While applying the rational formula for computing the design discharge the rainfall duration is stipulated at the time of concentration because.
The pressure gradient in the direction of flow is equal to the
-
Solution
\(\frac{\partial \tau }{\partial y}=\frac{\partial p}{\partial x}\)
A 50 KN force of applied on a cantilever beam as shown.The moment about the fixed end A is _______ Nm.
-
Solution
Taking moments about A
MA= 50 cos 30° (0.25) – 50 sin 30° (3 + 0.5)
= 10.825 – 87.5
= – 76.675 Nm
If the settlement of a single pile in sand is denoted by S and that of a group of N identical piles (each pile carrying the same load) by Sg, then the ratio sg/s/ will.
The given figure shows the arrow diagram for a particular project. The arrow A is known as
-
Solution
Here B-E make critical path as critical duration comes out to be 16 days.Hence activities B and E are critical activities.Activity (2) - (3) is a dummy activity.All activities come under logic arrows.So, Activity'A' is a sub- critical activity.
A triangle in the xy-plane is bounded by the straight lines 2x= 3y, y= 0 and x= 3. The volume above the triangle and under the plane x+ y+ z= 6 is __________
-
Solution
Volume =\(\int \int zdxdy=\int_{x=0}^{3}\int_{y=0}^{\frac{2}{3}x}(6-x-y)dxdy\)
=\(\int_{0}^{3}\left [ 6y-xy-\frac{y^{2}}{2} \right ]_{0}^{\frac{2}{3}x}\, dx\int_{0}^{3}\left ( 4X-\frac{8}{9}x^{2} \right )dx \\ =10\)
The Laplace transform of the causal periodic square wave of period T shown in the figure below is
-
Solution
F(S)=\(\frac{1}{1-e^{-sT}}\int_{0}^{T/2}e^{-sT}dt\)
=\(\frac{1}{1-e^{-sT}}\left [ \frac{e^{-sT}}{-s} \right ]_{0}^{\frac{T}{2}}=\frac{1}{s(1-e^{-sT})}(1-e^{^{-\frac{s}{2}}})\)
=\(\frac{\left ( 1-e^{-\frac{sT}{2}} \right )}{s(1-e^{-sT/2})(1+e^{-sT/2})}=\frac{1}{s(1-e^{-sT/2})}\)
The region specified by {(ρ,φ,z):3≤ρ≤5,π⁄8≤φ≤π⁄4,3≤Z≤4.5}in cylindrical coordinates has volume of_________
-
Solution
=\(\int_{\rho =3}^{5}\int_{\phi =\frac{\pi }{8}}^{\pi/4}\int_{Z=3}^{4.5}\rho d\rho d\phi dz=\int_{3}^{4.5}\int_{\pi /8}^{\pi 4}\left ( \frac{\rho ^{2}}{2} \right )_{3}^{5}d\phi \, dz\)
=\(\int_{3}^{4.5}\int_{\pi /8}^{\pi /4}\left ( \frac{25}{2}-\frac{9}{2} \right )d\phi \, dz\)
=\(\int_{3}^{4.5}\int_{\pi /8}^{\pi /4}8.d\phi \, dz\)
=\(8\int_{3}^{4.5}[\phi ]_{X/8}^{X/4}dz\)
=\(8\left [ \frac{\pi }{4}-\frac{\pi }{8} \right ]\int_{3}^{4.5}dz\)
=\(8\frac{\pi }{8}[z]_{3}^{4.5}\)
=\(8\frac{\pi }{8}(1.5)\)
= 4.712
The integral \(\frac{1}{2\pi }\int \int _{D}\)(x+y+10)dxdy,where D denotes the disc: x2+y 2≤4, evaluates to____
-
Solution
\(\frac{1}{2\pi }\int \int _{D}(x+y+10)dxdy\)
Put x = r cosθ; y = r sin θ; dxdy = r dr dθ
=\(\frac{1}{2\pi }\int_{0}^{2\pi}\int_{0}^{2}(r\, cos\theta +r\, sin\theta +10)r\, dr\, d\theta\)
=\(\frac{1}{2\pi }\int_{0}^{2\pi}\int_{0}^{2}[r(cos\theta +sin\theta )+10]r\, dr\, d\theta\)
=\(\frac{1}{2\pi }\int_{0}^{2\pi}\int_{0}^{2}[r^{2}(cos\theta +sin\theta )+10r]dr\, d\theta\)
=\(\frac{1}{2\pi }\left ( \int_{0}^{2\pi}(cos\theta+sin\theta )\left ( \frac{r^{3}}{3} \right )_{0}^{2}\, d\theta \\ + 10\int_{0}^{2\pi }\left ( \frac{r^{2}}{2} \right )_{0}^{2}d\theta \right)\)
=\(\frac{1}{2\pi }\int_{0}^{2\pi }\frac{8}{3}(cos\theta +sin\theta )d\theta +\frac{1}{2\pi }\int_{0}^{2\pi }5.(4)d\theta\)
=\(\frac{4}{3\pi }[sin\theta-cos\theta ]_{0}^{2\pi }+\frac{10}{\pi }[\theta ]_{0}^{2\pi }\)
=\(\frac{4}{3\pi }[sin\theta-cos\theta ]_{0}^{2\pi }+\frac{10}{\pi }[\theta ]_{0}^{2\pi }\)
= 0 + 20 = 20
How many distinct values of x satisfy the equation sin(x) = x/2,where x is in radians ?
-
Solution
Let y = sin(x) = x/2
⇒y =x/2 and y =sin(x)The curves y = sin x and y=x⁄2touch each other at 3 points (as shown in the figure).