From a box containing 4 white and 6 black balls. 3 balls are transferred to another empty box. From new box a ball is drawn and it is black. What is the probability that out of 3 balls transferred 2 are white and one black?
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Solution
Let A, B, C, D be the event when 0(W) and 3(B); 1(W)and 2(B); 2(W) and 1(B); 3(W) and 0(B) are transferred to the second box.Let E be the event of drawing a black ball from the second box
P(A)=\(\frac{^{6}C_{3}}{^{10}C_{3}}=\frac{1}{6},P(B)=\frac{^{4}C_{1}\times ^{6}C_{2}}{^{10}C_{3}}=\frac{1}{2}\)
P(C)=\(\frac{3}{10},P(D)=\frac{^{4}C_{3}}{^{10}C_{3}}=\frac{1}{30}\)
P = P(A) + P(B) + P(C) +P(D)
P=1⁄6+1⁄2+P3⁄10+1⁄30=1
∴ Events,A, B, C, Dare mutual exclusive and also exhaustive
P(E/A)=3/3 =1; P(E/B) = 2/3;
P(E/C)=1/3; P(E/D) = 0Required probability
P(C/E)=\(\frac{P(EC)\times P(C)}{P(E/A)\times P(A)P(E/B)\times P(B)+P(EC)\times P(C)+P(E/D)\times P(D)}\)
=\(\frac{\frac{1}{3}\times \frac{3}{10}}{1\times \frac{1}{6}+\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{3}{10}+0\times \frac{1}{30}}\)
=\(\frac{\frac{1}{10}}{\frac{1}{6}+\frac{1}{3}+\frac{1}{10}}=\frac{1}{6}\)
There are 9 objects and 9 boxes. Out of 9,5 objects can not fit to be put in 3 small boxes. How many arrangements can be made, such that each object can be put in one box only?
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Solution
Since, 5 objects can not fit to be put 3 small boxes, these 5 objects can be put in the remaining 6 boxes in 6P5ways.
The remaining 4 objects can be put in 4P4ways.
∴Total number of arrangement = 6P5×4P4
=(6×5×4×3×2)×(4×3×2×1)
=720×24
=17280
The number of vertices of odd degree is always
Matrix \(\begin{bmatrix} 8\, x\, 0 & & \\ 4\, 0\, 2 & & \\ 12\, 6\, 0 & & \end{bmatrix}\) will become singular, if value of x is
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Solution
The given matrix will become singular, if
\(\begin{vmatrix} 8\, x\, 0 & & \\ 4\, 0\, 2 & & \\ 12\, 6\, 0 & & \end{vmatrix}=0\)
8 (–12) – x(– 24)=0
– 96 + 24x=0
x=4
The z-transform of an cos nθ is
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Solution
Since, z(cos nθ) =\(\frac{z(z-cos\, \theta )}{z^{2}-2zcos\theta +1}=\overline{u}(z)\)
∴By change of scale property,we have
z (an cos nθ)=u(z/a)
\(=\frac{z/a(z/a-cos\, \theta )}{\frac{z^{2}}{a^{2}}-2\frac{z}{a}cos\, \theta +1}=\frac{z(z-a\, cos\, \theta )}{z^{2}-2za\, cos\, \theta +a^{2}}\)
The capacitor in the circuit as shown below is initially charged to 12 V with S1 and S2 open S1 is closed at t =0 while S2 is closed at t = 3. The waveform of the current in the capacitor is represented by
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Solution
At t= 3, since 2Ω is shorted, Ic jumps to a new valueand then decreases with reduced time constant.
If the rotor power factor of a 3-phase induction motor is 0.866, the displacement between the stator magnetic field and the rotor magnetic field in degree is _________.
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Solution
Rotor power factor, cos φ2= 0.866
∴Angle between stator and rotor magnitude axis
= 90° + 30° = 120°
An analog oscilloscope is calibrated for 10 volt per div and its sweep circuit is set for 50 ms. Two and half cycle of sine wave with a distance of 3 div between upper and low erextremities appear on the screen. What is the rms voltage and frequency of sine wave?
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Solution
An analog oscilloscope is calibrated for 10 volt per div and its sweep circuit is set for 50 ms.
(ii) Two and half cycle of sine wave with a distance of 3 div between upper and lower extremities appear on the screen.We know, VPP= NV×\(\frac{volt}{div}\)
and \(v_{rms}=\frac{V_{PP}}{2\sqrt{2}}for\, sine\,wave=\frac{N_{v}\times volt}{2\sqrt{2}\times div}\)
∴\(v_{rms}=\frac{1}{2\sqrt{2}}\times 3div\times 10\frac{volt}{div}\)
= 10.61 V
We known =fsignal × Tsweep
where n is number of cycles of signal displayed
\(f_{signal}=\frac{n}{T_{sweep}}=\frac{2.5}{50ms}=50Hz\)
The percentage resistance and reactance of a50 kVA, 400/200V, single phase transformer are 3% and 5% respectively.If the constant losses in the transformer are 1%, the maximum possible efficiency of the transformer is ________
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Solution
The percentage resistance and reactance of 50 KVA. 400/200 V, a single phase transformer are 3%and 5% respectively.X of full load at maximum efficiency
\(=\sqrt{\frac{Iron\,loss}{Full\, load\, Cu\, loss}}=\sqrt{\frac{1}{3}}=0.57\)
Maximum efficiency
\(=\frac{1\times 0.577}{0.577+2(0.01)}\times 100=96.6%\)
A 3kV, 750 A power electronic circuit has thyristors with 800V and 175 A rating. Using a derating of 25%, find the number of thyristors in series and parallel.
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Solution
0.25=\(1-\frac{750}{n_{P}\times 175}\, or\, n_{P}=5.71\, or\, 6\)
and 0.25=1-\(\frac{3000}{n_{s}\times 800}or\, n_{s}=5\)