All India Civil Engineering (CE) Group

In a shear test on cohesionless soils, if the initial void ratio is less than the critical void ratio, the sample will:
  • a)
    Increase in volume
  • b)
    Decrease in volume
  • c)
    Initially decrease and then increase in volume
  • d)
    Initially increase in volume and then remain constant
Correct answer is option 'C'. Can you explain this answer?

Nidhi Tiwari answered  •  8 hours ago
Understanding Shear Test on Cohesionless Soils
In the context of a shear test on cohesionless soils, the behavior of the soil sample is influenced by its initial void ratio relative to the critical void ratio.
Initial Void Ratio vs. Critical Void Ratio
- Initial Void Ratio: This represents the ratio of the volume of voids to the volume of solids in the soil.
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For I think cylinder of diameter D wall thickness T and subjected to internal feud pressure P the longitudinals stress would be?

Nidhi Tiwari answered  •  8 hours ago
Introduction
When a cylinder with a diameter D and wall thickness T is subjected to internal fluid pressure P, it experiences stress due to the internal pressure. The longitudinal stress is particularly critical in assessing the structural integrity of the cylinder.
Longitudinal Stress Formula
The longitudinal stress (σ_long) in a thin-walled cylinder can be approximat
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For the function f(t) = e-t/τ ; the Taylor series approximation for t < τ="" />
  • a)
    1 + t/ τ
  • b)
    1 - t/ τ
  • c)
    1 - t/2 τ2
  • d)
    1 + t
Correct answer is option 'B'. Can you explain this answer?

Prashanth Mehra answered  •  9 hours ago
Understanding the Function f(t)
The function f(t) = e^(-t/τ) represents an exponential decay, where τ is a time constant. To approximate this function using a Taylor series around t = 0, we need to expand it.
Taylor Series Expansion
The Taylor series expansion of a function f(t) around t = 0 is given by:
- f(t) = f(0) + f'(0)t + f''(0)t²/2 + ...
For the funct
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In the analysis of thick cylinders subjected to internal pressure, radial and hoop stresses vary
  • a)
    has a constant value
  • b)
    Linearly
  • c)
    Parabolically
  • d)
    Hyperbolically
Correct answer is option 'D'. Can you explain this answer?

Athul Das answered  •  13 hours ago
Understanding Thick Cylinder Stress Analysis
Thick cylinders, such as pressure vessels, experience varying stresses due to internal pressure. The analysis of these stresses is crucial for ensuring structural integrity and safety.
Types of Stresses in Thick Cylinders
- Radial Stress: This stress acts perpendicular to the radius of the cylinder and varies from the
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In an open channel with a rectangular cross-section, the flow depth is doubled while keeping the channel width and slope constant. Assuming the flow remains subcritical and the Manning's roughness coefficient remains unchanged, the new discharge will:
  • a)
    Double
  • b)
    Quadruple
  • c)
    Octuple
  • d)
    Increase by a factor of eight
Correct answer is option 'C'. Can you explain this answer?

Nilanjan Rane answered  •  13 hours ago
Understanding Flow in Open Channels
In an open channel with a rectangular cross-section, the relationship between flow depth and discharge can be analyzed using hydraulic principles.
Discharge Formula
The discharge (Q) in an open channel can be expressed as:
Q = A * V
Where:
- A = Cross-sectional area of flow
- V = Velocity of flow
For rectangu
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Depth-Area-Duration curves of precipitations are drawn as :
  • a)
    Minimizing envelopes through appropriate data points
  • b)
    Maximizing envelopes through appropriate data points
  • c)
    best fit mean waves through the appropriate data points
  • d)
    best fit straight liens through the appropriate data points
Correct answer is option 'B'. Can you explain this answer?

Rounak Saini answered  •  14 hours ago
Understanding Depth-Area-Duration Curves
Depth-Area-Duration (DAD) curves are essential tools in hydrology, particularly in the analysis of precipitation events. These curves illustrate the relationship between the depth of precipitation, the area over which it occurs, and the duration of the event.
Maximizing Envelopes
The correct answer is option 'B' – maximizing env
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If the heights of a lighthouse and an observer in a ship above MSL respectively are 165 m and 10 m, then the horizontal distance from the ship to the lighthouse is _________ m.
Correct answer is between '56000,58000'. Can you explain this answer?

Aarav Chauhan answered  •  14 hours ago
Understanding the Problem
The goal is to determine the horizontal distance from a ship to a lighthouse given their respective heights above Mean Sea Level (MSL). The heights are:
- Lighthouse height = 165 m
- Observer height (ship) = 10 m
Applying the Line of Sight Formula
To find the distance to the horizon from a height, we can use the formula:
- Dista
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फोनोग्राफ का आविष्कार किस वैज्ञानिक ने किया था?
  • a)
    अलेक्जेंडर ग्राहम बेल
  • b)
    थॉमस एडीसन
  • c)
    जगदीश चंद्र बोस
  • d)
    जॉर्ज ईस्टमैन
Correct answer is option 'B'. Can you explain this answer?

Gowri Singh answered  •  19 hours ago
फोनोग्राफ का आविष्कार
फोनोग्राफ का आविष्कार 1877 में थॉमस एडिसन द्वारा किया गया था। यह एक महत्वपूर्ण आविष्कार था जिसने ध्वनि रिकॉर्डिंग और पुन: उत्पादन के क्षेत्र में क्रांति ला दी।
थॉमस एडिसन का योगदान
- थॉमस एडिसन एक प्रसिद्ध अमेरिकी आविष्कारक थे।
- उन्होंने फोनोग्राफ का निर्माण करते समय ध्वनि तरंगों को एक सिलेंडर पर रिकॉर्ड करने की तकनीक विकसित की।
फोन
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- फोनोग्राफ में एक नुकीला सुई और एक धातु का सिलेंडर होता था।
- जब कोई आवाज़ सुई के पास आती, तो वह सुई को हिला देती, जिससे सिलेंडर पर उभरे हुए ट्रैक बनते थे।
- इस तरह से ध्वनि को रिकॉर्ड किया जाता था, जिसे बाद में पुन: उत्पन्न किया जा सकता था।
फोनोग्राफ का महत्व
- यह पहला उपकरण था जो ध्वनि को रिकॉर्ड और पुन: प्ले करने में सक्षम था।
- इस आविष्कार ने संगीत उद्योग में एक नया युग शुरू किया, जिससे संगीत को व्यापक दर्शकों तक पहुँचाने में मदद मिली।
निष्कर्ष
इस प्रकार, फोनोग्राफ का आविष्कार थॉमस एडिसन द्वारा किया गया था, और यह तकनीकी प्रगति का एक महत्वपूर्ण मील का पत्थर था। यह न केवल मनोरंजन के क्षेत्र में, बल्कि ध्वनि विज्ञान में भी एक महत्वपूर्ण योगदान था।

A 5 m high retaining wall is as shown in figure determine ranking active earth pressure (in kN/m) per unit length of the retaining wall after the formation of tension crack:( answer up to 1 decimal point)
Correct answer is '46.9'. Can you explain this answer?

Engineers Adda answered  •  23 hours ago

Ka = (1 – sin φ)/ (1 + sin φ) = 0.33
At z = 0, σ’v = 0
pa = σv ka – 2c √ka = – 2 × 5 × √0.33
= – 5.77 kPa
Depth of tension crack
(zv) = 2c/(r √ka) = (2 × 5)/(17.5 × √0.33) = 0.99m
At bottom,
σ'v = (17.55 x 5) = 87.5 kPa
Pa = 87.5 × 0.33 – 5.77 = 23.392 kPa
After formation of tension crack = 1/2 × 23.393 × 4.01 = 46.9 kN/m 

If one were to use a piezometer to measure fluid pressure, which of the following disadvantages are they subject to?
  • a)
    It has a complicated construct
  • b)
    It is expensive to build
  • c)
    It is not suitable for medium-pressure measurement
  • d)
    It cannot be used to measure vacuum pressure range
Correct answer is option 'D'. Can you explain this answer?

Arshiya Roy answered  •  yesterday
Understanding Piezometers
A piezometer is a device used to measure fluid pressure at a specific point in a fluid system. While it has many advantages, it also comes with certain limitations.
Disadvantage of Measuring Vacuum Pressure
The correct answer to the disadvantage of a piezometer is option 'D':
Inability to Measure Vacuum Pressure
- Piezometers
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1cu.msteek calculated kaise kre?

Tanishq Rane answered  •  yesterday
1. Introduction to 1 Cubic Meter Steel
Calculating the quantity of steel in a cubic meter is essential in civil engineering for accurate material estimation and cost planning.
2. Understanding Cubic Meter of Steel
- Cubic Meter (cu.m): It is a unit of volume that measures the amount of space occupied by a material.
- Steel Weight: The weight of stee
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1cu.m m steel kaise kre?

Jhanvi Choudhary answered  •  yesterday
1. Steel Ka Pariman Kaise Nikaalein
Steel ka pariman nikaalne ke liye, humein uski density aur volume ki jankari honi chahiye. 1 cubic meter steel ka weight nikaalne ke liye, hum density ka istemal karte hain.
2. Steel Ki Density
- Steel ki average density lagbhag 7850 kg/m³ hoti hai.
3. Volume Ka Formula
- Volume = Length × Width × Height
- Agar
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Siva Gopal asked a question

Once surrounded and protected by vast wilderness, many of the national parks are adversely affected by activities outside their boundaries. The National Park Organic Act established the national park system and empowered the Secretary of the Interior to manage activities within the parks. Conditions outside park boundaries are not subject to regulation by the Park Service unless they involve the direct use of park resources.
 Several approaches to protecting the national parks from external degradation have been proposed, such as one focusing on enacting federal legislation granting the National Park Service broader powers over lands adjacent to the national parks. Legislation addressing external threats to the national parks twice passed the House of Representatives but died without action in the Senate. Also brought to the table as a possible remedy is giving the states bordering the parks a significant and meaningful role in developing federal park management policy.  
Because the livelihood of many citizens is linked to the management of national parks, local politicians often encourage state involvement in federal planning. But, state legislatures have not always addressed the fundamental policy issues of whether states should protect park wildlife.  
Timber harvesting, ranching and energy exploration compete with wildlife within the local ecosystem. Priorities among different land uses are not generally established by current legislation. Additionally, often no mechanism exists to coordinate planning by the state environmental regulatory agencies. These factors limit the impact of legislation aimed at protecting park wildlife and the larger park ecosystem. 
Even if these deficiencies can be overcome, state participation must be consistent with existing federal legislation. States lack jurisdiction within national parks themselves, and therefore state solutions cannot reach activities inside the parks, thus limiting state action to the land adjacent to the national parks. Under the supremacy clause, federal laws and regulations supersede state action if state law conflicts with federal legislation, if Congress precludes local regulation, or if federal regulation is so pervasive that no room remains for state control. Assuming that federal regulations leave open the possibility of state control, state participation in policy making must be harmonized with existing federal legislation.  
The residents of states bordering national parks are affected by park management policies. They in turn affect the success of those policies. This interrelationship must be considered in responding to the external threats problem. Local participation is necessary in deciding how to protect park wildlife. Local interests should not, however, dictate national policy, nor should they be used as a pretext to ignore the threats to park regions.
Direction: Read the above Paragraph and answer the follownig Quetions
Q.The passage provides support for which of the following assertions? 
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