Question 1 |
The product of matrices (PQ)^{-1}P is
P^{-1} | |
Q^{-1} | |
P^{-1}Q^{-1}P | |
PQP^{-1} |
Question 1 Explanation:
\begin{aligned} \left ( PQ \right )^{-1}P&=\left (Q^{-1}P^{-1} \right )P \\ &=\left ( Q^{-1}\ \right )\left ( P^{-1}P \right ) \\ &=\left ( Q^{-1} \right )\left ( I \right )=Q^{-1} \end{aligned}
Question 2 |
The general solution of \frac{d^{2}y}{dx^{2}}+y=0 is
y = P \cos x + Q \sin x | |
y = P \cos x | |
y = P \sin x | |
y = P \sin^2 x |
Question 2 Explanation:
\begin{aligned} \frac{d^{2}y}{dx^{2}}+y&=0 \\ D^{2}+1&=0 \\ D=\pm i &=0\pm 1i \\ &\text{General solution is,} \\ y&=e^{0x}\left [ C_{1}\cos \left ( 1\times x \right )+C_{2}\sin \left ( 1\times x \right ) \right ] \\ &=C_{1}\cos x+C_{2}\sin x \\ &=P\cos x+Q\sin x \end{aligned}
where P and Q are some constants.
where P and Q are some constants.
Question 3 |
A mild steel specimen is under uniaxial tensile stress. Young's modulus and yield
stress for mild steel are 2 \times 10^{5} MPa and 250 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen
without permanent set is
156 Nmm/mm^{3} | |
15.6 Nmm/mm^{3} | |
1.56 Nmm/mm^{3} | |
0.156 Nmm/mm^{3} |
Question 3 Explanation:
The strain energy per unit volume may be given as
\begin{aligned} u &=\frac{1}{2} \times \frac{\sigma_{y}^{2}}{E}=\frac{1}{2} \times \frac{(250)^{2}}{2 \times 10^{5}} \\ &=0.156 \mathrm{N}\; \mathrm{mm} / \mathrm{mm}^{3} \end{aligned}
\begin{aligned} u &=\frac{1}{2} \times \frac{\sigma_{y}^{2}}{E}=\frac{1}{2} \times \frac{(250)^{2}}{2 \times 10^{5}} \\ &=0.156 \mathrm{N}\; \mathrm{mm} / \mathrm{mm}^{3} \end{aligned}
Question 4 |
A reinforced concrete structure has to be constructed along a sea coast. The
minimum grade of concrete to be used as per IS : 456-2000 is
M15 | |
M20 | |
M25 | |
M30 |
Question 4 Explanation:
As per clause 8.2.8 of 1S 456: 2.000 concrete in
sea water or exposed directly along the sea coast
shall be atleast M20 grade in the case of plain
concrete and M30 in case of reinforced concrete.
Question 5 |
In the design of a reinforced concrete beam the requirement for bond is not
getting satisfied. The economical option to satisfy the requirement for bond is by
bounding of bars | |
providing smaller diameter bars more in number | |
providing larger diameter bars less in number | |
providing same diameter bars more in number |
Question 5 Explanation:
Bond stress \left(\tau_{b d}\right)=\frac{\text { Tensile force }}{(n \pi \phi) \sigma_{s t}}
\tau_{b d} should be less than permissible value, if it is greater than \left(\tau_{b d}\right)_{\text {permissible }} then best economical solution is to reduce the diameter of bar and increase its number.
\tau_{b d} should be less than permissible value, if it is greater than \left(\tau_{b d}\right)_{\text {permissible }} then best economical solution is to reduce the diameter of bar and increase its number.
There are 5 questions to complete.