[ 2 Marks Questions ]

Question 12 Marks

Give the uses of polar satellites.

Answer

Polar satellites are uses in weather and environment monitoring.
They are used in spying work for military purposes.
They are used to study topography of Moon, Venus and Mars.

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Question 22 Marks

Give some uses of geostationary satellites.

Answer

In communicating radio, T.V and telephone signals across the world.
In studying upper regions of the atmosphere.
In forecasting weather.
In deter ming the exact shape and dimensions of the earth.
In studying solar radiations and cosmic rays.

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Question 32 Marks

Define orbital velocity.

Answer

Orbital velocity is the velocity required to put the satellite into its orbit around the earth.

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Question 42 Marks

What do you mean by weight of a body? Is it a scalar or vector?

Answer

Weight of a body is defined as the gravitational force with which a body is attracted towards the centre of the earth. Hence the weight of a body is given by $w=m g$ (or) $\vec{W}=m \vec{g}$

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Question 52 Marks

What is meant by acceleration due to gravity? Is is a scalar or a vector?

Answer

The acceleration produced in a freely falling body under the gravitational pull of the earth. It is a vector having direction towards the centre of the earth.

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Question 62 Marks

What is meant by the term free fall?

Answer

The motion of a body under the influence of gravity alone is called a free fall.

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Question 72 Marks

Why is G called the universal gravitational constant?

Answer

The value of G does not depend on the nature and size of the bodies. It also does not depend on the nature of the medium between the two bodies. That is why G is called universal gravitational constant.

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Question 82 Marks

Distinguish between the terms gravitation and gravity.

Answer

Gravitation: It is the force of attraction between any two bodies in the universe.
Gravity: It is the force of attraction between the earth and any object lying on or near its surface.

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Question 92 Marks

A body weight 500 N on the surface of the Earth. How much would it weight half way below the surface of Earth.

Answer

Weight on surface of Earth, $mg =500 N$
Weight below the surface of Earth at $d =\frac{ R }{2}$
From variation of ' $g$ ' with depth
$
\begin{aligned}
g ^{\prime} & =g\left(1-\frac{d}{ R }\right) \text {} \\
m g^{\prime} & =m g\left[1-\frac{\left(\frac{ R }{2}\right)}{ R }\right]=500\left[1-\frac{1}{2}\right]=500\left[\frac{1}{2}\right] \\
W ^{\prime} & =250 N
\end{aligned}
$

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Question 102 Marks

An object weight $72 N$ on the Earth. What its weight at a height $\frac{ R }{2}$ from Earth.

Answer

We know that
$
\begin{aligned}
g & =\underline{ GM } \\
g^{\prime} & = g \left(\frac{ R }{ R +h}\right)^2=g\left(\frac{ R }{ R +\frac{ R }{2}}\right)^2=g\left(\frac{2 R }{3 R }\right)^2 \\
g^{\prime} & =\frac{4}{9} g \quad \text {} \\
\text { Weight } W ^{\prime} & =\frac{4}{9} W =\frac{4}{9}(72) \\
W ^{\prime} & =32 N
\end{aligned}
$

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Question 112 Marks

A body weight $700 g$ on the surface of Earth. How much it weight on the surface of planet whose mass is $\frac{1}{7}$ and radius is half that of the Earth.

Answer

Acceleration due to gravity $g =\frac{ GM }{ R ^2}$
$
M _{ c }=\frac{ M }{7} \text { and } R _e=\frac{ R }{2}
$
On the planet $g_p=\frac{ G \left(\frac{ M }{7}\right)}{\left(\frac{ R }{2}\right)^2}=\frac{4}{7}\left(\frac{ GM }{ R ^2}\right)=\frac{4}{7} g$
Hence weight on the planet $W_p=700 \times \frac{4}{7}$
$
W _{ p }=400 \text { gram }
$

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Question 122 Marks

At what distance from the centre of Earth, the value acceleration due to gravity ‘g’ will be half that of the surface?

Answer

According to acceleration due to gravity
$
\begin{aligned}
\frac{g^{\prime}}{g} & =\left(\frac{ R }{ R +h}\right)^2 \text { } \\
\frac{1}{2} & =\left(\frac{ R }{ R +h}\right)^2 \Rightarrow \frac{1}{\sqrt{2}}=\frac{ R }{ R +h} \\
R +h & =\sqrt{2} R \Rightarrow h=(\sqrt{2}-1) R \\
h & =0.414 R
\end{aligned}
$
Hence distance from centre $= R +0.414 R =1.414 R$

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Question 132 Marks

If the mass of Earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the Earth is 9.8 ms-2. Calculate the value of ‘g’ on that planet?

Answer

Acceleration due to gravity $g=\frac{ GM }{ R ^2}$
$
\begin{aligned}
g_p & =\frac{ GM _{ P }}{ R _{ P }^2} \text { and } g_e=\frac{ GM _e}{ R _e^2} \\
\frac{g_p}{g_e} & =\frac{ GM _{ P }}{ R _{ P }^2} \times \frac{ R _e^2}{ GM _e}=\frac{ M _{ P }}{ M _e}\left(\frac{ R _e}{ R _{ P ^{\prime}}}\right)^2 \\
g_{ P } & =g_e\left(\frac{ M _p}{ M _e}\right)\left(\frac{ R _e}{ R _p}\right)=9.8\left(\frac{1}{80}\right)(2)^2=9.8 \times \frac{1}{20} \\
g_{ P } & =0.49 ms ^{-2}
\end{aligned}
$

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Question 142 Marks

If R is the radius of the Earth and g is the acceleration due to gravity on the Earth’s surface. Find the mean density of the Earth.

Answer

Acceleration due to gravity $g=\frac{G M}{R^2}$
Acceleration due to gravity $g=\frac{ GM }{ R _{\text {Sa }}{ }_{\text {Sa }}}$
Mass of the Earth $M =\frac{4}{3} \pi R^3 \rho$
$
\begin{aligned}
\therefore \quad g & =\frac{G}{ R ^2}\left[\frac{4}{3} \pi R ^3 \rho\right] \\
\rho & =\frac{3 g}{4 \pi RG }
\end{aligned}
$

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Question 152 Marks

The escape velocity of a body from Earth’s surface is ve. What will be the escape velocity of the same body from a height equal to 7R from Earth’s surface.

Answer

$v_e \propto \frac{1}{\sqrt{r}}$ where $r$ is the position of body from the surface.$\begin{array}{c}\frac{v_e}{v^{\prime}}=\sqrt{\frac{r^{\prime}}{r}}=\sqrt{\frac{ R +7 R }{ R }}=\sqrt{\frac{8 R }{ R }}=\sqrt{8}=2 \sqrt{2} \\\therefore v^{\prime}=\frac{v_e}{2 \sqrt{2}}\end{array}$

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Question 162 Marks

If the mean radius of the Earth is R, its angular velocity is ω, and the acceleration due to gravity at the surface of the Earth is g, then what will be the cube of the radius of the orbit of a geostationary satellite.

Answer

Let $r$ be the radius of the geostationary orbit. Angular velocity of revolution of a geostationary satellite is same as the angular velocity of rotation of the Earth.
$
\begin{aligned}
m r \omega^2 & =\frac{ GM m}{r^2} \\
r^3 & =\frac{ GM m}{m \omega^2} \times \frac{ R ^2}{ R ^2}=\frac{ GM }{ R ^2}\left(\frac{ R ^2}{\omega^2}\right) \quad\left[g=\frac{ GM }{ R ^2}\right] \\
r^3 & =\frac{g R ^2}{\omega^2} \quad \text { }
\end{aligned}
$

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Question 172 Marks

The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms-2 and 6400 km respectively. The required energy for this work will be.

Answer

Potential energy $U=-m g R_e+m g h$(The first term is independent of the height, so it can be taken to zero.)
$W=U=m g h[h \approx R]$
$=1000 \times 10 \times 6400 \times 10^3=64 \times 10^9$
$W=6.4 \times 10^{10} J$

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Question 182 Marks

A particle of mass $10$ g is kept on the surface of a uniform sphere of mass $100$ kg and radius $10$ cm. Find the work to done against the gravitational force between them to take the particle is away from the sphere.

Answer

$ U =\frac{- GMm m}{ R }=\frac{-6.67 \times 10^{-11} \times 100 \times 10 \times 10^{-3}}{10 \times 10^{-2}}$
$U =6.67 \times 10^{-10} J \text { }$
So, the amount of work done to take the particle upto infinite will be $6.67 \times 10^{-10} J$

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Question 192 Marks

A body of mass $‘m’$ kg starts falling from a point $2R$ above the Earth’s surface. What is its $K.E.$ When it has fallen to a point $‘R’$ above the Earth’s surface.

Answer

Potential energy
$ U =\frac{- GM m}{r}=\frac{ GM m}{ R +h}$
$U _{\text {initial }}=-\frac{ GM m}{r} \text { and } U _{\text {final }}=-\frac{ GM m}{2 R }$
$\text { Loss of P.E. }=\text { gain in K.E. }=\frac{ GM m}{2 R }-\frac{ GM m}{3 R }=\frac{ GM m}{6 R }$
$\therefore \quad \text { K.E. }=\frac{ GMm }{6 R } $

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Question 202 Marks

An infinite number of bodies, each of mass 2 kg, are situated on x-axis at distances lm, 2m, 4m, 8m from the origin. What will be the resultant gravitational potential due to this system at the origin.

Answer

Gravitational potential at the origin is
$
\begin{aligned}
& \mathrm{V}=-\mathrm{G}\left[\frac{m}{r_1}+\frac{m}{r_2}+\frac{m}{r_3}+\frac{m}{r_4}+\ldots .\right]=-\mathrm{G}\left[\frac{2}{1}+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\ldots .\right] \\
& =-2 \mathrm{G}\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots . .\right]=-2 \mathrm{G}\left[\frac{1}{1-\left(\frac{1}{2}\right)}\right] \\
& V=-4 G \\
&
\end{aligned}
$

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Question 212 Marks

Two satellites A and B of the same mass are revolving around the Earth in circular orbits such that the distance of B from the centre of the Earth is thrice as compared to the distance of A from the centre. What will be the ratio of centripetal force on B to that on A.

Answer

The necessary centripetal force is provided by the gravitational force of attraction, for circular orbit
Force on $A , \frac{m v^2}{r}=\frac{ GM m}{r^2}$
Force on $B , \frac{m v^2}{r}=\frac{ GMm }{(3 r)^2}$
The ratio of centripetal force,
$\frac{ F _{ B }}{ F _{ A }}=\frac{ GM m}{9 r^2} \times \frac{r^2}{ GM m} $
$\frac{ F _{ B }}{ F _{ A }}=\frac{1}{9}$

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Question 222 Marks

Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational force between them is proportional to.

Answer

The gravitational force $ F=\frac{G(M)(M)}{(2 R)^2}=\frac{G\left(\frac{4}{3} \pi R^3 \rho\right)^2}{4 R^2} $ Mass of the solid sphere $M=\frac{4}{3} \pi R^3 \rho$ $ F=\frac{G\left(16 \pi^2 R^6 \rho^2\right.}{9 \times 4 \times R^2} $ $ F=\frac{4}{9} G \pi^2 \rho^2 R^4 ; F \propto \frac{4}{9} \pi^2 \rho^2 R^4 $ $\therefore$ The gravitational force between them is proportional to 4 th power of radius.

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Question 232 Marks

A satellite moves in a circle around the Earth, the radius of this circle is equal to one half of the radius of the Moon’s orbit. The satellite completes one revolution in…… lunar month.

Answer

Time period of revolution of moon around the Earth $T_m=1$ lunar month$\begin{aligned}\frac{ T _e}{ T _m} & =\left(\frac{a_e}{a_m}\right)^{\frac{3}{2}} \\T _e & = T _m \times\left(\frac{1}{2}\right)^{\frac{3}{2}}=1 \times(2)^{\frac{-3}{2}} \\T _e & =2^{\frac{-3}{2}} \text { lunar month }\end{aligned}$

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Question 242 Marks

The figure shows elliptical orbit of a planet ‘M’ about the Sun ‘S’, the shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C and D and t2 is the time to move from A to B then.

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Question 252 Marks

The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period will become.

Answer

$T _2= T _1\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}}= T _1\left[\frac{4 a_1}{a_1}\right]^{\frac{3}{2}}=5(4)^{\frac{3}{2}}=5(8)$$T _2=40$ hours

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Question 262 Marks

A geo-stationary satellite is orbiting the Earth of a height of $6R$ above the surface of Earth R being the radius of the Earth calculate the time period of another satellite at a height of $2.5R$ from the surface of Earth. Distance of satellite from the center are $7$R and $3.5$ R respectively.

Answer

$\frac{ T _2}{ T _1}=\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}}$
$ T _2= T _1\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}}=24\left(\frac{3.5 R }{7 R }\right)^{\frac{3}{2}} $
$T _2=8.49 \text { (or) } 6 \sqrt{2} \text { hours }$

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Physics [ 2 Marks Questions ]

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