Given below are the class 9 physics gravitation ,thrust,pressure,relative density numericals along with Archimedes principle numericals
a. Calculation problems
b. short answer question questions
d. Long answer questions
Question 1.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 2.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 3.
What do you understand by 1 Pascal and 1 Newton?
Answer
Pascal is the SI unit of Pressure and it is defined as 1 Newton force or thrust on $1 \; m^2$ of Surface area
Newton is the SI unit of Force. it is the amount of force the will accelerate a mass of 1kg to $1 \; m/s^2$.
Question 4.
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 5.
If a fresh egg is put into a beaker filled with water, it sinks. On dissolving a lot of salt in the water, the egg begins to rise and then floats. Why?
Answer
The salt solution has much more density then the plain water solution.Now the density of salt solution is more than egg,so egg begins to float
Question 6.
Define pressure? Give its mathematical expression and its SI unit?
Answer
The thrust on unit area is called pressure.
$Pressure = \frac {thrust}{Surface \; Area}$
Its SI unit is Pascal
Question 7.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer
Upward direction
Question 8.
Differentiate between density and relative density?
Answer
Density: Mass per unit volume of a substance is called density. SI unit is kg/m^{3}
$ density= \frac {mass}{volume}$
Relative Density: 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
$Relative \; density = \frac {density \; of \; the object}{density \; of \; water}$
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 10.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) bothof 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 11.How do snow shoes stop you from sinking into snow?
Answer
Snow shoes are very large and have large surface area.Since the surface area is large than the pressure exerted by the shoes will be less on ground. So due to this person will not sink into snow.
Question 12.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/s^{2}
$v^2= u^2 + 2a s$
$0 = u^2 + 2 \times (-9.8 ) \times 10$v
u = 14 m/s
(ii) $v = u + a t$
$0 = 14 - 9.8 \times t$
t = 1.43 s.
Question 13.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 14.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 15.Define relative density. Relative density of mercury is 13.6. The density of water is 1000 kg/m
^{3}. 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 16.You are provide with a hollow iron ball of volume 20cm
^{3}, 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 17.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/m
^{3})?
Answer
Volume of the dining room= $50 \;m \times 15 \; m \times 3.5 \; m= 2625 m^3$
Now Density of air= 1.30kg/m^{3}
Now
$density = \frac {mass}{volume}$
or $mass = density \times volume= 1.30 \times 2625=3412.5 kg$
Question 18.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 19.A force of 100N acts on a surface of area 25m
^{2}? 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
Question 20.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 21.Define thrust? Name the factors on which the pressure depends. If a man standing on a sand surface lies down on the sand, how will the pressure exerted on sand by the man be affected in two positions?
Answer
The force acting on an object perpendicular to the surface is called thrust
Pressure depends on thrust and contact surface area. When the person is standing, the surface area is less hence more pressure. When the person is lying the surface area is more hence less pressure.
Question 22.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 23.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 24.A force of 250N acts on a surface of area $15cm^2$. Calculate thrust and pressure?
Answer
Thrust= Force=250 N
$Pressure = \frac {Thrust}{area} = \frac {250}{15 \times 10^{-4}} = 166666.7 Pa$
Question 25.A force of 400N acts on a surface of area 25cm
^{2}. 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 26. 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.
$H=R(\sqrt {2} -1)$
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