Step 1: Define shrinkage factor.
Shrinkage factor = \(\frac{\text{Shrunk length}}{\text{Original length}}\).
Step 2: Substitution.
Original length = 10 cm, Shrunk length = 9.5 cm.
\[
\text{Shrinkage Factor} = \frac{9.5}{10} = 0.95
\]
Step 3: Conclusion.
Therefore, the shrinkage factor is 0.95. Correct answer is (C).
The solution(s) of the ordinary differential equation $y'' + y = 0$, is:
(A) $\cos x$
(B) $\sin x$
(C) $1 + \cos x$
(D) $1 + \sin x$
Choose the most appropriate answer from the options given below:
For the matrix, $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the eigenvalues ($\lambda$) and eigenvectors ($X$) respectively are:
The value of $\iint_S \vec{F} \cdot \vec{N} \, ds$ where $\vec{F} = 2x^2y \hat{i} - y^2 \hat{j} + 4xz^2 \hat{k}$ and $S$ is the closed surface of the region in the first octant bounded by the cylinder $y^2 + z^2 = 9$ and the planes $x = 0, x = 2, y = 0, z = 0$, is:
The value of the integral $\displaystyle \oint_C \frac{z^3 - 6}{2z - i} \, dz$, where $C: |z| \leq 1$, is: