Textbook - GRAVITATION Questions in English

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force.
The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For two objects of masses $m_1$ and $m_2$ and the distance between them $r$,the force $(F)$ of attraction acting between them is given by the universal law of gravitation as:
$F = \frac{G m_1 m_2}{r^2}$
Where,$G$ is the universal gravitation constant and its value is $6.67 \times 10^{-11} \text{ Nm}^2 \text{ kg}^{-2}$.

(N/A) Let $M_E$ be the mass of the Earth and $m$ be the mass of an object on its surface. If $R$ is the radius of the Earth,then according to the universal law of gravitation,the gravitational force $(F)$ acting between the Earth and the object is given by the relation:
$F = \frac{G M_E m}{R^2}$
Where:
$G$ is the universal gravitational constant,
$M_E$ is the mass of the Earth,
$m$ is the mass of the object,
$R$ is the distance between the center of the Earth and the object (which is equal to the radius of the Earth).

(N/A) The gravity of the Earth attracts every object towards its centre. When an object is released from a height,it falls towards the surface of the Earth solely under the influence of gravitational force. This motion of the object is known as free fall.

(N/A) When an object falls towards the ground from a height,its velocity changes during the fall. This change in velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity,denoted by $g$. Its standard value on the surface of the Earth is approximately $9.8 \, m/s^2$.

(N/A)

$S.No.$	Mass vs Weight
$I.$	Mass is the quantity of matter contained in the body; Weight is the force of gravity acting on the body.
$II.$	Mass is the measure of inertia of the body; Weight is the measure of gravity.
$III.$	Mass is a constant quantity; Weight is not constant and varies at different places.
$IV.$	Mass is a scalar quantity (only magnitude); Weight is a vector quantity (magnitude and direction).
$V.$	The $SI$ unit of mass is kilogram $(kg)$; The $SI$ unit of weight is Newton $(N)$.

(N/A) Let $M_E$ be the mass of the Earth and $m$ be the mass of an object on the surface of the Earth. Let $R_E$ be the radius of the Earth. According to the universal law of gravitation,the weight $W_E$ of the object on the surface of the Earth is given by:
$W_E = \frac{G M_E m}{R_E^2}$
Let $M_M$ and $R_M$ be the mass and radius of the moon,respectively. Then,according to the universal law of gravitation,the weight $W_M$ of the object on the surface of the moon is given by:
$W_M = \frac{G M_M m}{R_M^2}$
Now,taking the ratio of the two weights:
$\frac{W_M}{W_E} = \frac{M_M R_E^2}{M_E R_M^2}$
Given values:
$M_E = 5.98 \times 10^{24} \text{ kg}$
$M_M = 7.36 \times 10^{22} \text{ kg}$
$R_E = 6.4 \times 10^6 \text{ m}$
$R_M = 1.74 \times 10^6 \text{ m}$
Substituting these values:
$\frac{W_M}{W_E} = \frac{7.36 \times 10^{22} \times (6.4 \times 10^6)^2}{5.98 \times 10^{24} \times (1.74 \times 10^6)^2} \approx 0.165 \approx \frac{1}{6}$
Therefore,the weight of an object on the moon is $\frac{1}{6}^{th}$ of its weight on the Earth.

(B) It is difficult to hold a school bag with a thin strap because the pressure exerted on the shoulders is very high.
Pressure is defined as force per unit area,given by the formula $P = F/A$.
Since pressure is inversely proportional to the surface area $(A)$ on which the force $(F)$ acts,a smaller surface area results in a higher pressure.
In the case of a thin strap,the contact surface area is very small,which leads to a significantly larger pressure on the shoulder,causing pain and discomfort.

(B) The upward force exerted by a liquid on an object immersed in it is known as buoyancy.
When an object is immersed in a fluid,the fluid exerts an upward force on the object.
This force is called the buoyant force or upthrust.
As the object is pushed deeper into the liquid,the volume of the displaced liquid increases,which leads to an increase in the buoyant force.

(C) Whether an object floats or sinks depends on the relative density of the object compared to the liquid.
If the density of an object is greater than the density of the liquid,the weight of the object (force of gravity) exceeds the buoyant force,causing it to sink.
Conversely,if the density of the object is less than the density of the liquid,the buoyant force acting on the object is greater than the force of gravity,causing it to float on the surface.

(A) When you stand on a weighing machine,the machine measures the normal force exerted by your body on it.
Due to the presence of air,a buoyant force acts on your body in the upward direction.
This buoyant force reduces the net downward force exerted by your body on the weighing machine.
Therefore,the reading shown by the weighing machine is slightly less than your actual mass.
Thus,your actual mass is more than $42\, kg$.

(A) The bag of cotton is heavier than the iron bar.
This is because the volume of the cotton bag is much larger than that of the iron bar.
According to Archimedes' principle,an object immersed in a fluid (air) experiences an upward buoyant force equal to the weight of the fluid displaced.
Since the cotton bag occupies more space,it displaces a larger volume of air,resulting in a greater buoyant force acting on it compared to the iron bar.
This buoyant force reduces the apparent weight of the cotton bag on the weighing machine.
Therefore,for the weighing machine to show $100\, kg$ for both,the actual mass of the cotton bag must be greater than that of the iron bar to compensate for the higher buoyant force.

(D) According to the universal law of gravitation,the gravitational force $(F)$ acting between two objects is inversely proportional to the square of the distance $(r)$ between them,i.e.,
$F = G \frac{m_{1} m_{2}}{r^{2}}$
If the distance $r$ is reduced to $r/2$,the new gravitational force $(F')$ will be:
$F' = G \frac{m_{1} m_{2}}{(r/2)^{2}}$
$F' = G \frac{m_{1} m_{2}}{r^{2}/4}$
$F' = 4 \times (G \frac{m_{1} m_{2}}{r^{2}})$
$F' = 4F$
Therefore,if the distance is reduced to half,the gravitational force becomes four times the original value.

(B) According to Newton's second law of motion,$F = ma$. The gravitational force acting on an object is $F = G \frac{Mm}{r^2}$. Equating these,$ma = G \frac{Mm}{r^2}$,which simplifies to $a = G \frac{M}{r^2}$. Here,$a$ is the acceleration due to gravity $(g)$,$G$ is the gravitational constant,$M$ is the mass of the Earth,and $r$ is the distance from the center of the Earth. Since $g$ depends only on the mass of the Earth and the distance from its center,it is independent of the mass of the falling object $(m)$. Therefore,in the absence of air resistance,all objects fall with the same acceleration,and heavy objects do not fall faster than light objects.

(B) According to the universal law of gravitation,the gravitational force $F$ exerted on an object of mass $m$ is given by the formula:
$F = \frac{G M m}{r^2}$
Where:
Mass of Earth,$M = 6 \times 10^{24} \,kg$
Mass of object,$m = 1 \,kg$
Universal gravitational constant,$G = 6.67 \times 10^{-11} \,Nm^2 \,kg^{-2}$
Radius of the Earth,$r = R = 6.4 \times 10^6 \,m$
Substituting the values into the formula:
$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2}$
$F = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}}$
$F \approx 9.8 \,N$
Thus,the magnitude of the gravitational force is approximately $9.8 \,N$.

(C) According to Newton's universal law of gravitation,every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This force is mutual. Therefore,the Earth attracts the moon with a force that is exactly the same as the force with which the moon attracts the Earth. This is in accordance with Newton's third law of motion,which states that for every action,there is an equal and opposite reaction.

(N/A) According to Newton's third law of motion,the Earth and the moon exert equal and opposite gravitational forces on each other.
However,according to Newton's second law of motion,$F = ma$,the acceleration produced in a body is inversely proportional to its mass $(a = F/m)$.
The mass of the Earth is significantly larger than the mass of the moon.
Due to this massive difference in mass,the acceleration produced in the Earth is negligible compared to the acceleration produced in the moon.
Therefore,the Earth does not show any observable movement towards the moon.

(N/A) According to the universal law of gravitation,the force of gravitation between two objects is given by $F = \frac{G M m}{r^{2}}$.
$(i)$ $F$ is directly proportional to the masses of the objects. If the mass of one object is doubled,then the gravitational force will also get doubled.
$(ii)$ $F$ is inversely proportional to the square of the distance between the objects. If the distance is doubled,the gravitational force becomes one-fourth of its original value. Similarly,if the distance is tripled,the gravitational force becomes one-ninth of its original value.
$(iii)$ $F$ is directly proportional to the product of the masses of the objects. If the masses of both objects are doubled,the new force $F' = \frac{G(2M)(2m)}{r^{2}} = 4 \times \frac{G M m}{r^{2}} = 4F$. Thus,the gravitational force becomes four times the original value.

(D) The universal law of gravitation is significant because:
$1$. It binds us to the Earth.
$2$. It is responsible for the motion of the Moon around the Earth.
$3$. It is responsible for the motion of planets around the Sun.
$4$. It causes the occurrence of tides due to the Moon and the Sun.

(A) When objects fall towards the Earth under the effect of gravitational force alone,they are said to be in free fall.
The acceleration produced in an object due to the Earth's gravitational force is known as the acceleration due to gravity or acceleration of free fall.
Its value is approximately $9.8\, ms^{-2}$,which is constant for all objects regardless of their mass.

(B) The gravitational force exerted by the earth on an object is defined as the weight of that object.
It is calculated by the formula $W = m \times g$,where $m$ is the mass of the object and $g$ is the acceleration due to gravity.

(B) The weight of a body on the Earth is given by the formula $W = mg$.
Where:
$m$ = Mass of the body.
$g$ = Acceleration due to gravity.
The value of $g$ is greater at the poles than at the equator because the Earth is flattened at the poles and bulges at the equator.
Since $W = mg$,the weight of the same mass of gold will be greater at the poles and less at the equator.
Therefore,Amit's friend will not agree with the weight of the gold bought because the weight of the gold decreases when moved from the poles to the equator.

(B) When a sheet of paper is crumpled into a ball,its surface area exposed to the air decreases.
Since air resistance depends on the surface area of the object moving through it,the crumpled ball experiences significantly less air resistance compared to the flat sheet of paper.
As a result,the crumpled ball falls faster than the flat sheet of paper.

(C) Weight of an object on the Earth is calculated as: $W_e = m \times g$,where $m = 10\, kg$ and $g = 9.8\, m/s^2$.
So,$W_e = 10\, kg \times 9.8\, m/s^2 = 98\, N$.
The gravitational force on the moon is $\frac{1}{6}$ of that on the Earth.
Therefore,the weight of the object on the moon is: $W_m = \frac{1}{6} \times W_e$.
$W_m = \frac{1}{6} \times 98\, N \approx 16.3\, N$.
Thus,the weight on the moon is $16.3\, N$ and the weight on the Earth is $98\, N$.

(D) Given:
Initial velocity $u = 49\, m/s$
Final velocity at maximum height $v = 0\, m/s$
Acceleration due to gravity $g = -9.8\, m/s^2$ (acting downwards)
$(i)$ To find the maximum height $(h)$:
Using the third equation of motion: $v^2 - u^2 = 2gh$
$0^2 - (49)^2 = 2 \times (-9.8) \times h$
$-2401 = -19.6 \times h$
$h = 2401 / 19.6 = 122.5\, m$
$(ii)$ To find the total time of flight $(T)$:
Using the first equation of motion for ascent: $v = u + gt$
$0 = 49 + (-9.8) \times t$
$9.8t = 49$
$t = 5\, s$
Since the time of ascent equals the time of descent,the total time $T = 2 \times t = 2 \times 5 = 10\, s$.

(A) According to the equation of motion under gravity $v^2 - u^2 = 2gs$.
Where,
$u = \text{Initial velocity of the stone} = 0\, ms^{-1}$.
$v = \text{Final velocity of the stone}$.
$s = \text{Height of the tower} = 19.6\, m$.
$g = \text{Acceleration due to gravity} = 9.8\, ms^{-2}$.
Substituting the values in the equation:
$v^2 - 0^2 = 2 \times 9.8 \times 19.6$.
$v^2 = 19.6 \times 19.6 = (19.6)^2$.
$v = 19.6\, ms^{-1}$.
Hence,the velocity of the stone just before touching the ground is $19.6\, ms^{-1}$.

(B) Using the equation of motion $v^2 - u^2 = 2as$,where $a = -g$ for upward motion.
Given:
Initial velocity $u = 40 \, m/s$
Final velocity at maximum height $v = 0 \, m/s$
Acceleration $g = 10 \, m/s^2$
Using $v^2 = u^2 - 2gh$:
$0^2 = (40)^2 - 2(10)h$
$20h = 1600$
$h = 80 \, m$ (Maximum height).
Total distance covered = distance upward + distance downward = $80 \, m + 80 \, m = 160 \, m$.
Net displacement = final position - initial position = $0 \, m$ (since the stone returns to the starting point).

(C) According to the universal law of gravitation,the force of attraction between the Earth and the Sun is given by:
$F = \frac{G \times M_{\text{Sun}} \times M_{\text{Earth}}}{R^2}$
Where:
$M_{\text{Sun}} = 2 \times 10^{30} \, kg$
$M_{\text{Earth}} = 6 \times 10^{24} \, kg$
$R = 1.5 \times 10^{11} \, m$
$G = 6.7 \times 10^{-11} \, Nm^2 \, kg^{-2}$
Substituting the values:
$F = \frac{6.7 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{11})^2}$
$F = \frac{80.4 \times 10^{43}}{2.25 \times 10^{22}}$
$F = 35.73 \times 10^{21} \, N = 3.57 \times 10^{22} \, N$
Thus,the gravitational force is $3.57 \times 10^{22} \, N$.

(D) Let the two stones meet after a time $t$.
For the stone dropped from the tower:
Initial velocity,$u = 0\, m/s$.
Let the displacement of the stone in time $t$ from the top of the tower be $s$.
Acceleration due to gravity,$g = 9.8\, m/s^2$.
Using the equation of motion,$s = ut + \frac{1}{2}at^2$:
$s = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$s = 4.9t^2$ ............. $(1)$
For the stone thrown upwards:
Initial velocity,$u = 25\, m/s$.
Let the displacement of the stone from the ground in time $t$ be $s'$.
Acceleration due to gravity,$g = -9.8\, m/s^2$.
Using the equation of motion,$s' = ut + \frac{1}{2}at^2$:
$s' = 25t - \frac{1}{2} \times 9.8 \times t^2$
$s' = 25t - 4.9t^2$ ............. $(2)$
The sum of the displacements of both stones at the meeting point is equal to the total height of the tower,$100\, m$:
$s + s' = 100$
$4.9t^2 + 25t - 4.9t^2 = 100$
$25t = 100$
$t = 4\, s$
In $4\, s$,the distance covered by the falling stone from the top is:
$s = 4.9 \times (4)^2 = 4.9 \times 16 = 78.4\, m$.
Therefore,the stones will meet after $4\, s$ at a height of $(100 - 78.4) = 20.6\, m$ from the ground.

(A) Time of ascent is equal to the time of descent. The ball takes a total of $6\, s$ for its upward and downward journey.
Hence,it has taken $3\, s$ to attain the maximum height.
Final velocity of the ball at the maximum height,$v = 0\, m/s$.
Acceleration due to gravity,$g = -9.8\, m/s^2$.
Using the equation of motion,$v = u + at$,we have:
$0 = u + (-9.8 \times 3)$
$u = 9.8 \times 3 = 29.4\, m/s$.
Hence,the ball was thrown upwards with a velocity of $29.4\, m/s$.
$(b)$ Let the maximum height attained by the ball be $h$.
Initial velocity during the upward journey,$u = 29.4\, m/s$.
Final velocity,$v = 0\, m/s$.
Acceleration due to gravity,$g = -9.8\, m/s^2$.
Using the equation of motion,$s = ut + \frac{1}{2}at^2$:
$h = 29.4 \times 3 - \frac{1}{2} \times 9.8 \times 3^2 = 88.2 - 44.1 = 44.1\, m$.
Hence,the maximum height is $44.1\, m$.
$(c)$ The ball attains the maximum height after $3\, s$. After attaining this height,it starts falling downwards.
In this case,initial velocity,$u = 0\, m/s$.
The position of the ball after $4\, s$ of the throw is given by the distance traveled by it during its downward journey in $4\, s - 3\, s = 1\, s$.
Using the equation of motion,$s = ut + \frac{1}{2}at^2$:
$s = 0 \times 1 + \frac{1}{2} \times 9.8 \times 1^2 = 4.9\, m$.
Now,the total height is $44.1\, m$.
This means the ball is $39.2\, m$ $(44.1\, m - 4.9\, m)$ above the ground after $4\, s$.

(B) When an object is immersed in a liquid,the liquid exerts an upward force on the object due to the pressure difference between the top and bottom surfaces of the object. This upward force is known as the buoyant force or upthrust.

(B) Two forces act on an object immersed in water.
One is the gravitational force,which pulls the object downwards.
The other is the buoyant force,which pushes the object upwards.
If the upward buoyant force is greater than the downward gravitational force,the object will rise to the surface.
Since the density of plastic is less than the density of water,the buoyant force exerted by the water on the plastic block is greater than the gravitational force acting on it.
Therefore,the block of plastic rises to the surface when released under water.

(B) If the density of an object is more than the density of a liquid,then it sinks in the liquid. On the other hand,if the density of an object is less than the density of a liquid,then it floats on the surface of the liquid.
Here,the density of the substance $= \frac{\text{Mass of the substance}}{\text{Volume of the substance}} = \frac{50}{20} = 2.5\, g\, cm^{-3}$.
The density of the substance $(2.5\, g\, cm^{-3})$ is greater than the density of water $(1\, g\, cm^{-3})$. Therefore,the substance will sink in water.

(N/A) Density of the packet = $\frac{\text{Mass}}{\text{Volume}} = \frac{500\, g}{350\, cm^3} \approx 1.43\, g/cm^3$.
Since the density of the packet $(1.43\, g/cm^3)$ is greater than the density of water $(1\, g/cm^3)$,the packet will sink in water.
According to Archimedes' principle,the volume of water displaced by a fully submerged object is equal to the volume of the object.
Mass of displaced water = $\text{Volume of packet} \times \text{Density of water} = 350\, cm^3 \times 1\, g/cm^3 = 350\, g$.