datavorous/entrance-exam-dataset huggingface dataset:
A message signal is super-imposed with a carrier signal. The resulting modulating signal $C_m(t)$ is given by
$C_m(t)=A_1 \sin \left(\omega_1 t\right)+A_2 \sin \left(\omega_2 t\right)-A_2 \sin \left(\omega_3 t\right) \text {, }$
where $\omega_2 < \omega_1 < \omega_3$. The modulation index and the angular frequency fo the message signal respectively, are

Options:
A $\frac{A_2}{A_1}, \frac{\omega_3-\omega_2}{2}$
B $\frac{2 A_2}{A_1}, \omega_3-\omega_2$
C $\frac{A_1}{2 A_2}, \frac{\omega_3-\omega_2}{2}$
D $\frac{2 A_2}{A_1}, \frac{\omega_3-\omega_2}{2}$

The correct answer is:
$\frac{2 A_2}{A_1}, \frac{\omega_3-\omega_2}{2}$

Solution:
Modulating signal is given as
$C_m(t)=A_1 \sin \left(\omega_1 t\right)+A_2 \sin \left(\omega_2 t\right)-A_3 \sin \left(\omega_3 t\right)$ $\ldots(\mathrm{i})$
where, $\quad \omega_1>\omega_2$ and $\omega_3>\omega_1$
Standard equation of modulating signal is given as
$C_m(t)=A_c \sin \left(\omega_c t\right)+\frac{\mu A_c}{2} \cos \left(\omega_c-\omega_m\right) t-\frac{\mu A_c}{2}$
$\cos \left(\omega_c+\omega_m\right) t$ ...(ii)
Clearly, $\omega_c>\omega_c-\omega_m$ and $\omega_c+\omega_m>\omega_c$
Eq. (i) can also be written as,
$C_m(t)=A_1 \sin \left(\omega_1 t\right)+A_2 \cos \left(\frac{\pi}{2}+\omega_2 t\right)-A_3 \cos$ $\left(\frac{\pi}{2}+\omega_3 t\right) \ldots$ (iii)
Comparing Eq. (ii) and Eq. (iii), we get
$A_1=A_c$
$\omega_c=\omega_1$
$\omega_c-\omega_m=\omega_2$ $\ldots(\mathrm{iv})$
and $\quad \omega_c+\omega_m=\omega_3$ $\ldots(\mathrm{v})$
Also, $\quad \frac{\mu A_c}{2}=A_2 \Rightarrow \mu=\frac{2 A_2}{A_1}$ $\ldots(\mathrm{vi})$
Substracting Eq. (iv) from Eq. (v), we have $\omega_3-\omega_2=\left(\omega_c+\omega_m\right)-\left(\omega_c-\omega_m\right)$
$\Rightarrow \quad 2 \omega_m=\omega_3-\omega_2$
$\Rightarrow \quad \omega_m=\frac{\omega_3-\omega_2}{2}$ $\ldots(\mathrm{vii})$
From Eq. (vi) and (vii), option (d) is correct.

Topic: Physics, Communication System, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 1)
A metallic shell has a point charge q kept inside its cavity. Which one of the following diagrams correctly represents the electric field lines?

Options:
A
B
C
D

The correct answer is:

Solution:
Electric field is zero everywhere inside a metal (conductor) i.e., field lines do not enter a metal plus these are perpendicular to a metal surface.

Topic: Physics, Electrostatics, NEET
A metallic wire with tension $T$ and at temperature $30^{\circ} \mathrm{C}$ vibrates with its fundamental frequency of $1 \mathrm{kHz}$. The same wire with the same tension but at $10^{\circ} \mathrm{C}$ temperature vibrates with a fundamental frequency of $1.001 \mathrm{kHz}$. The coefficient of linear expansion of the wire is

Options:
A $2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
B $1.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
C $1 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
D $0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

Solution:
Frequency of wire $n=\frac{1}{2 l} \frac{\sqrt{T}}{m}$ $m=$ mass per unit length
In both the condition same wire is kept same tension.
So,
$$
\begin{aligned}
\frac{n_1}{n_2}=\frac{l_2}{l_1} \Rightarrow \frac{1}{1.001} & =\frac{l_2}{l_1} \\
l_2 & =\frac{l_1}{1.001}
\end{aligned}
$$
Fall in temperature $\Delta t=30-10=20^{\circ} \mathrm{C}$
$$
\begin{aligned}
l_2 & =l_1(1-\alpha \Delta t) \\
\frac{l_1}{1.001} & =l_1(1-\alpha \times 20) \\
1-\alpha \times 20 & =\frac{1}{1.001}=\frac{0.001}{1.001} \\
\alpha=\frac{1}{1001 \times 20} & =0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2002
A metallic wire with tension $T$ and at temperature $30^{\circ} \mathrm{C}$ vibrates with its fundamental frequency of $1 \mathrm{kHz}$. The same wire with the same tension but at $10^{\circ} \mathrm{C}$ temperature vibrates with a fundamental frequency of $1.001 \mathrm{kHz}$. The coefficient of linear expansion of the wire is

Options:
A $2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
B $1.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
C $1 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
D $0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

Solution:
Frequency of wire $n=\frac{1}{2 l} \frac{\sqrt{T}}{m}$ $m=$ mass per unit length
In both the condition same wire is kept same tension.
So,
$$
\begin{aligned}
\frac{n_1}{n_2}=\frac{l_2}{l_1} \Rightarrow \frac{1}{1.001} & =\frac{l_2}{l_1} \\
l_2 & =\frac{l_1}{1.001}
\end{aligned}
$$
Fall in temperature $\Delta t=30-10=20^{\circ} \mathrm{C}$
$$
\begin{aligned}
l_2 & =l_1(1-\alpha \Delta t) \\
\frac{l_1}{1.001} & =l_1(1-\alpha \times 20) \\
1-\alpha \times 20 & =\frac{1}{1.001}=\frac{0.001}{1.001} \\
\alpha=\frac{1}{1001 \times 20} & =0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2002
A meter bridge is set-up as shown, to determine an unknown resistance $X$ using a standard $10 \mathrm{ohm}$ resistor. The galvanometer shows null point when tapping-key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is

Options:
A $10.2 \mathrm{ohm}$
B $10.6 \mathrm{ohm}$
C $10.8 \mathrm{ohm}$
D $11.1 \mathrm{ohm}$

The correct answer is:
$10.6 \mathrm{ohm}$

Solution:
Applying the condition of balanced Wheatstone bridge, we get
$\begin{aligned} & \frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(48+2) \mathrm{cm}}=\frac{53}{50}, \\ & X=10 \Omega \times \frac{53}{50}=10.6 \Omega\end{aligned}$

Topic: Physics, Current Electricity, AIIMS, AIIMS 2017
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency $340 \mathrm{~Hz}$ ) when the tube length is $25.5 \mathrm{~cm}$ or $79.3 \mathrm{~cm}$. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Options:

None

Solution:
[QUESTION]
One end of the tube is open and the other end is closed by the piston, so it behaves as a closed organ pipe, which produces only odd harmonics.
$\therefore \quad$ The pipe is in resonance with the fundamental mode and the 3rd harmonic $(\because 79.3 \approx 3 \times 25.5)$
For fundamental mode, $\frac{\lambda}{4}-l_1-25.5$ $\therefore \lambda=4 \times 25.5=102 \mathrm{~cm}=1.02 \mathrm{~m}$
Speed $=v=v \lambda=340 \times 1.02=346.8 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound
A moving boat is observed from the top of a cliff of $150 \mathrm{~m}$ height. The angle of depression of the boat changes from $60^{\circ}$ to $45^{\circ}$ in 2 minutes. What is the speed of the boat in metres per hour?

Options:
A $\frac{4500}{\sqrt{3}}$
B $\frac{4500(\sqrt{3}-1)}{\sqrt{3}}$
C $4500 \sqrt{3}$
D $\frac{4500(\sqrt{3}+1)}{\sqrt{3}}$

The correct answer is:
$\frac{4500(\sqrt{3}-1)}{\sqrt{3}}$

Solution:
$\tan 60^{\circ}=\frac{150}{x} \Rightarrow x=\frac{150}{\sqrt{3}}$
Also, $\tan 45^{\circ}=\frac{150}{x+y}$
$\Rightarrow \quad x+y=150$
$\Rightarrow \quad y=150-x=150-\frac{150}{\sqrt{3}}$
$\Rightarrow \quad y=150\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=$ distance travelled
$\quad$ Speed $(\operatorname{in} \mathrm{m} / \mathrm{hr})=\frac{150(\sqrt{3}-1)}{\sqrt{3}} \times \frac{60}{2}=4500 \frac{(\sqrt{3}-1)}{\sqrt{3}}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2016 (Phase 2)
A number in binary system is 110001 . It is equal to which one of the following numbers in decimal system

Options:
A 45
B 46
C 48
D 49

The correct answer is:
49

Solution:
$(110001)_{2}=1 \times 2^{5}+1 \times 2^{4}+0 \times 2^{3}+0 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0}$
$=32+16+0+0+0+1=(49)_{10}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
A number of elements are available in earth's crust but most abundant elements are.

Options:
A
$\mathrm{Al}$ and $\mathrm{Fe}$

B
$\mathrm{Al}$ and $\mathrm{Cu}$

C
Fe and $\mathrm{Cu}$

D
$\mathrm{Cu}$ and $\mathrm{Ag}$

The correct answer is:
$\mathrm{Al}$ and $\mathrm{Fe}$

Solution:
The most abundant elements are aluminium $(8.3 \%$ by weight) and iron ( $4.2 \%$ by weight).

Topic: Chemistry, General Principles and Processes of Isolation of Metals
A number of elements available in earth's crust but most abundant elements are

Options:
A Al and Fe
B Al and Cu
C Fe and Cu
D Cu and Ag

The correct answer is:
Al and Fe

Solution:
Aluminium is the third most abundant element by mass and iron is the second most abundant metal in the earth's crust.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JIPMER, JIPMER 2013
A pair of fair dice is rolled. What is the probability that the second dice lands on a higher value than does the first ?

Options:
A $\frac{1}{4}$
B $\frac{1}{6}$
C $\frac{5}{12}$
D $\frac{5}{18}$

The correct answer is:
$\frac{5}{12}$

Solution:
Possibilities of having higher number on second dice. $\begin{array}{ll}\text { First dice } & \text { Second dice }\end{array}$
$1 \quad 2,3,4,5,6 \rightarrow 5$ possibilities
$2 \quad 3,4,5,6 \rightarrow 4$ possibilities
$3 \quad 4,5,6 \rightarrow 3$ possibilities
$4 \quad 5,6 \rightarrow 2$ possibilities $5 \quad 6 \rightarrow 1$ possibility Total number of possibilities $=15$ Total number of events $=36$
$\therefore$ Probability $=\frac{15}{36}=\frac{5}{12}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first maximum of the diffraction pattern, the phase difference between the rays coming from the edges of the slit is

Options:
A $\frac{\pi}{4}$
B $\pi$
C $\frac{\pi}{2}$
D $3 \pi$

The correct answer is:
$3 \pi$

Solution:
For nth secondary maxima, the path difference $x$ is the odd multiple of the half of the wavelength, i.e.,
$x_n=(2 n+1) \frac{\lambda}{2}$
So, for the first maxima, the phase difference would be:
$\phi_1=\frac{2 \pi}{\lambda} x_1$
$\therefore \phi_1=\frac{2 \pi}{\lambda}\left(\frac{3 \lambda}{2}\right)=3 \pi$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2022 (06 Aug Shift 1)
A particle is acted upon by following forces:
(i) $\quad 2 \hat{i}+3 \hat{j}+5 \hat{k}$,
(ii) $-5 \hat{i}+4 \hat{j}-3 \hat{k}$ and (iii) $3 \hat{i}-7 \hat{k}$
In which plane does it move?

Options:
A $\mathrm{x}$ -yplane
B $\mathrm{y}$ -z plane
C $\quad z$ -x plane
D any arbitrary plane

The correct answer is:
$\mathrm{y}$ -z plane

Solution:
Three forces are given by, say, $\mathrm{F}_{1}, \mathrm{~F}_{2}$ and $\mathrm{F}_{3}$
$\overrightarrow{\mathrm{F}}_{1}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}, \overrightarrow{\mathrm{F}}_{2}=-5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
and $\vec{F}_{3}=3 \hat{i}-7 \hat{k}$
Total resultant force, $\overrightarrow{\mathrm{F}}$ is given by
$\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\overrightarrow{\mathrm{F}}_{3}$
$=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}-5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+3 \hat{\mathrm{i}}-7 \hat{\mathrm{k}}=7 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$
This show that the resultant force is in the $\mathrm{y}-\mathrm{z}$ plane. Thus, it moves in the $\mathrm{y}-\mathrm{z}$ plane.

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10âm sâ1 to 20âm sâ1 while passing through a distance 135 m in t seconds. The value of t is:

Options:
A 10
B 1.8
C 12
D 9

The correct answer is:
9

Solution:
Initial velocity, uâ=â10 m sâ1
Final velocity , vâ=â20 m sâ1
Distance , sâ=â135 m
Let , acceleration = a
Using the formula , v 2 â=â u 2 +2as
ââââaâ=â v 2 â u 2 2s
orââa=â (20) 2 â (10) 2 2Ã135 â=â 400â100 2Ã135
orââaâ=â3002Ã135â=â150135â=â109 m sâ2
Now , using the relation , vâ=âu+at
tâ=âvâuaâ=â20â10109â=â1010Ã9=â9 s.

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Topic: Physics, Motion In One Dimension, NEET
A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ moving with velocity $\mathrm{v}$ enters region-b form region-a along the normal to the boundary as shown is the figure. Region-b has a uniform magnetic field $B$ perpendicular to the plane of the paper. Also, region-b has length L. Choose the correct statements :

Options:
A The particle enters region-c only if $\mathrm{v}>\frac{\mathrm{qLB}}{\mathrm{m}}$
B The particle enters region-c only if $\mathrm{v} < \frac{\mathrm{qLB}}{\mathrm{m}}$
C Path of the particle is a circle in region-b
D Time spent in region -b is independent of velocity v

The correct answers are:
The particle enters region-c only if $\mathrm{v}>\frac{\mathrm{qLB}}{\mathrm{m}}$, Path of the particle is a circle in region-b, Time spent in region -b is independent of velocity v

Solution:
Particle enter region 'c' if
Radius is greater than ' $L$ ' $\Rightarrow \frac{m V}{q B}>L \Rightarrow V>\frac{q B L}{m}$
$\because$ B is uniform and $\overrightarrow{\mathrm{V}} \perp \overrightarrow{\mathrm{B}}$ so.
Path is circular in region b
$$
\text { Time spent }=\frac{\text { Angle cov ered }}{\omega} \quad \omega=\frac{\mathrm{qB}}{\mathrm{m}}
$$
Now angle covered depens on whether particle goes to region c or comes back to region 'a'

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2021
A particle with charge +Q and mass m enters a magnetic field of magnitude B, existing only to the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of B. Then, the time spent by the particle in the field will be

Options:
A 2πmQB
B πmQB
C mvQB
D Infinite

The correct answer is:
πmQB

Solution:
Fâ=QVâÃBâ
F=QVB
Since this force is constant and is perpendicular to particle's direction of motion. Particle will move in a circle.
â´ QVB=mV2R
R=mVQB
â´ Time required to travel ÏR distance
t=ÏRV
â t=ÏÃmVQBv=ÏmQB
t=ÏmQB

Topic: Physics, Magnetic Effects of Current, NEET
A passenger in an aeroplane shall

Options:
A
never see a rainbow

B
may see a primary and a secondary rainbow as concentric circles

C
may see a primary and a secondary rainbow as concentric arcs

D
shall never see a secondary rainbow

The correct answer is:
may see a primary and a secondary rainbow as concentric circles

Solution:
When a passenger in an aeroplane then he may see primary and seoondary rainbow such as concentric circles.

Topic: Physics, Ray Optics
A patient comes to the clinic with a genetic disorder characterized by a webbed neck and the presence of rudimentary ovaries. This person most likely has

Options:
A two X and one Y sex chromosomes
B only one sex chromosome which is X
C three 21st chromosomes
D three X chromosomes

The correct answer is:
only one sex chromosome which is X

Solution:
Turner's Syndrome is a disorder is caused due to the absence of one of the X chromosomes. These patients are females with only 45 chromosomes. Such females are sterile as ovaries are rudimentary. They lack other secondary sexual characters. They have a short neck with a webbed appearance, low hairline at the back of the neck, low-set ears, hands, and feet that are swollen or puffy at birth.

Topic: Biology, Principles of Inheritance and Variation, NEET
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite of bank is $60^{\circ}$. When he retires $40 \mathrm{~m}$. from the bank, he finds the angle to be $30^{\circ} .$ What is the breadth of the river?

Options:
A $60 \mathrm{~m}$
B $40 \mathrm{~m}$
C $30 \mathrm{~m}$
D $20 \mathrm{~m}$

The correct answer is:
$20 \mathrm{~m}$

Solution:
$\operatorname{In} \Delta \mathrm{ABC}, \tan 60^{\circ}=\frac{\mathrm{h}}{\mathrm{x}}$
$\Rightarrow \sqrt{3}=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{h}=\sqrt{3} \mathrm{x}$
$\operatorname{In} \Delta \mathrm{ABD}, \tan 30^{\circ}=\frac{\mathrm{h}}{\mathrm{x}+40}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{x}+40} \Rightarrow \mathrm{x}+40=\sqrt{3} \mathrm{~h}$
Putting value of h from equation (1), we get $x+40=3 x$
$\mathrm{x}=20 \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2013 (Phase 2)
A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93 \mathrm{~mm}$ of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is

Options:
A $+3 \%$
B $-3 \%$
C $+3.2 \%$
D $-3.2 \%$

The correct answer is:
$-3.2 \%$

Solution:
Given, diameter, $d_1=0.9 \mathrm{~mm}=0.9 \times 10^{-3} \mathrm{~m}$
$$
\begin{aligned}
& =9 \times 10^{-4} \mathrm{~m} \\
d_2=0.93 \mathrm{~mm} & =9.3 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
Frequency of fundamental tone is given as
$$
f=\frac{1}{2 l} \sqrt{\frac{T}{m}}
$$
where, $m=$ mass per unit length.
$$
=\frac{M}{l}=\frac{V \cdot \rho}{l}=\frac{A \cdot l}{l} \cdot \rho=A \rho
$$
$$
\Rightarrow \quad m=\frac{\pi d^2 \rho}{4}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& f=\frac{1}{2 l} \sqrt{\frac{T}{\pi d^2 \rho / 4}}=\frac{1}{2 l} \sqrt{\frac{4 T}{\pi d^2 \rho}}=\frac{1}{l} \sqrt{\frac{T}{\pi d^2 \rho}} \\
\Rightarrow & f=\frac{1}{l d} \sqrt{\frac{T}{\pi \rho}} \Rightarrow f \propto \frac{1}{d} \\
\therefore & \frac{f_1}{f_2}=\frac{d_2}{d_1}=\frac{9.3 \times 10^{-4}}{9 \times 10^{-4}}=\frac{9.3}{9} \Rightarrow \frac{f_2}{f_1}=\frac{9}{9.3}
\end{aligned}
$$
$\therefore$ Percentage change in frequency
$$
\begin{aligned}
& =\frac{f_2-f_1}{f_1} \times 100=\left(\frac{f_2}{f_1}-1\right) \times 100 \\
& =\left(\frac{9}{9.3}-1\right) \times 100=-3.2 \%
\end{aligned}
$$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
A pipe $20 \mathrm{~cm}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of $1237.5$ $\mathrm{Hz}$ ? (sound velocity in air $=330 \mathrm{~ms}^{-1}$ )

Options:

None

Solution:
As given that,
Length of pipe, $l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$$
v=1237.5 \mathrm{~Hz}, v=330 \mathrm{~m} / \mathrm{s}
$$
$l=\frac{\lambda}{4}$ or $\lambda=4 l$
$$
\begin{aligned}
&\left(v_{\text {funda }}\right)_1=\frac{v}{4 l}=\frac{330}{4 \times 20 \times 10^{-2}} \\
&\left(v_{\text {funda }}\right)_1=\frac{330 \times 100}{80}=412.5 \mathrm{~Hz} \\
&\frac{\left(v_{\text {given }}\right)}{\left(v_{\text {funda }}\right)_1}=\frac{1237.5}{412.5}=\frac{3}{1}
\end{aligned}
$$
The frequency in Ist, IInd, IIIrd .... harmonic are in the ratio $1: 2: 3: 4 \ldots \ldots$ at one end open organ pipe.

Topic: Physics, Waves and Sound
A pipe closed at one end has lengh $0 \cdot 8 \mathrm{~cm}$. At its open end a $0.5 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~m} / \mathrm{s}$, the mass of the string is

Options:
A 8 gram
B 2 gram
C 10 gram
D 4 gram

The correct answer is:
10 gram

Solution:
No Solution

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (15 Oct Shift 2)
A pipe closed at one end has length $0.8 \mathrm{~m}$. At its open end $0.5 \mathrm{~m}$ long uniform string is vibrating in its $2^{\text {nd }}$ harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~m} / \mathrm{s}$, the mass of the string is

Options:
A 20 gram
B 10 gram
C 5 gram
D 15 gram

The correct answer is:
10 gram

Solution:
$\begin{aligned} & 2 \times\left[\frac{1}{2 \ell_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\right]=\frac{v}{4 \ell_2} \\ & \frac{1}{0.5} \sqrt{\frac{50}{\mathrm{~m}}}=\frac{320}{4 \times 0.8} \\ & \therefore \mathrm{m}=0.02 \mathrm{~kg} / \mathrm{m} \\ & \therefore \text { Total mass of the string } \\ & =0.02 \times 0.5 \mathrm{~kg}=10 \mathrm{gm}\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (21 Sep Shift 2)
A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (velocity of sound in air =332 m s-1)

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
Fundamental frequency
n= V 4L = 332 4Ã83Ã 10 â2 =100 .
Hence possible frequencies are n=100 .
n 1 =300.
n 2 =500.
n 3 =700.
n 5 =900.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2018
A pipe closed at one end length $0.8 \mathrm{~m}$. At its open end a $0.5 \mathrm{mlong}$ uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~m} / \mathrm{s}$, the mass of the string used is

Options:
A $8 \mathrm{~g}$
B $4 \mathrm{~g}$
C $12 \mathrm{~g}$
D $10 \mathrm{~g}$

The correct answer is:
$10 \mathrm{~g}$

Solution:
For fundamental harmonics of a closed end pipe, $\frac{\lambda}{4}=L$
Velocity of sound is in air is given by, $v=\lambda f$
For second harmonics of a string, $\lambda^{\prime}=L^{\prime}$ the wavelength is the length of the string.
Speed of sound on the wave is related as, $v_{\mathrm{S}}=\lambda^{\prime} f$.
Speed of sound is given by, $v_{\mathrm{S}}=\sqrt{\frac{T}{\mu}}, T$ is the tension in the wire and $\mu$ the mass per unit length of the string.
Since, the string in second harmonics is in resonance with the closed end pipe in fundamental mode,
$\begin{aligned} & f=\frac{v}{4 L}=\frac{v_{\mathrm{S}}}{L^{\prime}}=\frac{1}{L^{\prime}} \sqrt{\frac{T}{\mu}} \\ & \Rightarrow \frac{320 \mathrm{~m} / \mathrm{s}}{4 \times 0.8 \mathrm{~m}}=\frac{1}{0.5 \mathrm{~m}} \sqrt{\frac{50 \mathrm{~N}}{\mu}} \\ & \Rightarrow \mu=\frac{1}{50} \mathrm{kgm}^{-1} \\ & \Rightarrow m=\mu L^{\prime}=\frac{0.5 \mathrm{~m}}{50 \mathrm{~kg}^{-1} \mathrm{~m}}=10 \mathrm{~g}\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (06 Aug Shift 2)
A pipe of 1 m length is closed at one end. Taking the speed of sound in air as 320 m s-1, the air column in the pipe cannot resonate for the frequency in Hz

Options:
A 80
B 160
C 240
D 560

The correct answer is:
160

Solution:
Given, v=320 m s-1, l=1 m
For an organ pipe whose one end is closed, only odd harmonics containing odd multiples of fundamental frequency are present.
Resonate frequency of the first mode, n1=v4l′
Second mode, n2=3n1,
Third mode, n3=5n1,
Fourth mode, n4=7n1,
Fifth mode, n5=9n1
So, n1=3204=80 Hz
n2=3×80=240 Hz
n3=5×80=400 Hz
n4=7×80=560 Hz
n5=9×80=720 Hz
So 160 Hz is not possible

Topic: Physics, Waves and Sound, NEET
A pipe of length 1.5 m closed at one end is filled with gas and resonates in its fundamental mode with a tuning fork. Another open organ pipe of same dimensions filled with air resonates in its fundamental mode with the same tuning fork. If this experiment is performed at 30Â° C, then the speed of sound at 0 Â°C in the same gas is (Speed of sound in air is 360 m/s)

Options:
A 580 m/s
B 683 m/s
C 880 m/s
D 743 m/s

The correct answer is:
683 m/s

Solution:
For open pipe f=v2â=3602Ã15=120 Hz

For the closed at one end pipe, let the velocity of sound in gas at 30 Â°C be v'

f'=v'4â=v'4Ã1.5=v'6

Since same tuning fork is used frequency will be same and thus,

f'=fâv'=720 m/s

Also

vâTâv=720Ã273303=683 m/s.

Topic: Physics, Waves and Sound, NEET