Q: An object is projected vertically upwards from the surface of the earth with a velocity 3 times the escape velocity v_{e} from earth's surface. What will be its final velocity after escape from the earth's gravitational pull?

We know, V_e\textrm{ (on surface of earth)}= \sqrt{\dfrac{2GM}{R}} Given, v_i=3v_e Let mass of object= m ,Apply energy conservation, \dfrac{1}{2}mv_i + \dfrac{-GMm}{R} = \dfrac{1}{2}mv_f^2 +0 (as R_{final}=\infty) \dfrac{1}{2}m\times 9\dfrac{2GM}{R} - \dfrac{GMm}{R}=\dfrac{1}{2} mv_f^2 \dfrac{8GMm}{R}=\dfrac{1}{2}mv_f^2 v_f=\sqrt{\dfrac{16GM}{R}} v_f=\sqrt 8\times \sqrt{\dfrac{2GM}{R}}=\sqrt 8 v_e Hence option \textbf B is the correct answer.

Q: A particle of mass m is placed at a distance of 4R from the centre of a huge uniform sphere of mass M and radius R . A spherical cavity of diameter R is made in the sphere as shown in the figure. If the gravitational interaction potential energy of the system of mass m and the remaining sphere after making the cavity is \displaystyle -\frac{\lambda GMm}{28R} Find the value of \displaystyle {\lambda }

Since the radius of the cavity is half the radius of the sphere it was made in, the mass removed to make it is one-eighth of the mass of the sphere. Gravitational Potential Energy =PE_{sys} \Rightarrow{PE}_{sys}={PE}_{due\ to\ sphere} - {PE}_{due\ to\ removed\ mass} { PE }_{ due\ to\ sphere }=-\dfrac { GMm }{ 4R } \\ { PE }_{ due\ to\ removed\ mass }=-\dfrac { Gm(\dfrac { M }{ 8 } ) }{ 4R-\dfrac { R }{ 2 } } =-\dfrac { GMm }{ 28R } . So, { PE }_{ sys }=-\dfrac { GMm }{ 4R } +\dfrac { GMm }{ 28R } =-\dfrac { 6GMm }{ 28R } Therefore, \lambda =6

Q: An artificial satellite is launched into a circular orbit around the Earth with velocity v relative to the reference frame moving translationally and fixed to the Earth's rotation axis. The distance from the satellite to the Earth's surface. Find x . The radius of the Earth and the free-fall acceleration on its surface are supposed to be known.

Let the mass of satellite be ' m ' and the mass and radius of earth be M,R respectively. V is the velocity of the satellite relative to earth In an earth's orbit, centripetal force is equal to gravitational pull of the earth. \Rightarrow \dfrac { GMm }{ { \left( R+h \right) }^{ 2 } } =\dfrac { m{ v }^{ 2 } }{ R+h } \\ \Rightarrow \left( R+h \right) =\dfrac { GM }{ { v }^{ 2 } } \\ \Rightarrow h=\dfrac { GM }{ { v }^{ 2 } } -R But, g=\dfrac { GM }{ { R }^{ 2 } } Therefore, \Rightarrow h=\dfrac { g{ R }^{ 2 } }{ { v }^{ 2 } } -R\\ \therefore \quad h=R\left( \dfrac { g{ R } }{ { v }^{ 2 } } -1 \right) Hence, x=1

Q: Gravitational acceleration on the surface of a planet is \dfrac{\sqrt{6}}{11}g , where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is \dfrac{2}{3} times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms ^{-1} , the escape speed on the surface of the planet in kms ^{-1} will be

let the gravitational acceleration on the surface be g{'} and desity be \rho{'} \displaystyle \dfrac{g'}{g}=\dfrac{\sqrt{6}}{11}; \dfrac{\rho '}{\rho }=\dfrac{2}{3} Hence, \displaystyle \dfrac{R'}{R}=\dfrac{3\sqrt{6}}{22} \displaystyle \dfrac{v'_{esc}}{v_{esc}}\propto\sqrt{\dfrac{R'^{2}\rho '}{R^{2}\rho }}=\dfrac{3}{11} v_{esc}'=3km/s.

Question 1

Two metal spheres of the same material and radius  r  are in contact with each other. The gravitational force of attraction between the spheres is given by:

Accepted Answer

Mass of each metal sphere is given by: m= \dfrac{4}{3}\pi r^{3}\rho The force of attraction between two metal sphere is given by: F = \dfrac{Gmm}{(2r)^2} F = \dfrac{G(\dfrac{4}{3}\pi r^{3}\rho)(\dfrac{4}{3}\pi r^{3}\rho)}{4r^2} F = \dfrac{G(\frac{4}{3}\pi \rho)(\dfrac{4}{3}\pi \rho)}{4}\dfrac{(r^3)(r^3)}{r^2} F = K r^4

Question 2

From a sphere of mass  M  and radius  R , a smaller sphere of radius  \dfrac{R}{2}  is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is  3R , the gravitational force between the two spheres is :

Accepted Answer

If Mass of the sphere is M, Mass of the carved out sphere is  \dfrac{M}{R^3} 	imes ( \dfrac{R}{2}) ^3 = \dfrac{M}{8}  Consider the larger sphere to be made of positive mass M and negative mass  \dfrac{-M}{8}  Net force on the   + \dfrac{M}{8}  sphere is  \dfrac { GM\dfrac { M }{ 8 }  }{  (3R) ^{ 2 } } + \dfrac { G(\dfrac { -M }{ 8 } )(\quad \dfrac { M }{ 8 } ) }{ (3R-\dfrac{R}{2}) ^2 }   = \dfrac { GM^2}{ R^2 } ( \dfrac{1}{72} - \dfrac{1}{400}) = \dfrac { GM^2}{ R^2 } ( \dfrac{50-9}{400}) = \dfrac { 41GM^2}{ 3600R^2 }

Question 3

A particle is moving with speed  v = b\sqrt{x}  along positive x-axis. Calculate the speed of the particle at time  t = 	au  (assume that the particle is at origin at  t = 0 )

Accepted Answer

v = b\sqrt{x} \dfrac{dv}{dt} = \dfrac{b}{2\sqrt{x}} \dfrac{dx}{dt} a = \dfrac{bv}{2\sqrt{x}} a = \dfrac{dv}{dt} = a = \dfrac{b^2}{2} v = \dfrac{b^2}{2}	au

Question 4

The position of a particle as a function of time t, is given by  x(t)=at+bt^2-ct^3  where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be?

Accepted Answer

x=at+bt^2-ct^3 v=\dfrac{dx}{dt}=a+2bt-3ct^2 a=\dfrac{dv}{dt}=2b-6ct=0\Rightarrow t=\dfrac{b}{3c} v_{\left(at t =\dfrac{b}{3c}\right)}=a+2b\left(\dfrac{b}{3c}\right)-3c\left(\dfrac{b}{3c}\right) =a+\dfrac{b^2}{3c} .

Question 5

A satellite is seen after each  8 h  over the equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of the earth and the satellite is the same will be

Accepted Answer

The time period is given by, T=\dfrac{2\pi}{\omega} Angular Velocity is given by, \omega=\dfrac{2\pi}{T} and then find relative angular velocityGiven  8=\dfrac {2\pi}{\omega_1+\omega_2}=\frac {2\pi}{\dfrac {2\pi}{T_1}+\dfrac {2\pi}{T_2}} \Rightarrow T_1=24 h  for the earthSubstituting this value in the above equation we get T_2=12 h (T_2  being the time period of the satellite, it will remain the same as the distance from the centre of the earth remains constant)Now when the sense of rotation of the earth and the satellite is same we get \Rightarrow T=\dfrac {2\pi}{\omega_2-\omega_1}=\dfrac {2\pi}{\dfrac {2\pi}{T_1}-\dfrac {2\pi}{T_2}}=24 h

Question 6

Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds is

2 : 1

. If one of them is at a height of

100 k m

, the height of the other satellite is

(

Radius of earth

R = 6400

km

)

Accepted Answer

Orbital velocity is given by:  V=\sqrt{\dfrac{GM}{R+h}} Where  G  is the Universal Gravitational constant,  M  is the mass of the Earth,  h  is the height of the satellite from the earth's surface and  R  is the radius  of the earth.  V \propto \dfrac{1}{\sqrt{R+h}} \dfrac{V_1}{V_2} = \sqrt{\dfrac{R+h_2}{R+h_1}} Given:  \dfrac{V_1}{V_2} =\dfrac{2}{1} ,  h_1=100  km and  R = 6500   km   \Rightarrow \dfrac{2}{1} = \sqrt{\dfrac{6400+h_2}{6400+100}} \Rightarrow 6400+h_2 = 4 	imes 6500 = 26000   km  Hence, the height above the earth's surface,  h_2=26000-6400 = 19600   km

Question 7

An object is projected vertically upwards from the surface of the earth with a velocity  3  times the escape velocity  v_{e}  from earth's surface. What will be its final velocity after escape from the earth's gravitational pull?

Accepted Answer

We know, V_e	extrm{   (on surface of earth)}= \sqrt{\dfrac{2GM}{R}} Given, v_i=3v_e Let mass of object= m ,Apply energy conservation, \dfrac{1}{2}mv_i + \dfrac{-GMm}{R} = \dfrac{1}{2}mv_f^2 +0      (as   R_{final}=\infty) \dfrac{1}{2}m	imes 9\dfrac{2GM}{R} - \dfrac{GMm}{R}=\dfrac{1}{2} mv_f^2 \dfrac{8GMm}{R}=\dfrac{1}{2}mv_f^2 v_f=\sqrt{\dfrac{16GM}{R}} v_f=\sqrt 8	imes \sqrt{\dfrac{2GM}{R}}=\sqrt 8 v_e Hence option  	extbf B  is the correct answer.

Question 8

A particle of mass m is placed at a distance of  4R  from the centre of a huge uniform sphere of mass  M  and radius  R . A spherical cavity of diameter  R  is made in the sphere as shown in the figure. If the gravitational interaction potential energy of the system of mass  m  and the remaining sphere after making the cavity is  \displaystyle -\frac{\lambda GMm}{28R}  Find the value of  \displaystyle {\lambda }

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Since the radius of the&#10;cavity is half the radius of the sphere it was made in, the mass removed to&#10;make it is one-eighth of the mass of the sphere.&#10;&#10;Gravitational Potential&#10;Energy  =PE_{sys} &#10;&#10; \Rightarrow{PE}_{sys}={PE}_{due\ to\ sphere}&#10;- {PE}_{due\ to\ removed\ mass} &#10;&#10; { PE }_{&#10;due\ to\ sphere }=-\dfrac { GMm }{ 4R } \ { PE }_{ due\ to\ removed\ mass }=-\dfrac&#10;{ Gm(\dfrac { M }{ 8 } ) }{ 4R-\dfrac { R }{ 2 }  } =-\dfrac { GMm }{ 28R&#10;}  .&#10;&#10;So,  { PE }_{ sys }=-\dfrac&#10;{ GMm }{ 4R } +\dfrac { GMm }{ 28R } =-\dfrac { 6GMm }{ 28R }  &#10;&#10;Therefore,  \lambda&#10;=6 &#10;&#10;

Question 9

An artificial satellite is launched into a circular orbit around the Earth with velocity v relative to the reference frame moving translationally and fixed to the Earth's rotation axis. The distance from the satellite to the Earth's surface. Find  x .  The radius of the Earth  and the free-fall acceleration on its surface are supposed to be known.

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Let the mass of&#10;satellite be ' m ' and the mass and radius of earth be  M,R  respectively.&#10;&#10; V  is the velocity&#10;of the satellite relative to earth&#10;&#10;In an earth's orbit,&#10;centripetal force is equal to gravitational pull of the earth.&#10;&#10; \Rightarrow \dfrac {&#10;GMm }{ { \left( R+h ight)  }^{ 2 } } =\dfrac { m{ v }^{ 2 } }{ R+h } \&#10;\Rightarrow \left( R+h ight) =\dfrac { GM }{ { v }^{ 2 } } \ \Rightarrow h=\dfrac&#10;{ GM }{ { v }^{ 2 } } -R &#10;&#10;But,  g=\dfrac { GM }{&#10;{ R }^{ 2 } }  &#10;&#10;Therefore,&#10; \Rightarrow h=\dfrac { g{ R }^{ 2 } }{ { v }^{ 2 } } -R\ 	herefore \quad&#10;h=R\left( \dfrac { g{ R } }{ { v }^{ 2 } } -1 ight)  &#10;&#10;Hence,  x=1 &#10;&#10;

Question 10

Gravitational acceleration on the surface of a planet is

\frac{6}{1 1} g

, where

g

is the gravitational acceleration on the surface of the earth. The average mass density of the planet is

\frac{2}{3}

times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms

^{- 1}

, the escape speed on the surface of the planet in kms

^{- 1}

will be

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;let the gravitational&#10;acceleration on the surface be  g{'}  and desity be  \rho{'} &#10;&#10; \displaystyle \dfrac{g'}{g}=\dfrac{\sqrt{6}}{11};&#10;\dfrac{\rho '}{\rho }=\dfrac{2}{3} &#10;&#10;Hence,&#10; \displaystyle \dfrac{R'}{R}=\dfrac{3\sqrt{6}}{22} &#10;&#10; \displaystyle \dfrac{v'_{esc}}{v_{esc}}\propto\sqrt{\dfrac{R'^{2}\rho&#10;'}{R^{2}\rho }}=\dfrac{3}{11} &#10;&#10; v_{esc}'=3km/s. &#10;&#10;

Question 11

A solid sphere of uniform density and radius  R  applies a gravitational force of attraction equal to  F_1  on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius  \dfrac{R}{2}  is now made in the sphere, as shown in the figure. The sphere with cavity now applies a gravitational force  F_2  on the same particle. The ratio  \dfrac{F_2}{F_1}  is

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;From super position&#10;principle , &#10;&#10; F_{1}=F_{r}+F_{c} &#10;&#10;Here  F_{r}=  force&#10;due to remaining part  =F_{2}  &#10;&#10; F_{c}= &#10;force due to mass on the cavity&#10;&#10; F_{1}=\dfrac{GMm}{9R^{2}} &#10;&#10; \displaystyle&#10;{F_{c}=\dfrac{G(\dfrac{M}{8})m}{(\dfrac{5}{2}R)^{2}}=\dfrac{GMm}{50R^{2}}} &#10;&#10; =\displaystyle \succ&#10;F_{2}=F_{1}-F_{c}=\dfrac{GMm}{9R^{2}}-\dfrac{GMm}{50R^{2}}=\dfrac{41GMm}{450R^{2}} &#10;&#10; \Rightarrow  \displaystyle \dfrac{F_{2}}{F_{1}}=\dfrac{41}{50} &#10;&#10;

Question 12

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line, joining the Sun and the Earth. The Sun is  3	imes 10^5  times heavier than the Earth and is at a distance  2.5	imes 10^4  times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is  v_e = 11.2 km s^{-1} . The minimum initial velocity  (v_s)  required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

Accepted Answer

Given: M_s = 3	imes10^5\ M_e d = 2.5 	imes 10^4\ R_e Also   V_e = 11.2 Km/s  =  \sqrt{\dfrac{2GM_e}{R_e}} Now for the sun earth system,  \dfrac{1}{2}mV_s^2 - \dfrac{GM_em}{R_e}-\dfrac{GM_sm}{R_e+d}=0 Assuming  R_e<<d and taking the value of  \dfrac{GM_e}{R}=\dfrac{11.2^2}{2} V_s \approx\ 42\ Km/s

Question 13

Two bodies, each of mass  M,  are kept fixed at a separation  2L . A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is  G.  The correct statement(s) is (are)

Accepted Answer

\displaystyle \dfrac{-2\mathrm{G}\mathrm{M}\mathrm{m}}{\mathrm{L}}+\dfrac{1}{2}\mathrm{m}\mathrm{v}^{2}=0 \Rightarrow \mathrm{v}=2\sqrt{\dfrac{\mathrm{G}\mathrm{M}}{\mathrm{L}}} Note:The energy of mass m means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each)  which is the potential energy of the system.

Question 14

Three particles, each of mass  m  are situated at the vertices of an equilateral triangle of side  a . The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle should move in a circle while maintaining the original mutual separation  a . Then their time period of revolution is :

Accepted Answer

As per the figure on the left hand side. Resultant Gravitational force acting on each particle is: F=\sqrt{(\dfrac{Gm^2}{a^2})^2+(\dfrac{Gm^2}{a^2})^2+2(\dfrac{Gm^2}{a^2})(\dfrac{Gm^2}{a^2})\cos 60}=\sqrt{3}\dfrac{Gm^2}{a^2} As per figure on right side, as the particles are expected to move in a circle of radius  r  (say) while maintaining original separation  a . \dfrac{a/2}{r}=\cos 30 r=\dfrac{a}{\sqrt{3}} Because of circular motion, the centripetal force is balanced by resultant gravitational force. If  v  is the velocity: \dfrac{mv^2}{r}=\sqrt{3}\dfrac{Gm^2}{a^2} \dfrac{\sqrt{3}mv^2}{a}=\sqrt{3}\dfrac{Gm^2}{a^2} v=\sqrt{\dfrac{Gm}{a}} Time period,  T=\dfrac{2\pi r}{v} =\dfrac{2\pi a}{\sqrt{3}}\sqrt{\dfrac{a}{Gm}} =2\pi\sqrt{\dfrac{a^3}{{3}Gm}}

Question 15

A satellite is seen after each  8 h  over the equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of the earth and the satellite is the same will be

Accepted Answer

The time period is given by, T=\dfrac{2\pi}{\omega} Angular Velocity is given by, \omega=\dfrac{2\pi}{T} and then find relative angular velocityGiven  8=\dfrac {2\pi}{\omega_1+\omega_2}=\frac {2\pi}{\dfrac {2\pi}{T_1}+\dfrac {2\pi}{T_2}} \Rightarrow T_1=24 h  for the earthSubstituting this value in the above equation we get T_2=12 h (T_2  being the time period of the satellite, it will remain the same as the distance from the centre of the earth remains constant)Now when the sense of rotation of the earth and the satellite is same we get \Rightarrow T=\dfrac {2\pi}{\omega_2-\omega_1}=\dfrac {2\pi}{\dfrac {2\pi}{T_1}-\dfrac {2\pi}{T_2}}=24 h

Question 16

Two planets A, B have their radii in the ratio   2: 5  and densities in the ratio   1: 6  respectively. Choose the correct statement(s)

Accepted Answer

R_A:R_B=2:5, D_A:D_B=1:6 \displaystyle g_A:g_B=\dfrac {\dfrac {GM_A}{R_A^2}}{\dfrac {GM_B}{R_B^{2}}}=\dfrac {4}{3}\pi \dfrac {R_A^3D_A}{R_A^2}	imes \dfrac {R_B^2}{\dfrac {4}{3}\pi R_B^3d_B}=\dfrac {R_Ad_A}{R_Bd_B} \dfrac {g_A}{g_B}=\dfrac {2}{5}	imes \dfrac {1}{6}=\dfrac {1}{15} W_B=mg_B=15mg_A=15W_A For the same volume of planets, mass of planet A is  1/6  the mass of planet B.Planet B has a greater volume than planet A.

Question 17

An object is weighed at the North pole by a beam balance and a spring balance, giving readings of

W_{B}

and

W_{S}

respectively. It is again weighed in the same manner at the equator, giving reading of

W_{B}^{'}

and

W_{S}^{'}

respectively. Assume that the acceleration due to gravity is the same everywhere and that the balances are quite sensitive.

Accepted Answer

Options A,C,D.Due to rotation of Earth, weight is lower at the equator.Beam balance measures mass of the body. Its reading does not change at pole and Equator.

Question 18

Inside a uniform spherical shell

Accepted Answer

Since there is no mass enclosed inside the spherical shell ,gravitational field is 0 from gauss law for gravitation. and same everywhere inside the spherical shell.Gravitational potential is given by  V=-\int gdr V=constant  g=0 So,  Gravitational potential is also same everywhere inside the spherical shell.Thus  B,C and D are correct .

Question 19

A bullet is fired vertically upwards with velocity  \upsilon   from the surface of a spherical planet. When it reaches its maximum heights, its acceleration due to the planet's gravity is  1/4 th of its value at the surface of the planet. If the escape velocity from the planet is  \upsilon_{esc} = \upsilon \sqrt N , then the value of N is (ignore energy loss due to atmosphere)

Accepted Answer

We know that at the surface of the earth, value of   g = \dfrac{G M}{R^2}  At height  h  above the Earth's surface, the value of acceleration due to gravity  g' = \dfrac{GM}{(R+h)^2} So it is given that  g' = \dfrac{g}{4}  When the bullet reaches maximum height, acceleration due to gravity is  \dfrac{1}{4} th of that at planet's surface. That implies  \dfrac{GM}{4R^2} = \dfrac{GM}{(R+h)^2} \ \ \rightarrow h = R  By conservation of mechanical energy,  \dfrac{-GMm}{R} + \dfrac{1}{2}mv^2 = \dfrac{-GMm}{h+R } + 0  since velocity is zero at max height.   \Rightarrow \dfrac{1}{2}mv^2 = \dfrac{GMm}{2R}    v = \sqrt{\dfrac{GM}{R}} = \sqrt{\dfrac{2GM}{2R}} = \dfrac{1}{\sqrt{2}} \sqrt{\dfrac{2GM}{R}}= \dfrac{1}{\sqrt{2}}v_{esc}     \Rightarrow v_{esc} = \sqrt{2}v   \Rightarrow N = 2

Question 20

lmagine a vertical mine shaft of depth x metres located near the equator (of earth). A plumb line is hanging in the shaft and a stone is dropped from rest next to it. The stone will not fall parallel to the plumb line, but will deviate in easterly direction by an amount of y metres.
Given

ω

is angular speed of earth and

g

is acceleration due to gravity)
If y

= \frac{ω g}{n} (\frac{2 x}{g})^{\frac{3}{2}}

find n.

View solution

Accepted Answer

Let, 'v' be the velocity attained by stone at time 't'. Thus, acceleration of the body in east direction due to coriolis effect is  2\omega v . Integrating this acceleration twice w.r.t time, we get the displacement of stone 'y'. Hence,  y\quad =\quad 2\omega v\ v\quad =\quad gt\ x\quad =\quad \dfrac { g{ t }^{ 2 } }{ 2 } \ \Rightarrow \quad y\quad =\quad 2\omega gt\ \iint { ydt } \quad =\quad \dfrac { \omega g{ t }^{ 3 } }{ 3 } \ \Rightarrow \quad y\quad =\quad \dfrac { \omega g }{ 3 } { \left( \dfrac { 2x }{ g }  ight)  }^{ \dfrac { 3 }{ 2 }  }\quad \Rightarrow \quad n\quad =\quad 3

Question 21

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M. The point masses are connected by a rigid massless rod of length  l  and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance  r = 3l  from M, the tension in the rod is zero for   m = k \displaystyle ( \frac{M}{288})  . The value of k is

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Force on left mass m&#10; F_1 = \dfrac{GMm}{(3l)^2} - \dfrac{Gmm}{(l)^2} &#10;&#10;Force on right mass m&#10; F_2 = \dfrac{GMm}{(4l)^2} + \dfrac{Gmm}{(l)^2} &#10;&#10;When tension in the&#10;light rod is zero,  F_1 =F_2 &#10;&#10;  	herefore \dfrac{GMm}{(3l)^2}&#10;- \dfrac{Gmm}{(l)^2} = \dfrac{GMm}{(4l)^2} + \dfrac{Gmm}{(l)^2} &#10;&#10;  \Rightarrow \dfrac{7GMm}{144l^2}&#10;= \dfrac{2Gm^2}{l^2} &#10;&#10;  \Rightarrow m =\dfrac{7M}{288} &#10;&#10; \Rightarrow k =7 &#10;&#10;

Question 22

A particle of mass m is placed at a distance of  4R  from the centre of a huge uniform sphere of mass  M  and radius  R . A spherical cavity of diameter  R  is made in the sphere as shown in the figure. If the gravitational interaction potential energy of the system of mass  m  and the remaining sphere after making the cavity is  \displaystyle -\frac{\lambda GMm}{28R}  Find the value of  \displaystyle {\lambda }

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Since the radius of the&#10;cavity is half the radius of the sphere it was made in, the mass removed to&#10;make it is one-eighth of the mass of the sphere.&#10;&#10;Gravitational Potential&#10;Energy  =PE_{sys} &#10;&#10; \Rightarrow{PE}_{sys}={PE}_{due\ to\ sphere}&#10;- {PE}_{due\ to\ removed\ mass} &#10;&#10; { PE }_{&#10;due\ to\ sphere }=-\dfrac { GMm }{ 4R } \ { PE }_{ due\ to\ removed\ mass }=-\dfrac&#10;{ Gm(\dfrac { M }{ 8 } ) }{ 4R-\dfrac { R }{ 2 }  } =-\dfrac { GMm }{ 28R&#10;}  .&#10;&#10;So,  { PE }_{ sys }=-\dfrac&#10;{ GMm }{ 4R } +\dfrac { GMm }{ 28R } =-\dfrac { 6GMm }{ 28R }  &#10;&#10;Therefore,  \lambda&#10;=6 &#10;&#10;

Question 23

A planet of mass M, has two natural satellites with masses

m_{1}

and

m_{2}

. The radii of their circular orbits are

R_{1}

and

R_{2}

respectively. Ignore the gravitational force between the satellites. Define

v_{1}, L_{1}, K_{1}

and

T_{1}

to be respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite

1

; and

v_{2}, L_{2}, K_{2}

and

T_{2}

to be the corresponding quantities of satellite

2

. Given

m_{1} / m_{2} = 2

and

R_{1} / R_{2} = 1 / 4

, match the ratios in List - I to the numbers in List-II.

List - I	List - II
P. $\frac{v _{1}}{v _{2}}$	1. $\frac{1}{8}$
Q. $\frac{L _{1}}{L _{2}}$	2. $1$
R. $\frac{K _{1}}{K _{2}}$	3. $2$
S. $\frac{T _{1}}{T _{2}}$	4. $8$

Accepted Answer

\dfrac{GMm_1}{R^{2}_{1}}  =   \dfrac{m_1 v_1^2}{R_1}    v_{1}^{2} = \dfrac{GM}{R_1}, v_{2}^{2} = \dfrac{GM}{R_2}    \dfrac{ v_{1}^{2}}{ v_{2}^{2}} = \dfrac{R_2}{R_1} = 4           (P)  \dfrac{v_1}{v_2} = 2            (Q) L = mvR   \dfrac{L_1}{L_2} = \dfrac{m_1v_1R_1}{m_2v_2R_2} = 2 	imes 2 	imes \dfrac{1}{4} = 1      (R)  K =  \dfrac{1}{2} mv^2   \dfrac{K_1}{K_2} = \dfrac{m_1v^{2}_{1}}{m_2v^{2}_{2}} = 2 	imes (2)^2 = 8    (S) T =   2 \pi R / v \dfrac{T_1}{T_2} = \dfrac{R_1}{v_1} 	imes \dfrac{v_2}{R_2} = \dfrac{R_1}{R_2} 	imes \dfrac{v_2}{v_1} = \dfrac{1}{4} 	imes \dfrac{1}{2} = \dfrac{1}{8}

Question 24

Match the following.

Accepted Answer

extbf{Part A} We know, 	extrm{Escape Velocity}=\sqrt{\dfrac{2GM_p}{R_p}}         ( M_p=  mass of planet).Here Escape Velocity is independent of mass of body, direction of propagation,and temperature of planet.Hence  	extbf{A}ightarrow 1,3,5 	extbf{Part B} Atmosphere exists when the air molecules can't escape theplanets gravitational force.Hence, existence of atmosphere on a planet also depends upon theescape velocity of the planet, but is independent of the mass of the planet.And the temperature of the planet.And existence of atmosphere has no reference with mass of body or mass of body, so we have nothing to do with these options.Hence  	extbf{B}ightarrow 2,4,5 	extbf{Part C} 	extrm{Acceleration due to gravity(g)}=\dfrac{GM_p}{R^2}      (M_p=  mass of planet.)It is independent of the mass of body upon which the acceleration is acting,Whatever may be the direction of propagation but the value of  g  is same,also it is independent of temperature of the planet.But  g  depends on the escape velocity of the planet,as  g=\dfrac{v^2}{R} Hence  	extbf Cightarrow 1,3,5 	extbf{Part D} 	extrm{Potential Energy of planet mass system}=\dfrac{GM_pm}{R} It is clear that it us independent of direction of propagation of the body,and the temperature of the planet.Hence  	extbf Dightarrow 1,5

Question 25

Question 26

Question 27

Question 28

The value of  h  will be

Accepted Answer

Gravity is related to the angular speed is given by g^{'}=g-\omega^2R So the weight at the equator is  mg^{'}=m(g-\omega^2R) Gravity above the earth's surface is given by g{'}=g\left(1-\dfrac {2h}{R}\right) Implies the weight at the equator above the earth's surface is  mg^{'}=mg\left(1-\dfrac {2h}{R}\right) It is given that these two weights are equal, \Rightarrow m(g-\omega^2R)=mg\left(1-\dfrac {2h}{R}\right) \Rightarrow g-\omega^2R=g-\dfrac {2gh}{R} \Rightarrow \omega^2R=\dfrac {2gh}{R} \Rightarrow h=\dfrac {\omega^2R^2}{2g}

Question 29

What is the gravitational field at Q?

Accepted Answer

Let us calculate field at Q in case the cavity weren't there, since mass is proportional to  { R }^{ 3 } Gravitational field  { g }_{ 0 }=GM(\dfrac { { 2 }^{ 3 } }{ { 4 }^{ 3 } } )\dfrac { 1 }{ { 2 }^{ 2 } } =\dfrac { GM }{ 32 }  Also, field due to mass previously in right cavity will be 0 since the point is at its centre. Gravitational field due to left sphere, g'=GM(\dfrac { { 1 }^{ 3 } }{ { 4 }^{ 3 } } )\dfrac { 1 }{ { 4 }^{ 2 } } =\dfrac { GM }{ 1024 }  Thus, actual magnetic field at Q is g={ g }_{ 0 }-g'=GM(\dfrac { 1 }{ 32 } -\dfrac { 1 }{ 1024 } )=\dfrac { 31GM }{ 1024 }  Answer is option A.

Question 30

Inside a uniform sphere of density  \rho  there is a spherical cavity whose centre is at a distance  l  from the centre of the sphere. Find the strength  G  of the gravitational field inside the cavity.

Accepted Answer

Treating the cavity as negative mass of density  -\rho  in a uniform sphere density  +\rho  and using the superposition principle, the sought field strength is, \vec{G}=\vec{G_1}+\vec{G_2} or  \displaystyle\vec{G}=-\frac{4}{3}\pi\gamma\rho\vec{r_{\displaystyle+}}+-\frac{4}{3}\gamma\pi(-\rho)\vec{r_{\displaystyle-}} (where  \vec{r_{\displaystyle+}}  and  \vec{r_{\displaystyle-}}  are the position vectors of an arbitrary point  P  inside the cavity with respect to centre of sphere and cavity respectively.)Thus  \displaystyle\vec{G}=-\frac{4}{3}\pi\gamma\rho\left(\vec{r_{\displaystyle+}}-\vec{r_{\displaystyle-}}\right)=-\frac{4}{3}\pi\gamma\rho\vec{l}