datavorous/entrance-exam-dataset huggingface dataset:
A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it, but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope? g = 1 0 m s - 2

Options:
A 2.5 m ​ s−2
B 5âm ââsâ2
C 10âm ââsâ2
D 25âm ââsâ2

The correct answer is:
2.5 m ​ s−2

Solution:
For monkey, we can write
Mg - T = Ma
or M(g + a) = T ....(i)
for suspended weight,
T â¤ mg ....(ii)
T â¤ 25 Ã 10 or 250 N [breaking strength]
â
â´ Equation (i) becomes

20.(10 + a) â¤ 250
â a â¤ 50 2 0 or a â¤ 2 . 5 m s - 2
â´ amax = 2.5 m s-2

Topic: Physics, Laws of Motion, JEE Main
A monkey of mass \( 40 \mathrm{~kg} \) climbs on a massless rope which can stand a maximum tension of \( 500 \mathrm{~N} \). In which of the following cases will the rope break? (Take \( g=10 \mathrm{~ms}^{-2} \) ). \( \|\backslash\| \backslash \| \)

Options:
A The monkey climbs up with an acceleration of \( 5 \mathrm{~m} \mathrm{~s}^{-2} \).
B The monkey climbs down with an acceleration of \( 5 \mathrm{~m} \mathrm{~s}^{-2} \).
C The monkey climbs up with a uniform speed of \( 5 \mathrm{~m} \mathrm{~s}^{-2} \).
D The monkey falls down the rope freely under gravity.

The correct answer is:
The monkey climbs up with an acceleration of \( 5 \mathrm{~m} \mathrm{~s}^{-2} \).

Solution:
Here, mass of monkey, m=40 kg
Maximum tension the rope can stand, T=500 N
Tension in the rope will be equal to apparent weight of the monkey.
The rope will break when R exceeds T.
(A) When the monkey climbs up with an acceleration a=5 ms-2
R=m g+a= 40 10+5= 600 N ∴R>T
Hence, the rope will break
(B) When the monkey climbs down with an acceleration a=5 ms-2
R=m g-a= 40 10-5=200 N∴R<T
Hence, the rope will not break.
(C) When the monkey climbs up with a uniform speed v=5 ms-1,its acceleration a=0 m/s2
∴R=mg=40×10=400 N ∴R<T
Hence, the rope will not break.
(D) When the monkey falls down the rope freely under gravity
a=g∴R=m g-a= m g-g=0
Hence, the rope will not break.

Topic: Physics, Laws of Motion, JEE Main
A monoatomic gas at pressure $p_1$ and $V_1$ is compressed adiabatically to $\frac{1}{8}$ th its original volume. What is the final pressure of the gas?

Options:
A $64 \mathrm{p}_1$
B $\mathrm{p}_1$
C $16 \mathrm{p}_1$
D $32 \mathrm{p}_1$

The correct answer is:
$32 \mathrm{p}_1$

Solution:
$\quad \mathrm{p}_1 \mathrm{~V}_1^{5 / 3}=\mathrm{p}^{\prime}\left(\frac{\mathrm{V}_1}{8}\right)^{5 / 3}$
$$
\begin{aligned}
p^{\prime} & =p_1(8)^{5 / 3} \\
& =p_1 \times 2^5 \\
p^{\prime} & =32 p_1
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A monoatomic gas at pressure $p_1$ and $V_1$ is compressed adiabatically to $\frac{1}{8}$ th its original volume. What is the final pressure of the gas?

Options:
A $64 \mathrm{p}_1$
B $\mathrm{p}_1$
C $16 \mathrm{p}_1$
D $32 \mathrm{p}_1$

The correct answer is:
$32 \mathrm{p}_1$

Solution:
$\quad \mathrm{p}_1 \mathrm{~V}_1^{5 / 3}=\mathrm{p}^{\prime}\left(\frac{\mathrm{V}_1}{8}\right)^{5 / 3}$
$$
\begin{aligned}
p^{\prime} & =p_1(8)^{5 / 3} \\
& =p_1 \times 2^5 \\
p^{\prime} & =32 p_1
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2010 (Mains)
A monoatomic gas at pressure P and volume V is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be

Options:
A P
B 8P
C 32P
D 64P

The correct answer is:
32P

Solution:
Change in entropy is given by dS=dQT. Since, the process is dS=0, dQ=0. Therefore, constant entropy means the process is adiabatic.

PVγ=constant

⇒P1V1γ=P2V2γ

For a monatomic gas γ=53

⇒P1V153=P2V1853

⇒P1V153=P2 V15332

⇒P2=32P1

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
A monoatomic gas is compressed from a volume of 2 m3 to a volume of 1 m3 at a constant pressure of 100 N m2. Then it is heated at constant volume by supplying 150 J of energy. As a result, the internal energy of the gas

Options:
A Decreases by 50 J
B Increases by 250 J
C Decreases by 250 J
D 0 J

The correct answer is:
Increases by 250 J

Solution:
Isobaric process with Q=0
∴ 0=W+ΔU
∴ 0=PΔV+ΔU
∴ ΔU=-PΔV
=-1001-2 J
ΔU=100 J
Constant volume with Q=150 J
Q=0+ΔU
∴ ΔU2=150J
∴ ΔU=ΔU1+ΔU2
= + 2 5 0 J
Hence, internal energy increases by 250 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2014 (19 Apr Online)
A monoatomic gas is suddenly compressed to $(1 / 8)^{\text {th }}$ of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is ( $\gamma=5 / 3$ )

Options:
A 32
B 8
C $\frac{40}{3}$
D $\frac{24}{5}$

The correct answer is:
32

Solution:
$$
\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{8}, \gamma=\frac{5}{3}
$$
For adiabatic process,
$$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{5 / 3}=(2)^5=32
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
A monoatomic gas molecule has

Options:
A Three degrees of freedom
B Four degrees of freedom
C Five degrees of freedom
D Six degrees of freedom

The correct answer is:
Three degrees of freedom

Solution:
For monoatomic gas molecules, it has 3 degrees of freedom. All 3 are translation al degrees of freedom.

Topic: Physics, Kinetic Theory of Gases, JEE Main
A monoatomic gas of mass 4.0 u is kept in an insulated container. The container is moving with velocity 30 m s-1. If the container is suddenly stopped then a change in temperature of the gas (R=gas constant) is x3R. Value of x is,

Options:

The correct answer is:
3600

Solution:
Given that mass of gas is 4u hence its molar mass M is 4 g mol-1

∴12mv2=nCvΔT

⇒12 m×302=mM×3R2×ΔT

∴ ΔT=36003R

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A monoatomic gas of molar mass M0 is filled into an insulated container which is moving along a horizontal surface with velocity a v. If the container is stopped suddenly, then the change in temperature of the gas is

Options:
A M0v22R
B M0v23R
C 2M0v23R
D 3M0v22R

The correct answer is:
M0v23R

Solution:
If the container is suddenly stopped loss in kinetic energy of the gas is

ΔK=12nM0v2, where n is the number of moles of the gas.

If ΔT is the change in the temperature of the gas, then

32nRΔT=12nM0v2

ΔT=M0v23R

Topic: Physics, Thermodynamics, JEE Main
A monoatomic gas of molar mass $m$ is kept in a insulated container. Container is moving with velocity $v$. If the container is suddenly stopped, then the change in the temperature of the gas is

Options:
A
$m v^2 / 4 R$

B
$m v^2 / 2 R$

C
$m v^2 / R$

D
$m v^2 / 3 R$

The correct answer is:

$m v^2 / 3 R$

Solution:
Let the insulated container has $n$ moles of the monoatomic gas. The loss of kinetic energy of the gas,
$$
\Delta E_K=\frac{1}{2}(m n) v^2
$$
If the change in temperature of the gas is $\Delta T$, then heat gained by the gas is,
$$
\Delta Q=\frac{3}{2} n R \Delta T
$$
Now, according to question,
$$
\begin{aligned}
\Delta Q & =\Delta E_K \\
\frac{3}{2} n R \Delta T & =\frac{1}{2} m n v^2 \\
\Rightarrow \quad \Delta T & =\frac{m v^2}{3 R}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A monoatomic gas of $n$-moles is heated from temperature $T_1$ to $T_2$ under two different conditions
(i) at constant volume and (ii) at constant pressure. The change in internal energy of the gas is

Options:
A More for (i)
B More for (ii)
C Same in both cases
D Independent of number of moles

The correct answer is:
Same in both cases

Solution:
Change in internal energy $\Delta U=\mu C_V \Delta T$
it doesn't depend upon type of process. Actually it is a state function

Topic: Physics, Thermodynamics, JEE Main
A monoatomic gas of presuure 'P' having volume 'V' expands isothermally to a volume '2V' and then adiabatically to a volume '16V'. The final pressure of the gas is (ratio of specific heats $=\frac{5}{3}$)

Options:
A $\frac{P}{16}$
B $\mathrm{P}$
C $\frac{\mathrm{P}}{32}$
D $\frac{\mathrm{P}}{64}$

The correct answer is:
$\frac{\mathrm{P}}{64}$

Solution:
$\gamma=\frac{5}{3}$
Case I: $\quad P_{1} V_{1}=P_{2} V_{2}$
$\mathrm{PV}=\mathrm{P}_{2} \times 2 \mathrm{~V}$
$\therefore \quad P_{2}=\frac{P}{2}$
Case II : $\quad P_{2} V_{2}^{\gamma}=P_{3} V_{3}^{\gamma}$
$$
\begin{array}{l}
\left(\frac{P}{2}\right)(2 V)^{\gamma}=P_{3}(16 V)^{\gamma} \\
P_{3}=\frac{P}{2} \frac{(2 V)^{\gamma}}{(16 V)^{\gamma}}=\frac{P}{2}\left(\frac{1}{8}\right)^{\gamma} \\
=\frac{P}{2}\left(\frac{1}{2^{3}}\right)^{5 / 3}=\frac{P}{2}\left(\frac{1}{2}\right)^{5} \\
=\frac{P}{2 \times 32}=\frac{P}{64}
\end{array}
$$

Topic: Physics, Thermodynamics, JEE Main
A monoatomic gas performs a work of Q4 where Q is the heat supplied to it. The molar heat capacity of the gas will be _____ R during this transformation.

Where R is the gas constant.

Options:

The correct answer is:
2

Solution:
Using first law of thermodynamics,

Q=∆U+W

⇒Q=ΔU+Q4

⇒ΔU=3Q4

⇒∆U=32nRΔT=3Q4

Therefore, molar heat capacity, C=Qn∆T

⇒C=2R

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A monoatomic gas undergoes a process in which the pressure (P) and the volume (V) of the gas are related as PV-3= constant. What will be the molar heat capacity of gas for this process?

Options:
A 5R3
B 35R
C 74R
D R4

The correct answer is:
74R

Solution:
PV-3=constant n=-3
C=Rγ-1-Rn-1 γ=53
C=R53-1-R-3-1
=3R2+R4
=74R

Topic: Physics, Thermodynamics, JEE Main
A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If l1 and l2 are the lengths of the gas column, before and after the expansion respectively, then the value of T1 T2 will be:

Options:
A l1l223
B l2l123
C l2l1
D l1l2

The correct answer is:
l2l123

Solution:
PVr= constant

TVr-1= constant

T(ℓ)53-1= constant

T1 T2=ℓ2ℓ12/3

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A monoatomic ideal gas, initially at temperature $T_1$, is enclosed in a cylinder fitted with a mmminmin frictionless piston. The gas is allowed to expand adiabatically to a temperature. ${ }_2$ by releasing the piston suddenly. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively, then $T_1 / T_2$ is given by

Options:
A $\left(\frac{L_1}{L_2}\right)^{2 / 3}$
B $\frac{L_1}{L_2}$
C $\frac{L_2}{L_1}$
D $\left(\frac{L_2}{L_1}\right)^{2 / 3}$

The correct answer is:
$\left(\frac{L_2}{L_1}\right)^{2 / 3}$

Solution:
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \Rightarrow \frac{T_1}{T_2}=\left(\frac{V_2}{V_1}\right)^{\gamma-1}=\left(\frac{L_2 A}{L_1 A}\right)^{\frac{5}{3}-1}=\left(\frac{L_2}{L_1}\right)^{\frac{2}{3}}$

Topic: Physics, Thermodynamics, JEE Main
A monochromatic beam of light falls on Young's double slit experiment apparatus as shown in figure. A thin sheet of glass is inserted in front of lower slit S2(Î»=600 nm is wavelength of source)

If the central bright fringe is obtained on screen at 'O' then

Options:
A μ-1t=dsin⁡θ
B μ-1t=dcos⁡θ
C μt=dθ
D Not possible

The correct answer is:
μ-1t=dsin⁡θ

Solution:
Total path difference ∆x=μ-1t-dsin⁡θ . For central maxima ∆x=0 hence μ-1t=dsin⁡θ

Topic: Physics, Wave Optics, JEE Main
A monochromatic beam of light has a frequency $v=\frac{3}{2 \pi} \times 10^{12} \mathrm{~Hz}$ and is propagating along the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$. It is polarized along the $\hat{k}$ direction. The acceptable form for the magnetic field is:

Options:
A
$\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]$

B
$\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]$

C
$\frac{E_0}{C} \hat{k} \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \vec{r}+\left(3 \times 10^{12}\right) t\right]$

D
$\frac{E_0}{C} \frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \vec{r}+\left(3 \times 10^{12}\right) t\right]$

The correct answer is:
$\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]$

Solution:
No Solution Available

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A monochromatic beam of light has a frequency v=32π×1012 Hz and is propagating along the direction i^+j^ 2 . It is polarized along the k^ direction. The acceptance form the magnetic field is:

Options:
A E0Ci^+j^Â 2Â Cosâ¡104i^+j^2.râÂ -3Ã1012t
B E0CÂ k^Â Cosâ¡104i^+j^2.râ+3Ã1012t
C E0Ci^-j^2 Cosâ¡104i^+j^2.râ+3Ã1012t
D Â E0Ci^+j^+k^3Cosâ¡104i^+j^2.râ+3Ã1012t

The correct answer is:
E0Ci^-j^2 Cosâ¡104i^+j^2.râ+3Ã1012t

Solution:
E^×B^ should have given direction of wave propagation,⇒ K^×B^∥ i^+j^ 2
K^×i^+j^ 2=j^--i^2=i^+j^ 2 ∥i^+j^ 2 .
Wave propagation vector K^ should be along i^+j^ 2.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2018 (15 Apr)
A monochromatic beam of light is used for the formation of fringes on the screen by illuminating the two slits in the Young's double slit interference experiment. When a thin film of mica is interposed in the path of one of the interfering beams then

Options:
A the fringe width increases
B the fringe width decreases
C the fringe width remains the same but the pattern shifts
D the fringe pattern disappears.

The correct answer is:
the fringe width remains the same but the pattern shifts

Solution:
The presence of thin mica in the path of one of the interfering beams changes only the path difference between the beams due to which the pattern shifts upwards. But the fringe width remains the same. As mochromatic light is used, one cannot make out which is the central shift as all the fringes are having the same colour and intensity.

Topic: Physics, Wave Optics, AIIMS, AIIMS 2004
A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation.

If the refractive index of the material of the prism is $\sqrt{3}$,

then the angle of incidence is :

Options:
A $90^{\circ}$
B $30^{\circ}$
C $60^{\circ}$
D $45^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
For minimum deviation:

$\mathrm{r}_{1}=\mathrm{r}_{2}=\frac{\mathrm{A}}{2}=30^{\circ}$

by Snell's law $\mu_{1} \sin \mathrm{i}=\mu_{2} \sin \mathrm{r}$

$1 \times \sin i=\sqrt{3} \times \frac{1}{2}=\frac{\sqrt{3}}{2} \Rightarrow i=60$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A monochromatic light of wavelength 6000 Ao is incident on the single slit of width 0.01 mm. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 cm, the linear width of the central maximum is :

Options:
A 60 mm
B 24 mm
C 120 mm
D 12 mm

The correct answer is:
24 mm

Solution:
The linear width of central maxima is given by,

W=2λDd=2×6×10−7×0.21×10−5

=2.4×10−2=24 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A monochromatic light source $\mathrm{S}$ of wavelength $440 \mathrm{~nm}$ is placed slightly above a plane mirror $\mathrm{M}$ as shown. Image of $\mathrm{S}$ in $\mathrm{M}$ can be used as a virtual source to produce interference fringes on the screen. The distance of source $S$ from $O$ is $20.0 \mathrm{~cm}$, and the distance of screen from $\mathrm{O}$ is $100.0 \mathrm{~cm}$ (figure is not to scale). If the angle $\theta=0.50 \times 10^{-3}$ radians, the width of the interference fringes observed on the screen is $-$

Options:
A $2.20 \mathrm{~mm}$
B $2.64 \mathrm{~mm}$
C $1.10 \mathrm{~mm}$
D $0.55 \mathrm{~mm}$

The correct answer is:
$2.64 \mathrm{~mm}$

Solution:
$\mathrm{S}$ and $\mathrm{S}_{\mathrm{I}}$ are source of $\mathrm{YDSE}$ $\theta=0.5 \times 10^{-3}$ radian (very small)
$$
\begin{array}{l}
\mathrm{D}=\mathrm{SO} \cos \theta+100 \\
=20 \times 1+100 \\
=120 \mathrm{~cm} \\
\mathrm{~d}=2 \times \mathrm{SO} \sin \theta \\
\Rightarrow \quad 2 \times 20 \times \theta \\
\Rightarrow \quad 40 \times 0.5 \times 10^{-3} \mathrm{~cm} \\
2 \times 10^{-2} \mathrm{~cm} \\
\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{440 \times 10^{-6} \times 120 \times 10^{2}}{2 \times 10^{-2} \times 10^{2}} \\
264 \times 10^{-2} \\
\Rightarrow 2.64 \mathrm{~mm}
\end{array}
$$

Topic: Physics, Wave Optics, JEE Main
A monochromatic light wave with wavelength λ1 and frequency ν1 in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength λ2 and frequency ν2 of the refracted wave are:

Options:
A λ2=2λ1,ν2=ν1
B λ2=λ1,ν2=12ν1
C λ2=λ1,ν2=2ν1
D λ2=12λ1,ν2=ν1

The correct answer is:
λ2=12λ1,ν2=ν1

Solution:
By using Snell's law, n1sinθ1=n2sinθ2, the refractive index is

μ=sin45°sin30°=1212=2

Also using the relation n=cv where n is the refractive index, c is the velocity of light in air and v is the velocity in the medium

⇒cair v =2=λ1 λ2

The velocities are given by

v=νλ2c=νλ1

⇒λ1=2λ2 and ν1=ν2.

The frequency of light does not change when the medium changes.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm ?

Options:
A 0.24 V
B 1.5 V
C 1.25 V
D 0.96 V

The correct answer is:
1.25 V

Solution:
eVs=hcλ-ϕ

eV'-V=hc1λ'-1λ

V'=hce1λ'-1λ+V

=12401474-1670+0.48=1.25eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A monochromatic plane wave of speed c and wavelength Î» is diffracted at a small aperture. The diagram illustrates successive wavefronts (vibrating in the same phase). After what time will some portion of the wavefront XY reach P?

Options:
A 3 λ 2 c
B 2 λ c
C 3 Î» c
D 4 λ c

The correct answer is:
3 λ c

Solution:
This distance between the 2 successive lines (wavefronts) is λ . This is for both the straight and curved lines.
So, wavefront XY is separated by a distance of 3λ from the wavefront that has reached point P.
Distance that needs to be traveled =3λ
Speed of light =c
Speed = Distance/Time
Time = distance/speed = 3λ c

Topic: Physics, Wave Optics, JEE Main
A monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is
[Use ch $=1242 \mathrm{eV}-\mathrm{m}$ ]

Options:
A 80 nm
B 85.5 nm
C 97.4 nm
D 100.2 nm

The correct answer is:
97.4 nm

Solution:
Number of line emitted $=6, n_i=1, n_f=4$
$\begin{aligned}
\frac{n(n-1)}{2} & =6 \\
n^2-n-12 & =0 \\
(n-4)(n+3) & =0 \\
n=4, n & =-3 \quad \text{(Not possible)}
\end{aligned}$
Using, $\frac{1}{\lambda}=R \sqrt{\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
$\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{4^2}\right)=R \frac{15}{16}$
where, $R=1.09 \times 10^7 \mathrm{~m}^{-1}$
$\lambda=\frac{16}{15} \times \frac{1}{1.09 \times 10^7}=97.4 \mathrm{~nm}$

Topic: Physics, Atomic Physics, JEE Main
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \mathrm{G}$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \mathrm{~cm}\left(\mathrm{~m}_{\mathrm{e}}=9.11 \times 10^{-19} \mathrm{C}\right)$. [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth's magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Options:

None

Solution:
Kinetic energy of electron $=18 \mathrm{keV}$
$\frac{1 \mathrm{~m}_e v^2}{2}=18 \times 10^3 \times 1.6 \times 10^{-19}$ joule
velocity $\mathrm{v}=8 \times 10^7 \mathrm{~ms}^{-1}$
velocity in $\mathrm{x}$ direction remain constant, hence time to cross $30 \mathrm{~cm}$
$t=\frac{30 \times 10^{-2}}{8 \times 10^7}$
$t=\frac{3}{8} \times 10^{-9} \mathrm{sec}$
A magnetic force $F=e v B$ is acting in vertical direction, which provides acceleration in vertical direction $a_y=\frac{e v B}{m_e}=5.63 \times 10^{14} \mathrm{~ms}^{-2}$
Vertical deflection $S_y=u_y t+\frac{1}{2} a_y t^2$
$$
\begin{aligned}
S_y &=0+\frac{1}{2} \times 5.63 \times 10^{14} \times\left[\frac{3}{8} \times 10^{-9}\right]^2 \\
S_y &=\frac{1}{2} \times \frac{9}{64} \times 5.63 \times 10^{-4} \\
&=395.85 \times 10^{-7} \text { meter. }
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter
A monoprotic acid in $1.00 \mathrm{M}$ solution is $0.01 \%$ ionised. The dissociation constant of this acid is

Options:
A $1 \times 10^{-8}$
B $1 \times 10^{-4}$
C $1 \times 10^{-6}$
D $10^{-5}$

The correct answer is:
$1 \times 10^{-8}$

Solution:
$K=\frac{\alpha^2 C}{1-\alpha} ; \alpha=\frac{0.01}{100} \approx 1 \therefore K=\alpha^2 C=\left[\frac{0.01}{100}\right]^2 \times 1$
$=1 \times 10^{-8}$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A mosquito is moving with a velocity v→=0.5t2i^+3tj^+9k^ m s-1 and accelerating in uniform conditions. What will be the direction of mosquitoes after 2 s ?

Options:
A tan-123 from x-axis
B tan-1856 from y-axis
C tan-152 from y-axis
D tan-152 from x-axis

The correct answer is:
tan-1856 from y-axis

Solution:
Given :
v→=0.5t2i^+3tj^+9k^

v→at t=2=2i^+6j^+9k^

∴ Angle made by direction of motion of mosquito will be,

cos-1211 (from x-axis) =tan-11172

cos-1611 (from y-axis) =tan-1856

cos-1911 (from z-axis) =tan-1409

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A mother is afflicted by Down's syndrome caused by an extra copy of chromosome 21. The father is normal. Thus, the percentage of offspring affected by the disorder would be

Options:
A 100%
B 75%
C 50%
D 25%

The correct answer is:
50%

Solution:
Down's Syndrome is case of trisomy of 21st chromosome. In which 21st pair carries 3 chromosomes. During gamete formation 50% gametes carry normal number of chromosomes i.e. 23, but remaining 50% gametes carry 24 chromosomes (one more chromosome in 21st pair). So if gamete with 24 chromosomes is fused with male gemete (sperm) then again case of Down's syndrome is generated.

Topic: Biology, Principles of Inheritance and Variation, NEET