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Explanation:

As per given data:

The distance traveled by each vehicle

Vehicle (1) ends at about 100 meters.

Vehicle (2) ends at about 250 meters.

Vehicle (3) ends at about 400 meters.

Vehicle (4) ends at about 400 meters.

Vehicle (5) ends at about 350 meters.

Vehicle (6) ends at about 150 meters.

The mean speed is given by

\(\text { Mean speed } =\frac{\text { Total distance travelled }}{\text { Total time taken }} \)

Mean Speed \(=\frac{100+250+400+400+350+150}{10+15+30+25+15+10}\)

Mean Speed = 15.714 m/sec = 56.57 km/h ≃ 57 km/h

Explanation:

As per given data:

Initial cost = 100000

Salvage value = 10,000

Accounting life = 5 years

(i) Depreciation using straight-line method

Annual depreciation 

\(=\frac{\text { Initial cost }- \text { Salvage value }}{\text { Accounting life or Service life }} \)

\(=\frac{100000-10000}{5}=18000\)

(ii) Depreciation using double declining balance method

In this method, the asset is assumed to loose its value by a fixed factor of the book value i.e. double the straight line rate.

Fixed factor for double declining balance method = FDDB

\(\rm \text { FDDB }=\frac{2}{n} \), here n = 5 years

\(\rm \therefore \quad F D D B=\frac{2}{5}=0.4\)

Depreciation for the first year = FDDB × Initial cost

= 0.4 × 100000 = 40000

Book value at the end of 1^st year = Initial cost – Depreciation

= 100000 – 40000 = 60000

Depreciation for 2^nd year = 60000 × 0.4 = 24000

Difference in depreciation between double declining balance and straight-line method

= 24000 – 18000 = 6000

Explanation:

As per given data:

Discharge = 7.5 m³/sec/m

Flow depth at downstream horizontal apron (y₁) = 0.5 m.

\(\rm v_{1}=\frac{q}{y_{1}}=\frac{7.5}{0.5}=15 \mathrm{~m} / \mathrm{s}\)

Initial Froude number (Fr₁)

= \(\rm \frac{v_{1}}{\sqrt{g y_{1}}}=\frac{15}{\sqrt{9.81 \times 0.5}}\)

= 6.7729 > 1

Let y₂ be the sequent depth corresponding to initial depth y₁ = 0.5 m

\(\frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=\frac{1}{2}\left[\sqrt{1+8 \mathrm{Fr}_{1}^{2}-1}\right] \)

or \(\frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=\frac{1}{2}\left[\sqrt{1+8 \times 6.7729^{2}}-1\right] \)

or \(\frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=9.0914\) 

∴y₂ = 0.5 × 9.0914 = 4.5457

So, tail water depth after rounding off 2 decimal places = 4.55 m.

Explanation:

Given Data:

Initial height of the bird at point P = 8 m above MSL.

Final height of the bird at point Q = 3 m above MSL.

Ground surface gradient = 1 in 2.

Step 1: Determine the change in height

Calculate the height difference between points P and Q:

Height difference = Height at P - Height at Q = 8 m - 3 m = 5 m.

Step 2: Calculate the horizontal distance using the gradient

Using the gradient of 1 in 2, where for every 1 m drop in height, the horizontal distance is 2 m:

Horizontal distance = Height difference × Gradient = 5 m × 2 = 10 m.

Conclusion: The horizontal distance between points P and Q where the bird flies parallel to the ground is 10 meters.

Explanation:

Explanation:

As per Given data:

\(\rm \vec{v}=3 \hat{i}+(5 x) \hat{j}\)

Here, u = 3, v = (5x)

Check for incompressibility 

\(\rm \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=\frac{\partial(3)}{\partial x}+\frac{\partial(5 x)}{\partial y}=0\)

Since, \(\rm \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\)

Fluid is incompressible.

Check for rotationality

\(\rm \omega_{z}=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right]\)

\(\rm =\frac{1}{2}\left[\frac{\partial(5 x)}{\partial x}-\frac{\partial(3)}{\partial y}\right] \)

\( =\frac{1}{2}[5-0]=2.5 \neq 0\)

So, flow is rotational.

Solution:

To verify the given statements about vector fields and scalar fields:

Given Data:

Scalar field \( r \) and vector field \( \mathbf{p} \) are defined as follows:

\( r = 6x^2 + 4y^2 - z^2 - 9yz - 2xy + 3xz - yz \)

\( \mathbf{p} = (2x^2 - 3xy + z^2) \mathbf{i} + (2y^2 - 3yz + x^2) \mathbf{j} + (2z^2 - 3xz + x^2) \mathbf{k} \)

Step 1: Analyze Statement P (Curl of the gradient of the scalar field \( r \) is a null vector)

Compute the gradient of scalar field \( r \):

\( \nabla r = (12x - 2y + 3z) \mathbf{i} + (8y - 10z - x) \mathbf{j} + (-2z - 9y + 3x) \mathbf{k} \)

Compute the curl of this gradient vector:

\( \nabla \times (\nabla r) = (0, 0, 0) \) (Since the curl of any gradient is always zero.)

Statement P is confirmed true.

Step 2: Analyze Statement Q (Divergence of curl of the vector field \( \mathbf{p} \) is zero)

Compute the curl of vector field \( \mathbf{p} \):

\( \nabla \times \mathbf{p} = (3y, 5z, 4x) \)

Compute the divergence of this curl vector:

\( \nabla \cdot (\nabla \times \mathbf{p}) = 0 \) (As the divergence of any curl is always zero.)

Statement Q is confirmed true.

Conclusion: Both statements P and Q are true based on vector calculus identities.

Correct Option: Both P and Q are true.

Explanation:

Given data

Emission standard (based on 11% oxygen)

For HCl = 50 mg/Nm³

The measured concentration of HCl and oxygen in flue gas are 42 mg/Nm³ and 13% respectively.

As per CPCB,

The calculated emission concentration at the standard percentage oxygen concentration (ES)

\(\mathrm{ES}=\frac{21-\mathrm{OS}}{21-\mathrm{OM}} \times \mathrm{EM}\)

EM = Measured emission concentration at the standard percentage oxygen concentration

OS = Standard oxygen concentration

OM = Measured oxygen concentration

So, 

\(\mathrm{ES}=\frac{21-11}{21-13} \times 42=52.5 \mathrm{mg} / \mathrm{Nm}^{3}>50 \mathrm{mg} / \mathrm{Nm}^{3}\)

Here, no compliance as the corrected HCl emission is greater than the emission standard.

Explanation:

The resultant stress at the bottom fibre in beam B,

\(\rm =\frac{P}{A}+\frac{P e}{Z_{b}}-\frac{M_{1}}{Z_{b}}\)

\(\rm =\frac{1000 \times 10^{3}}{A}+\frac{1000 \times 10^{3}}{Z_{b}}-\frac{M_{1}}{Z_{b}}\)

For no crack, stress at bottom = f_cr

\(\rm \Rightarrow \frac{1000 \times 10^{3}}{A}+\frac{1000 \times 10^{3} \times 100}{Z_{b}}-\frac{M_{1}}{Z_{b}}=-f_{c r} \quad \ldots \text { (i) }\)

The resultant stress at the bottom fibre in beam B₂ i.e.

when no prestressing force is here = \(\rm -\frac{M_2}{Z_b}\)

For no crack, stress at bottom = f_cr

\(\Rightarrow-\frac{\mathrm{M}_{2}}{\mathrm{Z}_{\mathrm{b}}}=-\mathrm{f}_{\mathrm{cr}} \ldots \text { (ii) }\)

Also, for beam B₁, without the moment M₁

\(\rm \frac{P}{A}-\frac{P e}{Z_{t}}=0 \)

\(\rm \Rightarrow \frac{1}{A}-\frac{e}{Z_{t}}=0\)

\(\rm \Rightarrow \frac{1}{A}=\frac{e}{Z_{t}} \)

\(\rm \Rightarrow A=\frac{Z_{t}}{e} \)

\(\rm \frac{Z_{b}}{Z_{t}}=\frac{\left(l / y_{b}\right)}{\left(1 / y_{t}\right)}=\frac{y_{t}}{y_{b}}=\frac{2 H}{3(H / 3)}=2 \)

\(\rm \Rightarrow Z_{t}=0.5 Z_{b} \)

\(\rm \Rightarrow A=\frac{0.5 Z_{b}}{e}\)

Adding equation (i) and (ii) 

\(\rm \Rightarrow \frac{1000 \times 10^{3}}{A}+\frac{1000 \times 10^{3} \times 100}{Z_{b}}-\frac{M_{1}}{Z_{b}}+\frac{M_{2}}{Z_{b}}=0 \)

\(\rm \Rightarrow \frac{1000 \times 10^{3} \times 100}{0.5 Z_{b}}+\frac{1000 \times 10^{3}}{Z_{b}}-\frac{\left(M_{1}-M_{2}\right)}{Z_{b}}=0\)

⇒ M₁ – M₂ = 300 kN.m

Explanation:

As per given data

cot θ = 1.5

θ = 0.588 rad

Area (A) = y²(θ + cot θ)

⇒ A = 2.088y²

p = 2y(θ + cot θ)

⇒ R = \(\rm \frac{y}{2}\)

The chezy’s equation is given by: 

\(\rm Q =A V=A \times C \sqrt{R S} \)

⇒ \(\rm 20 =2.088 y^{2} \times 150 \sqrt{\frac{y}{2} \times \frac{1}{200}} \)

⇒ \(\rm y^{5 / 2} =\frac{20 \times 2 \times 10}{2.088 \times 150}\)

⇒ y = 1.102m