Gravitation Test

Result

Gravitation Test - 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The value of acceleration due to gravity at a depth of $$1600 km$$ is equal to [Radius of earth $$= 6400 km$$]

A
$$9.8{ ms }^{ -2 }$$

B
$$4.9{ ms }^{ -2 }$$

C
$$7.35{ ms }^{ -2 }$$

D
$$19.6{ ms }^{ -2 }$$

Solution

Given : $$d = 1600 km$$ $$R = 6400 km$$
Acceleration due to gravity at a depth $$d$$, $$g_d = g_s\bigg(1 -\dfrac{d}{R}\bigg)$$
where $$g_s = 9.8 ms^{-2}$$ is acceleration due to gravity at earth's surface.
$$\therefore$$ $$g_d = 9.8 \bigg(1 -\dfrac{1600}{6400}\bigg) = 7.35 ms^{-2}$$
Question 2
1 / -0

You are to examine these two statements carefully and select the answers to these items using the code given below :
Statement I : The acceleration due to gravity decreases with increase in height from the surface of the Earth.
Statement II : The acceleration due to gravity is inversely proportional to the square of the distance from the centre of the Earth.

A
Both the statements are individually true and Statement II is the correct explanation of Statement I

B
Both the statements are individually true but Statement II is not the correct explanation of Statement I

C
Statement I is true but Statement II is false

D
Statement I is false but Statement II is true

Solution

The acceleration due to gravity decreases with increases in height from the surface of the earth because it is universally proportional to the square of the distance from the center of the earth.
Question 3
1 / -0

The depth $$'d'$$ at which the value of acceleration due to gravity becomes $$\dfrac {1}{n}$$ times the value at the earth's surface is $$(R =$$ radius of earth)

A
$$d = R \left (\dfrac {n}{n - 1}\right )$$

B
$$d = R \left (\dfrac {n - 1}{n}\right )$$

C
$$d = R \left (\dfrac {n - 1}{2n}\right )$$

D
$$d = R^{2} \left (\dfrac {n - 1}{n}\right )$$

Solution

Acceleration due to gravity at a depth $$d$$ under the earth surface $$g' = g_s\bigg(1-\dfrac{d}{R}\bigg)$$
Given : $$g' = \dfrac{g_s}{n}$$
$$\therefore$$ $$\dfrac{g_s}{n} = g_s\bigg(1-\dfrac{d}{R}\bigg)$$
Or $$\dfrac{1}{n} = \bigg(1-\dfrac{d}{R}\bigg)$$
$$\implies$$ $$d =R\bigg(\dfrac{n-1}{n}\bigg)$$
Question 4
1 / -0

Let '$${ g }_{ h }$$' and '$${ g }_{ d }$$' be the acceleration due to gravity at height '$$h$$' above the earth's surface and at depth '$$d$$' below the earth's surface respectively. IF $${ g }_{ h }={ g }_{ d }$$ then the relation between '$$h$$' and '$$d$$' is

A
$$d=h$$

B
$$d=\dfrac { h }{ 2 } $$

C
$$d=\dfrac { h }{ 4 } $$

D
$$d=2h$$

Solution

Acceleration due to gravity at height $$h$$ above the earth surface $$g_h = g(1-\dfrac{2h}{R})$$
where $$R$$ is the radius of earth.
Acceleration due to gravity at depth $$d$$ below the earth surface $$g_d = g(1-\dfrac{d}{R})$$
But $$g_h = g_d$$ implies $$ g(1-\dfrac{2h}{R})=$$ $$g(1-\dfrac{d}{R})$$
$$\implies$$ $$d = 2h$$
Question 5
1 / -0

If the escape speed of a projectile on Earth's surface is $$11.2{ kms }^{ -1 }$$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth :

A
$$56.63{ kms }^{ -1 }$$

B
$$33{ kms }^{ -1 }$$

C
$$39{ kms }^{ -1 }$$

D
$$31.7{ kms }^{ -1 }$$

Solution

According to the principle of conservation of energy, we have
Initial kinetic energy $$+$$ Initial potential energy $$=$$ Final kinetic energy $$+$$ Final potential energy
$$\Rightarrow \dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { GMm }{ R } =\dfrac { 1 }{ 2 } m{ v }^{ { \prime }^{ 2 } }+0$$
$$\Rightarrow \dfrac { 1 }{ 2 } m{ v }^{ { \prime }^{ 2 } }=\dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { GMm }{ R } $$ ....(i)
Also consider, $${ v }_{ e }=$$ escape velocity
$$\dfrac { 1 }{ 2 } m{ v }_{ e }^{ 2 }=\dfrac { GMm }{ R } $$ .....(ii)
$$\therefore$$ From equations (i) and (ii), we get
$$\dfrac { 1 }{ 2 } m{ v }^{ { \prime }^{ 2 } }=\dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { 1 }{ 2 } m{ v }_{ e }^{ 2 }$$ .....(iii)
$$\Rightarrow { v }^{ { \prime }^{ 2 } }={ v }^{ 2 }-{ v }_{ e }^{ 2 }$$
Now, $${ v }_{ e }=11.2{ kms }^{ -1 } $$ and $$v=3{ v }_{ e } $$ .....(iv)
From equations (iii) and (iv), we get
$${ v }^{ { \prime }^{ 2 } }={ \left( 3{ v }_{ e } \right) }^{ 2 }-{ v }_{ e }^{ 2' }$$
$${ v }^{ { \prime }^{ 2 } }=9{ v }_{ e }^{ 2 }-{ v }_{ e }^{ 2 }=8{ v }_{ e }^{ 2 }=8\times { \left( 11.2 \right) }^{ 2 }$$
$$v'^2=8\times { \left( 11.2 \right) }^{ 2 }$$
$$\Rightarrow { v }^{ \prime }=\sqrt { 8\times { \left( 11.2 \right) }^{ 2 } } =\sqrt { 8 } \times 11.2$$
$${ v }^{ \prime }=2\times 1.414\times 11.2=31.68{ kms }^{ -1 }$$
$$\therefore $$ Speed of the body far away from the Earth $${ v }^{ \prime }=31.68{ kms }^{ -1 }=31.7{ kms }^{ -1 }$$
Question 6
1 / -0

A man weighs $$60kg$$ at earth surface. At what height above the earth's weight become $$30kg$$, Given radius of earth is $$6400km$$:

A
$$2624km$$

B
$$3000km$$

C
$$2020km$$

D
None of these

Solution

Let acceleration due to gravity at a height $$h$$ above the surface of earth be $$g'$$
So, $$mg'=m\dfrac{gR^2}{(R+h)^2}$$
or, $$30=60\times\dfrac{(6400)^2}{(6400+h)^2}$$
So, $$h=6400\sqrt{2}-6400=2624km$$
Question 7
1 / -0

The earth's radius is R and acceleration due to gravity at its surface is g. If a body of mass m is sent to a height $$\displaystyle h=\frac { R }{ 5 } $$ from the earth's surface, the potential energy increases by

A
$$\displaystyle mgh$$

B
$$\displaystyle \frac { 4 }{ 5 } mgh$$

C
$$\displaystyle \frac { 5 }{ 6 } mgh$$

D
$$\displaystyle \frac { 6 }{ 7 } mgh$$

Solution

Increase in potential energy
$$\displaystyle \Delta U={ U }_{ 2 }-{ U }_{ 1 }$$
$$\displaystyle =\left[ -\frac { GMm }{ \left( R+\frac { R }{ 5 } \right) } \right] -\left[ -\frac { GMm }{ R } \right] $$
$$\displaystyle =\frac { GMm }{ R } -\frac { 5GMm }{ 6R } $$
$$\displaystyle =\frac { GMm }{ 6R } =\frac { g{ R }^{ 2 }m }{ 6R } =\frac { mgR }{ 6 } $$
$$\displaystyle =mg.\frac { 5h }{ 6 } =\frac { 5 }{ 6 } mgh$$
Question 8
1 / -0

Weight of a body of mass m decreases by $$1\%$$ when it is raised to height h above the Earth's surface. If the body is taken to a depth h in a mine, then its weight will:

A
Decreases by $$0.5\%$$

B
Decreases by $$2\%$$

C
Increases by $$0.5\%$$

D
Increase by $$1\%$$

Solution

Hint :- Gravitational acceleration g varies decreasingly with height as well as depth.

Step 1: Calculate height h
g = $$\dfrac{GM}{r^{2}}$$

Then g$$_{h}$$ = $$\dfrac{GM}{(r + h)^{2}}$$

$$\dfrac{g}{g_{h}}$$ = $$\dfrac{100}{99}$$ = ($$\dfrac{r + h}{r}$$)$$^{2}$$

$$\dfrac{r + h}{r}$$ = 1.005

r + h = 1.005r
h= 0.005r

Step 2: Calculate % variation in g with given depth
g$$_{d}$$ = g(1 - $$\dfrac{d}{r}$$)

d =h= 0.005r
g$$_{d}$$ = 1 - 0.005 = 0.995
Thus the weight will be decreased by 0.5%.
Question 9
1 / -0

Which one of the following statements is correct?

A
Acceleration due to gravity decreases with the increase of altitude

B
Acceleration due to gravity increase with the increase of depth (assuming earth to be a sphere of uniform density)

C
Acceleration due to gravity decreases with the increase of latitude

D
Acceleration due to gravity is independent of the mass of the earth

Solution
Question 10
1 / -0

The ratio of acceleration due to gravity at a height h above the surface of the earth and at a depth h below the surface of the earth h < radius of earth

A
Is constant

B
Increases linearly with h

C
Decreases linearly with h

D
Decreases parabolically with h

Solution

Acceleration due to gravity height h,
$$\displaystyle g_1= g \left( 1-\frac{2h}{R} \right)$$
Acceleration due to gravity at depth h,
$$\displaystyle g_2= g \left( 1-\frac{h}{R} \right)$$
$$\therefore$$ $$\displaystyle \frac{g_1}{g_2} =\frac{1-2h/R}{1-h/R}$$
$$\displaystyle = \left( 1-\frac{2h}{R} \right) \left( 1-\frac{h}{R} \right)^{-1}$$
$$\displaystyle =\left( 1-\frac{h}{R} \right)$$
[neglecting higher power of $$\displaystyle \frac{h}{R}$$]
$$\displaystyle \therefore \frac{g_1}{g_2}$$ decreases linearly with h.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Gravitation Test - 46

Result

Gravitation Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

$$9.8{ ms }^{ -2 }$$

$$4.9{ ms }^{ -2 }$$

$$7.35{ ms }^{ -2 }$$

$$19.6{ ms }^{ -2 }$$

Solution

Question 2

Both the statements are individually true and Statement II is the correct explanation of Statement I

Both the statements are individually true but Statement II is not the correct explanation of Statement I

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Solution

Question 3

$$d = R \left (\dfrac {n}{n - 1}\right )$$

$$d = R \left (\dfrac {n - 1}{n}\right )$$

$$d = R \left (\dfrac {n - 1}{2n}\right )$$

$$d = R^{2} \left (\dfrac {n - 1}{n}\right )$$

Solution

Question 4

$$d=h$$

$$d=\dfrac { h }{ 2 } $$

$$d=\dfrac { h }{ 4 } $$

$$d=2h$$

Solution

Question 5

$$56.63{ kms }^{ -1 }$$

$$33{ kms }^{ -1 }$$

$$39{ kms }^{ -1 }$$

$$31.7{ kms }^{ -1 }$$

Solution

Question 6

$$2624km$$

$$3000km$$

$$2020km$$

None of these

Solution

Question 7

$$\displaystyle mgh$$

$$\displaystyle \frac { 4 }{ 5 } mgh$$

$$\displaystyle \frac { 5 }{ 6 } mgh$$

$$\displaystyle \frac { 6 }{ 7 } mgh$$

Solution

Question 8

Decreases by $$0.5\%$$

Decreases by $$2\%$$

Increases by $$0.5\%$$

Increase by $$1\%$$

Solution

Question 9

Acceleration due to gravity decreases with the increase of altitude

Acceleration due to gravity increase with the increase of depth (assuming earth to be a sphere of uniform density)

Acceleration due to gravity decreases with the increase of latitude

Acceleration due to gravity is independent of the mass of the earth

Solution

Question 10

Is constant

Increases linearly with h

Decreases linearly with h

Decreases parabolically with h

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.