Online Electronics and Communication Engineering Test - Previous Exam Papers Test 4
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- Maximum energy of electrons liberated photoelectrically is independent of light intensity
- Maximum energy of electrons liberated photoelectrically varies nonlinearly with frequency of incident light.
Statement (ii) is false because maximum energy of photons varies linearly with frequency of incident light. It can be shown as follows :
F1 > F2 > F3
j = light intensity = constant
Photocurrent Vs Anode voltage with frequency and incident light as a parameter.
The light intensity is constant.
VT(new) = VT(odd) +
= 6.903 x 10-8
∴ 
∴ 
fB = - 8.6 x 1011
The threshold voltage is always negative for p-channel and hence implant is of p-type.
The given digits are 2, 2, 3, 3, 3, 4, 4, 4, 4 we have to find the numbers that are greater than 300
∴ The first digit can be 3 or 4 but not 2.
Now, let us fix the first, second and third digits as 3, 2, 2, then the fourth place can be filled in 3 ways.
∴ The number of ways is 3 similarly, we fix first third and fourth place as 3, 2 and 2 respectively (4) so the second place can be filled in 3 ways again,
The number of ways is 3
Now, we fix first, second and fourth, previous cases and we obtain the same result.
∴ The number of ways is 3 so, the total number of ways is 9 similarly this can done by fixing the numbers as 3 and 4 (instead of 2) and thereby we obtain the a ways each
The number of numbers starting with 3 is 27
Similarly by taking 4 as the first digit we get 27 numbers
∴ The number of numbers that are greater than 3000 is 27 + 27 = 54
But, 3222, 4222, is not possible as there are only two 2's, 3333 is not possible as there are only three 3's
∴ The total number of numbers that are greater than 3000 is 54 - 3 = 51.
With odd number of variables (n) terms with even number of 0's and terms with odd number of 1 's are the same.
Thus the k-map for the above condition with minterms represented by terms with even number of 0's is
for n = 3 
∴ The expression can be represented by equivalence (EX NOR) or (EX OR) expressions.
Source a : Power : 8 x (4) = - 32 W. Source A has delivered the power as it is. - ve.
Source b: Source B has absorbed power of + 12 W as power is positive
Source c: Source C has absorbed power of 5 x 16 = 80 W
Source d: Delivers power of 3 x ix x (5) = 3 x 4 x (- 5) = -60 W
Total power of the network is - 32 + 12 + 80 - 60 = 0 Watts.
VGS = VG - IDRS
- 2 = 5.44 - 2.22 X 10-3 X RS
RS = 3.35 KΩ.
Then for
x(2t) the Fourier transform will be:
Using scaling property
.
In case of capacitor, the charges on two plates of capacitor are of opposite sign.
We know that electric field due to infinite sheet charge with density rs is 
Outside the capacitor, electric field intensities due to upper plate and lower plate cancel each other.
∴ Eoutside = 0
Inside the capacitor, the electric field intensities due to upper and lower plate add together
∴
X = 10 cm, ZL = 0, βl = 
Thus Zin is Z0
which is inductive.
(V/m). The maximum value of 
will occur; if and only if
To find value of y at which E will be maximum, differentiable ET with respect to y and set it equal to zero.
- Registers are made of edge-triggered FFs, whereas latches are made from level-triggered FFs.
- Registers are temporary storage devices whereas latches are not.
- A latch employs cross-coupled feedback connections.
- A register stores a binary word whereas a latch does not.
- Bus is a group of wires carrying information.
- Bus is needed to achieve reasonable speed of operation.
- Bus can carry data or address.
- A bus can be shared by more than one device.
and 
. The behaviour of the communication system is modelled byP(Y = 1 |x = 1|) =
and P(Y = 0|x = 0) = 
Find P(Y = 1)
