Gravitation Test

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Gravitation Test - 25

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Question 1
1 / -0

If R$$=$$radius of the earth and g $$=$$acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r<R) from the centre of the earth is proportional to

A
$$r$$

B
$$r^{2}$$

C
$$r^{-2}$$

D
$$r^{-1}$$

Solution

$$F(d) = GM'm/r^2$$

$$ M' = Mr^3/R^3$$

$$F(d)= GMmr/R^3$$

$$r = R-d$$

$$g(d) = GM(R-d)/R^3$$

$$g(d) = g[(R-d)/R]$$

$$g(d) = g(1-\dfrac {d}{R})$$

Therefore, the gravitational field varies proportional to $$r$$.
Question 2
1 / -0

If R$$=$$radius of the earth and g$$=$$acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to

A
$$r$$

B
$$r^{2}$$

C
$$r^{-2}$$

D
$$r^{-1}$$

Solution

For $$r>R$$, earth can be considered as a point mass.
Therefore, $$ F = \dfrac{GMm}{r^2}$$
From the equation, $$F$$ is inversely proportional to $$r^2$$.
Question 3
1 / -0

The height at which the value of acceleration due to gravity becomes $$50$$% of that at the surface of the earth (Radius of the earth $$ = 6400$$ km) is nearly

A
$$2650$$ km

B
$$2430$$ km

C
$$2250$$ km

D
$$2350$$ km

Solution

Acceleration due to gravity a height $$h$$ from earth's surface is given by:
$$g_h = g(\dfrac {R}{R+h})^2$$

Here, $$g_h = \dfrac{g}{2}$$

$$\Rightarrow \dfrac {g}{2}=g(\dfrac {R}{R+h})^2$$

$$\Rightarrow \dfrac {R+h}{R}=1.414$$

$$\Rightarrow \dfrac {h}{R}=0.414$$
$$\Rightarrow h=0.414\times 6400=2649.6\ km$$
Question 4
1 / -0

The acceleration due to gravity on the surface of moon is $$1/6$$ that on the earth and the diameter of the earth is $$4$$ times the diameter of the moon. The rough ratio of the escape velocity of the moon to that of the earth is

A
$$1 : 4$$

B
$$4 : 1$$

C
$$5 : 1$$

D
$$1 : 5$$

Solution

Acceleration due to gravity on the surface of a celestial body is given by:
$$g = \dfrac{GM}{R^2}$$
It is given that $$g_e:g_m = 1:6$$
and $$R_e:R_m = 4:1$$

Escape velocity for a planet is given by $$v = \sqrt {2gR}$$
Ratio of escape velocity is given by:
$$\dfrac{v_m}{v_e} = \dfrac{\sqrt{2g_mR_m}}{\sqrt{2g_eR_e}}$$
$$=\dfrac{1}{\sqrt{6 \times 4}} \approx \dfrac{1}{5}$$
Question 5
1 / -0

If the radius of earth decreases by $$10\%$$, the mass remaining unchanged, then 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

The acceleration due to gravity is given by:
$$g=\dfrac {GM}{R^2}$$

$$\Rightarrow g\propto \dfrac {1}{R^2}$$

$$\Rightarrow \dfrac {\Delta g}{g}=-2\dfrac {\Delta R}{R}=-2(-0.1)=0.2$$
So, the value of $$g$$ increases by $$20\%$$.
Question 6
1 / -0

What is the escape velocity from the surface of the earth of radius $$R$$ and density $$\rho $$ ?

A
$$2R\sqrt{\dfrac{2\pi \rho G}{3}}$$

B
$$2\sqrt{\dfrac{2\pi \rho G}{3}}$$

C
$$2\pi \sqrt{\dfrac{R}{g}}$$

D
$$\sqrt{\dfrac{2\pi G\rho }{R^{2}}}$$

Solution

Escape speed is given by $$v_e=\sqrt {\dfrac {2GM}{R}}$$

But $$M=Volume\times density \Rightarrow M=\dfrac{4}{3} \pi r^3 \rho$$

$$\Rightarrow v_e=\sqrt {2G\dfrac {\dfrac {4}{3}\pi R^3\rho }{R}}$$

$$\Rightarrow v_e=2R\sqrt {\dfrac {2\pi \rho G}{3}}$$
Question 7
1 / -0

The ratio of the radii of planets A and B is $$K_{1}$$ and ratio of accelerations due to gravity on them is $$K_{2}$$ The ratio of escape velocities from them will be:

A
$$K_{1}K_{2}$$

B
$$\sqrt{K_{1}K_{2}}$$

C
$$\sqrt{\dfrac{K_{1}}{K_{2}}}$$

D
$$\sqrt{\dfrac{K_{2}}{K_{1}}}$$

Solution

Escape speed is, $$v_e=\sqrt {2gR}$$
$$\therefore \dfrac {v_1}{v_2}=\sqrt{\dfrac {g_1R_1}{g_2R_2}}$$
It is given, $$\dfrac {R_1}{R_2}=K_1$$ and $$\dfrac {g_1}{g_2}=K_2$$
$$\Rightarrow \dfrac {v_1}{v_2}=\sqrt {K_1K_2}$$
Question 8
1 / -0

A spaceship is launched into a circular orbit of radius $$R$$ close to the surface of earth. The additional velocity to be imparted to the spaceship in the orbit to overcome the earth's gravitational pull is : $$($$ $$g =$$ acceleration due to gravity $$)$$

A
$$1.414Rg$$

B
$$1.414\sqrt{Rg}$$

C
$$0.414Rg$$

D
$$0.414\sqrt{gR}$$

Solution

Velocity of the spaceship when in a circular orbit of radius $$R =\sqrt{gR}$$
where g is the acceleration due to gravity and $$R$$ is the radius of the earth .
Velocity needed to escape earth's gravitational pull $$=\sqrt{2gR}$$
$$\therefore$$ Additional velocity to be imparted $$= \sqrt{2gR} - \sqrt{gR}$$ $$ =(\sqrt{2} -1)\sqrt{gR}$$
Question 9
1 / -0

If the radius of earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth's surface would

A
decrease

B
remain unchanged

C
increase

D
nothing will happen

Solution

The acceleration due to gravity is given by:
$$g=\dfrac {GM}{R^2}$$
$$\Rightarrow g\propto \dfrac {1}{R^2}$$
$$\Rightarrow \dfrac {\Delta g}{g}=-2\dfrac {\Delta R}{R}=-2(-0.01)=0.02$$
So, the value of $$g$$ increases by 2 percent.
Question 10
1 / -0

A tunnel is dug along a diameter of the Earth. The force on a particle of mass $$m$$ placed in the tunnel at a distance $$x$$ from the centre is:

A
$$\dfrac{GM_{e}m}{R^{3}}x$$

B
$$\dfrac{GM_{e}m}{R^{2}}x$$

C
$$\dfrac{GM_{e}m}{R^{3}x}$$

D
$$\dfrac{GM_{e}mR^{3}}{x}$$

Solution

The value of $$g^{'}$$ at depth $$d$$ is given by,
$$g^{'}=g \left( 1-\dfrac{d}{R} \right)$$

The value of $$g^{'}$$ at a distance $$x$$ form Earth's centre is
$$g^{'}= g \left( 1-\dfrac{R-x}{R} \right) = \dfrac{gx}{R}$$

But $$g = \dfrac{GM}{R^2}$$

$$\Rightarrow g^{'}=\dfrac{\left(\dfrac {GM}{R^2} \right)}{R} x=\dfrac {GMx}{R^3}$$

Force on a particle of mass $$m$$ is $$mg=\dfrac {GMmx}{R^3}$$

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Gravitation Test - 25

Result

Gravitation Test - 25

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SHARING IS CARING

Question 1

$$r$$

$$r^{2}$$

$$r^{-2}$$

$$r^{-1}$$

Solution

Question 2

$$r$$

$$r^{2}$$

$$r^{-2}$$

$$r^{-1}$$

Solution

Question 3

$$2650$$ km

$$2430$$ km

$$2250$$ km

$$2350$$ km

Solution

Question 4

$$1 : 4$$

$$4 : 1$$

$$5 : 1$$

$$1 : 5$$

Solution

Question 5

decreases by $$19\%$$

increases by $$19\%$$

decreases by more than $$19\%$$

increases by more than $$19\%$$

Solution

Question 6

$$2R\sqrt{\dfrac{2\pi \rho G}{3}}$$

$$2\sqrt{\dfrac{2\pi \rho G}{3}}$$

$$2\pi \sqrt{\dfrac{R}{g}}$$

$$\sqrt{\dfrac{2\pi G\rho }{R^{2}}}$$

Solution

Question 7

$$K_{1}K_{2}$$

$$\sqrt{K_{1}K_{2}}$$

$$\sqrt{\dfrac{K_{1}}{K_{2}}}$$

$$\sqrt{\dfrac{K_{2}}{K_{1}}}$$

Solution

Question 8

$$1.414Rg$$

$$1.414\sqrt{Rg}$$

$$0.414Rg$$

$$0.414\sqrt{gR}$$

Solution

Question 9

decrease

remain unchanged

increase

nothing will happen

Solution

Question 10

$$\dfrac{GM_{e}m}{R^{3}}x$$

$$\dfrac{GM_{e}m}{R^{2}}x$$

$$\dfrac{GM_{e}m}{R^{3}x}$$

$$\dfrac{GM_{e}mR^{3}}{x}$$

Solution

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