NEET
- Exam
- NEET
- Subject
- Physics
- Chapter
- Mechanical Properties of Solids
NEET · Physics · Mechanical Properties of Solids
A copper wire ((Y=10^ 11 Nm^ -2 )) of length 8 m and steel wire...
A copper wire (\left(\mathrm{Y}=10^{11} \mathrm{Nm}^{-2}\right)) of length 8 m and steel wire (\left(\mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}\right)) of length 4 m and each of cross-sectional area (0.5 \mathrm{ cm}^2) are fastened end to end and stretched with a tension of 500 N. Match the given columns and select the correct option from the codes given below.
(
)
| Column-I | Column-II |
| (i) Elongation in copper wire (in mm) | (p) 0.2 |
| (ii) Elongation in steel wire (in mm) | (q) 1.0 |
| (iii) Total elongation (in mm) | (r) 0.8 |
Options
- A, ii , iii
- B(\mathrm{i}-\mathrm{p}), ii -q , iii -r
- C, ii , iii
- D, ii , iii
Correct answer
C. , ii , iii
Step-by-step solution
The pressure inside a soap bubble is inversely proportional to its radius ((P_{ex} \propto \frac{1}{r})). This means the smaller bubble (A) has a higher internal pressure than the larger bubble (B). When the valve V is opened, air will flow from the region of higher pressure (bubble A) to the region of lower pressure (bubble B).
As air flows out of A, its size will decrease, which further increases its internal pressure. As air flows into B, its size will increase, which decreases its internal pressure. This process will continue until bubble A becomes smaller and smaller, and bubble B becomes larger and larger.
Therefore, A will become smaller and B will become larger. This corresponds to option (3).