datavorous/entrance-exam-dataset huggingface dataset:
A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then least count of the travelling microscope is _____ ×10-6 m.

Options:

The correct answer is:
5

Solution:
Given here, main scale reading, MS=140 cm

And, 50×VS=49×MS

Least count of travelling microscope is LC=MS-VS

=MS-4950 MS

=150×MS

=150×140 cm=12000×10-2 m

=5×10-6 m

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A travelling nucleus A having a de-Broglie wavelength λA spontaneously splits into two nuclei B and C of equal masses. After the split, B travels in the same direction while C travels in the opposite direction with a speed equal to half of B. The values of the de-Broglie wavelengths λB and λC will be

Options:
A λA, 2λA
B λA2, λA
C λA, λA2
D 2λA, λA

The correct answer is:
λA2, λA

Solution:
Aâu âv/2C Bâv2m m m

Applying conservation of linear momentum

2mu = mv - mv2

So,

Î»A=h2mu Î»B=hmv=Î»A2 Î»C=hmv2=Î»A

Topic: Physics, Atomic Physics, JEE Main
A travelling wave is described by the equation yx,t=0.05sin8x-4t m. The velocity of the wave is:

[All the quantities are in SI unit]

Options:
A 4 m s-1
B 2 m s-1
C 0.5 m s-1
D 8 m s-1

The correct answer is:
0.5 m s-1

Solution:
The given wave equation is of the form:

y=Asinkx-ωt

with k=8 and ω=4

Since, k=2πλ and ω=2πf

Then, ωk=fλ=v

⇒v=48=0.5 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A trend which is common to elements of both the group 1st and group 17th, on going from top to bottom

Options:
A Boiling point increases
B Electorn affinity increases
C Oxidising power increases
D Ionization energy decreses.

The correct answer is:
Ionization energy decreses.

Solution:
A trend which is common to elements of both the group 1st and group 17th, on going from top to bottom is that the ionization energy decreases since the atomic size increases.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3) . Then the centroid of this triangle is:

Options:
A 13,1
B 1, 73
C 13, 2
D 13, 53

The correct answer is:
13, 2

Solution:
Let co-ordinates of B=x1,y1
∴ x1+12= -1 & y1+2 2=1
⇒ x1=-3, y1=0
∴B=(-3, 0)
Let co-ordinates of C=x2,y2
∴ x2+12=2 & y2+22=3
∴ C=3, 4
Let centroid G=h,k
h=1+x1+x23=1-3+33=13
k=2+y1+y23=2+0+43=2
∴ Centroid, G=13,2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A triangle has two of its vertices at 0,1 and 2,2 in the Cartesian plane. Its third vertex lies on the x-axis. If the area of the triangle is 2 square units, then the sum of the square of the possible values of abscissa of the third vertex is

Options:
A 40
B 10
C 52
D 61

The correct answer is:
40

Solution:
Let, the third vertex be a,0

Area of triangle, Δ=12011221a01=2

∴001211a-11=4
⇒a+2=4

i.e. a=2 or a=-6
Hence, the sum of the square of the possible values of abscissa =40

Topic: Mathematics, Straight Lines, JEE Main
A triangle with vertices (4, 0), (-1, -1), (3, 5) is

Options:
A
isosceles and right angled

B
isosceles but not right angled

C
right angled but not isosceles

D
neither right angled nor isoceles

The correct answer is:
isosceles and right angled

Solution:
$A B=\sqrt{(4+1)^2+(0+1)^2}=\sqrt{26} ; B C=\sqrt{(3+1)^2+(5+1)^2}=\sqrt{52}$
$C A=\sqrt{(4-3)^2+(0-5)^2}=\sqrt{26} ;$ So, in isosceles triangle $A B=C A$
For right angled triangle $\mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2$
So, here $B C=\sqrt{52}$ or $B C^2=52$ or $(\sqrt{26})^2+(\sqrt{26})^2=52$
So, given triangle is right angled and also isosceles

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2002
A triangular lamina of area $A$ and height $h$ is immersed in a liquid of density ${ }^\rho$ in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is

Options:
A $\frac{1}{2}$ Apgh
B $\frac{1}{3} A \rho g h$
C $\frac{1}{6} A \rho g h$
D $\frac{2}{3} A \rho g h$

The correct answer is:
$\frac{1}{3} A \rho g h$

Solution:
Thrust on lamina $=$ pressure at centroid $\times$ Area
$=\frac{h \rho g}{3} \times A=\frac{1}{3} A \rho g h$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is

Options:
A
$\frac{3}{2} x^2$

B
$\sqrt{\frac{x^3}{8}}$

C
$\frac{1}{2} x^2$

D
$\pi \mathrm{x}^2$

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

Solution:
Area $=\frac{1}{2} x^2 \sin \theta$
$A_{\max }=\frac{1}{2} x^2\left(\right.$ at $\left.\sin \theta=1, \quad \theta=\frac{\pi}{2}\right)$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2006
A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length \( x \). The maximum area enclosed by the park is:

Options:
A \( \frac{3}{2} x^{2} \)
B \( \sqrt{\frac{x^{3}}{8}} \)
C \( \pi x^{2} \)
D \( \frac{1}{2} x^{2} \)

The correct answer is:
\( \frac{1}{2} x^{2} \)

Solution:
AT = x sin α and BT = x cos α .

Area of triangle ABC=12×base×height
=122BTAT
=122x2cos α sin α=12x2sin2α

Now, -1≤sin2α≤1 ⇒-x22≤x2sin2α2≤x22
∴ Maximum area of Δ ABC = 1 2 x 2

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
A triangular plate is shown. A force F→=4i^-3j^ is applied at point P. The torque at point P with respect to point O and Q are:

Options:
A -15-203,15-203
B 15+203,15-203
C 15-203,15+203
D -15+203,15+203

The correct answer is:
-15-203,15-203

Solution:
F→=4i^-3j^

r→1=5i^+53j^ and r→2=-5i^+53j^

Torque about 'O'

τ→O=r→1×F→=-15-203k^

Torque about 'Q'

τ→Q=r→2×F→=15-203k^

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A triangular prism of mass \( M \) with a block of mass \( m \) placed on it, is released from rest on a smooth inclined plane of
inclination \( \theta \). The block does not slip on the prism. Then

Options:
A The acceleration of the prism is \( g \cos \theta \).
B The acceleration of the prism is \( g \tan \theta \).
C The minimum coefficient of friction between the block and the prism is \( \mu_{\min }=\cot \theta \).
D The minimum coefficient of friction between the block and the prism is \( \mu_{\min }=\tan \theta \).

The correct answer is:
The minimum coefficient of friction between the block and the prism is \( \mu_{\min }=\tan \theta \).

Solution:
As the small block does not slip over the prism,
We can consider the mass of small block into prism.
Where, N is normal reaction
and here (m + M)g sin θ is the only force in the line of inclination acting downwards

Fnet=ma⇒m+Mgsinθ=m+Ma⇒a=gsinθ
∴ g sin θ will be its acceleration
∴ option 1 and 2 are wrong

Now FBD of block with respect to prism
Where,
mg sin θ is the pseudo force
f is force of friction
N is normal reaction
Now writing equations of forces
⇒ N + mg sin θ sin θ = mg
⇒ N = mg - mg sin2θ ...(1)
and f = mg sin θ cos θ
f = μN
⇒μ(mg - mg sin2θ)=mg sin θ cos θ
⇒ μ(1 - sin2θ) = sin θ cos θ ⇒ μ cos2θ = sin θ cos θ
⇒ μ = tan θ
⇒ μmin = tan θ

Topic: Physics, Laws of Motion, JEE Main
A trolley of mass $300 \mathrm{~kg}$ carrying a sand bag of $25 \mathrm{~kg}$ is moving uniformly with a speed of $27 \mathrm{~km} / \mathrm{h}$ on a frictionless track. After a while, sand starts leaking out of a hole on the trolley's floor at the rate of $0.05 \mathrm{~kg} /$ s. What is the speed of the trolley after the entire sand bag is emptied?

Options:

None

Solution:
Since the trolley carrying the sand bag is moving uniformly, so the external force on the system is zero. When the sand leaks out, it doesn't apply any external force on the trolley. So the speed of the trolley doesn't change i.e it remains $27 \mathrm{kmh}^{-1}$

Topic: Physics, Work Power Energy
A truck accelerates from speed $v$ to $2 v$. Work done in during this is

Options:
A three times as the work done in accelerating it from rest to $v$
B same as the work done in accelerating it from rest to $\mathrm{V}$
C four times as the work done in accelerating it from rest to $v$
D less than the work done in accelerating it from rest to $v$

The correct answer is:
three times as the work done in accelerating it from rest to $v$

Solution:
Work done to accelerate from 0 to $v$ is, $W_{1}=\frac{1}{2} m v^{2}$ Work done to accelerate from $v$ to $2 v$ is,
$\begin{aligned}
W_{2} &=\frac{1}{2} m(2 v)^{2} \\
\Delta W &=W_{2}-W_{1}=\frac{1}{2} m(2 v)^{2}-\frac{1}{2} m v^{2} \\
\Rightarrow \quad & 4 W_{1}-W_{1}=3 W_{1}
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A truck is moving with a constant velocity of 54 km/h. In which direction (angle with the direction of motion of track) should a stone be projected up with a velocity of 20 m/s, from the floor of the truck, so as to appear at right angles to the truck, for a person standing on the earth ?

Options:
A cos -1 - 3 4
B cos -1 - 1 4
C cos -1 2 3
D cos -1 3 4

The correct answer is:
cos -1 - 3 4

Solution:
Let the stone be projected at an angle α to the direction of motion of truck with a speed of v = 20 m/s.
Since the resultant displacement along horizontal is zero.
The velocity along horizontal = 0
1 5 + 2 0 cos α = 0 ⇒ cos α = - 3 4
⇒ α = cos -1 - 3 4

Topic: Physics, Motion In Two Dimensions, JEE Main
A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck, suddenly moves to the right with an acceleration of a. The pendulum will tilt

Options:
A to the left and the angle of inclination of the pendulum with the vertical is $\sin ^{-1}\left(\frac{g}{a}\right)$
B to the left and angle of inclination of the pendulum with the vertical is $\tan ^{-1}\left(\frac{a}{g}\right)$
C to the left and angle of inclination of the pendulum with the vertical is $\sin ^{1-}\left(\frac{a}{g}\right)$
D to the left and angle of inclination of the pendulum with the vertical is $\tan ^{-1}\left(\frac{g}{a}\right)$

The correct answer is:
to the left and angle of inclination of the pendulum with the vertical is $\tan ^{-1}\left(\frac{a}{g}\right)$

Solution:
As the truck move to the right, so the bob will move to the left due to inertia of rest with acceleration a.
Thus, the given situation can be drawn as
From the above diagram (b) as the string moves by an angle of $\theta$ with the vertical then the tangent angle is
$$
\tan \theta=\frac{\mathrm{ma}}{\mathrm{mg}} \Rightarrow \theta=\tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{g}}\right)
$$

Topic: Physics, Oscillations, JEE Main
A truck starts from rest and accelerates uniformly at $2.0$ $\mathrm{ms}^{-2}$. At $t=10 \mathrm{~s}$, a stone is dropped by a person standing on the top of the truck ( $6 \mathrm{~m}$ high from the ground). What are the (a) velocity and (b) acceleration of the stone at $t=11 s$ ? (Neglect air resistance.)

Options:

None

Solution:
Given, $u=0, t=10 \mathrm{~s}, a=2 \mathrm{~ms}^{-2}$
The velocity of the truck when the stone is dropped, $v=u$ $+a t=0+2 \times 10=20 \mathrm{~ms}^{-1}$.
(a) When the stone is dropped, horizontal velocity, $v_x=v=20 \mathrm{~ms}^{-1} ; v_x$ remains constant as air resistance is neglected
In vertical direction, $a=g=9.8 \mathrm{~ms}^{-2}, u=0$, $t=11-10=1 \mathrm{~s}$.
$v_y=v=u+\mathrm{g} t=0+10 \times 1=10 \mathrm{~m} / \mathrm{s}$.
Resultant velocity,
$$
v=\sqrt{v_x^2+v_y^2}=\sqrt{20^2+10^2}=22.3 \mathrm{~ms}^{-1}
$$
Let $\theta$ be angle which $v_x$ makes with resultant velocity, $34^{\prime}$
(b) When the stone is dropped from the car, horizontal force on the stone is zero. Only acceleration of the stone $=$ acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ in the vertically downward direction. The path of the stone would be parabolic.

Topic: Physics, Laws of Motion
A tube A of radius r is connected to two other tubes B and C with the help of a junction valve. The tubes B and C have radii r2 and r3 and the flow velocity in each is v and 3v respectively. If the flow velocity in tube A is nmv, where n and m are integers, then what is the value of n+m?

Options:

The correct answer is:
19.00

Solution:
The volume flowrate in tube A will be equal to the total volume flowrate through B and C.

πr2vA=πr24v+πr293v

⇒vA=712v

n=7, m=12

⇒n+m=19

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A tube of length 50 cm is filled completely with an incompressible liquid of mass 250 g and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity xF rad s-1. If F be the force exerted by the liquid at the other end then the value of x will be _____ .

Options:

The correct answer is:
4

Solution:
Considering an element of tube of length dx having mass dm at a distance of x from right end as shown in figure below.

Now, force acting of that element is F=∫dmω2x

Integrating the above relation for whole length of tube, we have

F=∫0LmLdxω2x

=mLω2L22 =mω2L2

⇒ω=2mLF =20.25×0.5F

=16F=4F rad s-1

Hence, the value of x=4.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A tube of length l and radius R carries a steady flow of fluid whose density is ρ and velocity η . The fluid flow velocity depends on the distance r from the axis of the tube as v=v01-r2R2. Find the friction force exerted on the tube by the fluid.

Options:
A 4πηlV0
B 2πηlV0
C πηlV0
D 23πηlV0

The correct answer is:
4πηlV0

Solution:
v=v01-r2R2
dvdr= -2v0rR2
f=-Î·Advdr
=Î·A.2v0rR2
=Î·. 2ÏRl.2v0.RR2
=4ÏÎ·lv0

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A tuning fork A of unknown frequency produces 5 beats s-1 with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats s-1. What is the frequency of fork A ?

Options:
A 335 Hz
B 338 Hz
C 345 Hz
D 342 Hz

The correct answer is:
335 Hz

Solution:
Initially beat frequency =5 Hz

so, ρA=340±5=345 Hz, or 335 Hz

after filing frequency increases slightly so, new value of frequency of A>ρA

Now, beat frequency =2 Hz

⇒ new ρA=340±2=342 Hz, or 338 Hz

hence, original frequency of A is ρA=335 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A tuning fork and a sonometer wire were sounded together and produce 4 beats per second. When the length of sonometer wire is $95 \mathrm{~cm}$ or $100 \mathrm{~cm}$, the frequency of the tuning fork is

Options:
A 156 Hz
B 152 Hz
C 148 Hz
D 160 Hz

The correct answer is:
156 Hz

Solution:
Probable frequencies of tuning fork be $n+4$ or $n-4$
Frequency of sonometer wire $n \propto \frac{1}{l}$
$\begin{aligned} & \therefore \frac{n+4}{n-4}=\frac{100}{95} \text { or } 95(n+4)=100(n-4) \\ & \text { or } 95 n+380=100 n-400 \text { or } 5 n=780 \text { or } n=156\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
A tuning fork arrangement (pair) produces 4 beats / sec with one fork of frequency $288 \mathrm{cps}$. A little wax is placed on the unknown fork and it then produces 2 beats /sec. The frequency of the unknown fork is

Options:
A
286 cps

B
292 cps

C
294 cps

D
288 cps

The correct answer is:
292 cps

Solution:
The fact that placing wax decreases the frequency of the unknown fork and also the beat frequency states that the unknown fork is of higher frequency.
$$
n-288=4 \Rightarrow n=292 \mathrm{cps}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2002
A tuning fork having a frequency of 170 Hz is vibrating just above a cylindrical tube. The height of the tube is 110 cm. Water is slowly poured in it. What is the minimum height (in cm) of water column required for resonance to occur? (Velocity of sound in air = 340 m s-1)

Options:

The correct answer is:
60

Solution:
Î»=2 m
Resonance will occur at Î»4, 3Î»4, 5Î»4
â 50, 150, 250 cm
Minimum height of water =110-Î»4=110-50=60 cm

Topic: Physics, Waves and Sound, JEE Main
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be _____cm. (Take speed of sound in air as 340 m s-1 )

Options:

The correct answer is:
34

Solution:
For closed organ pipe

f=(2n+1)v4l

for minimum length, n=0

f0=v4l

⇒l=v4f0

=3404×250=34 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A tuning fork of known frequency $256 \mathrm{~Hz}$ makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options:
A
$256+2 \mathrm{~Hz}$

B
$256-2 \mathrm{~Hz}$

C
$256-5 \mathrm{~Hz}$

D
$256+5 \mathrm{~Hz}$

The correct answer is:
$256+2 \mathrm{~Hz}$

Solution:
No Solution Available

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2003
A tuning fork produces waves in a medium. If the temperature of the medium changes, then which of the following will change

Options:
A Amplitude
B Frequency
C Wavelength
D Time-period

The correct answer is:
Wavelength

Solution:
If the temperature changes then velocity of wave and its wavelength changes. Frequency amplitude and time period remains constant.

Topic: Physics, Waves and Sound, JEE Main
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ Hz.

Options:

The correct answer is:
6

Solution:
Frequency of sonometer wire is given by, f=12 LTμ

Therefore, f1=126μ f2=1254μ

Hence, f1f2=13

Also it is given that, f2-f1=12

So, f1=6 Hz.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A tuning fork vibrates with a frequency of 256 Hz. Taking the speed of sound to be 345.6 m s-1 in the air, find the wavelength and the distance, which the sound travels during the time, fork makes 60 vibrations.

Options:
A 1.35 m, 81 m

B 1.40 m, 84 m

C 1.45 m, 87 m

D 1.55 m, 93 m

The correct answer is:
1.35 m, 81 m

Solution:
Here, f= 256 Hz;  υ = 345.6 m s−1
∴ λ =υf=345.6256=1.35 m
In one cycle, sound advances by distance λ
Therefore, the required distance travelled by sound,
S = 60. λ =81 m

Topic: Physics, Waves and Sound, JEE Main
A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (speed of sound in air is 340 m s-1)

Options:
A 220 cm
B 200 cm
C 190 cm
D 180 cm

The correct answer is:
200 cm

Solution:
Organ pipe will have frequency either 255 or 257 Hz. Using 255 Hz,
255= 3v2l,⇒l=3×3402×255 m⇒l=200 cm. Using 257 Hz,257= 3v2l'⇒l'=3×3402×257 m=198 cm.

Topic: Physics, Waves and Sound, JEE Main
A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (speed of sound in air is 340 m s-1)

Options:
A 220 cm
B 200 cm
C 190 cm
D 180 cm

The correct answer is:
200 cm

Solution:
Organ pipe will have frequency either 255 or 257 Hz. Using 255 Hz,
255= 3v2l,⇒l=3×3402×255 m⇒l=200 cm. Using 257 Hz,257= 3v2l'⇒l'=3×3402×257 m=198 cm.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2018 (15 Apr)
A tuning fork vibrates with frequency $256 \mathrm{~Hz}$ and gives one best per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound of air is $340 \mathrm{~ms}^{-1}$ )

Options:
A
(a) $190 \mathrm{~cm}$

B
$180 \mathrm{~cm}$

C
$220 \mathrm{~cm}$

D
$200 \mathrm{~cm}$

The correct answer is:
$200 \mathrm{~cm}$

Solution:
According to question, tuning fork gives 1 beat/second with (N) $3^{\text {rd }}$ normal mode. Therefore, organ pipe will have frequency $(256 \pm 1) \mathrm{Hz}$. In open organ pipe, frequency $\mathrm{n}=\frac{\mathrm{NV}}{2 \ell}$
or, $255=\frac{3 \times 340}{2 \times \ell} \Rightarrow \ell=2 \mathrm{~m}=200 \mathrm{~cm}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period

Options:
A 42 minutes
B 1 day
C 1 hour
D 84.6 minutes

The correct answer is:
84.6 minutes

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}=2 \pi \sqrt{\frac{64 \times 10^{6}}{9.8}}$

$=2 \times \frac{22}{7} \times \frac{8 \times 10^{3}}{7 \times \sqrt{2}}$

$=\frac{\sqrt{2} \times 22 \times 8 \times 1000}{49 \times 60} \min =84.6 \mathrm{~min}$

Topic: Physics, Oscillations, JEE Main
A tunning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is _____ cm.
(Velocity of sound in air is 340 ms-1)

Options:

The correct answer is:
50

Solution:
We know that, v=λf, then, λ=vf=340 m s-1340 Hz=340 m s-1340 s-1=1 m

First resonating length, l1=λ4=1004=25 cm

Second resonating length, l2=3λ4=3×1004=75 cm

Third resonating length, l3=5λ4=5×1004=125 cm

So third resonance is not possible since the length of the tube is 125 cm

Thus, minimum height of water necessary for resonance =125−75=50 cm

Topic: Physics, Waves and Sound, JEE Main
A tunning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is _____ cm.
(Velocity of sound in air is 340 ms-1)

Options:

The correct answer is:
50

Solution:
We know that, v=λf, then, λ=vf=340 m s-1340 Hz=340 m s-1340 s-1=1 m

First resonating length, l1=λ4=1004=25 cm

Second resonating length, l2=3λ4=3×1004=75 cm

Third resonating length, l3=5λ4=5×1004=125 cm

So third resonance is not possible since the length of the tube is 125 cm

Thus, minimum height of water necessary for resonance =125−75=50 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A tunning fork of frequency 480 Hz is used in an experiment for measuring speed of sound v in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1=30 cm and l2=70  cm. Then, ν is equal to:

Options:
A 384 m s-1
B 332 m s-1
C 338 m s-1
D 379 m s-1

The correct answer is:
384 m s-1

Solution:
Speed of sound, v=2f l2-l1
=2×4800.70-0.30
=384 m/s

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A T.V. tower has a height of 100 m. The population covered by T.V. broadcast is, if the average population density around the tower is 1000 people km-2.

Options:
A 39.5× 10 5  units
B 19.5× 10 5  units
C 29.5× 10 5  units
D 9.4× 10 5  units

The correct answer is:
39.5× 10 5  units

Solution:
d= 2Rh
N=π d 2 σ=2πRhσ
=2×3.14×6400×0.1×1000
=2×3.14×6.4× 10 5 =39.5× 10 5  units

Topic: Physics, Communication System, JEE Main
A TV tower has a height of 150 m. The area of the region covered by the TV broadcast is (radius of earth=6.4Ã106 m)

Options:
A 9.6ÏÃ108Â km2
B 19.2ÏÃ108Â km2
C 19.2ÏÃ107Â km2
D 1.92π×103 km2

The correct answer is:
1.92π×103 km2

Solution:
Area of region covered =πd2=π(2hR)
=π×2×150×6.4×106
=1.92 π×109 m2=1.92 π×103km2

Topic: Physics, Communication System, JEE Main
A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given: radius of earth =6.4×106 m).

Options:
A 48Â km
B 40Â km
C 80 km
D 65 km

The correct answer is:
65 km

Solution:
D=2hTR+2hRR
D=2Ã140Ã64Ã105+2Ã40Ã64Ã105
D=8Ã103[28+8]
D=8× 10 3 [ 2 7 +2 2 ]
D=16× 10 3 [ 7+ 2 ]
D=16Ã1032.6+1.4
=64000m=64kmâ65km

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (10 Jan Shift 1)