A steel rod of radius 20 mm and length of 2 m is acted upon by a force of 400 kN along the length. The values of stress and strain are respectively (Y = 2 10 Nm )
1
Nm ,
2
Nm ,
3
Nm ,
4
Nm ,
Official Solution
Correct Option: (2)
The radius of the steel rod is mm m.
The length of the steel rod is m.
The applied force is kN N.
Young's modulus of steel is Nm . First, calculate the cross-sectional area of the rod:
$ , . Next, calculate the stress :
\) using Young's modulus :
\) $ The values of stress and strain are approximately Nm and respectively.
02
PYQ 2025
medium
physicsID: ap-eapce
If the pressure on a body is increased from kPa to kPa, the volume of the body decreases by . The compressibility of the material of the body (in ) is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Compressibility Definition Compressibility is defined as: where:
- = Percentage volume change = ,
- kPa = Pa. Step 2: Calculating Conclusion Thus, the correct answer is:
03
PYQ 2025
hard
physicsID: ap-eapce
The elastic potential energy stored in a copper rod of length one meter and area of cross-section when stretched by 1 mm is:
1
2
3
4
Official Solution
Correct Option: (1)
The elastic potential energy stored in a stretched material is given by:
where is the force, is the length of the rod, is the area of cross-section, and is the Young’s modulus. First, calculate the force using the elongation formula:
where . After solving for the energy, we get:
04
PYQ 2025
medium
physicsID: ap-eapce
When a wire made of material with Young's modulus Y is subjected to a stress S, the elastic potential energy per unit volume stored in the wire is
1
2
3
4
Official Solution
Correct Option: (3)
Young's modulus .
Let Stress be and Strain be .
So, .
Elastic potential energy per unit volume (Energy Density, U) stored in a stretched wire is given by:
Substitute the expressions for Stress and Strain:
Substitute :
This matches option (3). Other forms of energy density:
Using : .
05
PYQ 2025
medium
physicsID: ap-eapce
If two soap bubbles each of radius combine in vacuum under isothermal conditions, then the radius of the new bubble formed is
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: In vacuum and under isothermal conditions, volume is conserved when two bubbles combine. Step 2: Volume of each bubble = Total volume of two bubbles:
Let be the radius of the new bubble. Then:
This is correct for general isothermal processes. However, in this particular question, **surface energy in vacuum** is also conserved (assuming ideal condition), and **pressure difference** inside bubble is considered. Hence, **pressure-volume** product is conserved due to **isothermal process**:
For small soap bubbles under vacuum, the relation reduces to:
Step 3: Since , we get:
But in case of isothermal expansion of **gas** inside ideal soap bubbles (in vacuum), using pressure relation:
Final Calculation:
% Final Answer
06
PYQ 2025
medium
physicsID: ap-eapce
When a wire of length clamped at one end is pulled by a force from the other end, its length increases by . If the radius and the applied force are halved, then the increase in its length is
1
2
3
4
Official Solution
Correct Option: (4)
Extension in a wire:
Radius halved Area becomes
Force halved
New extension:
07
PYQ 2025
medium
physicsID: ap-eapce
If the pressure on a body is increased from kPa to kPa, the volume of the body decreases by . The compressibility of the material of the body (in ) is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Compressibility Definition Compressibility is defined as: where:
- = Percentage volume change = ,
- kPa = Pa. Step 2: Calculating Conclusion Thus, the correct answer is:
08
PYQ 2025
medium
physicsID: ap-eapce
As shown in the figure, a light uniform rod PQ of length 150 cm is suspended from the ceiling horizontally using two metal wires A and B tied to the ends of the rod. The ratios of the radii and the Young’s modulus of the materials of the two wires A and B are respectively 2 : 3 and 3 : 2. The position at which a weight should be suspended from the rod such that the elongations of the two wires become equal is
1
90 cm from P
2
100 cm from P
3
40 cm from Q
4
45 cm from Q
Official Solution
Correct Option: (1)
Step 1: Understand the problem setup A rod of length 150 cm is suspended horizontally by two wires A and B attached at ends P and Q respectively. A weight is hung at some point along PQ so that the elongations of both wires are equal. Step 2: Given ratios - Radius ratio: - Young’s modulus ratio: Step 3: Relationship between elongation, force, and material properties Elongation in a wire is given by: where is force, length, cross-sectional area ( ), and Young’s modulus. Step 4: Express elongations for both wires Let the weight be suspended at distance from P. The tension in wire A = , tension in wire B = due to lever principle. Area of wire A: Area of wire B: Step 5: Equal elongation condition cancels out. Substitute and : Simplify denominators: Cross multiply: Step 6: Conclusion The weight should be suspended 90 cm from point P for equal elongations.
