Question:
Which one is an example of branched polymer?
Which one is an example of branched polymer?
Show Hint
LDPE is softer and more flexible due to its branched structure, unlike HDPE which is stronger and linear.
Updated On: Sep 24, 2025
- HDPE
- LDPE
- Nylon-66
- PET
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding branched polymers.
Branched polymers have a main linear chain with side branches. This branching reduces packing efficiency and lowers density compared to linear polymers.
Step 2: Analysis of options.
- (A) HDPE: High Density Polyethylene is a linear polymer, not branched.
- (B) LDPE: Low Density Polyethylene is a branched polymer with side chains, making it the correct example.
- (C) Nylon-66: This is a condensation polymer but not branched.
- (D) PET: Polyethylene terephthalate is a linear polyester, not branched.
Step 3: Conclusion.
The correct answer is (B) LDPE, as it is a branched polymer.
Was this answer helpful?
0
0
Top Questions on Polymer Science
- Buna-S is a polymer of:
- Which one is appropriate for the poly-ethyleneterephthalate?
- The functionality of the Phenol in the preparation of phenol-formaldehyde resin is :
- Carothers equation may define gel formation for the:
- ABS is a:
View More Questions
Questions Asked in CUET PG exam
- Match List-I with List-II:
\[ \begin{array}{|l|l|} \hline \textbf{List-I (Equation)} & \textbf{List-II (Roots)} \\ \hline (A)\; (z + 1)^{2n} + (z - 1)^{2n} = 0 & (I)\; \cos\!\left(\tfrac{\pi}{5}\right) \pm i\sin\!\left(\tfrac{\pi}{5}\right), -1, \cos\!\left(\tfrac{3\pi}{5}\right) \pm i\sin\!\left(\tfrac{3\pi}{5}\right) \\ \hline (B)\; z^5 + z^4 + z^3 + z^2 + z + 1 = 0 & (II)\; \text{purely imaginary number} \\ \hline (C)\; (z-1)^5 + z^5 = 0 & (III)\; -1,\; \pm\frac{1}{2} \pm i\frac{\sqrt{3}}{2} \\ \hline (D)\; z^5 + 1 = 0 & (IV)\; \tfrac{1}{2}\left[1+i\cot\!\left(\tfrac{p\pi}{10}\right)\right], \; p = 1,3,5,7,9 \\ \hline \end{array} \]
Choose the correct answer from the options given below:- CUET (PG) - 2025
- Complex numbers
- Which of the following statements are correct?
If \[ A = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix}, \] then
(A) \[ B^n = \begin{pmatrix} 1 & nq \\ 0 & 1 \end{pmatrix} \]
(B) \[ A^n = \begin{pmatrix} p^n & q\frac{p^n-1}{p-1} \\ 0 & 1 \end{pmatrix}, \; \text{if } p \neq 1 \]
(C) \[ AB = \begin{pmatrix} p & pq+q \\ 0 & 1 \end{pmatrix} \]
(D) \[ B^{n-1} = \begin{pmatrix} 1 & (n+1)q \\ 0 & 1 \end{pmatrix} \]
(E) \[ AB^n = \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix} \]
Choose the correct answer from the options given below:- CUET (PG) - 2025
- Linear Algebra
- If the matrix \[ A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \] satisfies the matrix equation:
- CUET (PG) - 2025
- Linear Algebra
- If the rank of matrix \[ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & \lambda \end{pmatrix} \] is 2, then the value of \(\lambda\) is:
- CUET (PG) - 2025
- Linear Algebra
- Match List-I with List-II
\[ \begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A) \; \lim_{x \to 0} \frac{\ln(1+x)}{\sin x} & (I) \; 1 \\ (B) \; \lim_{x \to \infty} 2x \tan\left(\tfrac{1}{x}\right) & (II) \; 0 \\ (C) \; \lim_{x \to \infty} \frac{x^2}{e^x} & (III) \; 2 \\ (D) \; \lim_{x \to 1} x^{\tfrac{1}{x-1}} & (IV) \; e \\ \hline \end{array} \]
Choose the correct answer from the options given below:
- CUET (PG) - 2025
- Calculus
View More Questions