Gravitation Test

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Gravitation Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

'Weight' of a body may have the following attributes.
I. It is the gravitational force acting on a body at the earth's surface
II. It is independent of the mass of the body
III. The body feels weightlessness during free fall
IV. It is different at different places on earth's surface.

A
I, II and III only

B
I, III and IV only

C
I and III only

D
I, II, III and IV

Solution

Weight on a body is defined as the force acting on it due to gravity.

$$W = mg$$

Since, the value of $$g$$ is different at different places on the earth. Hence, weight of a body also differs at different places.

Since, in a state of free fall, there is no reaction force acting on a body, hence the body feel weightlessness in a state of free fall.

Hence, cases I, III and IV are True.
Question 2
1 / -0

An aluminium sphere is dipped into water. Which of the following is true?

A
Buoyancy will be less in water at $$0^oC$$ than that in water at $$4^oC$$.

B
Buoyancy will be more in water at $$0^oC$$ than that in water at $$4^oC$$.

C
Buoyancy in water at $$0^oC$$ will be same as that in water $$4^o C$$.

D
Buoyancy may be more or less in water at $$4^o$$ depending on the radius of the sphere.

Solution

$$\textbf{Explanation:}$$
$$\bullet$$Buoyant force is force exerted by liquid on solid block that depends on volume of submerged block and density of liquid.

$$\bullet$$Force F = Vdg, where V is volume of submerged aluminum sphere, d is density of liquid and g is acceleration due to gravity.

$$\bullet$$So, buoyancy will be maximum if density is maximum. For water its density is maximum at 4° C.

$$\bullet$$So, buoyancy will be less in water at 0° C than that in water at 4° C.

Hence, option A is the correct answer.
Question 3
1 / -0

The value of universal gravitational constant on earth for a particle of mass 5 kg is

A
$$6.67 \times 10^{-11}$$

B
$$6.67 \times 10^{-7}$$

C
$$5 \times 6.67 \times 10^{-11}$$

D
$$6.67 \times 10^{-23}$$

Solution

Gravitational constant depends only on the mass of the earth and not on the mass of the particle on the earth. Hence its values remains constant = $$6.67 \times 10^{-11}$$

The correct option is (a)
Question 4
1 / -0

Two identical particles of mass $$m$$ are placed at a distance $$r$$ from each other. If their separation is doubled, then the effect on gravitational constant will be

A
Gravitational constant remains same

B
Gravitational constant becomes quadrupled

C
Gravitational constant becomes 1/4th of the actual one

D
Gravitational constant becomes doubled

Solution

According to the Newton's law of gravitation, gravitational force between two objects is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

$$ F \propto \dfrac{m_1m_2}{r^2}$$

Hence, $$ F = \dfrac{Gm_1m_2}{r^2}$$, where $$G$$ is the proportionality constant , known as the Universal Gravitational constant.

Gravitational constant does not depend on masses or the distance between them

Changing the distance between the masses, changes the force between them.

Hence, The correct option is (A)
Question 5
1 / -0

Which of the following statements is correct regarding the universal gravitational constant G?

A
G has same value in all systems of units.

B
The value of G is same everywhere in the universe.

C
The value of G was first experimentally determined by Johannes Kepler.

D
G is a vector quantity

Solution

G has different value in different system of units.
In SI system the value of G is $$6.67\times 10^{-11}N m^{2} kg^{-2}$$ whereas in CGS its value is $$6.67\times 10^{-8}dyne cm^{2} g^{-2}$$.
The value of G is same throughout the universe.
The value of G was first experimentally determined by English scientist Henry Cavendish.
G is a scalar quantity.
Question 6
1 / -0

If the gravitational constant is expressed in terms of $$dynes\ m^{-2} kg^{2}$$, how will the value of G change:

A
$$6.67 \times 10^{-11} dynes \ kg^2 m^{-2}$$

B
$$6.67 \times 10^{-8} dynes \ kg^2 m^{-2}$$

C
$$6.67 \times 10^{-6} dynes \ kg^2 m^{-2}$$

D
$$6.67 \times 10^{-3} dynes \ kg^2 m^{-2}$$

Solution

'G' is expressed as
$$G=6.67\times { 10 }^{ -11 }N.{ m }^{ -2 }.{ kg }^{ +2 }\\ \therefore 1N={ 10 }^{ 5 }dyne\\ 1m={ 10 }^{ 2 }cm\\ 1kg={ 10 }^{ 3 }g.$$
$$G=6.67\times { 10 }^{ -11 }\times { 10 }^{ 5 }dynes.kg.{ m }^{ -2 }$$ [As,$$1N={ 10 }^{ 5 }$$dyne]
Question 7
1 / -0

If the radius of the earth decreases by $$10\%$$, the mass remaining unchanged, what will happen to the acceleration due to gravity?

A
Decreases by $$19\%$$

B
Increases by $$19\%$$

C
Decreases by more than $$19\%$$

D
Increases by more than $$19\%$$

Solution

Acceleration due to gravity is given by $$g=\cfrac { GM }{ { R }^{ 2 } } $$
R becomes $${ R }^{ \prime }=R-\cfrac { R\times 10 }{ 100 } =\cfrac { 9R }{ 10 } $$

Acceleration due to gravity becomes
$$\therefore { g }^{ \prime }=\cfrac { GM }{ { R }^{ \prime 2 } } =\cfrac { GM }{ \left( \cfrac { 9R }{ 10 } \right) ^{ 2 } } =\cfrac { 100 }{ 81 } \cfrac { GM }{ { R }^{ 2 } } =\cfrac { 100 }{ 81 } g\\ \therefore \cfrac { { g }^{ \prime }-g }{ g } \times 100%=\cfrac { g\left( \cfrac { 100 }{ 81 } -1 \right) }{ g } \times 100%=23.456%$$

So, increase is:
$$=\dfrac{\dfrac{100g}{81}-g}{g}\times100\approx24\%$$

So, option D is correct
Question 8
1 / -0

Which of the following combination is true about the mass of a body according to Newton?
I. Mass is the quantity of matter contained in a body
II. Mass is represented by the relation: mass $$=\displaystyle\frac{force}{velocity}$$
III. Mass is always conserved
IV. Mass of a body is the ratio of its weight to acceleration due to gravity at a particular place.

A
I, II and IV only

B
I, III and IV only

C
I, II and III only

D
I, II, III and IV

Solution

I. It states the definition of mass which is true.

II. According to Newton's second law of motion, $$F=ma$$
$$\Rightarrow$$ Mass $$=\dfrac{force}{acceleration} $$
So, the given statement is incorrect.

III. Mass is always conserved, i.e. law of conservation of mass.

IV. $$W=mg \Rightarrow m = \dfrac{W}{g}$$
So, the given statement is incorrect.

$$\therefore$$ Option B is correct.
Question 9
1 / -0

What is the range of gravitational force. ?

A
$$10^{-2}$$m

B
$$10^{-15}$$m

C
infinite

D
$$10^{-10}$$m

Solution

The range of gravitational force is infinite.
Question 10
1 / -0

The radii of two planets are $${ R }_{ 1 }$$ and $${ R }_{ 2 }$$ respectively and their densities are $${ \rho }_{ 1 }$$ and $${ \rho }_{ 2 }$$ respectively. The ratio of the accelerations due to gravity $$\dfrac{{ g }_{ 1 }}{{ g }_{ 2 }}$$ at their surfaces is

A
$$\dfrac { { { R }_{ 1 }{ \rho }_{ 2 } }}{ { R }_{ 2 }{ \rho }_{ 1 } }$$

B
$$\dfrac { { { R }_{ 1 }{ \rho }_{ 1 } }}{ { R }_{ 2 }{ \rho }_{ 2 } }$$

C
$$\dfrac { { \rho }_{ 1 }{ R }_{ 2 }^{ 2 } }{ { \rho }_{ 2 }{ R }_{ 1 }^{ 2 } }$$

D
$$\dfrac { { { R }_{ 1 }{ R } }_{ 2 } }{ { \rho }_{ 1 }{ \rho }_{ 2 } }$$

Solution

Using the relation $$g= \dfrac{GM}{R^2}$$
Also $$M=\rho \times V=\rho \times \dfrac43 \pi R^3$$

Hence, $$g=\dfrac 43 G\rho\pi R$$

$$\dfrac{g_1}{g_2}=\dfrac {G\rho_1 \pi R_1} {G \rho_2 \pi R_2}$$

$$\dfrac{g_1}{g_2}=\cfrac{R_1 \rho_1}{R_2 \rho_2}$$

Hence, B is the correct option.

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Gravitation Test - 55

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Gravitation Test - 55

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Question 1

I, II and III only

I, III and IV only

I and III only

I, II, III and IV

Solution

Question 2

Buoyancy will be less in water at $$0^oC$$ than that in water at $$4^oC$$.

Buoyancy will be more in water at $$0^oC$$ than that in water at $$4^oC$$.

Buoyancy in water at $$0^oC$$ will be same as that in water $$4^o C$$.

Buoyancy may be more or less in water at $$4^o$$ depending on the radius of the sphere.

Solution

Question 3

$$6.67 \times 10^{-11}$$

$$6.67 \times 10^{-7}$$

$$5 \times 6.67 \times 10^{-11}$$

$$6.67 \times 10^{-23}$$

Solution

Question 4

Gravitational constant remains same

Gravitational constant becomes quadrupled

Gravitational constant becomes 1/4th of the actual one

Gravitational constant becomes doubled

Solution

Question 5

G has same value in all systems of units.

The value of G is same everywhere in the universe.

The value of G was first experimentally determined by Johannes Kepler.

G is a vector quantity

Solution

Question 6

$$6.67 \times 10^{-11} dynes \ kg^2 m^{-2}$$

$$6.67 \times 10^{-8} dynes \ kg^2 m^{-2}$$

$$6.67 \times 10^{-6} dynes \ kg^2 m^{-2}$$

$$6.67 \times 10^{-3} dynes \ kg^2 m^{-2}$$

Solution

Question 7

Decreases by $$19\%$$

Increases by $$19\%$$

Decreases by more than $$19\%$$

Increases by more than $$19\%$$

Solution

Question 8

I, II and IV only

I, III and IV only

I, II and III only

I, II, III and IV

Solution

Question 9

$$10^{-2}$$m

$$10^{-15}$$m

infinite

$$10^{-10}$$m

Solution

Question 10

$$\dfrac { { { R }_{ 1 }{ \rho }_{ 2 } }}{ { R }_{ 2 }{ \rho }_{ 1 } }$$

$$\dfrac { { { R }_{ 1 }{ \rho }_{ 1 } }}{ { R }_{ 2 }{ \rho }_{ 2 } }$$

$$\dfrac { { \rho }_{ 1 }{ R }_{ 2 }^{ 2 } }{ { \rho }_{ 2 }{ R }_{ 1 }^{ 2 } }$$

$$\dfrac { { { R }_{ 1 }{ R } }_{ 2 } }{ { \rho }_{ 1 }{ \rho }_{ 2 } }$$

Solution

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