Engineering

ISBT Reasoning for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams

Q1.
In a basic transmission line, the voltage at the receiving end without load is 660 V; and it is 420 V with full load. What is the percentage of voltage regulation ?
1) 47% 2) 57%
3) 67% 4) 77%
5)None of these
Answer : 57%
Explanation :
Voltage regulation =(VNL - VFL) / VFL
= (600 - 420)/420 = 0.57 or 57%
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Q2.
A disk unit has 24 recording surfaces. It has a total of 14000 cylinders. There is an average of 400 sectors per track. Each sector contains 512 bytes of data. What is the data transfer rate at a rotational speed of 7200 r.p.m.?
1) 68.80 ×10^6 bytes/s 2) 24.58 ×10^6 bytes/s
3) 68.80 ×10^3 bytes/s 4) 24.58 ×10^3 bytes/s
5)None of these
Answer : 24.58 ×10^6 bytes/s
Explanation :
The data transfer rate = (No. of sector/track)×(Bytes in each sector)×(Rotational speed (revolution per sec.))
Rotational Speed = 7200 rpm × (1/60) = 120 rev. per sec.
= 400×512×120 = 24.58×106
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Q3.
Consider the following processes which arrived in the order P1, P2 and P3:
Process            Burst Time
  P1                    24 ms
  P2                     3 ms
  P3                     3 ms

What is the average waiting time by FCFS scheduling ?
1) 17 ms 2) 19 ms
3) 21 ms 4) 23 ms
5)None of these
Answer : 17 ms
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Q4.
Consider the following statements regarding the functions of an operating system in a computer:
1. It controls hardware access
2. It manages files and folders
3. It provides a user interface
4. It manages applications

Which of the above statements are correct ?
1) 1, 2 and 3 only 2) 1, 2 and 4 only
3) 3 and 4 only (d) 4) 1, 2, 3 and 4
5)None of these
Answer : 1, 2, 3 and 4
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Q5.
The noise factor of an attenuator pad that has an insertion loss of 6 dB is - 
1) 0.25 2) 0.5
3) 2 4) 4
5)None of these
Answer : 4
Explanation :
Insertion loss= 6dB
Insertion loss (IL) is given by
I.L. = 10 log (PT/PR), where PT is the transmitted power PR is the power received by the load after insertion 10 log (PT/PR) = 6 dB
So, (PT/PR)  = 100.6 = 4
Noise factor = (PT/PR) = 4 
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Q6.
A weighted complete graph with n vertices has weights 2|i – j| at edges (vj, vj). The weight of a minimum spanning tree is - 
1) n2/2 2) n/2
3) 2n – 2 4) n – 1
5)None of these
Answer : 2n – 2
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Q7.
An asynchronous link between two computers uses the start-stop scheme, with one start bit and one stop bit, and a transmission rate of 48.8 kbits/s. What is the effective transmission rate as seen by the two computers ?
1) 480 bytes/s 2) 488 bytes/s
3) 4800 bytes/s 4) 4880 bytes/s
5)None of these
Answer : 4880 bytes/s
Explanation :
In 8 bit of a single character total bit = 8 + start bit + stop bit = 8 + 1 + 1 = 10 bit
For every 10 bit wastage of bits = 2
For 48.8 × 103 bit wastage of bits.
(2 /10) x 48.8 x 103 = 9760 bit
Effective Transmission rate = 48.8 x 103 - 9760 = 39040 bits/s
39040/8 bytes/ s = 4880 bytes/s
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Q8.
A cache line has 128 bytes. The main memory has latency 64 ns and bandwidth 1 GB/s. The time required to fetch the entire cache line is - 
1) 32 ns 2) 64 ns
3) 96 s 4) 192 ns
5)None of these
Answer : 192 ns
Explanation :
Bandwidth 1GB/s means 109 bytes can be loaded online in 1 second. To load 128 bytes
= 128 x (1/109 ) = 128 nsec.
Main memory latency = 64 nsec.
Time required to fetch cacheline = 128 + 64 = 192 nsec.
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Q9.
A 32 kB RAM is formed by 16 numbers of a particular type of SRAM IC. If each IC needs 14 address bits, what is the IC capacity ?
1) 4 kbits 2) 8 kbits
3) 16 kbits 4) 32 kbits
5)None of these
Answer : 16 kbits
Explanation :
Given, total capacity = 32 kB = 32k × 8 bits = (256 /16) k bits
Since,  IC’s are used.
So each IC provides = 256 kbits 16
14 bits address gives 16 k location. Each chip has 16 k location.
Capacity of each IC = 16 k bits.
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Q10.
For derivative control action, the actuating signal consists of proportional error signal with addition of -
1) derivative of the error signals 2) integral of the error signals
3) steady-state error 4) a constant which is a function of the system type
5)None of these
Answer : derivative of the error signals
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