Mass and Weight Questions in English

(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.

(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 $42\, kg$.

(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.

(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.

(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) 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.

(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.

(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.

(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.

(B) Potential energy $(PE)$ is defined by the formula $PE = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height of the object from the reference level.
Since the car is moving on a levelled road,the height $(h)$ remains constant throughout the motion.
Because $m$,$g$,and $h$ do not change,the potential energy of the car remains constant.
Therefore,the potential energy does not change.

(D) The weight of a person is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
Since the weight on planet $A$ is half that on Earth,the acceleration due to gravity on planet $A$ $(g_A)$ is half of that on Earth $(g_E)$,i.e.,$g_A = \frac{1}{2} g_E$.
When a person jumps,the work done by their muscles is converted into potential energy $(PE = mgh)$. Assuming the muscular force and work done remain constant,the potential energy at the maximum height is constant.
Therefore,$m g_E h_E = m g_A h_A$.
Substituting the values: $g_E \times 0.4 = (\frac{1}{2} g_E) \times h_A$.
$0.4 = \frac{1}{2} h_A$.
$h_A = 0.4 \times 2 = 0.8 \, m$.
Thus,the person can jump to a height of $0.8 \, m$ on planet $A$.

(N/A) $1\, kg$ weight is defined as the gravitational force exerted by the Earth on an object with a mass of $1\, kg$.
It is calculated using the formula $F = m \times g$,where $m = 1\, kg$ and $g \approx 9.8\, m/s^2$.
Therefore,$1\, kg\, wt = 1\, kg \times 9.8\, m/s^2 = 9.8\, N$.

(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.

(A) The gravitational force exerted by the earth on any object is specifically known as the weight of that object. It is defined as the force with which the earth attracts a body towards its center.

(A) The mass of an object is the quantity of matter contained in it.
Mass is a scalar quantity and remains constant regardless of the object's location in the universe.
Unlike weight,which depends on the acceleration due to gravity $(g)$,mass does not change with position.
Therefore,if the mass of the body is $10 \,kg$ on the surface of the earth,it will remain $10 \,kg$ at the centre of the earth.

(N/A) $(i)$ The weight on Mars is $38 \%$ of the weight on Earth.
Weight on Mars $= 0.38 \times 750 \, N = 285 \, N$.
$(ii)$ The relationship between weight $(W)$,mass $(m)$,and acceleration due to gravity $(g)$ is given by $W = m \times g$.
Given $W = 750 \, N$ and $g = 10 \, m s^{-2}$.
Mass $m = W / g = 750 / 10 = 75 \, kg$.
Therefore,Mona's mass on Earth is $75 \, kg$.

(N/A) Mass is defined as the quantity of matter contained in an object and is an intrinsic property that remains constant regardless of location.
Weight is defined as the force with which an object is attracted towards the center of a planet or celestial body,calculated as $W = m \times g$,where $m$ is mass and $g$ is the acceleration due to gravity.
Since the value of $g$ varies from place to place (e.g.,it is different on the Earth's surface compared to the Moon),the weight of an object changes accordingly,even though its mass remains constant.

(A) The weight of the man on the Earth's surface is $W_{E} = 600\, N$.
The mass of the man is calculated as $m = \frac{W_{E}}{g_{E}} = \frac{600\, N}{10\, m s^{-2}} = 60\, kg$.
Mass is a constant quantity and does not change with location,so the mass on the Moon remains $60\, kg$.
The weight of the man on the Moon's surface is $W_{M} = \frac{W_{E}}{6}$,because the acceleration due to gravity on the Moon is $\frac{1}{6}$ of that on the Earth.
Therefore,$W_{M} = \frac{600\, N}{6} = 100\, N$.

(B) The weight of an object is defined as the force with which it is attracted towards the center of a planet,calculated as $W = m \times g$.
Since the mass $(m)$ of the man remains constant everywhere,his weight depends on the acceleration due to gravity $(g)$.
The acceleration due to gravity on the moon is approximately one-sixth $(1/6)$ of the acceleration due to gravity on the earth.
Therefore,the weight of the man will decrease on the surface of the moon compared to his weight on the earth.

(A) The effective acceleration due to gravity $(g')$ is given by the formula $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angular velocity of the earth,$R$ is the radius of the earth,and $\lambda$ is the latitude.
If the earth stops rotating,$\omega = 0$,which means the term $\omega^2 R \cos^2 \lambda$ becomes zero.
Consequently,the effective gravity $(g')$ increases at all points except at the poles (where $\lambda = 90^\circ$ and $\cos 90^\circ = 0$).
Since weight $(W = mg)$ is directly proportional to the effective acceleration due to gravity,the weight of a body will increase at all locations on the earth's surface except at the poles,where it remains unchanged.

(B) The weight of the object is the force required to lift it against gravity.
Given mass $m = 100 \,g = 0.1 \,kg$.
Gravitational acceleration on Earth $g \approx 10 \,m/s^2$.
The force $F$ required is equal to the weight $W = m \times g$.
$F = 0.1 \,kg \times 10 \,m/s^2 = 1 \,N$.
Therefore,a force of at least $1 \,N$ is required to lift the object.

(A) The weight of an object is given by the formula $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
The value of $g$ depends on the radius of the Earth $(R)$ as $g = \frac{GM}{R^2}$.
The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. Therefore,the radius of the Earth at the equator is greater than the radius at the poles.
Since $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,the value of $g$ is lower at the equator and higher at the poles (like Antarctica).
Consequently,the weight of the bag of sugar will be more when taken to Antarctica compared to its weight at the equator.

(N/A)

Acceleration due to gravity $(g)$	Universal gravitational constant $(G)$
$(i)$ It is the acceleration acquired by a body due to the earth's gravitational pull on it.	$(i)$ It is numerically equal to the force of attraction between two masses of $1\, kg$ each separated by a distance of $1\, m$.
$(ii)$ The value of $g$ is different at different places on the surface of the earth and varies from one celestial body to another.	$(ii)$ The value of $G$ is a universal constant,i.e.,its value is the same $(6.67 \times 10^{-11}\, Nm^2 kg^{-2})$ everywhere in the universe.
$(iii)$ It is a vector quantity.	$(iii)$ It is a scalar quantity.

(A) The acceleration due to gravity $g$ at the earth's surface is given by $g = \frac{GM}{R^2}$.
The weight of an object at the earth's surface is $W = mg = 20 \, N$.
At a height $h = 6400 \, km$ above the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.
The acceleration due to gravity at this height $g_h$ is given by $g_h = \frac{GM}{(R+h)^2} = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{1}{4} g$.
Therefore,the weight of the object at this height $W_h$ is $W_h = m \cdot g_h = m \cdot (\frac{1}{4} g) = \frac{1}{4} (mg)$.
Substituting the value of $W = 20 \, N$,we get $W_h = \frac{1}{4} \times 20 \, N = 5 \, N$.

(N/A) The distinction is shown in the table below:

Mass	Weight
$1.$ It is the amount of matter contained in a body.	$1.$ It is a force equal to the gravitational pull exerted by a planet.
$2.$ It is a constant quantity and does not change with respect to position or place.	$2.$ It is a variable quantity and changes with the change in acceleration due to gravity of a place.
$3.$ Mass of a body can never be zero.	$3.$ Weight of a body can be zero during free fall.
$4.$ It is measured by using a physical balance.	$4.$ It is measured by using a spring balance.
$5.$ It is a scalar quantity.	$5.$ It is a vector quantity.
$6.$ It is measured in kilograms $(kg)$.	$6.$ It is measured in newtons $(N)$.

When an object is moved from the equator to the poles, its weight increases because the acceleration due to gravity $(g)$ is greater at the poles than at the equator.
Weight of an object becomes zero at the center of the Earth or when a body is in a state of free fall.

(N/A) Let the mass of an object be $m$. Let its weight on the moon be $W_{m}$ and its weight on the earth be $W_{E}$.
Using the universal law of gravitation,the weight of an object on the moon is given by $W_{m} = \frac{GM_{m}m}{R_{m}^{2}}$,where $M_{m}$ is the mass of the moon and $R_{m}$ is its radius.
Similarly,the weight of the object on the earth is $W_{E} = \frac{GM_{E}m}{R_{E}^{2}}$,where $M_{E}$ is the mass of the earth and $R_{E}$ is its radius.
Taking the ratio of the two weights:
$\frac{W_{m}}{W_{E}} = \frac{GM_{m}m}{R_{m}^{2}} \times \frac{R_{E}^{2}}{GM_{E}m} = \frac{M_{m}}{M_{E}} \times \left( \frac{R_{E}}{R_{m}} \right)^{2}$
Substituting the given values:
$\frac{W_{m}}{W_{E}} = \frac{7.36 \times 10^{22}}{5.98 \times 10^{24}} \times \left( \frac{6.37 \times 10^{6}}{1.74 \times 10^{6}} \right)^{2}$
$\frac{W_{m}}{W_{E}} \approx 0.0123 \times (3.66)^{2} \approx 0.0123 \times 13.4 = 0.165 \approx \frac{1}{6}$
Thus,the weight of an object on the moon is one-sixth of its weight on the earth.

(N/A) Mass is the quantity of matter contained in an object. The $SI$ unit of mass is $kg$.
Weight is the force with which an object is attracted towards the center of a planet or celestial body. The $SI$ unit of weight is the newton $(N)$.
Since mass is a constant property of an object and does not change with location,the mass of the object on the Moon remains $20 \, kg$.
The weight of the object on the Moon is calculated using the formula $W = m \times g$,where $m = 20 \, kg$ and $g = 1.6 \, m s^{-2}$.
$W = 20 \times 1.6 = 32 \, N$.
Therefore,the mass on the Moon is $20 \, kg$ and the weight on the Moon is $32 \, N$.

(A) The weight of an object on Earth is calculated using the formula $W = m \times g$,where $m$ is the mass and $g$ is the acceleration due to gravity $(g \approx 10\, m/s^2)$.
Weight on Earth: $W = 50\, kg \times 10\, m/s^2 = 500\, N$.
On the Moon,the mass of an object remains constant because it is an intrinsic property of the body.
However,the weight on the Moon changes because the acceleration due to gravity on the Moon is approximately $1/6$th of that on the Earth. Therefore,her weight on the Moon will be $500/6 \approx 83.33\, N$.

(B) The weight of a body is given by the formula $W = m \times g$,where $m$ is the mass and $g$ is the acceleration due to gravity.
The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator.
Because the radius of the Earth is smaller at the poles compared to the equator,the value of $g$ is greater at the poles.
Since $W$ is directly proportional to $g$,a body will weigh more at the poles than at the equator.

(D) Let $M$ be the mass and $R$ be the radius of the earth.
The weight of an object of mass $m$ on the surface of the earth is given by $W = mg = \frac{GMm}{R^2}$.
If the diameter becomes two times,the radius $R$ also becomes two times,i.e.,$R' = 2R$.
The new weight $W'$ will be $W' = \frac{GMm}{(R')^2} = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$.
Thus,$W' = \frac{1}{4} W$.
Therefore,the weight of the object will become one-fourth of its original value.

(97.2 N) Weight on Earth $(W) = 450 N$
Acceleration due to gravity on Earth $(g) = 10 m s^{-2}$
Therefore,the mass of the body $(m) = \frac{W}{g} = \frac{450}{10} = 45 kg$.
Now,the acceleration due to gravity on the surface of Mars $(g_{m})$ is calculated using the formula $g_{m} = \frac{GM_{m}}{R_{m}^{2}}$.
Substituting the values: $g_{m} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{23}}{(4.3 \times 10^{6})^{2}}$.
$g_{m} = \frac{40.02 \times 10^{12}}{18.49 \times 10^{12}} \approx 2.16 m s^{-2}$.
Therefore,weight on Mars $= m \times g_{m} = 45 \times 2.16 = 97.2 N$.

(A) Given: Weight on Earth $W_{E} = 600\,N$,Weight on Moon $W_{M} = 100\,N$,acceleration due to gravity on Earth $g = 10\,m s^{-2}$.
First,we calculate the mass $(m)$ of the man using the formula $W = m \times g$:
$m = \frac{W_{E}}{g} = \frac{600}{10} = 60\,kg$.
Since mass remains constant everywhere,the mass of the man on the Moon is also $60\,kg$.
Now,we calculate the acceleration due to gravity on the Moon $(g_{m})$ using the formula $W_{M} = m \times g_{m}$:
$g_{m} = \frac{W_{M}}{m} = \frac{100}{60} = 1.67\,m s^{-2}$.

(N/A) Given: Weight $(W) = 392 \, N$,Acceleration due to gravity $(g) = 9.8 \, m s^{-2}$.
Mass $(m) = W / g = 392 / 9.8 = 40 \, kg$.
$(b)$ $(i)$ The mass of an object remains constant regardless of location. Therefore,his mass on the moon will be $40 \, kg$.
$(ii)$ The weight of an object on the moon is $1/6$ of its weight on the earth. Therefore,weight on the moon $= 392 / 6 \approx 65.33 \, N$.
$(iii)$ Acceleration due to gravity on the moon $(g_m) = \text{Weight on moon} / \text{Mass} = 65.33 / 40 \approx 1.63 \, m s^{-2}$.

(N/A) Given: Mass $(m) = 30 \ kg$,$g_{earth} = 10 \ m/s^2$.
$(i)$ Weight on the moon $(W_m)$:
$g_{moon} = 1/6 \times g_{earth} = 1/6 \times 10 = 5/3 \ m/s^2$.
$W_m = m \times g_{moon} = 30 \times (5/3) = 50 \ N$.
$(ii)$ Weight on the planet $(W_p)$:
$g_{planet} = 3 \times g_{earth} = 3 \times 10 = 30 \ m/s^2$.
$W_p = m \times g_{planet} = 30 \times 30 = 900 \ N$.

(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 $80\, kg$.

(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 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.

(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.