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Explanation:

• Swell potential of soil decreases with an increase in the shrinkage limit.
• Correct: The shrinkage limit is the maximum water content at which further reduction in water content does not lead to a reduction in the volume of the soil. A higher shrinkage limit indicates that the soil can retain more water without additional volume change, thus leading to a decrease in swell potential as the soil's capacity to absorb water decreases.
• In electrical resistivity tomography, the depth of current penetration is half of the spacing between the electrodes.
• Incorrect: Typically, the depth of current penetration in electrical resistivity tomography is approximately equal to the spacing between the electrodes, not half.
• Both loose and dense sands with different initial void ratios can attain similar void ratio at large strain during shearing.
• Correct: Both initially loose and initially dense sands can approach a constant void ratio, known as the critical void ratio, when subjected to large shearing strains. This behavior is a characteristic of granular materials like sand under shearing stress.
• Among the several corrections to be applied to the SPT-N value, the dilatancy correction is applied before all other corrections.
• Incorrect: Dilatancy correction is generally applied after corrections for overburden pressure have been made to the SPT-N values.

Explanation:

Given data:

Effective depth = 250 mm

0.2% main reinforcement using 8 mm diameter steel bars

Now, Area of reinforcement considering the width of slab as 1 m.

\(\rm A_{s t}=\frac{0.2}{100} b d=\frac{0.2}{100} \times 1000 \times 250=500 \mathrm{~mm}^{2}\)

\(\rm \text { Spacing }=\frac{1000}{\left(\frac{\mathrm{~A}_{\mathrm{st}}}{\frac{\pi}{4} \times 8^{2}}\right)}=100.53 \mathrm{~mm}\)

Rounded upto the nearest integer considering safety is 100 mm.

Explanation:

As per given data:

\(\text { Scale } \Rightarrow \frac{1}{1000}\)

Horizontal distance on map = 10 mm

Now, Horizontal distance between two contour on ground

\(=\frac{10}{\text { Scale }}=10 \times \frac{1000}{1} \mathrm{~mm}=10 \mathrm{~m} \)

Hence,\(\text { Slope }=\frac{\text { Vertical }}{\text { Horizontal }} \times 100=\frac{1 \mathrm{~m}}{10 \mathrm{~m}} \times 100=10 %\)

Explanation:

As per given data:

(ϕ) friction angle = 30°

For active state

Failure plane inclination with major principal plane (horizontal)

\(\Rightarrow \theta_{c}=45^{\circ}+\frac{\phi}{2}=45^{\circ}+\frac{30^{\circ}}{2}\)

θc = 60° 

For passive state

Failure plane inclination with major principal plane

\(\left(\theta_{\mathrm{C}}\right)=45^{\circ}+\frac{\phi}{2}\)

\(\theta_{\mathrm{c}}=45^{\circ}+\frac{30^{\circ}}{2}=60^{\circ}\)

Hence inclination of failure planes with major principal plane for both active & passive conditions is 60°

Explanation:

As per the given details:

Bending Moment at hinge = 0 

\(\Rightarrow \mathrm{R}_{\mathrm{A}} \ell-\frac{\mathrm{q} \ell^{2}}{2}=0\)

\(\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{q} \ell}{2} \) upward i.e. (+)ve

Now, equilibrium in the vertical direction

∑F_V = 0 ⇒ R_A + R_B + R_C = 2qℓ

⇒ R_B + R_C = 2qℓ - R_A

\(=2 \mathrm{q} \ell-\frac{\mathrm{q} \ell}{2}=1.5 \mathrm{q} \ell \) …(A)

Moment equilibrium at B

∑M_B = 0

⇒ RA(2ℓ) - q(2ℓ)ℓ - Rcℓ = 0

\(\frac{\mathrm{q} \ell}{2}(2 \ell)-2 \mathrm{q} \ell^{2}-\mathrm{R}_{\mathrm{C}} \ell=0\)

-qℓ² - R_Cℓ = 0

R_C = qℓ

⇒ RC = qℓ downward

⇒ From (A), R_B = 1.5qℓ + qℓ = 2.5qℓ

⇒ R_B = 2.5 qℓ upward i.e. (+)ve

Explanation:

The analogy works based on the increasing intensity of the actions described by the words. Here's the breakdown:

1. Simmer → Seethe → Smolder:

• Simmer suggests a mild or low-level intensity.
• Seethe indicates a higher intensity, where things are boiling or agitated.
• Smolder refers to something burning slowly with little flame, implying an intense state of heat, though not yet fully destroyed or burnt.

Now, for the analogy of [break → raze → _______]:

• Break suggests a low-level action, where something is damaged or disrupted.
• Raze involves completely destroying or leveling something, a much stronger action.
• The final word must indicate a complete or intense form of destruction.

The appropriate word to complete the analogy is obliterate, which means to completely destroy or wipe out, just as smolder represents a highly intense state of burning.

So, the solution is: break → raze → obliterate

Explanation:

Minute hand covers 6° angle in 1 min.

Second hand covers 360° angle in 1 min.

Means 354° angle gained by second hand over min. hand in 1 min. of time

So, 360° angle gained in \( \frac{360}{354}\) min.

(which is required time for one crossing)

But first crossing taking \(\left(\frac{30}{354}\right)\) min. only

So, total time taken in 50

Crossings = \(1 \times \frac{30}{354}+49 \times \frac{360}{354}=49.91 \) min.

So, no. of crossings = 50

Explanation:

Let no. of houses on each side = n

The sum of odd-numbered houses

= (301 + 303 + 305, …, n)

= \(\frac{n}{2}[2 \times 301+(n-1) 2] \)

= \(\frac{n}{2}[2 n+600]=n(n+300)\)

The sum of even-numbered houses

= 302 + 304 + 306, …, n

= \(\frac{n}{2}[2 \times 302+(n-1) 2] \)

= \(\frac{n}{2}[2 n+602]\)

= n[n + 301]

According to question

n[n + 301] – n[n + 300] = 27

n[n + 30] – n – 300] = 27

n = 27

n[n + 301] – n[n + 300] = 27

n[n + 30] – n – 300] = 27

n = 27

Alternate method:

If one house each side difference = 302 – 301 = 1

If two house each side difference = (302 + 304) – (301 + 303) = 2

If three houses each side difference = (302 + 304 + 306) – (301 + 303 + 39) = 3

So, to make difference equal to 27 number of houses on each side must be 27.

Explanation:

7 – 3 = 4 = 4 × 1

15 – 7 = 8 = 4 × 2

x – 15 = 16 = 8 × 2

63 – x = 32 = 16 × 2

127 – 63 = 64 = 32 × 2

255 – 127 = 128 = 64 × 2

x – 15 = 16 ⇒ x = 31

Explanation:

\(\rm \left(\frac{p}{q}\right)^{p / q}=p^{\left(\frac{p}{q}-1\right)}\)

\(\rm \frac{p^{p / q}}{q^{p / q}}=p^{p / q} \times p^{-1} \)

\(\rm \left(\frac{p^{p / q}}{q^{p / q}}-\frac{p^{p / q}}{p}\right)=0\)

\(\rm p^{p / q}\left(\frac{1}{q^{p / q}}-\frac{1}{p}\right)=0 \)

\(\rm \left(\frac{1}{q^{p / q}}-\frac{1}{p}\right)=0 \)

(∵ pp/q ≠ 0)

\(\rm \frac{p-q^{p / q}}{p \cdot q^{p / q}}=0\)

⇒ p = q^p/q

p = (q^1/q)^p

p^1/p = q^1/q

q^p = p^q