CE

div(\phi \mathrm{u})=\phi div(\mu)+ugrad(\phi)

Option (B) is correct.

Since 30 \% of the total vehicles are without pollution clearance certificate.
Out of the 100 chosen vehicle, 30 \% i.e. 100 \times 0.3=30 vehicle are expected to be without pollution clearance certificate.

\phi_{\mathrm{BA}}=\frac{\text { T.L. }}{\mathrm{GJ}}=\frac{(100)^{*} 5}{50000}=0.01 \mathrm{rad}
\Rightarrow Torsional strain energy (U)
\begin{aligned} & U=\frac{T^{2} L}{2 G J}=\frac{1}{2} \times T * \phi_{B A} \\ & U=\frac{1}{2} * 100 * 0.01=0.5 \mathrm{kN}-\mathrm{m} \end{aligned}
Hence, statement (i) correct and statement (ii) is incorrect.

In M20, M refers to mix and 20 to characteristic cube strength. As per clause no. 6.1.1, IS456: 2000 characteristic strength is defined as the strength below which not more than 5 percent of the test results are expected to fall.
Hence, correct option is (C).

\begin{aligned} I & =\int_{-1}^{1} \frac{1}{x^{2}} d x \\ & =2 \int_{0}^{1} \frac{1}{x^{2}} d x \quad(\because \quad f(-x)=f(x)) \\ & =2 \lim _{t \rightarrow 0^{+}} \int_{t}^{1} \frac{d x}{x^{2}} \\ & =2 \lim _{t \rightarrow 0^{+}}\left(\frac{-1}{x}\right)_{t}^{1} \\ & =-2 \lim _{t \rightarrow 0^{+}}\left(1-\frac{1}{t}\right) \\ & =2 \lim _{t \rightarrow 0^{+}}\left(\frac{1}{t}-1\right) \\ & =2 \lim _{h \rightarrow 0^{+}}\left(\frac{1}{0+h}-1\right)\\ & =2(\infty-1) \\ & =\infty \;\;\; (Divergent) \end{aligned}

\sigma_{A B}=\frac{P_{A B}}{A_{A B}}=\frac{250 \times 10^{3}}{100 \times 100}=25 \mathrm{~N} / \mathrm{mm}^{2}
\sigma_{B C}=\frac{P_{B C}}{A_{B C}}=\frac{50 \times 10^3}{50 \times 50}=20 \mathrm{N} / \mathrm{mm}^{2}
\sigma_{\max }=\sigma_{\mathrm{AB}}=25 \mathrm{~N} / \mathrm{mm}^{2}

Under sustained compressive loading, deformation in concrete increases with time even through the applied stress level is not changed. The time dependent component of strain is called creep.
Hence, correct answer is (A).

Given data,
Grade of concrete M-20
Grade of steel Fe-415
Balanced section, singly reinforced beam.
As per Clause No. 38.1, IS 456: 2000,
Maximum strain in concrete at the outermost compression fibre =0.0035
and strain in the tension reinforcement for balanced section at ultimate state under flexure
\begin{aligned} & =0.002+\frac{f_{y}}{1.15 E_{s}} \\ & =0.002+\frac{415}{1.15 \times 2 \times 10^{5}}=0.0038 \end{aligned}
Hence, correct answer is (A).

As the water-cement ratio increases, the porosity in the hardened concrete increases and hence the strength decreases.
Also, as water-cement ratio increases, ducts higher water availability, the workability increases.
Hence, correct answer is (A).

As the water-cement ratio increases, the porosity in the hardened concrete increases and hence the strength decreases.
Also, as water-cement ratio increases, ducts higher water availability, the workability increases.
Hence, correct answer is (A).

Given that,
Coarse aggregate = 260 gms
Fine aggregate = 240 gms
G_{CA}=2.6
G_{FA}=2.4
Bulk specific gravity G_m=2.3
Percentage air voids in the mix = ?
G_t (Theoretical specific gravity)

\begin{aligned} &=\frac{\Sigma W}{\Sigma \frac{W}{G}}\\ &=\frac{260+240}{\frac{260}{2.6}+\frac{240}{2.4}}\\ &=2.5 \end{aligned}
\begin{aligned} % \text{ air voids} (V_V)&=\frac{G_t-G_m}{G_t} \times 100\\ &=\frac{2.5-2.3}{2.3}\times 100\\ V_V&=8% \end{aligned}

C_3S- Responsible for early age strength
C_2S - Responsible for later age strength and lowest heat of hydration
C_3A- Flash setting and highest heat of hydration

Pozzolanic materials have no cementing properties itself but have the property of combining with lime to produce stable compound.
Pozzolana is considered as siliceous and aluminous materials and when added in cement it have SiO_2 and Al_2O_3 form.
So, pozzolanic reaction :
H_2O + Reactive slilica-di-oxide + H_2O \rightarrow C-S-H gel or tobermonite gel

Minimum force for sliding
(P_{min})_{sliding}=(f_s)_{max} ...(i)
Applying equilibrium equation in vertical direction
Normal reaction = Weight
N=mg=25 kN ...(ii)
Using equation (i) and (ii)
(P_{min})_{sliding}=\mu N =0.3 \times 25=7.5 kN
Minimum force for overturning

At the verge of overturning
(P_{min})_{oberturning} \times 2=W \times 2
(P_{min})_{oberturning}=\frac{25 \times 0.5} {2}=6.25 kN
Here, (P_{min})_{oberturning} \lt (P_{min})_{sliding}
First overtuning will take place.
Sliding will not take place.

According to given condition given function f(x,y) is nothing but constant function i.e. f(x,y)=3 because this is the only function whose value is 3 at any point on the boundary of unit circle and it is also satisfying Laplace equation, so
f(0,0)=3

MTA- Marks to All
I=\int \left ( x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\infty \right )dx
I=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+...
Option (A)
\frac{1}{1+x}=(1+x)^{-1}=1-x+x^2-x^3...\infty
So, its incorrect.
Option (B)
\frac{1}{1+x^2}=(1+x^2)^{-1}=1-x^2+x^4-x^6...\infty
So, its incorrect.
Similarly option (C) and (D) both are incorrect.
No-correct choice given.

E=2G(1+\mu )
G= Shear modulas
\mu =Poission's ratio
E= Young's modulus

The question is based on LSM design principle as it is describing different conditions related to strain
Depending on amount of reinforcement in a cross- section, here ca be three types of sections viz. balanced, under reinforced and over reinforced.
Balanced section is a section that is expected to result in a balanced failure. It means at the ultimate limit state in flexure, the concrete will attain a limiting compressive strain of 0.0035 and steel will attain minimum specified tensile strain of 0.002+\frac{0.87f_y}{E_s}
Under reinforced section is a section in which steel yield before collapse. Over reinforced section is a section in which crushing of concrete in compression i.e. attainment of compressive strain of 0.0035 occurs prior to yielding of steel.
In case of over reinforced section the deflection, crack width remain relatively low and failure occurs without any sign of warning and hence over reinforced flexural members are not recommended by IS code.
Based on the above information:
Statement I is true.
Statement II is true.
Statement III is false.
Statement IV is false.

C_3S- Responsible for early age strength
C_2S - Responsible for later age strength and lowest heat of hydration
C_3A- Flash setting and highest heat of hydration

\begin{aligned} z&=\sin(y+it)+ \cos (y-it)\\ \frac{\partial z}{\partial y}&=\cos (y+it)-\sin (y-it)\\ \frac{\partial ^2 z}{\partial ^2 y^2}&=-\sin(y+it)- \cos (y-it)\\ \frac{\partial ^2 z}{\partial ^2 y^2}&=-z \;\;...(i)\\ \frac{\partial z}{\partial t}&=i \cos (y+it)+i\sin (y-it)\\ \frac{\partial ^2 z}{\partial ^2 t^2}&=+\sin(y+it)+ \cos (y-it)\\ \frac{\partial ^2 z}{\partial ^2 t^2}&=z\;\;...(ii)\\ &\text{Adding (i) and (ii)}\\ &\frac{\partial ^2 z}{\partial ^2 y^2}+\frac{\partial ^2 z}{\partial ^2 t^2}=0 \end{aligned}

\frac{d^3y}{dx^3}+x\left ( \frac{dy}{dx} \right )^{3/2}+x^2y=0
Power of \left ( \frac{dy}{dx} \right ) is fractional, make it integer.
\frac{d^3y}{dx^3}+x^2y=-x\left ( \frac{dy}{dx} \right )^{3/2}
\left (\frac{d^3y}{dx^3}+x^2y \right )^2=x^2\left ( \frac{dy}{dx} \right )^{3}
Now order = 3 and degree = 2

Given,
Hoop stress (\sigma _h)=\frac{pd}{2t}=30MPa
Maximum shear stress in plane (\tau _{max})_{\text{in plane}}=\frac{\frac{pd}{2t}-\frac{pd}{4t}}{2}=7.5MPa

This is a question of working stress method i.e. elastic theory.
Modular ratio
=\frac{E_s}{E_c}=\frac{\text{Young's modulus of steel}}{\text{Young's modulus of concrete}}

Pozzolanic materials have no cementing properties itself but have the property of combining with lime to produce stable compound.
Pozzolana is considered as siliceous and aluminous materials and when added in cement it have SiO_2 and Al_2O_3 form.
So, pozzolanic reaction :
H_2O + Reactive slilica-di-oxide + H_2O \rightarrow C-S-H gel or tobermonite gel

Velocity, V=\omega r.
Here, \omega= constant.
Hence, more is the distance from the axis of rotation more will be the velocity.
\therefore \quad \mathrm{V}_{\mathrm{P}} \gt \mathrm{V}_{\mathrm{R}} \gt \mathrm{V}_{\mathrm{Q}}

4^{8^{x}}=256
\rightarrow Taking log to the base 4 . on both side.
8^{x}=\log _{4} 256=4
Taking lot to the base 8 on both sides, we get
\begin{aligned} x & =\log _{8} 4 \\ & =\log _{2^{3}} 2^{2} \\ x & =\frac{2}{3} \end{aligned}

Let us form the pairs of Husband-wife. Now these pairs can be arranged around circular table in
\begin{aligned} & =(3-1) ! \text { ways } \\ & =2 \text { ways } \end{aligned}
Their possible internal arrangement \mathrm{s}
\begin{aligned} & =2 ! \times 2 ! \times 2 ! \\ & =8 \end{aligned}
Hence, total seating arrangement.
=2 \times 8=16

Often and seldom are antonyms. Hence, correct option will be the antonym of kind i.e. Cruel.

Velocity, V=\omega r.
Here, \omega= constant.
Hence, more is the distance from the axis of rotation more will be the velocity.
\therefore \quad \mathrm{V}_{\mathrm{P}} \gt \mathrm{V}_{\mathrm{R}} \gt \mathrm{V}_{\mathrm{Q}}

4^{8^{x}}=256
\rightarrow Taking log to the base 4 . on both side.
8^{x}=\log _{4} 256=4
Taking lot to the base 8 on both sides, we get
\begin{aligned} x & =\log _{8} 4 \\ & =\log _{2^{3}} 2^{2} \\ x & =\frac{2}{3} \end{aligned}

Let us form the pairs of Husband-wife. Now these pairs can be arranged around circular table in
\begin{aligned} & =(3-1) ! \text { ways } \\ & =2 \text { ways } \end{aligned}
Their possible internal arrangement \mathrm{s}
\begin{aligned} & =2 ! \times 2 ! \times 2 ! \\ & =8 \end{aligned}
Hence, total seating arrangement.
=2 \times 8=16

Using R_{2} \rightarrow R_{2} \rightarrow 3 R_{1}, R_{3} \rightarrow R_{3}-5 R_{1}, R_{4} \rightarrow R_{4}-7 R_{1}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{array}\right]
Using R_{3} \rightarrow R_{3}-2 R_{2}, R_{4} \rightarrow R_{4}-3 R_{2}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]
So, \rho(A)= No. of non-zero rows = 2.

\begin{array}{l} \quad P Q=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \\ (P Q)^{\top}=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \end{array}
Now using Reversal law
Q^{\top} P^{\top}=(P Q) T=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]

PDF:f(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-(x-\mu )^2/(2\sigma ^2)}
CDF:F(x)=\frac{1}{2}\left [ 1+eff\left ( \frac{x-\mu }{\sigma \sqrt{2}} \right ) \right ]

As per Muller Breslau principle ILD for stress function (shear -V_{G}) will be a combination of curves (3^{\circ} curves).

The Gypsum is added to cement at the end of grinding clinker it is added to prevent quick setting.