Question 1 |
The sum of two normally distributed random variables X and Y is
always normally distributed | |
normally distributed, only if X and Y are independent | |
normally distributed, only if X and Y have the same standard deviation | |
normally distributed, only if X and Y have the same mean |
Question 1 Explanation:
\begin{aligned} X_{1} &\sim N\left(\mu_{1}, \sigma_{1}\right)\\ \text{and }\quad X_{2} &\sim N\left(\mu_{2}, \sigma_{2}\right)\\ \text{then }\quad X_{1}+X_{2} &\sim N\left(\mu_{1}+\mu_{2}, \sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}\right) \end{aligned}
Always normally distributed.
Always normally distributed.
Question 2 |
A matrix P is decomposed into its symmetric part S and skew symmetric part V. If
S=\begin{pmatrix} -4 &4 &2 \\ 4& 3 & 7/2\\ 2& 7/2 & 2 \end{pmatrix}, V=\begin{pmatrix} 0 &-2 &3 \\ 2& 0 & 7/2\\ -3& -7/2 & 0 \end{pmatrix},
Then matrix P is
S=\begin{pmatrix} -4 &4 &2 \\ 4& 3 & 7/2\\ 2& 7/2 & 2 \end{pmatrix}, V=\begin{pmatrix} 0 &-2 &3 \\ 2& 0 & 7/2\\ -3& -7/2 & 0 \end{pmatrix},
Then matrix P is
A | |
B | |
C | |
D |
Question 2 Explanation:
\begin{array}{l} S=\frac{P+P^{T}}{2} \\ V=\frac{P-P^{T}}{2} \\ P=S+V=\left(\begin{array}{ccc} -4 & 2 & 5 \\ 6 & 3 & 7 \\ -1 & 0 & 2 \end{array}\right) \end{array}
Question 3 |
Let I=\int_{x=0}^{1}\int_{y=0}^{x^2}xy^2dydx then, I may also be expressed as
I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}xy^2dxdy | |
I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}yx^2dxdy | |
I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}xy^2dxdy | |
I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}yx^2dxdy |
Question 3 Explanation:
I=\int_{0}^{1} \int_{0}^{x^{2}} x y^{2} d y d x
Change on rules, I=\int_{y=0}^{1} \int_{x=\sqrt{y}}^{1} x y^{2} d x d y
Change on rules, I=\int_{y=0}^{1} \int_{x=\sqrt{y}}^{1} x y^{2} d x d y
Question 4 |
The solution of
\frac{d^2y}{dt^2}-y=1,
which additionally satisfies y|_{t=0}=\left.\begin{matrix} \frac{dy}{dt} \end{matrix}\right|_{t=0}=0 in the Laplace s-domain is
\frac{d^2y}{dt^2}-y=1,
which additionally satisfies y|_{t=0}=\left.\begin{matrix} \frac{dy}{dt} \end{matrix}\right|_{t=0}=0 in the Laplace s-domain is
\frac{1}{s(s+1)(s-1)} | |
\frac{1}{s(s+1)} | |
\frac{1}{s(s-1)} | |
\frac{1}{s-1} |
Question 4 Explanation:
\begin{aligned} y^{\prime\prime}-y &=1 \\ y(0) &=1 \\ y^{\prime}(0) &=1 \\ L\left\{y^{\prime\prime}-y\right\} &=L\{1\} \\ s^{2} Y(s)-s y(0)-y^{\prime}(0)-y(s) &=\frac{1}{s} \\ y(s) &=\frac{1}{s\left(s^{2}-1\right)}\\ & =\frac{1}{s(s+1)(s-1)} \end{aligned}
Question 5 |
An attempt is made to pull a roller of weight W over a curb (step) by applying a horizontal
force F as shown in the figure.
The coefficient of static friction between the roller and the ground (including the edge of the step) is \mu. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.
The coefficient of static friction between the roller and the ground (including the edge of the step) is \mu. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.
A | |
B | |
C | |
D |
Question 5 Explanation:
Weigh = W
Note:
(i) When the cylinder is about to make out of the curb, it will loose its contact at point A, only contact will be at it B.
(ii) At verge of moving out of curb, Roller will be in equation under W, F and contact force from B and these three forces has to be concurrent so contact force from B will pass through C.
(iii) Even the surfaces are rough but there will be no friction at B for the said condition.
FBD
There are 5 questions to complete.