# GATE Mechanical Engineering 2020 SET-2

 Question 1
The sum of two normally distributed random variables X and Y is
 A always normally distributed B normally distributed, only if X and Y are independent C normally distributed, only if X and Y have the same standard deviation D normally distributed, only if X and Y have the same mean
Engineering Mathematics   Probability and Statistics
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.
 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
 A A B B C C D D
Engineering Mathematics   Linear Algebra
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
 A $I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}xy^2dxdy$ B $I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}yx^2dxdy$ C $I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}xy^2dxdy$ D $I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}yx^2dxdy$
Engineering Mathematics   Calculus
Question 3 Explanation:
$I=\int_{0}^{1} \int_{0}^{x^{2}} x y^{2} d y d x$

Change on rules, $I=\int_{y=0}^{1} \int_{x=\sqrt{y}}^{1} x y^{2} d x d y$
 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
 A $\frac{1}{s(s+1)(s-1)}$ B $\frac{1}{s(s+1)}$ C $\frac{1}{s(s-1)}$ D $\frac{1}{s-1}$
Engineering Mathematics   Differential Equations
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.
 A A B B C C D D
Engineering Mechanics   FBD, Equilirbium, Plane Trusses and Virtual work
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.