JEE Main Previous Year Questions (2025): Mechanical Properties of Fluids

 Page 1


JEE Main Previous Year Questions 
(2025): Mechanical Properties of 
Fluids 
 
Q1: Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each 
other. The radius of curvature of the common surface, in cm , is ____ . 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 4 
Solution: 
To find the radius of curvature of the common surface between two soap bubbles, we use the 
formula: 
?? =
?? 1
· ?? 2
?? 1
- ?? 2
 
Given that ?? 1
= 2 cm and ?? 2
= 4 cm , we substitute these values into the formula: 
?? =
2 · 4
2 - 4
=
8
-2
= 4 cm 
Thus, the radius of curvature of the common surface is 4 cm . 
Q2: An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free 
surface of a liquid having surface tension ?? . ?????? ?? /?? ?? and density ????
?? ???? /?? ?? . The 
difference between pressure inside the bubble and atmospheric pressure is ____ 
?? /?? ?? . (Take ?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 2190 
Solution: 
To find the difference between the pressure inside the air bubble and the atmospheric pressure, 
we use the fact that at depth h in a liquid of density ?? , the external (hydrostatic) pressure 
exceeds atmospheric by ???? h. Additionally, for a spherical bubble in a liquid, the internal 
pressure exceeds the external pressure by 
2?? ?? , where ?? is the surface tension and ?? is the radius 
of the bubble. 
Hence, the internal pressure of the bubble is 
?? inside 
= ?? atm 
+ ???? h +
2?? ?? . 
Therefore, the difference between the bubble's internal pressure and the atmospheric pressure 
is 
Page 2


JEE Main Previous Year Questions 
(2025): Mechanical Properties of 
Fluids 
 
Q1: Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each 
other. The radius of curvature of the common surface, in cm , is ____ . 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 4 
Solution: 
To find the radius of curvature of the common surface between two soap bubbles, we use the 
formula: 
?? =
?? 1
· ?? 2
?? 1
- ?? 2
 
Given that ?? 1
= 2 cm and ?? 2
= 4 cm , we substitute these values into the formula: 
?? =
2 · 4
2 - 4
=
8
-2
= 4 cm 
Thus, the radius of curvature of the common surface is 4 cm . 
Q2: An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free 
surface of a liquid having surface tension ?? . ?????? ?? /?? ?? and density ????
?? ???? /?? ?? . The 
difference between pressure inside the bubble and atmospheric pressure is ____ 
?? /?? ?? . (Take ?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 2190 
Solution: 
To find the difference between the pressure inside the air bubble and the atmospheric pressure, 
we use the fact that at depth h in a liquid of density ?? , the external (hydrostatic) pressure 
exceeds atmospheric by ???? h. Additionally, for a spherical bubble in a liquid, the internal 
pressure exceeds the external pressure by 
2?? ?? , where ?? is the surface tension and ?? is the radius 
of the bubble. 
Hence, the internal pressure of the bubble is 
?? inside 
= ?? atm 
+ ???? h +
2?? ?? . 
Therefore, the difference between the bubble's internal pressure and the atmospheric pressure 
is 
??? = ?? inside 
- ?? atm 
= ???? h +
2?? ?? . 
Substituting the given values: 
?? = 10
3
 kg/m
3
?? = 10 m/s
2
h = 0.20 m
?? = 0.095 J/m
2
?? = 1.0 × 10
-3
 m
 
we compute each term: 
Hydrostatic term: 
???? h = ( 10
3
 kg/m
3
)· ( 10 m/s
2
)· ( 0.20 m)= 2000 N/m
2
. 
Surface tension term: 
2?? ?? =
2 × 0.095 J/m
2
1.0 × 10
-3
 m
=
2 × 0.095
10
-3
= 2 × 95 = 190 N/m
2
 
Hence, 
??? = 2000 N/m
2
+ 190 N/m
2
= 2190 N/m
2
 
Thus, the required pressure difference between the inside of the bubble and the atmospheric 
pressure is 
2190 N/m
2
. 
Q3: In a measurement, it is asked to find modulus of elasticity per unit torque applied 
on the system. The measured quantity has dimension of [?? ?? ?? ?? ?? ?? ]. If ?? = ?? , the value 
of ?? is ____ ? 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 0 
Solution: 
Given, measured quantity =
 Modulus of elasticity 
 torque 
 
=
?? ????
=
?? ?????? where, ?? = stress, ?? = strain, ?? = torque 
=
?? ???????? 
So, dimension =
1
[?? 2
][?? ]
 ( ?? is dimensionless) 
= [?? -3
] = [?? 0
?? -3
?? 0
] = [?? ?? ?? ?? ?? ?? ] 
? ?? = 0 
Q4: A vessel with square cross-section and height of ???? is vertically partitioned. A 
small window of ?????? ????
?? with hinged door is fitted at a depth of 3 m in the partition 
Page 3


JEE Main Previous Year Questions 
(2025): Mechanical Properties of 
Fluids 
 
Q1: Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each 
other. The radius of curvature of the common surface, in cm , is ____ . 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 4 
Solution: 
To find the radius of curvature of the common surface between two soap bubbles, we use the 
formula: 
?? =
?? 1
· ?? 2
?? 1
- ?? 2
 
Given that ?? 1
= 2 cm and ?? 2
= 4 cm , we substitute these values into the formula: 
?? =
2 · 4
2 - 4
=
8
-2
= 4 cm 
Thus, the radius of curvature of the common surface is 4 cm . 
Q2: An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free 
surface of a liquid having surface tension ?? . ?????? ?? /?? ?? and density ????
?? ???? /?? ?? . The 
difference between pressure inside the bubble and atmospheric pressure is ____ 
?? /?? ?? . (Take ?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 2190 
Solution: 
To find the difference between the pressure inside the air bubble and the atmospheric pressure, 
we use the fact that at depth h in a liquid of density ?? , the external (hydrostatic) pressure 
exceeds atmospheric by ???? h. Additionally, for a spherical bubble in a liquid, the internal 
pressure exceeds the external pressure by 
2?? ?? , where ?? is the surface tension and ?? is the radius 
of the bubble. 
Hence, the internal pressure of the bubble is 
?? inside 
= ?? atm 
+ ???? h +
2?? ?? . 
Therefore, the difference between the bubble's internal pressure and the atmospheric pressure 
is 
??? = ?? inside 
- ?? atm 
= ???? h +
2?? ?? . 
Substituting the given values: 
?? = 10
3
 kg/m
3
?? = 10 m/s
2
h = 0.20 m
?? = 0.095 J/m
2
?? = 1.0 × 10
-3
 m
 
we compute each term: 
Hydrostatic term: 
???? h = ( 10
3
 kg/m
3
)· ( 10 m/s
2
)· ( 0.20 m)= 2000 N/m
2
. 
Surface tension term: 
2?? ?? =
2 × 0.095 J/m
2
1.0 × 10
-3
 m
=
2 × 0.095
10
-3
= 2 × 95 = 190 N/m
2
 
Hence, 
??? = 2000 N/m
2
+ 190 N/m
2
= 2190 N/m
2
 
Thus, the required pressure difference between the inside of the bubble and the atmospheric 
pressure is 
2190 N/m
2
. 
Q3: In a measurement, it is asked to find modulus of elasticity per unit torque applied 
on the system. The measured quantity has dimension of [?? ?? ?? ?? ?? ?? ]. If ?? = ?? , the value 
of ?? is ____ ? 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 0 
Solution: 
Given, measured quantity =
 Modulus of elasticity 
 torque 
 
=
?? ????
=
?? ?????? where, ?? = stress, ?? = strain, ?? = torque 
=
?? ???????? 
So, dimension =
1
[?? 2
][?? ]
 ( ?? is dimensionless) 
= [?? -3
] = [?? 0
?? -3
?? 0
] = [?? ?? ?? ?? ?? ?? ] 
? ?? = 0 
Q4: A vessel with square cross-section and height of ???? is vertically partitioned. A 
small window of ?????? ????
?? with hinged door is fitted at a depth of 3 m in the partition 
wall. One part of the vessel is filled completely with water and the other side is filled 
with the liquid having density ?? . ?? × ????
?? ?? ?? /?? ?? . What force one needs to apply on 
the hinged door so that it does not get opened ? 
(Acceleration due to gravity = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 150 
Solution: 
 
in equilibrium 
?? ?????? + ?? ?? = ?? l
 ? ?? ?????? = ?? l
- ?? ?? = ( ?? 0
+ ?? l
?? h) ?? - ( ?? 0
+ ?? ?? ?? h) ?? = ( ?? l
- ?? ?? ) ?? h?? = ( 1500- 1000 )× 10 × 3 × ( 100 × 10
-4
)
 = 150 m
 
Q5: A steel wire of length 2 m and Young's modulus ?? . ?? × ????
????
 ?? ?? -?? is stretched by 
a force. If Poisson ratio and transverse strain for the wire are 0.2 and ????
-?? 
respectively, then the elastic potential energy density of the wire is ____ × ????
?? (in SI 
units). 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 25 
Solution: 
To find the elastic potential energy density of the steel wire, we need to use the given 
information and formulae for strain and energy density. 
Given: 
The length of the wire, l = 2 m 
Young's modulus, ?? = 2.0 × 10
11
 N/m
2
 
Poisson's ratio, ?? = 0.2 
Transverse strain, 
??? ?? = 10
-3
 
The formula for Poisson's ratio is: 
Page 4


JEE Main Previous Year Questions 
(2025): Mechanical Properties of 
Fluids 
 
Q1: Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each 
other. The radius of curvature of the common surface, in cm , is ____ . 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 4 
Solution: 
To find the radius of curvature of the common surface between two soap bubbles, we use the 
formula: 
?? =
?? 1
· ?? 2
?? 1
- ?? 2
 
Given that ?? 1
= 2 cm and ?? 2
= 4 cm , we substitute these values into the formula: 
?? =
2 · 4
2 - 4
=
8
-2
= 4 cm 
Thus, the radius of curvature of the common surface is 4 cm . 
Q2: An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free 
surface of a liquid having surface tension ?? . ?????? ?? /?? ?? and density ????
?? ???? /?? ?? . The 
difference between pressure inside the bubble and atmospheric pressure is ____ 
?? /?? ?? . (Take ?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 2190 
Solution: 
To find the difference between the pressure inside the air bubble and the atmospheric pressure, 
we use the fact that at depth h in a liquid of density ?? , the external (hydrostatic) pressure 
exceeds atmospheric by ???? h. Additionally, for a spherical bubble in a liquid, the internal 
pressure exceeds the external pressure by 
2?? ?? , where ?? is the surface tension and ?? is the radius 
of the bubble. 
Hence, the internal pressure of the bubble is 
?? inside 
= ?? atm 
+ ???? h +
2?? ?? . 
Therefore, the difference between the bubble's internal pressure and the atmospheric pressure 
is 
??? = ?? inside 
- ?? atm 
= ???? h +
2?? ?? . 
Substituting the given values: 
?? = 10
3
 kg/m
3
?? = 10 m/s
2
h = 0.20 m
?? = 0.095 J/m
2
?? = 1.0 × 10
-3
 m
 
we compute each term: 
Hydrostatic term: 
???? h = ( 10
3
 kg/m
3
)· ( 10 m/s
2
)· ( 0.20 m)= 2000 N/m
2
. 
Surface tension term: 
2?? ?? =
2 × 0.095 J/m
2
1.0 × 10
-3
 m
=
2 × 0.095
10
-3
= 2 × 95 = 190 N/m
2
 
Hence, 
??? = 2000 N/m
2
+ 190 N/m
2
= 2190 N/m
2
 
Thus, the required pressure difference between the inside of the bubble and the atmospheric 
pressure is 
2190 N/m
2
. 
Q3: In a measurement, it is asked to find modulus of elasticity per unit torque applied 
on the system. The measured quantity has dimension of [?? ?? ?? ?? ?? ?? ]. If ?? = ?? , the value 
of ?? is ____ ? 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 0 
Solution: 
Given, measured quantity =
 Modulus of elasticity 
 torque 
 
=
?? ????
=
?? ?????? where, ?? = stress, ?? = strain, ?? = torque 
=
?? ???????? 
So, dimension =
1
[?? 2
][?? ]
 ( ?? is dimensionless) 
= [?? -3
] = [?? 0
?? -3
?? 0
] = [?? ?? ?? ?? ?? ?? ] 
? ?? = 0 
Q4: A vessel with square cross-section and height of ???? is vertically partitioned. A 
small window of ?????? ????
?? with hinged door is fitted at a depth of 3 m in the partition 
wall. One part of the vessel is filled completely with water and the other side is filled 
with the liquid having density ?? . ?? × ????
?? ?? ?? /?? ?? . What force one needs to apply on 
the hinged door so that it does not get opened ? 
(Acceleration due to gravity = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 150 
Solution: 
 
in equilibrium 
?? ?????? + ?? ?? = ?? l
 ? ?? ?????? = ?? l
- ?? ?? = ( ?? 0
+ ?? l
?? h) ?? - ( ?? 0
+ ?? ?? ?? h) ?? = ( ?? l
- ?? ?? ) ?? h?? = ( 1500- 1000 )× 10 × 3 × ( 100 × 10
-4
)
 = 150 m
 
Q5: A steel wire of length 2 m and Young's modulus ?? . ?? × ????
????
 ?? ?? -?? is stretched by 
a force. If Poisson ratio and transverse strain for the wire are 0.2 and ????
-?? 
respectively, then the elastic potential energy density of the wire is ____ × ????
?? (in SI 
units). 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 25 
Solution: 
To find the elastic potential energy density of the steel wire, we need to use the given 
information and formulae for strain and energy density. 
Given: 
The length of the wire, l = 2 m 
Young's modulus, ?? = 2.0 × 10
11
 N/m
2
 
Poisson's ratio, ?? = 0.2 
Transverse strain, 
??? ?? = 10
-3
 
The formula for Poisson's ratio is: 
?? = -
(
??? ?? )
(
?l
l
)
 
From this, we solve for the longitudinal strain 
?l
l
 : 
?l
l
=
1
?? × (
??? ?? ) 
Substitute the given values: 
?l
l
=
1
0.2
× 10
-3
= 5 × 10
-3
 
The elastic potential energy density ?? is given by: 
?? =
1
2
?? ?? l
2
 
where ?? l
=
?l
l
. Plug in the values: 
?? =
1
2
× 2 × 10
11
× ( 5 × 10
-3
)
2
 
Simplify further: 
?? =
1
2
× 2 × 10
11
× 25 × 10
-6
?? = 25 × 10
5
( in SI units )
 
Thus, the elastic potential energy density of the wire is 25 × 10
5
 SI units. 
Q6: The length of a light string is 1.4 m when the tension on it is 5 N . If the tension 
increases to 7 N , the length of the string is 1.56 m . The original length of the string is 
____ m. 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 1 
Solution: 
To find the original length of the string, we use the relationship between tension, elasticity 
constant ( ?? ), and the change in length of the string. 
Given the equation for tension: 
T = K( l - l
0
) 
where l is the length of the string under tension and l
0
 is the original length. 
When the tension is 5 N , the equation becomes: 
5 = K( 1.4 - l
0
) 
When the tension increases to 7 N , the equation is: 
Page 5


JEE Main Previous Year Questions 
(2025): Mechanical Properties of 
Fluids 
 
Q1: Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each 
other. The radius of curvature of the common surface, in cm , is ____ . 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 4 
Solution: 
To find the radius of curvature of the common surface between two soap bubbles, we use the 
formula: 
?? =
?? 1
· ?? 2
?? 1
- ?? 2
 
Given that ?? 1
= 2 cm and ?? 2
= 4 cm , we substitute these values into the formula: 
?? =
2 · 4
2 - 4
=
8
-2
= 4 cm 
Thus, the radius of curvature of the common surface is 4 cm . 
Q2: An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free 
surface of a liquid having surface tension ?? . ?????? ?? /?? ?? and density ????
?? ???? /?? ?? . The 
difference between pressure inside the bubble and atmospheric pressure is ____ 
?? /?? ?? . (Take ?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 2190 
Solution: 
To find the difference between the pressure inside the air bubble and the atmospheric pressure, 
we use the fact that at depth h in a liquid of density ?? , the external (hydrostatic) pressure 
exceeds atmospheric by ???? h. Additionally, for a spherical bubble in a liquid, the internal 
pressure exceeds the external pressure by 
2?? ?? , where ?? is the surface tension and ?? is the radius 
of the bubble. 
Hence, the internal pressure of the bubble is 
?? inside 
= ?? atm 
+ ???? h +
2?? ?? . 
Therefore, the difference between the bubble's internal pressure and the atmospheric pressure 
is 
??? = ?? inside 
- ?? atm 
= ???? h +
2?? ?? . 
Substituting the given values: 
?? = 10
3
 kg/m
3
?? = 10 m/s
2
h = 0.20 m
?? = 0.095 J/m
2
?? = 1.0 × 10
-3
 m
 
we compute each term: 
Hydrostatic term: 
???? h = ( 10
3
 kg/m
3
)· ( 10 m/s
2
)· ( 0.20 m)= 2000 N/m
2
. 
Surface tension term: 
2?? ?? =
2 × 0.095 J/m
2
1.0 × 10
-3
 m
=
2 × 0.095
10
-3
= 2 × 95 = 190 N/m
2
 
Hence, 
??? = 2000 N/m
2
+ 190 N/m
2
= 2190 N/m
2
 
Thus, the required pressure difference between the inside of the bubble and the atmospheric 
pressure is 
2190 N/m
2
. 
Q3: In a measurement, it is asked to find modulus of elasticity per unit torque applied 
on the system. The measured quantity has dimension of [?? ?? ?? ?? ?? ?? ]. If ?? = ?? , the value 
of ?? is ____ ? 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 0 
Solution: 
Given, measured quantity =
 Modulus of elasticity 
 torque 
 
=
?? ????
=
?? ?????? where, ?? = stress, ?? = strain, ?? = torque 
=
?? ???????? 
So, dimension =
1
[?? 2
][?? ]
 ( ?? is dimensionless) 
= [?? -3
] = [?? 0
?? -3
?? 0
] = [?? ?? ?? ?? ?? ?? ] 
? ?? = 0 
Q4: A vessel with square cross-section and height of ???? is vertically partitioned. A 
small window of ?????? ????
?? with hinged door is fitted at a depth of 3 m in the partition 
wall. One part of the vessel is filled completely with water and the other side is filled 
with the liquid having density ?? . ?? × ????
?? ?? ?? /?? ?? . What force one needs to apply on 
the hinged door so that it does not get opened ? 
(Acceleration due to gravity = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 150 
Solution: 
 
in equilibrium 
?? ?????? + ?? ?? = ?? l
 ? ?? ?????? = ?? l
- ?? ?? = ( ?? 0
+ ?? l
?? h) ?? - ( ?? 0
+ ?? ?? ?? h) ?? = ( ?? l
- ?? ?? ) ?? h?? = ( 1500- 1000 )× 10 × 3 × ( 100 × 10
-4
)
 = 150 m
 
Q5: A steel wire of length 2 m and Young's modulus ?? . ?? × ????
????
 ?? ?? -?? is stretched by 
a force. If Poisson ratio and transverse strain for the wire are 0.2 and ????
-?? 
respectively, then the elastic potential energy density of the wire is ____ × ????
?? (in SI 
units). 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 25 
Solution: 
To find the elastic potential energy density of the steel wire, we need to use the given 
information and formulae for strain and energy density. 
Given: 
The length of the wire, l = 2 m 
Young's modulus, ?? = 2.0 × 10
11
 N/m
2
 
Poisson's ratio, ?? = 0.2 
Transverse strain, 
??? ?? = 10
-3
 
The formula for Poisson's ratio is: 
?? = -
(
??? ?? )
(
?l
l
)
 
From this, we solve for the longitudinal strain 
?l
l
 : 
?l
l
=
1
?? × (
??? ?? ) 
Substitute the given values: 
?l
l
=
1
0.2
× 10
-3
= 5 × 10
-3
 
The elastic potential energy density ?? is given by: 
?? =
1
2
?? ?? l
2
 
where ?? l
=
?l
l
. Plug in the values: 
?? =
1
2
× 2 × 10
11
× ( 5 × 10
-3
)
2
 
Simplify further: 
?? =
1
2
× 2 × 10
11
× 25 × 10
-6
?? = 25 × 10
5
( in SI units )
 
Thus, the elastic potential energy density of the wire is 25 × 10
5
 SI units. 
Q6: The length of a light string is 1.4 m when the tension on it is 5 N . If the tension 
increases to 7 N , the length of the string is 1.56 m . The original length of the string is 
____ m. 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 1 
Solution: 
To find the original length of the string, we use the relationship between tension, elasticity 
constant ( ?? ), and the change in length of the string. 
Given the equation for tension: 
T = K( l - l
0
) 
where l is the length of the string under tension and l
0
 is the original length. 
When the tension is 5 N , the equation becomes: 
5 = K( 1.4 - l
0
) 
When the tension increases to 7 N , the equation is: 
7 = K( 1.56 - l
0
) 
By setting up a ratio from the two equations, we have: 
5
1.4 - l
0
=
7
1.56 - l
0
 
Solving this proportion gives us the original length l
0
 : 
l
0
= 1 m 
Thus, the original length of the string is 1 meter. 
Q7: The excess pressure inside a soap bubble ?? in air is half the excess pressure inside 
another soap bubble ?? in air. If the volume of the bubble ?? is ?? times the volume of 
the bubble ?? , then, the value of ?? is ____ . 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 8 
Solution: 
The excess pressure inside a soap bubble is determined by the formula: 
??? =
4?? ?? 
where ?P is the excess pressure, T is the surface tension of the soap film, and R is the radius of 
the bubble. 
Given that the excess pressure inside bubble A is half that inside bubble B , we have: 
?P
A
=
1
2
?P
B
 
This implies: 
?? A
?? B
=
?P
B
?P
A
= 2 
Now, considering the volumes of the bubbles, the relationship between volume and radius for a 
sphere is given by: 
?? =
4
3
?? ?? 3
 
Thus, the ratio of the volumes of bubbles A and B is: 
?? A
?? B
= (
?? A
?? B
)
3
= 2
3
= 8 
Therefore, the volume of bubble A is ?? = 8 times the volume of bubble B . 
Q8: Two slabs with square cross section of different materials ( ?? , ?? ) with equal sides 
( ?? ) and thickness ?? ?? and ?? ?? such that ?? ?? = ?? ?? ?? and ?? > ?? ?? 
. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing