? = \(L\left [ \frac{2\delta (t-3)+3\delta (t-2)}{t} \right ]\)
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Solution
\(L\left [ \frac{f(t)}{t} \right ]=\int_{s}^{\infty }F(s)ds\)
\(L\left [ \frac{2\delta (t-3)+3\delta (t-2)}{t} \right ]\)
⇒\(L\left [ \frac{2\delta (t-3)}{t} \right ]+L\left [ \frac{3\delta (t-2)}{t} \right ]\)
\(2\int_{s}^{\infty }e^{-3s}ds+3\int_{s}^{\infty }c^{-2s}ds\)
\(2\left [ \frac{e^{-3s}}{-3} \right ]_{s}^{\infty }+3\left [ \frac{e^{-2s}}{-2} \right ]_{s}^{\infty }\)
\(\frac{2}{-3}\left [ 0-e^{-3s} \right ]-\frac{3}{2}\left [ 0-e^{-2s} \right ]\)
\(\frac{2}{-3}\left [-e^{-3s} \right ]-\frac{3}{2}\left [-e^{-2s} \right ]\)
\(\frac{2}{-3}e^{-3s}+\frac{3}{2}e^{-2s}\)
If f(a + b– x) = f(x), then\(\int_{a}^{b}\)xf(x)dx=
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Solution
Let I =\(\int_{a}^{b}xf(x)dx\)......(1)
Now,I =\(\int_{a}^{b}xf(x)dx\)
\(\int_{a}^{b}(a+b-x)f(x)dx\)......(2)
By adding (1) & (2)
2I=\(\int_{a}^{b}(a+b)f(x)dx\)
I=\(\left ( \frac{a+b}{2} \right )\int_{a}^{b}f(x)dx\)
On evaluating \(\underset{x\rightarrow0}{lim}\frac{x\, tan\, 2x-2x\, tan\, x}{(1-cos2x)^{2}}\),the value we get is
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Solution
\(\underset{x\rightarrow0}{lim}\frac{x\, tan\, 2x-2x\, tan\, x}{(1-cos2x)^{2}}\)
\(\underset{x\rightarrow0}{lim}\frac{x\left \{ 2x+\frac{8x^{3}}{3}+\frac{64x^{5}}{15} \right \}-2x\left \{ x+\frac{x^{3}}{3}+\frac{2x^{5}}{15}+... \right \}}{4sin^{4}x}\)
=\(\underset{x\rightarrow0}{lim}\)x4{8⁄3-2⁄3+ terms containing positive powers}
⇒\(\frac{1}{4}\underset{x\rightarrow0}{lim}\frac{\left ( \frac{8}{3}-\frac{2}{3} \right )+0}{\left ( \frac{sin^{4}x}{x^{4}} \right )}=\frac{1}{4}\times 2=\frac{1}{2}\times 2=\frac{1}{2}=0.5\)
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Solution
Here, h = 1
By Simpson’s rule:\(\int_{0}^{4}e^{x}dx\)
=1⁄3[1 + 4 (2.72 + 20.09) + 2 (7.39) + 54.6]
= 53.873
If A is 3 × 3 with eigen values 0, 1 and –1, then |I + A46| is
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Solution
Eigen values of A are 0,1, –1
So, λ(λ– 1) (λ+ 1)= 0
λ(λ2– 1) = 0
⇒λ3– λ= 0
⇒λ3= l
A3= A ⇒ A2. A-1= A.A-1 ⇒ A2= I ⇒ A3= A
A45= (A3)15⇒ A45= A15 ⇒ A45 = (A3)5
⇒ A45= A3.A2= AI
A46= A.A45= A.AI = I2 = I
∴|I + A46| = |I + I| =|2I| = 23|I| = 8
A thick cylinder with internal diameter d and out side diameter 2d is subjected to internal pressure P.Then the maximum hoop stress developed in the cylinder is:
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Solution
Writing Lame’s equations
⇒ Hoop stress,σk=\(A+\frac{B}{r^{2}}\)............ (1)
⇒Radial stress, σr=A-\(A-\frac{B}{r^{2}}\)............ (2)
Given: Internal diameter = d
Outside diameter = 2d
In equation (2),
p=σr=A-\(\frac{B}{(d)^{2}}\)............ (3)
0 = A-\(\frac{B}{(2d)^{2}}\)............ (4)
On solving equations (3) and (4) we get,
A=\(\frac{P}{3},B=4d^{2}\times \frac{p}{3}\)
Now,maximum hoop stress,
\(\sigma _{max}=A+\frac{B}{r^{2}}=\frac{p}{3}+\frac{4pd^{2}}{3r^{2}}\)
=p⁄3+\(\frac{4pd^{2}}{3d^{2}}=\frac{p}{3}+\frac{4p}{3}=\frac{5}{3}p\)
Consider a two dimensional state of stress given for an element as shown in figure given below:
Then the coordinates of the centre of the Mohr’s circle will be ………………..
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Solution
Given: σ1 (in x-direction) = 200 MPa
σ2 (in y-direction) = 100 MPa
The location of co-ordinates of centre of Mohr’s circle
x-cordinates =\(\frac{\sigma _{1}+\sigma _{2}}{2}=\frac{200-100}{2}=50\)
y-cordinate = 0
Hence, (50, 0)
If α= helix angle and Pc= circular pitch, then which one of the following expresses the axial pitch of a helical gear?
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Solution
Horizontal component of normal force = Pc cos a and, axial pitch of helical gear
\(\frac{P_{C}}{tan\, \alpha }\)
A body of mass ‘m’ and radius of gyration ‘k’ is to be replaced by two masses m1 and m2 located at distances h1 and h2 from the C.G.of original body.An equivalent dynamic system will result:
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Solution
An equivalent dynamic system will result as :Radius of Gyration,k=\(\sqrt{h_{1}h_{2}}\)
A shaft of 50 mm diameter and in length carrier a disc which has mass eccentricity equal to 190 micron. The displacement of shaft at a speed which is 90% of critical speed in micron is ………………..
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Solution
Given, eccentricity (e) = 190 micron Speed of shaft (ws) = 90% of critical speed (wc)
wc=\(\frac{90}{100}\times w_{c}= 0.9w_{c}\Rightarrow \frac{w_{c}}{w_{s}}=0.9\)
Now,using the following relation,
\(\frac{x}{e}=\frac{1}{\left ( \frac{w_{c}}{w_{e}x} \right )^{2}-1}=\frac{1}{\left ( \frac{1}{0.9} \right )^{2}-1}=\frac{x}{190}\)
X = 808.5 micron.