb Shift 1)
An experiment has 10 equally likely outcomes. Let $A$ and $B$ be two non-empty events of the experiment. If $A$ consists of 4 outcomes, the number of outcomes that $B$ must have so that $A$ and $B$ are independent, is

Options:
A
2,4 or 8

B
3,6 or 9

C
4 or 8

D
5 or 10

The correct answer is:
5 or 10

Solution:
$P(A)=\frac{2}{5}$
For independent events,
$$
\begin{aligned}
& P(A \cap B)=P(A) P(B) \Rightarrow P(A \cap B) \leq \frac{2}{5} \\
& \Rightarrow P(A \cap B)=\frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10} \\
& \text { (i) } P(A \cap B)=\frac{1}{10} \\
& \Rightarrow P(A) \cdot P(B)=\frac{1}{10} \\
& \Rightarrow \quad P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4}, \text { not possible. } \\
& \text { (ii) } P(A \cap B)=\frac{2}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{2}{10} \\
& \Rightarrow \quad P(B)=\frac{5}{10}, \text { outcomes of } B=5
\end{aligned}
$$
$$
\text { (iii) } \begin{aligned}
& P(A \cap B)=\frac{3}{10} \Rightarrow P(A) \cdot P(B)=\frac{3}{10} \\
& \Rightarrow \quad \frac{2}{5} \times P(B)=\frac{3}{10} \\
& P(B)=\frac{3}{4}, \text { not possible }
\end{aligned}
$$
$$
\text { (iv) } \begin{aligned}
& P(A \cap B)=\frac{4}{10} \\
\Rightarrow & P(A) \cdot P(B)=\frac{4}{10} \\
\Rightarrow & P(B)=1, \text { outcomes of } B=10 .
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main
An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distance are measured by

Options:
A
a vernier scale provided on the microscope

B
a standard laboratory scale

C
a meter scale provided on the microscope

D
a screw gauage provided on the microscope

The correct answer is:
a vernier scale provided on the microscope

Solution:
No Solution Available

Topic: Physics, Ray Optics, JEE Main, JEE Main 2008
An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :

Options:
A 4.4%
B 2.27%
C 1.7%
D 2.7%

The correct answer is:
2.7%

Solution:
T = 2 π L g
∴ T 2 = 4 π 2 L g
∴ g = 4 π 2 L T 2
% Error in g = Δ L L 0 × 1 0 0 + 2 Δ T T 0 × 1 0 0
= 1 × 1 0 - 3 0.20 × 1 0 0 + 2 1 9 0 × 1 0 0
= 0.5 + 2 1 0 9
= 0.5 + 2 0 9
= 0.5 + 2.22
= 2.7%

Topic: Physics, Mathematics in Physics, JEE Main
An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :

Options:
A 4.4%
B 2.27%
C 1.7%
D 2.7%

The correct answer is:
2.7%

Solution:
T = 2 π L g
∴ T 2 = 4 π 2 L g
∴ g = 4 π 2 L T 2
% Error in g = Δ L L 0 × 1 0 0 + 2 Δ T T 0 × 1 0 0
= 1 × 1 0 - 3 0.20 × 1 0 0 + 2 1 9 0 × 1 0 0
= 0.5 + 2 1 0 9
= 0.5 + 2 0 9
= 0.5 + 2.22
= 2.7%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2014 (09 Apr Online)
An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is

Options:
A 496729
B 192729
C 240729
D 256729

The correct answer is:
256729

Solution:
Let p be the probability of successes and q be the probability of failure.

p=2q and p+q=1

gives p=23 and q=13

For having at least 5 successes in 6 trials,

Required probability=C5623513+C66236

=C1623513+C06236

=6.25+6436=256729

Topic: Mathematics, Probability, JEE Main, JEE Main 2016 (10 Apr Online)
An experiment takes 10 min to raise the temperature of water in a container from 0oC to 100oC and another 55 min to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of the water to be 1 cal g∘C-1 , the heat of vaporization according to this experiment will come out to be:

Options:
A 530 cal g-1
B 550 cal g-1
C 540 cal g-1
D 560 cal g-1

The correct answer is:
550 cal g-1

Solution:
Let the constant state of supply of heat per unit time =λ. So according to the first condition. λt1=mc×100 ⇒λm=100Ct1 ...(i), and according to the second condition, λt2=mLf⇒λm=Lft2 ...(ii). By (i) and (ii) 100Ct1=Lft2⇒1cal×10010=Lf55⇒Lf=550 cal g-1

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2015 (11 Apr Online)
An experiment yields 3 mutually exclusive and exhaustive events A,B and C . If PA=2P(B)=3PC, then PA is equal to

Options:
A 111
B 211
C 311
D 611

The correct answer is:
611

Solution:
Given, events A, B and C are mutually exclusive and exhaustive, hence PAâªBâªC=1
â PA+PB+PC=1
âPA+12PA+13PA=1
â 116PA=1
âPA=611

Topic: Mathematics, Probability, JEE Main
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, $1 \mathrm{~kg}$ first part moving with a velocity of $12 \mathrm{~ms}^{-1}$ and $2 \mathrm{~kg}$ second part moving with a velocity of $8 \mathrm{~ms}^{-1}$. If the third part flies off with a velocity of $4 \mathrm{~ms}^{-1}$, its mass would be

Options:
A $5 \mathrm{~kg}$
B $7 \mathrm{~kg}$
C $17 \mathrm{~kg}$
D $3 \mathrm{~kg}$

The correct answer is:
$5 \mathrm{~kg}$

Solution:
Key Idea Apply law of conservation of linear momentum.
Momentum of first part $=1 \times 12=12 \mathrm{~kg} \mathrm{~ms}^{-1}$
Momentum of the second part
$$
=2 \times 8=16 \mathrm{~kg} \mathrm{~ms}^{-1}
$$
$\therefore$ Resultant momentum
$$
=\sqrt{(12)^2+(16)^2}=20 \mathrm{~kg} \mathrm{~ms}^{-1}
$$
The third part should also have the same momentum.
Let the mass of the third part be, then
$$
\begin{aligned}
4 \times M & =20 \\
M & =5 \mathrm{~kg}
\end{aligned}
$$
$$
\text { Alternative : }
$$
$$
\begin{aligned}
\mathrm{Mv} \cos \theta & =12 \\
\mathrm{Mv} \sin \theta & =16 \\
\tan \theta & =\frac{16}{12}=\frac{4}{3} \\
M & =\frac{12 \times 5}{4 \times 3}=\frac{60}{12}=5 \mathrm{~kg}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An explosion blows a rock into three parts Two parts go off at right angles to each other. These two are, $1 \mathrm{~kg}$ first part moving with a velocity of $12 \mathrm{~ms}^{-1}$ and $2 \mathrm{~kg}$ second part moving with a velocity of $8 \mathrm{~ms}^{-1}$. If the thirds part files off with a velocity of $4 \mathrm{~ms}^{-1}$, its mass would be :

Options:
A $3 \mathrm{~kg}$
B $5 \mathrm{~kg}$
C $7 \mathrm{~kg}$
D $17 \mathrm{~kg}$

The correct answer is:
$5 \mathrm{~kg}$

Solution:
$$
\begin{aligned}
\overrightarrow{\mathrm{P}}_1+\overrightarrow{\mathrm{P}}_2+\overrightarrow{\mathrm{P}}_3 & =0 \\
\left|\overrightarrow{\mathrm{P}}_3\right| & =\left|\overrightarrow{\mathrm{P}}_1+\overrightarrow{\mathrm{P}}_2\right| \\
\mathrm{m} \times 4 & =\sqrt{\mathrm{P}_1^2+\mathrm{P}_2^2} \\
& =\sqrt{12^2+16^2} \\
\mathrm{~m} & =5 \mathrm{~kg}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2009 (Mains)
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1 \mathrm{~kg}$ moves with a speed of $12 \mathrm{~ms}^{-1}$ and the second part of mass $2 \mathrm{~kg}$ moves with $8 \mathrm{~ms}^{-1}$ speed. If the third part flies off with $4 \mathrm{~ms}^{-1}$ speed, then its mass is

Options:
A $3 \mathrm{~kg}$
B $5 \mathrm{~kg}$
C $7 \mathrm{~kg}$
D $17 \mathrm{~kg}$

The correct answer is:
$5 \mathrm{~kg}$

Solution:
We have $p_1+p_2+p_3=0 [\because p=m v]$
$\begin{aligned}
& \therefore 1 \times 12 i+2 \times 8 j+p_3=0 \\
& \Rightarrow 12 \mathrm{i}+16 \mathrm{j}+\mathrm{p}_3=0 \\
& \Rightarrow \mathrm{P}_3=-(12 \mathrm{i}+16 \mathrm{j}) \\
& \therefore \mathrm{p}_3=\sqrt{(12)^2+(16)^2} \\
& =\sqrt{144+256} \\
& =20 \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}$
Now, $p_3=m_3 v_3$
$\Rightarrow m_3=\frac{p_3}{V_3}=\frac{20}{4}=5 \mathrm{~kg}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An expression for a dimensionless quantity P is given by P=αβlogekTβx; where α and β are constants, x is distance; k is Boltzmann constant and T is the temperature. Then the dimensions of α will be

Options:
A M0L-1T0
B ML0T-2
C MLT-2
D ML2T-2

The correct answer is:
MLT-2

Solution:
Given p=αβlnkTβx

We know that expression inside log is dimensionless.

Therefore, kTβx is dimensionless

⇒β=kTx=ML2T-2K-1×K×L-1=MLT-2

Since p is dimensionless ∴ p=αβ=M0L0T0 ⇒α=β ⇒α=MLT-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2022 (26 Jun Shift 1)
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

Options:
A 3PKα
B P3αK
C PαK
D 3αPK

The correct answer is:
P3αK

Solution:
B=-PdVV
dVV=-PB
dV=-PVB
ΔV=VγΔTΔVV=γΔT
ΔVV=3αΔT
K=PΔVV=P3αΔT
ΔT=P3αk

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2017 (02 Apr)
An H atom and a Li++ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta and EH and ELi their respective energies, then

Options:
A lH>lLi and EH>ELi
B lH=lLi and EH<ELi
C lH<lLi and EH>ELi
D lH>lLi and EH<<ELi

The correct answer is:
lH=lLi and EH<ELi

Solution:
In the second excited state, n=3
â´lH=lLi=3h2Ï
As EâZ2and ZH=1, and ZLi=3
â´ ELi=9EH or EH<ELi

Topic: Physics, Atomic Physics, JEE Main
An H-atom moving with speed v makes a head-on collision with another H-atom at rest. Both the atoms are in the ground state. The minimum value of velocity v for which one of the atoms may excite is

Options:
A 6.25×104 m s-1
B 8×104 m s-1
C 7.25×104 m s-1
D 13.6×104 m s-1

The correct answer is:
6.25×104 m s-1

Solution:
For vmin collision should be completely inelastic.
According to energy conservation principle,
â´12mvmin2=12mv2+12mv2+âE ...(i))
According to momentum conservation principle.
mvmin=mv+mv
â´v=mmin2 ...(ii)
After solving eq. (i) an d(ii), we get
12mvmin2=2âE
â´vmin=4âEm
Here, âE= minimum excitation energy
=10.2 eV
m=1.67Ã10-27 kg
â´vmin=4Ã10.2Ã1.6Ã10-191.67Ã10-27
=6.25×104 m s-1

Topic: Physics, Atomic Physics, JEE Main
An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is ν- , m is its mass and kB is Boltzmann's constant, then its temperature will be:

Options:
A mÎ½-Â 25kB
B mν- 26kB
C mν- 23kB
D mÎ½-Â 27kB

The correct answer is:
mν- 23kB

Solution:
Vrms=3kBTm
v-=3kBTm
T=mv-23kB

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (09 Apr Shift 1)
An He+ ion in the excited state, on the transition to the ground state, emits two photons in succession with wavelengths 108.5 and 30.4 nm. What is the quantum number n corresponding to the excited state of He+ ion.

Options:

The correct answer is:
5

Solution:
The value of hc=1242 nm eV
∴ The energy of the first photon is
Δ E⁡ 1 = hc λ 1 = 1 2 4 2 n m eV 1 0 8 . 5 n m = 1 1 . 4 5 eV
The energy of the second photon is
Δ E⁡ 2 = hc λ 2 = 1 2 4 2 n m eV 3 0 . 4 n m = 4 0 . 8 6 eV
The total energy emitted by the electron during jumping from n2 to n1 is ΔE=ΔE1+ΔE2
=(11.45+40.86) eV=52.31 eV
or Δ E⁡ = 1 3 . 6 Z⁡ 2 1 n 1 2 - 1 n 2 2 eV
or Δ E⁡ = 1 3 . 6 Z⁡ 2 1 n 1 2 - 1 n 2 eV
or 5 2 . 3 1 eV = 1 3 . 6 Z⁡ 2 1 n 1 2 - 1 n 2 eV
But the final state of the electron is the ground state
∴ n1=1
∴ 5 2 . 3 1 eV = 1 3 . 6 Z⁡ 2 1 - 1 n 2 eV
or 1 - 1 n 2 Z 2 = 5 2 . 3 1 1 3 . 6
or 1 - 1 n 2 Z 2 = 3 . 8 5
or 1 - 1 n 2 = 3 . 8 5 2 2 = 3 . 8 5 4 = 0 . 9 6 For He + Z = 2
∴ 1 n 2 = 1 - 0 . 9 6 = 0 . 0 4
or n 2 = 1 0 . 0 4 = 2 5
∴ n=5

Topic: Physics, Atomic Physics, JEE Main
An ice berg of density $900 \mathrm{Kg} / \mathrm{m}^3$ is floating in water of density $1000 \mathrm{Kg} / \mathrm{m}^3$. The percentage of volume of ice-cube outside the water is

Options:
A $20 \%$
B $35 \%$
C $10 \%$
D $25 \%$

The correct answer is:
$10 \%$

Solution:
Let the total volume of ice-berg is $V$ and its density is $\rho$. If this ice-berg floats in water with volume $V / n$
inside it then $V_{in} \sigma g=V \rho g \Rightarrow V_{i n}=\left(\frac{\rho}{\sigma}\right) V$
or $V_{o u t}=V-V_m=\left(\frac{\sigma-\rho}{\sigma}\right) V$
$\Rightarrow \frac{V_{\text {out }}}{V}=\left(\frac{\sigma-\rho}{\sigma}\right)=\frac{1000-900}{1000}=\frac{1}{10}$
$\therefore V_{\text {out }}=10 \%$ of $V$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
An ice cube of dimensions 60 cm×50 cm×20 cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0°C of temperature is brought to a room of temperature 40°C. The rate of melting of ice is approximately: (Latent heat of fusion of ice is 3.4×105 J kg-1 and thermal conducting of insulation wall is 0.05 W m-1 °C-1)

Options:
A 61×10-1 kg s-1
B 61×10-5 kg s-1
C 208 kg s-1
D 30×10-5 kg s-1

The correct answer is:
61×10-5 kg s-1

Solution:
Using the equation of conduction,

dQdt=KAΔTl

Now the total area of the box will be,

A=20.6×0.5+0.5×0.2+0.2×0.6

A=20.3+0.1+0.12

A=20.52=1.04 m2

Therefore,

dQdt=KAΔTl=0.05×1.04×400.01=208 J s-1

This heat will melt the ice. Therefore,

dQdt=dmdtL⇒208=dmdt×3.4×105⇒dmdt=61×10-5 kg s-1

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (26 Jul Shift 2)
An ideal capacitor of capacitance 0.2 μF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V is:

Options:
A 0.15 A
B 0.17 A
C 0.34 A
D 0.25 A

The correct answer is:
0.17 A

Solution:
The energy stored in the capacitor is given by, Uc=12CV2. The energy stored in the inductor is given by. UL=12LI2. In LC circuit there is no loss of energy in the form of heat so, using energy conservation, Uci+ULi=Ucf+ULf

⇒ 12×0.2×10-6×102+0=12×0.2×10-6×52+12×0.5×10-3I2
I= 3×10-1A=0.17 A

Topic: Physics, Alternating Current, JEE Main
An ideal capacitor of capacitance 0.2 μF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V is:

Options:
A 0.15 A
B 0.17 A
C 0.34 A
D 0.25 A

The correct answer is:
0.17 A

Solution:
The energy stored in the capacitor is given by, Uc=12CV2. The energy stored in the inductor is given by. UL=12LI2. In LC circuit there is no loss of energy in the form of heat so, using energy conservation, Uci+ULi=Ucf+ULf

⇒ 12×0.2×10-6×102+0=12×0.2×10-6×52+12×0.5×10-3I2
I= 3×10-1A=0.17 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2018 (15 Apr)
An ideal capacitor of capacitance $0.2 \mu \mathrm{F}$ is charged to a potential difference of $10 \mathrm{~V}$. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $0.5 \mathrm{mH}$. The current at a time when the potential difference across the capacitor is $5 \mathrm{~V}$, is:

Options:
A
$0.17 \mathrm{~A}$

B
$0.15 \mathrm{~A}$

C
$0.34 \mathrm{~A}$

D
$0.25 \mathrm{~A}$

The correct answer is:
$0.17 \mathrm{~A}$

Solution:
Given: Capacitance, $C=0.2 \mu \mathrm{F}=0.2 \times 10^{-6}$ $\mathrm{F}$
Inductance $\mathrm{L}=0.5 \mathrm{mH}=0.5 \times 10^{-3} \mathrm{H}$
Current $\mathrm{I}$ = ?
Using energy conservation
$$
\frac{1}{2} C V^2=\frac{1}{2} C V_1^2+\frac{1}{2} L I^2
$$
$\frac{1}{2} \times 0.2 \times 10^{-6} \times 10^2+0$
$=\frac{1}{2} \times 0.2 \times 10^{-6} \times 5^2+\frac{1}{2} \times 0.5 \times 10^{-3} I^2$
$\therefore \quad I=\sqrt{3} \times 10^{-1} \mathrm{~A}=0.17 \mathrm{~A}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
An ideal Carnot's engine works between 227 Â°C and 57 Â°C. The efficiency of the engine will be

Options:
A 22%
B 34%
C 55%
D 13.5%

The correct answer is:
34%

Solution:
T1=227 Â°C=(227+273) K=500 K
T2=57 Â°C=(57+273) K=330 K
Efficiency of Carnot's engine,
Î·=1-T2T1=1-330500=0.34 or Î·=34%

Topic: Physics, Thermodynamics, JEE Main