11: Gravitation : Neet-xii-physics

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NEET-XII-Physics

11: Gravitation

with Solutions - page 5

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Qstn #12
A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mass m when it is at a distance x from the centre of the tunnel.
Ans : Let d be distance of the particle from the centre of the Earth.

`` \,\mathrm{\,Now\,},{d}^{2}={x}^{2}+\left(\frac{{R}^{2}}{4}\right)=\frac{4{x}^{2}+{R}^{2}}{4}``
`` \Rightarrow d=\left(\frac{1}{2}\right)\sqrt{4{x}^{2}+{R}^{2}}``
Let M be the mass of the Earth and M' be the mass of the sphere of radius d.
Then we have:
`` M=\left(\frac{4}{3}\right)\pi {R}^{3}\,\mathrm{\,\rho \,}``
`` {M}^{1}=\left(\frac{4}{3}\right)\,\mathrm{\,\pi \,}{d}^{3}\,\mathrm{\,\rho \,}``
`` \therefore \frac{{M}^{\mathit{1}}}{M}=\frac{{d}^{3}}{{R}^{3}}``
Gravitational force on the particle of mass m is given by
`` F=\frac{G{M}^{1}m}{{d}^{2}}``
`` \Rightarrow F=\frac{\,\mathrm{\,G\,}{d}^{3}Mm}{{R}^{3}{d}^{2}}``
`` \Rightarrow F=\frac{GMm}{{R}^{3}}d``
∴ Normal force exerted by the wall, F_N = F cos θ`` =\frac{\,\mathrm{\,GM\,}md}{{\,\mathrm{\,R\,}}^{3}}\times \frac{\,\mathrm{\,R\,}}{2d}=\frac{\,\mathrm{\,GM\,}md}{2{\,\mathrm{\,R\,}}^{2}}``
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Qstn #13
A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure (11-E1). A particle of mass m‘ is placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if

#13-a
r < x < 2r,
Ans : Consider that the particle is placed at a distance x from O.
Here, r < x < 2r
Let us consider a thin solid sphere of radius (x `` -`` r).

Mass of the sphere, `` dm=\frac{m}{\left({\displaystyle \frac{4}{3}}\right)\,\mathrm{\,\pi \,}{r}^{3}}\times \frac{4}{3}\,\mathrm{\,\pi \,}(x-r{)}^{3}=\frac{m(x-r{)}^{3}}{{r}^{3}}``
Then the gravitational force on the particle due to the solid sphere is given by
`` \,\mathrm{\,F\,}=\frac{\,\mathrm{\,G\,}m\text{'}dm}{(x-r{)}^{2}}``
`` =\frac{\,\mathrm{\,G\,}{\displaystyle \frac{m(x-r{)}^{3}}{{r}^{3}}}m\text{'}}{(x-r{)}^{2}}=\frac{\,\mathrm{\,G\,}mm\text{'}(x-r)}{{r}^{3}}``
Force on the particle due to the shell will be zero because gravitational field intensity inside a shell is zero.

#13-b
2r < x < 2R and
Ans : If 2r < x < 2R,
Force on the body due to the shell will again be zero as particle is still inside the shell.
then F is only due to the solid sphere.
`` \therefore F=\frac{\,\mathrm{\,G\,}mm\text{'}}{{\left(x-r\right)}^{2}}``

#13-c
x > 2R.
Figure
Ans : If x > 2R, then the gravitational force is due to both the sphere and the shell.
Now, we have:
Gravitational force due to shell, `` F=\frac{\,\mathrm{\,GM\,}m\text{'}}{{\left(x-R\right)}^{2}}``
Gravitational force due to the sphere`` =\frac{\,\mathrm{\,G\,}mm\text{'}}{{\left(x-r\right)}^{2}}``
As both the forces are acting along the same line joining the particle with the centre of the sphere and shell so both the forces can be added directly without worrying about their vector nature.
∴ Resultant force`` =\frac{\,\mathrm{\,G\,}mm\text{'}}{{\left(x-r\right)}^{2}}+\frac{GMm\text{'}}{{\left(x-R\right)}^{2}}``
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Qstn #14
A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a (figure 11-E2). The centre of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P₁ and P₂ shown in the figure
Figure
Ans : At point P₁, the gravitational field due to the sphere and the shell is given by
F`` =\frac{GM}{{\left(3a+a\right)}^{2}}+0=\frac{GM}{16{a}^{2}}``
At point P₂, the gravitational field due to the sphere and the shell is given by
`` F=\frac{\,\mathrm{\,GM\,}}{{\left(a+4a+a\right)}^{2}}+\frac{\,\mathrm{\,GM\,}}{{\left(4a+a\right)}^{2}}``
`` \Rightarrow F=\frac{\,\mathrm{\,GM\,}}{36{a}^{2}}+\frac{\,\mathrm{\,GM\,}}{25{a}^{2}}``
`` \Rightarrow F=\frac{\,\mathrm{\,GM\,}}{{a}^{2}}\left(\frac{1}{36}+\frac{1}{25}\right)``
`` \Rightarrow F=\frac{\,\mathrm{\,GM\,}}{{a}^{2}}\left(\frac{25+36}{900}\right)``
`` \Rightarrow F=\left(\frac{61}{900}\right)\frac{\,\mathrm{\,GM\,}}{{a}^{2}}``
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Qstn #15
A thin spherical shell having uniform density is cut in two parts by a plane and kept separated as shown in figure (11-E3). The point A is the centre of the plane section of the first part and B is the centre of the plane section of the second part. Show that the gravitational field at A due to the first part is equal in magnitude to the gravitational field at B due to the second part.
Figure
Ans : We know that in a thin spherical shell of uniform density, the gravitational field at its internal point is zero. So, at points A and B, the gravitational fields are equal and opposite and, thus, cancel each other. So the net field is zero.
Hence, E_A = E_B

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Qstn #16
Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level?
Ans : Let the mass of 0.10 kg be at a distance x from the 2 kg mass and at a distance of (2 - x) from the 4 kg mass.
Force between 0.1 kg mass and 2 kg mass = force between 0.1 kg mass and 4 kg mass
`` \therefore \frac{2\times 0.1}{{x}^{2}}=-\frac{4\times 0.1}{{\left(2-x\right)}^{2}}``
`` \Rightarrow \frac{0.2}{{x}^{2}}=\frac{0.4}{{\left(2-x\right)}^{2}}``
`` \Rightarrow \frac{1}{{x}^{2}}=\frac{2}{{\left(2-x\right)}^{2}}``
`` \Rightarrow {\left(2-x\right)}^{2}=2{x}^{2}``
`` \Rightarrow 2-x=\sqrt{2}x``
`` \Rightarrow x\left(\sqrt{2}+1\right)=2``
`` \Rightarrow x=\frac{2}{2.414}``
`` =0.83\,\mathrm{\,m\,}\,\mathrm{\,from\,}\,\mathrm{\,the\,}2\,\mathrm{\,kg\,}\,\mathrm{\,mass\,}``
Now, gravitation potential energy of the system is given by
`` U=-G\left[\frac{0.1\times 2}{0.83}+\frac{0.1\times 4}{1.17}+\frac{2\times 4}{2}\right]``
`` \Rightarrow U\mathit{=}-6.67\times {10}^{11}\left[\frac{0.1\times 2}{0.83}+\frac{0.1\times 4}{1.17}+\frac{2\times 4}{2}\right]``
`` \Rightarrow U=-3.06\times {10}^{-10}\,\mathrm{\,J\,}``
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Qstn #17
Three particles of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.
Ans : The work done in increasing the side of the triangle from a to 2a is equal to the difference of the potential energies of the system.
i.e., work done = final potential energy of the system `` -`` initial potential energy of the system
`` \therefore W=-3\frac{G{m}^{2}}{2a}-\left(-3\frac{G{m}^{2}}{a}\right)``
`` \Rightarrow W=3\frac{G{m}^{2}}{2a}``
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Qstn #18
A particle of mass 100 g is kept on the surface of a uniform sphere of mass 10 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle away from the sphere.
Ans : The work done against the gravitational force to take the particle away from the sphere to infinity is equal to
the difference between the potential energy of the particle at infinity and potential energy of the particle at the surface of the sphere.

`` \therefore W=0-\left(-\frac{\,\mathrm{\,G\,}\times 10\times 0.1}{1\times 0.1}\right)``
`` =\frac{6.67\times {10}^{-11}\times 1}{1\times 0.1}``
`` =6.67\times {10}^{-10}\,\mathrm{\,J\,}``
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Qstn #19
The gravitational field in a region is given by
E→=5 N kg-1 i→+12 N kg-1 j→.
Ans : Gravitational field, `` \stackrel{\to }{E}=\left(5\,\mathrm{\,N\,}/\,\mathrm{\,kg\,}\right)\stackrel{‸}{i}+\left(12\,\mathrm{\,N\,}/\,\mathrm{\,kg\,}\right)\stackrel{‸}{j}``

#19-a
Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin.
Ans : `` \stackrel{\to }{\,\mathrm{\,F\,}}=m\stackrel{\to }{\,\mathrm{\,E\,}}``
`` =2\,\mathrm{\,kg\,}\left[\left(5\,\mathrm{\,N\,}/\,\mathrm{\,kg\,}\right)\stackrel{‸}{i}+\left(12\,\mathrm{\,N\,}/\,\mathrm{\,Kg\,}\right)\stackrel{‸}{j}\right]``
`` =\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{i}+\left(24\,\mathrm{\,N\,}\right)\stackrel{‸}{j}``
`` \therefore \left|\stackrel{\to }{\,\mathrm{\,F\,}}\right|=\sqrt{100+576}=\sqrt{676}=26\,\mathrm{\,N\,}``

#19-b
Find the potential at the points (12 m, 0) and (0, 5 m) if the potential at the origin is taken to be zero.
Ans : `` V=-\stackrel{\to }{\,\mathrm{\,E\,}.}\stackrel{\to }{r}``
Potential at (12 m, 0)`` =-60\,\mathrm{\,J\,}/\,\mathrm{\,Kg\,}``
Potential at (0, 5 m) = `` -60\,\mathrm{\,J\,}/\,\mathrm{\,kg\,}``

#19-c
Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12 m, 5 m).
Ans : change in potential=final potential -initial potential
initial potential=potential at the origin=0
final potential=potential at (12,5)
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{V}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\stackrel{\to }{E}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{.}\stackrel{\to }{r}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{10}\stackrel{‸}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{+}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{24}\stackrel{⏜}{j}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{.}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{12}\stackrel{⏜}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{+}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{5}\stackrel{⏜}{j}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{120}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{+}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{120}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{\text{J}}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{240}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{\text{J}}``

#19-d
Find the change in potential energy if the particle is taken from (12 m, 0) to (0, 5 m).
Ans : `` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{∆}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{V}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\stackrel{\to }{E}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{.}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{∆}\stackrel{\to }{r}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{∆}\stackrel{\to }{r}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{12}\stackrel{⏜}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{+}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{0}\stackrel{⏜}{j}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{0}\stackrel{⏜}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{+}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{5}\stackrel{⏜}{j}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{12}\stackrel{⏜}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{5}\stackrel{⏜}{j}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{∆}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{V}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{10}\stackrel{⏜}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{+}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{12}\stackrel{⏜}{j}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{.}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{(}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{12}\stackrel{⏜}{i}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{-}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{5}\stackrel{⏜}{j}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{)}``
`` \textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{=}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{0}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{}\textcolor[rgb]{0.09803921568627451,0.09803921568627451,0.09803921568627451}{\text{J}}``
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