datavorous/entrance-exam-dataset huggingface dataset:
A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field $2^{5 / 4} \mathrm{~s}$. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is :

Options:
A $2^{1 / 4}$
B $2^{1 / 2}$
C 2
D $2^{5 / 4}$

The correct answer is:
2

Solution:
Time period of magnet in vibration magnetometer,
$T=2 \pi \sqrt{\frac{I}{M H}}$
First case, $T_1=2 \pi \sqrt{\frac{I_1+I_2}{M^{\prime} H}}$
where, $M=$ resultant magnetic moment of two magnets
Here, two identical magnets are placed perpendicular to each other.
$\therefore \quad I_1=I_2=I$ (Let)
and $\quad M^{\prime}=\sqrt{M^2+M^2}=M \sqrt{2}$
$\therefore \quad T_1=2 \pi \sqrt{\frac{2 I}{\sqrt{2} M H}}$
$2^{5 / 4}=2 \pi \sqrt{\frac{2 I}{\sqrt{2} M H}}$ $\ldots$ (i)
When one magnet is removed, then time period,
$T_2=2 \pi \sqrt{\frac{I}{M H}}$ ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{2^{5 / 4}}{T_2}=\sqrt{\frac{2}{\sqrt{2}}}=\sqrt{\sqrt{2}}$
$\frac{2^{5 / 4}}{T_2}=2^{1 / 4}$
$T_2=\frac{2^{5 / 4}}{2^{1 / 4}}=2 \mathrm{~s}$

Topic: Physics, Oscillations, JEE Main
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \mathrm{~s}$ in earth's horizontal magnetic field of $24 \mu \mathrm{T}$. When a horizontal field of $18 \mu \mathrm{T}$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

Options:
A $1 \mathrm{~s}$
B $2 \mathrm{~s}$
C $3 \mathrm{~s}$
D $4 \mathrm{~s}$

The correct answer is:
$4 \mathrm{~s}$

Solution:
The time period T of oscillation of a magnet is given by
\(T=2 \pi \sqrt{\frac{I}{M B}}\)
where,
I = Moment of inertia of the magnet about the axis of rotation
\(M=\) Magnetic moment of the magnet
\(B=\) Uniform magnetic field As I, B remains the same
\(\therefore \quad T \propto \frac{1}{\sqrt{B}}\) or \(\frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}\)
According to given problem,
\(\begin{aligned}
B_1 & =24 \mu \mathrm{T} \\
B_2 & =24 \mu \mathrm{T}-18 \mu \mathrm{T}=6 \mu \mathrm{T} \\
T_1 & =2 \mathrm{~s} \\
\therefore \quad T_2 & =(2 \mathrm{~s}) \sqrt{\frac{(24 \mu \mathrm{T})}{(6 \mu \mathrm{T})}}=4 \mathrm{~s}
\end{aligned}\)

Topic: Physics, Magnetic Properties of Matter, JEE Main
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \mathrm{~s}$ in earth's horizontal magnetic field of $24 \mu \mathrm{T}$. When a horizontal field of $18 \mu \mathrm{T}$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

Options:
A $1 \mathrm{~s}$
B $2 \mathrm{~s}$
C $3 \mathrm{~s}$
D $4 \mathrm{~s}$

The correct answer is:
$4 \mathrm{~s}$

Solution:
The time period T of oscillation of a magnet is given by
\(T=2 \pi \sqrt{\frac{I}{M B}}\)
where,
I = Moment of inertia of the magnet about the axis of rotation
\(M=\) Magnetic moment of the magnet
\(B=\) Uniform magnetic field As I, B remains the same
\(\therefore \quad T \propto \frac{1}{\sqrt{B}}\) or \(\frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}\)
According to given problem,
\(\begin{aligned}
B_1 & =24 \mu \mathrm{T} \\
B_2 & =24 \mu \mathrm{T}-18 \mu \mathrm{T}=6 \mu \mathrm{T} \\
T_1 & =2 \mathrm{~s} \\
\therefore \quad T_2 & =(2 \mathrm{~s}) \sqrt{\frac{(24 \mu \mathrm{T})}{(6 \mu \mathrm{T})}}=4 \mathrm{~s}
\end{aligned}\)

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2010 (Screening)
A village has 10 players. A team of 6 players is to be formed. 5 members are chosen out of these 10 players and then the captain is chosen from the remaining players. Then total number of ways of choosing such teams is

Options:
A 1260
B 210
C $\left.{ }^{10} \mathrm{C}_6\right) 5$ !
D $\left({ }^{10} \mathrm{C}_5\right) 6$

The correct answer is:
1260

Solution:
Total number of ways of choosing such terms is
$$
\begin{aligned}
& { }^{10} C_5 \cdot{ }^5 C_1 \\
& =\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times 5 \\
& =10 \times 3 \times 7 \times 6=1260 \\
&
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2017
A viscous fluid is flowing through a cylindrical tube. The velocity distribution of the fluid is best represented by the diagram

Options:
A

B

C

D None of these

The correct answer is:

Solution:
Correct option is 3.
The velocity distribution of fluid is (A) , as due to friction between layers it will varies as $v=\frac{R^2}{4 \mu}(-d p / d x)$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A viscous fluid is flowing through a cylindrical tube. The velocity distribution of the fluid is best represented by the diagram

Options:
A
B
C
D None of these

The correct answer is:

Solution:
Figure shows the flow speed profile for laminar flow of a viscous fluid in a long cylindrical pipe. The speed is greatest along the axis and zero at the pipe walls,

We know that,v=R2μ-dPdx

where, R=radius of pipe

μ=viscosity

So, according to the above explanation solution 3 is correct.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A voltmeter having a resistance of 998 ohms is connected to a cell of e.m.f. 2 volt and internal resistance 2 ohm. The error in the measurement of e.m.f. will be

Options:
A $4 \times 10^{-1}$ volt
B $2 \times 10^{-3}$ volt
C $4 \times 10^{-3}$ volt
D $2 \times 10^{-1}$ volt

The correct answer is:
$4 \times 10^{-3}$ volt

Solution:
Error in measurement $=$ Actual value Measured value
$\begin{aligned} & \text { Actual value }=2 \mathrm{~V} \\ & i=\frac{2}{998+2}=\frac{1}{500} \mathrm{~A}\end{aligned}$
Since $E=V+i r \Rightarrow V=E-i r=2-\frac{1}{500} \times 2=\frac{998}{500} \mathrm{~V}$
$\begin{aligned} & \therefore \text { Measured value }=\frac{998}{500} \mathrm{~V} \\ & \Rightarrow \text { Error }=2-\frac{998}{500}=4 \times 10^{-3} \mathrm{volt}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
A voltmeter of variable ranges 3 V, 15 V, 150 V is to be designed by connecting resistances R1, R2, R3 in series with a galvanometer of resistance G = 20 Î©, as shown in the figure. The galvanometer gives full pass through its coil. Then, the resistances R1, R2 and R3 (in kilo ohms) should be, respectively

Options:
A 3,Â 12,Â 135
B 2.98, 12, 135
C 2.98,Â 14.98,Â 149.98
D 3.86, 12.55, 122.97

The correct answer is:
2.98, 12, 135

Solution:
R =VIg â Rg
R1 =3Igâ Rg = 2980 Î© = 2.98 kÎ©
R1 + R2=15Igâ Rg = 14980Î© = 14.98 kÎ©
R1 + R2 + R3= 150Ig â Rg = 149980 Î©
= 149.98 kÎ©
âR1 = 2.98 kÎ©
R2 = 14.98 â 2.98 = 12 kÎ©
R3 = 149.98 â 14.98 = 135 kÎ©

Topic: Physics, Current Electricity, JEE Main
A voltmeter with a resistance 50Ã103 Î© is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is

Options:
A 9 ×106 Ω
B 105 Ω
C 1.5 ×105 Ω
D 9 ×105 Ω

The correct answer is:
105 Ω

Solution:
V=igRÂ Â Â Â Â Â Â Â andÂ Â V′=igR′Â Â Â Â Â orÂ R′R=V′V
Or R′=V′VR=3VVÃ50Ã103=1.5Â Ã105Î©
â´Â Â Â Â Â Additional resistance
=1.5Ã105-0.5Â Ã105=105Â Î©

Topic: Physics, Current Electricity, JEE Main
A wall is moving with a velocity u and a source of sound is moving with a velocity u2, in the same direction, as shown in the figure. Assume that the sound travels with velocity 10u, the ratio of incident sound wavelength on the wall to the reflected sound wavelength by the wall is equal to

Options:
A 9 :11
B 11Â :9
C 4Â :5
D 5Â :4

The correct answer is:
9 :11

Solution:
Î»i= Wavelength of the incident sound
=10u-u2f=19u2f
fi= Frequency of the incident sound
=10u-u10u-u2f=1819f
=f=fr= Frequency of the reflected sound
Î»r= Wavelength of the reflected sound
=10u+ufr=11u18fÃ19
=11Ã1918.uf
Î»iÎ»r=19u2fÃ18f11Ã19u=911

Topic: Physics, Waves and Sound, JEE Main
A watch shows time as $3: 25$ when seen through a mirror, time appeared will be

Options:
A $8: 35$
B $9: 35$
C $7: 35$
D $8: 25$

The correct answer is:
$8: 35$

Solution:
The correct option is 1 $8: 35$
For solving this type of problems remembe Actual time $=11: 60$ - given time So here Actual time $=11: 60-3: 25=8: 35$

Topic: Physics, Ray Optics, JEE Main
A water drop is divided into eight equal droplets then the pressure difference between the inner and outer side of the big drop will be

Options:
A same as for smaller droplet
B 1/2 of that for smaller droplet
C 1/4 of that for smaller droplet
D Twice that for smaller droplet

The correct answer is:
1/2 of that for smaller droplet

Solution:
When break into eight drop radius of the small drop is r=Rn1/3⇒ r=R81/3=R2
For large drop ΔP1=2TR
for small drop ΔP2=2Tr=2TR/2= 22TR

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N m-1. In this process the gain in surface energy will be

Options:
A 2.8×10-4 J
B 1.5×10-3 J
C 1.9×10-4 J
D 9.4×10-5 J

The correct answer is:
2.8×10-4 J

Solution:
By conserving the total volume 43πR3=64×43πr3⇒R=4r

Change in surface energy is

ΔU=T×ΔA, =T64×4πr2-4πR2

=4πTR264×116-1=12×3.14×0.075×10-4

≈2.8×10-4 J

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A water drop of radius 1μm falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8×10-5 N s m-2 and its density is negligible as compared to that of water 106 g m-3. Terminal velocity of the water drop is

(Take acceleration due to gravity =10 m s-2)

Options:
A 145.4×10-6 m s-1
B 123.4×10-6 m s-1
C 118.0×10-6 m s-1
D 132.6×10-6 m s-1

The correct answer is:
123.4×10-6 m s-1

Solution:
Terminal velocity is given by vT=2r2ρ-σg9η, where, r is radius of falling body, ρ is density of falling body and σ is density of fluid.

⇒vT=2×10-12×106×10-3×109×1.8×10-5

⇒vT=123.4×10-6 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A water drop of radius ' $r$ ' and volume ' $\mathrm{V}$ ' is kept in between the two identical glass plates such that it forms a thin layer of area ' $\mathrm{A}$ ' between the plates. A force ' $\mathrm{F}$ ' is applied such that the two plates separate from each other. The surface tension ' $\mathrm{T}$ ' of the liquid is

Options:
A $\frac{F V}{2 A^2}$
B $\frac{\mathrm{A}^2}{\mathrm{FV}}$
C $\frac{\mathrm{AV}}{\mathrm{F}^2}$
D $\frac{\mathrm{FV}}{4 \mathrm{~A}^2}$

The correct answer is:
$\frac{F V}{2 A^2}$

Solution:
Consider complete wetting.
The Laplace pressure jump is given by:
On plugging into equation (1), $\frac{\mathrm{T}}{\left(\frac{\mathrm{V}}{2 \mathrm{~A}}\right)}=\frac{\mathrm{F}}{\mathrm{A}}$
$$
\therefore \mathrm{T}=\frac{\mathrm{FV}}{2 \mathrm{~A}^2}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $\mathrm{v}$, the total area around the fountain that gets wet is :

Options:
A
$\pi \frac{v^4}{g^2}$

B
$\frac{\pi}{2} \frac{v^4}{g^2}$

C
$\pi \frac{v^2}{g^2}$

D
$\pi \frac{v^4}{g}$

The correct answer is:
$\pi \frac{v^4}{g^2}$

Solution:
Max. range $=\frac{u^2}{\mathrm{~g}}$ i.e., $\frac{v^2}{\mathrm{~g}}$ (radius of circle)
$$
\text { Area occupied }=\pi\left(\frac{\mathrm{v}^2}{\mathrm{~g}}\right)^2=\frac{\pi \mathrm{v}^4}{\mathrm{~g}^2}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2011
A water heater of power 2000 W is used to heat water. The specific heat capacity of water is 4200 J kg-1 K-1. The efficiency of heater is 70%. Time required to heat 2 kg of water from 10°C to 60°C is _____ s.

(Assume that the specific heat capacity of water remains constant over the temperature range of the water).

Options:

The correct answer is:
300

Solution:
Heat provided to water will be responsible for increase in temperature. Therefore, we can write

η×P×t=M×s×ΔT

Now, η=70100=0.7 and energy=power×time.

Therefore,

t=2×4200×60°-10°0.7×2000 s=300 s

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A water sample has ppm level concentration of following anions, F-=10;SO42-=100;NO3-=50 . The anion/anions that make/makes the water sample unsuitable for drinking is/are

Options:
A only F-
B only SO42-
C only NO3-
D both SO42- and NO3-

The correct answer is:
only F-

Solution:
Above 500 ppm of SO42- ions in drinking water, can cause laxative effect otherwise lesser ppm value is permissible for drinking.
Maximum limit of NO3- ions in drinking water is 50 ppm, above this limit it can cause the disease like methemoglobinemia.
More than 1 ppm F- ions in drinking water are not fit for drinking, it can cause decay of bones and teeth.

Topic: Chemistry, Environmental Chemistry, JEE Main
A water sample has ppm level concentration of following anions
F-=10; SO42-=100; NO3-=50
The anion/anions that make/makes the water sample unsuitable for drinking is/are

Options:
A Both SO42- and NO3-
B Only F-
C Only SO42-
D Only NO3-

The correct answer is:
Only F-

Solution:
Ions Maximum prescribed
concentration in drinking water
Fluoride 1.5 ppm
Sulphate 500 ppm
Nitrate 50 ppm
Fluoride ion concentration is more than the limit, so it makes water unsuitable for drinking

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2017 (02 Apr)
A water sample has ppm level concentration of the following metals:
Fe=0.2; Mn=5.0; Cu=3.0; Zn=5.0 .
The metal that makes the water sample unsuitable for drinking is

Options:
A Zn
B Fe
C Mn
D Cu

The correct answer is:
Mn

Solution:
The concentration of the ions for drinking water are,
Fe≤0.2 ppm
Mn≤0.05 ppm
Cu≤3.0 ppm
Zn≤5.0 ppm
Here, Mn concentration is greater than the allowed limit. So, it makes the water sample unsuitable for drinking.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semivertical angle is tan-134. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is _______.

Options:

The correct answer is:
5

Solution:
Plotting the diagram we have,

Now given, tanθ=34=rh and let water is poured at constant rate dVdt=6 cubic meter per hour,

Now we know that volume of cone is given by V=13πr2 h=13πh3tan2θ=9π48 h3=3π16 h3 by using tanθ=34=rh

Now differentiating above equation we get,

⇒dVdt=3π16·3h2·dhdt=6

⇒dhdt=23πm/hr

Now, we know that curved surface area of cone is given by,

S=πrl=1516πh2 as sinθ=35=rl⇒l=53r and r=34h

Now differentiating both side we get,

⇒dSdt=15π16·2hdhdt

⇒dSdt=5 m2/hr as dhdt=23πm/hr

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan-112. Water is poured into it at a constant rate of 5 cubic m/min. Then the rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is 10 m; is:

Options:
A 110π
B 115π
C 15π
D 2π

The correct answer is:
15π

Solution:
The given water tank is of the shape shown by the diagram.

The semi-vertical angle θ=tan-112, ⇒tanθ=12

Let at any time t min, height of water level is h m and radius of cone filled with water be r m.

Also, we have tanθ=rh

⇒rh=12

⇒r=h2 ...i

Now, the volume of the water at time t min in the cone is V=13πr2h

On putting the value of r from the equation i, we get

V=π3h34=π12h3

Now, differentiating with respect to t, we get

dVdt=π12⋅3h2dhdt

Put the given value of dVdt & h

⇒5=π4⋅100dhdt

⇒dhdt=15π m/min

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A water tank is filled upto height h and is being emptied through a small hole at the bottom. Ratio of time taken for the level of water to fall from h to h2 and from h2to zero is

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

The correct answer is:
2 - 1

Solution:
Time taken for the level to fall from H to H' is given by
t = A A 0 2 g H - H â²
According to problem the time taken for the level to fall from h to h 2
t 1 = A A 0 2 g [ h - h 2 ]
Similarly, time taken for water level to fall from h2 to 0 is
t 2 = A A 0 2 g [ h 2 - 0 ]
â´ t 1 t 2 = 1 - 1 2 1 2 - 0 = 2 - 1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A wave in a string has an amplitude of $2 \mathrm{~cm}$. The wave travels in the +ve direction of $\mathrm{x}$-axis with a speed of $128 \mathrm{~m} / \mathrm{s}$ and it is noted that 5 complete waves fit in $4 \mathrm{~m}$ length of the string. The equation describing the wave is

Options:
A $y=(0.02) m \sin (7.58 x-1005 t)$
B $y=(0.02) \mathrm{m} \sin (7.85 x+1005 \mathrm{t})$
C $y=(0.02) m \sin (15.7 x-2010 t)$
D $y=(0.02) m \sin (15.7 x+2010 t)$

The correct answer is:
$y=(0.02) m \sin (7.58 x-1005 t)$

Solution:
$5 \lambda=4$
$$
\lambda=\frac{4}{5}
$$
$$
\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{10 \pi}{4}=7.85
$$
wave moves along positive $\mathrm{X}$-direction

Topic: Physics, Waves and Sound, NEET, NEET 2009 (Mains)
A wave in a string has an amplitude of $2 \mathrm{~cm}$. The wave travels in the +ve direction of $x$ axis with a speed of $128 \mathrm{~ms}^{-1}$ and it is noted that 5 complete waves fit in $4 \mathrm{~m}$ length of the string. The equation describing the wave is

Options:
A $y=(0.02) \mathrm{m} \sin (7.85 x+1005 \mathrm{t})$
B $\mathrm{y}=(0.02) \mathrm{m} \sin (15.7 \mathrm{x}-2010 \mathrm{t})$
C $y=(0.02) \mathrm{m} \sin (15.7 \mathrm{x}+2010 \mathrm{t})$
D $y=(0.02) m \sin (7.85 x-1005 \mathrm{t})$

The correct answer is:
$y=(0.02) m \sin (7.85 x-1005 \mathrm{t})$

Solution:
Key Idea Find the parameters and put in the general wave equation.
Here,
$$
\begin{aligned}
A & =2 \mathrm{~cm} \\
\text { direction } & =+ \text { ve direction } \\
\mathrm{v} & =128 \mathrm{~ms}^{-1}
\end{aligned}
$$
and
$$
5 \lambda=4
$$
Now,
$$
\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi \times 5}{4}=7.85
$$
and
$$
\mathrm{v}=\frac{\omega}{\mathrm{k}}=128 \mathrm{~ms}^{-1}
$$
$$
\begin{aligned}
\Rightarrow \quad \omega & =\mathrm{v} \times \mathrm{k}=128 \times 7.85 \\
& =1005
\end{aligned}
$$
$$
\begin{array}{lll}
\text { As, } & y & =\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}) \\
\therefore & & \mathrm{y}=2 \sin (7.85 \mathrm{x}-1005 \mathrm{t}) \\
& & =(0.02) \mathrm{m} \sin (7.85 \mathrm{x}-1005 \mathrm{t})
\end{array}
$$

Topic: Physics, Waves and Sound, JEE Main
A wave is reflected from a rigid support. The change in phase on reflection will
be

Options:
A $\pi / 4$
B $\pi / 2$
C $\pi$
D $2 \pi$

The correct answer is:
$\pi$

Solution:
After reflection from rigid support, a wave suffers a phase change of $\pi$.

Topic: Physics, Waves and Sound, JEE Main
A wave of frequency =3 GHz, strikes a particle of size 1100th of λ, then this phenomenon is called as

Options:
A Diffraction
B Scattering
C Reflection
D Refraction

The correct answer is:
Scattering

Solution:
Rayleigh scattering applies to particles that are small with respect to wavelengths of light.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A wave of the wavelength 5900ââ« emitted by an atom or molecule must have some finite total length which is known as coherence length. For the mentioned wave, this length is 2.4 cm. The number of oscillations in this length will be

Options:
A 4.068×108
B 4.068×104
C 4.068×106
D 4.068×105

The correct answer is:
4.068×104

Solution:
Number of oscillations in coherence length lλ=0.0245900×10-10
=40677.9
=4.068×104

Topic: Physics, Wave Optics, JEE Main
A wave represented by the equation $y_1=a \cos$ $(k x-\omega t)$ is superimposed with another wave to form a stationary wave such that the point $x-0$ is node. The equation for the other wave is

Options:
A
$a \cos (k x-\omega t+\pi)$

B
$a \cos (k x+\omega t+\pi)$

C
$a \cos \left(k x+\omega t+\frac{\pi}{2}\right)$

D
$a \cos \left(k x-\omega t+\frac{\pi}{2}\right)$

The correct answer is:
$a \cos (k x+\omega t+\pi)$

Solution:
Since the point $x=0$ is a node and reflection is taking place from point $x=0$. This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of $\pi$ or a path change of $\frac{\lambda}{2}$.
So, if $y_{\text {incident }}=a \cos (k x-\omega t)$
$$
\begin{aligned}
\Rightarrow y_{\text {incident }} & =a \cos (-k x-\omega t+\pi) \\
& =-a \cos (\omega t+k x)
\end{aligned}
$$
Hence equation for the other wave
$$
y=a \cos (k x+\omega t+\pi)
$$

Topic: Physics, Waves and Sound, JEE Main
A wave represented by the equation $y_1=a \cos$ $(k x-\omega t)$ is superimposed with another wave to form a stationary wave such that the point $x-0$ is node. The equation for the other wave is

Options:
A
$a \cos (k x-\omega t+\pi)$

B
$a \cos (k x+\omega t+\pi)$

C
$a \cos \left(k x+\omega t+\frac{\pi}{2}\right)$

D
$a \cos \left(k x-\omega t+\frac{\pi}{2}\right)$

The correct answer is:
$a \cos (k x+\omega t+\pi)$

Solution:
Since the point $x=0$ is a node and reflection is taking place from point $x=0$. This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of $\pi$ or a path change of $\frac{\lambda}{2}$.
So, if $y_{\text {incident }}=a \cos (k x-\omega t)$
$$
\begin{aligned}
\Rightarrow y_{\text {incident }} & =a \cos (-k x-\omega t+\pi) \\
& =-a \cos (\omega t+k x)
\end{aligned}
$$
Hence equation for the other wave
$$
y=a \cos (k x+\omega t+\pi)
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2012 (12 May Online)
A wave travelling along the $x$-axis is described by the equation $y(x, t)=0.005 \cos (\alpha x-\beta t)$. If the wavelength and the time period of the wave are $0.08 \mathrm{~m}$ and $2.0 \mathrm{~s}$, respectively, then $\alpha$ and $\beta$ in appropriate units are

Options:
A
$\alpha=25.00 \pi, \beta=\pi$

B
$\alpha=\frac{0.08}{\pi}, \frac{2.0}{\pi}$

C
$\alpha=\frac{0.04}{\pi}, \beta=\frac{1.0}{\pi}$

D
$\alpha=12.50 \pi, \beta=\frac{\pi}{2.0}$

The correct answer is:
$\alpha=25.00 \pi, \beta=\pi$

Solution:
$$
y=0.005 \cos (\alpha x-\beta \mathrm{t})
$$
comparing the equation with the standard form,
$$
\begin{aligned}
& y=A \cos \left[\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right) 2 \pi\right] \\
& 2 \pi / \lambda=\alpha \text { and } 2 \pi / T=\beta \\
& \alpha=2 \pi / 0.08=25.00 \pi \\
& \beta=\pi
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2008
A wave travelling in the positive $x$-direction with amplitude $\mathrm{A}=0.2 \mathrm{~m}$, velocity $v=360 \mathrm{~ms}^{-1}$ and wavelength $\lambda$ $=60 \mathrm{~m}$, then the correct expression for the wave is:

Options:
A $y=0.2 \sin \left[2 \pi\left(6 t+\frac{x}{60}\right)\right]$
B $y=0.2 \sin \left[\pi\left(6 t+\frac{x}{60}\right)\right]$
C $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
D $y=0.2 \sin \left[\pi\left(6 t-\frac{x}{60}\right)\right]$

The correct answer is:
$y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$

Solution:
The general equation of wave travelling in $x$-direction is given as
$$
y=A \sin \left[\frac{2 \pi}{\lambda}(V t-x)\right]
$$
Where $A=0.2 \mathrm{~m}, v=360 \mathrm{~ms}^{-1}, \lambda=$ $60 \mathrm{~m}$
$\therefore$ The equation becomes
$$
\begin{aligned}
& y=0.2 \sin \left[\frac{2 \pi}{60}(360 t-x)\right] \\
& y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
A wave $y=a \sin (\omega t-k x)$ on a string meets with another wave producing a node at $x=0$. Then the equation of the unknown wave is

Options:
A
$y=a \sin (\omega t+k x)$

B
$y=-a \sin (\omega t+k x)$

C
$\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$

D
$\mathrm{y}=-\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$

The correct answer is:
$y=-a \sin (\omega t+k x)$

Solution:
$y_1+y_2=a \sin (\omega t-k x)-a \sin (\omega t+k x)$
$=-2 a \cos \omega t \times \sin k x \Rightarrow y_1+y_2=0$ at $x=0$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2002