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Karnataka PUC I Kinetic Theory & Oscillations — practice questions

Explore 50 free Kinetic Theory & Oscillations questions in Karnataka PUC I Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Kinetic Theory & Oscillations questions and exercises with answers

Showing questions 1–30 of 50.

The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:Read the answer and worked solution →A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:Read the answer and worked solution →The time period of a spring-mass system with mass $m$ and spring constant $k$ is:Read the answer and worked solution →The total mechanical energy of a particle of mass $m$ in SHM with amplitude $A$ and angular frequency $\omega$ is:Read the answer and worked solution →The root-mean-square (rms) speed of gas molecules is given by:Read the answer and worked solution →The mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:Read the answer and worked solution →If the absolute temperature of an ideal gas is doubled, the rms speed of its molecules:Read the answer and worked solution →The mean free path $\lambda$ of a gas molecule of diameter $d$ at number density $n$ is approximately:Read the answer and worked solution →For an ideal gas, the relationship between $C_p$ and $C_v$ is (Mayer's relation):Read the answer and worked solution →The ideal gas equation is:Read the answer and worked solution →The relation between universal gas constant $R$, Avogadro number $N_A$, and Boltzmann constant $k_B$ is:Read the answer and worked solution →The **mean free path** of a gas molecule is:Read the answer and worked solution →The pressure of an ideal gas in kinetic theory is given by:Read the answer and worked solution →The **rms speed** of gas molecules at absolute temperature $T$ is:Read the answer and worked solution →Average translational kinetic energy of a gas molecule at temperature $T$ is:Read the answer and worked solution →The **law of equipartition of energy** states that, at temperature $T$, each quadratic degree of freedom contributes:Read the answer and worked solution →Number of degrees of freedom (DOF) of a monatomic gas (e.g. helium) molecule:Read the answer and worked solution →Mayer's relation for an ideal gas is:Read the answer and worked solution →**Adiabatic ratio** $\gamma$ of a gas is defined as:Read the answer and worked solution →Number of degrees of freedom of a **rigid diatomic molecule** (e.g. O₂ at room temperature, vibrations not excited):Read the answer and worked solution →If vibrations are excited, a diatomic molecule has DOF:Read the answer and worked solution →For a monatomic ideal gas, $C_V$ and $C_P$ per mole are:Read the answer and worked solution →For a **rigid diatomic** ideal gas, $\gamma = C_P/C_V$ equals:Read the answer and worked solution →The rms speed of N₂ at 300 K (M = 28 g/mol) is approximately (use $R = 8.314$ J/(mol·K)):Read the answer and worked solution →If the temperature of a gas is doubled (at constant volume), the rms speed changes by a factor of:Read the answer and worked solution →Average translational KE of a He atom at 300 K is approximately ($k_B = 1.38 \times 10^{-23}$ J/K):Read the answer and worked solution →Mean free path of nitrogen at STP (d = 324 pm) is approximately:Read the answer and worked solution →If P is doubled at constant T (compress gas isothermally), the mean free path:Read the answer and worked solution →An ideal gas is held in a container at temperature T. The total internal energy of one mole of a monatomic gas is:Read the answer and worked solution →For a diatomic gas with all vibrational modes excited, the molar specific heat at constant volume is:Read the answer and worked solution →

Worked practice questions

The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:

  1. $T = 2\pi \sqrt{\dfrac{L}{g}}$
  2. $T = 2\pi \sqrt{\dfrac{g}{L}}$
  3. $T = 2\pi L g$
  4. $T = \sqrt{\dfrac{L}{g}}$
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Correct answer: A. $T = 2\pi \sqrt{\dfrac{L}{g}}$

For small oscillations the pendulum's equation of motion reduces to SHM with $\omega = \sqrt{g/L}$. So period $T = 2\pi/\omega = 2\pi\sqrt{L/g}$. Independent of the mass of the bob (a key contrast with the spring-mass system, where $T \propto \sqrt{m}$). Note: only valid for small-amplitude swings. At larger amplitudes the small-angle approximation fails and the period depends slightly on amplitude.

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A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:

  1. $A\omega$
  2. $A\omega^2$
  3. $A/\omega$
  4. $\omega/A$
Show answer and solution

Correct answer: A. $A\omega$

Position $x(t) = A \sin(\omega t + \phi)$. Velocity is its time derivative: $v(t) = A\omega \cos(\omega t + \phi)$. Maximum value is $v_{\max} = A\omega$, achieved when the particle passes through the mean position ($x = 0$). Maximum acceleration is at the extremes: $a_{\max} = A\omega^2$ (option B is the acceleration max, the common trap).

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The time period of a spring-mass system with mass $m$ and spring constant $k$ is:

  1. $T = 2\pi \sqrt{\dfrac{k}{m}}$
  2. $T = 2\pi \sqrt{\dfrac{m}{k}}$
  3. $T = 2\pi k m$
  4. $T = 2\pi (k + m)$
Show answer and solution

Correct answer: B. $T = 2\pi \sqrt{\dfrac{m}{k}}$

Spring-mass system: Newton's second law gives $m\ddot x = -kx$, hence $\omega = \sqrt{k/m}$. Period: $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{m/k}$. A stiffer spring (larger $k$) gives a shorter period. A heavier mass gives a longer period. Option A inverts the ratio, the classic trap.

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