3 Marks Question

Question 13 Marks

The geostationary orbit of the earth is at a distance of about $36000\ km$ from the earth's surface. Find the weight of a $120\ kg $ equipment placed in a geostationary satellite. The radius of the earth is $6400\ km.$

Answer

The geostationary orbit of the Earth is at a distance of about $36000\ km.$
We know that the value acceleration due to gravity above the surface of the Earth is given by $\text{g}'=\frac{\text{Gm}}{(\text{R + h})^2}.$
At h = 36000km, we have:$\text{g}'=\frac{\text{Gm}}{(36000+6400)^2}$
At the surface, we have:$\text{g}=\frac{\text{Gm}}{(6400)^2}$
$\therefore\frac{\text{g}'}{\text{g}}=\frac{6400\times6400}{42400\times42400}$
$=\frac{256}{106\times106}=0.0228$
$\Rightarrow\text{g}'=0.0227\times9.8=0.223 [$Taking $g = 9.8m/s^2$ at the surface of the earth$]$
For a $120\ kg$ equipment placed in a geostationary satellite, its weight will be mg' $= 120 \times 0.233$
$\Rightarrow26.76\approx27\text{N}$

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Question 23 Marks

The gravitational force acting on a particle of $1g$ due to a similar particle is equal to $6.67 \times 10^{-17}N$. Calculate the separation between the particles.

Answer

Mass of the particle $m = 1 \text{gm}=\frac{1}{1000}\text{kg}$
Let the distance between the two particles be $r.$
Gravitational force between the particle, $F = 6.67 \times 10^{-17}N$
Also, $\text{F}=\frac{\text{Gm}_1\text{m}_2}{\text{r}^2}$
Substituting the respective values in the above formula, we get:$6.67\times10^{-17}=\frac{6.67\times10^{-11}\times\big(\frac{1}{10000}\big)\times\big(\frac{1}{10000}\big)}{\text{r}^2}$
$\Rightarrow\text{r}^2=\frac{6.67\times10^{-6}\times10^{-11}}{6.67\times10^{-17}}$
$\Rightarrow\frac{10^{-17}}{10^{-17}}=1$
$\Rightarrow\text{r}=\sqrt{1}=1\text{m}$

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Question 33 Marks

When you hold a pen and write on your notebook, what kind of force is exerted by you on the pen? By the pen on the notebook? By you on the notebook?

Answer

Yes, In case of moving frame of reference a uniformly moving particle can undergo curved path.
For example if you roll a ball out of marry go round while it is rotating the ball as viewed from marry go round would be going in straight direction outward of Mary go round. but as seen from ground the ball will be undergoing semicircle like shape.

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Question 43 Marks

At what distance should two charges, each equal to $1C,$ be placed so that the force between them equals your weight?

Answer

Given: $q_1 = q_2 = 1C$ By Coulomb's law, the force of attraction between the two charges is given by
$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$=\frac{9\times10^9\times1\times1}{\text{r}^2}$
However, the force of attraction is equal to the weight $(F = mg).$
$\therefore\text{mg}=\frac{9\times10^9}{\text{r}^2}$
$\Rightarrow\text{r}^2=\frac{9\times10^9}{\text{m}\times10}=\frac{9\times10^8}{\text{m}} ($Taking $g = 10\ m/s^2)$
$\Rightarrow\text{r}^2=\frac{9\times10^8}{\text{m}}$
$\Rightarrow\text{r}=\frac{3\times10^4}{\sqrt{\text{m}}}$
Assuming that $m = 81\ kg,$
we have:$\text{r}=\frac{3\times10^4}{\sqrt{81}}$
$=\frac{3}{9}\times10^4\text{m}$
$=3333.3\text{m}$
$\therefore$ The distance $r$ is $3333.3m.$

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Question 53 Marks

The average separation between the proton and the electron in a hydrogen atom in ground state is $5.3 \times 10^{-11} m.$

Calculate the Coulomb force between them at this separation.
When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?

Answer

Average separation between the proton and the electron of a Hydrogen atom in ground state $, r = 5.3 \times 10^{−11}m$

Coulomb force when the proton and the electron in a hydrogen atom in ground state

$\text{F}=9\times10^9\times\frac{\text{q}_1\text{q}_2}{\text{r}_2}$
$=\frac{9\times10^9\times(1.6\times10^{-19})^2}{(5.3\times10^{-11})^2}=8.2\times10^{-8}\text{N}$

Coulomb force when the average distance between the proton and the electron becomes $4 $ times that of its ground state

Coulomb force, $\text{F}=\frac{1}{4\pi\in_0}=\frac{\text{q}_1\text{q}_2}{(4\text{r})^2}$
$=\frac{9\times10^9\times(1.6\times10^{-19})^2}{16\times(5.3)^2\times10^{-22}}$
$=\frac{9\times(1.6)^2}{10\times(5.3)^2}\times10^{-7}$
$=0.0512\times10^{-7}$
$=5.1\times10^{-9}\text{N}$

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Question 63 Marks

A lawyer alleges in court that the police has forced his client to issue a statement of confession. What kind of force is this?

Answer

In car the coin would fall directly downward in case viewed from car. in case viewed from outside the coin would be under projectile trajectory as if coin was projected horizontally with the speed of car.
In case of freely falling elevator the coin would retain its position where it was left as it will not feel any gravity.

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Question 73 Marks

Figure shows a cart. Complete the table shown.

Answer

Force on	Force by	Nature	Direction
cart	gravity(weight)	gravitational	downward
	friction(road)	mechanical	backward
	pull (horse)	mechanical	forward
horse	cart(weight)	gravitational	downward
horse	pull(cart)	mechanical	backward
driver	pull (cart)	mechanical	forward

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