Gravitation Test

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Gravitation Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

A spring balance is calibrated at sea level. If this balance is used to measure the weight of a body at successive increasing heights from the surface of the earth, then the weight indicated by spring balance will

A
decrease continuously

B
increase continuously

C
first decrease, then increase

D
remains constant

Solution

The acceleration due to gravity on or above earth is given by g = $$ \dfrac {GM}{r^2} $$, where r is distance from earth's center.
i.e. $$g \alpha \dfrac{1}{r^2} $$
now as u increase the heights, r values takes $$ r_e, r_e + h , r_e + 2h $$ etc.
We can observe that, r is increasing and this implies that $$g$$ is decreasing accordingly.
Question 2
1 / -0

An object lying at the equator of the earth will fly off the surface (will feel weightlessness), if the length of the day becomes

A
$$1.00 h$$

B
$$1.41 h$$

C
$$17.0 h$$

D
$$17.0 min$$

Solution

Escape velocity at earth radius $$R$$ is given by $$V = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR} $$
Now, time period $$T=\dfrac{2\pi R}{\sqrt{2gR}} = \pi \sqrt{\dfrac{2R}{g}} = 3.14 \times \sqrt{\dfrac{2 \times 6.4 \times 10^6}{9.8}} = 1\ hr(approx.) $$
A is the right answer.
Question 3
1 / -0

At a height $$h$$ above the surface of earth, the change in value of $$g$$ is same as that at a depth $$x$$ below the surface on the earth. Both $$x$$ and $$h$$ are very small in comparison to the radius of the earth. Then,

A
$$x = h$$

B
$$x = 2h$$

C
$$2x = h$$

D
$$x= h^2$$

Solution

Let, $$g_h$$ be the acceleration due to gravity at height $$h.$$
$$g_x$$ be the acceleration due to gravity at depth $$x.$$
The acceleration due to gravity on the surface of Earth is
$$g = \dfrac{GM}{R^2}$$
where$$, R$$ is the radius o Earth$$, M$$ is mass of Earth and $$G$$ is gravitational constant.
$$\therefore g \propto \dfrac{1}{R^2}$$
$$\Rightarrow g_h \propto \dfrac{1}{(R+h)^2}$$
Therefore
$$\dfrac{g_h}{g} = \dfrac{R^2}{(R+h)^2}$$
$$\dfrac{g_h}{g} = \dfrac{1}{(1+\frac{h}{R})^2}$$
$$\dfrac{g_h}{g} = \left(1+\dfrac{h}{R}\right)^{-2}$$
$$\dfrac{g_h}{g} = \left(1-\dfrac{2h}{R}\right)$$
$$g_h = g - \dfrac{2gh}{R}$$
$$g - g_h = \dfrac{2gh}{R}$$
Also, the acceleration due to gravity at depth $$x,$$
$$g_x = \dfrac{4}{3}G\rho (R-x)$$
$$\Rightarrow g_x \propto (R-x)$$
Therefore
$$\dfrac{g_x}{g} = \dfrac{R-x}{R}$$
$$\dfrac{g_x}{g} = 1 - \dfrac{x}{R}$$
$$g_x = g - \dfrac{gx}{R}$$
$$g - g_x = \dfrac{gx}{R}$$
If the change in the value of $$g$$ at height $$h$$ above earth surface is the same as that at depth $$x$$ i.e.
$$g - g_h = g - g_x$$
$$\dfrac{2gh}{R} = \dfrac{gx}{R}$$
$$\therefore x = 2h$$
Question 4
1 / -0

If the change in the value of $$g$$ at height $$h$$ above earth surface is the same as that at depth $$x (x \quad or\quad h < R_e),$$ then

A
$$x = h^2$$

B
$$x = h$$

C
$$x = \dfrac{h}{2}$$

D
$$x = 2h$$

Solution

Let $$g_h$$ be the acceleration due to gravity at height $$h$$.
$$g_x$$ be the acceleration due to gravity at depth $$x$$
The acceleration due to gravity on the surface of Earth is
$$g = \dfrac{GM}{R^2}$$
where, $$R$$ is the radius o Earth$$, M$$ is mass of Earth and $$G$$ is gravitational constant.
$$\therefore g \propto \dfrac{1}{R^2}$$
$$\Rightarrow g_h \propto \dfrac{1}{(R+h)^2}$$
Therefore
$$\dfrac{g_h}{g} = \dfrac{R^2}{(R+h)^2}$$
$$\dfrac{g_h}{g} = \dfrac{1}{\left(1+\dfrac{h}{R}\right)^2}$$
$$\dfrac{g_h}{g} = \left(1+\dfrac{h}{R}\right)^{-2}$$
$$\dfrac{g_h}{g} = \left(1-\dfrac{2h}{R}\right)$$
$$g_h = g - \dfrac{2gh}{R}$$
$$g - g_h = \dfrac{2gh}{R}$$
Also, the acceleration due to gravity at depth $$x,$$
$$g_x = \dfrac{4}{3}G\rho (R-x)$$
$$\Rightarrow g_x \propto (R-x)$$
Therefore
$$\dfrac{g_x}{g} = \dfrac{R-x}{R}$$
$$\dfrac{g_x}{g} = 1 - \dfrac{x}{R}$$

$$g_x = g - \dfrac{gx}{R}$$

$$g - g_x = \dfrac{gx}{R}$$
If the change in the value of $$g$$ at height $$h$$ above earth surface is the same as that at depth $$x$$ i.e.
$$g - g_h = g - g_x$$
$$\dfrac{2gh}{R} = \dfrac{gx}{R}$$

$$\therefore x = 2h$$
Question 5
1 / -0

At some planet, $$g = 1.96\ ms^{-2}$$. If it is safe to jump from a height of $$2\ m$$ on earth, then the corresponding safe height on that planet is

A
$$2\ m$$

B
$$5\ m$$

C
$$10\ m$$

D
$$20\ m$$

Solution

Gravitation potential on the planet$$, P.E._P = mgh_P$$
$$\Rightarrow P.E._P = m\times 1.96h_P$$
Gravitation potential on the earth$$, P.E._E = mgh_E$$
$$\Rightarrow P.E._E = m\times 9.8\times 2$$
$$\Rightarrow P.E._E = P.E._E $$
$$\Rightarrow m\times 1.96h_P=m\times 9.8\times 2$$
$$\Rightarrow h_P = \dfrac{m\times 9.8\times 2}{m\times 1.96} = 10\ m$$
Question 6
1 / -0

If $${ W }_{ 1 }$$, $${ W }_{ 2 }$$ and $${ W }_{ 3 }$$ represent the work done in moving a particle from $$A$$ to $$B$$ along three different paths $$1$$, $$2$$ and $$3$$, respectively, (as shown in the figure) in the gravitational field of a point mass $$m$$, find the correct relation between $${ W }_{ 1 }$$, $${ W }_{ 2 }$$ and $${ W }_{ 3 }$$

A
$${ W }_{ 1 }>{ W }_{ 2 }>{ W }_{ 3 }$$

B
$${ W }_{ 1 }={ W }_{ 2 }={ W }_{ 3 }$$

C
$${ W }_{ 1 }<{ W }_{ 2 }<{ W }_{ 3 }$$

D
$${ W }_{ 2 }>{ W }_{ 1 }>{ W }_{ 3 }$$

Solution

In a conservative field work done does not depend on the path, and gravitational field is a conservative field.
$$\therefore W_1=W_2=W_3$$
Question 7
1 / -0

If $$g$$ is same at a height $$h$$ and at a depth $$d$$, then

A
$${ R }={ 2d } $$

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

C
$${ h }={ d } $$

D
none

Solution

We know, $$g_h=g(1-\dfrac{2h}{R})$$ and $$g_d=(1-\dfrac{d}{R})$$
When $$g_d=g_h$$ then,
$$(1-\dfrac{2h}{R})=1-\dfrac{d}{R}$$
$$\therefore (\dfrac{-2h}{R})=-\dfrac{d}{R}$$
$$d=2h$$
Question 8
1 / -0

A missile is launched with a velocity less than the escape velocity from earth. The sum of its kinetic energy and potential energy is

A
positive

B
negative

C
zero

D
may be positive or negative depending on the direction of projection

Solution

The potential at infinite distance from earth is taken to be zero. Thus if it is given energy to give it just the escape velocity, it has no velocity and hence no kinetic energy after reaching infinity and hence zero total energy.
Thus if the given velocity is less than escape velocity of earth, the total energy is negative.
Question 9
1 / -0

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth.

A
The acceleration of S is always directed towards the centre of earth.

B
The angular momentum of S about the centre of earth changes in direction but its magnitude remains constant.

C
The total mechanical energy of S varies periodically with time.

D
The linear momentum of S remains constant in magnitude.

Solution

Force on the satellite is always towards the earth; therefore acceleration of satellite $$S$$ is always directed towards the centre of the earth. Net torque of this gravitational force $$N$$ about the centre of the earth is zero. Therefore, angular momentum (both in magnitude and direction) of $$S$$ about the centre of the earth is constant throughout. Since force $$F$$ is conservative in nature, therefore mechanical energy of the satellite remains constant. Speed of $$S$$ is maximum when it is nearest to the earth and minimum when it is farthest.
Question 10
1 / -0

If the escape velocity from the surface of a spherical planet is given by $$2 \displaystyle \sqrt{\frac{GM}{K}}$$, where $$M$$ is the mass of the planet and $$K$$ is a constant, then the radius of the planet is

A
$$\dfrac{K}{2}$$

B
$$K$$

C
$$2K$$

D
$$4K$$

Solution

Escape velocity of planet with radius $$R,\quad v_e = \sqrt{\dfrac{2GM}{R}}$$
given, $$v_e = 2\sqrt{\dfrac{GM}{K}}$$
So, $$v_e = 2\sqrt{\dfrac{GM}{K}} =\sqrt{\dfrac{2GM}{R}}$$ and solving it we get
$$ R = \dfrac{K}{2}$$

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 32

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Gravitation Test - 32

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Question 1

decrease continuously

increase continuously

first decrease, then increase

remains constant

Solution

Question 2

$$1.00 h$$

$$1.41 h$$

$$17.0 h$$

$$17.0 min$$

Solution

Question 3

$$x = h$$

$$x = 2h$$

$$2x = h$$

$$x= h^2$$

Solution

Question 4

$$x = h^2$$

$$x = h$$

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

$$x = 2h$$

Solution

Question 5

$$2\ m$$

$$5\ m$$

$$10\ m$$

$$20\ m$$

Solution

Question 6

$${ W }_{ 1 }>{ W }_{ 2 }>{ W }_{ 3 }$$

$${ W }_{ 1 }={ W }_{ 2 }={ W }_{ 3 }$$

$${ W }_{ 1 }<{ W }_{ 2 }<{ W }_{ 3 }$$

$${ W }_{ 2 }>{ W }_{ 1 }>{ W }_{ 3 }$$

Solution

Question 7

$${ R }={ 2d } $$

$${ d }={ 2h } $$

$${ h }={ d } $$

none

Solution

Question 8

positive

negative

zero

may be positive or negative depending on the direction of projection

Solution

Question 9

The acceleration of S is always directed towards the centre of earth.

The angular momentum of S about the centre of earth changes in direction but its magnitude remains constant.

The total mechanical energy of S varies periodically with time.

The linear momentum of S remains constant in magnitude.

Solution

Question 10

$$\dfrac{K}{2}$$

$$K$$

$$2K$$

$$4K$$

Solution

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