NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Have you seen a swing going back and forth or a ball bouncing? These are oscillations. In this chapter, you will study how objects move in repeating patterns, such as a pendulum or the vibration of a guitar string. Oscillations are everywhere, from the ticking to sound waves' vibrations.

Oscillations is a crucial topic in the Class 11 NCERT syllabus, covering fundamental concepts. The Oscillations chapter in Class 11 Physics is crucial as it forms the foundation for wave motion, SHM (Simple Harmonic Motion), and resonance. It is widely applied in pendulums, springs, sound waves, AC circuits, and quantum mechanics. If you struggle to grasp the concept of Oscillations or solve NCERT textbook questions, Careers360 provides detailed NCERT solutions to assist you in your studies.

Through these problems, the students gain a robust concept of wave motion, mechanical vibrations, and oscillatory systems. All of these topics are crucial for CBSE exams, JEE Main, and NEET since oscillation is an important chapter.

NCERT Solutions for Class 11 Physics Chapter 13: Oscillations

Searching for a simple and free solution to crack your Class 11 Physics exams? You can download the Chapter 13 NCERT Solutions PDF. It's full of step-by-step solutions to help you master waves and vibrations and make complex material easy and simple to understand. Ideal for last-minute revisions and to shine in your exam.

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NCERT Solutions For Class 11 Physics Chapter 13 Oscillations: Exercise Solution

Oscillation Class 11 Physics

NCERT answers for oscillations allow students to have a thorough understanding of the various factors that affect an object's motion. Period, frequency, simple harmonic motion, and uniform circular motion are a few of these factors. Chapter 13 of Physics in Class 11 NCERT Solutions offers a thorough explanation of every important concept involved, such as force laws, velocity, forced oscillations, systems that undergo harmonic motion (such as spring- mass systems), etc.

NCERT Solutions for Class 11 Physics Chapter-Wise

Have you ever wondered why gases appear to expand or how they produce pressure? The Kinetic Theory of Gases explains it by imagining that gases consist of small particles moving about rapidly. Scientists such as Boyle, Newton, Maxwell, and Boltzmann contributed significantly to this theory, which explains gas behavior, their response to changes in temperature and pressure, and even how they flow and mix together.

NCERT solutions for important chapter kinetic theory class 11 chapter 12 physics are created by subject matter experts to offer accurate and clear answers to each and every NCERT exercise problem. Students can build a strong concept by practicing these Kinetic Theory Class 11 Physics solutions. Step-by-step solutions make learning convenient and are provided below.

Also Read

• Kinetic Theory Class 11th Notes - Free NCERT Class 11 Physics Chapter 12 Notes Download PDF
• NCERT Exemplar Class 11 Physics Solutions Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12: Download PDF

Free download of Kinetic Theory of Gases Class 11 Solutions PDF for CBSE exams and strengthen your understanding with step-by-step explanations and solved exercises.

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NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Exercise Questions

Q12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V _actual is

$\\V_{actual}=N_{A}\frac{4}{3}\pi \left ( \frac{d}{2} \right )^{3}$ $ \\ $
$V_{actual}=6.023\times 10^{23}\times \frac{4}{3}\pi \times \left ( \frac{3\times 10^{-10}}{2} \right )^{3}\\$
$ V_{actual}=8.51\times 10^{-6}m^{3}\\ $
$V_{actual}=8.51\times 10^{-3}litres$

The volume occupied by a mole of oxygen gas at STP is V _molar = 22.4 litres

$\\\frac{V_{actual}}{V_{molar}}=\frac{8.51\times 10^{-3}}{22.4}\\ $

$\frac{V_{actual}}{V_{molar}}=3.8\times 10^{-4}$

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, $0^{0}C$ ). Show that it is $22.4 Litres$ .

Answer:

As per the ideal gas equation

$\\PV=nRT\\ V=\frac{nRT}{P}$

For one mole of a gas at STP we have

$\\V=\frac{1\times 8.314\times 273}{1.013\times 10^{5}}\\$
$ V=0.0224 m^{3}\\ $

$V=22.4\ litres$

Q12.3 Figure 13.8 shows plot of $PV/T$ versus P for $1.00\times 10^{-3}$ kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: $T_{1}> T_{2}\: \: orT_{1}< T_{2}$ ?
(c) What is the value of $PV/T$ where the curves meet on the y-axis?
(d) If we obtained similar plots for $1.00\times 10^{-3}$ kg of hydrogen, would we get the same
value of $PV/T$ at the point where the curves meet on the y-axis? If not, what mass
of hydrogen yields the same value of $PV/T$ (for low pressure high temperature
region of the plot) ? (Molecular mass of $H_{2}=2.02\mu$ , of $O_{2}=32.0\mu$ ,
$R=8.31J mol^{-1}K^{-1}$ .)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T ₁ is closer to the horizontal line that the one corresponding to T ₂ we conclude T ₁ is greater than T ₂.

(c) As per the ideal gas equation

$\frac{PV}{T}=nR$

The molar mass of oxygen = 32 g

$n=\frac{1}{32}$

R = 8.314

$\\nR=\frac{1}{32}\times 8.314\\ nR=0.256JK^{-1}$

(d) If we obtained similar plots for $1.00\times 10^{-3}$ kg of hydrogen we would not get the same
value of $PV/T$ at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

$m=\frac{PV}{T} \frac{M}{R}={0.256}\times \frac{2}{8.314}=5.48\times10^{-5}Kg$

Q12.4 An oxygen cylinder of volume $30$ litres has an initial gauge pressure of $15$ atm and a temperature of $27^{0}C$ . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11$ atm and its temperature drops to $17^{0}C$ . Estimate the mass of
oxygen taken out of the cylinder ( $R=8.31Jmol^{-1}K^{-1}$ , molecular mass of $O_{2}=32\mu$ ).

Answer:

Initial volume, V ₁ = Volume of Cylinder = 30 l

Initial Pressure P ₁ = 15 atm

Initial Temperature T ₁ = 27 ^o C = 300 K

The initial number of moles n ₁ inside the cylinder is

$\\n_{1}=\frac{P_{1}V_{1}}{RT_{1}}\\ n_{1}=\frac{15\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 300}\\ n_{1}=18.28$

Final volume, V ₂ = Volume of Cylinder = 30 l

Final Pressure P ₂ = 11 atm

Final Temperature T ₂ = 17 ^o C = 290 K

The final number of moles n ₂ inside the cylinder is

$\\n_{2}=\frac{P_{2}V_{2}}{RT_{2}}\\ n_{2}=\frac{11\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 290}\\ n_{2}=13.86$

Moles of oxygen taken out of the cylinder = n ₂ -n ₁ = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

$\\m=4.42\times 32\\ m=141.44g$

Q12.5 An air bubble of volume $1.0 cm^{3}$ rises from the bottom of a lake $40m$ deep at a temperature of $12^{0}C$ . To what volume does it grow when it reaches the surface, which is at a temperature of $35^{0}C$ ?

Answer:

Initial Volume of the bubble, V ₁ = 1.0 cm ³

Initial temperature, T ₁ = 12 ^o C = 273 + 12 = 285 K

The density of water is $\rho _{w}=10^{3}\ kg\ m^{-3}$

Initial Pressure is P ₁

Depth of the bottom of the lake = 40 m

$\\P_{1}=Atmospheric\ Pressure+Pressure\ due\ to\ water\\ P_{1}=P_{atm}+\rho _{w}gh\\ P_{1}=1.013\times 10^{5}+10^{3}\times 9.8\times 40\\ P_{1}=4.93\times 10^{5}Pa$

Final Temperature, T ₂ = 35 ^o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure $=1.013\times 10^{5}Pa$

Let the final volume be V ₂

As the number of moles inside the bubble remains constant, we have

$\\\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\\$

$ V_{2}=\frac{P_{1}T_{2}V_{1}}{P_{2}T_{1}}\\ $

$V_{2}=\frac{4.93\times 10^{5}\times 308\times 1}{1.013\times 10^{5}\times 285}\\ $

$V_{2}=5.26\ cm^{3}$

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0m^{3}$ at a temperature of $27^{0}C$ and $1atm$ pressure.

Answer:

The volume of the room, V = 25.0 m ³

Temperature of the room, T = 27 ^o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

$\\n=\frac{PV}{RT}\\ n=\frac{1.013\times 10^{5}\times 25}{8.314\times 300}\\ n=1015.35$

Avogadro's Number, $N_{A}=6.022\times 10^{23}$

The number of molecules inside the room is N

$\\N=nN_{A}\\ N=1015.35\times 6.022\times 10^{23}\\ N=6.114\times 10^{26}$

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( $27^{0}C$ ), (ii) the temperature on the surface of the Sun ( $6000K$ ), (iii) the temperature of $10$ million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as $\frac{3kT}{2}$ since it is monoatomic

(i)

$\\E=\frac{3kT}{2}\\ $

$E=\frac{3\times 1.38\times 10^{-23}\times 300}{2}\\ $

$E=6.21\times 10^{-21}\ J$

(ii)

$\\E=\frac{3kT}{2}\\ $

$E=\frac{3\times 1.38\times 10^{-23}\times 6000}{2}\\ $

$E=1.242\times 10^{-19}\ J$

(iii)

$\\E=\frac{3kT}{2}\\$

$ E=\frac{3\times 1.38\times 10^{-23}\times10^{7}}{2}\\ $

$E=2.07\times 10^{-16}\ J$

Q12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v _rms the largest?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

$v_{rms}=\sqrt{\frac{3kT}{m}}$

As we can see, v_rms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.

Q12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-20^{0}C$ ? (atomic mass of $Ar=39.9\mu$ , of $He=4.0\mu$ ).

Answer:

As we know root mean square velocity is given as $v_{rms}=\sqrt{\frac{3RT}{M}}$

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at $-20^{0}C$

$\\\sqrt{\frac{3R\times T}{39.9}}=\sqrt{\frac{3R\times 253}{4}}\\ T=2523.7\ K$

Q12.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0 atm$ and temperature $17^{0}C$ . Take the radius of a nitrogen molecule to be roughly $1.0A$ . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of $N_{2}=28.0\mu$ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 ^o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N ₂ = 28 u

The molar mass of N ₂ = 28 g

From the ideal gas equation

$\\PV=nRT\\ $

$\frac{n}{V}=\frac{P}{RT}\\$

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

$\\n'=\frac{N_{A}n}{V}=\frac{6.022\times 10^{23}\times 2\times 1.013\times 10^{5}}{8.314\times (17+273)}\\ $

$n'=5.06\times 10^{25}$

The mean free path $\lambda$ is given as

$\\\lambda =\frac{1}{\sqrt{2}\pi n'd^{2}}\\$

$ \lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times 10^{25}\times (2\times 1\times 10^{-10})^{2}}\\ $

$\lambda =1.11\times 10^{-7}\ m$

The root mean square velocity v _rms is given as

$\\v_{rms}=\sqrt{\frac{3RT}{M}}\\ $

$v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times 10^{-3}}}\\ $

$v_{rms}=508.26\ m\ s^{-1}$

The time between collisions T is given as

$\\T=\frac{1}{Collision\ Frequency}\\ $

$T=\frac{1}{\nu }\\$

$ T=\frac{\lambda }{v_{rms}}\\ $

$T=\frac{1.11\times 10^{-7}}{508.26}\\ $

$T=2.18\times 10^{-10}s$

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

$\\T'=\frac{d}{v_{rms}}\\ $

$T'=\frac{2\times 1\times 10^{-10}}{508.26}\\ $

$T'=3.935\times 10^{-13}s$

The ratio of the average time between collisions to the collision time is

$
\begin{aligned}
\frac{T}{T^{\prime}} & =\frac{2.18 \times 10^{-10}}{3.935 \times 10^{-13}} \\
\frac{T}{T^{\prime}} & =554
\end{aligned}
$
Thus, we can see that the time between collisions is much larger than the collision time.

Additional Questions

Q 1) A metre-long narrow bore held horizontally (and closed at one end) contains a $76 cm$ long mercury thread, which traps a $15 cm$ column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P ₁ = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm ²

The initial volume of the air column, V ₁ = 15x cm ³

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P ₂ = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V ₂ = (24 + y)x cm ³

Since the temperature of the air column does not change

$\\P_{1}V_{1}=P_{2}V_{2}\\ 76\times 15x=y\times (24+y)x\\ 1140=y^{2}+24y\\ y^{2}+24y-1140=0$

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7cm^{3}s^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2cm^{3}s^{-1}$. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M ₁ and M ₂ diffuse with rates R ₁ and R ₂ respectively, their diffusion rates are related by the following equation

$\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}$

In the given question

R ₁ = 28.7 cm ³ s ^-1

R ₂ = 7.2 cm ³ s ^-1

M ₁ = 2 g

$\\M_{2}=M_{1}\left ( \frac{R_{1}}{R_{2}} \right )^{2}\\ M_{2}=2\times \left ( \frac{28.7}{7.2} \right )^{2}\\ M_{2}=31.78g$

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres $n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]$

where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column:

$n_{2}=n_{1}exp\left [ -mg N_{A}(\rho -\rho ^{'})(h_{2}-h_{1})/ (\rho RT)\right ]$
where $\rho$ is the density of the suspended particle, and $\rho ^{'}$ , that of the surrounding medium. [ $N_{A}$ s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

$n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]$ $(i)$

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

$F_{G}=\frac{4}{3}\pi r^{3}\rho g$

The buoyant force acting on them would be

$F_{B}=\frac{4}{3}\pi r^{3}\rho' g$

The net force acting on the particles become

$\\F_{net}=F_{G}-F_{B}\\ F_{net}=\frac{4}{3}\pi r^{3}\rho g-\frac{4}{3}\pi r^{3}\rho' g\\F_{net}=\frac{4}{3}\pi r^{3}g(\rho -\rho ')$

Replacing mg in equation (i) with the above equation, we get

$\\n_{2}=n_{1}exp\left [ -\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1}) /k_{b}T\right ]\\ n_{2}=n_{1}exp\left [ \frac{-\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1})}{\frac{RT}{N_{A}}} \right ]\\ n_{2}=n_{1}exp\left [ \frac{-mgN_{A}(\rho -\rho ')(h_{2}-h_{1})}{RT\rho '} \right ]$

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Answer:

Let one mole of a substance of atomic radius r and density $\rho$ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is $N_{A}=6.022\times 10^{23}$

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3M}{4N_{A}\pi \rho } \right )^{\frac{1}{3}}$

For Carbon

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 12.01}{4\times 6.022\times 10^{23}\times \pi\times 2.22\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.29$ Å.

For gold

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 197.00}{4\times 6.022\times 10^{23}\times \pi\times 19.32\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.59$ Å.

For Nitrogen

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 14.01}{4\times 6.022\times 10^{23}\times \pi\times 1.00\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.77$ Å.

For Lithium

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 6.94}{4\times 6.022\times 10^{23}\times \pi\times 0.53\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.73$ Å.

For Fluorine

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times19.00}{4\times 6.022\times 10^{23}\times \pi\times 1.14\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.88 $ Å.

Class 11 Physics Chapter 12 NCERT solutions help students understand how tiny particles in matter behave. This chapter explains gases and their thermal properties, linking to thermodynamics. Studying it helps in exams and builds a strong foundation in physics.

Kinetic Theory Chapter 12 Physics NCERT Solutions: Concepts and Important Formulas

12.1 Introduction

The kinetic theory explains the behaviour of gases by assuming they consist of a large number of tiny particles (atoms or molecules) in constant random motion. This theory helps derive gas laws and explain properties like pressure, temperature, and specific heat.

12.2 Molecular Nature of Matter

• Matter is made up of tiny molecules that are in continuous motion.
• The molecular interactions vary in solids, liquids, and gases.
• Gases have negligible interatomic forces, allowing them to expand and fill any container.

12.3 Behavior of Gases

• Gases obey different laws like Boyle's Law and Charles's Law at low pressure and high temperature.
• The Ideal Gas Equation is given by:

$
P V=n R T
$

where $\mathbf{P}=$ pressure, $\mathbf{V}=$ volume, $\mathbf{n}=$ number of moles, $\mathbf{R}=$ universal gas constant, $\mathbf{T}=$ temperature (Kelvin).

12.4 Kinetic Theory of an Ideal Gas

• Gas molecules move in random directions, colliding elastically with each other and with the walls of the container.
• Pressure of an Ideal Gas:

$
P=\frac{1}{3} \rho v_{\mathrm{rms}}^2
$

where $\rho=$ density of gas, $\mathrm{vamx}_{\mathrm{am}}=$ root mean square velocity of gas molecules.

• Root Mean Square Speed ( $\mathrm{v}_{\mathrm{mas}}$ ):

$
v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}
$

where $\mathbf{k}=$ Boltzmann constant, $\mathbf{T}=$ absolute temperature, $\mathbf{m}=$ mass of a gas molecule.

12.5 Law of Equipartition of Energy

• Energy is equally distributed among all degrees of freedom in thermal equilibrium.
• Energy per molecule in an ideal gas:

$
E=\frac{f}{2} k T
$

where $f=$ degrees of freedom.

12.6 Specific Heat Capacity

• Monoatomic gas: $C_v=\frac{3}{2} R, \quad C_p=\frac{5}{2} R$
• Diatomic gas: $C_v=\frac{5}{2} R, \quad C_p=\frac{7}{2} R$
• Relation between Cp and Cv :

$
C_p-C_v=R
$

12.7 Mean Free Path

• The mean free path $(\lambda)$ is the average distance a molecule travels before colliding with another molecule.

$
\lambda=\frac{k T}{\sqrt{2} \pi d^2 P}
$

where $\mathbf{d}=$ diameter of the gas molecule, $\mathbf{P}=$ pressure .

Importance of NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

• Helps in understanding key concepts and equations in the chapter.
• Useful for class exams and competitive exams like NEET & JEE Mains (1 question is usually asked).
• Covers important formulas, making problem-solving easier.
• This chapter holds 2-3% weightage in competitive exams, so mastering it is beneficial.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT Solutions for Class 11 Physics Chapter-Wise

Importance of NCERT Solutions for Class 11 Physics Chapter 13 Oscillations:

For JEE Mains, oscillation and wave questions represent 6.67 percent of the total questions. The majority of the earlier oscillation-related JEE mains problems were from the subjects of SHM and basic pendulum. There will likely be two oscillation-related questions in the NEET test. To perform well in Class 11 and competitive exams, use the CBSE NCERT answers for Class 11 Physics Chapter 13 Oscillations.

Smart Tips to learn Class 11 Oscillations:

1. Understand Basics First- Understand what oscillations and SHM are before diving into formulas. Focus on SHM Concepts. Learn words such as amplitude, frequency, period, and phase correctly.

2. Use Graphs & Diagrams- Visualize SHM using displacement, velocity, and acceleration vs time graphs.

3. Memorize Formulas with Meaning- Don't just memorize—understand what each formula means and when to apply it.

4. Practice NCERT Numericals- Complete all examples and exercises, even for solid conceptual use.

5. Revise Regularly- Create a formula sheet and regularly revise it.

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