Question 1 |
Two coins are tossed simultaneously. The probability (upto two decimal points accuracy) of getting at least one head is _______
0.5 | |
0.65 | |
0.75 | |
0.8 |
Question 1 Explanation:
Total four possibilities {HH. HT, TH, TT}
The probability of getting at least one head is \frac{3}{4}.
The probability of getting at least one head is \frac{3}{4}.
Question 2 |
The divergence of the vector -yi+xj______
0 | |
1 | |
2 | |
0.5 |
Question 2 Explanation:
\begin{aligned} \vec { F } & = - y \bar { i } + x \bar { j } \\ \nabla \cdot \bar { F } & = \frac { \partial } { \partial x } ( - y ) + \frac { \partial } { \partial y } ( x ) \\ & = 0 + 0 = 0 \end{aligned}
Question 3 |
The determinant of a 2x2 matrix is 50. If one eigenvalue of the matrix is 10, the other eigenvalue is _____.
5 | |
6 | |
7 | |
8 |
Question 3 Explanation:
The product of eigen value of always equal to
the determinant value of the matrix.
\begin{aligned} \lambda_{1} &=10 \quad \lambda_{2}=\text { unknown } \quad|A|=50 \\ \lambda_{1} \cdot \lambda_{2} &=50 \\ 10\left(\lambda_{2}\right) &=50\\ \therefore \qquad \lambda_{2}&=5 \\ \end{aligned}
the determinant value of the matrix.
\begin{aligned} \lambda_{1} &=10 \quad \lambda_{2}=\text { unknown } \quad|A|=50 \\ \lambda_{1} \cdot \lambda_{2} &=50 \\ 10\left(\lambda_{2}\right) &=50\\ \therefore \qquad \lambda_{2}&=5 \\ \end{aligned}
Question 4 |
A sample of 15 data is follows: 17, 18, 17, 17, 13, 18, 5, 5, 6, 7, 8, 9, 20, 17, 3. The mode of the data is
4 | |
13 | |
17 | |
20 |
Question 4 Explanation:
Mode means highest number of observations
or occurrence of data most of the time as data
17, occurs four times, i.e., highest time. So
mode is 17.
Question 5 |
The Laplace transform of te^{t} is
\frac{s}{(s+1)^{2}} | |
\frac{1}{(s-1)^{2}} | |
\frac{1}{(s+1)^{2}} | |
\frac{s}{s-1} |
Question 5 Explanation:
f ( t ) = t e ^ { t }
L ( t ) = \frac { 1 } { s ^ { 2 } }
By first shifting rule
L\left( t e ^ { t } \right) = \frac { 1 } { ( s - 1 ) ^ { 2 } }
L ( t ) = \frac { 1 } { s ^ { 2 } }
By first shifting rule
L\left( t e ^ { t } \right) = \frac { 1 } { ( s - 1 ) ^ { 2 } }
Question 6 |
A mass m is attached to two identical springs having spring constant k as shown in the figure. The natural frequency \omega
of this single degree of freedom system is


\sqrt{\frac{2k}{m}} | |
\sqrt{\frac{k}{m}} | |
\sqrt{\frac{k}{2m}} | |
\sqrt{\frac{4k}{m}} |
Question 6 Explanation:

Equivalent stiffness
k_{e q}=k+k=2 k
Natural frequency is
\omega_{n}=\sqrt{\frac{k_{\mathrm{eq}}}{m}}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}
Question 7 |
The state of stress at a point is \sigma _{x}=\sigma _{y}=\sigma _{z}=t_{xz}=t_{zx}=t_{yz}=t_{zy}=0 and t_{xy}=t_{yx}=50MPa . The maximum normal stress (in MPa) at that point is_____.
49 | |
50 | |
55 | |
60 |
Question 7 Explanation:
Given state of stress condition indicates pure shear state of stress.
For pure shear state of stress,
Max. tensile stress = Max. comp. stress = Max. Shear stress
=\tau_{X Y}=50 \mathrm{MPa} Hence, Max. normal stress =50 \mathrm{MPa}
For pure shear state of stress,
Max. tensile stress = Max. comp. stress = Max. Shear stress
=\tau_{X Y}=50 \mathrm{MPa} Hence, Max. normal stress =50 \mathrm{MPa}
Question 8 |
For a loaded cantilever beam of uniform cross-section, the bending moment (in N.mm) along the length is M(x)=5x^{2}+10x , where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x=10 mm is
100 | |
105 | |
110 | |
115 |
Question 8 Explanation:
(\mathrm{BM})_{x-x}=M_{x-x}=5 x^{2}+10 x

\text { (S.F) }_{x-x}=F_{x-x}=?
We know that at section X-X
\begin{aligned} F_{X-X} &=\frac{d}{d x}\left[5 x^{2}+10 x\right] \\ F_{X-X} &=10 x+10 \\ \left(F_{X-X}\right)_{x=10 \mathrm{mm}} &=10(10)+10=110 \mathrm{N} \end{aligned}

\text { (S.F) }_{x-x}=F_{x-x}=?
We know that at section X-X
\begin{aligned} F_{X-X} &=\frac{d}{d x}\left[5 x^{2}+10 x\right] \\ F_{X-X} &=10 x+10 \\ \left(F_{X-X}\right)_{x=10 \mathrm{mm}} &=10(10)+10=110 \mathrm{N} \end{aligned}
Question 9 |
A cantilever beam of length L and flexural modulus EI is subjected to a point load P at the free end. The elastic strain energy stored in the beam due to bending (neglecting transverse shear) is
\frac{P^{2}L^{3}}{6EI} | |
\frac{P^{2}L^{3}}{3EI} | |
\frac{PL^{3}}{3EI} | |
\frac{PL^{3}}{6EI} |
Question 9 Explanation:
Let U is the S.E. in the beam due to B.M. (M)

U=\int_{0}^{L}\frac{(M_{x-x})^2 dx}{2(EI)_{x-x}}=\int_{0}^{L}\frac{(Px)^2 dx}{2EI_{NA}}
U=\frac{P^2}{2EI_{NA}}\int_{0}^{L}x^2dx=\frac{P^2}{2EI}\left ( \frac{L^3}{3} \right )
U=\frac{P^2L^3}{6EI_{NA}}

U=\int_{0}^{L}\frac{(M_{x-x})^2 dx}{2(EI)_{x-x}}=\int_{0}^{L}\frac{(Px)^2 dx}{2EI_{NA}}
U=\frac{P^2}{2EI_{NA}}\int_{0}^{L}x^2dx=\frac{P^2}{2EI}\left ( \frac{L^3}{3} \right )
U=\frac{P^2L^3}{6EI_{NA}}
Question 10 |
A steel bar is held by two fixed supports as shown in the figure and is subjected to an increases of temperature \Delta T=100^{\circ}C.If the coefficent of thermal exapnasion and young's modules of elasticity of steel are 11\, \times \, 10^{-6} \, /^{\circ}C and 200GPa, respectively, the magnitude of thermal stress (in MPa) induced in the bar is _____


220 | |
225 | |
230 | |
235 |
Question 10 Explanation:
We know that for completely restricted expansion, thermal stress developed in bar is given by
\begin{aligned} \sigma_{t h} &=\alpha(\Delta T) E \\ &=11 \times 10^{-6} \times 100 \times 200 \times 10^{3}\\ &=220 \mathrm{MPa}\text{(comp. in native)} \end{aligned}
\begin{aligned} \sigma_{t h} &=\alpha(\Delta T) E \\ &=11 \times 10^{-6} \times 100 \times 200 \times 10^{3}\\ &=220 \mathrm{MPa}\text{(comp. in native)} \end{aligned}
There are 10 questions to complete.