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Explanation:

The sum of the interior angle is given by

= (2n – 4) × 90

For 3 sides, the sum of the interior angle is

= (2×3 – 4) × 90 = 180°

For 4 sides, the sum of the interior angle is

= (2×4 – 4) × 90 = 360°

Hence, the sum of the interior angles increases by 

= 360° - 180° = 180°

Explanation:

As per given equation

\(\rm \frac{\partial^{2} u}{\partial x^{2}}=2\) …(i)

\(\rm \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)=2\)

Integrating both sides with respect to x

\(\rm \frac{\partial u}{\partial x}=2 x+f(y)\)

Again integrating with respect to x

u = x² + xf(y) + f(y)

u = x² + xf(y) + h(y)

Explanation:

As per given data:

From the inextensibility of BF & BA,

ΔB_x = 0 ΔB_y = 0

From the inextensibility of BE and EF,

ΔΕ_x = 0 ΔΕ_y =0

From the inextensibility of CE and BC

ΔC_x = 0 ΔC_y = 0

From the inextensibility of BD and CD,

ΔD_x = 0 ΔD_y = 0

From the inextensibility of DG,

ΔG_x = 0

From the inextensibility of EH and HF,

ΔΗ_x = 0 ΔΗ_y = 0

From inextensibility FI,

ΔI_x = 0

⇒ If at joint (F), FB, FE, FH, FI are rigidly connected then possible displacements at F = θ_F.

If at joint I, FI and IH are rigidly connected, possible displacements at I = θ_I. 

Hence unknown joint displacements are

θ_B, θ_C, θ_D, θ_E, θ_F, θ_G, Δ_Gy, θ_H, θ_I

⇒ D_k = 9

However, if somebody assumes all members at joint (F) to be connected with pin then at (F) we have unknown joint displacements as θ_FB, θ_FE, θ_FH, θ_FI.

Similarly, if at joint I if somebody assumes FI and HI to be pin connected then unknown joint displacements at I are θ_IF, θ_IH.

Hence, D_k will increase by 4.

⇒ Dk = 9 + 4 = 13

Explanation:

The degrees of freedom (DOF) in an open channel flow system refer to the number of independent variables required to describe the flow conditions.

For a natural open channel flow with a mobile bed, the key parameters influencing the system are:

1. Flow depth (y)
2. Velocity (V)
3. Slope of the bed (S)
4. Sediment concentration (C)

Since these four variables are independently adjustable in a mobile bed channel, the number of degrees of freedom is 4.

Explanation:

Ductility Test (True)

• The ductility test measures the ability of bitumen to stretch under tensile stress, assessing adhesiveness and flexibility.

Flash Point (True)

• The flash point is the minimum temperature at which bitumen vapors ignite momentarily when exposed to a flame.

Kinematic Viscosity (True)

• Kinematic viscosity represents the resistance to flow under gravity, crucial for evaluating bitumen’s workability at different temperatures.

Softening Point of Bitumen (False)

• Softer grade bitumen has a lower softening point, meaning it softens at a lower temperature compared to harder grades.
• Harder bitumen has a higher softening point, making it more resistant to deformation at high temperatures.
• Thus, this statement is incorrect.

Explanation:

Carbon (C) (Increases Strength)

• Strengthens steel by increasing hardness and tensile strength.
• However, excessive carbon reduces ductility and weldability.

Manganese (Mn) (Increases Strength)

• Improves strength, toughness, and wear resistance.
• Helps in deoxidation and grain refinement.

Chlorine (Cl) (Does Not Increase Strength)

• Harmful impurity in steel.
• Causes corrosion and brittleness, reducing strength.

Sulphur (S) (Does Not Increase Strength)

• Weakens steel by making it brittle and reducing ductility.
• Increases machinability but is considered an undesirable impurity in structural steel.

Explanation:

Zenith angle (Z) is measured from the vertical (0° at zenith downward to 180° at nadir).

Vertical angle (V) is measured from the horizontal (0° at the horizon, +ve above, -ve below).

Conversion Formula: V=90^∘−Z

Given Data:

Zenith Angle (Z) = 93°00'00"

Using the formula: 

V=90^∘−93^∘=−3^∘00′00"

Explanation:

• Speed of the car: 60 km/hr
• Downward gradient of the highway: 2%
• Distance to the stopped vehicle: 130 m
• Brake efficiency: 60%
• Coefficient of friction: 0.7
• Driver's reaction time: 2.5 seconds
• Acceleration due to gravity: 9.81 m/s²

The distance (in meters) required by the driver to bring the car to a safe stop is

\(\mathrm{SSD} =\mathrm{vt}_{\mathrm{r}}+\frac{\mathrm{v}^{2}}{2 \mathrm{~g}(\mu-0.01 \mathrm{n})} \)

SSD = \(\left(\frac{5}{18} \times 60\right) \times 2.5+\frac{\left(\frac{5}{18} \times 60\right)^{2}}{2 \times 9.81(0.6 \times 0.7-0.01 \times 2)}\)

SSD = 41.67 + 35.4 = 77.07 m

Explanation:

The given equation is

x5 – 5x4 – 10x3 + 50x2 + 9x – 45 = 0

Now, 

f(0) = 0 – 45 = Negative

f(2) = 2⁵ – 5 × 2⁴ – 10 × 2³ + 50 × 2² + 9 × 2 – 45

= 32 – 65 – 80 + 200 + 18 – 45

= 60 = Positive

Hence, If f(x) is continuous on [a, b] and f(a) and f(b) are of opposite sign then there exist atleast one point c on [a, b] such that f(c) = 0 so atleast one root lying between (0, 2).

Explanation:

The most general case of second-order partial differential equation

\(\rm A \frac{\partial^{2} u}{\partial x}+B \frac{\partial^{2} u}{\partial x \partial y}+C \frac{\partial^{2} u}{\partial y^{2}}+D \frac{\partial u}{\partial x}+E \frac{\partial u}{\partial y}+f u=G\) …(i)

Given that the partial differential equation is

\(\rm x \frac{\partial^{2} f}{\partial x^{2}}+y \frac{\partial^{2} f}{\partial y^{2}}=\frac{x^{2}+y^{2}}{2}\) …(ii)

Now, Comparing (i) and (ii)

A = x and B = 0, C = y

Discrimanant = B² – 4AC = 0 – 4 × x × y = –4xy

–4yx > 0 if x < 0 and y > 0

So, the given equation is hyperbolic if x < 0 and y > 0.

Similarly, we can prove if x > 0, y > 0 given partial differential equation is elliptic also.