datavorous/entrance-exam-dataset huggingface dataset:
A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of λ. An alpha particle having certain kinetic energy has the same de-Brogle wavelength λ. The ratio of kinetic energy of proton and that of alpha particle is :

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

The correct answer is:
4 : 1

Solution:
As we know, de Broglie wavelength is given by,

λ=hp

Now kinetic energy can be written as,

K=12mv2=p22m=h22mλ2

Therefore,

K∝1m

Hence, required ratio

KpKα=mαmp=4⇒mα:mp=4:1

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the proton will move in a circular path of radius

Options:
A $\frac{2 E m}{q B}$
B $\frac{\sqrt{2 E m}}{q B}$
C $\frac{\sqrt{E m}}{2 q B}$
D $\sqrt{\frac{2 E q}{m B}}$

The correct answer is:
$\frac{\sqrt{2 E m}}{q B}$

Solution:
Given, Kinetic energy $=E$
Mass $=m$
Magnetic field $=B$
Charge $=q$ We know that
$$
F=q v B \sin \theta
$$
(motion of a charged particle in a uniform magnetic field) If
$$
\theta=90^{\circ}
$$
Then
$$
F=q v B
$$
We know that also (centripetal force)
$$
F=\frac{m v^{2}}{r}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
q v B &=\frac{m v^{2}}{r}, r=\frac{m v}{q B}\left[\because E=\frac{1}{2} m v^{2}, v=\sqrt{\frac{2 E}{m}}\right] \\
\therefore \quad r &=\frac{m \sqrt{\frac{2 E}{m}}}{q B} \Rightarrow r=\frac{\sqrt{2 E m}}{q B}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton of mass $\mathrm{m}$ collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^{\circ}$ with respect to each other. The mass of unknown particle is:

Options:
A
$\frac{\mathrm{m}}{\sqrt{3}}$

B
$\frac{\mathrm{m}}{2}$

C
$2 \mathrm{~m}$

D
$\mathrm{m}$

The correct answer is:
$\mathrm{m}$

Solution:
Apply principle of conservation of momentum along $x$-direction,
$$
\begin{aligned}
m u &=m v_1 \cos 45^{\circ}+M v_2 \cos 45^{\circ} \\
m u &=\frac{1}{\sqrt{2}}\left(m v_1+M v_2\right)
\end{aligned}
$$
Along $y$-direction,
$$
\begin{aligned}
&o=m v_1 \sin 45^{\circ}-M v_2 \sin 45^{\circ} \\
&o=\left(m v_1-M v_2\right) \frac{1}{\sqrt{2}}
\end{aligned}
$$
Coefficient of restution $e=1$
$$
=\frac{v_2-v_1 \cos 90}{u \cos 45}
$$
( $\because$ Collision is elastic)
$$
\begin{aligned}
&\Rightarrow \frac{v_2}{\frac{u}{\sqrt{2}}}=1 \\
&\Rightarrow u=\sqrt{2} v_2
\end{aligned}
$$
Solving eqs (i), (ii), \& (iii), we get mass of unknown particle, $M=m$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A proton of mass $m$ moving with a speed $v$ ( $< < $ c. velocity of light in vacuum) completes circular orbit in time $T$ in a uniform Whetic ficld. If the speed of the proton is increased to $\sqrt{2}$ v, what will be time needed to plete the circular orbit?

Options:
A $\sqrt{T}$
B $T$
C $\frac{T}{\sqrt{2}}$
D $\frac{T}{2}$

The correct answer is:
$T$

Solution:
$\because$ When a charged particle moves at a circular path in uniform magnetic field, it time period is independent from its speed.
Time period of proton will not change with change in its speed.

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton, when accelerated through a potential difference of V= 29.6 V, has a wavelength Î» associated with it. An Î± - particle, in order to have the same Î», must be accelerated through a potential difference of how many volts?

Options:

The correct answer is:
3.7

Solution:
λ P = λ α
orh2mpQPVp=h2mαQαVα
∴ mpQpVp=mαQαVα
∴ Vα=mpmαQpQαVp

⇒Vα=1412×29.6 V=3.7 V

Topic: Physics, Electromagnetic Waves, JEE Main
A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude π2×10-3 T. The angle between the direction of magnetic field and velocity of proton is 60o. The pitch of the helical path taken by the proton is ____ cm. (Take, mass of proton =1.6×10-27 kg and charge on proton =1.6×10-19 C).

Options:

The correct answer is:
40

Solution:
The given data is

K.E=2 eV

B=π2×10-3 T

θ=60o

The pitch of the proton is given by

P=2πmqB×vcosθ ...(i)

By using

K=mv22⇒v=2Km

Substituting the values in equation (i)

=2π×2mKE×12×21.6×10-19×π×10-3

=2×2×1.6×10-27×2×1.6×10-19×1031.6×10-19

=2×2×10–1=0.4 m

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A pulley of radius 1.5 m is rotated about its axis by a force F=12t-3t2 N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m2, the number of rotations made by the pulley before its direction of motion is reversed, will be Kπ. The value of K is _____ .

Options:

The correct answer is:
18

Solution:
Torque of pulley about its axis is τ=Iα, where, I is moment of inertia and α is angular acceleration.

Torque of a force about its axis is τ=FR, here, R is radius of pulley.

Then, 12t-3t21.5=4.5α

⇒α=4t-t2

Angular acceleration in terms of angular velocity is

⇒α=dωdt=4t-t2

⇒ω=∫0t4t-t2dt

⇒ω=2t2-t33

For ω=0, 2t2-t33=0⇒t22-t3=0

⇒t=0 or 6 s

Now, using ω=dθdt, we have

dθdt=2t2-t33⇒θ=∫062t2-t33dt

=2t33-t41206

=6323-612=638-612=636=36

Number of revolutions =362π=18π

∴K=18

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A pulley of radius 2 m is rotated about its axis by a force =20t-5t2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotation made by the pulley before its direction of motion is reversed is

Options:
A more than 3 but less than 6
B more than 6 but less than 9
C more than 9
D less than 3

The correct answer is:
more than 3 but less than 6

Solution:
We have,

Torque =r×F=lα

or 220t-5t2=10α

α=4t-t2

But α=dωdt, so

dωdt=4t-t2

dω=4t-t2dt

Integrating ω=2t2-t33

ω will be zero at t=6 s

So, ω=dθdt=2t2-t33

or dθ=2t2-t33dt

Again, integrating both sides
θ=2t33-t412

since. t=6 s

so, θ=2×633-6412=36

Number of turns, n=θ2π=362π=5.73

Topic: Physics, Rotational Motion, JEE Main
A pulley of radius $2 \mathrm{~m}$ is rotated about its axis by a force $\mathrm{F}=\left(20 \mathrm{t}-5 \mathrm{t}^2\right)$ Newton (where $\mathrm{t}$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation made by the pulley before its direction of motion if reversed, is :

Options:
A
more than 3 but less than 6

B
more than 6 but less than 9

C
more than 9

D
less than 3

The correct answer is:
more than 3 but less than 6

Solution:
$\mathrm{r} \times \mathrm{F}=\mathrm{l} \times \alpha$
$2\left(20 \mathrm{t}-5 \mathrm{t}^2\right)=10 \alpha \Rightarrow \alpha=4 \mathrm{t}-\mathrm{t}^2$
$\frac{\mathrm{d} \omega}{\mathrm{dt}}=4 \mathrm{t}-\mathrm{t}^2$
$\mathrm{~d} \omega=\left(4 \mathrm{t}^2-\mathrm{t}^2\right) \mathrm{dt}$
$\omega=2 \mathrm{t}^2-\frac{\mathrm{t}^3}{3}$ (on integration)
$\omega=0 \Rightarrow \mathrm{t}=6 \mathrm{~s}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2011
A pulse or a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with

Options:
A The same phase as the incident pulse but with velocity reversed
B A phase change of $180^{\circ}$ with no reversal of velocity
C The same phase as the incident pulse with no reversal of velocity
D A phase change of $180^{\circ}$ with velocity reversed

The correct answer is:
A phase change of $180^{\circ}$ with velocity reversed

Solution:
On reflection from fixed end (denser medium) a phase difference of $\pi$ is introduced.

Topic: Physics, Waves and Sound, JEE Main
A purse contains 10 ten rupee coins and 5 five rupee coins. Two coins are randomly drawn. If the expected value of 2 drawn coins is λ, then 9λ4 is equal to

Options:

The correct answer is:
37.5

Solution:
2 coins can be both 10 rupee coins, both 5 rupee coins or one 5 rupee coin and one 10 rupee coin
So, expected value =10C215C2×20+10C215C2×10+10C15C115C2×15
=37×20+221×10+1021×15
=35021=503

Topic: Mathematics, Probability, JEE Main
A purse contains three 10 paise, three 50 paise and ten 1 rupee coins. If three coins are selected at random, then the probability that the total amount is 2 rupees is

Options:
A 25
B 356
C 14
D 18

The correct answer is:
356

Solution:
Favourable cases= There is only one possibility as 3 coins must be two 50 paise coin and one 1 rupee coin
Number of favourable ways =C32ÃC101
Total number of ways =C163
Hence, the required probability = C 3 2 × C 10 1 C 16 3 = 3×10×6 16×15×14 = 3 56

Topic: Mathematics, Probability, JEE Main
A quantity of electricity required to reduce 12.3 g of nitrobenzene to aniline arising 50% current efficiency is

Options:
A 115800 C
B 57900 C
C 23160 C
D 28950 C

The correct answer is:
115800 C

Solution:
Molar mass of nitrobenzene = 123 Equivalent mass of nitrobenzene
=  Molecular mass 6
=1236=20.5
Number of equivalents of nitrobenzene
= Mass Equivalent mass
=12.320.5=0.6
Amount of charge required =0.6×96500=57900 Coulomb
Amount of charge required when current efficiency is 50%
=2×57900
= 115800 C

Topic: Chemistry, Redox Reactions, JEE Main
A quick supply of $\mathrm{Cl}_2$ gas may be made by reacting crystals of $\mathrm{KMnO}_4$ with a concentrated solution of

Options:
A Potassium chloride
B Sodium chloride
C Bleaching powder
D Hydrochloric acid

The correct answer is:
Hydrochloric acid

Solution:
$2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}_2+5 \mathrm{Cl}_2+8 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A radar has a power of $1 \mathrm{~Kw}$ and is operating at a frequency of $10 \mathrm{~GHz}$. It is located on a mountain top of height $500 \mathrm{~m}$. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth $=6.4 \times 10^6 \mathrm{~m}$ ) is

Options:
A
$80 \mathrm{~km}$

B
$16 \mathrm{~km}$

C
$40 \mathrm{~km}$

D
$64 \mathrm{~km}$

The correct answer is:
$80 \mathrm{~km}$

Solution:
Maximum distance on earth where object can be detected is $\mathrm{d}$, then
$\begin{array}{ll}
& (h+R)^2=d^2+R^2 \\
\Rightarrow & d^2=h^2+2 R h \\
\text { since } & h < < R, \quad \Rightarrow d^2=2 h R \\
\Rightarrow & d=\sqrt{2(500)\left(6.4 \times 10^6\right)}=80 \mathrm{~km}
\end{array}$

Topic: Physics, Communication System, JEE Main, JEE Main 2012 (Offline)
A radar sends an electromagnetic signal of electric field E0=2.25 V m-1 and magnetic field B0=1.5×10-8 T which strikes a target on line of sight at a distance of 3 km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t=0 from radar, then after how much time echo will reach to the radar?

Options:
A 2.0×10-5 s
B 4.0×10-5 s
C 1.0×10-5s
D 8.0×10-5s

The correct answer is:
4.0×10-5 s

Solution:
The speed of light is given by,

c=E0 B0=2.251.5×10-8=1.5×108 m s-1

Therefore, time taken for the echo will be,

t=2dc=6×1031.5×108=4×10-5 s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A radar sends waves towards a distant object and receives the signal reflected by the object. These waves are

Options:
A sound waves
B light waves
C radio waves
D microwaves

The correct answer is:
microwaves

Solution:
Microwaves are used to locate the flying objects by radar.

Topic: Physics, Communication System, JEE Main
A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Options:
A
$E / c$

B
$2 \mathrm{E} / \mathrm{c}$

C
Ec

D
$\mathrm{E} / \mathrm{c}^2$

The correct answer is:
$2 \mathrm{E} / \mathrm{c}$

Solution:
$\Delta \mathrm{P}_{\text {surface }}=-\Delta \mathrm{P}=\frac{2 \mathrm{E}}{\mathrm{C}}$.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2004
A radio active material is reduced to 18 of its original amount in 3 days. If 8×10-3 kg of the material is left after 5 days the initial amount of the material is

Options:
A 40 g
B 32 g
C 64 g
D 256 g

The correct answer is:
256 g

Solution:
The initial and final amount of the radioactive substance are related by the formula

NN0=12n ...1

where, n is the number of half lives.

Substitute 18 for NN0 into equation (1) and solve to calculate the value of n.

18=12n⇒123=12n⇒n=3

The half-life of the radioactive element can be calculated as follows:

tt12=n ...2

Substitute the values of the known parameters into equation (2) and solve to calculate the half-life.

3 dayst12= 3⇒t12= 1 day

The initial and the final mass of the radioactive element are related by

mm0=12t't12 ...3

Substitute the values of the known parameters into equation (3) and solve to calculate the initial mass of the substance.

8×10-3 kgm0= 1251⇒m0= 25×8×10-3 kg= 256 g

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A radio can tune to any station in 6MHz to 10MHz band. The value of corresponding wavelength bandwidth will be

Options:
A 4 m
B 20 m
C 30 m
D 50 m

The correct answer is:
20 m

Solution:
Given: Frequency f1=6MHz

Frequency f2=10MHz

Wavelength can be calculated using λ=cf.

⇒λ1=cf1=3×1086×106=50 m

⇒λ2=cf2=3×10810×106=30 m

Wavelength bandwidth is the difference of upper lower wavelengths =λ1-λ2=20 m

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (26 Jul Shift 1)
A radio isotope ‘X’ with a half life 1.4×109 years decays to ‘Y’ which is stable. A sample of the rock from a cave was found to contain ‘X’ and ‘Y’ in the ratio 1:7. The age of the rock is

Options:
A 1.96×109 years
B 3.92×109 years
C 4.20×109 years
D 8.40×109 years

The correct answer is:
4.20×109 years

Solution:
Given a radioactive isotope decays into a stable element,

X→Y

At time t=0, x= N0, y=0

At time t=t, x=N0-x, y=x

N0-xx=17⇒x=7N08
So, remaining nuclei of X=N0-x=N08=N023
So, three half-life periods would have been passed,
⇒t=3t12=3×1.4×109=4.2×109 years.

Topic: Physics, Nuclear Physics, NEET, NEET 2014
A radio transmitter transmits at $830 \mathrm{kHz}$. At a certain distance from the transmitter magnetic field has amplitude $4.82 \times 10^{-11} \mathrm{~T}$. The electric field and the wavelength are respectively

Options:
A
$0.014 \mathrm{~N} / \mathrm{C}, 36 \mathrm{~m}$

B
$0.14 \mathrm{~N} / \mathrm{C}, 36 \mathrm{~m}$

C
$0.14 \mathrm{~N} / \mathrm{C}, 360 \mathrm{~m}$

D
$0.014 \mathrm{~N} / \mathrm{C}, 360 \mathrm{~m}$

The correct answer is:
$0.014 \mathrm{~N} / \mathrm{C}, 360 \mathrm{~m}$

Solution:
Frequency of EM wave $v=830 \mathrm{KHz}$ $=830 \times 10^3 \mathrm{~Hz}$.
Magnetic field, $B=4.82 \times 10^{-11} \mathrm{~T}$
As we know, frequency, $v=\frac{c}{\lambda}$
$$
\begin{aligned}
\text { or } \lambda & =\frac{c}{v}=\frac{3 \times 10^8}{830 \times 10^3} \\
\lambda & \simeq 360 \mathrm{~m} \\
\text { And, } E & =B C=4.82 \times 10^{-11} \times 3 \times 10^8 \\
& =0.014 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2012 (26 May Online)
A radioactive element forms its own isotope after 3 consecutive disintegrations. The particles emitted are

Options:
A $3 \beta$-particles
B $2 \beta$-particles and $1 \alpha$-particle
C $2 \beta$-particles and $1 \gamma$-particle
D $2 \alpha$-particles and $1 \beta$-particle

The correct answer is:
$2 \beta$-particles and $1 \alpha$-particle

Solution:
To form its own isotope atomic number (Z) should remain same.
So, the emission of one $\alpha$-particle and two $\beta$-particles will maintain the $Z$ same.
where $\alpha$-particle $={ }_{2} \mathrm{He}^{4}$; b-particle $={ }_{-1} \beta^{0}$

Topic: Physics, Nuclear Physics, KCET, KCET 2007
A radioactive element gets spilled over the floor of a room. Its half-life period is $30$ days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

Options:
A
$1000$ days

B
$300$ days

C
$10$ days

D
$100$ days

The correct answer is:
$100$ days

Solution:
Activity $\left(-\frac{\mathrm{dN}}{\mathrm{dt}}\right) \propto \mathrm{N}$
$\mathrm{N}=\mathrm{N}_{\circ}\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\frac{\mathrm{N}}{\mathrm{N}_{\circ}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\frac{1}{10}=\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 10=2^{\mathrm{n}}$
$\log 10=\mathrm{nlog} 2$
$\Rightarrow \mathrm{n}=\frac{1}{0.301}=3.32$
$\mathrm{t}=\mathrm{n} \times \mathrm{t}_{112}$
$=3.32 \times 30=99.6 \text { days }$
Hence, (D) is correct.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2007
A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is____(Nearest integer)
(Given : antilog 0.125=1.333, antilog 0.693=4.93)

Options:

The correct answer is:
75

Solution:
For first order reaction,

K=2.303tlogNoNt ... 1

Also, K=0.693t12 ... 2

Equating 1 and 2

Hence, 2.303tlogNoNt=0.693t12

2.30383logNoNt=0.693200

NoNt=43

%NtN0=0.75×100

=75%

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A radioactive element X92242 emits two α-particles, one electron and two positrons. The product nucleus is represented by YP234. The value of P is ______.

Options:

The correct answer is:
87

Solution:
When a radioactive nucleus emits 1 α-particle, its mass number decreases by 4 and the atomic number decreases by 2. During positron emission, the mass number remains unchanged but the atomic number decreases by 1. Similarly, during electron emission, the mass number remains unchanged but the atomic number decreases by 1.

So, the value of P=92-2-2+1-1-1

⇒P=92-5

⇒P=87.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is

Options:
A 2 hours
B 4 hours
C between 4 hours and 6 hours
D 6 hours

The correct answer is:
between 4 hours and 6 hours

Solution:
Initially at $t=0$, amount of $X$ in
sample $=5 X$
After time ' $t$ ' amount of $X$ remained in sample $=X$
From, $\frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{t}{T}}$
We have, $\frac{X}{5 X}=\left(\frac{1}{2}\right)^{\frac{t}{2}} \Rightarrow\left(\frac{1}{2}\right)^{\frac{t}{2}}=\frac{1}{5}$
$\therefore T=2 \mathrm{~h}$
As, $\frac{1}{2^2} < \frac{1}{5} < \frac{1}{2^3} \Rightarrow \frac{1}{2^2} < \frac{1}{2^{t / 2}} < \frac{1}{2^3}$
$2 < t / 2 < 3 \Rightarrow 4 < t < 6$

Topic: Physics, Nuclear Physics, JEE Main
A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is

Options:
A 2 hours
B 4 hours
C between 4 hours and 6 hours
D 6 hours

The correct answer is:
between 4 hours and 6 hours

Solution:
Initially at $t=0$, amount of $X$ in
sample $=5 X$
After time ' $t$ ' amount of $X$ remained in sample $=X$
From, $\frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{t}{T}}$
We have, $\frac{X}{5 X}=\left(\frac{1}{2}\right)^{\frac{t}{2}} \Rightarrow\left(\frac{1}{2}\right)^{\frac{t}{2}}=\frac{1}{5}$
$\therefore T=2 \mathrm{~h}$
As, $\frac{1}{2^2} < \frac{1}{5} < \frac{1}{2^3} \Rightarrow \frac{1}{2^2} < \frac{1}{2^{t / 2}} < \frac{1}{2^3}$
$2 < t / 2 < 3 \Rightarrow 4 < t < 6$

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 1)
A radioactive isotope, A undergoes simultaneous decay to different nuclei as :
Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining ?

Options:
A \( 6.0 \mathrm{~h} \)
B \( 7.0 \mathrm{~h} \)
C \( 8.0 \mathrm{~h} \)
D \( 5.0 \mathrm{~h} \)

The correct answer is:
\( 6.0 \mathrm{~h} \)

Solution:
t 1/2 (P)  = 2 t 1/2  (Q)
â Q is formed at twice the rate of formation of P i.e., at any time t; [Q] = 2[P]
Rate constant for decay of A(k) = kP + kQ
â k = ln 2 9 + ln 2 4 . 5
= ln 2 3 h -1
Let after t-hours, [Q] = 2 [A] and [P] = x%

Given, 2x = 2 (100 - 3x)
â x = 2 5
â t 3 ln 2 = ln( 1 0 0 1 0 0 - 3 x ) = ln 4
â t = 6 . 0 h

Topic: Chemistry, Chemical Kinetics, JEE Main
A radioactive isotope decays at such a rate that after 192 minutes only $\frac{1}{16}$ of the original amount remains. The half-life of the radioactive isotope is

Options:
A $12 \mathrm{~min}$
B $24 \mathrm{~min}$
C $32 \mathrm{~min}$
D $48 \mathrm{~min}$

The correct answer is:
$48 \mathrm{~min}$

Solution:
For radioactive decay
$t_{1 / 2}=\frac{0.693}{k}$
For concentration to reduce to $\frac{1}{16}$ of original amount
$\begin{aligned}
t &=t_{1 / 2} \times 4 \\
\therefore \quad t_{1 / 2} &=\frac{192}{4}=48 \mathrm{~min}
\end{aligned}$

Topic: Chemistry, Chemical Kinetics, COMEDK, COMEDK 2016
A radioactive isotope having a half - life period of $3$ days was received after $12$ days. If $3 \mathrm{~g}$ of the isotope is left in the container, what would be the initial mass of the isotope ?

Options:
A
$12 \mathrm{~g}$

B
$36 \mathrm{~g}$

C
$48 \mathrm{~g}$

D
$24 \mathrm{~g}$

The correct answer is:
$48 \mathrm{~g}$

Solution:
Given $t_{1 / 2}=3$
Total time $\mathrm{T}=12$
No. of half lives $(n)=\frac{12}{3}=4$
$\left(\frac{1}{2}\right)^n=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}$
$\therefore \left(\frac{1}{2}\right)^4=\frac{3}{\mathrm{~N}_{\mathrm{o}}}$
$\frac{3}{\mathrm{~N}_{\mathrm{o}}}=\frac{1}{16}$
$\mathrm{~N}_{\mathrm{o}}=48 \mathrm{~g}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2013 (25 Apr Online)
A radioactive isotope having $t_{1 / 2}=3$ days was read after 12 days. If $3 g$ of isotope is contained in container, the initial weight of isotope was

Options:
A $12 \mathrm{~g}$
B $24 \mathrm{~g}$
C $36 \mathrm{~g}$
D $48 \mathrm{~g}$

The correct answer is:
$48 \mathrm{~g}$

Solution:
$\begin{aligned} T &=n \times t_{1 / 2} \\ 12 &=n \times 3 \\ n &=4 \\ N &=N_{0} \times\left(\frac{1}{2}\right)^{n} \\ 3 &=N_{0} \times\left(\frac{1}{2}\right)^{4} \\ 3 &=N_{0} \times \frac{1}{16} \\ N_{0} &=16 \times 3=48 \mathrm{~g} \end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main
A radioactive material decays by simultaneous emission of two particles with half-lives 1620 years and 810 years respectively. The time in years after which one - fourth of material remains, is

Options:
A 1080 years
B 2340 years
C 4860 years
D 3240 years

The correct answer is:
1080 years

Solution:
Since, from Rutherford - Soddy law, the number of atoms left after half-lives is given by
N=N012n
Where, N0 is the original number of atoms.
The number of half-lives, n=time of decayeffective half-life
Relation between effective disintegration constant λ and half life T
λ=ln⁡2T
∴ λ1+λ2=ln⁡2T1+ln⁡2T2
Effective half life,
1T=1T1+1T2=11620+1810
1T=1+21620⇒T=540 years
∴ n=T540
∴ N=N012t540⇒NN0=122=12t540
⇒ t540=2⇒t=2×540=1080 years

Topic: Physics, Nuclear Physics, JEE Main
A radioactive material decays by simultaneous emission of two particles with half-lives 1620 yr and 810 yr respectively. The time in year after which one-fourth of the material remains, is

Options:
A 4860 yr

B 3240 yr
C 2340 yr
D 1080 yr

The correct answer is:
1080 yr

Solution:
From Rutherford-Soddy law, the number of atoms left after n half-lives is given by
N=N012n
Where,N0 is original number of atoms.
The number of half-life
n=time of decayeffective half-life
Relation between effective disintegration constant (Î») and half-life (T) is
Î»=ln â¡2T
â´ Î»1+Î»2=ln â¡2T1+ln â¡2T2
Effective half -life
1T=1T1+1T2=11620+1810
1T=1+21620âTâ540 yr
â´ n= t540
â´ N=N012t/540
â NN0=122=12t/540
â t540=2
â t= 2 Ã 540 = 1080 yr

Topic: Physics, Nuclear Physics, JEE Main