Answer

Consider the Universal law of gravitation,

According to that law, the force of attraction between two bodies is

$\mathrm{F} =\frac{(\mathrm{Gm_1m_2})}{r^2}$

Where,

$\mathrm{m_1}$ and $\mathrm{m2}$ are the masses of the two bodies.

$\mathrm{G}$ is the gravitational constant.

$\mathrm{r}$ is the distance between the two bodies.

Given that the distance is reduced to half then,

$\mathrm{r}= 1/2 ~r$

$\mathrm{F} =\frac{(\mathrm{Gm_1m_2})}{r^2}$

$\mathrm{F} =\frac{(\mathrm{Gm_1m_2})}{(\frac{r}{2})^2}$

$\mathrm{F} =\frac{4(\mathrm{Gm_1m_2})}{r^2}$

$\mathrm{F} = 4\mathrm{F}$

Therefore once the space between the objects is reduced to half, then the force of gravitation will increase by fourfold the first force.

Answer

All objects fall from the top with a constant acceleration called acceleration due to gravity (g). This is constant on earth and therefore the value of ‘g’ doesn’t depend on the mass of an object. Hence, heavier objects don’t fall quicker than light-weight objects provided there’s no air resistance.

Answer

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by

$$ F=\frac{\left(G m_1 m_2\right)}{r^2} $$

Here

$ \begin{aligned} & m_1=\text { mass of Earth }=6.0 \times 10^{24} \mathrm{~kg} \\ & m_2=\text { mass of the body }=1 \mathrm{~kg} \end{aligned} $

$r=$ distance between the two bodies

Radius of Earth $=6.4 \times 10^6 \mathrm{~m}$

$\mathrm{G}=$ Universal gravitational constant $=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$

By substituting all the values in the equation

$$ \begin{aligned} & F=\frac{\left(G m_1 m_2\right)}{r^2} \\ & F=\frac{6.67 \times 10^{-11}\left(6.0 \times 10^{24} \times 1\right)}{\left(6.4 \times 10^6\right)^2} \\ & \mathrm{~F}=9.8 \mathrm{~N} \\ \end{aligned} $$

This shows that Earth exerts a force of $9.8 \mathrm{~N}$ on a body of mass $1 \mathrm{~kg}$. The body will exert an equal force of attraction of $9.8 \mathrm{~N}$ on the Earth.

Answer

The earth attracts the moon with a force same as the force with which the moon attracts the earth. However, these forces are in opposite directions. By universal law of gravitation, the force between moon and also the sun can be

$$ F=\frac{\left(G m_1 m_2\right)}{d^2} $$

Where,

$\mathrm{d}=$ distance between the earth and moon.

$\mathrm{m}_1$ and $\mathrm{m}_2=$ masses of earth and moon respectively.

Answer

According to the universal law of gravitation and Newton’s third law, we all know that the force of attraction between two objects is the same, however in the opposite directions. So the earth attracts the moon with a force same as the moon attracts the earth but in opposite directions. Since earth is larger in mass compared to that of the moon, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. Therefore, for this reason the earth does not move towards the moon.

Answer

(i)

According to universal law of gravitation, the force between 2 objects $\left(\mathrm{m}_1\right.$ and $\left.\mathrm{m}_2\right)$ is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

$$ F=\frac{\left(G 2 m_1 m_2\right)}{R^2} $$

If the mass is doubled for one object.

$\mathrm{F}=2 \mathrm{~F}$, so the force is also doubled.

(ii)

If the distance between the objects is doubled and tripled

If it’s doubled

Hence,

$ \begin{aligned} & \mathrm{F}=\left(\mathrm{Gm}_1 \mathrm{~m}_2\right) /(2 \mathrm{R})^2 \\ & \mathrm{~F}=1 / 4\left(\mathrm{Gm}_1 \mathrm{~m}_2\right) / \mathrm{R}^2 \\ & \mathrm{~F}=\mathrm{F} / 4 \end{aligned} $

Force thus becomes one-fourth of its initial force.

Now, if it’s tripled

Hence,

$ \begin{aligned} & \mathrm{F}=\left(\mathrm{Gm}_1 \mathrm{~m}_2\right) /(3 \mathrm{R})^2 \\ & \mathrm{~F}=1 / 9\left(\mathrm{Gm}_1 \mathrm{~m}_2\right) / \mathrm{R}^2 \\ & \mathrm{~F}=\mathrm{F} / 9 \end{aligned} $

Force thus becomes one-ninth of its initial force.

(iii)

If masses of both the objects are doubled, then

$ F=\frac{\left(G 2 m_1 2 m_2\right)}{R^2} $

$F=4 F$, Force will therefore be four times greater than its actual value.

Answer

The universal law of gravitation explains many phenomena that were believed to be unconnected:

(i) The motion of the moon round the earth

(ii) The responsibility of gravity on the weight of the body which keeps us on the ground

(iii) The tides because of the moon and therefore the Sun

(iv) The motion of planets round the Sun

Answer

Acceleration due to gravity is the acceleration gained by an object due to gravitational force. On Earth, all bodies experience a downward force of gravity which Earth’s mass exerts on them. The Earth’s gravity is measured by the acceleration of the freely falling objects. At Earth’s surface, the acceleration of gravity is $9.8 \mathrm{~ms}^{-2}$ and it is denoted by ’ $\mathrm{g}$ ‘. Thus, for every second an object is in free fall, its speed increases by about 9.8 metres per second.

Answer

The gravitation force between the earth and an object is called weight. Weight is equal to the product of acceleration due to the gravity and mass of the object.

Answer

The weight of a body on the earth’s surface;

W = mg (where m = mass of the body and g = acceleration due to gravity)

The value of g is larger at poles when compared to the equator. So gold can weigh less at the equator as compared to the poles.

Therefore, Amit’s friend won’t believe the load of the gold bought.

Answer

A sheet of paper has a larger surface area when compared to a crumpled paper ball. A sheet of paper will face a lot of air resistance. Thus, a sheet of paper falls slower than the crumpled ball.

Answer

Given data:

Acceleration due to earth’s gravity $=\mathrm{g}_{\mathrm{c}}$ or $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$

Object’s mass, $\mathrm{m}=10 \mathrm{~kg}$

Acceleration due to moon gravity $=\mathrm{g}_{\mathrm{m}}$

Weight on the earth $=\mathrm{W}_{\mathrm{c}}$

Weight on the moon $=\mathrm{W}_{\mathrm{m}}$

Weight $=$ mass $\times$ gravity

$\mathrm{g_m}=(1 / 6) \mathrm{g_c}$ (given)

So $\mathrm{W_m}=\mathrm{mg_m}=\mathrm{mx}(1 / 6) \mathrm{g_c}$

$\mathrm{W}_{\mathrm{m}}=10 \times(1 / 6) \times 9.8=16.34 \mathrm{~N}$

$\mathrm{W_c}=\mathrm{m} \mathrm{x} \mathrm{g_c}=10 \times 9.8$

$\mathrm{W}_{\mathrm{e}}=98 \mathrm{~N}$

Answer

Given data:

Initial velocity $\mathrm{u}=49 \mathrm{~m} / \mathrm{s}$

Final speed $\mathrm{v}$ at maximum height $=0$

Acceleration due to earth gravity $\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^2$ (thus negative as ball is thrown up).

By third equation of motion,

$ \begin{aligned} & 2 \mathrm{gH}=\mathrm{v}^2-\mathrm{u}^2 \\ & 2 \times(-9.8) \times \mathrm{H}=0-(49)^2 \\ & -19.6 \mathrm{H}=-2401 \\ & \mathrm{H}=122.5 \mathrm{~m} \end{aligned} $

Total time $\mathrm{T}=$ Time to ascend $\left(\mathrm{T_a}\right)+$ Time to descend $\left(\mathrm{T_d} \right)$

$ \begin{aligned} & \mathrm{v}=\mathrm{u}+\mathrm{gt} \\ & 0=49+(-9.8) \times \mathrm{T}_{\mathrm{a}} \\ & \mathrm{Ta}=(49 / 9.8)=5 \mathrm{~s} \end{aligned} $

Also, $\mathrm{T}_{\mathrm{d}}=5 \mathrm{~s}$

Therefore $\mathrm{T}=\mathrm{T_a} +\mathrm{T_d}$

$ \begin{aligned} & \mathrm{T}=5+5 \\ & \mathrm{~T}=10 \mathrm{~s} \end{aligned} $

Answer

Given data:

Initial velocity

$ \mathrm{u}=0 $

Tower height $=$ total distance $=19.6 \mathrm{~m}$

$ \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2 $

Consider third equation of motion

$ \begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gs} \\ & \mathrm{v}^2=0+2 \times 9.8 \times 19.6 \\ & \mathrm{v}^2=384.16 \\ & \mathrm{v}=\sqrt{ }(384.16) \\ & \mathrm{v}=19.6 \mathrm{~m} / \mathrm{s} \end{aligned} $

Answer

Given data:

Initial velocity $u=40 \mathrm{~m} / \mathrm{s}$

$ \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 $

Max height final velocity $=0$

Consider third equation of motion

$\mathrm{v}^2=\mathrm{u}^2-2 \mathrm{gs}$ [negative as the object goes up]

$ \begin{aligned} & 0=(40)^2-2 \times 10 \times s \\ & \mathrm{~s}=(40 \times 40) / 20 \end{aligned} $

Maximum height $\mathrm{s}=80 \mathrm{~m}$

Total Distance $=\mathrm{s}+\mathrm{s}=80+80$

Total Distance $=160 \mathrm{~m}$

Total displacement $=0$ (The first point is the same as the last point)

Answer

Given data:

Mass of the sun $\mathrm{m}_{\mathrm{s}}=2 \times 10^{30} \mathrm{~kg}$

Mass of the earth $\mathrm{m}_{\mathrm{c}}=6 \times 10^{24} \mathrm{~kg}$

Gravitation constant $\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 / \mathrm{kg}^2$

Average distance $\mathrm{r}=1.5 \times 10^{11} \mathrm{~m}$

Consider Universal law of Gravitation

$ \begin{aligned} & F=\frac{\left(G m_1 m_2\right)}{d^2} \\ & F=\frac{\left(6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}\right)}{\left(1.5 \times 10^{11}\right)^2} \\ & \mathrm{~F}=3.56 \times 10^{22} \mathrm{~N} \end{aligned} $

Answer

Given data:

(i) When the stone from the top of the tower is thrown, Initial velocity $u^{\prime}=0$

Distance travelled $=\mathrm{x}$

Time taken $=\mathrm{t}$

Therefore,

$ \begin{aligned} & s=u t+\frac{1}{2} g t^2 \\ & x=0+(1 / 2) g^2 \\ & x=5 t^2—–(a) \end{aligned} $

(ii) When the stone is thrown upwards,

Initial velocity $u=25 \mathrm{~m} / \mathrm{s}$

Distance travelled $=(100-\mathrm{x})$

Time taken $=\mathrm{t}$

$ \begin{aligned} & s=u t-\frac{1}{2} g t^2 \\ & (100-\mathrm{x})=25 \mathrm{t}-(1 / 2) \times 10 \mathrm{xt}^2 \\ & \mathrm{x}=100-25 \mathrm{t}+5 \mathrm{t}^2—– \text { (b) } \end{aligned} $

From equations (a) and (b)

$ \begin{aligned} & 5 \mathrm{t}^2=100-25 \mathrm{t}+5 \mathrm{t}^2 \\ & \mathrm{t}=(100 / 25)=4 \mathrm{sec} . \end{aligned} $

After 4sec, two stones will meet

From (a)

$ \mathrm{x}=5 \mathrm{t}^2=5 \mathrm{x} 4 \times 4=80 \mathrm{~m} . $

Putting the value of $x$ in (100-x)

$ =(100-80)=20 \mathrm{~m} \text {. } $

This means that after $4 \mathrm{sec}, 2$ stones meet a distance of $20 \mathrm{~m}$ from the ground.

Answer

Given data:

$ \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 $

Total time $\mathrm{T}=6 \mathrm{sec}$

$ \mathrm{T_a}=\mathrm{T_d}=3 \mathrm{sec} $ (a) Final velocity at maximum height $\mathrm{v}=0$

From first equation of motion:-

$ \begin{aligned} & \mathrm{v}=\mathrm{u}-\mathrm{gt_a} \\ & \mathrm{u}=\mathrm{v}+\mathrm{gt_a} \\ & =0+10 \times 3 \\ & =30 \mathrm{~m} / \mathrm{s} \end{aligned} $

The velocity with which stone was thrown up is $30 \mathrm{~m} / \mathrm{s}$.

(b) From second equation of motion

$ \begin{aligned} & s=u t_a-\frac{1}{2} g\left(t_a\right)^2 \\ & =30 \times 3-(1 / 2) \times 10 \times(3)^2 \\ & =90-45=45 \mathrm{~m} \end{aligned} $

The maximum height stone reaches is $45 \mathrm{~m}$.

(c) In $3 \mathrm{sec}$, it reaches the maximum height.

Distance travelled in another $1 \mathrm{sec}=\mathrm{s}$,

$ \begin{aligned} & s=u t_a-\frac{1}{2} g\left(t_a\right)^2 \\ & \mathrm{~s}=0+10 \times 1 \times 1 \\ & \mathrm{~s}=5 \mathrm{~m} . \end{aligned} $

The distance travelled in another $1 \mathrm{sec}=5 \mathrm{~m}$.

Therefore in $4 \mathrm{sec}$, the position of point $\mathrm{p}(45-5)$ $=40 \mathrm{~m}$ from the ground.

Answer

The buoyant force on an object that is immersed in a liquid will be in a vertically upward direction.

Answer

The density of plastic is lesser than that of water. Therefore, the force of buoyancy on plastic block will be greater than the weight of plastic block. Hence, the acceleration of plastic block is going to be in the upward direction. So, the plastic block comes up to the surface of water.

Answer

To find the Density of the substance the formula is

Density $=($ Mass/Volume $)$

Density $=(50 / 20)=2.5 \mathrm{~g} / \mathrm{cm}^3$

Density of water $=1 \mathrm{~g} / \mathrm{cm}^3$

Density of the substance is greater than density of water. So the substance will sink.

Answer

Density of sealed packet $=500 / 350=1.42 \mathrm{~g} / \mathrm{cm}^3$

Density of sealed packet is greater than density of water

Therefore the packet will sink.

Considering Archimedes Principle,

Displaced water volume $=$ Force exerted on the sealed packet.

Volume of water displaced $=350 \mathrm{~cm}^3$

Therefore displaced water mass $=\rho \times \mathrm{V}$

$ =1 \times 350 $

Mass of displaced water $=350 \mathrm{~g}$.