Question 1 |
The Fourier cosine series for an even function f(x) is given by
f(x)=a_{0}+\sum_{n=1}^{\infty}a_{n}\cos(nx)
The value of the coefficient a_{2} for the function f(x)=\cos ^{2}(x) in \left [ 0 , \pi \right ] is
f(x)=a_{0}+\sum_{n=1}^{\infty}a_{n}\cos(nx)
The value of the coefficient a_{2} for the function f(x)=\cos ^{2}(x) in \left [ 0 , \pi \right ] is
-0.5 | |
0 | |
0.5 | |
1 |
Question 1 Explanation:
\begin{aligned} \cos ^{2} x &=\frac{1+\cos 2 x}{2} \\ f(x) &=\frac{1}{2}+\frac{\cos 2 x}{2} \\ f(x) &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cdot \cos n x \\ a_{0} &=1 \\ a_{1} &=0 \\ a_{2} &=\frac{1}{2} \end{aligned}
Question 2 |
The divergence of the vector field \overrightarrow{u}=e^{x}\left ( \cos y\hat{i} + \sin y\hat{j} \right) is
0 | |
e^{x}\cos y + e^{x}\sin y | |
2e^{x}\cos y | |
2e^{x}\sin y |
Question 2 Explanation:
\begin{aligned} \vec{u} &=e^{x} \cos y \hat{i}+e^{x} \cdot \sin y \hat{j} \\ \nabla \cdot \vec{u} &=\frac{\partial}{\partial x}\left(u_{1}\right)+\frac{\partial}{\partial y}\left(u_{2}\right) \\ &=\frac{\partial}{\partial x}\left(e^{x} \cdot \cos y\right)+\frac{\partial}{\partial y}\left(e^{x} \cdot \sin y\right) \\ &=e^{x} \cos y+e^{x} \cos y \\ \nabla \cdot \vec{u} &=2 e^{x} \cdot \cos y \end{aligned}
Question 3 |
Consider a function u which depends on position x and time t. The partial differential
equation
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}
is known as the
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}
is known as the
Wave equation | |
Heat equation | |
Laplace's equation | |
Elasticity equation |
Question 3 Explanation:
\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}is known as heat equation
Question 4 |
If y is the solution of the differential equation
y^{3}\frac{d y}{dx}+x^{3}=0,y(0)=1, the value of y\left ( -1 \right ) is
y^{3}\frac{d y}{dx}+x^{3}=0,y(0)=1, the value of y\left ( -1 \right ) is
-2 | |
-1 | |
0 | |
1 |
Question 4 Explanation:
\begin{aligned} y^{3} \frac{d y}{d x} &=-x^{3} \\ y^{3} d y &=-x^{3} d x \\ \int y^{3} d y &=-\int x^{3} d x \\ \frac{y^{4}}{4} &=\frac{-x^{4}}{4}+C \\ \frac{x^{4}+y^{4}}{4} &=C \\ y(0) &=1 \\ \frac{0+1}{4} &=C \\ C &=\frac{1}{4} \\ x^{4}+y^{4} &=1 \\ y^{4} &=1-x^{4} \\ y &=\sqrt[4]{1-x^{4}} \\ \text{When,}\quad x &=-1 \\ y &=0 \end{aligned}
Question 5 |
The minimum axial compressive load, P, required to initiate buckling for a pinned-pinned slender column with bending stiffness EI and length L is
P=\frac{\pi ^{2}EI}{4L^{2}} | |
P=\frac{\pi ^{2}EI}{L^{2}} | |
P=\frac{3\pi ^{2}EI}{4L^{2}} | |
P=\frac{4\pi ^{2}EI}{L^{2}} |
Question 5 Explanation:
For both ends hinged buckling load,
P=\frac{\pi^{2} E I}{L^{2}}
P=\frac{\pi^{2} E I}{L^{2}}
Question 6 |
A frictionless gear train is shown in the figure. The leftmost 12-teeth gear is given a torque of 100 N-m. The output torque from the 60-teeth gear on the right in N-m is
5 | |
20 | |
500 | |
2000 |
Question 6 Explanation:
\begin{aligned} \tau_{1}&=100 \mathrm{Nm}\\ \text{Let speed of 1 is }N_{1} \\ (1,2):\qquad N_{2}&=N_{1} \times \frac{T_{1}}{T_{2}}=N_{1} \times \frac{12}{48}=\frac{N_{1}}{4} \\ N_{3}=N_{2}&=\frac{N_{1}}{4}\\ (3,4) \qquad N_{4}&=N_{3} \times \frac{T_{3}}{T_{4}}=\frac{N_{1}}{4} \times \frac{12}{60} \\ N_{4}&=\frac{N_{1}}{20} \end{aligned}
By Power conservation
(Assume \eta (efficiency) =1)
\begin{aligned} \tau_{1} \times N_{1} &=\tau_{4} \times N_{4} \\ 100 \times N_{1} &=\tau_{4} \times \frac{N_{1}}{20} \\ \tau_{4} &=2000 \mathrm{N}-\mathrm{m} \end{aligned}
Question 7 |
In a single degree of freedom underdamped spring-mass-damper system as shown in the figure, an additional damper is added in parallel such that the system still remains underdamped. Which one of the following statements is ALWAYS true?
Transmissibility will increase. | |
Transmissibility will decrease. | |
Time period of free oscillations will increase. | |
Time period of free oscillations will decrease. |
Question 7 Explanation:
\text { Tansmissibility, } \quad \in=\frac{\sqrt{1+\left(\frac{2 \xi \omega}{\omega_{n}}\right)^{2}}}{\sqrt{\left\{1-\left(\frac{\omega}{\omega_{n}}\right)^{2}\right\}^{2}+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}}}
Additional damper is added into parallel.
Then damping will increase.
But it is still under damped.
But \xi will increase.
Here no unbalance force is there.
But \omega_{d}=\sqrt{1-\xi^{2}} \cdot \omega_{n} \quad\left\{\omega_{n}=\sqrt{\frac{K}{M}}\right\}
But as \xi increases
then \xi^{2} will increase
Then \omega_{d} will decrease
Then T_{d}=\frac{2 \pi}{\omega_{d}} \text { will increase }
Additional damper is added into parallel.
Then damping will increase.
But it is still under damped.
But \xi will increase.
Here no unbalance force is there.
But \omega_{d}=\sqrt{1-\xi^{2}} \cdot \omega_{n} \quad\left\{\omega_{n}=\sqrt{\frac{K}{M}}\right\}
But as \xi increases
then \xi^{2} will increase
Then \omega_{d} will decrease
Then T_{d}=\frac{2 \pi}{\omega_{d}} \text { will increase }
Question 8 |
Pre-tensioning of a bolted joint is used to
strain harden the bolt head | |
decrease stiffness of the bolted joint | |
increase stiffness of the bolted joint | |
prevent yielding of the thread root |
Question 8 Explanation:
Pretension increase stiffness of system..
Question 9 |
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 \mum. If the peak wavelength of emitted radiation changes to 2.90 \mum, then the temperature (in K) of the black body is
500 | |
1000 | |
4000 | |
8000 |
Question 9 Explanation:
From Wein's displacement law
For black body,
\begin{aligned} \lambda_{M} T &=\text { constant } \\ \lambda_{M 1} T_{1} &=\lambda_{M 2} T_{2} \\ 1.45 \times 2000 &=\lambda_{M 2} T_{2}=2.90 \times T_{2} \\ \therefore \quad T_{2} &=\left(\frac{1.45}{2.90} \times 2000\right)=1000 \mathrm{K} \end{aligned}
For black body,
\begin{aligned} \lambda_{M} T &=\text { constant } \\ \lambda_{M 1} T_{1} &=\lambda_{M 2} T_{2} \\ 1.45 \times 2000 &=\lambda_{M 2} T_{2}=2.90 \times T_{2} \\ \therefore \quad T_{2} &=\left(\frac{1.45}{2.90} \times 2000\right)=1000 \mathrm{K} \end{aligned}
Question 10 |
For an ideal gas with constant properties undergoing a quasi-static process, which one of
the following represents the change of entropy (\Delta s) from state 1 to 2?
\Delta s=C_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )-R ln\left ( \frac{p_{2}}{p_{1}} \right ) | |
\Delta s=C_{v}ln\left ( \frac{T_{2}}{T_{1}} \right )-C_{p} ln\left ( \frac{v_{2}}{v_{1}} \right ) | |
\Delta s=C_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )-C_{v} ln\left ( \frac{p_{2}}{p_{1}} \right ) | |
\Delta s=C_{v}ln\left ( \frac{T_{2}}{T_{1}} \right )+R ln\left ( \frac{v_{1}}{v_{2}} \right ) |
Question 10 Explanation:
\begin{array}{l}\mathrm{Tds}=\mathrm{dH}-\mathrm{VdP}\\ \mathrm{dS}=\frac{\mathrm{dH}}{\mathrm T}-\frac{\mathrm V}{\mathrm T}\mathrm{dP}\\ \mathrm{For}\;\mathrm{an}\;\mathrm{ideal}\;\mathrm{gas}\\ \mathrm{PV}=\mathrm{RT}\\ \frac{\mathrm V}{\mathrm T}=\frac{\mathrm R}{\mathrm P}\\ \mathrm{dS}={\mathrm C}_{\mathrm P} \frac{\mathrm{dT}}{\mathrm T}-\frac{\mathrm R}{\mathrm p}\mathrm{dP}\\ {\mathrm S}_2-{\mathrm S}_1={\mathrm C}_\mathrm P\ln\frac{{\mathrm T}_2}{{\mathrm T}_1}-\mathrm{Rln}\frac{{\mathrm P}_2}{{\mathrm P}_1}\\\\\end{array}
There are 10 questions to complete.