A closed system undergoes a process 1-2 for which the values of Q1–2 and W1–2 are+ 20 kJ and + 50 kJ respectively. If the system is returned to state 1 and Q2–1 is– 10 kJ,what is the value of the work W2–1?
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Solution
Writting the following equation,
U1– U2= W1–2– Q1–2 = + 50 – (+20) = 50 – 20
U1– U2= 30 k J
U2– U1= – 30 kJ
Now,U2+ Q2–1 = U1– W2–1
W2–1= U2– U1 + Q2–1= – 30 – 10 = – 40 k
A particle is projected under gravity (g = 9.8 m/s2) with a velocity of 30 m/s at an elevation of 40°. The times of flight to a height of 10 mare ……………….. .
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Solution
Given: Initial velocity (u) = 30 m/s, Elevation(q) = 40°
Height (h) = 10 m, g= 9.8 m/s2
Using the following equation,h= 4 sin θ × T –1⁄2gT2
10 = (30 × sin 40) T– 12× 9.8 (T)2
⇒10 = 19.3 T – 4.9T2
4.9 T2– 19.3T + 10 = 0
On solving, we get
T = 3.33 sec., 0.61 sec.
If the probability of an acceptance of a 1% defective lot is 0.95, then AOQ is ………………..
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Solution
Given: defective lot = 1%
Probability of acceptance = 0.95
Then, Average outgoing quality (AOQ)=\(\frac{Defective \times Probability}{100}\)
=\(\frac{1\times 0.95}{100}=0.0095\)
It is required to cut screw threads with double start and 2 mm pitch on a lathe having lead screw pitch of 6 mm.The spindle ratio between lathe spindle and lead screw will be……………….. .
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Solution
Given: Pitch (P) = 2 mm,
Screw thread = double start = N= 2,
lead screw pitch (Ps) = 6 mm
Then, lead of screw (L) = Pitch (P) × N
= 2 × 2= 4 mmSpindle ratio=\(\frac{lathe\,spindle}{lead\,screw}=\frac{4mm}{6mm}\)
=2⁄3= 2 : 3 = 0.67
The hardness of lathe bed material should be measured by:
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Solution
As we know that the lathe bed cannot be arranged on a hardness testing machine for testing purposes.So, shore scleroscope test is used for this type of situations in which the concept of measurement of height of bounces of ball is being calculated.
A box contains 20 defective items and 80 non-defective items.If two items are selected at random without replacement, what will be the probability that both items are defective?
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Solution
Total number of items = 100
Number of defective items = 20
Number of Non-defective items = 80
Then the probability that both items are defective,when 2 items are selected at random is,p=\(\frac{^{20}C_{2}\times ^{80}C_{0}}{^{100}C_{2}}=\frac{\frac{20\times 19}{2\times 1}}{\frac{100\times 99}{2\times 1}}=\frac{19}{495}\)
If φ(x, y) and Ψ(x, y) are functions with continuous second derivatives, then φ(x, y)+ i Ψ(x, y) can be expressed as an analytic function of x + iy (i =\(\sqrt{-1}\)) when
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Solution
Given φ(x, y) and ψ(x, y) are continuous second derivative function. So, it can be expressed as an analytic function.
Since, Cauchy– Riemann equation satisfies analytic function.
When f(z) = w = u+ ivThen cauchy–Riemann equation is
\(\left ( \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}and\frac{\partial u}{\partial y}=\frac{-\partial v}{\partial x} \right )\)
But, w = φ(x, y)+ i ψ(x, y) (Given)Then by Cauchy Riemann equation
\(\left ( \frac{\partial \phi }{\partial x}=\frac{\partial \Psi }{\partial y}and\frac{\partial \phi }{\partial y}=\frac{-\partial \Psi }{\partial x} \right )\)
The partial differential equation \(\frac{\partial u}{\partial x}+u\frac{\partial u}{\partial x}=\frac{\partial^{2} u}{\partial x^{2}}\) is a
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Solution
Order is highest index of a derivative present in a partial differential equation.
Here,\(\frac{\partial^{2} u}{\partial x^{2}}\)⇒highest index 2 ⇒ order = 2
The minimum value of y =x2 in the interval [1,5] is ______.
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Solution
y =x2
At x =0, y = 0
x= 1, y = 1
x =5, y = 25
As the function is increasing in the first quadrant. So,minimum value in interval[1, 5]is at 1.
Minimum y =1 at x =1
A is a 3 × 4 real matrix and AX = B is an inconsistent system of equations. The highest possible rank of A is ____.
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Solution
By Echelon form: We know that, the rank of any matrix is equal to number of non-zero rows.
Here, the given matrix is of order 3 × 4, i.e. only three rows. So, the highest possible rank should be 3.