Gravitation Test

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Gravitation Test - 36

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Question 1
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 in its weight will

A
decrease by $$0.5%$$

B
decrease by $$2%$$

C
increase by $$0.5%$$

D
increase by $$1%$$

Solution

As,$$\dfrac{g'-g}{g}=-\dfrac{2h}{R_e}=-1%$$

$$\dfrac{g''-g}{g}=-\dfrac{h}{R_e}=-\dfrac{1}{2}%=-0.5%$$

so, Weight in mne will decrease by $$0.5%$$.
Question 2
1 / -0

The gravitational potential energy of a body at a distance $$r$$ from the centre of earth is $$U$$. Its weight at a distance $$2r$$ from the centre of earth is

A
$$\displaystyle\frac{U}{r}$$

B
$$\displaystyle\frac{U}{2r}$$

C
$$\displaystyle\frac{U}{4r}$$

D
$$\displaystyle\frac{U}{\sqrt{2}r}$$

Solution

$$U=\displaystyle\frac{-GMm}{r}$$

$$\therefore -GM=\displaystyle\frac{Ur}{m}$$

$$E=-\displaystyle\frac{GM}{(2r)^{2}}=\displaystyle\dfrac{\left(\dfrac{Ur}{m}\right)}{4r^{2}}=\dfrac{U}{4mr}$$

$$F=mE=\displaystyle\dfrac{U}{4r}$$
Question 3
1 / -0

Which of the following statement is wrong for acceleration due to gravity.

A
$$g$$ decreases on going above the surface of earth

B
$$g$$ increases on going below the surface of earth

C
$$g$$ is maximum at pole

D
$$g$$ increases on going from equator to poles

Solution

Acceleration due to gravity at a depth x from the surface, $$g'=g\bigg(1-\dfrac{x}{R}\bigg)$$
Thus, g decreases on going below the surface of earth.
Question 4
1 / -0

The value of '$$g$$' reduces to half of its value at surface of earth at a height '$$h$$', then

A
$$h = R$$

B
$$h = 2R$$

C
$$h \equiv ( \sqrt 2 +1)R$$

D
$$h \equiv ( \sqrt 2 -1)R$$

Solution

$$g'=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$$
As $$g'=g/2$$

$$\dfrac{g}{2}=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$$

$$\implies 1+ \dfrac{h}{R}= \sqrt{2}$$
Hence $$h= (\sqrt{2}-1)R$$
Question 5
1 / -0

Acceleration due to gravity at earth's surface is '$$g$$' m/s$$^2$$. Find the effective value of acceleration due to gravity at a height of $$32 km$$ from sea level :($$R_e = 6400 km$$).

A
$$0.5 g ms^{2}$$

B
$$0.99 g ms^{2}$$

C
$$1.01 g ms^{2}$$

D
$$0.90 g ms^{2}$$

Solution

Acceleration due to gravity at a height h from the earth surface, $$g'=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$$
Given $$h= 32 km \quad R= 6400 km$$
Using bionomial expansion, $$(1+x)^n= 1+ nx$$ for $$x<< 1$$

Thus as $$h<< R$$ $$\implies g'=g\bigg[1-\dfrac{2h}{R}\bigg]$$

$$ g'=g\bigg[1-\dfrac{2(32)}{6400}\bigg]$$
$$\implies g'= .99 g m/s^2$$
Question 6
1 / -0

A body of mass $$m$$ is lifted up from the surface of earth to a height three times the radius of the earth $$R$$.The change in potential energy of the body is

A
$$3mgR$$

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

C
$$\displaystyle\frac{3}{4}mgR$$

D
$$2mgR$$

Solution

Hint: Use formula of change in potential energy
$$Step - 1: Put\ formula\ of\ change\ in\ potential\ energy\ with\ height$$
Change in potential energy is given by the formula:
$$∆U = \dfrac{m\ g\ h}{1\ +\ \dfrac{h}{R}}$$
Now, if body is lifted to height three times to the radius of earth, then h = 3 R.
$$Step - 2: Put\ values\ in\ formula\ $$
On putting value in formula,
$$∆U = \dfrac{m\ g\ (3\ R)}{1\ +\ \dfrac{3\ R}{R}} = \dfrac{3}{4} m g R$$
$$Answer:$$
Hence, option C is the correct answer.
Question 7
1 / -0

At some planet '$$g$$' is $$1.96 m/s^2$$. If it is safe to jump from a height of $$2m$$ on earth, then what should be corresponding safe height for jumping on that planet

A
$$5m$$

B
$$2m$$

C
$$10m$$

D
$$20m$$

Solution

At earth, $$g_e=9.8 m/s^2 \quad h_e=2 m$$
At other planet, $$g_p=1.96 \quad h=?$$
$$\implies \dfrac{g_p}{g_e}= \dfrac{1}{5}$$
Now at some planet g reduces by $$5$$ times, thus safe height will increases by $$5$$ times.
Hence $$h= 5\times 2= 10 m$$

Question 8

1 / -0

On the surface of earth acceleration due to gravity is g and gravitational potential is V.

Column-I	Column-II
a) At height h = R, value of g	(p) decreases by a factor 1/4
(b) At depth h = R/2 value of g	(q) decreases by a factor 1/2
(c) At height h = R, of V	(r) Increases by a factor 11/8
(d) At depth h = R/2 of V	(s) Increases by a factor 2 (t) None

(a - p), (b - q), (c - s), (d - t)

(a - q), (b - p), (c - t), (d - s)

(a - t), (b - s), (c - p), (d - q)

None of these

Solution

$$\text{Value of g at a distance 'r' from center of earth}=\dfrac{GM}{r^2}$$

$$\text{'g' at surface}=\dfrac{GM}{R^2}$$

$$'g' \text{at height}\ 'R' \text{from surface}=\dfrac{GM}{(2R)^2}$$

$$\text{Hence decreased by a factor of}\ \dfrac{1}{4}.$$

$$\text{Dependence of 'g' on the depth h is}\ g_h=\dfrac{4\pi}{3}G\rho h$$

$$\text{Hence decreases by a factor of}\ \dfrac{1}{2}\ \text{at}\ h=\dfrac{R}{2}.$$

$$\text{Gravitational potential varies with distance 'r' from center of earth as}-\dfrac{GM}{r}$$

$$\text{Hence at surface it has value}-\dfrac{GM}{R}$$

$$\text{At a height 'R' it has value}-\dfrac{GM}{2R}$$

$$\text{Hence it has 'increased' by a factor 2, since potential is negative.}$$

Question 9
1 / -0

A missile is lauched with a velocity less than the escape velocity. Sum of its kinetic energy and potential energy is.

A
positive

B
negative

C
may be negative or positive depending upon its

initial velocity

D
zero

Solution

At the surface, $$T.E= \dfrac{1}{2}mv^2 + (-\dfrac{GMm}{R})$$
For $$v= escape velocity (v_e)= \sqrt{\dfrac{2GM}{R}}$$ $$\implies T.E=0$$
Thus for $$v< v_e$$ $$T.E <0$$
Question 10
1 / -0

The escape velocity for a planet is $$\displaystyle v_{e}$$ A particle starts from rest at a large distance from the planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre Its speed at the centre of the planet will be

A
$$\displaystyle \sqrt{1.5}v_{e}$$

B
$$\displaystyle \frac{v_{e}}{\sqrt{2}}$$

C
$$\displaystyle v_{e}$$

D
zero

Solution

From mechanical energy conservation 0 + 0 = $$\displaystyle \frac{1}{2}mv^{2}-\frac{3GMm}{2R}\: \: \Rightarrow \: \: v=\sqrt{\frac{3GM}{R}}=\sqrt{1.5}v_{e}$$

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Gravitation Test - 36

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Gravitation Test - 36

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Question 1

decrease by $$0.5%$$

decrease by $$2%$$

increase by $$0.5%$$

increase by $$1%$$

Solution

Question 2

$$\displaystyle\frac{U}{r}$$

$$\displaystyle\frac{U}{2r}$$

$$\displaystyle\frac{U}{4r}$$

$$\displaystyle\frac{U}{\sqrt{2}r}$$

Solution

Question 3

$$g$$ decreases on going above the surface of earth

$$g$$ increases on going below the surface of earth

$$g$$ is maximum at pole

$$g$$ increases on going from equator to poles

Solution

Question 4

$$h = R$$

$$h = 2R$$

$$h \equiv ( \sqrt 2 +1)R$$

$$h \equiv ( \sqrt 2 -1)R$$

Solution

Question 5

$$0.5 g ms^{2}$$

$$0.99 g ms^{2}$$

$$1.01 g ms^{2}$$

$$0.90 g ms^{2}$$

Solution

Question 6

$$3mgR$$

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

$$\displaystyle\frac{3}{4}mgR$$

$$2mgR$$

Solution

Question 7

$$5m$$

$$2m$$

$$10m$$

$$20m$$

Solution

Question 8

(a - p), (b - q), (c - s), (d - t)

(a - q), (b - p), (c - t), (d - s)

(a - t), (b - s), (c - p), (d - q)

None of these

Solution

Question 9

positive

negative

may be negative or positive depending upon itsinitial velocity

zero

Solution

Question 10

$$\displaystyle \sqrt{1.5}v_{e}$$

$$\displaystyle \frac{v_{e}}{\sqrt{2}}$$

$$\displaystyle v_{e}$$

zero

Solution

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may be negative or positive depending upon its

initial velocity