datavorous/entrance-exam-dataset huggingface dataset:
Two Polaroids P1 and P2 are placed with their axis perpendicular to each other. Unpolarised light with intensity I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its axis makes an angle 45o with that of P1. The intensity of transmitted light through P2 is

Options:
A I02
B I04
C I08
D I016

The correct answer is:
I08

Solution:
By Malus law, the intensity of transmitted light via P1 is I1=I02
The intensity of light transmitted through P3 as obtained from the above figure is I2=I02cos2⁡ 45o
=I02×12
=I04
Intensity of light transmitted via P2 is I3=I04cos2⁡45o
I3=I08

Topic: Physics, Wave Optics, JEE Main
Two poles of height 10 meters and 20 meters stand at the centres of two circular plots which touch each other externally at a point and the two poles subtend angles 30o and 60o respectively at this point, then the distance between the centres of these circular plots is

Options:
A 30 meters
B 503 meters
C 703 meters
D 103+20 meters

The correct answer is:
503 meters

Solution:
Let, A and B be the centres of the two circles where the poles of height 10m and 20m respectively stand making angles 30o and 60o respectively at the point O where these circles touch each other externally.
The sum of the radii of the two circles = The distance between the centres of the two circles.
=10cotâ¡30o+20cotâ¡60o
=103+203=30+203 meters

Topic: Mathematics, Heights and Distances, JEE Main
Two poles of height a meters and b meters stand at the centres of two circular plots which touch each other externally at a point and the two poles subtend angles of 30 o and 60 o respectively at this point, then the distance (in meters) between the centres of these plots is

Options:
A a+b
B 3a+b3
C a+3b3
D a3+b

The correct answer is:
3a+b3

Solution:
Let A and B be the centres of the two circles where the poles of height a and b respectively stand making angles 30 o and 60 o respectively at the point O where these circles touch each other externally.
Sum of the radii of the two circles
=acot 30 o +bcot 60 o
=a3+b3=3a+b3
= distance between the centres of the two circles.

Topic: Mathematics, Heights and Distances, JEE Main
Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their tops makes an angle of 15° with the ground. Then the distance (in m) between the poles, is

Options:
A 103-1
B 522+3
C 52+3
D 53+1

The correct answer is:
52+3

Solution:
Let, DC be the pole of height 5 m and AB be the pole of height 10 m.

Also, let DM⊥AB, then ACDM is a rectangle and hence, AM=DC=5 m and BM=AB-AM=10-5=5 m and the distance between the poles is AC=MD=K (let).

In ΔBDM
tan15°=BMMD=5K

⇒K=5tan15°

⇒K=53+13-1×3+13+1

⇒K=53+1+233-1

⇒K=52+3 m.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Two poles standing on a horizontal ground are of heights x meters and 40 meters respectively. The line joining their tops makes an angle of 30 o with the ground and the distance between the foot of the poles is 303 meters, then the value of x can be

Options:
A 20
B 30
C 10
D 50

The correct answer is:
10

Solution:
Given AB=x, CD=40 and ∠DBE= 30 o
In ∆BDE,
DE BE =tan 30 o
DE=BEtan 30 o
40−x=303⋅13

⇒x=10 or 70

Topic: Mathematics, Heights and Distances, JEE Main
Two poles standing on horizontal ground are of heights 10 meters & 40 meters respectively. The line joining their tops makes an angle of 30o with the ground. Then the distance (in meters) between the foot of the poles is

Options:
A 20
B 30
C 203
D 303

The correct answer is:
303

Solution:
given, AB=10, CD=40
∠DBE=30o
in ΔBDE,

DEBE=tan⁡30o
BE=DEtan⁡30o=3013=303 meters

Topic: Mathematics, Heights and Distances, JEE Main
Two positive ions, each carrying a charge ' $\mathrm{q}$ ' are separated by a distance ' $\mathrm{d}$ '. If ' $\mathrm{F}$ ' is the force of repulsion between the ions, the number of electrons from each ion will be $(\mathrm{e}=$ charge on electron, $\varepsilon_0=$ permittivity of free space)

Options:
A $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{~d}^2}{\mathrm{e}^2}}$
B $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}}{\mathrm{e}^2}}$
C $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}}{\mathrm{e}}}$
D $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}$

The correct answer is:
$\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}$

Solution:
$\begin{aligned} & \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{~d}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(\mathrm{ne})^2}{\mathrm{~d}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{n}^2 \mathrm{e}^2}{\mathrm{~d}^2} \\ & \therefore \mathrm{n}=\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main
Two possible stereoisomers for
are

Options:
A enantiomers
B diastereomers
C conformers
D rotamers

The correct answer is:
enantiomers

Solution:
Enantionmers Formed because cyclohexa diene shows optical activity.

Topic: Chemistry, General Organic Chemistry, JEE Main
Two projectile thrown at 30° and 45° with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

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

The correct answer is:
2:1

Solution:
Time taken to reach maximum height is given by

t=usinθg, where, u is initial velocity.

Here, t1=t2

∴u1sinθ1g=u2sinθ2g

⇒u1sin30=u2sin45

⇒u1u2=1212=21=2:1

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Two projectiles A and B are thrown with initial velocities of 40 m s-1 and 60 m s-1 at angles 30° and 60° with the horizontal respectively. The ratio of their ranges respectively is (g = 10 m s-2)

Options:
A 4 : 9
B 2 : 3
C 3: 2
D 1 : 1

The correct answer is:
4 : 9

Solution:
The range R1 of the first projectile is given by

R1=u12sin2θ1g ...(1)

The range R2 of the second projectile is given by

R2=u22sin2θ2g ...(2)

Divide equation (1) by equation (2) and simplify to obtain the ratio of the ranges.

R1R2=u12sin2θ1gu22sin2θ2g= u12sin2θ1u22sin2θ2 ...3

Substitute the values of the known parameters into equation (3) to calculate the required ratio.

R1R2=402×sin60°602×sin120°=49

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Two projectiles are projected at 30° and 60°with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

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

The correct answer is:
1 : 3

Solution:
The maximum height is given by, H=u2sin2θ2g.

The given data is

θ1=30°θ2=60°u1=u2

Therefore,

H1maxH2max=u12sin2θ1u22sin2θ2=sin30°sin60°2=12322=13

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Two projectiles are thrown with same initial velocity making an angle of 45° and 30° with the horizontal respectively. The ratio of their respective ranges will be

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

The correct answer is:
2:3

Solution:
Let projection speed is u, The formula for the range of the projectile is given by, R=u2sin2θg. Therefore,

R1=u2sin90°g; R2=u2sin60°g

R1R2=sin90°sin60°=23

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Two pure inductors each of self inductance $L$ are connected in parallel but are well separated from each other. The total inductance is

Options:
A $2 L$
B $L$
C $)^{\frac{L}{2}}$
D $\frac{L}{4}$

The correct answer is:
$)^{\frac{L}{2}}$

Solution:
Inductors obey the laws of parallel and series combination of resistors.

Topic: Physics, Electromagnetic Induction, JEE Main
Two quantities A and B are related by the relation AB=m, where m is linear mass density and A is force. The dimensions of B will be

Options:
A Same as that of latent heat
B Same a that of pressure
C Same as that of work
D Same as that of momentum

The correct answer is:
Same as that of latent heat

Solution:
B=Am
B=forcemass/length
B=force×lengthmass=energymass=[latent heat]

Topic: Physics, Units and Dimensions, JEE Main
Two racing cars of masses $\mathrm{m}_1$ and $\mathrm{m}_2$ are moving in circles of radii $\mathrm{r}_1$ and $\mathrm{r}_2$ respectively. Their speeds are such that each makes a complete circle in the same duration of time $t$. The ratio of the angular speed of the first to the second car is

Options:
A $\mathrm{m}_1 \mathrm{r}_1: \mathrm{m}_2 \mathrm{r}_2$
B $1: 1$
C $\mathrm{m}_1: \mathrm{m}_2$
D $r_1: r_2$

The correct answer is:
$1: 1$

Solution:
Angular velocity is the rate of change of angular displacement with time.
Given that both the cars complete one circle in the same time ' $t$ '
Angular speed $=\omega=\frac{\text { angle subtended by the car for one revolution }(27}{\text { Time Taken }(\mathrm{t})}$
Angular speed of first car, $=\omega_1=\frac{2 \pi}{t}$
Angular speed of second car, $=\omega_2=\frac{2 \pi}{t}$
The ratio of angular speed of first car to that of second $=\frac{\omega_1}{\omega_2}=1: 1$

Topic: Physics, Rotational Motion, JEE Main
Two radiations of photon energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$, successively illuminate a photosensitive metallic surface of work function $0.5 \mathrm{eV}$. The ratio of the maximum speeds of the emitted electrons is

Options:
A $1: 4$
B $1: 2$
C $1: 1$
D $1: 5$

The correct answer is:
$1: 2$

Solution:
We have
$$
\frac{1}{2} m v_{\max }^2=E-\phi
$$
Here case (i)
$$
\begin{gathered}
\frac{1}{2} m v_{1 \max }^2=(1-0.5) \mathrm{eV} \\
\frac{1}{2} m v_{2 \max }^2=(2.5-0.5) \mathrm{eV}
\end{gathered}
$$
Hence, $\quad \frac{v_{1 \max }^2}{v_{2 \max }^2}=\frac{1}{4}$
$$
\frac{v_{1 \max }}{v_{2 \max }}=\frac{1}{2}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2012 (Mains)
Two radio antennas are located at points \( A \) and \( B, 200 \mathrm{~m} \) part as shown in the figure. The sources are coherent and are producing electromagnetic waves in phase at a frequency of \( 6 \mathrm{MHz} \). A radio receiver is moved out from point \( B \) along a line perpendicular to the line connecting \( A \) and \( B \) (line \( B C \) shown in figure). At what distances from \( B \) will there be destructive interference?

Options:
A \( 787.5,229 \mathrm{~m}, 97.5 \mathrm{~m}, 26.7 \mathrm{~m} \)
B \( 789.5,230 \mathrm{~m}, 98.5 \mathrm{~m}, 27.7 \mathrm{~m} \)
C \( 788.5,228 \mathrm{~m}, 98.5 \mathrm{~m}, 28.7 \mathrm{~m} \)
D \( 777.5,239 \mathrm{~m}, 87.5 \mathrm{~m}, 36.7 \mathrm{~m} \)

The correct answer is:
\( 787.5,229 \mathrm{~m}, 97.5 \mathrm{~m}, 26.7 \mathrm{~m} \)

Solution:
Speed of light, c=nλ, where n is frequency and λ is wavelength

⇒λ=cn
⇒λ=3×1086×106 ⇒λ=50 m
Path difference at P=AP-BP
Path difference at P=4λ2+x2-x ...1
For destructive interference at P
Path difference =(2n-1)λ2 ...2
From (1) and (2),
4λ2+x2-x=2n-1λ2
⇒4λ2+x2=2n-1λ2+x
For first minima, n=1,
x1=63λ4 ⇒x1=787.5 m
For second minima n=2,
x2=55λ12 ⇒x2=229 m
For third minima n=3,
x3=39λ20 ⇒x3=97.5 m

Topic: Physics, Waves and Sound, JEE Main
Two radioactive elements $A$ and $B$ initially have same number of atoms. The half life of $A$ is same as the average life of $B$. If $\lambda_A$ and $\lambda_B$ are decay constants of $A$ and $B$ respectively, then choose the correct relation from the given options.

Options:
A $\lambda_A=\lambda_B$
B $\lambda_A=2 \lambda_B$
C $\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ln 2$
D $\lambda_{\mathrm{A}} \ln 2=\lambda_{\mathrm{B}}$

The correct answer is:
$\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ln 2$

Solution:
It is given that
$\begin{aligned}
& \left(\frac{\mathrm{T}_1}{2}\right)_{\mathrm{A}}=\left(\mathrm{T}_{\mathrm{av}}\right)_{\mathrm{B}} \\
& \because \mathrm{T}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\lambda} \text { and } \mathrm{T}_{\mathrm{av}}=\frac{1}{\lambda} \\
& \therefore \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ell n 2
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
Two radioactive elements A and B initially have same number of atoms. The half life of A is same as the average life of B. If λA and λB are decay constants of A and B respectively, then choose the correct relation from the given options.

Options:
A λAln2=λB
B λA=λB
C λA=λBln2
D λA=2λB

The correct answer is:
λA=λBln2

Solution:
It is given that

T12(A)=Tav(B)

The half life of a substance is given by T12=ln2λ

So,

ln2λA=1λB

⇒λA=λBln2

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Two radioactive materials A and B have decay constants 10λ and λ , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time:

Options:
A 110λ
B 19λ
C 111λ
D 1110λ

The correct answer is:
19λ

Solution:
NA=N0e-10λt
NB=N0e-λt
NANB=1e=e-10λt+λt=e-9λt
1=9λt
t=19λ

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Two radioactive materials $R_1$ and $R_2$ have decay constants $6 \lambda$ and $\lambda$, respectively. The half life of $R_2$ is $1.4 \times 10^{17} \mathrm{~s}$. Initially they contain some number of nuclei.
The time at which the ratio of the remaining nuclei of $R_2$ to that of $R_1$ will be $e$ is $($ Let $\ln 2=0.7)$

Options:
A $2 \times 10^{16} \mathrm{~s}$
B $4 \times 10^{16} \mathrm{~s}$
C $3 \times 10^{16} \mathrm{~s}$
D $5 \times 10^{16} \mathrm{~s}$

The correct answer is:
$4 \times 10^{16} \mathrm{~s}$

Solution:
For two radioactive elements $R_1$ and $R_2$, decay constants,
$\lambda_1=6 \lambda$ and $\lambda_2=\lambda$
Half-life of $R_2,\left(t_{1 / 2}\right)_2=1.4 \times 10^{17} \mathrm{~s}$
If $N_0$ be the initial number of nuclei at $t=0$, in both $R_1$ and $R_2$ or $N_1$ and $N_2$ be the nuclei after time $t$.
Then,
$N_1=N_0\left(\frac{1}{2}\right)^{n_1}$
and
$N_2=N_0\left(\frac{1}{2}\right)^{n_2}$
$\therefore \quad \frac{N_2}{N_1}=\left(\frac{1}{2}\right)^{n_2-n_1}$
$e=\left(\frac{1}{2}\right)^{n_2-n_1} \quad\left[\because \frac{N_2}{N_1}=e\right]$
$\begin{aligned} \log _e e & =\left(n_2-n_1\right) \log \frac{1}{2} \\ \Rightarrow \quad 1 & =\left(n_2-n_1\right) \log 2^{-1} \\ 1 & =\left(n_1-n_2\right) \log 2 \\ 1 & =\left(n_1-n_2\right) 0.7 \\ \Rightarrow \quad n_1-n_2 & =\frac{1}{0.7} \\ \frac{t}{\left(t_{1 / 2}\right)_1} & -\frac{t}{\left(t_{1 / 2}\right)_2}=\frac{1}{0.7}\end{aligned}$
$\begin{aligned} & \text { Since, }\left(t_{1 / 2}\right)_2=\frac{0.693}{\lambda_2}=\frac{0.693}{\lambda} \\ & \Rightarrow \quad \lambda=\frac{0.693}{\left(t_{1 / 2}\right)_2}=\frac{0.693}{1.4 \times 10^{17}}=0.495 \times 10^{-17} \\ & \therefore \quad \lambda_1=6 \lambda=6 \times 0.495 \times 10^{-17} \\ & =2.97 \times 10^{-17} \mathrm{~m} \\ & \therefore \quad\left(t_{1 / 2}\right)_1=\frac{0.693}{\lambda_1}=\frac{0.693}{2.97 \times 10^{-17}} \\ & \left(t_{1 / 2}\right)_1=0.23 \times 10^{17} \mathrm{~s} \\ & \end{aligned}$
$\therefore$ From Eq. (i), we get
$\begin{aligned} \frac{t}{0.23 \times 10^{17}}-\frac{t}{1.4 \times 10^{17}}=\frac{1}{0.7} \\ t\left(\frac{1}{0.23}-\frac{1}{1.4}\right)=\frac{10^{17}}{0.7} \\ t(3.64)=\frac{10^{17}}{0.7} \\ t=\frac{10^{17}}{0.7 \times 3.64}=\frac{10}{2.56} \times 10^{16} \\ =3.9 \times 10^{16}=4 \times 10^{16} \mathrm{~s}\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
Two radioactive materials $R_1$ and $R_2$ have decay constants $6 \lambda$ and $\lambda$, respectively. The half life of $R_2$ is $1.4 \times 10^{17} \mathrm{~s}$. Initially they contain some number of nuclei.
The time at which the ratio of the remaining nuclei of $R_2$ to that of $R_1$ will be $e$ is $($ Let $\ln 2=0.7)$

Options:
A $2 \times 10^{16} \mathrm{~s}$
B $4 \times 10^{16} \mathrm{~s}$
C $3 \times 10^{16} \mathrm{~s}$
D $5 \times 10^{16} \mathrm{~s}$

The correct answer is:
$4 \times 10^{16} \mathrm{~s}$

Solution:
For two radioactive elements $R_1$ and $R_2$, decay constants,
$\lambda_1=6 \lambda$ and $\lambda_2=\lambda$
Half-life of $R_2,\left(t_{1 / 2}\right)_2=1.4 \times 10^{17} \mathrm{~s}$
If $N_0$ be the initial number of nuclei at $t=0$, in both $R_1$ and $R_2$ or $N_1$ and $N_2$ be the nuclei after time $t$.
Then,
$N_1=N_0\left(\frac{1}{2}\right)^{n_1}$
and
$N_2=N_0\left(\frac{1}{2}\right)^{n_2}$
$\therefore \quad \frac{N_2}{N_1}=\left(\frac{1}{2}\right)^{n_2-n_1}$
$e=\left(\frac{1}{2}\right)^{n_2-n_1} \quad\left[\because \frac{N_2}{N_1}=e\right]$
$\begin{aligned} \log _e e & =\left(n_2-n_1\right) \log \frac{1}{2} \\ \Rightarrow \quad 1 & =\left(n_2-n_1\right) \log 2^{-1} \\ 1 & =\left(n_1-n_2\right) \log 2 \\ 1 & =\left(n_1-n_2\right) 0.7 \\ \Rightarrow \quad n_1-n_2 & =\frac{1}{0.7} \\ \frac{t}{\left(t_{1 / 2}\right)_1} & -\frac{t}{\left(t_{1 / 2}\right)_2}=\frac{1}{0.7}\end{aligned}$
$\begin{aligned} & \text { Since, }\left(t_{1 / 2}\right)_2=\frac{0.693}{\lambda_2}=\frac{0.693}{\lambda} \\ & \Rightarrow \quad \lambda=\frac{0.693}{\left(t_{1 / 2}\right)_2}=\frac{0.693}{1.4 \times 10^{17}}=0.495 \times 10^{-17} \\ & \therefore \quad \lambda_1=6 \lambda=6 \times 0.495 \times 10^{-17} \\ & =2.97 \times 10^{-17} \mathrm{~m} \\ & \therefore \quad\left(t_{1 / 2}\right)_1=\frac{0.693}{\lambda_1}=\frac{0.693}{2.97 \times 10^{-17}} \\ & \left(t_{1 / 2}\right)_1=0.23 \times 10^{17} \mathrm{~s} \\ & \end{aligned}$
$\therefore$ From Eq. (i), we get
$\begin{aligned} \frac{t}{0.23 \times 10^{17}}-\frac{t}{1.4 \times 10^{17}}=\frac{1}{0.7} \\ t\left(\frac{1}{0.23}-\frac{1}{1.4}\right)=\frac{10^{17}}{0.7} \\ t(3.64)=\frac{10^{17}}{0.7} \\ t=\frac{10^{17}}{0.7 \times 3.64}=\frac{10}{2.56} \times 10^{16} \\ =3.9 \times 10^{16}=4 \times 10^{16} \mathrm{~s}\end{aligned}$

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2019 (06 May Shift 1)
Two radioactive materials X1 and X2 have decay constants 10Î» and Î» respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1e after a time

Options:
A 110λ
B 111λ
C 1110λ
D 19λ

The correct answer is:
19λ

Solution:
Here, Nx1(t)Nx2(t)=1e
or N0e-10λtN0e-λt=1e
(Because initially, both have the same number of nuclei, N0).
or e=e-λte-10λt=e9λt
9λt=1
t=19λ

Topic: Physics, Nuclear Physics, JEE Main
Two radioactive materials $X_1$ and $X_2$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $\mathrm{X}_1$ to that of $X_2$ will be $\frac{1}{e}$ after a time

Options:
A $\frac{e}{\lambda}$
B $\lambda$
C $\frac{1}{2} \lambda$
D $\frac{1}{4 \lambda}$

The correct answer is:
$\frac{1}{4 \lambda}$

Solution:
$$
\begin{aligned}
& \frac{N_{x_1}}{N_{x_2}}=\frac{e^{-5 \lambda t}}{e^{-\lambda t}}=\frac{1}{e} \\
& \Rightarrow \quad t=\frac{1}{4 \lambda}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2008 (Mains)
Two radioactive materials X 1 and X 2 have decay constants 10Î» and Î» respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X 1 to that of X 2 will be 1/e after time

Options:
A 1 10λ
B 1 11λ
C 11 10λ
D 1 9Î»

The correct answer is:
1 9λ

Solution:
N 1 = N 0 e â10Î»ât
and N 2 = N 0 e âÎ»ât

N 1 N 2 = e - 1 0 Î» t e - Î» t = 1 e (given)
â t = 1 9 Î»

Topic: Physics, Nuclear Physics, JEE Main
Two radioactive materials $\mathrm{Y}_1$ and $\mathrm{Y}_2$ initially contain same number of nuclei. Their decay constants are $9 \lambda \mathrm{s}^{-1}$ and $6 \lambda \mathrm{s}^{-1}$ respectively. The time after which the ratio of number of undecayed nuclei of $\mathrm{Y}_1$ and $\mathrm{Y}_2$ becomes $\frac{1}{e}$ is

Options:
A $\frac{1}{3 \lambda} \mathrm{s}$
B $\frac{1}{15 \lambda} \mathrm{s}$
C $\frac{1}{10 \lambda} \mathrm{s}$
D $\frac{1}{8 \lambda} \mathrm{s}$

The correct answer is:
$\frac{1}{3 \lambda} \mathrm{s}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Nuclear Physics, JEE Main
Two radioactive nuclei $P$ and $Q$, in a given sample decay into a stable nucleous $R$. At time $t=0$, number of $P$ species are $4 N_0$ and that of $Q$ are $N_0$. Half-life of $P$ (for conversion to $R$ ) is 1 min where as that of $Q$ is $2 \mathrm{~min}$. Initially there are no nuclei of $R$ present in the sample. When number of nuclei of $P$ and $Q$ are equal, the number of nuclei of $R$ present in the sample would be

Options:
A $3 N_0$
B $\frac{9 N_0}{2}$
C $\frac{5 N_0}{2}$
D $2 N_0$

The correct answer is:
$\frac{9 N_0}{2}$

Solution:
Initially $P \rightarrow 4 N_0$
$$
Q \rightarrow N_0
$$
Half life $T_p \rightarrow 1 \mathrm{~min}$
$$
T_Q \rightarrow 2 \min
$$
Let after time $t$ number of nuclei of $P$ and $Q$ are equalie, $\quad \frac{4 N_0}{2^{t / 1}}=\frac{N_0}{2^{t / 2}}$
$$
\begin{aligned}
4 & =2^{t / 2} \\
2^2 & =2^{t / 2} \\
\frac{t}{2} & =2 \\
t & =4 \mathrm{~min}
\end{aligned}
$$
Disactive nucleus or Nuclei of $R$
$$
\begin{aligned}
& =\left(4 N_0-\frac{4 N_0}{2^4}\right)+\left(N_0-\frac{N_0}{2^2}\right) \\
& =4 N_0-\frac{N_0}{4}+N_0-\frac{N_0}{4} \\
& =5 N_0-\frac{N_0}{2} \\
& =\frac{9}{2} N_0
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2011 (Mains)
Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0, the number of P species are 4N0 and that of Q is N0. The half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q is equal, the number of nuclei of R present in the sample would be

Options:
A 2N0
B 3N0
C 9N02
D 5N02

The correct answer is:
9N02

Solution:
PQNo. of nuclei, at t = 04N0N0Half-life1 min2 minNo. of nuclei afterNPNQtime t
Let after t min, the number of nuclei of P and Q are equal.
∴NP=4N012t/1andNQ=N012t/2
As NP = NQ
∴4N012t/1=N012t/2
42t/1=12t/2or4=2t2t/2
or 4 = 2t/2 or 22 = 2t/2
ort2=2ort=4 min
After 4 minutes, both P and Q have an equal number of nuclei.
∴ Number of nuclei of R
=4N0-N04+N0-N04
=15N04+3N04=9N02

Topic: Physics, Nuclear Physics, JEE Main
Two radioactive substances A and B have decay constants 5Î» and Î» respectively. At t=0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become 1e2 will be

Options:
A 1λ
B 12λ
C 2λ
D 14λ

The correct answer is:
12λ

Solution:
Using the decay equation, N=N0e-λt
For radioactive substance A, N1=N0e-5λt ...........(i)
For radioactive substance B, N2=N0e-λt ............(ii)
From equation (i) and (ii)
N1N2=e-5λte-λt
1e2=e-4λt
4λt=2
Time taken, t=12λ

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Two radioactive substances A and B have decay constants 5Î» and Î» respectively. At t=0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become 1e2 will be

Options:
A 1λ
B 12λ
C 2λ
D 14λ

The correct answer is:
12λ

Solution:
N=N0e-λt
N1=N0e-5λt ...........(i)
N2=N0e-λt ............(ii)
From equation (i) and (ii)
N1N2=e-5λte-λt
1e2=e-4λt
4λt=2
t=12λ

Topic: Physics, Atomic Physics, JEE Main
Two radioactive substances $A$ and $B$ have decay constants $5 \lambda$ and $\lambda$ respectively. At $t=0$ they have the same number of nuclei. The ratio of number of nuclei of $A$ to those of $B$ will be $\left(\frac{1}{e}\right)^2$ after a time interval

Options:
A $\frac{1}{4 \lambda}$
B $4 \lambda$
C $2 \lambda$
D $\frac{1}{2 \lambda}$

The correct answer is:
$\frac{1}{2 \lambda}$

Solution:
Number of nuclei remained after time $t$ can be written as
$N=N_0 e^{-\lambda t}$
where $N_0$ is initial number of nuclei of both the substances.
and
$\begin{aligned}
& N_1=N_0 e^{-5 \lambda t} \ldots(i)\\
& N_2=N_0 e^{-\lambda t} \ldots(ii)
\end{aligned}$
Dividing Eq. (i) by Eq. (ii), we obtain
$\frac{N_1}{N_2}=e^{(-5 \lambda+\lambda) t}=e^{-4 \lambda t}=\frac{1}{e^{4 \lambda t}}$
But, we have given
$\frac{N_1}{N_2}=\left(\frac{1}{e}\right)^2=\frac{1}{e^2}$
Hence, $\quad \frac{1}{e^2}=\frac{1}{e^{4 \lambda t}}$
Comparing the powers, we get
or
$\begin{aligned}
& 2=4 \lambda t \\
& t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}
\end{aligned}$

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2009