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Physics 4 Marks Question

PART - 1 CH - 7 Gravitation · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 9 questions.

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Question 14 Marks
A comet orbits the sun in a highly elliptical orbit. does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the sun.
Answer
(a) According to Kepler's second law, when the comet is near the sun, it moves fast. But when the planet
is far away, it moves slowly. Therefore, we can say that the linear motion of the comet doesn't remain constant.
(b) The angular speed of the comet also changes slightly.
(c) As per the rules, the angular momentum of the comet is conserved.
(d) The linear speed of the rotating comet is changing from time to time hence its kinetic energy also changes
continuously.
(e) The value of potential energy depends on the distance from the Sun to the Earth. This is the reason why
the potential energy keep changing in an elliptical orbit as the distance between the Sun and the comet changes continously.
(f) We know that mechanical energy i.e., total energy is equal to the sum of kinetic energy and potential energy. Hence this value always remains constant according to the law of conservation of energy. The total energy of the comet remains constant.
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Question 24 Marks
Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky way to be $10^5 ly$.
Answer
Given,
Radius of a star $(r)=$ Distance of the star from the center of galaxy
$
\begin{array}{l}
=50,000 ly \\
=50000 \times 9.46 \times 10^{15} m \\
=4.73 \times 10^{20} m
\end{array}
$
Let us consider the mass of the stars of the galaxy
$
M=2.5 \times 10^{11} \times 2 \times 10^{30} kg
$
$
\begin{aligned}
\because \quad \text { Mass of one sun } & =2 \times 10^{30} kg \\
& =5 \times 10^{41} kg \\
G & =6.67 \times 10^{-11} Nm^2 / kg^2
\end{aligned}
$
$T =$ Time period of one revolution $=$ ?
Diameter of galaxy $=10^5$ light years
We know,
$\begin{array}{llrl} & M _{ l } & =\frac{4 \pi^2 r^3}{ GT ^2} \\ \text { or } & T ^2 & =\frac{4 \pi^2 r^3}{ GM } \\ \therefore & T & =\sqrt{\frac{4 \pi^2 r^3}{ GM }}\end{array}$
Put the values
$\begin{array}{l}=\sqrt{\frac{4 \times(3.14)^2 \times\left(4.73 \times 10^{20}\right)^3}{6.67 \times 10^{-11} \times 5 \times 10^{41}}} \\ =\sqrt{\frac{4173.53 \times 10^{60} \times 10^{11}}{5 \times 6.67 \times 10^{41}}} \\ =\sqrt{125.14 \times 10^{30}} \\ =11.2 \times 10^{15} \\ =1.12 \times 10^{16} \text { seconds }\end{array}$
In converting in years,
$\begin{array}{l}=\frac{1.12 \times 10^{16}}{365 \times 24 \times 3600} \\ =3.55 \times 10^8 \text { years }\end{array}$
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Question 34 Marks
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Answer
Suppose A and B are different positions of two spheres and the midpoint of AB is P.
The gravitational field at P due to mass at A is
Image
$
\begin{aligned}
E_1 & =\frac{GM}{r^2} \\
& =\frac{G \times 100}{(0.5)^2}(Along PA)
\end{aligned}
$
Gravitational field at P due to mass B is
$
E_2=\frac{G \times 100}{(0.5)^2} \quad[\text { Along } P B]
$
$\because E _1$ and $E _2$ are equal in magnitude but being in the opposite directions the resultant gravitational field at P will be zero due to which the gravitational force will be zero. Gravitational potential is a scalar quantity, hence total gravitational potential
$\begin{aligned} V & = V _{ A }+ V _{ B } \\ & =-\frac{ GM }{r}-\frac{ GM }{r}=\frac{-2 GM }{r} \\ & =\frac{-2 \times 6.67 \times 10^{-11} \times 100}{0.5} \\ & =-2.668 \times 10^{-8} J / kg \\ & \cong-2.7 \times 10^{-8} J / kg \end{aligned}$
Thus, the object will be in equilibrium. But this will happen in a state of temporary equilibrium because even a slight displacement from position A and B will disturb the equilibrium that can't be restored.
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Question 44 Marks
Choose the correct alternative :
(a) Acceleration due to gravity increases/ decreases with increasing altitude.
(b) Acceleration due to gravity increases/ decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula - GMm $\left(1 / r_2-1 / r_1\right)$ is more/ less accurate than the formula $m g\left(r_2-r_1\right)$ for the difference of potential energy between two points $r_2$ and $r_1$ distance away from the centre of the earth.
Answer
(a) Gravitational acceleration decreases with increasing altitude.
$\because$ Acceleration due to gravity at height $h$
$
g_h=g\left(1-\frac{2 h}{R}\right)
$
Hence as h increases, the value of $g_h$ decreases.
(b) The acceleratrion due to gravity decreases with increasing depth (assuming the earth is a sphere of uniform density).
$\because \quad g_d=\left(1-\frac{d}{ R }\right)$
Hence as $d$ increases the value of gravitational acceleration $g_d$ decreases.
(c) $\because$ Gravitation acceleration $g=\frac{ GM }{ R ^2}$, where M is mass of Earth.
$\therefore$ Acceleration due to gravity does not depend on mass (m) of the object.
(d) The formula of potential energy difference between two points at distance $r_2$ and $r_1$ from the center of Earth is
-GM $m\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$ more accurate than the formula $m g\left(r_2-r_1\right)$.
The value of $g$ keep changing in different situations.
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Question 54 Marks
The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth ? Ignore the presence of the sun and other planets.
Answer
Suppose the initial speed of projectile on the surface of earth is v and the final speed of the projectile for away from earth is v'. Initial potential energy of the projectile (at the Earth's surface)
$=-\frac{ GM m}{ R }$
And final potential energy (at infinity) $=0$
Hence, from the law of conservation of energy, Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy.
$\Rightarrow \quad \frac{1}{2} m v ^2-\frac{ GM m}{ R }=\frac{1}{2} m v ^2+0$$\ldots$(1)
But the escape velocity of object
$
\begin{array}{rlrl}
v_e & =\sqrt{\frac{2 GM}{R}} \\
\therefore \quad & \frac{1}{2} m v_e^2 & =\frac{GM m}{R}
\end{array}
$
$\therefore$ From equation (1)
$
\frac{1}{2} m v^2-\frac{1}{2} m v_e^2=\frac{1}{2} m v^2
$
$\because$ According to question,
$
\begin{aligned}
v & =3 v_e \\
\therefore \quad \frac{1}{2} m\left(3 v_e\right)^2-\frac{1}{2} m v_e^2 & =\frac{1}{2} m v^{\prime 2} \\
\Rightarrow \quad 9 v_e^2-v_e^2 & =v^{\prime 2} \\
v^{\prime 2} & =8 v_e^2 \\
v^{\prime} & =\sqrt{8 v_e} \\
v^{\prime} & =2.83 \times 11.2 Kms^{-1} \\
v^{\prime} & =31.68 Kms^{-1}
\end{aligned}
$
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Question 64 Marks
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250N on the surface?
Answer
Let the weight of the object on the earth's surface is W and weight at depth d from the earth is $W _{ D }$. Given,
$W =m g=250 N$$\ldots$(1)
and
$W _d=m g_d$$\ldots$(2)
We know that,
$g_d=g\left(1-\frac{d}{ R }\right)$$\ldots$(3)
Given,
$d=\frac{1}{2} R , R =$ radius of earth.$\ldots$(4)
Put the values,
$\begin{aligned} g_d & =g\left(1-\frac{\frac{1}{2} R }{ R }\right) \\ & = g \left(1-\frac{1}{2}\right)=\frac{1}{2} g\end{aligned}$
$g_d=\frac{1}{2} g$$\ldots$(5)
Put this value in equation (2)
$\begin{aligned} W _d & =m \times \frac{1}{2} g=\frac{1}{2} m g \\ & =\frac{1}{2} W=\frac{1}{2} \times 250\end{aligned}$ [From equation(1)]
$W _d=125 N$
Thus, the weight of the object at half the distance twoards the center of earth is 125N.
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Question 74 Marks
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer
Given
Height from earth surface $h=\frac{1}{2} R$
Radius of earth $=R$
Gravitation force on this object due to earth $=$ ?
$
W=mg=63 N
$
We know that,
$
\begin{array}{l}
g_h=g\left(1+\frac{h}{R}\right)^{-2} \\
g_h=g\left(1+\frac{\frac{1}{2} R}{R}\right)^{-2}=g\left(1+\frac{1}{2}\right)^{-2}
\end{array}
$
or
$
=g\left(\frac{3}{2}\right)^{-2}=g\left(\frac{2}{3}\right)^2=\frac{4}{9} g
$
or
$
g_h=\frac{4}{9} g \ldots(1)
$
Let mass of object $=m$
If W and $W _{ h }$ are the weights on earth and weight at a height $h$ from earth is surface respectively, then,
W = mg = 63 N
And
$
\begin{aligned}
W_h & =m g_h=m \times \frac{4}{9} g \quad \because g_h=\frac{4}{9} h \\
W_h & =\frac{4}{9} m g \\
& =\frac{4}{9} \times 63=4 \times 7=28 N
\end{aligned}
$
$W _h=28 N$
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Question 84 Marks
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ?
Answer
We know from Kepler's third law that the square of the period of revolution of the planet $\propto$ (semi major axis).
That is $\quad T ^2 \propto r^3$
Here we will take $r_5$ for saturn and $r_{ E }$ for earth and Time for saturn is $T _{ S }$ and for earth it is $T _{ E }$.
$\therefore$ For saturn and saturn,
$T _{ S }^2 \propto r_{ S }^3$$\ldots$(1)
And $\quad T _{ E }^2 \propto r_{ E }^3$ $\ldots$(2)
By dividing equation (1) and (2),
$\left(\frac{ T _{ S }}{ T _{ E }}\right)^2=\left(\frac{r_{ S }}{r_{ E }}\right)^3$
It is given,
$T _{ S }=29.5 T_{ E }$
or
$
\frac{T_S}{T_E}=29.5
$
And, 
$r_{ E }=$ distance of Earth from Sun $=1.5 \times 10^8 km$
$r_{ S }=$ distance of Saturn from Sun $=$ ?
$
\begin{aligned}
(29.5)^2 & =\left(\frac{r_{S}}{1.5 \times 10^8}\right)^3=\frac{\left(r_{S}\right)^3}{3.75 \times 10^{24}} \\
\left(r_{S}\right)^3 & =(29.5)^2 \times 3.375 \times 10^{24} \\
& =2937 \times 10^{24} \\
r_{S} & =\left(2937 \times 10^{24}\right)^{1 / 3} \\
r_{c} & =14.3 \times 10^8 km
\end{aligned}
$
$=1.43 \times 10^9 km$
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Question 94 Marks
How will you ‘weigh the sun’, that is estimate its mass ? The mean orbital radius of the earth around the sun is 1.5 × 108 km.
Answer
It is given,
Radius of Earth's orbit $r=1.5 \times 10^8 km$
$
=1.5 \times 10^{11} m
$
Time of Earth's revolution around the Sun $=365$ days
$
\begin{aligned}
& =365 \times 24 \times 3600 sec \\
G & =6.67 \times 10^{-11} Nm^2 / kg^2
\end{aligned}
$
We know that, 
$\begin{aligned} & \frac{r}{T^2}=\frac{ GM _{ S }}{4 \pi^2 r^2} \\ \text { Here, } & M _{ S }=\text { Mass of Sun } \\ \therefore & M _{ S }=\frac{4 \pi^2 r^3}{T^2 G }\end{aligned}$
Put the values,
$\begin{array}{l}=\frac{4 \times 9.86 \times\left(1.5 \times 10^{11}\right)^3}{(365 \times 24 \times 3600)^2 \times 6.67 \times 10^{-11}} kg \\ =\frac{4 \times 9.86 \times 3.75 \times 10^{33}}{99.5 \times 10^{13} \times 6.67 \times 10^{-11}} \\ =\frac{147.9 \times 10^{31}}{663.7}\end{array}$
$\begin{array}{l}=0.223 \times 10^{31} \\ =2.23 \times 10^{30} kg\end{array}$
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