Q: Find the gravitational attraction between the two atoms in a hydrogen molecule. Given that G=6.67\times 10^{-11} Nm^2 kg^2 , mass of a hydrogen atom =1.67\times 10^{-27} kg and distance between the two atoms =1^oA.

Here, G=6.67\times 10^{-11} N m^2 kg^2 M_1=M_2=1.67\times 10^{-27}kg; r=1A^o=10^{-10}m Now, F=G\dfrac {M_1M_2}{r^2} =\dfrac {6.67\times 10^{-11}\times 1.67\times 10^{-27}\times 1.67\times 10^{-27}}{(10^{-10})^2} =1.86\times 10^{-44}N

Q: Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Velocity of a particle is given by.

{ F }_{ 1 }={ F }_{ 2 }=\cfrac { G{ m }^{ 2 } }{ 2{ r }^{ 2 } } \\ { F }_{ s }=\cfrac { G{ m }^{ 2 } }{ 4{ r }^{ 2 } } Centripetal force \cfrac { m{ v }^{ 2 } }{ r } =\cfrac { 2G{ m }^{ 2 } }{ 2{ r }^{ 2 } } \times \sin { 45° } +\cfrac { G{ m }^{ 2 } }{ 4{ r }^{ 2 } } \\ \Rightarrow \cfrac { m{ v }^{ 2 } }{ r } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } \times \cfrac { 1 }{ \sqrt { 2 } } +\cfrac { G{ m }^{ 2 } }{ 4{ r }^{ 2 } } \\ \Rightarrow { v }^{ 2 }=\cfrac { Gm }{ r } \left[ \cfrac { 1 }{ \sqrt { 2 } } +\cfrac { 1 }{ 4 } \right] \\ { v }^{ 2 }=\cfrac { GM }{ r } \left[ \cfrac { 2\sqrt { 2 } +1 }{ 4 } \right] \\ v=\left(\dfrac { GM }{ r } \left[ \cfrac { 2\sqrt { 2 } +1 }{ 4 } \right] \right)^{\dfrac{1}{2}}

Q: Let \omega be the angular velocity of the earth's rotation about its axis. Assume due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth a below earth's surface at a pole( d<<R ). The value of d is:

Lets \omega = the angular velocity of the earth's rotation about it's axis. g= acceleration due to gravity. R= radius of earth d= depth a below earth surface at which the acceleration due to gravity is same at the equator and the poles. g_{eff} due to rotation = g_{eff} due to depth g-\omega ^2R=g(1-\dfrac{d}{R}) g-R\omega ^2=g-\dfrac{gd}{R} -R\omega ^2=-\dfrac{gd}{R} d=\dfrac{\omega ^2R^2}{g} The correct option is A.

Q: Find the potential energy of the gravitational interaction of a mass point of mass m and a thin uniform rod of mass M and length l , if they are located along a straight line at a distance a from each other; also find the force of their interaction.

Choose the location of the point mass as the origin. Then the potential energy dU of an element of mass \displaystyle dM=\frac{M}{l}dx of the rod in the field of the point mass is \displaystyle dU=-\gamma m\frac{M}{l}dx\frac{1}{x} where x is the distance between the element and the point. (Note that the rod and the point mass are on a straight line.) If then a is the distance of the nearer end of the rod from the point mass. \displaystyle U=-\gamma\frac{mM}{l}\int_a^{\displaystyle a+l}{\frac{dx}{x}}=-\gamma m\frac{M}{l}\ln{\left(1+\frac{l}{a}\right)} The force of interaction is \displaystyle F=\frac{\partial U}{\partial a} \displaystyle=\gamma\frac{mM}{l}\times\frac{1}{\displaystyle1+\frac{l}{a}}\left(-\frac{l}{a^2}\right)=-\frac{\gamma mM}{a(a+l)} Minus sign means attraction.

Q: Mass M is distributed uniformly over a rod of length L . Find the gravitational field at P at a distance a on perpendicular bisector.

\displaystyle dm = \dfrac{M}{L} dx \displaystyle d E = \dfrac{G \dfrac{M}{L}dx}{(x^2 + a^2)} d E_{net} = d E cos \theta = \dfrac{GM a dx}{L (x^2 + a^2)^{3/2}} E = \displaystyle \int d E cos \theta = \dfrac{GM a}{L} \int_{-L/2}^{L/2} \dfrac{dx}{(x^2 + a^2)^{3/2}} Put x = a tan \theta then dx = a sec^2 \theta d \theta \displaystyle E = \dfrac{GM a}{L} \int \dfrac{a sec^2 \theta d \theta}{a^3 sec^3 \theta} = \dfrac{GM}{La} \int cos \theta d \theta = \displaystyle \dfrac{GM}{La} sin \theta = \dfrac{GM x}{La \sqrt{x^2 + a^2}} \int_{-L/2}^{L/2} = \displaystyle \dfrac{GM (2L)}{aL (L^2 + 4a^2)^{1/2}} = \dfrac{2 GM}{a (L^2 + 4a^2)^{1/2}}

Question 1

Find the gravitational attraction between the two atoms in a hydrogen molecule. Given that  G=6.67	imes 10^{-11} Nm^2 kg^2 , mass of a hydrogen atom  =1.67	imes 10^{-27}  kg and distance between the two atoms  =1^oA.

Accepted Answer

Here,  G=6.67	imes 10^{-11} N m^2 kg^2 M_1=M_2=1.67	imes 10^{-27}kg; r=1A^o=10^{-10}m Now,  F=G\dfrac {M_1M_2}{r^2} =\dfrac {6.67	imes 10^{-11}	imes 1.67	imes 10^{-27}	imes 1.67	imes 10^{-27}}{(10^{-10})^2} =1.86	imes 10^{-44}N

Question 2

Mass  m  is divided into two parts  X\ m  and  (1-X)\ m . For a given separation, the value of  X  for which the gravitational force of attraction between the two pieces becomes maximum is :

Accepted Answer

Gravitational Force between 2 bodies is given by  F = \dfrac{Gm_1 m_2}{R^2} where ,  m_1  and   m_2  are the masses of the bodies ,  G  is the universal Gravitational constant and  R  is the distance between them.For a given distance ,  F = \dfrac{Gm(1-X) m(X)}{R^2}  is maximum when  X(1-X)  is maximum . By differentiation, we get, \dfrac{dF}{dX} = \dfrac{Gm^2}{R^2}\dfrac{d}{dX}(X(1-X)) \dfrac{dF}{dX} = \dfrac{Gm^2}{R^2}\dfrac{d}{dX}(X-X^2) \dfrac{dF}{dX} = \dfrac{Gm^2}{R^2}(1-2X)=0 1-2X =0 X = \dfrac{1}{2} The gravitational force of attraction has a maximum value at  X =1/2

Question 3

The point at which the gravitational force acting on any mass is zero due to the Earth and the Moon system is (The mass of the Earth is approximately  81  times the mass of the Moon and the distance between the Earth and the Moon is  3,85,000 \;km .)

Accepted Answer

For zero gravitational force acting on the mass  m_1 ,   \dfrac{GMm_1}{(d-x)^2}  = \dfrac{Gmm_1}{x^2}  Here  x  is distance from moon,  M  is mass of earth,  m  is mass of moon and  d  is the distance between the earth and moon.  \Rightarrow \dfrac{M}{(d-x)^2}  = \dfrac{m}{x^2}   \Rightarrow \dfrac{81m}{(d-x)^2}  = \dfrac{m}{x^2}   \Rightarrow \dfrac{81}{(d-x)^2}  = \dfrac{1}{x^2}   \Rightarrow \dfrac{(d-x)^2}{81}  = x^2   \Rightarrow \dfrac{d-x}{9}  = x  Solving gives:  x  =  \dfrac{d}{10}  =  38,500   km

Question 4

Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Velocity of a particle is given by.

Accepted Answer

{ F }_{ 1 }={ F }_{ 2 }=\cfrac { G{ m }^{ 2 } }{&#10;2{ r }^{ 2 } } \ { F }_{ s }=\cfrac { G{ m }^{ 2 } }{ 4{ r }^{ 2 } }  &#10;&#10;Centripetal force  \cfrac { m{ v }^{ 2 } }{ r } =\cfrac { 2G{ m }^{ 2 } }{ 2{ r }^{ 2 } } 	imes \sin { 45&#176; } +\cfrac { G{ m }^{ 2 } }{ 4{ r }^{ 2 } } \ \Rightarrow \cfrac { m{ v }^{ 2 } }{ r } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } 	imes \cfrac { 1 }{ \sqrt { 2 }  } +\cfrac { G{ m }^{ 2 } }{ 4{ r }^{ 2 } } \ \Rightarrow { v }^{ 2 }=\cfrac { Gm }{ r } \left[ \cfrac { 1 }{ \sqrt { 2 }  } +\cfrac { 1 }{ 4 }  ight] \ { v }^{ 2 }=\cfrac { GM }{ r } \left[ \cfrac { 2\sqrt { 2 } +1 }{ 4 }  ight] \ v=\left(\dfrac { GM }{ r } \left[ \cfrac { 2\sqrt { 2 } +1 }{ 4 }  ight] ight)^{\dfrac{1}{2}} &#10;&#10;

Question 5

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, distant  2R  from the center of the sphere. A spherical cavity of radius  R/2  is now made on the sphere with cavity now applies a gravitational force F_{2} . Then  \dfrac{F_{2}}{F_{1}}  will be

Accepted Answer

Let, M is the mass of the sphere of uniform density  \rho  and m is the mass of the particle.The gravitational force of attraction  F_1  is given by, F_1 = \frac{GMm}{(2R)^2}     ........................(1)The mass of sphere of volume V, is given by M = \rho V M = \rho \frac{4}{3} \pi R^3  ......................(2)From equations (1) and (2), we get F_1 = \frac{G[\frac{4}{3} \pi R^3]m}{(2R)^2}  ...................(3)When the spherical cavity is made in sphere, the gravitational force  F_2  will be F_2 = \frac{G[M - \Delta M]m}{(2R)^2} F_2 = \frac{G[\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3]m}{(2R)^2} F_2 = \frac{G[\frac{4}{3} \pi R^3(1 - (\frac{1}{2})^3)]m}{(2R)^2}  .......................(4)Now, \frac{F_2}{F_1} = \frac{\frac{G[\frac{4}{3} \pi R^3(1 - (\frac{1}{2})^3)]m}{(2R)^2}}{\frac{G[\frac{4}{3} \pi R^3]m}{(2R)^2}}  ...............from equations (3) and (4) \frac{F_2}{F_1} = 1 - (\frac{1}{2})^3 \frac{F_2}{F_1} = 1 - (\frac{1}{8}) \frac{F_2}{F_1} = \frac{7}{8}

Question 6

The height at which the value of acceleration due to gravity becomes  50 % of that at the surface of the earth (Radius of the earth   = 6400  km) is nearly

Accepted Answer

Acceleration due to gravity a height  h  from earth's surface is given by: g_h = g(\dfrac {R}{R+h})^2 Here,  g_h = \dfrac{g}{2} \Rightarrow \dfrac {g}{2}=g(\dfrac {R}{R+h})^2 \Rightarrow \dfrac {R+h}{R}=1.414 \Rightarrow \dfrac {h}{R}=0.414 \Rightarrow h=0.414	imes 6400=2649.6\ km

Question 7

A tunnel is dug along a diameter of the Earth. The force on a particle of mass  m  placed in the tunnel at a distance  x   from the centre is:

Accepted Answer

The value of  g^{'}  at depth  d  is given by, g^{'}=g \left( 1-\dfrac{d}{R} \right) The value of  g^{'}  at a distance  x  form Earth's centre is  g^{'}= g \left( 1-\dfrac{R-x}{R} \right) = \dfrac{gx}{R} But  g = \dfrac{GM}{R^2} \Rightarrow  g^{'}=\dfrac{\left(\dfrac {GM}{R^2} \right)}{R} x=\dfrac {GMx}{R^3} Force on a particle of mass  m  is  mg=\dfrac {GMmx}{R^3}

Question 8

Let  \omega   be the angular velocity of the earth's rotation about its axis. Assume due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth a below earth's surface at a pole( d<<R ). The value of  d  is:

Accepted Answer

Lets  \omega =  the angular velocity of the earth's rotation about it's axis. g=  acceleration due to gravity. R=  radius of earth d=  depth a below earth surface at which the acceleration due to gravity is same at the equator and the poles. g_{eff}  due to rotation   = g_{eff}  due to depth g-\omega ^2R=g(1-\dfrac{d}{R}) g-R\omega ^2=g-\dfrac{gd}{R} -R\omega ^2=-\dfrac{gd}{R} d=\dfrac{\omega ^2R^2}{g} The correct option is A.

Question 9

Find the potential energy of the gravitational interaction of a mass point of mass  m  and a thin uniform rod of mass  M  and length  l , if they are located along a straight line at a distance  a  from each other; also find the force of their interaction.

Accepted Answer

Choose the location of the point mass as the origin. Then the potential energy  dU  of an element of mass  \displaystyle dM=\frac{M}{l}dx  of the rod in the field of the point mass is \displaystyle dU=-\gamma m\frac{M}{l}dx\frac{1}{x} where  x  is the distance between the element and the point. (Note that the rod and the point mass are on a straight line.) If then  a  is the distance of the nearer end of the rod from the point mass. \displaystyle U=-\gamma\frac{mM}{l}\int_a^{\displaystyle a+l}{\frac{dx}{x}}=-\gamma m\frac{M}{l}\ln{\left(1+\frac{l}{a}ight)} The force of interaction is  \displaystyle F=\frac{\partial U}{\partial a} \displaystyle=\gamma\frac{mM}{l}	imes\frac{1}{\displaystyle1+\frac{l}{a}}\left(-\frac{l}{a^2}ight)=-\frac{\gamma mM}{a(a+l)} Minus sign means attraction.

Question 10

Mass M is distributed uniformly over a rod of length  L . Find the gravitational field at P at a distance a on perpendicular bisector.

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10; \displaystyle dm = \dfrac{M}{L}&#10;dx &#10;&#10; \displaystyle d&#10; E = \dfrac{G \dfrac{M}{L}dx}{(x^2 + a^2)} d  E_{net} = d  E cos&#10; 	heta = \dfrac{GM a  dx}{L (x^2 + a^2)^{3/2}} &#10;&#10; E = \displaystyle&#10;\int d E  cos 	heta = \dfrac{GM a}{L} \int_{-L/2}^{L/2} \dfrac{dx}{(x^2 +&#10;a^2)^{3/2}} &#10;&#10;Put  x = a  tan&#10; 	heta  then  dx = a  sec^2 	heta d 	heta &#10;&#10; \displaystyle E = \dfrac{GM&#10;a}{L} \int \dfrac{a  sec^2 	heta d 	heta}{a^3 sec^3 	heta} = \dfrac{GM}{La}&#10;\int cos  	heta d 	heta &#10;&#10; = \displaystyle \dfrac{GM}{La}&#10;sin 	heta = \dfrac{GM x}{La \sqrt{x^2 + a^2}} \int_{-L/2}^{L/2} &#10;&#10; = \displaystyle \dfrac{GM&#10;(2L)}{aL (L^2 + 4a^2)^{1/2}} = \dfrac{2 GM}{a (L^2 + 4a^2)^{1/2}} &#10;&#10;

Question 11

A ring has mass  M , radius  R . A point mass  m  is placed at a distance  x  on the axial line as shown in the figure. Find  x , so that force experienced is maximum

Accepted Answer

All point on the circumference of the ring are at distance  \sqrt{R^2 + x^2}  from the center O of the ring.Force on the mass m at P :  F = \dfrac{ GMm}{R^2 +x^2} 	imes (2 \cos 	heta )  where  \cos 	heta = \dfrac{x}{\sqrt{R^2 + x^2} } Due to symmetry, vertical components of the forces from two symmetrical elements cancel out, the horizontal components add up, hence we get a factor of 2. 	herefore F =  \dfrac{ GMm}{R^2 +x^2} 	imes 2\dfrac{x}{\sqrt{R^2 + x^2} } When F is maximum,  \frac{dF}{dx} =0 \Rightarrow (R^2 + x^2) ^{-\dfrac{3}{2}} +x ( -\dfrac{3}{2} (R^2 + x^2) ^{-\dfrac{3}{2} -1} 	imes 2x =0  \Rightarrow (R^2 + x^2) = 3x^2 \Rightarrow x = \dfrac{R}{\sqrt{2}}

Question 12

The figure shows the elliptical orbit of a planet P about the sun S. The shaded area SCD is a twice shaded area SAB. if  {t}_{1}  is the the time for the planet to move from C to D and  {t}_{2}  is the time to move from A to B, then

Accepted Answer

We know,According to Kepler's law, time taken to complete a path in an elliptical orbit is directly proportional to the area swept from the focus as the centre. A\propto t A_{SAB}\propto t_2 A_{SCD}\propto t_1 Given, \dfrac{A_{SAB}}{A_{SCD}}=\dfrac{1}{2} \dfrac{t_2}{t_1}=\dfrac{1}{2} t_1=2t_2 Option  	extbf B  is the correct answer

Question 13

India's Mangalyan was sent to the Mars launching it into a transfer orbit  EOM  around the sun. It leaves the earth at  E  and meets mars at  M . If the semi-major axis of Earth's orbit is  a_{e}=1.5 	imes 10^{11}\ m  that of Mar's orbit  a_{m}=2.28 	imes 10^{11} , taken Kepler's laws given the estimate of time for mangalyan to reach Mars from Earth to be close to :

Accepted Answer

Solution: r = \dfrac{1.5 + 2.28}{2} = 1.89 \dfrac{T_m}{T_e} = (\dfrac{1.89}{1.5})^{\dfrac{3}{2}} t_m = \dfrac{T_m}{2} = \left(\dfrac{1.89}{1.5}\right)^{\dfrac{3}{2}} \Rightarrow \dfrac{365}{2}  x  1.41 = 257.4 \ days 257.4 is approximately equal to 260 days and close to option  CHence the correct option: C

Question 14

What is the escape velocity from the surface of the earth of radius  R  and density  \rho   ?

Accepted Answer

Escape speed is given by  v_e=\sqrt {\dfrac {2GM}{R}} But  M=Volume	imes density  \Rightarrow M=\dfrac{4}{3} \pi r^3 ho \Rightarrow v_e=\sqrt {2G\dfrac {\dfrac {4}{3}\pi R^3ho }{R}} \Rightarrow v_e=2R\sqrt {\dfrac {2\pi ho G}{3}}

Question 15

The escape velocity from the earth is  11 km/sec . The escape velocity from a planet having twice the radius and same density as earth is :

Accepted Answer

Escape speed is  v_e=\sqrt {\dfrac {2GM}{R}} If we express  Mass  in terms of  Volume  and  Density ,  v_e  in terms of density can be written as, v_e=2R\sqrt {\dfrac {2G\pi ho}{3}}=11 km/s ; on Earth's surface.For another planet, radius is twice that of Earth but density is same. 	herefore v^{'}_e=2(2R)\sqrt {\dfrac {2G\pi ho}{3}}=2	imes 11=22  km/s

Question 16

A satellite is revolving in a circular orbit at a height  h  from the earth surface (radius of earth  Rh < < R ). The minimum increased in its orbital velocity required, so that the satellite could escape from the earths gravitational field, is close to: (Neglect the effect of atmosphere.)

Accepted Answer

We know that,Orbital velocity  =v^2=\cfrac{GM}{R+h}=\cfrac{GM}{R}  as  h<<R Velocity required to escape \cfrac{1}{2} mv'^2=\cfrac{GM}{R}\ \implies v'^2=\sqrt{\cfrac{2GM}{R}} Increase in velocity =v'-v=\sqrt{\cfrac{2GM}{R}}-\sqrt{\cfrac{GM}{R}}\=\sqrt{2gR}-\sqrt {gR}\ =\sqrt gR(\sqrt 2 -1)

Question 17

If the mass of earth were  4  times the present mass, the mass of the moon were half the present mass and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be approximately.

Accepted Answer

Time period of moon is given by: T^2=\dfrac {4\pi^2}{GM}r^3 ,   where  M  is the mass of the Earth. T=27 daysWhen mass of the Earth is  4M , {T^{'}}^2=\dfrac {4\pi^2}{4MG}r^3=\dfrac {T^2}{4} \Rightarrow T^{'}=\dfrac{T}{2}=\dfrac{27}{2}=14  days