M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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16 questions · auto-graded multiple-choice test.

MCQ 11 Mark

As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from earth, this would.

A
Be similarly true.
B
Not be true because the force between earth and mercury is not inverse square law.
✓
Not be true because the major gravitational force on mercury is due to sun.
D
Not be true because mercury is influenced by forces other than gravitational forces.

Answer

Correct option: C.

Not be true because the major gravitational force on mercury is due to sun.

Force of attraction between any two objects obeys the inverse square law as its universal law. The relative motion between Earth, Mercury as observed from Earth will not be circular as the force on Mercury due to the sun is very large than due to Earth and due to the relative motion of Sun and Earth with Mercury.

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MCQ 21 Mark

Choose the wrong option.

A
Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass.
B
That the gravitational mass and inertial mass are equal is an experimental result.
C
That the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass.
✓
Gravitational mass of a particle like proton can depend on the presence of neighouring heavy objects but the inertial mass cannot.

Answer

Correct option: D.

Gravitational mass of a particle like proton can depend on the presence of neighouring heavy objects but the inertial mass cannot.

Key concept:

Inertial mass: It is the mass of the material body, which measures its inertia.
Gravitational Mass: It is the mass of the material body, which determines the gravitational pull acting upon it. According to the principle of equivalence, Gravitational mass of proton is equivalent to its inertial mass and is independent of presence of neighboring heavy objects.
Important point: Comparison between inertial and gravitational mass:

Both are measured in the same units.
Both are scalars.
Both do not depend on the shape and state of the body.
Inertial mass is measured by applying Newton’s second law of motion whereas gravitational mass is measured by applying Newton’s law of gravitation.
Spring balance measures gravitational mass and inertial balance measure inertial mass.

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MCQ 31 Mark

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon.

A
Will be elliptical.
✓
Will not be strictly elliptical because the total gravitational force on it is not central.
C
Is not elliptical but will necessarily be a closed curve.
D
Deviates considerably from being elliptical due to influence of planets other than earth.

Answer

Correct option: B.

Will not be strictly elliptical because the total gravitational force on it is not central.

Moon revolves around the earth in a nearly circular orbit. When it is observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon. So moon is moving under the combined gravitational pull acting on it due to the earth and the sun. Hence, total force on the moon is not central.

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MCQ 41 Mark

If the law of gravitation, instead of being inversesquare law, becomes an inversecube law.

Planets will not have elliptic orbits.
Circular orbits of planets is not possible.
Projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic.
There will be no gravitational force inside a spherical shell of uniform density.

✓
$a, b$ and $c$
B
$a, b$ and $d$
C
$a, c$ and $d$
D
$b, c$ and $d$

Answer

Correct option: A.

$a, b$ and $c$

If the law of gravitation becomes an inverse cube law instead of inverse square law, then for a planet of mass $m$ revolving around the sun of mass $M,$ we can write
$\text{F}=\frac{\text{GMm}}{\text{R}^3}=\frac{\text{mv}^2}{\text{R}} ($where $R$ is the radius of orbiting planet$)$
$\Rightarrow\ \text{Orbital speed}=\frac{\sqrt{\text{GM}}}{\text{R}}\Rightarrow\text{V}\propto\frac{1}{\text{ R}}$
Time period of revolution of a planet
$\text{T}=\frac{2\pi\text{R}}{\text{v}}=\frac{2\pi\text{R}}{\frac{\sqrt{\text{GM}}}{\text{R}}}=\frac{2\pi\text{R}^2}{\sqrt{\text{GM}}}$
$\Rightarrow\ \text{T}^2\propto\text{R}^4$
Hence, orbit will not be elliptical.
$[$For elliptical orbit $\text{T}^2\propto\text{R}^3]$
The circular orbits of the planets is not possible according to new law of gravitation.
As force $\text{F}=\Big(\frac{\text{GM}}{\text{R}^3}\Big)\text{m} = g'm$
where, $g' =\frac{\text{GM}}{\text{R}^3}$
As $g',$ acceleration due to gravity is constant, hence path followed by a projectile will be approximately parabolic, $(\text{as T}\propto\text{R}^2).$
Also, gravitational force inside a spherical shell of uniform density will have some value.
So, only option $(d)$ is incorrect.

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MCQ 51 Mark

Which of the following options are correct?

Acceleration due to gravity decreases with increasing altitude.
Acceleration due to gravity increases with increasing depth $($assume the earth to be a sphere of uniform density$).$
Acceleration due to gravity increases with increasing latitude.
Acceleration due to gravity is independent of the mass of the earth.

A
$a, c$
✓
$a, b$
C
$b, c$
D
$b, d$

Answer

Correct option: B.

$a, b$

The acceleration due to gravity at an altitude $($height$), gh =g(1-2).$

As value of $\cos$ decrease from $0^\circ$ to $90^\circ ($from $1$ to $0).$ The acceleration due to gravity increases from equator $(\lambda=0^0)$ to pole $(\lambda=90^0).$ Option $(c)$ is correct.
The acceleration due to gravity on surface of earth is $\text{g}=\frac{\text{GM}_\in}{\text{R}^2_\in}$ So $g$ on earth depends on mass of earth. Option $(d)$ is incorrect.
Increasing depth $(d)$ decreases the value of $ga$ . Option $(b)$ is incorrect.
If $\lambda$ is latitude on earth then $\text{g}\lambda=\text{g}-\omega^2\text{R}\cos^2\lambda$
Increasing height $(h)$ decreases the value of gh.option $(a)$ is correct.
Assuming the earth to be a sphere of uniform density, the acceleration due to gravity at a particular depth $(d), \text{gd}=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big).$

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MCQ 61 Mark

Particles of masses 2M, m and M are respectively at points A, B and C with AB = ½ (BC). m is muchmuch smaller than M and at time t = 0, they are all at rest. At subsequent times before any collision takes place:

A
M will remain at rest.
B
M will move towards M.
✓
M will move towards 2M.
D
M will have oscillatory motion.

Answer

Correct option: C.

M will move towards 2M.

Let FBC be the force experienced by mass m at point B due to mass M at point C, and FBA be the force experienced by mass m at point B due to mass 2M at point A.
$\text{F}_{\text{BC}}=\text{G}\frac{\text{mM}}{(\text{BC})^2}...(1)$
$\text{F}_\text{AB}=\text{G}\frac{\text{m2M}}{(\text{BA})^2}...(2)$
Suppose AB = x, then $\text{x}=\frac{1}{2}\text{ (BC)}\text{or}\text{(BC)}=2\text{x}$
Substituting the values of AB and BC in the above equations (1) and (2)
$\text{F}_{\text{BC}} =\text{G}\frac{\text{mM}}{(\text{2x})^2}$
$\text{F}_\text{BA}=\text{G}\frac{\text{m2M}}{(\text{x})^2}$
As $\text{F}_{\text{BA}}>\text{F}_{\text{BC}},$ m will move towards point A (position of particle with mass 2M).

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MCQ 71 Mark

In our solar system, the interplanetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They

A
Will not move around the sun since they have very small masses compared to sun.
B
Will move in an irregular way because of their small masses and will drift away into outer space.
C
Will move around the sun in closed orbits but not obey Kepler’s laws.
✓
Will move in orbits like planets and obey Kepler’s laws.

Answer

Correct option: D.

Will move in orbits like planets and obey Kepler’s laws.

Asteroids will move in orbits like planets and obey Kepler’s law because they are also being acted upon by central gravitational forces.

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MCQ 81 Mark

If the mass of sun were ten times smaller and gravitational constant $G$ were ten times larger in magnitudes.

Walking on ground would became more difficult.
The acceleration due to gravity on earth will not change.
Raindrops will fall much faster.
Airplanes will have to travel much faster.

✓
$a, c, d$
B
$b, c, d$
C
$a, b, c$
D
$a, b, d$

Answer

Correct option: A.

$a, c, d$

If the gravitational constant $G$ becomes $10$ times larger in magnitude.
$\text{G}'=10\text{G}$
Gravitational field due to the earth
$\text{g}'=\frac{\text{G}'\text{M}_\text{e}}{\text{r}^2}$
$=\frac{10\text{GM}_\text{e}}{\text{r}^2}=10\text{g}$
Weight of a person $=\text{mg}'=\text{m}\times10\text{g}=10\text{mg}$
Force on the man due to sun, $\text{F}=\frac{\text{GM}'_\text{s}\text{m}}{\text{r}^2}$
Mass of the sun $\text{M}'_\text{s}=\frac{1}{10}\text{M}_\text{s}\Rightarrow10\text{M}'_\text{s}=\text{M}_\text{s}$
$\text{F}=\frac{\text{GM}_\text{s}\text{m}}{\text{10r}^2}$
Weight of person becomes $10$ times larger so it will be more difficult to walk. Option $(a)$ is correct.
As $g’ = 10g$, the acceleration due to gravity changes. Option $(b)$ is incorrect.
The terminal velocity $\text{v}_\text{T}\propto\text{g}$ and $g', g' = 10g$, the terminal velocity increases $10$ times.
Hence the rain drops falls $10$ times faster. Option $(c)$ is correct.
As the $g’ = 10g$, to overcome the increase gravitational force and in order to maintain the speed the aeroplane will have to travel much faster. Option $(d)$ is correct.

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MCQ 91 Mark

If the sun and the planets carried huge amounts of opposite charges,

A
All three of Kepler’s laws would still be valid.
B
The first law will still be valid.
C
The second law will not change.
✓
All of the above

Answer

Correct option: D.

All of the above

Due to opposite charges the attractive electrostatic forces of attraction produced will be large, along with the gravitational forces. Both the forces will be added and would be radial in nature. Since both the forces are central forces and obey the inverse square law, all the three kepler's laws will be valid.

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MCQ 101 Mark

The centre of mass of an extended body on the surface of the earth and its centre of gravity.

A
Are always at the same point for any size of the body.
B
Are always at the same point only for spherical bodies.
C
Can never be at the same point.
✓
Is close to each other for objects, say of sizes less than $100m.$

Answer

Correct option: D.

Is close to each other for objects, say of sizes less than $100m.$

The centre of gravity $\text{(COG)}$ is based on weight of a body. The centre of mass $\text{(COM)}$ is based on mass of a body. The $\text{COG}$ is a point in a body over which the body can be perfectly balanced. The net torque due to gravity at $\text{COG}$ is zero. Whereas the $\text{COM}$ is the average location of the mass distribution of a body or it is a point where the whole mass of the body is supposed to be concentrated. If given some angular momentum the body would spin about the $\text{COM.}$ For small body, say of size less than $100m, << R_e$ when placed in uniform gravitational field, the $\text{COM}$ is very close to $\text{COG}$. If the size of the body increases i.e., very larger like lake, or mountain, its weight changes, and its $\text{COM}$ and $\text{COG}$ become far from each other. As in case of a spherical ball where the $\text{COM}$ and $\text{COG}$ are same, but in case of a mountain the $\text{COM}$ and the $\text{COG}$ are not aligned, where $\text{COM}$ lies a bit above its $\text{COG}$. Option $(d)$ is justified.

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MCQ 111 Mark

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity.

A
Will be directed towards the centre but not the same everywhere.
B
Will have the same value everywhere but not directed towards the centre.
C
Will be same everywhere in magnitude directed towards the centre.
✓
Cannot be zero at any point.

Answer

Correct option: D.

Cannot be zero at any point.

Acceleration due to gravity g = 0, at the centre if we assume the earth as a sphere of uniform density, then it can be treated as point mass placed at its centre.
But on surface of the earth the acceleration due to gravity cannot be zero at any point.

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MCQ 121 Mark

Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,

A
The solar cells and batteries in satellites run out.
B
The laws of gravitation predict a trajectory spiralling inwards.
✓
Of viscous forces causing the speed of satellite and hence height to gradually decrease.
D
Of collisions with other satellites.

Answer

Correct option: C.

Of viscous forces causing the speed of satellite and hence height to gradually decrease.

The P.E. of satellite orbiting in orbit of radius r due to earth of mass M is $\Big(\frac{-\text{GM}}{2\text{r}}\Big)$ negative sign shows the force of attraction between satellite and earth. Energy (P.E.) is continuously reduced due to atmospheric friction, the radius of the orbit or height decreases gradually, and ultimately it comes back to earth with increasing speed and burns in the atmosphere.

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MCQ 131 Mark

Supposing Newton's law of gravitation for gravitation forces $F_1$ and $F_2$ between two masses $m_1$ and $m_2$ at positions $r_1$ and $r_2$ read $F_1=-F_2=-\frac{r^{12}}{r_{12}^3} G_0^2\left(\frac{m_1 m_2}{M_0^2}\right)^n$ where $M_0$ is a constant of dimension of mass, $r_{12}=r_1-r_2$ and $n$ is a number. In such a case,

The acceleration due to gravity on earth will be different for different objects.
None of the three laws of Kepler will be valid.
Only the third law will become invalid.
For $n$ negative, an object lighter than water will sink in water.

✓
$a, c$
B
$a, b$
C
$a, d$
D
$b, d$

Answer

Correct option: A.

$a, c$

According to the problem,
$F_1=-F_2=-\frac{r^{12}}{r_{12}^3} GM_0^2\left(\frac{m_1 m_2}{M_0^2}\right)^{n}$
$\Rightarrow \overrightarrow{r}_{12}=r_1-r_2$
$\text { (a) } F=\frac{GM_0^{2(1-n)}(M)^{n}}{r_{12}^2}\left(m_1 m_2\right)^{n}$
Take, $m_1=M ($mass of earth$), m_2=m ($mas of the object$), r_{12} R ($radius of earth$)$
Therefore, $F =\left(\frac{ GM _0^{(2-2 n )}( M )^{ n }}{ R ^2}\right) m ^{ n }= Km ^{ n }$
Where $K$ is the constant or the term in the bracket is regarded as constant.
As $F = mg$, so $g = K m ^{ n -1}$, hence $g$ depends upon the mass of object.
Since, $g$ depends upon position vector and mass of object, hence it will be different for different objects. As $g$ is not constant, hence constant of proportionality will not be constant in Kepler's third law.
Hence, Kepler's third law will not be valid.
As the force is of central nature,
$\left[\because \text { Force } \propto \frac{1}{r^2}\right]$
Hence, first two Kepler's laws will be valid. Hence option $(b)$ in incorrect and $(c)$ is correct.
$(d)$ When $n$ is negative, $F = K / M ^{ n }$ Or we can say that $F$ is inversely proportional to mass. This implies that lighter bodies will experience a greater force than the heavier bodies and vice versa. Hence, object lighter than water will sink in water.

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MCQ 141 Mark

There have been suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth,

A
Nothing will change.
B
We will become hotter after billions of years.
C
We will be going around but not strictly in closed orbits.
✓
After sufficiently long time we will leave the solar system.

Answer

Correct option: D.

After sufficiently long time we will leave the solar system.

We know that gravitational force exists between the earth and the sun.
$F _{ G }=\frac{ G \left( M _{ S } \times m _{ e }\right)}{ r ^2}$ Where $M _{ S }$ is mass of the sun and $m _{ e }$ is mass of the earth.
This provides the necessary centripetal force for the circular orbit of the earth around the sun. As $G$ decreases with time, the gravitational force $F_G$ will become weaker with time. As $F_G$ is changing with time due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker.
Hence, after long time the earth will leave the solar system.

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MCQ 151 Mark

Which of the following are true?

A polar satellite goes around the earth’s pole in northsouth direction.
A geostationary satellite goes around the earth in eastwest direction.
A geostationary satellite goes around the earth in westeast direction.
A polar satellite goes around the earth in eastwest direction.

✓
$a, c$
B
$b, c$
C
$a, d$
D
$b, d$

Answer

Correct option: A.

$a, c$

The satellite which appears stationary relative to earth is called the geostationary satellite. It revolves around the earth in the west$-$east direction with the same angular velocity as done by the earth about its own axis in the west$-$east direction. A polar satellite revolves around the earth's pole in north$-$south direction. It is independent of earth's rotation. Option $(a), (c)$ are correct.

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MCQ 161 Mark

Different points in earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the c.m. (centre of mass) causing translation and a net torque at the c.m. causing rotation around an axis through the c.m. For the earthsun system (approximating the earth as a uniform density sphere).

✓
The torque is zero.
B
The torque causes the earth to spin.
C
The rigid body result is not applicable since the earth is not even approximately a rigid body.
D
The torque causes the earth to move around the sun.

Answer

Correct option: A.

The torque is zero.

As the earth is revolving around the sun in a circular motion (approximately in actual the path of earth around the sun is elliptical) due to gravitational attraction. When we consider the earth-sun as a single system and we are taking earth as a sphere of uniform density. Then the gravitational force (F) will be of radial nature, i.e. angle between position vector r and force F is zero. So, torque
$|\vec{\tau}|=|\vec{\text{r}}\times\vec{\text{F}}|=\text{r F}\sin0^0=0.$

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M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip