Given below are the class 9 physics gravitation Numericals .It includes thrust,pressure,relative density numericals along with Archimedes principle numericals
a. short answer question questions
b. Long answer questions
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short answer question questions
Question 1.
A solid weighs 200 gf in air, 160 gf in water and 170 gf in a liquid. Calculate the relative density of the solid and that of the liquid? Answer
a. This can be calculated using the formula
$Relative \; density \; of \; solid = \frac {Weight \; of \; solid \; in \; air }{ Loss \; of \; weight \; of \; solid \; in \; water}$v
$= \frac {200}{40} = 5$
b. This can be calculated using the formula
$Relative \; density \; of \; liquid = \frac {Loss \; of \; weight \; of \; solid \; in \;liquid}{Loss \; of \; weight \; of \; solid \; in \; water} = \frac {30}{40} = .75$
Question 2.
Define thrust. A force of 150N acts on a surface of area $15 \; cm^2$. Calculate thrust and pressure? Answer
The force acting on an object perpendicular to the surface is called thrust
Thrust =150 N
Now $Pressure = \frac {thrust}{Surface \; Area} = \frac {150 }{15 \times 10^{-4}} = 10^5 \; N/m^2$
Question 9.
Relative density of gold is 19.3. The density of water is $1000 \; kg/m^3$. Find the density of gold in SI unit in g/cc? Answer
$1000 \; kg/m^3 =1 \; g/cm^3$
$Relative \; density = \frac {density \; of \; the gold}{density \; of \; water}$
or
$density \; of \; gold = (Relative \; density) \times (density \; of \; water) = 19.3 g/cc$
Question 3.
The volume of a 350g sealed tin is 200 cubic.cm. Find the density of the tin in g/cc. Also find the density in SI unit? Answer
$Density = \frac {mass}{volume}$
$Density = \frac {350}{200 }=1.75 g/cm^3$
Density in SI unit
$=1.75 \times 1000= 1750 kg/m^3$
Question 4.
Relative density of silver is 10.8? What is the density of silver in SI unit?. Given The density of water is $1000 \; kg/m^3$ Answer
$Relative \; density = \frac {density \; of \; the silver}{density \; of \; water}$
or
$density \; of \; silver= (Relative \; density) \times (density \; of \; water) = 10800 kg/m^3$
Question 5.
A force of 250N acts on a surface of area $15cm^2$. Calculate thrust and pressure? Answer
Question 6.
A force of 400N acts on a surface of area 25cm2. Calculate thrust and pressure? Calculate the changed pressure if the force now is made to 600N? Answer
Thrust= 400 N
$Pressure = \frac {Thrust}{Surface area} = \frac {400}{ 25 \times 10^{-4}} = 160000 Pa$
Now New pressure
$Pressure = \frac {Thrust}{Surface area} = \frac {600}{ 25 \times 10^{-4}} = 240000 Pa$
Change in Pressure = 240000- 160000=80000 Pa
Question 7.
Define relative density. Relative density of mercury is 13.6. The density of water is 1000 kg/m3. What is the density of mercury in SI unit? Answer
Relative Density is the ratio between density of a substance and density of water is called relative density.Relative density has no units because it is the ratio between the similar physical quantities
$Relative \; density = \frac {density \; of \; the object}{density \; of \; water}$
$Relative \; density of mercury= \frac {density \; of \; the mercury}{density \; of \; water}$
or
$density \; of \; mercury= (Relative \; density) \times (density \; of \; water) = 13600 kg/m^3$
Question 8.
You are provide with a hollow iron ball of volume 20cm3, mass of 15g and a solid iron ball of mass 20g. Both are placed on the surface of water contained in a large tub. What will float? Give reason for your answer? Given density of iron= 8 g/cc Answer
Density of hollow ball = 15/20 = .75 gm/cc
Density of solid ball = 8 gm/cc
Now Density of water is 1 gm/cc
So Hollow ball will float and Solid ball will sink
Question 9.
The dining has dimensions $50 \;m \times 15 \; m \times 3.5 \; m$. Calculate the mass of air in the hall? (Density of air= 1.30kg/m3)? Answer
Volume of the dining room= $50 \;m \times 15 \; m \times 3.5 \; m= 2625 m^3$
Now Density of air= 1.30kg/m3
Now
$density = \frac {mass}{volume}$
or $mass = density \times volume= 1.30 \times 2625=3412.5 kg$
Question 10. Mass of body 'A' is equal to mass of body 'B'. If the density of body 'A' is greater than the density of body 'B', which is easier to lift through air? Question 11. A force of 100N acts on a surface of area 25m2? Calculate thrust and pressure. Calculate the changed pressure if the force now is reduced to 25N? Answer
Thrust =100 N
Pressure = 100/25 =4 Pascal
New Pressure =25/25 =1 Pascal
Change in Pressure = -3
Long answer type
Question 12. Two objects of masses $m_1$ and $m_2$ having the same size are dropped simultaneously from heights $h_1$ and $h_2$ respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if
(i) one of the objects is hollow and the other one is solid
(ii) both of them are hollow, size remaining the same in each case. Give reason? Answer
For solid ball as u=0
$h_1 =\frac {1}{2}gt_1^2$
For hollow ball as u=0
$h_2 =\frac {1}{2}gt_2^2$
From above two, we can find
$\frac {t_1}{t_2} = \sqrt {\frac {h_1}{h_2}}$
Ratio will not change in either case because acceleration remains the same.
In case of free-fall acceleration does not depend upon mass and size.
Question 13. A piece of iron weighs 44.5gf in air, 39.5 gf in water and 40.3 gf in an oil. Find:
a.the relative density of iron.
b. the density of oil in SI unit. What assumption have you made in your calculation?
Given density of water=1 gm/cc = 1000 kg/cubic m Answer
Method -1
a. Let V be the volume of iron and $\rho _i$ be the density of Iron,$\rho _o$ be the density of oil
Upthrust in Water = 44.5 - 39.5 = 5 gf =.049 N
Now
$ V \times 1000 \times 9.8 = .049$
or $V= .000005 m^3 = 5 cm^3$
Now mass of Iron=44.5 gm
Density of Iron = 44.5/5 = 8.9 g/cc
Now relative density of the iron = Density of iron /Density of watter = 8.9/1 = 8.9
b. Now Upthrust in oil = 44.5 - 40.3= 3.2 gf= .03136 N
Now
$V \times \rho _o \times 9.8 = .03136$
or $ \rho _o= 640 kg/m^3$
Method-2
a. This can be calculated using the formula
$Relative \; density \; of \; solid = \frac {Weight \; of \; solid \; in \; air }{ Loss \; of \; weight \; of \; solid \; in \; water}$
$= \frac {44.5}{5} = 8.9$
b. This can be calculated using the formula
$Relative \; density \; of \; oil = \frac {Loss \; of \; weight \; of \; solid \; in \;oil}{Loss \; of \; weight \; of \; solid \; in \; water} = \frac {3.2}{5} = .64$
Density of Oil =.64 X 1000 = 640 kg/cubic mv
Question 14. A solid weighs 50 gf in air(where gf is gram force) and 44 gf when completely immersed in water. Calculate:
a.the upthrust
b.the volume of the solid
c. the relative density of the solid
Given Density of water=$1000 kg/m^3$ Answer
a. Upthrust = Weight in air - Weight in water = 50 -44 = 6 gf = .0588 N
b. Let V be the volume
$Upthrust= V \rho_w g = V \times 1000 \times 9.8 = 9800V$
Now
$9800V=.0588$
or $V= .000006 m^3 = 6 cm^3$
c. Mass of Solid = 50g
$Volume= 6 cm^3$
$Density =\frac {mass}{volume} = 8.33 gm/cm^3$
Now $\text {relative density of the solid} = \frac {\text{Density of Solid}}{\text{Density of water}} = \frac {8.33}{1} = 8.33$
Question 15. An ball is thrown vertically upwards and rises to a height of 10 m.
Calculate
(i) the velocity with which the object was thrown upwards
(ii) the time taken by the object to reach the highest point.? Answer
i.Here v=0,u=?, s=10 m ,a= -g= -9.8 m/s2
$v^2= u^2 + 2a s$
$0 = u^2 + 2 \times (-9.8 ) \times 10$v
u = 14 m/s
(ii) $v = u + a t$
$0 = 14 - 9.8 \times t$
t = 1.43 s.
Question 16. A block of wood of mass 10kg and dimensions $40 \; cm \times 25 \;cm \times 10 \;cm$ is placed on a table top. Find the pressure exerted if the block lies on the table top with sides of dimension:
a.$10 \; cm \times 25 \; cm$
b.$40 \; cm \times 10 \; cm$ Answer
Mass of the wooden block = m= 10 kg
Dimensions = $40 \; cm \times 25 \; cm \times 10 \; cm$
Thrust on the table top is due to the weight of the wooden block.Now,
Weight of the box = $F = mg= 10 kg \times 9.8 m/s^2= 98 N$
a. Area of a side = $length \times breadth
= 10 \times 25
= 250 cm^2
= 0.025 m^2$
$
Pressure = \frac {weight}{ area}= \frac {98}{0.025}= 3920 N/m^2$
b.
Area= $length \times breadth= 40 \times 20= 800= 0.08 m^2$
$
Pressure = \frac {weight}{ area}= \frac {98 }{0.08}= 1225 N/m^2$
Question 17. Find the ratio of the pressure exerted by a block of 400N when placed on a table top along its two different sides with dimensions $20 \; cm \times 10 \; cm$ and $10 \; cm \times 15 \; cm$. Answer
$P=\frac {F}{A}$
Now
$P_1= \frac {F}{A_1}$
and
$P_2= \frac {F}{A_2}$
Now
$P_1 : P_2 = A_2 : A_1 = 150 :200= 3:4$
Question 18. If two forces in the ratio 5:7 act on two areas in the ratio 4:5, Find the ratio of pressure exerted? Answer
Let two forces $F_1=5x,F_2=7x$
Let two areas $A_1=4y, A_2=5y$
$P=\frac {F}{A}$
Now
$P_1= \frac {F_1}{A_1}$
and
$P_2= \frac {F_2}{A_2}$
Now
$P_1 : P_2 = F_1 A_2 : F_2 A_1 = 25xy : 28 xy=25 :28$
Question 19. The mass of the earth is $6 \times 10^{24} kg$ and that of the moon is $7.4 \times 10^{22} kg$. If the distance between the earth and the moon is $3.84 \times 10^{5} km$,calculate the force exerted by the earth on the moon. $G = 6.7 \times 10^{-11} N m^2 kg^{-2}$.? Answer
The mass of the earth, M = $6 \times 10^{24} kg$
The mass of the moon,m = $7.4 \times 10^{22} kg$
The distance between the earth and the moon,d = $3.84 \times 10^{5} km$
$= 3.84 \times 10^5 \times 1000 m$
$= 3.84 \times 10^8 m$
$G = 6.7 \times 10^{-11} N m^2 kg^{-2}$
Now
$F= G \frac {M \times m}{d^2}$v
Substituting the values
$F=2.01 \times 10^{20} N$
Question 20. Acceleration due to gravity at Height H above earth surface is half of the value of Acceleration due to gravity at the surface. What is the value H in terms of R where R is radius of Earth Answer
$g= \frac {GM}{r^2}$
Now
$g_s=\frac {GM}{R^2}$
$g_h=\frac {GM}{(R+H)^2}$
Now
$g_h= \frac {1}{2}g_s$
$\frac {GM}{(R+H)^2} = \frac {1}{2} \times \frac {GM}{R^2}$
$(R^2 + H^2 + 2RH)=2R^2$
$H^2 +2RH -R^2=0$
or
$H= \frac {-2R \pm \sqrt {4R^2 +4R^2}}{2}$
or
$H=R(\sqrt {2} -1)$ or $-R( \sqrt {2} +1)$
Ignorning negative.
$H=R(\sqrt {2} -1)$
Question 21.
A ball is dropped from a height of 20 m. a second ball is thrown downwards from the same height after one second with initial velocity u. if both the balls reach the ground at the same time, calculate the initial velocity of the second ball. (Take $g = 10 \ ms^{-2}$) Question 22.
What happens to the gravitational force between two objects if:
(i)The mass of one object is doubled?
(ii)The distance between the objects is doubled?
(iii)The masses of both the objects are doubled?
(iv)The distance between them is halved?
(v)Mass of one of the objects is halved?
(vi)the distance between them is reduced to $\frac{1}{4}$th Answer
We know that
$F=\frac {Gm_1 m_2}{r^2}$
(i) 2F
(ii) F/4
(iii) 4F
(iv) 4F
(v) F/2
(vi) 16F
Question 23.
Find the gravitational force between the sun and the earth. The mass of the sun is $2.0 \times 10^{30}$ kg, and the mass of the earth is $6.0 \times 10^{24}$ kg. the distance between the sun and the earth is 1.5 \times 1011 m. ($G = 6.67 \times\ {10}^{-11}\ {Nm}^2\ {kg}^{-2}$). Question 24
A ball thrown vertically up returns to the thrower after 6 s. find:
(i) The velocity with which it was thrown up.
(ii) The maximum height it reaches.
(iii) Its position after 4 s. Question 25
At what height above the earth’s surface would the value of acceleration due to gravity be half of what is on the surface? Calculate. Question 26
Communication satellites move in orbits of radius 44,400 km around the earth. Find the acceleration of such a satellite assuming that the only force acting on it is that due to the earth. Mass of the earth = $6 \times\ {10}^{24}$ kg. (G = $6.67 \times\ {10}^{-11}\ {Nm}^2\ {kg}^{-2}$). Answer
Radius of the orbit of the satellite, r = 44400 km = 444 * 105
Mass of the earth, Me =$6 \times\ {10}^{24}$ kg
Universal gravitational constant, G = $6 \times\ {10}^{24}$ kg
Now, the acceleration of the satellite,
$a = \frac {G \times M_e }{ r^2}$
Substituting the above values
$a= 0.203 m/s^2$ .
Question 27
Find the weight of an object at the height 6400 km above the earth’s surface. The weight of the object at the surface of the earth is 20 N and the radius of the earth is 6400 km. Question 28
A fire cracker is fired and it rises to a height of 1000 m. find the
(i) velocity by which it was released.
(ii) time taken by it to reach the highest point (take $g = 10 \ ms^{-2}$) Question 29
A car falls off a ledge and drops to the ground in 0.5 s. let $g = 10 \ ms^{-2}$.
(i) What is its speed on striking the ground?
(ii) What is its average speed during the 0.5 s?
(iii) How high is the edge from the ground? Answer
Given,
time taken=0.5s
g=10m/s2
u=0 as free fall
(i)
u=0
a=g
From first law of motion
v=u+at
v=0+10*0.5
v=5m/s
(ii) Average speed = $\frac {u+v}{2}=\frac {0+5}{2}=2.5$ m/s
(iii) From newton's equation of motion, As u=0,
$h=ut+ \frac {1}{2}gt^2$
$h= \frac {1}{2}gt^2= .5 \times 10 \times (0.5)^2 =1.25$ m
Question 30
The value of 'g' on earth’s surface is $9.8 \ m/s^2$. Suppose the earth suddenly shrinks to one – third of its present size without losing any mass. What is the value of g on the surface of shrinked earth?
Summary
This gravitation class 9 numericals with solutions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.
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