Class 11th

8.5 Let us assume that our galaxy consists of 2.5 * 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milk Way to be 10⁵ ly.

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Mass of our galaxy Milky Way, M = 2.5 $* 10^{11}$ solar mass

Solar mass = Mass of the Sun = 2.0 $* 10^{30}$ kg

Mass of our galaxy = 2.5 $* 10^{11} *$ 2.0 $* 10^{30}$ = 5 $* 10^{41}$ kg

Diameter of the Milky Way, d = 10⁵ ly

Radius of the Milky Way = 5 $* 10^{4}$ ly

1 ly = 9.46 $* 10^{15}$ m

r = 5 $* 10^{4} *$ 9.46 $* 10^{15}$ m = 4.73 $* 10^{20}$ m

As a star revolves around the galactic centre of the Milky Way, its time period is given by the relation: $τ$ = ( ${\frac{4 π^{2} r^{3}}{G M})}^{\frac{1}{2}}$ = ( ${\frac{4 * {3.14}^{2} {* 4.73}^{3} * 10^{60}}{6.67 * 10^{- 11} * 5 * 10^{41}})}^{\frac{1}{2}}$ = 1.12 $* 10^{16} s$ = 3.55 x $10^{8}$ years

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8.4 IO, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 * 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.

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The rotation period of the satellite Io, $T_{j u}$ = 1.769 days = 1.769 x 24 x 60 x 60s

Radius of the Orbit is given by the relation, $R_{j u}$ = 4.22 $* 10^{8}$ m

Mass is given by the relation: $M_{j}$ = $\frac{4 π^{2} R_{j u}^{3}}{G T_{j u}^{2}}$ ……(i)

Where $M_{j}$ = mass of Jupiter and G = Universal gravitational constant

Orbital period of the Earth, $T_{e}$ = 365.25 days = 365.25 x 24 x 60 x 60 s

Orbital radius of the Earth, $R_{e}$ = 1 AU = 1.496 x $10^{11}$ m

Mass of Sun is given as : $M_{s}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ …(ii)

$\frac{M_{s}}{M_{j}}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ $*$ $\frac{G T_{j u}^{2}}{4 π^{2} R_{j u}^{3}}$ = $\frac{R_{e}^{3}}{T_{e}^{2}}$ $*$ $\frac{T_{j u}^{2}}{R_{j u}^{3}}$ = ( ${\frac{1.496 x 10^{11}}{4.22 * 10^{8}})}^{3}$ $* ({\frac{1.769 x 24 x 60 x 60}{365.25 x 24 x 60 x 60})}^{2}$ = 1045.04

Hence it can be inferred that the mass of Jupiter is about one – thousandth that

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The rotation period of the satellite Io, $T_{j u}$ = 1.769 days = 1.769 x 24 x 60 x 60s

Radius of the Orbit is given by the relation, $R_{j u}$ = 4.22 $* 10^{8}$ m

Mass is given by the relation: $M_{j}$ = $\frac{4 π^{2} R_{j u}^{3}}{G T_{j u}^{2}}$ ……(i)

Where $M_{j}$ = mass of Jupiter and G = Universal gravitational constant

Orbital period of the Earth, $T_{e}$ = 365.25 days = 365.25 x 24 x 60 x 60 s

Orbital radius of the Earth, $R_{e}$ = 1 AU = 1.496 x $10^{11}$ m

Mass of Sun is given as : $M_{s}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ …(ii)

Hence it can be inferred that the mass of Jupiter is about one – thousandth that of the Sun

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8 months ago

8.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

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Time taken by the Earth to complete one revolution around the Sun, $T_{e}$ = 1 year

Orbital radius of the Earth in its orbit, $R_{e}$ = 1 AU

Time taken by the planet to complete one revolution around the Sun, $T_{p}$ = $\frac{1}{2} T_{e}$ = $\frac{1}{2}$ year

Orbital radius of the planet = $R_{p}$

From Kepler's 3rd law of planetary motion, we can write:

( ${\frac{R_{p}}{R_{e}})}^{3}$ = ( ${\frac{T_{p}}{T_{e}})}^{2}$

$\frac{R_{p}}{R_{e}}$ = ( ${\frac{T_{p}}{T_{e}})}^{2 / 3}$ = ( ${\frac{1}{2})}^{2 / 3}$ = 0.63

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth

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8 months ago

8.2 Choose the correct alternative:

(a) Acceleration due to gravity increases/decreases with increasing altitude

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density)

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body

(d) The formula –G Mm(1/r₂ – 1/r₁) is more/less accurate than the formula mg(r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth

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(a) Decreases - Acceleration due to gravity at depth h is given by $g_{h}$ = (1 – $\frac{2 h}{R_{e}}$ )g, where $R_{e} = R a d i u s o f t h e e a r t h,$ g = acceleration due to gravity on the surface of the Earth. From this equation, it is clear that acceleration due to gravity decreases with increase in height

(b) Decreases – Acceleration due to gravity at depth d is given by $g_{d}$ = (1- $\frac{d}{R_{e}}$ )g. So the acceleration due to gravity decreases with increase in depth.

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(b) Decreases – Acceleration due to gravity at depth d is given by $g_{d}$ = (1- $\frac{d}{R_{e}}$ )g. So the acceleration due to gravity decreases with increase in depth.

(c) Mass of the body – Acceleration due to gravity of body mass m is given by the relation g = $\frac{G M}{R^{2}}$ , where G = Universal gravitation constant, M = mass of the Earth and R = radius of the Earth. Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More – Gravitational potential energy of two points $r_{1}$ and $r_{2}$ distance away from the centre of the Earth is respectively given by:

V( $r_{1}$ ) = $-$ - $\frac{G m M}{r_{1}}$ and V( $r_{2}$ ) = $-$ - $\frac{G m M}{r_{2}}$

Difference in potential energy, V = V( $r_{2}$ ) $-$ V( $r_{1}$ ) = $- G m M (\frac{1}{r_{2}}$ $- \frac{1}{r_{1}}$ )

Hence this formula is more accurate than mg( $r_{2}$ - $r_{1}$ )

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8 months ago

15.27 A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

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Ultrasonic beep frequency emitted by bat, $ν$ = 40 kHz

Velocity of the bat, $v_{b}$ = 0.03v, where v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

$ν^{'} = (\frac{v}{v - v_{b}}$ ) $ν$ = $(\frac{v}{v - 0.03 v}$ ) $* 40$ = $\frac{40}{0.97}$ kHz = 41.24 kHz

This frequency is reflected by the stationary wall ( $v_{s} = 0)$ towards the bat

The frequency ( $ν'') o f t h e r e c e i v e d s o u n d i s g i v e n b y t h e r e l a t i o n :$

$ν^{''} = (\frac{v + v_{b}}{v}$ ) $ν'$ = ( $\frac{v + 0.03 v}{v}$ ) $* 41.24 k H z$ = 1.03 $* 41.24$ = 42.47 kHz

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8 months ago

15.26 Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^–1, and that of P wave is 8.0 km s^–1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur ?

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Let $v_{s}$ and $v_{p}$ be the velocities and $t_{s}$ and $t_{p}$ be the time taken to reach the seismograph from the epicentre of S and P waves respectively.

Let L be the distance between the epicentre and the seismograph.

We have:

L = $v_{s} t_{s}$ …. (i)

L = $v_{p} t_{p}$ …. (ii)

It is given, $v_{s}$ = 4 km/s and $v_{p}$ = 8 km/s

From equation (i) and (ii), we get

4 $t_{s}$ = 8 $t_{p}$ or $t_{s} =$ 2 $t_{p}$ …. (iii)

It is also given,

$t_{s} - t_{p} = 240 s$ so $t_{p} = 240 s a n d t_{s} = 480 s$

From equation (ii), we get, L = 8 $* 240$ = 1920 km

Hence, the earthquake occurred at a distance of 1920 km from the seismograph.

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8 months ago

15.25 A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^–1. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 m s^–1

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Operating frequency of the SONAR system, $ν$ = 40 kHz

Speed of enemy submarine, $v_{e} =$ 360 km/h = 100 m/s

Speed of sound in water, v = 1450 m/s

The source is at rest and the observer (enemy submarine) is moving towards it. Hence, the apparent frequency (f') received and reflected by the submarine is given by the relation:

f = ( $\frac{v + v_{e}}{v}) ν =$ ( $\frac{1450 + 100}{1450}) * 40$ = 42.76 kHz

The frequency (f') received by the enemy submarine is given by the relation:

f' = ( $\frac{v}{v - v_{e}}$ )f = ( $\frac{1450}{1450 - 100}$ ) $* 42.76$ = 45.93 kHz

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8 months ago

15.24 One end of a long string of linear mass density 8.0 * 10^–3 kg m^–1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

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The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $ω t - k x)$ ……… (i)

Linear mass density $μ =$ 8.0 $* 10^{- 3}$ kg/m and frequency of the tuning fork, $ν$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $* 9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ = $\sqrt{\frac{882}{8.0 * 10^{- 3}}}$ = 332 m/s

Angular frequency, $ω$ = 2 $π ν$ = 2 $π * 256$ = 1608.5 rad/s = 1.6 $* 10^{3}$ rad/s

Wavelength, $λ$ = $\frac{v}{ν}$ = $\frac{332}{256}$ &

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The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $ω t - k x)$ ……… (i)

Linear mass density $μ =$ 8.0 $* 10^{- 3}$ kg/m and frequency of the tuning fork, $ν$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $* 9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ = $\sqrt{\frac{882}{8.0 * 10^{- 3}}}$ = 332 m/s

Angular frequency, $ω$ = 2 $π ν$ = 2 $π * 256$ = 1608.5 rad/s = 1.6 $* 10^{3}$ rad/s

Wavelength, $λ$ = $\frac{v}{ν}$ = $\frac{332}{256}$ = 1.297 m

Propagation constant, k = $\frac{2 π}{λ}$ = $\frac{2 π}{1.297}$ = 4.844 /m

Substituting these values in the displacement equation, we get

y (x, t) = a sin ( $ω t - k x)$

y (x, t) = 0.05 sin (1.6 $* 10^{3}$ t – 4.844 x) m

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15.23 A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) Frequency (ii) Wavelength (iii) Speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is the whistle is blown for a split of second after every 20 s) is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz ?

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The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed in that medium. The short pip produced after 20 s does not mean that the frequency of the whistle is $\frac{1}{20} o r$ 0.05 Hz. It means that 0.05 Hz is the frequency of repetition of the pip of the whistle. So the answers are

(a)

(i) No

(ii) No

(iii) Yes

(b) No

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15.22 A travelling harmonic wave on a string is described by

y(x, t) = 7.5 sin (0.0050x +12t + $π$ /4)

(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.

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