If the Eigen values of a 3 × 3 real matrix P are 1, –2 and 3, then
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Solution
The characteristic equation is
(λ– λ1) (λ– λ2) (λ– λ3) = 0
(λ– 1) (λ+ 2) (λ– 3) = 0
(λ– 1) (λ2 – λ –6) = 0
λ3– λ2– 6λ– λ2+ λ +6 =0
⇒ λ3 – 2λ2– 5λ + 6 = 0
By Cayley Hamiltion theorem,
we have
P3– 2P2– 5P + 6
I = 06I = –P3+ 2P2+ 5P
6p–1= p2+ 2p + 5I
P–1= 1⁄6 [5I + 2P – P2]
The partial differential equation \(\frac{\partial ^{2}u}{\partial x^{2}}-7\frac{\partial ^{2}u}{\partial x\partial y }+2\frac{\partial ^{2}u}{\partial y^{2}}=0\) is a/an
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Solution
The convergence of which of the following method is sensitive to starting value?
For |z| = 1,\(\int_{z}\frac{z-3}{z^{2}+2z+5}dz\) where c is the circle, is _____.
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Solution
The interval in which the lagrange’s theorem is applicable for the function f(x)=1⁄x is
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Solution
From option (a), Since f(x) = 1⁄x is not define in interval[–3, 3]
f(0)= 1⁄0 → not define
From option (b), since,f(x) = 1⁄xis not define in interval[–2, 3]
f(0)= 1⁄0 → not define
From option (c), since, f(x)=1⁄x is define in interval[2, 3] because there does not zero lie in the interval [2, 3]
When the pressure on a given mass of liquid is increased from 3MPa to 3.5 MPa, the density of liquid increases from 500 kg/m3. Then the average value of bulk modulus of the liquid over the given pressure range will be ________.
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Solution
Given: Initial pressure (Pi) = 3MPa
Final pressure(Pf) = 3.5 MPa
Initial density ρi= 500 kg/m3
Final density (ρf) = 504 kg/m3
Bulk modulur ](K) = \(\rho \frac{\delta P}{\delta P}\)
where, ρ= density, δP = change in pressure
= Pf – Pi = 3.5 – 3
= 0.5 MPa
δP= 504 – 500 = 4Kg/m3
K=0.5⁄4 × 500 = 62.5 MPa
A heat engine is supplied with 2700 KJ/min of heat at 750°C.Heat rejection with 1000 KJ/min takes place at 99°C. Then the type of heat engine will be:
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Solution
The velocity of a particle is given as :
v = 9 – 0.06 x m/s
If x= 0, t = 0, then the distance travelled (when it comes to rest) and acceleration (at t =0) will be ________.
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Solution
A truss structure is shown in figure given below which supports a load 18 KN at point ‘S’. Then the load is the member PQ will be ________.
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Solution
Using method of sections, cut three members PQ, QT and TU by reaction (1) - (1) and considering the equilibrium of right portion of truss.
In a multiple disc clutch, the axial intensity of pressure is not to exceed 0.2 MPa. The inner radius of the discs is 100 mm and is half the outer radius. Then the axial force per pair of contact surfaces in N will be ________.
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Solution
Given : Pressure (P) = 0.2 MPa
Inner radius(R1) = 100mm
Outer radius (R2) = 200mm
Axial force (F)=Pressure × Area
=\(0.2\times \pi (R_{2}^{2}-R_{1}^{2})\)
=0.2 × 3.14((200)2-(100)2)
= 0.2 × 3.14((100)(300))= 18840 N