Mix Example - GRAVITATION Questions in English

(A) When an object is in free fall,its acceleration is equal to the acceleration due to gravity $(g)$,which is independent of the mass of the object.
Since both objects are falling freely near the surface of the moon,they will experience the same acceleration due to gravity $(g_{moon})$.
Because they start from the same initial conditions (assuming they are dropped from the same height at the same time),their velocities at any given instant will be identical.
Therefore,they will have the same velocities at any instant.

(B) The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator.
As a result,the distance from the center of the Earth to the surface is greater at the equator than at the poles.
The formula for acceleration due to gravity is $g = \frac{GM}{R^2}$,where $g$ is inversely proportional to the square of the radius $(R)$.
Since the radius $R$ is largest at the equator,the value of $g$ is least at the equator.

(C) According to Newton's law of universal gravitation,the force $F$ between two objects of masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
When the masses of both objects are halved,the new masses become $m_1' = \frac{m_1}{2}$ and $m_2' = \frac{m_2}{2}$.
The new gravitational force $F'$ is $F' = G \frac{(\frac{m_1}{2})(\frac{m_2}{2})}{r^2} = G \frac{m_1 m_2}{4r^2} = \frac{1}{4} F$.
Therefore,the gravitational force becomes $F/4$.

(D) When an object is in circular motion,it experiences a centripetal force directed towards the center,which keeps it on the circular path.
At any given instant,the velocity of the object is directed along the tangent to the circular path.
When the string breaks,the centripetal force is no longer acting on the stone.
According to Newton's first law of motion,the object will continue to move in the direction of its velocity at that instant.
Therefore,the stone will move along a straight line tangential to the circular path at the point where the string broke.

$(A)$ According to the law of flotation, the weight of the object is equal to the weight of the displaced liquid.
Let $V$ be the total volume of the object and $d$ be the density of the object.
Weight of object = $V \cdot d \cdot g$.
If a fraction $f$ of the volume is outside the liquid, then the volume inside the liquid is $V_{in} = V(1 - f)$.
Weight of displaced liquid = $V(1 - f) \cdot d_{liquid} \cdot g$.
Equating the two: $V \cdot d \cdot g = V(1 - f) \cdot d_{liquid} \cdot g$, which simplifies to $d = (1 - f) \cdot d_{liquid}$, or $d_{liquid} = \frac{d}{1 - f}$.
For $d_1$, $f_1 = \frac{1}{9}$, so $d_1 = \frac{d}{1 - 1/9} = \frac{d}{8/9} = 1.125d$.
For $d_2$, $f_2 = \frac{2}{11}$, so $d_2 = \frac{d}{1 - 2/11} = \frac{d}{9/11} = 1.222d$.
For $d_3$, $f_3 = \frac{3}{7}$, so $d_3 = \frac{d}{1 - 3/7} = \frac{d}{4/7} = 1.75d$.
Comparing the values, we get $1.125d < 1.222d < 1.75d$, which implies $d_1 < d_2 < d_3$.

(B) The relation $F = G \frac{Mm}{d^2}$ represents Newton's Law of Universal Gravitation.
In this formula,$G$ is known as the Universal Gravitational Constant.
It is called 'universal' because its value remains constant throughout the universe,regardless of the location,the nature of the masses,or the medium between them.
Its value is approximately $6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ everywhere.
Therefore,option $B$ is the correct statement.

(C) Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically,it is expressed as $F = G \frac{m_1 m_2}{r^2}$,where $F$ is the gravitational force,$G$ is the universal gravitational constant,$m_1$ and $m_2$ are the masses of the two bodies,and $r$ is the distance between them.
Since this law applies to all objects with mass,it describes the gravitational force between any two bodies having some mass in the universe.

(D) The quantity $G$ represents the Universal Gravitational Constant.
By definition,a universal constant has the same value throughout the universe,regardless of the location,the mass of the objects,or the distance between them.
Therefore,the value of $G$ is independent of the mass and radius of the earth.

(A) According to Newton's law of universal gravitation,the force $F$ between two objects of masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
If the mass of each particle is doubled,the new masses become $m_1' = 2m_1$ and $m_2' = 2m_2$.
The new gravitational force $F'$ is given by $F' = G \frac{(2m_1)(2m_2)}{r^2} = 4 \times (G \frac{m_1 m_2}{r^2}) = 4F$.
Therefore,the gravitational force becomes $4$ times the original value.

(B) The atmosphere is a layer of gases surrounding the Earth.
These gas molecules have mass and are subject to the Earth's gravitational pull.
Gravity acts as the force that keeps these gases close to the Earth's surface,preventing them from escaping into outer space.
Therefore,the atmosphere is held to the Earth by gravity.

(C) According to Newton's law of universal gravitation,the force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$,where $G$ is the universal gravitational constant.
If $m_1 = 1 \text{ unit}$,$m_2 = 1 \text{ unit}$,and $r = 1 \text{ unit}$,then $F = G \frac{1 \times 1}{1^2} = G$.
Therefore,the force of attraction between two unit point masses separated by a unit distance is numerically equal to the universal gravitational constant $G$.

(D) The acceleration due to gravity $(g)$ is zero at the centre of the earth.
We know that weight $(W)$ is the product of mass $(m)$ and acceleration due to gravity $(g)$,given by the formula $W = m \times g$.
Since the acceleration due to gravity at the centre of the earth is $0$,the weight of the object at the centre of the earth will be $m \times 0 = 0$.

(A) The weight of the liquid displaced by an object is equal to the loss in weight of the object when immersed in the liquid.
Loss in weight = Weight in air - Weight in water.
Loss in weight = $10 \, N - 8 \, N = 2 \, N$.
According to Archimedes' principle,the buoyant force acting on an object is equal to the weight of the liquid displaced by the object,which is equal to the apparent loss in weight.

(B) Pressure $(P)$ is defined as force $(F)$ divided by the area $(A)$ over which it acts,given by the formula $P = F/A$.
Since the weight of the box (force) remains constant in all three cases,the pressure is inversely proportional to the surface area $(P \propto 1/A)$.
To obtain the maximum pressure,the surface area of the base must be the minimum.
The three possible base areas are:
$1$. Length $\times$ Breadth = $60 \,cm \times 40 \,cm = 2400 \,cm^2$
$2$. Breadth $\times$ Width = $40 \,cm \times 20 \,cm = 800 \,cm^2$
$3$. Width $\times$ Length = $20 \,cm \times 60 \,cm = 1200 \,cm^2$
Comparing these,the minimum area is $800 \,cm^2$,which occurs when the breadth and width form the base.
Therefore,the pressure is maximum when the breadth and width form the base.

(C) According to Newton's Third Law of Motion,for every action,there is an equal and opposite reaction.
Since the gravitational force is a mutual interaction between two bodies,the force exerted by the Earth on the apple $(F_1)$ is equal in magnitude to the force exerted by the apple on the Earth $(F_2)$.
Therefore,$F_1 = F_2$.

(N/A) The source of the centripetal force required for a planet to revolve around the Sun is the gravitational force exerted by the Sun on the planet.
According to Newton's law of universal gravitation,this force $F$ is given by the formula $F = G \frac{M \times m}{r^2}$,where $M$ is the mass of the Sun,$m$ is the mass of the planet,$r$ is the distance between them,and $G$ is the universal gravitational constant.
Therefore,the force depends on the product of the masses of the planet and the Sun $(M \times m)$ and is inversely proportional to the square of the distance $(r^2)$ between them.

(NONE) Both stones will reach the ground at the same time. This is because the vertical motion of both stones is governed by the same gravitational acceleration $(g)$ and they are released from the same initial height $(h)$. The horizontal velocity imparted to the first stone does not affect its vertical motion or the time taken to fall under gravity,as described by the equation of motion $h = \frac{1}{2}gt^2$,where $t = \sqrt{\frac{2h}{g}}$.

(B) The $Moon$ revolves around the $Earth$ due to the centripetal force provided by the gravitational pull of the $Earth$.
If the gravitational force of the $Earth$ suddenly becomes zero,the centripetal force will also become zero.
According to $Newton's$ first law of motion,an object in motion will continue to move in a straight line unless acted upon by an external force.
Therefore,the $Moon$ will begin to move in a straight line along the tangent to its circular orbit at the instant the gravity becomes zero.

(B) The acceleration due to gravity,denoted by $g$,is not uniform across the Earth's surface.
Due to the Earth's rotation and its equatorial bulge,the radius of the Earth is greater at the equator than at the poles.
Since $g = GM/R^2$,where $R$ is the radius of the Earth,the value of $g$ is smaller at the equator and larger at the poles.
Using the equation of motion $h = (1/2)gt^2$,we get $t = \sqrt{2h/g}$.
Since $g$ is smaller at the equator,the time $t$ taken to reach the surface will be greater at the equator compared to the poles.
Therefore,the packets will not take the same time; the packet dropped at the equator will take a longer time to reach the surface.

(D) The weight of an object on the moon is $1/6$ of its weight on the earth because the acceleration due to gravity on the moon $(g_m)$ is $1/6$ of the acceleration due to gravity on the earth $(g_e)$.
Since the force applied by the person remains the same,the relationship between mass $(m)$ and force $(F)$ is given by $F = m \times g$.
On earth: $F = 15\, kg \times g_e$.
On the moon: $F = m_{moon} \times g_m = m_{moon} \times (g_e / 6)$.
Equating the forces: $15 \times g_e = m_{moon} \times (g_e / 6)$.
Solving for $m_{moon}$: $m_{moon} = 15 \times 6 = 90\, kg$.
Therefore,the person can lift a maximum mass of $90\, kg$ on the moon.

(A) The acceleration due to gravity $g$ on the surface of the earth is given by the formula: $g = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $R$ is its radius.
From this,we can express the mass $M$ as: $M = \frac{g R^2}{G}$.
The density $D$ of the earth is defined as the mass per unit volume: $D = \frac{M}{V}$.
Assuming the earth is a sphere,its volume $V$ is given by: $V = \frac{4}{3} \pi R^3$.
Substituting the expressions for $M$ and $V$ into the density formula:
$D = \frac{g R^2 / G}{4/3 \pi R^3} = \frac{g R^2}{G} \times \frac{3}{4 \pi R^3}$.
Simplifying the expression,we get: $D = \frac{3g}{4 \pi G R}$.

(C) The gravitational force of the Sun provides the necessary centripetal force to keep the Earth in its orbit.
This force acts perpendicular to the velocity of the Earth,constantly changing its direction to maintain a circular path.
Because the gravitational pull is exactly balanced by the requirements of the circular motion,the Earth neither falls into the Sun nor escapes into space.

(N/A) The weight of an object is directly proportional to the mass of the Earth $(M)$ and inversely proportional to the square of the radius of the Earth $(R)$.
Weight of a body $\propto \frac{M}{R^2}$.
The original weight is $W_0 = mg = G \frac{Mm}{R^2}$.
In the hypothetical case,the new mass $M' = 4M$ and the new radius $R' = \frac{R}{2}$ (since the diameter is halved,the radius is also halved).
The new weight $W_n$ is given by:
$W_n = G \frac{M'm}{(R')^2} = G \frac{(4M)m}{(\frac{R}{2})^2} = G \frac{4Mm}{\frac{R^2}{4}} = 16 \times (G \frac{Mm}{R^2}) = 16 W_0$.
Therefore,the weight of the object will become $16$ times its original value.

(N/A) According to Newton's law of universal gravitation,the force of attraction $(F)$ between two bodies is directly proportional to the product of their masses ($m_{1}$ and $m_{2}$) and inversely proportional to the square of the distance $(d)$ between them: $F \propto m_{1} m_{2}$ and $F \propto \frac{1}{d^{2}}$.
This hypothesis is incorrect. Two bricks tied together will fall at the same rate as a single brick under the action of gravity in a vacuum (free-fall). This is because the acceleration due to gravity $(g)$ is independent of the mass of the falling object. Since both objects experience the same acceleration $(g \approx 9.8 \ m/s^{2})$,they will reach the ground at the same time.

(N/A) For an object dropped from height $h$,the time $t$ taken to reach the ground is given by the equation of motion $h = \frac{1}{2} g t^2$,where $g$ is the acceleration due to gravity.
For the first object: $h_1 = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2h_1}{g}}$.
For the second object: $h_2 = \frac{1}{2} g t_2^2 \implies t_2 = \sqrt{\frac{2h_2}{g}}$.
The ratio of time taken is $\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}}$.
$(i)$ If one object is hollow and the other is solid,the ratio remains the same because the acceleration due to gravity $g$ is independent of the mass and density of the object.
(ii) If both are hollow,the ratio still remains the same for the same reason. In free-fall,the acceleration of an object is independent of its mass,shape,or size.

(N/A) $(i)$ The cube will experience a greater buoyant force in the saturated salt solution because the buoyant force is directly proportional to the density of the fluid $(F_b = \rho V g)$. Since the density of the salt solution is greater than that of water,the buoyant force in the salt solution is higher.
$(ii)$ The smaller cube ($4\, cm$ side) will experience a lesser buoyant force because the buoyant force is directly proportional to the volume of the displaced liquid. Since the volume of the smaller cube is less than the initial cube,the volume of water displaced is less,resulting in a lower buoyant force.
$(b)$ The volume of the ball $V = \frac{\text{mass}}{\text{density}} = \frac{4\, kg}{4000\, kg\, m^{-3}} = 10^{-3}\, m^3$.
Buoyant force $(F_b)$ = weight of the liquid displaced = $\text{density of water} \times \text{volume of ball} \times g$.
$F_b = 1000\, kg\, m^{-3} \times 10^{-3}\, m^3 \times 10\, m\, s^{-2} = 10\, N$.

(N/A) Free fall is defined as the motion of an object falling solely under the influence of gravitational force,with no other forces such as air resistance acting upon it. In this state,the object experiences an acceleration equal to the acceleration due to gravity,denoted by $g$.

(B) An object falling freely towards the Earth's surface under the influence of gravitational force is a classic example. For instance,when a ball is dropped from a height,it accelerates towards the Earth due to the gravitational pull exerted by the Earth on the object.

(N/A) The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. The acceleration due to gravity is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth. Since the radius $R$ is smaller at the poles compared to the equator,the value of $g$ is inversely proportional to the square of the radius,making $g$ greater at the poles.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses $(M_1 \times M_2)$ and inversely proportional to the square of the distance $(r^2)$ between their centres.
Mathematically, this is expressed as $F = G \frac{M_1 M_2}{r^2}$, where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M_1$ and $M_2$ are the masses of the two objects, and $r$ is the distance between them.

(B) The acceleration due to gravity is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the distance from the center of the Earth.
Since the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ is smaller at the poles compared to the equator.
Because $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,a smaller radius at the poles results in a higher value of $g$.
Conversely,the larger radius at the equator results in a lower value of $g$.

(C) During free fall,an object is influenced only by the gravitational force of the Earth. According to Newton's second law of motion,the acceleration produced in a body due to gravity is independent of the mass of the object. This acceleration is known as the acceleration due to gravity $(g)$,which is approximately $9.8 \ m/s^2$ near the surface of the Earth. Therefore,for all bodies,whether large or small,undergoing free fall,the acceleration remains constant.

(C) Mass is defined as the quantity of matter contained in a body.
It is an intrinsic property of an object and does not change with location,altitude,or depth.
Therefore,the mass of a body at the centre of the earth remains the same as it is at any other place on the earth.

(A) Yes,gravitational forces obey Newton's third law of motion.
According to Newton's third law,for every action,there is an equal and opposite reaction.
In the case of gravitation,if object $A$ exerts a gravitational force on object $B$,then object $B$ simultaneously exerts an equal and opposite gravitational force on object $A$.

(A) The gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by Newton's Law of Gravitation: $F = G \frac{m_1 m_2}{r^2}$.
This force depends only on the masses of the objects and the distance between them.
The gravitational constant $G$ is a universal constant and does not depend on the medium between the masses.
Therefore,the gravitational force remains the same regardless of whether the objects are in air or any other medium.
Thus,the ratio of the force in air to the force in any other medium is $1: 1$.

(N/A) The acceleration due to gravity $(g)$ at the surface of the earth is given by the formula:
$g = \frac{GM}{R^2}$
Where:
$G$ is the universal gravitational constant.
$M$ is the mass of the earth.
$R$ is the radius of the earth.

(C) No,it is not possible to shield a body from a gravitational field.
This is because the gravitational force is a fundamental interaction that acts between all masses in the universe.
Unlike electric or magnetic fields,which can be blocked by certain materials (like Faraday cages for electric fields),the gravitational force is independent of the nature of the medium present between two bodies.
Therefore,no material can act as a shield against gravity.

(C) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{4}{3} \pi G \rho R$,where $\rho$ is the density and $R$ is the radius of the planet.
Since $G$,$\pi$,and $R$ are constant,we have $g \propto \rho$.
If the density $\rho$ is doubled,the new acceleration due to gravity $g'$ becomes $g' = \frac{4}{3} \pi G (2\rho) R = 2g$.
Therefore,the value of $g$ on the planet is doubled.

(B) The acceleration due to gravity $(g)$ experienced by a body falling towards the Earth is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
Since this formula does not include the mass of the falling body $(m)$,the acceleration is independent of the mass of the body.
Therefore,all objects fall with the same acceleration regardless of their mass in the absence of air resistance.

(C) Inertial mass is an intrinsic property of an object that measures its resistance to acceleration when a force is applied. According to the principle of equivalence in physics,inertial mass is equivalent to gravitational mass,but gravity itself does not change or influence the value of the inertial mass of an object. Therefore,gravity has no effect on inertial mass.

(D) The value of acceleration due to gravity $(g)$ depends on the following factors:
$1$. Shape of the Earth: The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. Due to this,the radius at the poles is less than at the equator,making $g$ greater at the poles.
$2$. Rotation of the Earth: The rotation of the Earth creates a centrifugal force that reduces the effective gravity,especially at the equator.
$3$. Latitude: Since the Earth's radius varies with latitude,the value of $g$ changes accordingly.
$4$. Altitude or Height: As we move away from the surface of the Earth (increasing altitude),the value of $g$ decreases.
$5$. Depth: As we move inside the Earth (increasing depth),the value of $g$ decreases.
Therefore,the value of $g$ depends on the shape,rotation,latitude,altitude,and depth.

(B) The acceleration due to gravity is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the distance from the center of the Earth.
Since the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ at the poles is less than the radius $R$ at the equator.
Because $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,a smaller radius at the poles results in a higher value of $g$ compared to the equator.

(N/A) The acceleration due to gravity $(g)$ is given by the formula $g = GM/R^2$,where $R$ is the radius of the Earth.
Because the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius at the poles $(R_p)$ is less than the radius at the equator $(R_e)$.
Since $g$ is inversely proportional to the square of the radius $(g \propto 1/R^2)$,a smaller radius at the poles results in a higher value of $g$ $(g_p > g_e)$.
Weight is defined as $W = mg$. Since the mass $(m)$ of the body remains constant and $g_p > g_e$,it follows that $mg_p > mg_e$.
Therefore,a body weighs more at the poles than at the equator.

(B) The acceleration due to gravity $g$ at a height $h$ above the surface is given by $g_h = g(1 - 2h/R)$.
The acceleration due to gravity $g$ at a depth $d$ below the surface is given by $g_d = g(1 - d/R)$.
For the same distance $h = d = 1 \, km$,we compare the reduction in gravity.
At height $h$,the reduction is $2h/R$,while at depth $d$,the reduction is $d/R$.
Since $2h/R > d/R$,the value of $g$ decreases more at height $h$ than at depth $d$.
Therefore,the body will weigh more at a depth of $1 \, km$ below the surface of the earth.

(C) The weight of a body is determined by the gravitational force exerted by the Earth on it $(W = mg)$. The revolution of the Earth around the Sun is an orbital motion that does not change the gravitational pull of the Earth on objects located on its surface. Therefore,the revolution of the Earth has no effect on the weight of a body.

(B) The statement is false.
The value of acceleration due to gravity $(g)$ is not constant on the surface of the earth.
Because the earth is not a perfect sphere (it is flattened at the poles and bulges at the equator),the distance from the center of the earth to the surface is less at the poles and more at the equator.
Since $g = GM/R^2$,the value of $g$ is inversely proportional to the square of the radius $(R)$.
Therefore,$g$ is maximum at the poles and minimum at the equator.

(A) In projectile motion,the horizontal component of velocity $(u_x = u \cos \theta)$ remains constant throughout the flight.
However,the vertical component of velocity $(u_y)$ changes due to gravity.
$(i)$ The speed is maximum at the point of projection because the vertical component is at its maximum value here.
$(ii)$ The speed is minimum at the highest point of the trajectory because the vertical component of velocity becomes zero,leaving only the constant horizontal component $(u \cos \theta)$.

(B) The path followed by an object projected into the air,which is subject only to the acceleration of gravity,is called a $Trajectory$. This path is typically parabolic in shape.

(C) projectile is an object projected into the air,subject only to the acceleration of gravity.
When an object is thrown at an angle with the horizontal,it moves under the influence of gravity in a vertical plane.
The horizontal velocity remains constant (ignoring air resistance),while the vertical velocity changes due to gravity.
The combination of these two motions results in a curved path known as a $Parabola$.

(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the centre of the Earth is given by the formula $g' = g(1 - d/R)$,where $d$ is the depth from the surface.
At the centre of the Earth,the depth $(d)$ is equal to the radius of the Earth $(R)$.
Substituting $d = R$ into the formula,we get $g' = g(1 - R/R) = g(1 - 1) = 0$.
Therefore,the acceleration due to gravity at the centre of the Earth is $0 \ m/s^2$.