Mix Example - GRAVITATION Questions in English

(N/A) Given: Mass of the man $M = 80\, kg$,$g_{e} = 9.8\, m s^{-2}$,$g_{m} = 1.63\, m s^{-2}$.
Weight is calculated using the formula $W = m \times g$.
Weight on the moon $(W_{m})$ = $M \times g_{m} = 80 \times 1.63 = 130.40\, N$.
Mass is a fundamental property of an object and remains constant regardless of the location.
Therefore,the mass of the man on the Earth is $80\, kg$ and on the moon is also $80\, kg$.

(N/A) Given initial velocity $u = 40 \, m s^{-1}$,acceleration due to gravity $g = -10 \, m s^{-2}$ (acting downwards),and final velocity $v = 0 \, m s^{-1}$ at the maximum height.
Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - (40)^2 = 2 \times (-10) \times h$
$-1600 = -20 \times h$
$h = 80 \, m$.
Therefore,the maximum height reached is $80 \, m$.
The total distance covered is the sum of the upward and downward journey: $80 \, m + 80 \, m = 160 \, m$.
The net displacement is the difference between the final and initial position. Since the stone returns to the starting point,the net displacement is $0 \, m$.

(N/A) Given: height $h = 20\, m$,initial velocity $u = 0$,acceleration due to gravity $g = 10\, m s^{-2}$.
Using the second equation of motion $S = ut + \frac{1}{2}at^2$,where $S = h$ and $a = g$:
$20 = 0 \times t + \frac{1}{2} \times 10 \times t^2$
$20 = 5t^2$
$t^2 = 4$
$t = 2\, s$.
So,it will take $2\, s$ to reach the ground.
$(b)$ Using the first equation of motion $v = u + at$:
$v = 0 + 10 \times 2$
$v = 20\, m s^{-1}$.
Thus,the speed when it hits the ground is $20\, m s^{-1}$.

(A) According to Newton's law of universal gravitation,the force $F$ is inversely proportional to the square of the distance $r$ between the objects: $F \propto \frac{1}{r^2}$.
Given initial force $F_1 = 100 \, N$ and final force $F_2 = 50 \, N$.
We have the ratio: $\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}$.
Substituting the values: $\frac{50}{100} = \frac{r_1^2}{r_2^2} \implies \frac{1}{2} = \frac{r_1^2}{r_2^2}$.
Taking the square root on both sides: $\frac{1}{\sqrt{2}} = \frac{r_1}{r_2}$.
Therefore,$r_2 = \sqrt{2} \, r_1$.
Thus,the distance must be increased by a factor of $\sqrt{2}$.

(C) According to Newton's law of universal gravitation,the gravitational force $F$ between two bodies of masses $m_1$ and $m_2$ separated by a distance $r$ is given by the formula:
$F = G \frac{m_1 m_2}{r^2}$
Here,$G$ is the universal gravitational constant.
From this formula,it is clear that the force $F$ is inversely proportional to the square of the distance $r$ between the two bodies.
Therefore,$F \propto \frac{1}{r^2}$.

(D) The acceleration due to gravity $g$ 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.
Because the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ is smallest at the poles and largest at the equator.
Since $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,the value of $g$ is maximum where the radius $R$ is minimum.
The radius of the Earth is minimum at the poles (Antarctica is located near the South Pole).
Therefore,the value of $g$ is maximum at a camp site in Antarctica.

(A) Given:
Initial velocity $(u)$ = $0 \, m s^{-1}$ (as the stone is dropped).
Distance fallen $(s)$ = $20 \, m$.
Acceleration due to gravity $(g)$ = $10 \, m s^{-2}$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
Substituting the values: $v^2 - 0^2 = 2 \times 10 \times 20$.
$v^2 = 400$.
$v = \sqrt{400} = 20 \, m s^{-1}$.
Thus,the speed of the stone after falling $20 \, m$ is $20 \, m s^{-1}$.

(B) When a ball is thrown vertically upward,its velocity is directed upwards.
However,the acceleration due to gravity $(g)$ always acts vertically downwards towards the center of the Earth.
Since the motion of the ball is upwards and the acceleration due to gravity is downwards,the acceleration is in the direction opposite to the direction of motion.
Therefore,the correct option is $B$.

(C) When a projectile is at the highest point of its trajectory,its vertical velocity becomes $0 \ m/s$,but it still possesses a horizontal velocity component.
However,the acceleration due to gravity $(g)$ acts constantly on the projectile throughout its entire flight.
This acceleration is always directed vertically downwards towards the center of the Earth.
Therefore,even at the top of its path,the projectile experiences a constant downward acceleration equal to $g$.

(D) The acceleration due to gravity $(g)$ at a height $(h)$ above the Earth's surface is given by $g_h = g(1 - 2h/R)$,where $R$ is the radius of the Earth. As $h$ increases,$g_h$ decreases.
Similarly,the acceleration due to gravity at a depth $(d)$ below the Earth's surface is given by $g_d = g(1 - d/R)$. As $d$ increases,$g_d$ decreases.
Therefore,for both scientist $A$ (moving deep into a mine) and scientist $B$ (moving high up in a balloon),the measured acceleration due to gravity decreases.

(A) According to Newton's first law of motion,an object in motion will continue to move in a straight line at a constant speed unless acted upon by an external force.
Since the earth moves in a nearly circular orbit around the sun,its direction of motion is constantly changing at every point.
$A$ change in direction implies a change in velocity,which constitutes acceleration.
According to Newton's second law,acceleration requires a net force.
Therefore,the revolution of the earth around the sun serves as evidence that a centripetal force (gravitational force) must be acting on the earth,directed towards the sun.

(B) The effective acceleration due to gravity '$g'$ at the equator is given by the formula: $g' = g - R\omega^2$,where '$g$' is the acceleration due to gravity at the poles (or if the earth were stationary),'$R$' is the radius of the earth,and '$\omega$' is the angular velocity of the earth's rotation.
If the earth stops rotating,the angular velocity '$\omega$' becomes $0$.
Substituting this into the formula,we get: $g' = g - R(0)^2 = g$.
Since the original value was $g' = g - R\omega^2$ (which is less than '$g$'),the new value '$g$' is greater than the original value '$g'$ at the equator.
Therefore,the value of '$g$' at the equator will increase.

(C) The weight of an object is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
Since $g$ is greater at the poles than at the equator,the weight of an object is higher at the poles.
$A$ spring balance measures the weight $(W)$ of an object,which varies with the location due to the change in $g$.
If you purchase goods at the equator using a spring balance,you get a certain mass for a specific weight reading.
When you take these goods to the poles,the same mass will show a higher weight reading on a spring balance.
However,a beam balance compares the mass of the object with a standard mass (counterweight),which is unaffected by the change in $g$ because $g$ acts equally on both sides of the balance.
Therefore,to gain an advantage,one must use a spring balance to measure weight at the equator and sell it at the poles,as the spring balance will show a higher value at the poles for the same mass.

(D) The gravitational constant $(G)$ is a universal constant.
It is defined as the force of attraction between two unit masses separated by a unit distance.
Its value is $6.674 \times 10^{-11} \ N \ m^2/kg^2$.
Since it is a universal constant,it does not depend on the masses of the objects,the distance between them,the temperature of the medium,or any other physical conditions.
Therefore,the correct option is $D$.

(B) The acceleration due to gravity on the surface of the Earth is defined as the standard value $g$.
By definition,the gravitational acceleration experienced by an object at the Earth's surface is denoted by the symbol $g$,which is approximately $9.8 \ m/s^2$.
Therefore,the correct option is $B$.

(B) In a vacuum,there is no air resistance to oppose the motion of a falling object.
According to Newton's second law of motion,$F = ma$,and the gravitational force acting on an object is $F = mg$,where $m$ is the mass of the object and $g$ is the acceleration due to gravity.
Equating these,$ma = mg$,which simplifies to $a = g$.
Since $g$ is a constant value (approximately $9.8 \ m/s^2$ near the Earth's surface) and is independent of the mass of the object,all objects falling freely in a vacuum experience the same acceleration.

(C) The gravitational acceleration $g$ at a distance $r$ from the centre of the Earth is given by the formula $g' = \frac{GM}{r^2}$.
At the surface of the Earth,the distance $r = R$ (where $R$ is the radius of the Earth),so $g = \frac{GM}{R^2} = 9.8\, m s^{-2}$.
Given that the spaceship is at a distance $r = 2R$ from the centre of the Earth.
Substituting this into the formula: $g' = \frac{GM}{(2R)^2} = \frac{GM}{4R^2}$.
Since $g = \frac{GM}{R^2}$,we can write $g' = \frac{g}{4}$.
Substituting the value of $g = 9.8\, m s^{-2}$,we get $g' = \frac{9.8}{4} = 2.45\, m s^{-2}$.

(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
For Earth,$g_e = \frac{GM_e}{R_e^2} = 9.8 \, m s^{-2}$.
For the new planet,the mass $M_p = \frac{M_e}{2}$ and the radius $R_p = \frac{R_e}{2}$.
Substituting these values into the formula for the new planet's gravity $g_p$:
$g_p = \frac{G(M_e/2)}{(R_e/2)^2} = \frac{G M_e / 2}{R_e^2 / 4} = 2 \times \frac{GM_e}{R_e^2}$.
Since $\frac{GM_e}{R_e^2} = g_e$,we have $g_p = 2 \times g_e$.
$g_p = 2 \times 9.8 \, m s^{-2} = 19.6 \, m s^{-2}$.

(A) Given:
Initial velocity $u = 0 \, m s^{-1}$ (since the stone is dropped).
Distance covered $s = 100 \, m$.
Acceleration due to gravity $g = 9.8 \, m s^{-2}$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
Substituting the values: $v^2 - 0^2 = 2 \times 9.8 \times 100$.
$v^2 = 1960$.
$v = \sqrt{1960} \approx 44.27 \, m s^{-1}$.
Thus,the speed is approximately $44.2 \, m s^{-1}$.

(B) To find the initial velocity $(u)$ of the ball,we use the third equation of motion: $v^2 = u^2 + 2as$.
At maximum height,the final velocity $(v)$ is $0 \, m/s$.
The acceleration due to gravity $(a)$ is $-g = -9.8 \, m/s^2$.
The displacement $(s)$ is $100 \, m$.
Substituting these values into the equation: $0^2 = u^2 + 2(-9.8)(100)$.
$0 = u^2 - 1960$.
$u^2 = 1960$.
$u = \sqrt{1960} \approx 44.27 \, m/s$.
Rounding to the nearest option,the speed is $44.2 \, m/s$.

(C) Given:
Initial velocity $(u)$ = $0\, m/s$ (since the stone is dropped).
Time $(t)$ = $4\, s$.
Acceleration due to gravity $(g)$ = $9.8\, m/s^2$.
Using the second equation of motion:
$s = ut + \frac{1}{2}gt^2$
Substituting the values:
$h = (0 \times 4) + \frac{1}{2} \times 9.8 \times (4)^2$
$h = 0 + 4.9 \times 16$
$h = 78.4\, m$
Therefore,the height is $78.4\, m$.

(D) The weight of an object is defined as the force with which the earth attracts the object towards its center.
Mathematically,it is given by the formula $W = m \times g$,where $m$ is the mass of the object and $g$ is the acceleration due to gravity.
Unlike mass,which is the quantity of matter contained in an object and remains constant,weight is a force that varies depending on the value of $g$ at a specific location.
Therefore,option $D$ is the correct definition.

(A) The universal law of gravitation,expressed as $F = G \frac{m_1 m_2}{r^2}$,describes the gravitational force between two point masses.
For extended objects,this formula is strictly valid only when the objects are spherical and their mass distribution is uniform (spherically symmetric).
In such cases,the entire mass of the sphere can be considered to be concentrated at its center,allowing the distance $r$ to be measured between the centers of the two spheres.

(D) 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 the formula: $F = G \frac{m_1 m_2}{r^2}$.
Here,$G$ is the universal gravitational constant.
If the distance $r$ is doubled,the new distance becomes $r' = 2r$.
The new gravitational force $F'$ will be: $F' = G \frac{m_1 m_2}{(2r)^2} = G \frac{m_1 m_2}{4r^2}$.
This can be rewritten as: $F' = \frac{1}{4} \times (G \frac{m_1 m_2}{r^2}) = \frac{1}{4} F$.
Therefore,the gravitational force becomes one-fourth of its original value.

(C) When a bomb is released from an aircraft moving with a constant horizontal velocity, it possesses the same horizontal velocity as the aircraft at the moment of release.
Due to gravity, it experiences a constant downward acceleration $(g)$.
The combination of uniform horizontal motion and uniformly accelerated vertical motion results in a curved path known as a $parabola$.
Therefore, the trajectory of the bomb is a parabola.

(A) The initial velocity $u = 0 \, m/s$ and the acceleration due to gravity is $g$.
Using the equation of motion $s = ut + (1/2)gt^2$,for the total height $H = 100 \, m$ and time $T$:
$100 = 0 \cdot T + (1/2)gT^2 \implies 100 = (1/2)gT^2 \implies T^2 = 200/g$.
Now,let $h$ be the distance covered in time $t = T/2$:
$h = (1/2)g(T/2)^2 = (1/2)g(T^2/4) = (1/8)gT^2$.
Substituting $gT^2 = 200$ into the equation:
$h = (1/8) \cdot 200 = 25 \, m$.
Since $h$ is the distance covered from the top,the bottle will be $25 \, m$ from the top.

(A) Let the two bodies be $A$ and $B$. Body $A$ is $1 \, m$ above body $B$.
Both bodies are released simultaneously and fall freely under gravity,meaning their initial velocity $u = 0 \, m/s$ and acceleration $a = g = 9.8 \, m/s^2$ are the same for both.
The position of body $A$ at time $t$ is given by $y_A = y_{A,0} + ut + \frac{1}{2}gt^2$.
The position of body $B$ at time $t$ is given by $y_B = y_{B,0} + ut + \frac{1}{2}gt^2$.
The relative separation between the bodies at any time $t$ is given by $|y_A - y_B| = |(y_{A,0} + ut + \frac{1}{2}gt^2) - (y_{B,0} + ut + \frac{1}{2}gt^2)|$.
This simplifies to $|y_A - y_B| = |y_{A,0} - y_{B,0}|$.
Since the initial vertical separation $|y_{A,0} - y_{B,0}|$ is $1 \, m$,the relative separation remains constant at $1 \, m$ regardless of the time elapsed,as both bodies experience the same acceleration due to gravity.

(FALSE) The statement is False.
Explanation: The force of attraction between any two objects in the universe is known as gravitation. Gravity is specifically defined as the force of attraction exerted by the Earth on objects near its surface.

(FALSE) False.
The value of $G$ is the Universal Gravitational Constant, which is approximately $6.673 \times 10^{-11} \text{ N m}^2/\text{kg}^2$.
It is a fundamental constant of nature and does not depend on the mass, distance, or the medium between the two objects.

(TRUE) True. When a spring balance holding an object is released,it falls freely under the influence of gravity. In a state of free fall,both the spring balance and the object experience the same acceleration due to gravity $(g)$. Consequently,there is no net force exerted by the object on the spring of the balance,leading to a reading of zero weight.

(FALSE) False.
$G$ is the Universal Gravitational Constant. Its value is $6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.
It is a constant value throughout the universe and does not depend on the mass,radius,or shape of the bodies involved.

(TRUE) True.
For a small body,the gravitational field is considered uniform throughout its volume.
In a uniform gravitational field,the centre of mass (the point where the total mass is concentrated) and the centre of gravity (the point where the total gravitational force acts) coincide at the same location.

(TRUE) True.
When a body is thrown vertically upward,it experiences a constant downward acceleration due to gravity $(g)$,which is approximately $9.8 \ m/s^2$.
When a body falls downward,it also experiences the same constant downward acceleration due to gravity $(g = 9.8 \ m/s^2)$.
In physics,if we take the upward direction as positive $(+)$,then the downward acceleration is represented as negative $(-g)$.
Therefore,the magnitude remains the same $(9.8 \ m/s^2)$,but the sign is opposite depending on the chosen coordinate system.

(TRUE) The statement is True.
According to the formula for acceleration due to gravity at a depth $d$ below the surface of the Earth,$g_d = g(1 - d/R)$,where $g$ is the acceleration due to gravity at the surface,$d$ is the depth,and $R$ is the radius of the Earth.
At the centre of the Earth,the depth $d$ is equal to the radius $R$ $(d = R)$.
Substituting $d = R$ into the formula,we get $g_d = g(1 - R/R) = g(1 - 1) = g(0) = 0$.
Therefore,the acceleration due to gravity $(g)$ at the centre of the Earth is zero.

(A) The statement is True.
Inertia is defined as the inherent property of an object to resist any change in its state of rest or uniform motion.
According to Newton's first law of motion,mass is the measure of inertia.
Therefore,a heavier object has more inertia than a lighter object,confirming that inertia depends directly on the mass of the object.

(A) The statement is True.
According to Newton's Law of Universal Gravitation,every particle of matter in the universe attracts every other particle with a force that acts along the line joining their centres of mass. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

(TRUE) True.
The acceleration due to gravity $g$ on the surface of a planet is defined by the formula $g = GM/R^2$,where $G$ is the universal gravitational constant,$M$ is the mass of the planet,and $R$ is the radius of the planet.