Selina Concise Class 9 Physics Solution Chapter No. 5- ‘Upthrust in Fluids Archimedes Principle and Floatation’ For ICSE Board Students.
Exercise: A Solution
1.) Solution:
Ans:
Upthrust:
When a body is immersed or submerged completely or partially in a fluid then the upward force exerted on body by the fluid is called as the upthrust.
The following experiment describes the existence of upward thrust.
- We have taken a can whose open end is tighten with airtight stopper. When we immersed that can into the tub of water it floats on the water by deeping some part in the water.
- And if we pushed the can forcefully into the water we feel an upward force on the can, due to which it is difficult to push the can further into the water.
- And also it is observed that, more and more force will be required to push the can inside the water yet it may immersed completely in water.
- When the can is immersed completely inside the water, there will be the constant upward force acting on the can which makes it stationary. And if we again released the can it bounce back to the surface and again starts floating on the surface of water.
- Thus, this experiment proves there is existence of upward thrust by the water on the can.
2.) Solution:
Ans:
The buoyant force on a body due to liquid acts in upward direction and at the centre of buoyancy.
3.) Solution:
Ans:
Buoyancy is the property of liquid to exert an upward force on the body immersed in that liquid.
4.) Solution:
Ans:
When the body is immersed or submerged completely or partially in a fluid then the upward force exerted on the body by the fluid is called as the upward thrust.
The SI unit of upward thrust is N or kgf.
6.) Solution:
Ans:
We know that, when we immersed any body inside the liquid or fluid we feel an upward force which is the buoyancy force or upthrust. And this is because the body in order to immerse in the liquid displaces some amount of liquid and that amount of liquid apply force on it.
For example:
When a wooden block is immersed in water, initially it floats slowly as it’s density is less than the density of water but when we press it inside the water we feel some force from inside the water is opposing to us. And if we press more and more we feel the more and more upward pull and this is the force exerted by the water on the wooden block which is called as upthrust.
If we totally immersed the wooden block inside the water then the force required to immerse it is very large as upward thrust is also very large. And again if we have removed the force applying on the block it may comes to surface and again floats on the water.
7.) Solution:
Ans:
The floating and sinking of the objects in liquid or fluid depends on the density of the objects and the liquid or fluid.
When we immersed the wooden block in the water it comes to surface of water and floats. Because here the density of water is more than the density of wooden block. And due to which the upward thrust acting on the wooden block by the water is more which take it to surface and hence the wooden block floats on the surface of water.
8.) Solution:
Ans:
- When a body is immersed or submerged in a liquid or fluid the displaced fluid or liquid exerts the upward force on the body and that force is called as upward thrust. And due to this upward thrust the immersed body feels less weight than its actual weight.
- The following is the experiment which demonstrates the body immersed in a liquid feels light in weight than really is.
- We have taken an empty bucket to which a long rope is tied and we have immersed that empty bucket in the water of well and one end is in our hand.
- Now we are taking water from well to the empty bucket, here we observe that a bucket full of water we feels light in weight when it is totally inside the water but when it comes out from water slowly we required more force to lift that bucket.
- And from this experiment we can say that, the bucket full of water feels light in weight when it is inside the water than it is in air.
- Similar things happens when fisherman lift the fish from the water.
9.) Solution:
Ans:
We know that, when a body is immersed in the liquid, some of the liquid will be displaced which exerts an upward pull on the body called as upward thrust. And hence we feels the less weight of body inside the water and when we take out it then we feel more weight of body.
Similar things happens when a body is weighed by spring balance in air. In air there is some upward thrust acting on the body due to which its weight will be less but when we weighs same body in vacuum its weight will be more because there will be no upthrust acting on the body.
10.) Solution:
Ans:
- When a metal solid cylinder is tied to thread is hanging from the hook of the spring balance.
- When we gradually immersed the cylinder in in a jar full of water then we observe the following things.
- When the metal solid cylinder is in air it weighs more and hence spring balance shows more reading. But when we immersed it in the water contained in a jar, due to upward thrust by water on it we feel it light in weight and hence the reading shown by spring balance now will be less.
- When we taken out the cylinder then we again feel more weight and the spring balance shows the reading which was showed earlier.
- In this way, the weight of cylinder is more when it is in the air and its weight will be less in water. Because the upward thrust exerted by water is more than the upward thrust due to air. More the upward thrust we feel the less weight of the body.
11.) Solution:
Ans:
When a body is immersed in the liquid the liquid will be displaced the amount of liquid displaced will exerts an upward force on the body which is called as the upthrust.
The upthrust on the body mainly depends on the following factors:
- Volume of the body which is immersed in the liquid or fluid
- And the density of the liquid or fluid in which the body is immersed.
12.) Solution:
Ans:
The upthrust acting on the body of volume V submerged in a liquid of density q is given by,
Upthrust = V*q*g
Where, V- is the volume of body immersed
q- is the density of liquid in which body is immersed
g- is the acceleration due to gravity
Thus,we can say that the more is the volume of the body immersed more will be the upthrust acting on the body. Because, the body having more volume when immersed in the liquid it displaces more liquid and hence the upthrust acting on the body is more.
13.) Solution:
Ans:
Given that, the bunch of feathers and a stone of same mass are released simultaneously in air, then the stone will comes fast because the friction due to air acting on the feathers makes it slower, although their mass is same and hence acceleration due to gravity is acting is also same.
While when both stone and bunch of feathers are released in vacuum there will be no friction and nothing and hence both will comes to the ground at the same time although the acceleration due to gravity acting on them is same.
14.) Solution:
Ans:
- We know that, the upthrust exerted by the liquid of density q on the body of volume V immersed in it is given by,
- Upthrust = V*q*g
Where, V is the volume of body immersed in liquid
q- is the density of liquid in which the body is immersed
g- is the acceleration due to gravity
- Thus, more the density of the liquid more will be the thrust acting on it.
- As we know that, the sea water contains more salts it’s density is more than the river water.
- Hence, the upthrust exerted by sea water is more than the river water.
- Hence, here F2 is more than the F1.
- Because, sea water is more denser than the river water.
15.) Solution:
Ans:
- We know that, the upthrust acting on body depends on the volume of body, density of liquid and acceleration due to gravity.
- And as the density of glycerine is more than that of water hence the upthrust exerted by glycerine is more on the wooden block immersed in it.
- Hence, the wooden block floats easily on glycerine with less volume than the wooden block in water.
16.) Solution :
Ans:
Given that, V is the volume of the body and q is the density of body.
And qL is the density of liquid and g is the acceleration due to gravity.
a) then the weight of the body is the force acting on the body due to acceleration due to gravity.
Hence, weight of the body = V*q*g
b) the upthrust acting on the body depends on the volume of the body, density of the liquid and acceleration due to gravity.
Hence, upthrust is given by
Upthrust= V*qL*g
c) the apparent weight of the body in liquid is the difference between the weight of body in air and the weight of body in liquid.
Since, due to the upthrust the weight of body in liquid is less than in air.
Hence, apparent weight of body in liquid is given by,
Apparent weight= V*q*g – V*qL*g = V*g*(q – qL)
d) the loss in weight of the body is nothing but the upthrust acting on it.
Loss in weight of body= V*qL*g
17) Solution:
Ans:
- The following figure shows the direction of forces acting on the body held completely immersed in the liquid.
- The force F1 is the force due to acceleration due to gravity
- And the force F2 is the upthrust exerted by liquid on the body.
- It is observed that, when the force F1 is less than or equal to the force F2, the body will floats easily on the liquid.
- But, when the force F1 is more than the force F2, the body will sinks in the liquid.
18.) Solution:
Ans:
a) Two balls, one of iron and other of aluminium experiences the same upthrust when dipped completely in water if both have equal volumes.
b) An empty tin container with its mouth closed has an average density equal to that of liquid. The container taken is 2m below the surface of the liquid and is left there. Then the container will remain at the same position.
c) A piece of wood is held under water. The upthrust on it will be more than the weight of the wood piece.
19.) Solution:
Ans:
- We know that, when a body is immersed in the liquid, the liquid displaced by the body will exert the force on the body which is called as upthrust.
- And the upthrust is exactly equal to the weight of the liquid displaced by the body submerged in it.
We can prove this mathematically as follows.
- Let us consider the cylindrical shaped body ABCD as shown in figure whose area of cross section is A and which is immersed in the liquid of density q.
- Let the upper surface AB is at depth h1 and the lower surface of the body CD is at depth h2, both which are below the free surface of liquid.
- At depth h1, the pressure exerted on the surface AB is given by,
P1= h1*q*g
- And hence, the downward thrust on the upper surface AB will be,
F1= pressure*area= h1*q*g*A
- Similarly, the pressure exerted on the lower surface CD is given by,
P2= h2*q*g
- And hence, the upward thrust on the lower surface CD is given by,
F2= h2*q*g*A
- From above equations, we conclude that F1<F2 because h1<h2, due to change in pressure the body will exerts the net upward force which is called as upthrust or buoyant force and it is given by,
F= F2 – F1
F= (h2 – h1)*q*g*A
- But, (h2 – h1)*A = V which is the volume of the body immersed in the liquid.
Thus, F= V*q*g
- Upthrust= volume of body* density of liquid* acceleration due to gravity
- As the body immersed displaces the liquid equal to it’s volume, we can write it as,
- Upthrust= volume of liquid displaced*Density of liquid*acceleration due to gravity
Upthrust= mass of liquid displaced*acceleration due to gravity
Upthrust= weight of the liquid displaced
- Thus, we have proved here that the upthrust is equal to the weight of the liquid displaced when the body is immersed in it.
20.) Solution:
Ans:
- Given that, sphere of iron and another wood are having same radius hence their volume will be same.
- And we know that, the upthrust acting on the body depends on the volume of the body, acceleration due to gravity and density of the liquid.
- Here, both the spheres are immersed in the same liquid i.e. water and hence density of liquid will be same. Acceleration due to gravity g is remains same everywhere.
- So, we can say that the upthrust= V*q*g will be the same for both iron and wooden sphere immersed in the water having same radius.
- Thus, their ration of upthrust will be 1:1.
21.) Solution:
Ans:
We know that, the density of iron sphere is more than density of water. Hence, the weight of the iron sphere will be more which sinks inside the water.
While the density of wooden sphere is less than the water. And also the upthrust acting on wooden sphere will be balanced by the weight of the wooden sphere and hence it floats on the surface of water.
22.) Solution:
Ans:
- We know that, when we immersed the body in a liquid then some amount of liquid will be displaced and which exerts upward force called as upthrust.
- Now, if the upthrust is balanced by the weight of the body immersed then that body floats on the liquid.
- But, if upthrust is less than the downward force of the body then that body sinks in the liquid.
- In short, if the density of body is more than the liquid in which it is immersed then that body sinks in the liquid.
- For example:
- An iron balls immersed in water directly sink in the water.
- If the density of body is less than the density of liquid then that body floats on the liquid.
- For example:
- The wooden block floats on the water.
23.) Solution:
Ans:
- Given that, the density of body is q and the density of liquid in which the body is immersed is qL.
- We know that, when the density of body is less than or equal to the density of liquid then that body floats on the liquid.
- Hence, when q< or = qL, then body will floats on the liquid.
- Also, when the density of body is greater than the density of liquid in which it is immersed then that body sinks in the liquid.
- Hence, when qL< q, then the body will sinks in the liquid.
24.) Solution:
Ans:
As we already know that, when the body is immersed in the liquid it displaces some liquid and hence the liquid exerts some upward force on it which is called as upthrust.
When the whole body is in the water then the weight of body and the upthrust is balanced and the apparent weight of the body will be less. Hence we feel less weight of the body in liquid than air.
25.) Solution:
Ans:
Archimedes’s principle:
According to Archimedes’s principle, when we immerse the body partially or completely inside the liquid or fluid then that fluid or liquid exerts the upward force which we called as upthrust. And this upthrust is equal to the weight of the liquid which displaced by the body when immersed in the liquid or fluid.
26.) Solution:
Ans:
Archimedes’s principle:
According to Archimedes’s principle, when we immerse the body partially or completely inside the liquid or fluid then that fluid or liquid exerts the upward force which we called as upthrust. And this upthrust is equal to the weight of the liquid which displaced by the body when immersed in the liquid or fluid.
The following experiment verifies the Archimedes’s principle:
- We have taken two cylinders A and B of the same volume, cylinder A is the solid cylinder while cylinder B is the hollow cylinder.
- Now we have suspended this two cylinders in the left arm of the balance as shown in figure in a such way that, the hollow cylinder A will be in upper position and below which there is solid cylinder B.
- Now, we balance the beam by keeping some weights on the right arm and in this case both cylinders are in air.
- Now, we keep solid cylinder in beaker full of water such that it doesn’t touches the walls of beaker. And due to which the solid cylinder B loses its weight because there will be some upthrust acting on the solid cylinder B.
- Now when we balance the beam then we observe that, there will some weight loss in left arm of the beam balance.
- Now, we pour water into the hollow cylinder A till the beam will balanced to the weight when both cylinders are in air.
- After adding some water the beam balance again get balanced.
- This shows that, when the solid cylinder B is immersed in water there will be upthrust due to which the solid cylinder weighs less and the amount of water displaced by the solid cylinder when we add to hollow cylinder B the total beam balance is balanced.
- Hence, this experiment verifies Archimedes’s principle which states that the upthrust acting on the body is equal to the weight of the liquid displaced by that body.
: Multiple choice type:
1.) A body will experience minimum upthrust when it is completely immersed in
Ans: a) turpentine
2.) The SI unit of upthrust is
Ans: b) N
3.) A body of density q sinks in a liquid of density qL. The densities q and qL are related as
Ans: c) qL<q
Numerical:
1.) Solution:
Ans:
Given that,
Volume of the body = 100cm3= 100*10-6m3= 10-4m3
Density of liquid= 1.8*103kg/m3
Weight of body in air = 5kgf
Upthrust due to liquid on the body is given by,
Upthrust= volume of body*density of liquid*g
Upthrust= 10-4*1.8*103*g
= 0.18*g N
= 0.18kgf
Thus, upthrust due to liquid is given by 0.18 kgf.
Weight of the body in liquid is given by,
Weight of the body in liquid= Weight of the body in air – upthrust
= 5kgf – 0.18kgf
= 4.82kgf
Thus, the weight of the body in liquid is 4.82kgf.
Thus, here it is conclude that the weight of body in air is greater than the weight of body in liquid.
2.) Solution:
Ans:
Weight of the body in air = 450gf
Weight of the body when immersed in water= 310gf
a) The volume of the body is given by,
Volume of body= loss in weigh of body* density of water
Volume of body= (450-310)*1
Since, density of water is 1gm/cm3
Volume of body= 140cm3
b) The loss in weight of the body is given by,
Loss in weight = weight of body in air – weight of body in water
Loss in weight = 450 – 310
Loss in weight = 140gf
c) The upthrust on the body is given by,
Upthrust= loss in weight of the body = 140gf
3.) Solution:
Ans:
Volume of hollow iron ball A = 15cm3
Mass of hollow iron ball A = 12g
Mass of solid iron ball B= 12g
Density of iron = 8g/cm3
a) The upthrust on hollow iron ball A is given by,
Upthrust on ball A = volume of ball A*density of water*g
= 15*1*g
=15gf
Upthrust on ball B = volume of ball B* density of water*g
Since, volume of ball B= mass/density= 12/8= 1.5cm3
Upthrust on ball B =1.5*1*g= 1.5gf
b) Here, the upthrust on ball B is 1.5gf which is less than its weight and hence it sinks in the water.
While the upthrust on the ball A is 15gf which is greater than its weight 12g and hence when it is fully immersed in water it floats on the water.
4.) Solution:
Ans:
Given that,
Density of solid = 5000 kg/m3
Weight of solid in air = 0.5kgf
The apparent weight of the solid in water is given by,
Apparent weight= true weight – upthrust
The upthrust acting on the solid is given by,
Upthrust= volume of body*density of water*g
= (0.5/5000*g)*1000*g
= 500/5000= 1/10= 0.1kgf
Thus, apparent weight= real weight – upthrust
= 0.5 – 0.1= 0.4kgf
Thus, the apparent weight of the solid in water is 0.4kgf.
5.) Solution:
Ans:
Given that,
Volume of sphere A = 100cm3= 10-4m3
Density of sphere A = 0.3g/cm3
Volume of sphere B = 100cm3 = 10-4m3
Density of sphere B= 8.9g/cm3
Density of water =1g/cm3
Weight of sphere A= mass*g
= Volume of sphere A*density of sphere A*g= 100*0.3*g= 30gf
Weight of sphere A = 30gf
Similarly,
Weight of sphere B= 100*8.9*g= 890gf
Weight of sphere B= 890gf
Now, upthrust on sphere A is given by,
Upthrust on A = volume of A*density of water*g
= 100*1*g= 100gf
Upthrust on sphere A = 100gf
Similarly, upthrust on sphere B is given by,
Upthrust on B= 100*1*g= 100gf
Upthrust on sphere B is 100gf.
Hence, here we conclude that the upthrust acting on both the spheres is same as their volumes are also same.
As we know that, the density of iron is more than the density of water. Hence the iron sphere B sinks in water.
While the density of wood is less than the density of water hence the wood sphere A will floats on the water.
6.) Solution:
Ans:
Given that,
Mass of block = 13.5kg
Volume of block = 15*10-3m3
Density of water =1000kg/m3
- a) The upthrust on the block is given by,
Upthrust= volume of block*density of water*g
= 15*10-3*1000*g= 15kgf
b) As the upthrust on the block is 15kgf which is more than the weight of the block. Hence when block is completely immersed in water it will floats on the water.
c)
When the body floats on the water, then its weight is equal to the upthrust on the body.
Upthrust on block while floating = weight of the block= 13.5kgf
7.) Solution:
Ans:
Given that,
Weight of piece of brass in air= 175gf
Weight of brass in water = 150gf
Density of water=1g/cm3
The volume of brass piece is given by,
Since, volume of brass piece= weight loss= (175-150)= 25cm3
As we know that, when body immersed in water there is upthrust acting on the body due to which that body weighs less in water. And hence here the piece of brass also weighs less when immersed in water.
8.) Solution:
Ans:
Given that,
Edge of cube= 5cm
Volume of cube=5*5*5= 125cm3
Density of cube =9g/cm3
Mass of cube = volume of cube*density of cube= 125*9= 1125g
Thus, weight of the cube acting downwards is = 1125gf
Now, upthrust on cube is given by,
Upthrust = volume of cube* density of liquid*g
= 125*1.2*g= 150gf
Thus, tension in thread in downward direction is the net force acting in downward direction.
Tension in the thread = weight – upthrust
= 1125 – 150= 975gf = 9.75N
Thus, the tension in thread is 9.75N
9.) Solution:
Ans:
Given that,
Volume of block = 50*50*50cm3= 125000cm3= 0.125m3
g= 9.8N/kg
The buoyant force is nothing but the upthrust acting in the block and it is given by,
Upthrust = volume of block* density of water*g
= 0.125*1000*9.8
= 125*9.8 = 1225N
Thus, the buoyant force acting on the block is 1225N.
10.) Solution:
Ans:
Given that,
Mass of body = 3.5kg
Volume of water displaced by body = 1000cm3
Weight of body = 3.5kgf
As we know that,
According Archemedes principle, volume of body is equal to the volume of liquid displaced by that body.
Volume of body = volume of water displaced by body when immersed in water = 1000cm3
Thus, volume of body = 1000cm3
The upthrust on the body is given by,
Upthrust= volume of body *density of water*g
= 0.001*1000*g= 1kgf
Thus, upthrust on the body will be 1kgf.
The apparent weight of the body is given by,
Apparent weight= real weight – upthrust = 3.5 – 1= 2.5kgf
Thus, apparent weight of the body is 2.5 kgf.
Exercise-B Solution:
1.) Solution:
Ans:
- The density of any substance is defined as it is the mass of substance per unit volume.
- Hence, density of substance is given by
Density= mass of substance/volume of substance
- The SI unit of density of substance is kg/m3.
- For given substance the density remains constant.
- For example: The density of water is 1000kg/m3.
2.) Solution:
Ans:
- The CGS unit of density is g/cm3.
- The SI unit of density is kg/m3.
3.) Solution:
Ans:
The relationship between CGS and SI unit of density is given by,
1g/cm3 = 1000kg/m3
4.) Solution:
Ans:
The density of iron is 7800kg/m3.
As we know that, density is the mass of a substance per unit volume.
Hence, the given statement indicates the mass of iron per m3 is 7800kg.
5.) Solution:
Ans:
The density of water at 4°C is 1000kg/m3 in SI unit.
6.) Solution:
Ans:
- As we know that, there is no effect of temperature on mass of the substance.
- Hence, mass of the substance remains same at any temperature.
- We know that, most of the substances expands on heating that means if the temperature is increased the volume of the substances increases.
- And density is the mass per unit volume.
- Hence, as temperature increases the volume also increases and hence density of the substance decreases.
7.) Solution:
Ans:
On heating water from 0°C to 4°C the density of water increases and at 4°C it is 1000kg/m3.
Then after heating water above 4°C again density of water decreases.
8.) Solution:
Ans:
a) Mass= Volume*density
b) SI unit of density is kg/m3.
c) Density of water is 1000kg/m3.
d) Density in kg/m3 = 1000*density in g/cm3
9.) Solution:
Ans:
- Relative density of the substance is defined as it is the ratio of the density of that substance and the density of water at 4°C.
- Relative density can also be defined as it is the ratio of mass of specific volume of a substance to the mass of an equal volume of water at 4°C.
- It is the scalar quantity and has no unit.
10.) Solution:
Ans:
- As relative density is the ratio of two same quantities and hence it has no unit.
- It is the scalar quantity.
11.) Solution:
Ans:
Density:
- The density of the substance is defined as it is the ratio of mass of substance per unit volume.
- It is the scalar quantity.
- The SI unit of density is kg/m3.
Relative density:
- Relative density is defined as it is the ratio of density of that substance and the density of water at 4°C.
- It is the scalar quantity.
- As it is the ratio of two similar quantities, it has no unit.
12.) Solution:
Ans:
- We know that, the relative density of substance is given by,
- Relative density= mass of body/ mass of the water of volume equal to that of the body
- According to Archimedes’ principle, the mass of water of volume equal to that of body is equal to the mass of water displaced by the body.
- Hence, we can write,
Relative density= mass of body/mass of water displaced by the body
= Weight of the body/weight of the water displaced by the body
= Weight of body/ loss in weight of the body in water
Thus,
- Relative density= weight of body in air/ (weight of body in air – weight of body in water)
- Let, W1 be the weight of body in air and W2 be the weight of the body in water.
- Hence, relative density is given by,
R.D.= W1/(W1 – W2)
a) To find the relative density of solid denser than water and insoluble in water.
- Now, we have suspended the piece of solid from the left hook of physical balance and we measured its weight as W1.
- Now, we have placed the wooden bridge on the left of pan balance and a beaker is placed on the bridge which is two-third filled with water as shown in following figure.
- Now, we have immersed the solid in beaker in such way that it does not touches the walls and bottom of the beaker.
- Let, W2 be the weight of the solid we have measured in water.
- As weight of solid in air = W1 gf
- And weight of solid in water = W2 gf
- Thus, the loss in weight of solid when we immersed it in water is given by,
Loss in weight = (W1 – W2) gf
- Hence, relative density is given by,
R.D.= W1/(W1 – W2)
- Now, to find the relative density of solid which is denser than water and soluble in it we modify some steps.
- If the solid is soluble in water then we take another liquid of known relative density in which that solid must be insoluble and the solid should be sink in that liquid.
- And we repeat the above process again.
- Then the relative density of solid denser than water and soluble in water is given by,
- D.=( weight of solid in air/ loss in weight of solid in liquid)* R.D. of liquid
- In this way, we can modify the experiment to find the relative density of solid denser than water and soluble in water.
13.) Solution:
Ans:
Given that,
Wgf is the weight of body in air.
W1 gf is the weight of body when immersed in water
Then, volume of the body is given by,
- Volume of body= (W – W1) cm3
- Upthrust on the body= (W – W1) gf
- Relative density of material of body = W/(W – W1)
14.) Solution:
Ans:
We know that, the relative density of liquid is given by,
- D.= weight of the given volume of the liquid/ weight of the same volume of water
- But, according to Archimedes’ principle , when a body is immersed in water or liquid it displaces the liquid equal to it’s volume. Hence, we can write as
R.D.= weight of liquid displaced by body/ weight of water displaced by the same body
R.D.= ( weight of body in air – weight of body in liquid)/ weight body in air – weight of body in water)
- Thus, to find the relative density of liquid by using the Archimedes’ principle, we take a body which is heavier than both the liquid and water and that body must be insoluble in both water and liquid.
- Initially we measure the weight of body in air, then we measure it’s weight in liquid and then we wash it with water and we keep it for drying and then we measure it’s weight in water.
- Let us suppose, W1 gf be the weight of body in air, W2 gf be the weight of body in liquid and W3 gf be the weight of body in water.
- Then , the relative density of liquid is given by,
R.D. = (W1 – W2)/ (W1 – W3)
15.) Solution:
Ans:
Given that,
W1 gf be the weight of body in air
W2 gf be the weight of body in liquid
W3 gf be the weight of body in water.
Then, volume of the body is given by,
- Volume of body= (W1 – W3) cm3
Upthrust due to liquid is given by,
- Upthrust = (W1 – W2) gf
The relative density of solid is given by,
- D. of solid = W1/(W1 – W3)
The relative density of liquid is given by,
- D. of liquid= (W1 – W2)/ (W1 – W3)
Multiple choice type:
1.) Relative density of substance is expressed by comparing the density of that substance with the density of
Ans: c) water
2.) The unit of relative density is
Ans: d) no unit
3.) The density of water is
Ans: c) 1g/cm3
Numerical:
1.) Solution:
Ans:
Given that,
Density of copper is 8.83 g/cm3
8.83 g/cm3 = 8.83*10-3kg/10-6m3
Thus, 8.83g/cm3 = 8.83*103kg/m3
Hence, 8.83g/cm3 = 8830 kg/m3
2.) Solution:
Ans:
Given that,
Relative density of mercury is 13.6.
In CGS unit, it is 13.6 g/cm3.
In SI unit, it is 13.6*103 kg/m3.
3.) Solution:
Ans:
Given that,
Density of iron is 7.8*103kg/m3.
Hence, its relative density is given by,
R.D.= density of iron/ density of water= 7.8*103/ 103= 7.8
Thus, the relative density of iron was found to be 7.8
4.) Solution:
Ans:
Given that,
Relative density of silver is 10.8
We know that, the relative density is given by
R.D.= density of silver/ density of water
Density of silver = R.D.* density of water
Density of silver= 10.8*103 kg/m3
5.) Solution:
Ans:
Given that,
Relative density of body = 0.52
Hence, density of body = 0.52*103 kg/m3
Volume of body = 2m3
We know that, density is nothing but the mass per unit volume.
Density= mass/ volume
Hence, mass = density* volume = 0.52*103*2= 1.04*103 kg = 1040kg
Thus, the mass of body is found to be 1040kg.
6.) Solution:
Ans:
Given that,
Volume of room means the volume of air.
Volume of air= 4.5*3.5*2.5 m3
Density of air at NTP= 1.3kg/m3
And we know that,
Density= mass/ volume
Hence, mass of air = density of air*volume of air
Mass of air= 1.3*4.5*3.5*2.5= 51.18 kg
Thus, the mass of air in a given volume is 51.18 kg
7.) Solution:
Ans:
Given that,
Mass of stone = 113g
Water level rises from 30ml to 40ml.
Thus, rise in water level = 10ml
When the stone is immersed in water it’s level rises by 10ml, which means the volume of stone will be 10cm3.
Thus, density of stone is given by,
Density= mass/volume = 113/10= 11.3 g/cm3
Hence, the relative density of stone is 11.3
8.) Solution:
Ans:
Given that,
Volume of body = 100cm3
Weight of body in air = 1kgf =1000 gf
Mass of body = 1kg = 1000g
We know that,
Relative density of water= 1
And relative density of body = 10
We know that, relative density of solid is given by,
R.D. of body= W1/(W1 – W2)* R.D. of water
Where W2 is the weight of body in water.
Thus, 10= 1000/(1000 – W2)*1
1000 – W2 = 100
W2 = 900 gf
Thus, weight of body in water is 900 gf.
Relative density of body is given by,
R.D.= 1000/100=10
9.) Solution:
Ans:
Given that,
Mass of body = 70kg
Body displaces water when immersed= 20,000cm3= 0.02m3
We know that, according to Archimedes’s principle the mass of body is equal to the amount of water displaced by that body.
Hence, the mass of solid immersed in water= density of water*volume of water displaced
= 1000*0.02= 20kg
Thus, weight of body in water 20kgf.
Relative density of body is given by,
density = 70*1000/20,000= 3.5 g/cm3
Hence, R.D.= 3.5
10.) Solution:
Ans:
Given that,
Weight of solid in air = 120gf
Weight of solid when completely immersed in water= 105 gf
Then, relative density of solid is given by,
R.D. of solid= W1/(W1 – W2)*relative density of water
R.D. of solid= 120/(120 – 105)*1 = 8
Thus, relative density of solid is found to be 8.
11.) Solution:
Ans:
Given that,
Weight of solid in air= 32 gf
Weight of solid in water= 28.8 gf
Density of liquid= 0.9 g/cm3
The volume of solid is given by,
Volume= 32/10= 3.2 cm3
The relative density of solid is given by,
R.D. of solid= W1/(W1 – W2)*relative density of water
R.D. of solid= 32/(32 – 28.8)*2 = 10
Thus, relative density of solid is 10.
Let W3 be the weight of solid in liquid of density 0.9g/cm3.
Then,
R.D. of solid= W1/(W1 – W3)*relative density of liquid
10= 32/(32 – W3)*0.9
10= 28.8/(32-W3)
32-W3 = 2.88
W3= 32 – 2.88= 29.12 gf
12.) Solution:
Ans:
Given that,
Weight of body in air = 20 gf
Weight of body in water= 18gf
Then relative density of body is given by,
R.D. of body= W1/(W1 – W3)* relative density of water
R.D. of body= 20/(20-18)*1= 10
Thus, the relative density of body is found to be 10.
13.) Solution:
Ans:
Given that,
Weight of solid in air= 1.5 kgf
Weight of solid in liquid= 0.9 kgf
Density of liquid= 1.2*103 kg/m3
The relative density of solid is given by,
R.D. of solid= 1.5/(1.5 – 0.9)*(1.2*103/1*103)
R.D. of solid= 1.8/0.6= 3
Thus, the relative density of solid is found to be 3.
14.) Solution:
Ans:
Given that,
Relative density of pure gold is= 19.3
Weight of bangles in air= 25.25 gf
Weight of bangles in water= 25.075 gf
We find relative density of bangles and compared it with relative density of pure gold to check whether bangles are made of pure gold or not.
Now,
R.D. of bangles= W1/(W1 – W2)*relative density of water
R.D. of bangles= 25.25/(25.25 – 25.075)*1
R.D. of bangles= 11.6
But, the relative density of pure gold is 19.3.
Hence, we can say that bangles are not made from pure gold as it’s density is less than density of pure gold.
15.) Solution:
Ans:
Given that,
Weight of iron in air = 44.5 gf
Density of iron = 8900 kg/m3
Weight of iron in water can be determined as,
Weight of iron in water= 44.5*(1- 1000/8900)
= 44.5*(1 – 0.1123)
= 44.5*0.8877
= 39.50 kgf
Thus, weight of iron piece when immersed in water is 39.50 kgf
16.) Solution:
Ans:
Given that,
Mass of stone in air= 15.1g
Weight of stone when immersed in liquid= 10.9 gf
Weight of stone when immersed in water= 9.7 gf
a)
Weight of the piece of stone in air= 15.1 gf
b)
When stone is immersed in water there will be upthrust acting on it which is given by,
Upthrust= 15.1 – 9.7= 5.4 gf
Thus, mass of stone in water is 5.4 g.
Volume of piece of stone = mass/density= 5.4/1 = 5.4 cm3
Thus, volume of stone is 5.4 cm3
c)
Relative density of stone is given by,
R.D. of stone = 15.1/(15.1 – 9.7)= 15.1/5.4= 2.8
Thus, the relative density of stone is 2.8
d)
The relative density of liquid is given by,
R.D. of liquid= (15.1 – 10.9)/(15.1 – 9.7) = 4.2/5.4 = 0.78
Thus, relative density of liquid is 0.78
Exercise- C Solution
1.) Solution:
Ans:
We already know that, when the body is submerged in a liquid or fluid there will be an upward force acting on it at the same time the weight of the body acts vertically downwards. On the basis of these two forces principle of floatation is based which is explained as follows:
1) We know that, every body has specific mass on which acceleration due to gravity is acting due to which it feels weight. And there will be also a centre of gravity G for each body through which whole weight of the body acts downwards. And this force is responsible for sinking of the body in liquid.
2) When the body is submerged in liquid the liquid displaced, and that displaced liquid also has center of buoyancy B through which the upward force is acting called as upthrust. And this upthrust is always equal to the weight of the liquid displaced. And this force is responsible for the Floatation of the body on liquid.
2.) Solution:
Ans:
a)
We know that, according to principle of floatation when body is immersed in liquid there will be two forces acting on the body. One is the weight of the body acting vertically downwards and other is the upthrust which acts upwards on the body.
The following figure shows these two forces acting on the body immersed in water.
b)
We know that, the weight of body acting vertically downwards which is responsible for sinking of body in liquid.
And the upthrust acting on the body is responsible for the Floatation of the body in liquid.
Thus, if the weight acting downwards is more than upthrust then body will sinks in the liquid.
And if the upthrust acting in upward direction is more than weight acting downwards the body will floats on the liquid.
c)
When the body sinks in liquid then the net force acting on body will be it’s weight in downward direction.
When the body floats on the liquid then the net force acting on body will be upthrust in upward direction.
3.) Solution:
Ans:
- Given that, a piece of wood is suspended from the hook of spring balance then it shows reading 70 gf.
- When we lowered the wood into the water it obviously floats on water. Because density of wood is less than the density of water.
- And we know that, wood is floating on water means the weight acting downwards and upthrust acting upward both balances each other.
- Hence, here the apparent weight will be zero.
- Thus, the reading shown by spring balance is zero.
4.) Solution:
Ans:
a)
Here, the iron ball floats in mercury contained in beaker.
Because, the density of iron ball is less than the density of mercury.
b) As we already know that, when body floats on liquid it’s weight and upthrust are balances with each other.
Hence, the apparent weight of the iron ball here is also zero.
6.) Solution:
Ans:
As the density of iron nails is less than the density of mercury, the iron nails will floats on mercury.
While the density of iron nails is greater than the density of water and hence iron nails sinks in the water.
7.) Solution:
Ans:
- We are well known about, when weight of floating body is greater than upthrust, the body will sinks in liquid.
- And body will floats on liquid only when the weight of body is balanced by the upthrust.
- Thus, here body is floating on liquid it means the weight of body is equal to the upthrust.
8.) Solution:
Ans:
- When a homogeneous block floats on water which is partly immersed then the its centre of buoyancy B lies exactly below the centre of gravity G of the block.
- Similarly, when the block is completely immersed in water, at that time the centre of buoyancy B and the centre of gravity G of the block will coincide with each other.
10.) Solution:
Ans:
The following figure shows the forces acting on the body floating in water with its some part submerged.
There are two forces acting on body which are as follows:
- The weight of the body W acting downwards.
- The upthrust F acting upward on the body.
Here, the body is floating on water hence its centre of buoyancy B lies exactly below the centre of gravity G of the body as shown in figure.
As here the body is floating, then according to Archimedes’s principle
Weight of water displaced by floating body = weight of the body
11.) Solution:
Ans:
- When we immersed the body inside the water then buoyant force acts on the body and the point at which this resultant buoyant force acts is called as the centre of buoyancy.
- When the body is floating on liquid that means weight of body is balanced by the upthrust.
- And each body has center of gravity G and the centre of buoyancy when immersed in water.
- The floating body has center of buoyancy B which is at the centre of gravity of the submerged part and it is exactly below the centre of gravity G of the entire body.
12.) Solution:
Ans:
When there will be air inside the big closed jar, the balloon floats because density of helium gas is less than density of air.
But, when the air from the jar is pumped out there will less density of air in the jar remained. As air is removed form jar, there will be more density of helium gas due to which balloon starts to sink in that jar slowly.
13.) Solution:
Ans:
When a wooden block is immersed in water room temperature it floats on water as the density of wooden block is less than water.
a)
But, when we added salt to water then more part of the wooden block remains outside the water. This is because of the more upthrust exerted by the salty water. As salt is added the density of water increases which increases the upthrust acting on the wooden block and hence its more part lies above the water surface.
b)
When heat the water, the wooden block sinks in that heated water.
Because, when we heat the water it’s density decreases due to which the upthrust acting on the wooden block also decreases and hence weight of wooden block will be more which causes to sink in the heated water.
14.) Solution:
Ans:
Given that, V is the volume of body.
And q is the density of that body.
v be the volume of body which is immersed in the liquid of density qL.
We know that,
Weight=m*g
But, Density = mass/volume
Hence, mass m= volume*density
Thus, weight of body is given by,
Weight of body= volume of body*density of body*g
W= Vqg
Now, we know that according to Archimedes principle, the weight of liquid displaced is equal to upthrust acting on that body.
Hence, weight of liquid displaced= upthrust= volume of displaced liquid*density of liquid*g
Upthrust= vqLg
But, here the body floating on liquid, and hence the weight of body must be balanced by the upthrust acting on it.
Hence, Vqg= vqLg
We get, V/v= qL/q
Or v/V= q/qL
Hence proved.
15.) Solution:
Ans:
In general, we know that density of ice and wood is less than the density of water and hence they floats on water.
But, when we take comparison of density of ice with density of wood then we see that the density of wood (300kg/m3) is less than the density of ice (900kg/m3).
a)
As the density of ice is more there will be more part of ice is submerged in the water as compared to wood.
b) As the density of ice is more as compared to wood hence the upthrust acting on ice due to water is also more.
16.) Solution:
Ans:
Floating ice is less submerged in brine than water because of the greater density of brine than water.
As density of brine is more it’s upthrust acting on ice is more due to which ice floats on it and less part of ice will be submerged in the brine.
17.) Solution:
Ans:
a)
- Although the densities of sea water and river water are different but the weight of man swimming is same in both the water and hence he will displaced the same amount of water which is equal to his weight.
- And hence their ration will be 1:1 simply.
b)
- As we know that, the density of sea water is more due to salts in it than river water.
- More the sender water more will be the upthrust acting on the body and hence the man swimming in that water can swim easily without submerging his body inside the water.
- Hence, here the man find easier to swim in sea water.
18.) Solution:
Ans:
- The iron nail is very small in shape and having density greater than the density of water, hence when iron nail is immersed in water it sinks in the water.
- But, ships are also made up of iron but they are in huge size and hollow in weight. Also there will be more empty space in ships where air is continuously flowing.
- Due to which the average density of that ship will be less as compared to density of water and hence the ship floats on water.
- And very small part of ship submerged in water which displaces the water which is equal to the upthrust acting on it. And hence ship floats on water.
19.) Solution:
Ans:
Thus ships are floating on water only because of the average density of ship is less than the density of water.
20.) Solution:
Ans:
It’s very curious to know that when we add ice pieces in water after melting the ice the level of water does not changes.
Because, when ice floats on water the part which is not submerged in water that amount part of ice will be contracted after melting and hence the level of water after melting of ice also remains as it is.
21.) Solution:
Ans:
For a given condition, there will be three forces acting on the buoy.
- The weight of buoy acting vertically in downward direction.
- There will be upthrust acting on the buoy in upward direction due to water.
- And the tension in the thread which is also a ting vertically downwards.
- These three forces keep the buoy in equilibrium.
22.) Solution:
Ans:
When the loaded ship is in the sea water, as the density of sea water is more than the average density of ship it displaces small amount of water and floats on sea water.
While in case of river water, the density of river water is less than the sea water and hence when ship floats in river water most of the part of ship will be submerged in water and hence more amount of water will be displaced from the river water.
23.) Solution:
Ans:
a)
Icebergs are having less density than water due to which they floats on water. Also main thing is that, the more portion of the Icebergs lie inside the water which we can’t assume by seeing the Icebergs only.
Hence when ships collides with Icebergs it is more dangerous because the ship driver is not well known about how much portion of Icebergs is inside the water.
b)
An egg sinks in the fresh water but floats in a strong salty solution.
Because when the egg is in the normal water, the density of water is less than egg and hence it sinks in the normal water.
But, in strong salty water solution due addition of salt the density of water is increased which is more than the density of egg and hence the upthrust acting on the egg in upward direction is more. Due to which egg easily floats in strong salty water solution.
c)
When a balloon is filled with hydrogen, due to the density of hydrogen gas is less than the density of air the upthrust acting on the balloon is more and hence it rises in the air.
But, when the balloon is filled with carbon dioxide, due to more density of carbon dioxide the weight of balloon become more and hence the upthrust acting on it is less. Due to which it sinks and comes to the ground.
d)
When the ship in harbour being unloaded, it’s weight decrease and hence the weight acting vertically downwards also decreases. But, upthrust acting on the ship remains same and which is more than weight acting vertically downwards. Due to which the ship slowly rises in higher water.
e)
The balloon filled with hydrogen rises because the density of hydrogen gas is less than the density of air.
But after certain height the density of air decreases due to which the weight of balloon filled with hydrogen at that height is more. And hence from that height it will stops to further rise.
f)
When the loaded ship is in the sea water, as the density of sea water is more than the average density of ship it displaces small amount of water and floats on sea water.
While in case of river water, the density of river water is less than the sea water and hence when ship floats in river water most of the part of ship will be submerged in water and hence more amount of water will be displaced from the river water.
Multiple choice type:
1.) For a floating body, it’s weight W and upthrust F on it are related as
Ans: c) W= F
2.) A body of weight W is floating in a liquid, it’s apparent weight will be
Ans: d) zero
3.) A body floats on liquid A of density q1 with a part of it submerged inside liquid while in liquid B of density q2 totally submerged inside liquid. The densities q1 and q2 are related as
Ans: c) q2<q1
Numerical:
1.) Solution:
Ans:
Let us consider the volume of the rubber ball is V.
The volume of rubber ball which is outside the water= 1/3*V
Hence, the volume of rubber ball which is inside the water= V – 1/3*V= 2/3*V
But, we know that
Volume of rubber ball immersed/total volume of ball= density of rubber/density of water
2/3= density of rubber ball/1000
Hence, density of rubber ball= 2/3*1000 = 666.66 kg/m3
Thus, the density of the rubber ball is found to be 667 kg/m3.
2.) Solution:
Ans:
Given that,
Mass of wooden block = 24kg
Volume of wood = 0.032m3
Density of water = 1000kg/m3
a)
We know that,
Upthrust= weight of the block
24kgf= volume of block*density of water*g
24kgf= V*1000*g
Thus, V= 24/1000= 0.024m3
b)
Also we know that,
Volume of the submerged block/ total volume of block= density of wood/density of water
0.024/0.032= density of wood/1000
Thus, density of wood = 0.75*1000= 750kg/m3
Thus, the density of wooden block is 750kg/m3.
3.) Solution:
Ans:
Given that,
Side of wooden cube= 10cm
Mass of wooden cube= 700g
Hence, volume of wooden cube = 10*10*10= 1000cm3
Hence, we can find the density of wooden cube as,
Density of wooden cube= mass/volume= 700/1000= 0.7g/cm3
But, we know that,
the density of floating body/density of water= portion of wooden cube submerged
0.7/1= portion of wooden cube submerged in water
Thus, the portion of wooden cube submerged in water is 0.7.
But, total height of wooden cube is 10cm.
Thus, the height of wooden cube submerged in water = 10*0.7= 7cm
Hence, the remaining height 10-7=3cm
Hence, we can say that the height of the wooden cube which remain outside the water is 3cm.
4.) Solution:
Ans:
Given that,
Density of wax= 0.95g/cm3
Density of brine= 1.1g/cm3
We consider the entire volume of piece of wax is V and v be the volume of wax which is immersed in brine.
We know that,
v/V= density of wax/ density of brine
v/V= 0.95/1.1= 0.86
Volume of wax submerged in brine = 0.86V
5.) Solution:
Ans:
Given that,
Density of ice= 0.9g/cm3
Density of sea water= 1.1g/cm3
We consider v be the volume of iceberg submerged in the sea water.
And V be the total volume of Iceberg.
We know that,
v/V= density of ice/ density of sea water
v/V= 0.9/1.1= 9/11
Thus, v= 9/11*V
Thus, the portion of an iceberg will remain below the surface of water in sea is 9/11th of its total volume.
6.) Solution:
Ans:
Given that height of piece of wood= 15cm
Let A be the area of wood.
Hence, volume of wood is = A*15
Let q1, q2 and q3 be the densities of water, wood and spirit.
We know that, when wood is immersed in water then mass of wood is equal to the mass of water displaced by wood
Mass of wood= mass of water displaced
A*15*q2*g= A*10*q1*g
Thus, q2= 10*q1/15
Thus, q2= 10*1/15= 0.666g/cm3
Thus, the density of wood is found to be 0.67g/cm3.
Similarly, we can write
Mass of wood= mass of the spirit displaced
15*A*q2*g= A*12*q3*g
Thus, q3= 15*q2/12
q3= 15*0.67/12= 0.83g/cm3
Thus, the density of spirit is found to be 0.83g/cm3.
7.) Solution:
Ans:
a)
When the wooden block floats in water it’s two third volume is submerged in it.
Let, V be the total volume of wooden block.
Then, volume of wooden block submerged v= 2/3*V
Let q and q1 are the densities of water wood respectively.
Then, we can also write as
v/V= density of wood/density of water
2/3= density of wood/1000
Density of wood= 2/3*1000= 666.66 kg/m3
Thus, density of wood is found to be 667 kg/m3
b)
When the same block of totals volume V is submerged in oil three quarter of its volume is submerged in oil.
Volume of block submerged in oil v’= 3/4*V
Let q2 be the density of oil.
Then we can also write,
v’/V= density of wood/density of oil
3/4= 667/density of oil
Thus, density of oil= 667/0.75= 889.33 kg/m3.
Thus, the density of oil is found to be 889 kg/m3.
8.) Solution:
Ans:
Given that,
Density of ice = 0.92g/cm3
Density of sea water = 1.025g/cm3
Let V be the total volume of iceberg.
The volume of iceberg floating above water= 800cm3
Let v be the volume of iceberg floating inside water.
Then we can also write,
v/V= density of ice/density of sea water
v/V= 0.92/1.025= 0.8976
Thus, v= 0.8976*V
Total volume of iceberg – volume of iceberg inside water= volume of iceberg above water
V – 0.8976 V= 800
V*(1-0.8976)= 800
Thus, V= 800/0.1024= 7812.5cm3
Thus, the total volume of iceberg is found to be 7812.5 cm3.
9.) Solution:
Ans:
Given that,
Volume of plastic balloon= 15m3
Density of hydrogen= 0.09kg/m3
The mass of empty balloon= 7.15kg
Density of air = 1.3kg/m3
a)
The mass of hydrogen in balloon is given by,
Mass of hydrogen in balloon= volume of balloon*density of hydrogen
Mass of hydrogen in balloon= 15*0.09= 1.35kg
Thus, the mass of hydrogen in balloon is found to be 1.35kg
b)
The mass of hydrogen and balloon is given by,
Mass of hydrogen and balloon= mass of empty balloon+ mass of hydrogen in the balloon
= 7.15kg + 1.35kg= 8.5kg
Thus, the mass of hydrogen and balloon is found to be 8.5kg.
c)
Given that, xkg is the mass of equipment.
Hence, total mass of hydrogen, balloon and equipment is given by
= 8.5kg + xkg = (8.5 + X) kg
d)
The mass of air displaced by balloon is given by,
Mass of air displaced by balloon= volume of balloon*density of air
= 15*1.3= 19.5kg
Thus, the mass of air displaced by balloon is 19.5kg
e)
We know that, according to the law of floatation,
The mass of air displaced= total mass of hydrogen, balloon and equipment
19.5= 8.5+X
Thus, X= 19.5-8.5 = 11kg
Thus, the mass of equipment is found to be 11kg.