datavorous/entrance-exam-dataset huggingface dataset:
A sample of 0.125 g of an organic compound when analysed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759mmHg. The percentage of nitrogen in the given organic compound is____

(a) The vapour pressure of water at 280 K is 14.2mmHg

(b) R=0.082 L atm K-1 mol-1

Options:

The correct answer is:
22

Solution:
Let the mass of organic compound = m g

Volume of nitrogen collected = V1 mL

Room temperature = T1K

V=22.78ml,T=280 K

Ptotal=759mmHg

PN2= Atmospheric pressure – Aqueous tension

PN2=759-14.2=744.8mmHg

22400 mL N2 at STP weighs 28 g.

V mL N2 gas at STP weighs = 28×V22400g

nNz=744.8×22.78760×1000×0.082×280=0.00097

WNitrogen=0.02716g

% N=0.027160.125×1000=21.728

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A sample of 2 kg monoatomic helium gas (assumed ideal) is taken through the process ABC and another sample of 2 kg of the same gas is taken through the process ADC. Given that the molecular mass of helium = 4 amu, find the temperature of helium in the state D. [Take the universal gas constant R=253 J mol-1K-1]

Options:

The correct answer is:
240

Solution:
Number of gram moles of He, n=mM=2×1034=500

VA = 10m3, pA = 5 x 104 N/m2

TA=PAVAnR=(10)(5×104)(500)(253) K=120 K

Similarly, VB = 10m3, pB = 10 x 104 N/m2

TB=(10)(10×104)(500)(253) K=240 K

Similarly, VC = 20 m3, pC = 10 x 104 N/m2

TC=(20)(10×104)(500)(253) K=480 K

Similarly, VD = 20 m3, pD = 5x 104 N/m2

TD=(20)(5×104)(500)(253) K=240 K

Topic: Physics, Thermodynamics, JEE Main
A sample of 5 observations has mean 32 and median 33 . Later it is found that an observation was recorded incorrectly as 40 instead of 35 . If we correct the data, then which oneot the following is correct? $\quad$

Options:
A The mean and median remain the samd
B The median remains the same but the mean will decrease
C The mean and median both will decrease
D The mean remains the same but median will decrease

The correct answer is:
The median remains the same but the mean will decrease

Solution:
Mean $=32$, Median $=33$
It was recorded as 40 instead of $35 .$ So, sum will decrease and so mean. Median remains same.

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is: (atomic mass, Ba = 137 amu, Cl = 35.5 amu)

Options:
A BaCl2⋅H2O
B BaCl2⋅3H2O
C BaCl2⋅4H2O
D BaCl2⋅2H2O

The correct answer is:
BaCl2⋅2H2O

Solution:
BaCl2⋅xH2OâBaCl2+xÂ H2OÂ
(137+2Ã35.5+18x)
=208+18xÂ g/mole
208Â +Â 18Â x208=6152
10816+936Â x=12688
936Â x=1872
x=2
Formula is BaCl2⋅2H2O

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2015 (10 Apr Online)
A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is: (atomic mass, Ba = 137 amu, Cl = 35.5 amu)

Options:
A BaCl2⋅H2O
B BaCl2⋅3H2O
C BaCl2⋅4H2O
D BaCl2⋅2H2O

The correct answer is:
BaCl2⋅2H2O

Solution:
BaCl2⋅xH2OâBaCl2+xÂ H2OÂ
(137+2Ã35.5+18x)
=208+18xÂ g/mole
208Â +Â 18Â x208=6152
10816+936Â x=12688
936Â x=1872
x=2
Formula is BaCl2⋅2H2O

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A sample of a metal oxide has formula M0.83O1.00. The metal M can exist in two oxidation states +2 and +3. In the sample of M0.83O1.00, the percentage of metal ions existing in +2 oxidation state is %. (nearest integer)

Options:

The correct answer is:
59

Solution:
Let number of ions of M+2=x

Thus, as net charge has to be zero, M+3=0.83-x

⇒2x+3(0.83-x)+1(-2)=0

⇒2x+2.49-3x-2=0

∴x=0.49

%M2+ ions =0.490.83×100=59.03

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A sample of a radioactive element whose half-life is 30 s contains a million nuclei at a certain instant of time. How many nuclei will be left after $10 \mathrm{~s}$ ?

Options:
A $7.96 \times 10^{5}$
B $3.78 \times 10^{5}$
C $1.11 \times 10^{5}$
D $1.26 \times 10^{5}$

The correct answer is:
$7.96 \times 10^{5}$

Solution:
Half-life of radioactive sample,
$$
T_{1 / 2}=30 \mathrm{~s}, N_{0}=10^{6}
$$
If $N$ be the remaining number of nuclei after $t=10 \mathrm{~s}$, then according to radioactive decay law,
$$
\begin{aligned}
N &=N_{0}\left(\frac{1}{2}\right)^{n}=10^{6}\left(\frac{1}{2}\right)^{t / T_{1 / 2}} \\
&=10^{6}\left(\frac{1}{2}\right)^{\frac{10}{30}}=10^{6}\left(\frac{1}{2}\right)^{1 / 3} \\
&=10^{6}(0.796)=7.96 \times 10^{5}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, COMEDK, COMEDK 2015
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc . The work done by the gas along the path abc is:

Options:
A 130 J
B 100 J
C 120 J
D 140 J

The correct answer is:
130 J

Solution:
In a cyclic process, ΔU=0
From 1st law of thermodynamics
Q=ΔU+W
Q=W
Qab+Qbc+Qca=Wabc+Wca
Wabc=Qab+Qbc+Qca-Wca
Wabc=Qab+Qbc+Uca
Wabc=250+60-180=130 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A sample of atomic hydrogen gas is irradiated by a beam of ultraviolet radiation having a wavelength between 100 nm and 200 nm. Assuming that the atoms are in the ground state, the wavelength which will have low intensity in the transmitted beam is

Options:
A 104 nm
B 103 nm
C 105 nm
D 100 nm

The correct answer is:
103 nm

Solution:
The energy of a photon corresponding to λ=100 nm is,

1242eV nm100nm=12·42 eV

Similarly, the corresponding to λ = 200 nm is 6.21 eV.

The energy needed to take the atom from the ground state to the first excited state is

E2-E1=13.6 eV-3.4 eV=10.2 eV

The energy needed to take the atom from the ground state to the second excited state is

E3-E1=13.6 eV-1.5 eV=12.1 eV

The energy needed to take the atom from the ground state to the third excited state is

E4-E1=13.6 eV-0.85 eV=12.75 eV, and so on.

Thus, 10.2 eV photons and 12.1 eV photons have a large probability of being absorbed from the given range of 6.21 eV to 12.42 eV.

The corresponding wavelengths are

λ 1 = 1 2 4 2 eV nm 1 0 · 2 eV = 1 2 2 nm

λ 2 = 1 2 4 2 eV nm 1 2 · 1 eV = 1 0 3 nm
These wavelengths will have low intensity in the transmitted beam.

Topic: Physics, Atomic Physics, JEE Main
A sample ofÂ CaCO3 andÂ MgCO3 weighedÂ 2.21g is ignited to constant weight ofÂ 1.152g. The composition of the mixture is:

(Given molar mass inÂ gÂ molâ1, CaCO3:100,Â MgCO3:84)

Options:
A 1.187g CaCO3+1.023 g MgCO3
B 1.023 g CaCO3+1.023 g MgCO3
C 1.187 g CaCO3+1.187 g MgCO3
D 1.023 g CaCO3+1.187 g MgCO3

The correct answer is:
1.187g CaCO3+1.023 g MgCO3

Solution:
CaCO3s→ΔCaOs+CO2g

MgCO3s→ΔMgOs+CO2g

Let the weight of CaCO3 be x gm

∴ weight of MgCO3=2.21−x gm

Moles of CaCO3 decomposed = moles of CaO formed

x100= moles of CaO formed

∴ weight of CaO formed =x100×56
Moles of MgCO3 decomposed = moles of MgO formed

2.21−x84= moles of MgO formed

∴ weight of MgO formed =2.21−x84×40 ⇒2.21−x84×40+x100×56=1.152

∴x=1.187g= weight of CaCO3 and weight of MgCO3=1.023g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A sample of calcium carbonate $\left(\mathrm{CaCO}_3\right)$ has the following percentage composition : $\mathrm{Ca}=40 \% ; \mathrm{C}=$ $12 \% ; O=48 \%$
If the law of constant proportions is true, then the weight of calcium in $4 g$ of a sample of calcium carbonate obtained from another source will be

Options:
A $0.016 \mathrm{~g}$
B $0.16 \mathrm{~g}$
C $1.6 \mathrm{~g}$
D $16 \mathrm{~g}$

The correct answer is:
$1.6 \mathrm{~g}$

Solution:
In $100 \mathrm{~g} \mathrm{CaCO3,} 40 \mathrm{~g}$ Ca is present.
In 4 g CaCO3, Let x g Ca be present
$\begin{aligned} & x / 4=40 / 100 \\ & \Rightarrow x=1.6 \mathrm{~g}\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A sample of drinking water was found to be severely contaminated with chloroform \( \left(\mathrm{CHCl}_{3}\right) \) supposed to a carcinogen. The level of contamination was \( 15 \) ppm (by mass): (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.

Options:
A \( 1.25 \times 10^{-4} \mathrm{~m} \)
B \( 2.41 \times 10^{-4} \mathrm{~m} \)
C \( 4.85 \times 10^{-6} \mathrm{~m} \)
D \( 6.45 \times 10^{-8} \mathrm{~m} 6.45 \times 10^{-8} \mathrm{~m} \)

The correct answer is:
\( 1.25 \times 10^{-4} \mathrm{~m} \)

Solution:
i Percentage % by mass of CHCl3
= mass of CHCl 3 Mass of solution × 100 = 15.0 g 10 6 g × 100 = 1.5 × 10 - 3 %
ii Molality of solution (m) = Mass of CHCl 3 / Molar mass Mass of water in kg
= 1 5 / 1 2 + 1 + 106.5 10 6 / 1000 kg
= 15 g 119.5 g mol - 1 × 1000 10 6 kg
=1.25×10-4mol Kg-1
=1.25×10-4 m.

Topic: Chemistry, Solutions, JEE Main
A sample of gas at temperature $T$ is adiabatically expanded to double its volume. Adiabatic constant for the gas is $\gamma=3 / 2$. The work done by the gas in the process is:
$(\mu=1 \mathrm{~mole})$

Options:
A $R T[1-2 \sqrt{2}]$
B $R T[\sqrt{2}-2]$
C $R T[2-\sqrt{2}]$
D $R T[2 \sqrt{2}-1]$

The correct answer is:
$R T[2-\sqrt{2}]$

Solution:
$\begin{aligned} & \mathrm{W}=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma} \\ & \mathrm{TV}^{\gamma-1}=\text { constan } \mathrm{t}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\gamma-1} \\ & \mathrm{~T}_{\mathrm{f}}=\mathrm{T}\left(\frac{1}{2}\right)^{1 / 2}=\frac{\mathrm{T}}{\sqrt{2}} \\ & \mathrm{~W}=\frac{\mathrm{R}\left(\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right)}{1-\frac{3}{2}}=2 \mathrm{RT} \frac{(\sqrt{2}-1)}{\sqrt{2}} \\ & =\mathrm{RT}(2-\sqrt{2})\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A sample of gas expands from $V_1$ to $V_2$. In which of the following, the work done will be greatest?

Options:
A
Same in all processes

B
Isobaric process

C
Isothermal process

D
Adiabatic process

The correct answer is:
Isobaric process

Solution:
Work done $=$ Area bounded by PV graph and volume axis
Among the three process, bounded area by PV graph and volume axis is greatest for isobaric process, hence work done is greatest for isobaric process.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2013 (25 Apr Online)
A sample of gas is compressed by an average pressure of 0.50 atmosphere so as to decrease its volume from 400 cm3 to 200â cm 3 . During the process 8.00âJ of heat flows out to surroundings. The change in internal energy of the system is

Options:
A +2.13 J
B +10.13Â J
C â2.13Â J
D â10.13Â J

The correct answer is:
+2.13 J

Solution:
Here, ÎV=200-400=-200 cm3
A we know 1âLitreâ=1000 cm 3
âÎV=â0.2âLitre.
External Pressure P=0.50 then
Work done =-PÎV=+0.500.2
=+0.1âatmâLitre
and 1âlitre-atmosphereâ=â101.3âJoule
â´ W=+10.13 Joule
According to First law of thermodynamics
q=ÎE-W
then ÎE=q+W
Given q=-8.00 J
âÎE=â8+10.13=2.13âJoule

Topic: Chemistry, Thermodynamics (C), JEE Main
A sample of gas with γ=1.5 is taken through an adiabatic process in which the volume is compressed from 1200 cm3 to 300 cm3. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process =________ J.

Options:

The correct answer is:
480

Solution:
Vi=1200×10-6 m3, Pi=200 KPa
Vf=300×10-6 m3, Pf=?
γ=1.5
PiViγ=PfVfγ
200×1031200×10-6300×10-6γ=Pf
Pf=200×103×43/2
=200×103×8
=1600×103=16×105 Pa
W=PiVi-PfVfγ-1
W=2×105×1200×10-6-16×105×300×10-61.5-1=240-4800.5=-240×2=-480 J|W|=480 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A sample of H2SO4 (density 1.787 g mL-1 ) is labelled as 86% by weight. What volume (in mL) of this acid has to be used to make 1 L of 0.2 M H2SO4 ?
[Report your answer upto two decimal places.]

Options:

The correct answer is:
12.75

Solution:
H2SO4 is 86% by weight
So weight of H2SO4 = 86 g
Weight of solution = 100 g
So volume of solution =1001.787mL=1001.787×1000L
MH2SO4=8698×1001.787×1000=15.68
Let V mL of this H2SO4 are used to prepare 1 L of 0.2 M H2SO4 .
So mmoles of concentrated H2SO4= mmoles of dilute H2SO4
So V×15.68=1000×0.2
So V = 12.75 mL.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A sample of hydrogen gas was collected over water at 21°C and 685 mm Hg. The volume of the container was 7.80 L. Calculate the mass of H2(g) collected (vapour pressure of water = 18.6 mm Hg at 21°C )

Options:
A 0.283 g
B 0.570 g
C 0.589 g
D 7.14 g

The correct answer is:
0.570 g

Solution:
The volume of H2 at STP
V2=P1V1T1ÃT2P2
Here vapour pressure of H2 at 21Â°C
= 685 mm â 18.6 mm = 666.4 mm
So V 2 = 666.4âmmâHgâÃâ7.80âLâÃâ273âK 294âKâÃâ760âmm =6.38âL
At STP 22.4âLâ H 2 âweightâ=â2âg
So, at STP 6.38 L  H 2  weigh =  2 g × 6.38 L 22.4 L  = 0.570 g

Topic: Chemistry, States of Matter, JEE Main
A sample of Hydrogen peroxide solution labelled as \( 28 \) volume has density of \( 265 \mathrm{~g} / \mathrm{L} \). Mark the correct options(S)
representing concentration of same solution in other units.

Options:
A Molarity of \( \mathrm{H}_{2} \mathrm{O}_{2}=2.8 \)
B Percentage \( \mathrm{w} / \mathrm{v}=8.5 \)
C Mole fraction of \( \mathrm{H}_{2} \mathrm{O}_{2}=0.5 \)
D Molality of \( \mathrm{H}_{2} \mathrm{O}_{2}=12.53 \)

The correct answer is:
Percentage \( \mathrm{w} / \mathrm{v}=8.5 \)

Solution:
Given,

Volume strength is 28.

Then, Molarity is volume strength11.2=2811.2=2.5

Percentage wv is volume strength×6822.4 =28×68224=8.5.

Topic: Chemistry, Hydrogen, JEE Main
A sample of ideal gas \( (\gamma=1.4) \) is heated at constant pressure. If \( 100 \mathrm{~J} \) of heat is supplied to the gas the work done by the gas
is

Options:
A \( 28.57 \mathrm{~J} \)
B \( 56.54 \mathrm{~J} \)
C \( 38.92 \mathrm{~J} \)
D \( 65.38 \mathrm{~J} \)

The correct answer is:
\( 28.57 \mathrm{~J} \)

Solution:
dQ = n C P dT
dU = n C V dT
dW = dQ - dU = n C P - C V dT
dW dQ = n C P - C V dT n C P dT = C P C P - C V C P
dW 100 = 1 - C V C P = 1 - 1 1.4 = 0.4 1.4 = 4 14 = 2 7
dW = 100 × 2 7 = 28.57 J

Topic: Physics, Thermodynamics, JEE Main
A sample of ideal gas is at equilibrium. Which of the following quantity is zero?

Options:
A rms speed
B average speed
C average velocity
D most probable speed

The correct answer is:
average velocity

Solution:
The molecules move in all possible directions in an ideal gas at equilibrium. Since, momentum is a vector quantity for every direction of motion of the molecule, there exists an opposite direction of motion of the other. And we know that momentum is directly proportional to velocity. Thus, when the average momentum is zero therefore average velocity will also be zero for an ideal gas at equilibrium.

Topic: Physics, Kinetic Theory of Gases, JEE Main
A sample of milk splits after 60min . at 300K and after 40min . at 400K when the population of Iactobacillus acidophilus in it doubles. The activation energy (in kJ/mol ) for this process is closest to ______________.
(Given, R=8.3Jmol-1K-1,ln23=0.4,e-3=4.0 )

Options:

The correct answer is:
3.98

Solution:
lnk2k1=EaR1T11T2
ln6040=Ea8.3×100400×300
ln32×8.3×1200=Ea
⟹Ea=0.4×8.3×1200
⟹Ea=3984Jmol.
⟹Ea=3.984 k Jmol.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A sample of NaClO3 is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be: (Given: Molar mass of AgCl=143.5 g mol-1)

Options:
A 0.35
B 0.54
C 0.41
D 0.48

The correct answer is:
0.48

Solution:
2NaClO3→∆2NaCl+3O2
2 moles of NaCl=3 moles of oxygen
nNaCl=0.1632×23=1200×23=1300
The number of moles of NaCl=The number of moles of AgCl
1300=nAgCl
Weight of AgCl=1300×108+35.5=1300×143.5
=0.48 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2018 (15 Apr)
A sample of $\mathrm{NaClO}_3$ is converted by heat to $\mathrm{NaCl}$ with a loss of $0.16 \mathrm{~g}$ of oxygen. The residue is dissolved in water and precipitated as $\mathrm{AgCl}$. The mass of $\mathrm{AgCl}$ (in $\mathrm{g}$) obtained will be:
(Given: Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Options:
A
$0.35$

B
$0.54$

C
$0.41$

D
$0.48$

The correct answer is:
$0.48$

Solution:
No. of moles of oxygen in $0.16 \mathrm{~g}$ of oxygen molecule
$=\frac{0.16 \mathrm{~g}}{32 \mathrm{~g} / \mathrm{mol}}=0.005 \mathrm{~mole}$
$2 \mathrm{NaClO}_3 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaCl}+3 \mathrm{O}_2$
According to the reaction,
$3$ moles of $\mathrm{O}_2=2$ moles of $\mathrm{NaCl}$
$=2$ moles of $\mathrm{AgCl}$
Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} / \mathrm{mol}$
$0.005$ moles of $\mathrm{O}_2$ will ppt. $=0.005 \times \frac{2}{3}$ moles $\mathrm{AgCl}$
$=0.0033 \text { moles of } \mathrm{AgCl}$
$\therefore$ Mass of $\mathrm{AgCl}$ (in $\mathrm{g}$) obtained will be
$=143.5 \mathrm{~g} / \mathrm{mol} \times 0.0033$ moles $=0.48 \mathrm{~g}$.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A sample of radioactive element contains $8 \times 10^{16}$ active nuclei. The halt-life of the element is 15 days. The number of nuclei decayed after 60 days is

Options:
A $7.5 \times 10^{+16}$
B $2.0 \times 10^{16}$
C $0.5 \times 10^{16}$
D $4.0 \times 10^{16}$

The correct answer is:
$7.5 \times 10^{+16}$

Solution:
Half life, $\mathrm{T}=15$ days, time $\mathrm{t}=60$ days $=4 \mathrm{~T}$ The number of nuclei remaining is given by
$$
\begin{aligned}
& \mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}=\mathrm{N}_0\left(\frac{1}{2}\right)^4=\frac{1}{16} \times 8 \times 10^{16} \\
& =0.5 \times 10^{16} \\
& \therefore \text { No. of nuclei decayed }=\mathrm{N}_0-\mathrm{N} \\
& =8 \times 10^{16}-0.5 \times 10^{16}=7.5 \times 10^{16}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, MHT CET, MHT CET 2021 (20 Sep Shift 1)
A Sample of radioactive material A, that has an activity of 10 m Ci 1 Ci=3.7×1010 decays s-1, has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 m Ci. The correct choices for half-lives of A and B would then be, respectively:

Options:
A 5 days and 10 days
B 10 days and 40 days
C 20 days and 10 days
D 20 days and 5 days

The correct answer is:
20 days and 5 days

Solution:
Activity A=λN
10=λANA
20=λBNB
12=λAλB2
λAλB=14 ⇒TATB=41

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A sample of red ink (a colloidal suspension) is prepared by mixing eosine dye, egg white, HCHO and water. The component which ensures stability of the ink sample is :

Options:
A Egg white
B Water
C HCHO
D Eosin dye

The correct answer is:
Egg white

Solution:
Blue ink is a colloidal sol, so it can be stabilised by material like natural gum or Egg white/albumen.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A sample originally contained $10^{20}$ radioactive atoms, which emit $\alpha$-particles. The ratio of $\alpha$ particles emitted in the third year to that emitted during the second year is $0.3$. How many $\alpha$ particles were emitted in the first year?

Options:
A
$3 \times 10^{18}$

B
$3 \times 10^{19}$

C
$5 \times 10^{18}$

D
$7 \times 10^{19}$

The correct answer is:
$3 \times 10^{19}$

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2012 (07 May Online)
A sample space consists of \( 3 \) sample points with associated probabilities given as \( 2 p, p^{2}, 4 p-1 . \) Then the value of \( p \) is

Options:
A \( \sqrt{11}-3 \)
B \( \sqrt{10}-3 \)
C \( \frac{1}{4} < p < \frac{1}{2} \)
D none

The correct answer is:
\( \sqrt{11}-3 \)

Solution:
We know that sum of probabilities of all exhaustive events for a sample space is 1.

⇒p2+2p+4p-1=1

⇒p2+6p-2=0

⇒p=-6±62-41-221=-3±11 ...i

We also know that probability of an event belongs to 0,1.

⇒0≤p2≤1 & 0≤2p≤1 & 0≤4p-1≤1

⇒-1≤p≤1 & 0≤p≤12 & 14≤p≤12

⇒14≤p≤12 ...ii

From equations i & ii,

p=11-3

Topic: Mathematics, Probability, JEE Main
A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is :

Options:
A 1:6
B 1:3
C 1:2
D 3:4

The correct answer is:
1:6

Solution:
From angular momentum conservation

mV1r1=mV2r2

r1r2=v2v1=16

1:6

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A satellite is in an orbit around the earth. If its kinetic energy is doubled, then

Options:
A it will maintain its path
B it will fall on the earth
C it will rotate with a great speed
D it will escape out of earth’s gravitational field

The correct answer is:
it will escape out of earth’s gravitational field

Solution:
Kinetic energy, $K=\frac{1}{2} m v_0^2$
or $v_0=\sqrt{\frac{2 K}{m}}$
When kinetic energy is doubled, then
$v_0^{\prime}=\sqrt{\frac{2 \times 2 K}{m}}=\sqrt{2} \sqrt{\frac{2 K}{m}}=\sqrt{2} v_0$
But $\sqrt{2} v_0=v_e$ (escape velocity)
Hence satellite will escape out of earth's gravitational field.

Topic: Physics, Gravitation, AIIMS, AIIMS 2014
A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is:

Options:
A 1.5
B 2.0
C 0.7
D 3.0

The correct answer is:
3.0

Solution:
T2∝R3

T=kR3/2

dTT=32dRR

=32×0.02=0.03

%Change =3%

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01R . The period of the second satellite is larger than that of the first one by approximately

Options:

The correct answer is:
1.5

Solution:
In the problem orbital radius is increased by 1%
Time period of satellite T∝r3/2

⇒∆TT=32×∆rr
Percentage change in the time period
=32 [ % change in orbital radius]
=321%=1.5%

Topic: Physics, Gravitation, JEE Main
A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.02)R. The period of the second satellite is larger than the first one by approximately

Options:
A 0.7%
B 1.0%
C 1.5%
D 3.0%

The correct answer is:
3.0%

Solution:
from kepler's law,
T2∝R3 ⇒T∝R32
⇒T= k R 3 2 ;k k is some constant . for radius =R
T=k R 3 2
dT dR = 3 2 k R 1 2
when radius =1.02R;dR=1.02R−R=0.02R
dT dR = 3 2 k R 1 2
⇒dT= 3 2 k R 1 2 ×dR
⇒dT=1.5kR12×0.02R
% change = dT T ×100= 0.03T T ×100=3

Topic: Physics, Gravitation, JEE Main
A satellite is launched into a circular orbit, R4 above the surface of the earth. The time period of revolution is T=2Ïn32Rg. Where R is the radius of the earth and g is the acceleration due to gravity. Then what is value on n?

Options:

The correct answer is:
1.25

Solution:
T⁡=2πr⁡32gR⁡2 where r = R + h
Then, T⁡=2πR⁡+h⁡R⁡R⁡+h⁡g
Here, h=R4
T⁡=2πR⁡+R⁡4R⁡R⁡+R⁡4g=2π1.2532R⁡g

Topic: Physics, Gravitation, JEE Main
A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius Re. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become 32 times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is :

Options:
A 4Re
B 2.5Re
C 3Re
D 2Re

The correct answer is:
3Re

Solution:
V=GMRe

From angular momentum conversation

-GMmRe+12m32V2=-GMmRmax+12mVmin2

From angular momentum conversation

32VRe=VminRmax

Eliminating Vmin from equation (i) and (ii) we get

Rmax=3Re

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (03 Sep Shift 1)