datavorous/entrance-exam-dataset huggingface dataset:
When a metal is electroplated with silver (Ag)

Options:
A The metal is the anode
B Ag metal is the cathode
C The solution contains Ag+ ions
D The reaction at the anode is Ag+ + e–→ Ag

The correct answer is:
The solution contains Ag+ ions

Solution:
For electroplating of metal with silver
Cathode → Metal
Anode → Ag
Solution → Electrolyte containing Ag+

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
When a metal is to be extracted from its ore, if the gangue associated with the ore is silica, then

Options:
A An acidic flux is needed
B A basic flux is needed
C Both acidic and basic flux are needed
D Neither of them is needed

The correct answer is:
A basic flux is needed

Solution:
Correct Option is 2 A basic flux is needed
\(\underset{\text { Acidic impurity }}{\mathrm{SiO}_2}+\underset{\text { Basic flux }}{\mathrm{CaO}} \rightarrow \underset{\text { Slag }}{\mathrm{CaSiO}_3}\)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
When a metallic wire is held under a tensile force F1, the length of the wire is L1. If the force is changed to F2 , the length of the wire is L2. The original length L of the wire is

Options:
A L1F1+L2F2F1+F2
B L2-L1F1+F2
C F1L2-F2L1F1-F2
D F1L1-F2L2F1-F2

The correct answer is:
F1L2-F2L1F1-F2

Solution:
F∝dl
F1∝L1-l⇒ (1)
F2∝L2-l⇒ (2)
F1F2=L1-lL2-l
F1L2-F1l=F2L1-F2l
F1-F2l=F1L2-F2L1
l=F1L2-F2L1F1-F2

Topic: Physics, Mechanical Properties of Solids, JEE Main
When a missile is fired from a ship, the probability that it is intercepted is 13 and the probability that the missile hits the target, given that it is not intercepted, is 34. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:

Options:
A 38
B 127
C 18
D 34

The correct answer is:
18

Solution:
Probability of not getting intercepted =23

Probability of missile hitting target =34

∴ Probability that all 3 hit the target =23×343=18

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (25 Feb Shift 1)
When a mixture of methane and oxygen is passed through heated molybdenum oxide, the main product formed is

Options:
A methanol
B methanal
C methanoic acid
D ethanal

The correct answer is:
methanal

Solution:
When mixture of methane and oxygen is passed through heated molybdenum oxide, the main product formed is methanal.

Topic: Chemistry, Hydrocarbons, COMEDK, COMEDK 2016
When a neuron is in resting state i.e. not conducting any impulse, the axonal membrane is

Options:
A comparatively more permeable to Na+ ions and nearly impermeable to K+ ions
B equally permeable to both Na+ and K+ ions
C impermeable to K+ ions and Na+ ions
D comparatively more permeable to K+ ions and nearly impermeable to Na+ ions

The correct answer is:
comparatively more permeable to K+ ions and nearly impermeable to Na+ ions

Solution:
The main reason for conduction through the nerve is that it is polarised. This means the neuron is oppositely charged on the outside and inside the axoplasm. If a neuron is not conducting it still has charges and such a neuron is polarised or resting neuron. Such a neuron is positively charged outside and negatively charged inside the axoplasm. Positive charge is due to high concentration of sodium ions and negatively charged is due to high concentration of potassium and negatively charged proteins. The proteins are impermeable and hence only potassium ions can move from inside to out when the impulse will arrive. Thus the axonal membrane is more permeable to potassium ions and nearly impermeable to sodium ions.

Topic: Biology, Neural Control and Coordination, NEET
When a neutral atom is converted into cation, there is

Options:
A Decrease in the atomic number
B An increase in the atomic number
C A decrease in size
D An increase in size

The correct answer is:
A decrease in size

Solution:
During the conversion of neutral atom to cation size decreases because after removal one $\mathrm{e}^{-}$or more
(i) Nuclear charge per electron increases.
(ii) Outermost shell is completely removed.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom

Options:
A do not change for any type of radioactivity
B change for $\alpha$ and $\beta$ radioactivity but not for $\gamma$-
C change for $\alpha$-radioactivity but not for others
D change for $\beta$-radioactivity but not for others

The correct answer is:
change for $\alpha$ and $\beta$ radioactivity but not for $\gamma$-

Solution:
Alpha and beta decay eject charged particles from the nucleus, altering its charge and its interactions with electrons, which in turn alter the electronic energy levels.
Whereas Gamma radiation leaves the nucleus charge unchanged, and the electronic levels are not affected,

Topic: Physics, Nuclear Physics, JEE Main
When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom

Options:
A do not change for any type of radioactivity
B change for \( \alpha \) and \( \beta \) radioactivity but not for \( \gamma \)-radioactivity.
C change for \( \alpha \)-radioactivity but not for others.
D change for \( \beta \)-radioactivity but not for others.

The correct answer is:
change for \( \alpha \) and \( \beta \) radioactivity but not for \( \gamma \)-radioactivity.

Solution:
β- Particles carries one unit of negative charge, and α-particle carries 2 units of positive charge, and γ-particle carries no charge. So the electronic energy level of the atom changes in emission of α and β particle, but not in γ decay.

Topic: Physics, Nuclear Physics, JEE Main
When a oscillator completes 100 oscillations its amplitude reduces to $\frac{1}{3}$ of the initial value. What will be its amplitude when it completes 200 oscillations?

Options:
A $\frac{1}{8}$
B $\frac{2}{3}$
C $\frac{1}{6}$
D $\frac{1}{9}$

The correct answer is:
$\frac{1}{9}$

Solution:
Let $a_0$ be the initial amplitude and $b$ be the damping constant
The used formula is $a=a_0 e^{-b t}$
In Ist case : $t=100 T$ and $a=\frac{a_0}{3}$
$\therefore \frac{a_0}{3}=a_0 e^{-b(100 T)}$
$\begin{aligned}
& \Rightarrow e^{-b \times 100 T}=\frac{1}{3} \\
& \text {In 2nd case : } t=200 T \\
& a=a_0 e^{-200 b T} \\
& \Rightarrow a=a_0\left(e^{-100 b T}\right)^2 \\
& \Rightarrow a=a_0\left(\frac{1}{3}\right)^2 \\
& \Rightarrow a=\frac{a_0}{9}
\end{aligned}$

Topic: Physics, Oscillations, JEE Main
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:

Options:
A straight line
B elliptical
C circular
D parabolic

The correct answer is:
elliptical

Solution:
For a particle executing SHM,

x=Asinωt+ϕ

v=ωAcosωt+ϕ

⇒v2ω2A2+x2 A2=1⇒ equation of ellipse between v and x

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (24 Feb Shift 2)
When a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be

Options:
A Circular
B Ellipitical
C Sinusoidal
D Straight line

The correct answer is:
Ellipitical

Solution:
The displacement equation of a particle is given by

x=Asinωt+ϕ⇒sinωt+ϕ=xA

The velocity equation is given by,

v=ωAcosωt+ϕ⇒cosωt+ϕ=vωA

Since, sin2θ+cos2θ=1

⇒x2A2+v2ω2A2=1

The above equation represents an ellipse.

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (26 Jul Shift 1)
When a particle executing SHM oscillates with a frequency $v$, then the kinetic energy of the particle

Options:
A changes periodically with a frequency of $v$
B changes periodically with a frequency of $2 \mathrm{v}$
C changes periodically with a frequency of $\mathrm{v} / 2$
D remains constant

The correct answer is:
changes periodically with a frequency of $2 \mathrm{v}$

Solution:
We know that
Kinetic energy $K=\frac{1}{2} m u^{2}$
where $u=\frac{d y}{d t}=\omega a \cos \omega t$
So. $K=\frac{1}{2} m \omega^{2} a^{2} \cos ^{2} \omega t$
Hence, kinetic energy varies periodically with double the frequency of SHM. So, when a particle executing SHM oscillates with a frequency v, then the kinetic energy of particle changes periodically with a frequency of $2 v$.

Topic: Physics, Oscillations, JEE Main
When a particle executing SHM oscillates with a frequency $v$, then the kinetic energy of the particle

Options:
A changes periodically with a frequency of $v$
B changes periodically with a frequency of $2 \mathrm{v}$
C changes periodically with a frequency of $\mathrm{v} / 2$
D remains constant

The correct answer is:
changes periodically with a frequency of $2 \mathrm{v}$

Solution:
We know that
Kinetic energy $K=\frac{1}{2} m u^{2}$
where $u=\frac{d y}{d t}=\omega a \cos \omega t$
So. $K=\frac{1}{2} m \omega^{2} a^{2} \cos ^{2} \omega t$
Hence, kinetic energy varies periodically with double the frequency of SHM. So, when a particle executing SHM oscillates with a frequency v, then the kinetic energy of particle changes periodically with a frequency of $2 v$.

Topic: Physics, Oscillations, WBJEE, WBJEE 2014
When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion, is described by yt=y0sin2ωt, where 'y' is measured from the lower end of upstretched spring. Then ω is :

Options:
A 12gy0
B gy0
C g2y0
D 2gy0

The correct answer is:
g2y0

Solution:
y=y0sin2ωt

y=y02(1-cot2ωt)

y-y02=-y02 cos2ωt

Y=A cos2ωt

2ω=km

maximum displacemnet =y0=mgk

y0×(2ω)2=2g

ω=g2y0

Topic: Physics, Oscillations, JEE Main, JEE Main 2020 (06 Sep Shift 2)
When a person breathes out normally, then the amount of air which remains in the lung after a normal expiration is

Options:
A expiratory reserve volume.
B expiratory capacity.
C functional residual capacity (FRC).
D residual volume.

The correct answer is:
functional residual capacity (FRC).

Solution:
The additional volume of air a person can expire by forcible expiration is called Expiratory Reserve Volume (ERV). Average ERV is 1000 mL to 1100 mL. Inspiratory Capacity (IC) is the total volume of air a person inspires post a normal expiration. Inspiratory capacity includes tidal volume and inspiratory reserve volume (TV+IRV). The total volume of air expired after a normal inspiration is Expiratory Capacity (EC). Tidal volume and expiratory reserve volume (TV+ERV) are both included in EC. Functional Residual Capacity (FRC) is the volume of air remaining in the lungs after normal expiration. Includes both ERV+RV.

Topic: Biology, Breathing and Exchange of Gases, NEET
When a photon is absorbed, do all electrons come out as photoelectrons?

Options:
A Yes, all electrons are released as photoelectrons.
B No, most electrons are scattered into the metal.
C No, all electrons remain within the metal.
D Yes, but only a few electrons are released as photoelectrons.

The correct answer is:
No, most electrons are scattered into the metal.

Solution:
Correct Option is : (B)
No, most electrons are scattered into the metal.

Topic: Chemistry, Dual Nature of Matter, NEET
When a piece of metal is illuminated by monochromatic light of wavelength Î», then stopping potential is 3Vs. When the same surface is illuminated by the light of wavelength 2Î», then stopping potential becomes Vs. The value of threshold wavelength for photoelectric emission will be

Options:
A 4λ
B 8λ
C 43λ
D 6λ

The correct answer is:
4λ

Solution:
According to Einstein's photoelectric equation
eV=hc1Î»-1Î»0
Ist case 3eVs=hc1Î»-1Î»0 ...(i)
IInd case eVs=hc [12Î»-1Î»0] ...(ii)
Dividing Eq. (i)by Eq. (ii), we get
Î»0=4Î»

Topic: Physics, Dual Nature of Matter, JEE Main
When a plane polarised light is passed through an analyser and analyser is rotated through $90^{\circ}$, the intensity of the emerging light

Options:
A Varies between a maximum and minimum
B Becomes zero
C Does not vary
D Varies between a maximum and zero

The correct answer is:
Varies between a maximum and zero

Solution:
Varies between a maximum and zero

Topic: Physics, Wave Optics, JEE Main
When a planet revolves around the Sun, in general, for the planet

Options:
A linear momentum and linear velocity are constant.
B linear momentum and areal velocity are constant.
C kinetic and potential energy of the planet are constant.
D angular momentum about the Sun and areal velocity of the planet are constant.

The correct answer is:
angular momentum about the Sun and areal velocity of the planet are constant.

Solution:
When a planet revolves around the sun, net
external torque is zero. Therefore its angular momentum is conserved. As a consequence, its areal velotity remains the same/constant.

Topic: Physics, Gravitation, KCET, KCET 2023
When a plot between logk and 1/T is plotted we get the graph as shown.

What is the value of slope in the graph?

Options:
A EaRT
B -Ea2.303R
C -Ea2.303RTlog⁡A
D -Ea2.303RT

The correct answer is:
-Ea2.303R

Solution:
log⁡k=log⁡A-Ea2.303R1T
log⁡k=-Ea2.303RT+log⁡A
Slope =-Ea2.303R

Topic: Chemistry, Chemical Kinetics, JEE Main
When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :

Options:
A 60°
B 30°
C 90°
D 45°

The correct answer is:
45°

Solution:
Let I0 be intensity of unpolarised light incident on first polaroid.
I1= Intensity of light transmitted from 1st polaroid
=I02
If θ be the angle between 1st and 2nd polaroid and ϕ be the angle between 2nd and 3rd polaroid, then

θ+ϕ=90° ...1

If I2= Intensity from 2nd polaroid,

I2=I1cos2θ=I02cos2θ

If I3= Intensity from 3rd polaroid

I3=I2cos2ϕ=I1cos2θcos2ϕ=I02cos2θcos290°-θ=I02cos2θsin2θ=I022sinθcosθ22=I08sin22θ

Now, I3 will be maximum when

sin2θ=1

⇒2θ=90°

⇒θ=45°

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (27 Jan Shift 2)
When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:

Options:
A 14 W
B 12 W
C 2 W
D 4 W

The correct answer is:
4 W

Solution:
For first case:

V2R=W ...(i)

Now, equivalent resistance of the combination will be R2×R2R2+R2=R4

For second case:

W'=V2R4 ...(ii)

From (i) & (ii), we get
W'=4 W

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (30 Jan Shift 2)
When a pressure of 100 atmosphere is applied on a spherical ball, then its volume reduces to $0.01 \%$. The bulk modulus of the material of the rubber in dyne$/ \mathrm{cm}^2$ is

Options:
A $10 \times 10^{12}$
B $100 \times 10^{12}$
C $1 \times 10^{12}$
D $20 \times 10^{12}$

The correct answer is:
$1 \times 10^{12}$

Solution:
$K=\frac{100}{0.01 / 100}=10^6 \mathrm{~atm}=10^{11} \mathrm{~N} / \mathrm{m}^2=10^{12} \mathrm{dyne} / \mathrm{cm}^2$

Topic: Physics, Mechanical Properties of Solids, JEE Main
When a proton, anti-proton annihilate, the energy released is

Options:
A 3.0 Ã 10-10 J
B 28.8Â ÃÂ 10-10Â J
C 6Â ÃÂ 10-10Â J
D 9×1 0 - 1 0 J

The correct answer is:
3.0 Ã 10-10 J

Solution:
Assuming proton and antiproton are at rest and let them mutually annihilate.
Then the energy per photon:
Einstein's mass energy relation , the energy released is
ΔE=Δmc2
Where, c is the speed of light and Δm is mass.
Given,
Δm=1.67 × 10-27 kg , c = 3 × 108 m s-1
∴ ΔE=1.67 × 10-27 × 3 × 1082
ΔE≈ 1.5 × 10-10 J per photon.
Energy for two photons will be 3.0 ×10-10 J

Topic: Physics, Nuclear Physics, JEE Main
When a resistance of 5 Ω is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA, however when 1050 Ω resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is_________Ω.

Options:

The correct answer is:
50

Solution:
The formula to calculate the maximum current through the galvanometer, when a shunt resistance r1 is connected is given byiGmax=r1r1+RGimax ...1

When a series resistance r2 is connected, the potential difference across the galvanometer is given by

V=iGmaxRG+r2 ...2

From equations (1) and (2), it can be written that

V=r1r1+RGimaxRG+r2⇒Vr1+VRG=imaxr1RG+imaxr1r2⇒V-imaxr1RG=imaxr1r2-Vr1⇒RG=imaxr1r2-Vr1V-imaxr1 ...3

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

RG=0.250 A×5 Ω×1050 Ω-25 V×5 Ω25 V-0.250 A×5 Ω=50 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (13 Apr Shift 1)
When a rubber ball is taken to a depth of _______ m in deep sea, its volume decreases by 0.5%
(The bulk modulus of rubber =9.8×108 N m-2 Density of sea water =103 kg m-3,g=9.8 m s-2)

Options:

The correct answer is:
500

Solution:
-dVV=0.5100=0.005
B=-dPdVV=ρgh-dVV
9.8×108=103×9.8×h5×10-3⇒h=500 m

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (31 Aug Shift 1)
When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude
F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is :

Options:
A aL 2 + bL 3
B 1 2 aL 2 + bL 3
C aL 2 2 + bL 3 3
D 1 2 aL 2 2 + bL 3 3

The correct answer is:
aL 2 2 + bL 3 3

Solution:
W = &Integral; F d x
= &Integral; ax + bx 2 dx
= a 2 x 2 0 L + b 3 x 3 0 L
= a L 2 2 + b L 3 3

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2014 (06 Apr)
When a satellite moves around the earth in a certain orbit, the quantity which remains constant is :

Options:
A Angular velocity
B Kinetic energy
C Areal velocity
D Potential energy

The correct answer is:
Areal velocity

Solution:
From Kepler's law aerial velocity remains constant

Topic: Physics, Gravitation, NEET
When a semiconducting device is connected in series with a battery and a resistance, a current is found to flow in the circuit. If however, the polarity of the battery is reversed, practically no current flows in the circuit. The device may be

Options:
A a $p$ -type semiconductor
B a $n$ -type semiconductor
C an intrinsic semiconductor
D a $p-n$ junction

The correct answer is:
a $p-n$ junction

Solution:
When the battery and resistance is connected with $p n$ diode,

In case of forward bias that $p$ is connected to + ve terminal of battery, current flows. If one reverse the polarity i.e., $p$ -end is made negative, no current flows in semiconductor diode.

Topic: Physics, Semiconductors, JEE Main
When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

Options:
A Is less
B Is same
C Is more
D May be less or more depending on the nature of the material introduced

The correct answer is:
Is more

Solution:
When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to $Q C V$, the charge will increase.

Topic: Physics, Capacitance, JEE Main
When a small amount of $\mathrm{KMnO}_4$ is added to concentrated $\mathrm{H}_2 \mathrm{SO}_4$, a green oily compound is obtained which is highly explosive in nature. Compound may be :

Options:
A
$\mathrm{MnSO}_4$

B
$\mathrm{Mn}_2 \mathrm{O}_7$

C
$\mathrm{MnO}_2$

D
$\mathrm{Mn}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{Mn}_2 \mathrm{O}_7$

Solution:
$\mathrm{KMnO}_4$ reacts with $\mathrm{H}_2 \mathrm{SO}_4$ to form $\mathrm{Mn}_2 \mathrm{O}_7$ which is highly explosive substance.
$$
2 \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4 \underset{\mathrm{K}_2 \mathrm{SO}_4}{\longrightarrow}+\mathrm{Mn}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{O}
$$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2013 (23 Apr Online)
When a soap bubble is charged?

Options:
A Its radius increases
B Its radius decreases
C The radius remains the same
D Its radius may increase or decrease

The correct answer is:
Its radius increases

Solution:
When a soap bubble is charged, then the bubble expands because the charged particles uniformly distributed on it causes them to repel each other due to the electrostatic force. Hence, radius of soap bubble increases.

Topic: Physics, Mechanical Properties of Fluids, KCET, KCET 2020
When a solid is converted into liquid, entropy

Options:
A Becomes zero
B Remains the same
C Decreases
D Increases

The correct answer is:
Increases

Solution:
The correct option is $\mathbf{4}$ Increases
when solid is converted liquid, the particles are relatively more free to mov and randomness increases. Liquid state has more accessible microstates, sc more distribution of energy and hence entropy increases.
Theory:
Entropy ( $(\mathrm{C})$ :
- Entropy is defined as a measure of randomness or disorder of a system
- All-natural processes are spontaneous because of an increase in randomness
Suggest Corrections

Topic: Chemistry, Thermodynamics (C), JEE Main
When a solute is mixed, the water potential of the cell

Options:
A increases
B decreases
C first increases and then decreases
D remains unchanged

The correct answer is:
decreases

Solution:
Water molecules possess kinetic energy. In liquid and gaseous form they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. If some solute is dissolved in pure water, the solution has fewer free water and the concentration of water decreases, reducing its water potential.

Topic: Biology, Transport in Plants, NEET
When a spring is stretched by $2 \mathrm{~cm}$, it stores $100 J$ of energy. If it is stretched further by $2 \mathrm{~cm}$, the stored energy will be increased by

Options:
A 100 J
B 200 J
C 300 J
D 400 J

The correct answer is:
300 J

Solution:
$\begin{aligned} & 100=\frac{1}{2} k x^2 (given) \\ & W=\frac{1}{2} k\left(x_2^2-x_1^2\right)=\frac{1}{2} k\left[(2 x)^2-x^2\right] \\ & =3 \times\left(\frac{1}{2} k x^2\right)=3 \times 100=300 \mathrm{~J}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
When a standard solution of NaOH is left in air for a few hours

Options:
A a precipitate will form
B strength of solution will decrease
C the strength of solution will increase
D the concentration of Na+ ion in solution will remains same

The correct answer is:
strength of solution will decrease

Solution:
NaOH being deliquescent absorb water form atmosphere therefore strength of solution will decrease.

Topic: Chemistry, s Block Elements, NEET
When a stationary wave is formed then its frequency is

Options:
A Same as that of the individual waves
B Twice that of the individual waves
C Half that of the individual waves
D None of the above

The correct answer is:
Same as that of the individual waves

Solution:
If $y_{\text {incident }}=a \sin (\omega t-k x)$ and $y_{\text {stationary }}=a \sin (\omega t) \cos k x$ then it is clear that frequency of both is same $(\omega)$

Topic: Physics, Waves and Sound, JEE Main
When a string is divided into three segments of length $l_{1}, l_{2},$ and $l_{3}$, the fundamental frequencies of these three segments are $v_{1}, v_{2}$ and $v_{3}$ respectively. The original fundamental frequency

(v) of the string is

Options:
A $\quad \sqrt{v}=\sqrt{v_{1}}+\sqrt{v_{2}}+\sqrt{v_{3}}$
B $v=v_{1}+v_{2}+v_{3}$
C $\frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}$.
D $\frac{1}{\sqrt{v}}=\frac{1}{\sqrt{v_{1}}}+\frac{1}{\sqrt{v_{2}}}+\frac{1}{\sqrt{v_{3}}}$

The correct answer is:
$\frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}$.

Solution:
Fundamental frequency is given by

$$

v=\frac{1}{2 l} \sqrt{\frac{T}{\mu}} \Rightarrow v \propto \frac{1}{l} \Rightarrow P \propto \frac{1}{v}

$$

Since, $P$ divided into $l_{1}, l_{2}$ and $l_{3}$ segments Here $l=l_{1}+l_{2}+l_{3}$

So $\quad \frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}$

Topic: Physics, Waves and Sound, JEE Main
When a string is divided into three segments of lengths l1, l2 and l3 the fundamental frequencies of these three segments are Î½1, Î½2 and Î½3 respectively. The original fundamental frequency Î½ of the string is

Options:
A ν=ν1+ν2+ν3
B ν=ν1+ν2+ν3
C 1Î½=1Î½1+1Î½2+1Î½3
D 1ν=1ν1+1ν2+1ν3

The correct answer is:
1ν=1ν1+1ν2+1ν3

Solution:
Let l be the length of the string.
The fundamental frequency is given by
ν=12lT⁡μ
orν∝1l⁡ ( ∵ T and μ are constant )
orν∝kl⁡ ( where k is a constant )
Here, l1=kν1, l2=kν2, l3=kν3 and l=kν
But l=l1+l2+l3
∴ 1ν=1ν1+1ν2+1ν3

Topic: Physics, Oscillations, JEE Main