Physics Part Ii Chapter 13 Kinetic Theory
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    NCERT Solution For Class 11 Physics Physics Part Ii

    Kinetic Theory Here is the CBSE Physics Chapter 13 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Physics Kinetic Theory Chapter 13 NCERT Solutions for Class 11 Physics Kinetic Theory Chapter 13 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Physics.

    Question 1
    CBSEENPH110024919

    State Boyle's Law?

    Solution

    Boyle’s Law : Temperature remaining constant the volume of a given mass of dry gas is inversely proportional to its pressure.

    V α1p=T=Constant

    Question 2
    CBSEENPH110024921

    Calculate the degrees of freedom for monoatomic, diatomic and triatomic gas?

    Solution

    The degrees of freedom of the system is given by:- f=3 N-K

    where f = degrees of freedom

    N = Number of Particles in the system.

    K = Independent relation among the particles.

    Question 3
    CBSEENPH110024922

    what do you mean by mean free path and write the formula?

    Solution

    In kinetic theory the mean free path of a particle, such as a molecule, is the average distance the particle travels between collisions with other moving particles. The formula still holds for a particle with a high velocity relative to the velocities of an ensemble of identical particles with random locations.

     
    Question 8
    CBSEENPH11020505

    The mass of a hydrogen molecule is 3.32 x 10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:

    • 4.70 x 102 N/m2

    • 2.35 x 103 N /m2

    • 4.70 x 103 N/m2

    • 2.35 x 102 N /m2

    Solution

    B.

    2.35 x 103 N /m2

    Change in momentum

    P = P2J^  + P2J^ + P2i^ -P2i^P =  2P2J^  = IH molecule Iwalll = - 2P2J^  Pressure, P=FA = 2PAn (n = no. of particles)2 x 3.32 x 10-27x 103 x 10232 x 10-4= 2.35 x 103 N/m2

    Question 9
    CBSEENPH11020507

    A particle is moving in a circular path of radius a under the action of an attractive potential U = -k/2r2. its total energy is:

    • -32ka2

    • -k4a2

    • k2a2

    • zero

    Solution

    D.

    zero

    Given,

    U = - k2r2 Fr = - dUdr = - kr3Since the particle moves in a circular path of radius 'a'mv2a = ka3 mv2 = ka2 ....(1)Kinetic energy = 12mv2 = k2a2Total energy = - k2a2 + k2a2 = 0

    Question 10
    CBSEENPH11020838

    The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

    • KA < KB < KC

    • KA > KB > KC

    • KB > KA > KC

    • KB < KA < KC

    Solution

    B.

    KA > KB > KC

    Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant

    Point A is perihelion and C is aphelion.

    Clearly, VA> VB>VC

    So, KA>KB>KC

    Question 11
    CBSEENPH11020840

    A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy (Kt) as well as rotational kinetic energy (Kr) simultaneously. The ratio Kt: (Kt+ Kr) for the sphere is

    • 7:10

    • 5:7

    • 2:5

    • 10:7

    Solution

    B.

    5:7

    In rolling motion, rotational kinetic energy.

    Kt = 12mv2AndKt + Kr = 12mv2 + 12Iω2= 12mv2 + 1225mr2vr2 = 710mv2 KtKt +Kr = 12mv2710mv2 = 57

    Question 12
    CBSEENPH11020877

    A graph between pressure P (along with the y-axis) and absolute temperature, T (along with the x-axis) for equal moles of two gases have been drawn. Given that the volume of the second gas is more than the volume of first gas. Which of the following statement is correct?

    • Slope of gas 1 is less than gas 2

    • Slope og gas 1 is more than gas 2

    • Both have some slopes

    • None of the above

    Solution

    B.

    Slope og gas 1 is more than gas 2

    According to the ideal gas equation,

    pV = nRT .... (i)

    pV = nRTorpT = nRV

    P/T represents the slope of the graph. As the number of moles is the same for the two gases.

     PT  1V V2 >V1 (slope)2 < (slope)1

    Question 13
    CBSEENPH11026117

    A sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity u = 0.5 mls. The kinetic energy of the sphere is

    • 0.1 J

    • 0.3 J

    • 0.5 J

    • 0.42 J

    Solution

    D.

    0.42 J

    The kinetic energy of sphere while rolling on an inclined plane is given by

    EK =Tanslational kinetic energy  +  Rotational kinetic energy

    =12 mv2 × 2 × 0.52 + 12 × 23 × 2× 0.12 × 0.50.12EK  =  0.25 + 0.17EK  = 0.42 J

    Question 14
    CBSEENPH11026200

    Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?

    • increase

    • Decrease

    • No change

    • Can't be determined

    Solution

    C.

    No change

    Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased on decreased.

    Question 15
    CBSEENPH11026246

    Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?

    • Increase

    • Decrease

    • No change

    • Cannot be determined

    Solution

    C.

    No change

    The kinetic energy of an ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased or decreased.

    Question 16
    CBSEENPH11026287

    For a gas RCv= 0.67. This gas is made up of molecules which are

    • monoatomic

    • polyatomic

    • mixture of diatomic and polyatomic molecules

    • diatomic

    Solution

    A.

    monoatomic

    The molecule of a monoatomic gas has three translational degrees of freedom. Thus the average energy of a molecule at temperature T is ( 3/2 ) kBT.

    Given:- 

          RCV = 0.67

     So,   Cv = R0.67

                  = 1.5 R

                  = 32R

    This value corresponds to monoatomic

    Question 17
    CBSEENPH11026329

    A horizontal tube of length l closed at both ends, contains an ideal gas of molecular weight M. The tube is rotated at a constant angular velocity ω about a vertical axis passing through an end. Assuming the temperature to be uniform and constant. If p1 and p2 denote the pressure at free and the fixed end respectively, then choose the correct relation

    • p2p1 = e2 l22RT

    • p1p2 = eM ωl2RT

    • p1p2 = eω l M3RT

    • p2p1 = eM2 ω2 l23RT

    Solution

    A.

    p2p1 = e2 l22RT

    Consider the diagram

    Consider the elementary part of thickness dx at a distance x from the axis of rotation, then force on this part

              Adp = (dm)ω2 x              ......(i)

    where, dm = mass of element

               ω = angular velocity

    Now, from ideal gas equation

               pV = nRT 

       R - Avagadro constant

       V = Volume

        n = Number of moles of gas

               pA dx = dmMRT

    ⇒         dm = MpART dx                 ......(ii)

    From Eqs. (i) and (ii)

             Adp = M p AR Tω2 x dx

    0(minimum) to l (maximum) is the limit

           p1p2dpp = 0lM ω2R T x dx

    ⇒      ln p = 2RTx220l

    ⇒     lnp2p1 = 2 l22RT 

    ⇒       p2p1 = eM ω2 l22RT

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    Question 19
    CBSEENPH11026355

    The temperature of the cold junction of thermocouple is 0°C and the temperature of hot junction is To C. The emf is E = 16T - 0.04T2 µV. The inversion temperature Tf is

    • 300o C

    • 200o C

    • 500o C

    • 400o C

    Solution

    D.

    400o C

    The thermocouple is an electrical device consisting of two dissimilar electrical conductors forming electrical junctions at different temperatures.

    Given emf  E = 16 T - 0.04 T2 μV

    At inversion temperature, E = 0

    ⇒        16 T - 0.04T2 = 0

    ⇒        0.04 T2 = 16T

    ⇒              Tf160.04

    ⇒               Tf = 400o C

    Question 20
    CBSEENPH11026365

    The temperature of a gas is raised from 27° C to 927° C. The root mean square speed

    • gets halved

    • gets doubled

    • is 92727 times the earlier value

    • remains the same

    Solution

    B.

    gets doubled

    Given:-

    T1 = 27oC = 300 K

    T2 = 927oC = 1200 K

    According to Kinetic theory of ideal gas

          vrmsv2¯12 = 3 KTm

             vrms3kTm

    Where m - mass of the molecules 

               v2¯ - mean of squared speed

    ⇒     vrmsT

    ∴       vrms1vrms2 = T1T2

            vrms 1 vrms 2 = 3001200

            ( vrms )2 = 2 ( vrms )1

    Question 21
    CBSEENPH11026372

    Two different isotherms representing the relationship between pressure p and volume V at a given temperature of the same ideal gas are shown for masses m1 and m2 , then

    • Nothing can be predicted

    • m1 < m2

    • m1 = m2

    • m1 > m2

    Solution

    B.

    m1 < m2

      

       pV = nRT

                = mM RT

    For m1 

        p = m1M RTV1                        ....(i)

    For m2

        p = m2M RTV2                        .....(ii)

    From Eqs. (i) and (ii), we get

         m1MRTV1 = m2MRTV2

    Thus,   V ∝ m

            V2 > V1

            m2 > m1

    Question 22
    CBSEENPH11026451

    1 mole of H2 gas is contained in a box of volume V  = 1.00 m3  at  T = 300 K. The gas is heated to a temperature of  T = 3000 K  and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

    • same as the pressure initially

    • 2 times the pressure initially

    • 10 times the pressure initially

    • 20 times the pressure initially

    Solution

    D.

    20 times the pressure initially

    According to Ideal gas equation

                  PV  = nRT

                 P2 V1T2 = P1 V1T1

    ⇒           P2P1 = V1V2 T2T1

    Here T2 = 3000 K,  T1 = 300 K

    Since H2 splits into hydrogen atoms, therefore volume become half 

    i.e           V212 V1

    ∴               P2P1 = V112 V1

    ∴                 P2P1 = 20

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