physics ncert solutions class 11th

13.9 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 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

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13.9 Temperature of the helium atom, $T_{H e}$ = – 20 °C = 253 K and temperature of argon atom be = $T_{A r}$

Atomic mass of helium, $M_{H e}$ = 4.0 u

Atomic mass of Argon, $M_{A r}$ = 39.9 u

Let ${V_{r m s}}_{A r}$ be the rms speed of Argon and ${V_{r m s}}_{H e}$ be the rms speed of Helium

From the relation of $V_{r m s} = \sqrt{\frac{3 k T}{M}}$ we get

rms speed of Argon, ${V_{r m s}}_{A r} = \sqrt{\frac{3 k T_{A r}}{M_{A r}}}$

rms speed of Helium, ${V_{r m s}}_{H e} = \sqrt{\frac{3 k T_{H e}}{M_{H e}}}$

Since both the speeds are equal, we get

$\sqrt{\frac{3 k T_{A r}}{M_{A r}}}$ = $\sqrt{\frac{3 k T_{H e}}{M_{H e}}}$ or $\frac{T_{A r}}{M_{A r}}$ = $\frac{T_{H e}}{M_{H e}}$ or $T_{A r}$ = $\frac{T_{H e} * M_{A r}}{M_{H e}}$ = $\frac{253 * 39.9}{4}$ = 2523.675 K = 2.523 $* 10^{3}$ K

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13.8 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 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 vrms the largest?

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13.8 (a) According to Avogadro's law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro's number, N = 6.023 $* 10^{23}$

(b) The root mean square speed ( $V_{r m s})$ of a gas of mass m and temperature T is given by the relation $V_{r m s} = \sqrt{\frac{3 k T}{m}}$ . Where k is Boltzmann constant. For the given gases, k and T are constants. Hence $V_{r m s}$ depends only on the mass of the atoms

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among Neon, Chlorine and Uranium hexafluoride, the mass of the neon is the smallest, so Neon will have the highest root mean square sp

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13.8 (a) According to Avogadro's law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro's number, N = 6.023 $* 10^{23}$

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13.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

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physics ncert solutions class 11th Motion in a Plane

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13.7 (i) At room temperature, T = 27 $?$ = 300 K

$k_{B}$ is Boltzmann constant = 1.38 $* 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

Average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 300}{2}$ = 6.21 $* 10^{- 21}$ J

Hence, the average thermal energy of a helium atom at room temperature is 6.21 $* 10^{- 21}$ J

(ii) On the surface of the Sun, T = 6000 K

Hence average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 6000}{2}$ = 1.242 $* 10^{- 19}$ J

Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.242 $* 10^{- 19}$ J

(iii) Inside the core of a star, T = $10^{7}$ K

Hence average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 10^{7}}{2}$ = 2.07 $* 10^{- 16}$ J

Hence, the average thermal energy of a helium atom

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13.7 (i) At room temperature, T = 27 $?$ = 300 K

$k_{B}$ is Boltzmann constant = 1.38 $* 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

Average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 300}{2}$ = 6.21 $* 10^{- 21}$ J

Hence, the average thermal energy of a helium atom at room temperature is 6.21 $* 10^{- 21}$ J

(ii) On the surface of the Sun, T = 6000 K

Hence average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 6000}{2}$ = 1.242 $* 10^{- 19}$ J

Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.242 $* 10^{- 19}$ J

(iii) Inside the core of a star, T = $10^{7}$ K

Hence average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 10^{7}}{2}$ = 2.07 $* 10^{- 16}$ J

Hence, the average thermal energy of a helium atom inside the core of a star is 2.07 $* 10^{- 16}$ J

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4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 ? m/s and moves in the x-y plane with a constant acceleration of (8.0 ? + 2.0 ?) m s^-2.

(a) At what time is the x- coordinates of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

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4.21:

(a) Velocity , $\overset{?}{v}$ = 10.0 ? m/s

Acceleration, $\overset{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = 8.0 ? + 2.0 ?

$\overset{?}{d v}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$ ,

Where, $\overset{?}{u} =$ velocity vector of the particle at t =0

$\overset{?}{v} =$ velocity vector of the particle at time t

But $\overset{?}{v}$ = $\frac{\underset{d r}{\to}}{d t}$

$\overset{?}{d r}$ = $\overset{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\overset{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\overset{?}{r} =$ $\overset{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\overset{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\overset{?}{u}$ , we g

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4.21:

(a) Velocity , $\overset{?}{v}$ = 10.0 ? m/s

Acceleration, $\overset{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = 8.0 ? + 2.0 ?

$\overset{?}{d v}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$ ,

Where, $\overset{?}{u} =$ velocity vector of the particle at t =0

$\overset{?}{v} =$ velocity vector of the particle at time t

But $\overset{?}{v}$ = $\frac{\underset{d r}{\to}}{d t}$

$\overset{?}{d r}$ = $\overset{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\overset{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\overset{?}{r} =$ $\overset{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\overset{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\overset{?}{u}$ , we get

$\overset{?}{r} =$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x / 4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$

At t = 2 s,

$\overset{?}{v}$ (t) = 8.0 $*$ 2 ? + 2.0 $* 2$ ? + 10 $*$ ? = 16 ? + 14 ?

The magnitude of $\overset{?}{v}$ (t) is given by

$|\overset{?}{v}|$ = ( 162 + 142)1/2 = 21.26 m/s

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

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13.6 Volume of the room, V = 25.0 $m^{3}$

Temperature of the room, T = 27 $?$ = 300 K

Pressure of the room, P = 1 atm = 1 $*$ 1.013 $* 10^{5}$ Pa

The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) can be written as

PV = $k_{B}$ NT, where $k_{B}$ is Boltzmann constant = 1.38 $* 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

N is the number of air molecules in the room.

N = $\frac{P V}{k_{B} T}$ = $\frac{1 * 1.013 * 10^{5} * 25}{1.38 * 10^{- 23} * 300}$ = 6.117 $* 10^{26}$

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4.20 The position of a particle is given by

r = 3.0t ? − 2.0t 2 ? + 4.0 ? m

where t is in seconds and the coefficients have the proper units for r to be in meters.

(a) Find the v and a of the particle?

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

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4.20

(a) The position is given by $\overset{?}{r}$ = 3. 0t ? − 2.0t 2 ? + 4.0 ?

The velocity v is given by $\overset{?}{v}$ = $\frac{d \overset{?}{r}}{d t}$ = $\frac{d}{d t}$ (3.0t ? − 2.0t 2 ? + 4.0 ?)

$\overset{?}{v}$ = 3.0 ? – 4.0t?

Acceleration $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = $\frac{d}{d t}$ (3.0 ? – 4.0t?) = – 4.0?

(b) At t = 2.0s

$\overset{?}{v}$ = 3.0 ? – 4.0t?

The magnitude of velocity is given by $|\overset{?}{v}|$ = (3.02 +- 8.02)0.5 = 8.54 m/s

Direction, $θ$ = $\tan^{- 1} ? \frac{v_{y}}{v_{x}}$ = $\tan^{- 1} ? \frac{8}{3}$ = 69.44 $°$

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13.5 An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

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13.5 Volume of the air bubble $,$ $V_{1}$ = 1.0 ${c m}^{3}$ = 1 $* 10^{- 6}$ $m^{3}$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m, $T_{1}$ = 12 $?$ = 285 K

Temperature at the surface of the lake, $T_{2}$ = 35 $?$ = 308 K

The pressure at the surface of the lake, $P_{2}$ = 1 atm = 1.013 $* 10^{5}$ Pa

The pressure at 40m depth: $P_{1}$ = 1 atm + d $? g$

Where $?$ is the density of water = $10^{3}$ kg/ $m^{3}$

G = acceleration due to gravity = 9.8 m/ $s^{2}$

Hence $P_{1}$ = 1.013 $* 10^{5} + (40 * 10^{3} * 9.8)$ = 493300 Pa

From the relation $\frac{P_{1} V_{1}}{T_{1}}$ = $\frac{P_{2} V_{2}}{T_{2}}$ , where $V_{2}$ is the volume of bubble when it

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13.5 Volume of the air bubble $,$ $V_{1}$ = 1.0 ${c m}^{3}$ = 1 $* 10^{- 6}$ $m^{3}$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m, $T_{1}$ = 12 $?$ = 285 K

Temperature at the surface of the lake, $T_{2}$ = 35 $?$ = 308 K

The pressure at the surface of the lake, $P_{2}$ = 1 atm = 1.013 $* 10^{5}$ Pa

The pressure at 40m depth: $P_{1}$ = 1 atm + d $? g$

Where $?$ is the density of water = $10^{3}$ kg/ $m^{3}$

G = acceleration due to gravity = 9.8 m/ $s^{2}$

Hence $P_{1}$ = 1.013 $* 10^{5} + (40 * 10^{3} * 9.8)$ = 493300 Pa

From the relation $\frac{P_{1} V_{1}}{T_{1}}$ = $\frac{P_{2} V_{2}}{T_{2}}$ , where $V_{2}$ is the volume of bubble when it reaches the surface, we get, $V_{2}$ = $\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$ = $\frac{493300 * 1 * 10^{- 6} * 308}{285 * 1.013 * 10^{5}}$ $=$ 5.263 $*$ $10^{- 6} m^{3} = 5.263 {c m}^{3}$

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4.19 Read each statement below carefully and state, with reasons, if it is true or false:

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

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4.19

(a) The net acceleration of a particle in circular motion is always directed along the radius of the circle towards the centre (centripetal force), in the case of a circular uniform motion. Hence the statement is False.

(b) The centrifugal force direction is always along the tangent, hence the statement is True.

(c) In uniform circular motion, the direction of the acceleration vector points is always toward the centre of the circle. The average of these vectors over one cycle is a null vector. Hence the statement is True.

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4.18 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

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4.18 The radius of the loop, r = 1 km = 1000 m

Speed, v = 900 km/h = 250 m/s

Centripetal acceleration, AC = $\frac{v^{2}}{r}$ = 250 $* 250$ / 1000 m/s² = 62.5 m/s² = $\frac{62.5}{9.817}$ g = 6.37g

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4.17 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

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