Class 11th

Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Is linear speed, angular speed, kinetic and potential energy varies from point to point in the orbit.

(a) No

(b) No

(c) Yes

(d) No

(e) No

(f) Yes

(a) Escape velocity of a body from the Earth is given by the relation:

$v_{esc}$ = $\sqrt{2gR}$ …. (1)

Where g = acceleration due to gravity and R = radius of the Earth

So escape velocity is independent of the mass of the body.

(b) It does not depend on the location from where it is projected.

(c) Does not depend on the direction of projection

(d) Depends on the height of the location from where the body is launched.

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: $\tau$ = ( ${\frac{4{\pi }^2{r}^3}{GM})}^{\frac{1}{2}}$ = ( ${\frac{4*{3.14}^2{*4.73}^3*{10}^{60}}{6.67*{10}^{-11}*5*{10}^{41}})}^{\frac{1}{2}}$ = 1.12 $*{10}^{16}s$ = 3.55 x ${10}^8$ years

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

Ultrasonic beep frequency emitted by bat, $\nu$ = 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:

${\nu }^{\text{'}}=(\frac{v}{v-v_b}$ ) $\nu$ = $(\frac{v}{v-0.03v}$ ) $*40$ = $\frac{40}{0.97}$ kHz = 41.24 kHz

This frequency is reflected by the stationary wall ( $v_s=0)$ towards the bat

The frequency ( $\nu \text{'}\text{'})\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{y}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}:$

${\nu }^{\text{'}\text{'}}=(\frac{v+v_b}{v}$ ) $\nu \text{'}$ = ( $\frac{v+0.03v}{v}$ ) $*41.24\mathrm{k}\mathrm{H}\mathrm{z}$ = 1.03 $*41.24$ = 42.47 kHz

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,

$t_s-t_p=240s$ so $t_p=240sand{t}_s=480s$

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

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

Operating frequency of the SONAR system,  $\nu$ = 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 = ( $\frac{v+v_e}{v})\nu =$  ( $\frac{1450+100}{1450})*40$ = 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