Ratio of torque resisted by the upper one-third of the vane to lower two-third of the vane in vane shear test is _______(Take dia of vane blade = half of the vane length.)
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Solution
T=πd2τf\(\left ( \frac{H}{2}+\frac{d}{6} \right )\)
Given,H = 2d
Top 1/3rd portion T1/3=πd2τf\(\left ( \frac{H}{6}+\frac{d}{6} \right )\)
= πd2τf\(\left ( \frac{3d}{6} \right )\)
=\(\frac{\pi \tau fd^{3}}{2}\)
Bottom 2/3rd portion T2/3=\(\pi d^{2}\tau f\left ( \frac{2}{3}\left ( \frac{H}{2} \right )+\frac{d}{6} \right )\)
=πd2τf\(\left ( \frac{2}{3}d+\frac{d}{6} \right )\)
=πd2τf\(\left ( \frac{5d}{6} \right )\)
Ratio=\(\frac{\frac{1}{2}}{\frac{5}{6}}\Rightarrow \frac{1}{2}\times \frac{6}{5}\Rightarrow \frac{3}{5}=0.6\)
In normally consolidated soil, earth pressure at rest depends on which of the following
1.liquid limit
2.plastic limit
3.shrinkage limit
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Solution
In normally consolidated soil.
Ko= 0.19 + 0.233 log 10(IP)
So option (c) is correct
But if liquid limit changes and plastic limit does not changes in same proportion then also IP will change and vice versa so it depends on liquid and plastic limit also. So (a) option is also correct.
The capillary rise in soil A with D10= 0.06 mm is 60 cm.Estimate the capillary rise in soil B with D10= 0.1 mm.Assuming same void ratio in both soil _____ cm.
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Solution
size of void be d
\(V_{s}\alpha D_{10}^{3}\)
\(V_{s}\alpha d^{3}\)
e=\(\frac{V_{v}}{V_{s}}=\left ( \frac{d}{D_{10}} \right )^{3}\)
For Soil A,hc=\(\frac{0.3084}{d}\)
so d=0.3084⁄d=5.14+10-3cm
Now,\(\left ( \frac{d}{D_{10}} \right )_{A}=\left ( \frac{d}{D_{10}} \right )_{B}\) as void ratio is same
dB=\(\frac{5.14\times 10^{-3}}{0.06}\times 0.1\)
= 0.00857 cm
(hc)B=\(\frac{0.3084}{0.00857}=36\, cm\)
There is a free overall in horizontal frictionless rectangular channel. Assuming flow to be horizontal at section (1) with Froude number 6 and depth 1 m, and the pressure at the brink of section(2) to be atmospheric throughout the depth then depth at the brink will be equal to ______ m.
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Solution
y⁄1=\(\frac{2F_{0}^{2}}{2F_{0}^{2}+1}\)
y⁄1=\(\frac{(2\times 6)}{(2\times 6)+1}\)
y=\(\frac{12}{13}\Rightarrow y=0.923m\)
A hydraulic jump occurs in a horizontal 60° triangular channel. If the sequent depths in this channel are 0.5 m and 1.5 m respectively. The Froude’s no. after the jump will be_____ when Froude no before the jump is 25.
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Solution
For triangular channel section
F=\(\frac{Q}{A\sqrt{g\frac{A}{T}}}\)
⇒F2=\(\frac{Q^{2T}}{gA3}\)
⇒\(F^{2}=\frac{Q^{2}(2my)}{g\left ( \frac{1}{2}\times y\times 2my \right )^{3}}\)
⇒F2=\(\frac{2Q^{2}}{gm^{2}y^{2}}\)
so,Fα\(\frac{1}{y^{2.5}}\)
\(\frac{F_{2}}{F_{1}}=\left ( \frac{y_{1}}{y_{2}} \right )^{2.5}\)
F2=2.5\(\left ( \frac{0.5}{15} \right )^{2.5}\)
F2=\(\frac{2.5}{3^{2.5}}=0.16\)
The parallax bar reading for a control point B of the elevations 500 was 17 mm. The elevation of another point A which has the parallax reading of 12 mm is _______m. Take parallax bar constant C as 80 and flying height 1400 m
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Solution
Δp =ra– rb
= 12 – 17
= –5 mm
pa = C + ra
= 80 + 12 = 92
ha=hB+Δp⁄pa(H-hb)
=500+\(\frac{(-5)}{92}(1400-500)\)
= 451.08 m
The true bearing of a Tower T as observed from station A was 356° and the magnetic bearing of the same was 4°. The back bearing of the line AB when measured with prismatic compass were found to be 296°. Then the true forebearing of line AB will be _______.
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Solution
Declination = 8° W
Back bearing of AB = 296°
Fore bearing of AB = 296 –180= 116°
True bearing of AB = 116° – 8°= 108°
A four wheel tractor whose operating weight is 12000 kg is pulled along a road having a rising slope of 2% at a uniform speed. Assume grade resistance factor 10 kg/tonne. The tension in the two cable is 720 kg. The rolling resistance of the road will be ________ kg/tonne.
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Solution
Grade resistance = 10 ×2 × 12 = 240 kg
Total tractive effort required = 720 kg
∴Rolling resistance
\(=\frac{720-240}{12}=\frac{480}{12}=40kg/tonne\)
In a chemical reactor 20 m tall, the density of fluid mixture varies with z in metres above, the bottom of the reactor as
\(\delta =1000\left ( 1+\frac{z^{2}}{2}+\frac{z^{3}}{3} \right )\)
then, the pressure differance between top and bottom of the reactor will be ______ × 106N/m2.
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Solution
pdA – (p + dp) dA = weight of the fluid
⇒ (δg) dz dA
-dp=\(1000\left ( 1+\frac{z^{2}}{2}+\frac{z^{3}}{3} \right )dz\, g\)
P2– P1= 1000\(_{0}^{20}\left ( 1+\frac{z^{2}}{2}+\frac{z^{3}}{3} \right )_{0}^{20}\, g\)
=\(1000\left ( 1+\frac{\tau ^{3}}{6}+\frac{z^{4}}{12} \right )_{0}^{20}\, g\)
⇒\(1000\left ( 1+\frac{\tau ^{3}}{6}+\frac{z^{4}}{12} \right )_{0}^{20}\, g\)
⇒ 144.07 ×106N/m2.
The sides of a rectangle are (60 ±0.05 m) and (120 ± 0.03 m).The probable error in area will be
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Solution
error= \(ab\left ( \left ( \frac{e_{a}}{a} \right )^{2}+\left ( \frac{e_{b}}{b} \right )^{2} \right )^{\frac{1}{2}}\)
=\(60\times 120\left ( \left ( \frac{0.05}{60} \right )^{2}+\left ( \frac{0.03}{120} \right )^{2} \right )^{\frac{1}{2}}\)
= ± 6.26