Olympiad Science - Gravitation and Pressure
Exam Duration: 45 Mins Total Questions : 30
A body weighs 72 N on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth from the surface
- (a)
72 N
- (b)
28 N
- (c)
16 N
- (d)
32 N
W on the earth's surface = 72 N = mg ,
\(h=\frac { R }{ 2 } \\ { g }_{ h }\left( \frac { g }{ 1+\frac { h }{ R } } \right) ^{ 2 }\left( \frac { g }{ 1+\frac { 1 }{ 2 } } \right) ^{ 2 }=\frac { g }{ 9 } \left( 4 \right) \\ m\times g\left( \frac { 4 }{ 9 } \right) =72\times \frac { 4 }{ 9 } =32N\)
Where will it be profitable to purchase one kilogram sugar?
- (a)
At the poles
- (b)
At the equator
- (c)
At 45° latitude
- (d)
At 40° latitude
As the value of g is maximum at poles, for the same value, the weight of 1 kg sugar measures more at poles. Hence, it will be profitable to purchase one kilogram of sugar at the poles.
The figure given below shows a planet in elliptical orbit around the sun S.
At what position will the kinetic energy of the planet be maximum?
- (a)
P1
- (b)
P4
- (c)
P3
- (d)
P2
In the figure shown, at point P1 the kinetic energy of the planet is maximum as it has the least value of radius vector with greater forces of gravitational attraction.
The distances of two planets (Neptune and Saturn) from the sun are 1013 m and 1012 m respectively. Find the ratio of the time periods of the planets.
- (a)
\(100:1\)
- (b)
\(1:\sqrt { 10 } \)
- (c)
\(\sqrt { 10 } :1\)
- (d)
\(10\sqrt { 10 } :1\)
\({ d }_{ neptune }={ 10 }^{ 13 }m;{ d }_{ saturn }={ 10 }^{ 12 }m\\ { T }^{ 2 }\alpha { R }^{ 3 }\quad \quad (kepler's\quad { 3 }^{ rd }law)\\ \frac { { T }_{ n } }{ { T }_{ S } } =\left( \frac { { 10 }^{ 13 } }{ { 10 }^{ 12 } } \right) ^{ 3 }=\left[ 10 \right] ^{ 3 },\frac { { T }_{ neptune } }{ { T }_{ saturn } } =\sqrt { \frac { 1000 }{ 1 } } \\ \frac { { T }_{ neptune } }{ { T }_{ saturn } } =10\sqrt { 10 } :1\)
When does the value of g at a place increase?
- (a)
With decrease in the latitude of the place
- (b)
With increase in the latitude of the place
- (c)
With increase in the altitude of the place
- (d)
With increase in the angular rotation of the earth
The value of 9 at a place increases with increase in the latitude of the place.
Two planets A and 8 have their radii in the ratio of 2: 5 and densities in the ratio of 1: 6 respectively which of the following statements is NOT true regarding the given information?
- (a)
The ratio of acceleration due to gravity on them is 1 : 15
- (b)
For the same volume of planets, mass of planet A is greater than that of planet 8
- (c)
A body weighs 15 times more on planet 8 than on planet A
- (d)
Planet 8 has greater volume than planet A
\({ R }_{ A }:{ R }_{ B }=2:5,{ D }_{ A }:{ D }_{ B }=1:6,{ g }_{ A }:{ g }_{ B }\\ \frac { { \frac { { GM }_{ A } }{ { R }_{ A }^{ 2 } } } }{ \frac { { GM }_{ B } }{ { R }_{ B }^{ 2 } } } =\frac { 4 }{ 3 } \pi \frac { { R }_{ A }^{ 3 }{ D }_{ A } }{ { R }_{ A }^{ 2 } } \times \frac { { R }_{ B }^{ 2 } }{ \frac { 4 }{ 3 } \pi { R }_{ B }^{ 3 }{ D }_{ B } } =\frac { { R }_{ A }{ D }_{ A } }{ { R }_{ B }{ D }_{ B } } \\ \frac { { g }_{ A } }{ { g }_{ B } } =\frac { 2 }{ 5 } \times \frac { 1 }{ 6 } =\frac { 1 }{ 15 } \\ { W }_{ B }={ mg }_{ B }=15m{ g }_{ A }=15\quad { W }_{ A }\)
For the same volume of planets, mass of planet A is 1/6th the mass of planet B. Planet B has greater volume than planet A
Two spheres each of mass 105 kg and radius 10m are kept in contact. Find the force of gravitation acting between them.
- (a)
10-3 N
- (b)
6.67 x 10-3N
- (c)
6.67 x 10-11 N
- (d)
10-11 N
\({ F }_{ g }=\frac { { Gm }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } =\frac { 6.67\times { 10 }^{ -11 }\times { 10 }^{ 5 }\times { 10 }^{ 5 } }{ \left( 10 \right) ^{ 2 } } \\ =6.67\times { 10 }^{ -1 }\times { 10 }^{ -2 }\times 6.67\times { 10 }^{ -3 }N\)
A body weighs 700 N on the surface of the earth. What is its weight on the surface of a planet, whose mass is 1/7 th and radius is 1/2 as that of the earth?
- (a)
400 N
- (b)
300 N
- (c)
700 N
- (d)
500 N
\({ w }_{ earth }=700N={ mg }_{ earth }\)
\({ w }_{ earth }=\frac { { Gm }_{ planet } }{ { R }_{ planet }^{ 2 } } =\frac { G\left( \frac { 1 }{ 7 } { M }_{ earth } \right) }{ \left( \frac { 1 }{ 2 } { R }_{ earth } \right) ^{ 2 } } =\frac { 4 }{ 7 } { g }_{ earth }\)
\({ w }_{ planet }={ mg }_{ planet }=m\times \frac { 4 }{ 7 } { g }_{ earth }N\)
\(=\frac { 4 }{ 7 } =700N=400N\)
Find the ratio of the distances travelled by a body falling freely from rest in the first, second and third seconds respectively.
- (a)
1:4: 9
- (b)
1: 2: 3
- (c)
1:9:25
- (d)
1: 3: 5
\({ S }_{ n }=u+a\left[ n-\frac { 1 }{ 2 } \right] \\ \)
For a freely falling body,
u = 0 and a = +g,:. S1:S2:S3=
\(\frac { 1 }{ 2 } g\left( 1 \right) ^{ 2 }:\left( \frac { 1 }{ 2 } g\left( 2 \right) ^{ 2 }-\frac { 1 }{ 2 } g\left( 1 \right) ^{ 2 } \right) :\left( \frac { 1 }{ 2 } g\left( 3 \right) ^{ 2 }-\frac { 1 }{ 2 } g\left( 2 \right) ^{ 2 } \right) \\ \frac { g }{ 2 } :\frac { 3g }{ 2 } :\frac { 5g }{ 2 } 1:3:5\)
A stone is thrown upwards with a speed 'u' from the top of a tower. It reaches the ground with a velocity '3u'. Find the height of the tower.
- (a)
\(\sqrt { 3 } \)
- (b)
\(\frac { { 2 }u^{ 2 } }{ g } \)
- (c)
\(\frac { { 3u }^{ 2 } }{ g } \)
- (d)
\(\frac { { 4u }^{ 2 } }{ g } \)
v2 - u2 = 2as, v = 3u, a = - g, s = - h
:. (3u)2 - u2 = 2 (- g)(- h)
\(\Rightarrow 8{ u }^{ 2 }=2gh\Rightarrow h=\frac { { 4u }^{ 2 } }{ g } \\ \)
From the top of a building of height 40 m, a boy throws a stone vertically upwards with an initial velocity of 10ms-1 such that it eventually falls to the ground. At what time will the stone reach the ground? (g=10ms-2)
- (a)
4 s
- (b)
3 s
- (c)
2 s
- (d)
1 s
\(h=-ut+\frac { 1 }{ 2 } gt^{ 2 }\\ 40=-10t+\frac { 1 }{ 2 } \times 10\times { t }^{ 2 }\Rightarrow t=4s\\ \)
If it is safe to jump from a height of 2 m on to the earth, what would be the safe height on a planet where the value of 'g' is 1.96 m S-2?
- (a)
2 m
- (b)
4 m
- (c)
6 m
- (d)
10 m
\({ h }_{ 1 }=2m,{ g }_{ 1 }=9.8m{ s }^{ 12 },\quad \quad { g }_{ 2 }=1.96m{ s }^{ -2 }\\ { h }_{ 1 }{ g }_{ 1 }={ h }_{ 2 }g_{ 2 }\quad ,\quad { h }_{ 2 }=\frac { { h }_{ 1 }{ g }_{ 1 } }{ { g }_{ 2 } } =\frac { 2\times 9.8 }{ 1.96 } =10m\)
A body falling freely from rest covers \(\frac { 7 }{ 16 } \) of the total height in the last second of its fall. What is the height from which it falls?
- (a)
24.2 m
- (b)
38.4 m
- (c)
78.4 m
- (d)
46.8 m
\(Total\quad height\Rightarrow h=\frac { 1 }{ 2 } { gn }^{ 2 }\)
in the last second, the body covers \(\frac { 7 }{ 16 } \) th of the total height so, Sn =\(\frac { 7 }{ 16 } \)h
\(we\quad have\quad { S }_{ n }=u+a\left[ n-\frac { 1 }{ 2 } \right] \\ here\quad u=0\quad and\quad a=g\\ so,\quad { S }_{ n }=g\left[ n-\frac { 1 }{ 2 } \right] ,{ S }_{ n }=\frac { 7 }{ 16 } h\pi \frac { 7 }{ 16 } \left[ \frac { 1 }{ 2 } { gn }^{ 2 } \right] \\ \therefore \frac { 7 }{ 16 } \left[ \frac { 1 }{ 2 } g{ n }^{ 2 } \right] =g\left[ n-\frac { 2 }{ 2 } \right] \Rightarrow n=4s\\ so,h=\frac { 1 }{ 2 } \times 9.8\times { 4 }^{ 2 }=78.4m\)
If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first 3 seconds. Find the time of its travel
- (a)
3 s
- (b)
4 s
- (c)
5 s
- (d)
6 s
We have,the distance travelled by the body in the nth second equal to the distance travelled by it in 't' seconds. t = 3 s
So,sn = S
\(g\left[ n-\frac { 1 }{ 2 } \right] =\frac { 1 }{ 2 } { gt }^{ 2 }\Rightarrow n-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \times { 3 }^{ 2 }=\frac { 9 }{ 2 } \\ \Rightarrow 4.5\quad or\quad 5\quad seconds\\ \)
A stone is dropped from a tower. It was found that it covered a distance of 45 m during its last second of the fall. Calculate the time of the fall and the height of the tower? (9 = 70 m S-2)
- (a)
5 s,45 m
- (b)
10 s,90 m
- (c)
5 s, 125 m
- (d)
10 s, 100 m
During the last second it fell through 45 m
\({ S }_{ n }=\frac { g }{ 2 } (2n-1),45=\frac { 10 }{ 2 } (2n-1)\Rightarrow n=5\quad s\)
Time of flight is 5 s. Therefore, the height of the tower:
\({ S }=\frac { 1 }{ 2 } \times g\times { t }^{ 2 }=\frac { 1 }{ 2 } \times 10\times 25=125m\)
A liquid exerts a pressure of 100,999 Pa on an object when it is 73.5 cm below the liquid surface. What is the density of the liquid? (Take 9 = 70 m S-2)
- (a)
136 kg m-3
- (b)
272 kg m-3
- (c)
13,700 kg m-3
- (d)
27,200 kg m-3
pgh = 1,00,999 = (P ) (10)(0.735) p = 1,00,999 \(\div\) (tox 0.735) = 13,741 kg m-3 := 13,700 kg m-3
The diagram given below shows a special flask filled with water
Which of the followi ng statements is true?
- (a)
Point P has a higher pressure than point Q
- (b)
Point R has a higher pressure than point Q
- (c)
Point Q has the highest pressure as compared to point P and point R
- (d)
Point P,point Q and point R have the same pressure
The pressure at the same depth in the same liquid is the same.
A hollow metal sphere of mass 180.6 g contains a cavity of volume 2.5 cm3. This metal when placed in water, displaces 24 cc of water. Find the specific gravity of the metal.
- (a)
21.5 g cm-3
- (b)
6.3 g cm-3
- (c)
8.4 g cm-3
- (d)
24 g cm-3
Let the density of metal be p Mass of metal = 180.6 9
∴ Actual volume of metal = (180.6/ P ) cm-3
Volume of cavity = 2.5 cm-3 Volume of metal = volume of water displaced (d) = 24 cm-3 Volume of cavity = volume of liquid displaced - actual volume of metal
\(2.4=24-\frac { 180.6 }{ p } \Rightarrow \frac { 180.6 }{ p } \\ =24-2.5\Rightarrow \frac { 180.6 }{ p } =21.5\\ p=8.4g\quad c{ m }^{ -3 }\)
specific gravity of metal = 8.4 9 cm-3
A rectangular tank of 6 m long,2 m broad and 2 m deep is full of water, what is the thrust acting on the bottom of the tank?
- (a)
23.52 x 104 N
- (b)
23.52 N
- (c)
11.76 x 104
- (d)
3.92 X 104 N
Thrust = Total weight acting normally = mg But mass = volume x density
∴Thrust = volume of tank x density of water x 9 =(6 x 2 x 2) x (103) x (9.8)
= 23.52 x 104 N
When a body is immersed in a liquid, what is the loss in its weight equal to?
- (a)
The weight of the liquid
- (b)
The weight of the body
- (c)
The difference in the weights of body in air and in liquid
- (d)
The upthrust of liquid on the body
When a body is immersed in a liquid, the loss in its weight is equal to the upthrust of liquid on the body
Equal masses of water and a liquid of relative density 2 are mixed together, then what is the density of the mixture (in g cm-3)?
- (a)
2/3
- (b)
4/3
- (c)
3/2
- (d)
3
\(RD.\frac { Density\quad of\quad liquid }{ Density\quad of\quad water } =2\\ \Rightarrow Density\quad of\quad liquid=2g\quad { cm }^{ -3 }\\ Density\quad of\quad mixture(D)\\ =\frac { Total\quad mass }{ Total\quad volume } =\frac { 2m }{ { v }_{ 1 }+{ v }_{ 2 } } \\ \frac { m }{ { v }_{ 1 } } =2g{ cm }^{ -3 },\frac { m }{ { v }_{ 2 } } =1g{ cm }^{ -3 }\\ D=\frac { 2 }{ \frac { 1 }{ 2 } +1 } ;D=\frac { 4 }{ 3 } g{ cm }^{ -3 }\)
A stone is thrown vertically upwards with an initial velocity of 14m S-1. Find the time taken by the stone to strike the ground. (Take g=9.8 m S-2)
- (a)
2.86 s
- (b)
3.46 s
- (c)
3.86 s
- (d)
4.86 s
\(u=14m{ s }^{ -1 };g=9.8m{ s }^{ -2 }\\ t=\frac { 2u }{ g } =\frac { 12\times 14 }{ 9.8 } =2.86s\)
If suddenly the gravitational force of attraction between the earth and a satellite revolving around it becomes zero, what will happen?
- (a)
The satellite will continue to move in its orbit with the same velocity.
- (b)
The satellite will move tangentially and escape away from its orbit.
- (c)
The satellite will become stationary in its orbit.
- (d)
The satellite will move towards the earth
When gravitational force becomes zero, then centripetal force on the satellite also becomes zero. Therefore, the satellite moves tangentially and escapes away from its orbit.
Which of the following statements regarding gravitational force existing between two bodies is true?
- (a)
The first body exerts attractive force on the second body while the second body exerts repulsive force on the first body.
- (b)
The gravitational force is zero when they are kept in vacuum.
- (c)
Force exerted by the first body on the second is not equal to the force exerted by the second body on the first.
- (d)
Force exerted by the first body on the second body is equal to the force exerted by the second body on the first.
Force exerted by the first body on the second body is equal to the force exerted by the second body on the first body.
A body weighs 12 N on the surface of the moon. What is its weight on the surface of the earth?
- (a)
72 N
- (b)
2 N
- (c)
24 N
- (d)
Zero
\({ W }_{ earth }={ 6W }_{ moon }=6\times 12=72N\)
If F is the force between two bodies of masses m1 and m2 at certain separation, then what is the force between \(\sqrt { 5 } \) m1 and \(\sqrt { 3 } \) m2 at the same separation?
- (a)
\(\sqrt { 5 } F\)
- (b)
\(F/\sqrt { 15 } \)
- (c)
\(\\ \\ \sqrt { 15 } F\)
- (d)
F
\(Given\quad F=\frac { { m }_{ 1 }{ m }_{ 2 } }{ { d }^{ 2 } } \\ if\quad { m }_{ 2 }=\sqrt { 3 } { m }_{ 2 };{ m }_{ 1 }=\sqrt { 5 } m_{ 1 },\\ then\quad { F }^{ 1 }=\frac { g\sqrt { 5 } { m }_{ 1 }\times \sqrt { 3 } { m }_{ 2 } }{ { d }^{ 2 } } =\sqrt { 15 } F\)
A stone is dropped from a building and 2 seconds later another stone is dropped. How far apart are these two stones by the time the first one reaches a speed of 30 m s-1 (Take 9 = 10 m S-2)
- (a)
80 m
- (b)
100 m
- (c)
60 m
- (d)
40 m
For the first stone, u, = 0 Let the first stone pick up a speed of 30 m s1 after t1 see. Then, v1 = u1 + gt1 ⇒t, = 30/10 = 3 s.Total distance travelled by the first stone in 3 seconds is:\({ s }_{ 1 }={ u }_{ 1 }{ t }_{ 1 }+\frac { 1 }{ 2 } { gt }_{ 1 }\\ =0+\frac { 1 }{ 2 } \times 10\times \left( 3 \right) ^{ 2 }=45m\)The second stone is dropped after 2 seconds. This means that it only travelled for 1sec by the time the first stone reaches a speed of 30 m/second.
Distance travelled by the second stone in 1 second.\(\Rightarrow { s }_{ 2 }={ u }_{ 2 }{ t }_{ 2 }+\frac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }=0+\frac { 1 }{ 2 } \times 10\times \left( 1 \right) ^{ 2 }=5m\)
Distance between both the stones = S1 - S2= 45 - 5 = 40 m
Analyse the given statements and choose the correct option.
Statement I :The buoyant force of water on a submerged wooden cube is greater than the buoyant force of water on a steel cube of equal volume.
Statement II : The buoyant force on a body is equal to the weight of the liquid displaced by the body.
- (a)
Both statement 1 and statement /I are correct and statement /I is the correct explanation of statement I.
- (b)
Both statement 1 and statement /I are correct, but statement /I is not the correct explanation of statement I.
- (c)
Statement 1 iscorrect ,but statement /I is incorrect
- (d)
Statement 1is incorrect,but statement /I is correct
The buoyant force on a body is equal to the weight of the liquid displaced by the body
A block of wood of length 40 cm and area of cross section 15cm-2floats in water with 3/8 of its length above water. What is the density of wood? (density of water = 1 g cm-3)
- (a)
0.256 g cm-3
- (b)
0.526 g cm-3
- (c)
0.625 g cm-3
- (d)
0.650 g cm-3
Length of wood outside water\(=\frac { 3 }{ 8 } \times 40=15cm\)
Length of wood inside water=40-15=25cm
by laws of floatation\({ h }_{ wood }\times { d }_{ wood }={ h }_{ water }\times { d }_{ water }\)h and d stand for height and density
40 x dwood= 25 x 1 g cm-3
dwood =\(\frac { 25 }{ 40 } \)=0.625 g cm-3
A ball is released from the top of a tower of height 'h' metres. It takes 'T' seconds to reach the ground. What is the position of the ball in T/3 seconds
- (a)
h/9 metres from the ground
- (b)
7h/9 m from the ground
- (c)
8h/9 metres from the ground
- (d)
17h/18 m from the ground
\({ h }^{ 1 }=\frac { 1 }{ 2 } \times g\times \frac { { T }^{ 2 } }{ 9 } andT=\sqrt { \frac { 2h }{ g } } \)
From ground, the position of ball is
\(\\ \left( h-\frac { h }{ 9 } \right) \quad m=\frac { 8h }{ 9 } m\)