If tomato juice is having a pH of G-1, the hydrogen ion concentration will be
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Solution
pH = –log (H+)
pH = 4.1
∴H+= 7.94 × 10–5mol/L
During a 6 hour storm, the rainfall inensity was 0.75 km/hr. on a catchment runoff volume during this period was 2,880,00 m2. The total rainfall was lost due to infiltration,evaporation and transpiration in cm/hr is
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Solution
Water is pumped through a pipe line to a height of 10 mat a rate of 0.1m3/s. Frictional and other miner losses are 5m. Then the power of pump in kW required is______ .
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Solution
A rectangular channel of 4m width convergs water of 2m3/s under critical condition specific energy for this flow is (in m)
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Solution
Uniform flow in a channel is characterized by one of the following statements
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Solution
∵ Depth is constant (uniform flow) velocity is constant.
A strip footing 8m wide is designed for a total settlement of 40 mm. The sale bearing capacity (shear) was 150 kN/m2. and safe allowable soil pressure was 100 kN/m2. Due to importance of the structure, now the footing to be redesigned for total settlement of 25 mm. The new width of footing will be.
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Solution
Load, Q = (B × 1)qna
= 8 × 1× 100=800 kN
In a consolidated drained trianial test a specimens of clay fails at cell pressure of 60 kN/m2. The effective shear strength parameters are C’= 15 kNm2 and φ’= 20°.Find the compressive strength of the soil (in kPa)
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Solution
These are two footings resting on ground surface. One footing is square of dimension B. The other is strip footing of width B. Both of them are subjected to a loading intensity of q. The pressure intensity at any depth below the base of the footing along centre line would be
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Solution
Stresses:
Pstrip > Psquare > Pcircular
Data from a sieve analysis conducted on a given sample of soil showed that 67% of the particles passed through 75 marks. It sieve. The liquid limit and plastic limit of the fines fraction was found to be 45 and 33 percents respectively. The group symbol of soil is
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Solution
∴ 67% (> 50%) passed 75 µ since it is fine grained soil.
∴ WL(45%) > 35% it is intermediate in compressible soil(I)
Ip of soil= 45 – 33 = 12%
For WL= 45%, Ip= 0.73(WL– 20)
= 0.73(45 –20)= 18.25% (Equation of A line)
∴ soil is below A line
∴ it is M.I.
A sheet of water of thickness 1m is available to fill the voids of cohesion less soil two degree of saturation of 80%. The soil has a void ratio of 0.5. Find the thickness of soil layer required to accomodate this amount of water.
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Solution