Gravitation Test

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Gravitation Test - 39

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Question 1
1 / -0

A ball is dropped from a height $$h$$. As it bounces off the floor, its speed becomes $$80 \%$$ of what it was just before it hit the floor. The ball then rises to a height equal to

A
$$0.80h$$

B
$$0.75h$$

C
$$0.64h$$

D
$$0.50h$$

Solution

According to third equation of motion, $$v^2=u^2+2ax$$
Here, $$u= 0 \ m/s$$
$$a=-g$$ and $$x=-h$$
Negative sign indicates downward direction. Displacement and acceleration both are downwards.

So,
$$v=\pm \sqrt{2(-g)(-h)}$$
We take minus sign because it is downwards.
$$v=-\sqrt{2gh}$$

After bouncing. velocity becomes $$80 \%$$ of $$v,$$ i.e.,
$$v'=+0.8\sqrt{2gh}$$
(positive sign because the direction of ball has reversed after bouncing and is upwards.

Applying third equation of motion again, for $$u=v'$$, $$v = 0$$ and $$a=-g$$
$$v^{2} =u^{2}+2 \times a\times x$$
Thus,
$$0=0.64(2gh)+2(-g)x$$
or
$$x=0.64h$$
Question 2
1 / -0

A person jumps from the fifth storey of a building with load on his head. The weight experienced by him before reaching the earth will be:

A
$$zero$$

B
$$g \ kg-wt$$

C
$$m(g + a)$$

D
$$mg$$

Solution

Weight experienced by a body is equivalent to the reaction force acting on the same. When a person with a load on his head jumps from a building, he and the load are in a free fall. We know that there is no reaction force experienced by a body in free fall. Hence, the weight experienced by the person is also $$0$$.
Question 3
1 / -0

The value of G for two bodies in vacuum is $$6.67 \times 10^{-11} N-m^2/Kg^2$$. Its value in a dense medium of density $$10^{10} gm/cm^3$$ will be:

A
$$6.67 \times 10^{-11} N-m^2 /Kg$$

B
$$6.67 \times 10^{-31} N-m^2 /Kg$$

C
$$6.67 \times 10^{-21} N-m^2 /Kg$$

D
$$6.67 \times 10^{-10} N-m^2 /Kg$$

Solution

G is a Universal Gravitational constant and is same every where in the universe. Thus,
G $$ = 6.67 \times 10^{-11} N.m^{2}.kg^{-2}$$
Question 4
1 / -0

FIll in the blank.
When a solid floats in a liquid, the weight of _______ by its immersed part of the solid is equal to the weight of the solid.

A
less

B
buoyant force

C
weight

D
liquid displaced

Solution

Liquid displaced.
This is known as Archimedes principle.
Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
Question 5
1 / -0

Two balls of masses in ratio $$1 : 2$$ are dropped from the same height. Find the ratio of the forces acting on them during motion.

A
$$1 : 2$$

B
$$2 : 1$$

C
$$1 : 1$$

D
$$1 : 4$$

Solution

$$\dfrac{F_1}{F_2}=\dfrac{W_1}{W_2}=\dfrac{m_1g}{m_2g}=\dfrac{m_1}{m_2}=\dfrac{1}{2}$$
Question 6
1 / -0

A body of mass $$1.5 \ kg$$ is dropped from a height of $$12 \ m$$. What is the force acting on it during its fall? $$(g=9.8 \ ms^{-2})$$

A
$$1.5 \ N$$

B
$$9.8 \ N$$

C
$$14.7 \ N$$

D
$$147 \ N$$

Solution

Force acting on the body during its fall = weight of the body.
So, $$F=W=m \times g=1.5 \times 9.8=14.7\ N$$
And this force is independent of the height of fall.
Question 7
1 / -0

A mass $$m_1$$ is placed at the center of a shell of radius $$R$$. Another mass $$m_2$$ is placed at a distance $$R$$ from the surface of the shell and is immersed in oil of dielectric constant $$10$$. Find the gravitational force on $$m_1$$. (Assume that the shell and oil are practically massless).

A
zero

B
$$\displaystyle \dfrac{G(M+m_2)m_1}{10 R^2}$$

C
$$\displaystyle \dfrac{Gm_1 m_2}{4R^2}$$

D
$$\displaystyle \dfrac{Gm_1 (4M + m_2)}{4R^2}$$

Solution

Here, shell exerts no force as its gravitational field inside the shell is zero. Oil does not play any role.
So the gravitational force is usual i.e., $$F=Gm_1m_2/(2R)^2=Gm_1m_2/4R^2$$.
Question 8
1 / -0

The mass of the earth is $$80$$ times that of the moon. Their diameters are $$12800\ Km$$ and $$3200\ Km$$ respectively. The value of g on moon will be _______, if its value on earth is $$980\ cm/s^2$$.

A
$$98\ cm/s^2$$

B
$$196\ cm/s^2$$

C
$$100\ cm/s^2$$

D
$$294\ cm/s^2$$

Solution

Let $$ g_e, g_m $$ are accelarations due to gravity on earth and moon respectively.
Similarly let $$ M_e, M_m $$ are masses of earth and moon respectively, $$ r_e , r_m$$ are radiuses of earth and moon respectively.
Now, $$ g_e = \dfrac {GM_e}{{r}^{2}_{e}} $$

$$ g_m = \dfrac {GM_m}{{r}^{2}_{m}} $$

taking the ratio, $$ \dfrac {g_m}{g_e} = \left(\dfrac {GM_m}{{r}^{2}_{m}}\right) \times \left(\dfrac {{r}^{2}_{e}}{GM_e}\right) $$
simplifying and rearranging terms we get
$$ g_m = \left(\dfrac {M_m}{M_e}\right) \times \left(\dfrac {r_e}{r_m}\right)^2 \times g_e $$
Substituting the given values, $$ g_e =980\ cm s^{-2},$$

$$Given:\dfrac {M_e}{M_m} = 80 , \ \ 2r_e = 128 \times 10^7 cm, \ 2r_m = 32 \times 10^7 cm $$

$$ g_m = \left(\dfrac{1}{80}\right) \times \left(\dfrac{64 \times 10^7}{16 \times 10^7}\right)^2 \times 980 = 196\ cms^{-2} $$
Question 9
1 / -0

Two point masses each equal to $$1$$ kg attract one another with force of $$10^{-10} N$$. The distance between the two point masses is $$\displaystyle (G=6.6 \times 10^{-11} MKS$$ units$$)$$

A
$$\displaystyle 8cm$$

B
$$\displaystyle 0.8cm$$

C
$$\displaystyle 80cm$$

D
$$\displaystyle 0.08cm$$

Solution

According to newton's Gravitational Law
$$\displaystyle F_g=\frac {GM_1M_2}{r^2}$$ here $$\displaystyle F_g=10^{-10}$$ Newton,
$$\displaystyle m_1=m_2=1$$kg,
$$\displaystyle G=6.6 \times 10^{-11}$$
So $$\displaystyle r^2=\frac {GM_1M_2}{F_g} = \frac {6.6 \times 10^{-11} \times 1 \times 1}{10^{-10}}=0.66$$
or $$\displaystyle r=0.8125$$ metre $$\displaystyle = 81.25 cm = 80cm$$
Question 10
1 / -0

A sphere of iron and another of wood of the same radius are held under water. Let $$F_{A} $$ and $$F_{B}$$ are the uptrusts of iron and wood respectively.Compare the upthrust on the two spheres.

A
$$F_{A} $$=$$F_{B}$$

B
$$F_{A} $$>$$F_{B}$$

C
$$F_{A} $$<$$F_{B}$$

D
None of the above

Solution

As both have equal volume inside water, they will experience the same upthrust
$$F_{Thrust}=\rho Vg$$, where $$\rho$$ is density of water and $$V$$ is volume of spheres.
Hence, option A is correct.

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Gravitation Test - 39

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Gravitation Test - 39

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SHARING IS CARING

Question 1

$$0.80h$$

$$0.75h$$

$$0.64h$$

$$0.50h$$

Solution

Question 2

$$zero$$

$$g \ kg-wt$$

$$m(g + a)$$

$$mg$$

Solution

Question 3

$$6.67 \times 10^{-11} N-m^2 /Kg$$

$$6.67 \times 10^{-31} N-m^2 /Kg$$

$$6.67 \times 10^{-21} N-m^2 /Kg$$

$$6.67 \times 10^{-10} N-m^2 /Kg$$

Solution

Question 4

less

buoyant force

weight

liquid displaced

Solution

Question 5

$$1 : 2$$

$$2 : 1$$

$$1 : 1$$

$$1 : 4$$

Solution

Question 6

$$1.5 \ N$$

$$9.8 \ N$$

$$14.7 \ N$$

$$147 \ N$$

Solution

Question 7

zero

$$\displaystyle \dfrac{G(M+m_2)m_1}{10 R^2}$$

$$\displaystyle \dfrac{Gm_1 m_2}{4R^2}$$

$$\displaystyle \dfrac{Gm_1 (4M + m_2)}{4R^2}$$

Solution

Question 8

$$98\ cm/s^2$$

$$196\ cm/s^2$$

$$100\ cm/s^2$$

$$294\ cm/s^2$$

Solution

Question 9

$$\displaystyle 8cm$$

$$\displaystyle 0.8cm$$

$$\displaystyle 80cm$$

$$\displaystyle 0.08cm$$

Solution

Question 10

$$F_{A} $$=$$F_{B}$$

$$F_{A} $$>$$F_{B}$$

$$F_{A} $$<$$F_{B}$$

None of the above

Solution

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