CBSE Class 12 Marking Scheme 2022 for Computer Science Term 2

Marking Scheme
COMPUTER SCIENCE (Code : 083)
Maximum Marks: 35 Time: 2 hours

General Instructions

 The question paper is divided into 3 sections – A, B and C
 Section A, consists of 7 questions (1-7). Each question carries 2 marks.
 Section B, consists of 3 questions (8-10). Each question carries 3 marks.
 Section C, consists of 3 questions( 11-13). Each question carries 4 marks.
 Internal choices have been given for question numbers – 7, 8 and 12

Section -A
Each question carries 2 marks
Q. Part Question Marking Marks
No No. Instructions
1. Characteristics of Stacks: 1 mark for
 It is a LIFO data structure each point (2)
 The insertion and deletion happens at one
end i.e. from the top of the stack

2. (i) SMTP : Simple Mail Transfer Protocol ½ mark for
XML: Extensible Mark Up Language each correct (1)
expansion
(ii) Wired- optical fibre ½ mark for
Wireless – microwave each correct (1)
3. char(n): 1 mark for
 stores a fixed length string between 1 and each correct
255 characters difference
 if the value is of smaller length, adds blank ( minimum 2
spaces differences (2)
 some space is wasted to be given)
varchar(n) :
 stores a variable length string
 no blanks are added even if value is of
smaller length
 no wastage of space
[1]

4. (a) One record 1 mark for
(b) tuple each correct (2)
5.
(a) 29

(b) 19-Jul-2021
½ mark for (2)
(c) each correct
output
T006 Console 17-Nov- 15000 12
Table 2019

(d)
10-Mar- 2020
17-Nov-2019

6. (i) SHOW TABLES; 1 mark for
correct (1)
(ii) Equi- join: 1 mark for
 The join in which columns from two tables correct
are compared for equality difference
 Duplicate columns are shown (Any one (1)
Natural Join point may be
 The join in which only one of the identical given)
columns existing in both tables is present
 No duplication of columns

7. (a) Degree: 5 ½ mark each
Cardinality: 6 for correct
degree and
cardinality

(b) MOVIEID should be made the primary key as it ½ mark for
uniquely identifies each record of the table. correct field
and ½ mark (2)
for
justification

[2]

OR

(a) MOVIEID and TITLE ½ mark for
field name

(b) MOVIEID 1 mark for
correct
SECTION – B
Each question carries 3 marks

8. # Question No 8 (first option)
R={"OM":76, "JAI":45, "BOB":89, 1 mark for
"ALI":65, "ANU":90, "TOM":82} correct PUSH
def PUSH(S,N): operation
S.append(N)
def POP(S):
if S!=[]:
return S.pop() correct POP
else: operation
return None
ST=[] 1 mark for
for k in R: correct
if R[k]>=75: function calls
PUSH(ST,k) and (3)
while True: displaying
if ST!=[]: the output
print(POP(ST),end=" ")
else:
break

OR

# Question No 8 (second option) 1 mark for
correct PUSH
N=[12, 13, 34, 56, 21, 79, 98, 22, operation
35, 38]
def PUSH(S,N):

[3]

S.append(N) 1 mark for
def POP(S): correct POP
if S!=[]: operation
return S.pop()
else: 1 mark for
return None
correct
ST=[]
for k in N: function calls
if k%2==0: and
PUSH(ST,k) displaying
while True: the output
if ST!=[]:
print(POP(ST),end=" ")
else:
break

Note: Marks
to be
awarded for
any other
correct logic
given by the
student
9. (i) ALTER TABLE Item 1 mark for
ADD (Discount INT); correct (1)
command
(ii) DDL: DROP TABLE, ALTER TABLE ½ mark for (2)
DML: INSERT INTO, UPDATE...SET each correct
command
identified
10. CREATE DATABASE MYEARTH; 1 mark for
correctly
CREATE TABLE CITY creating
( database.
CITYCODE CHAR(5)PRIMARY KEY, (3)
CITYNAME CHAR(30),
2 marks for
SIZE INT,
AVGTEMP INT, correctly
POPULATIONRATE INT, creating the
POPULATION INT, table.
);

Section C
Each question carries 4 marks
11. (a) SELECT AVG(SALARY)

[4]

FROM EMPLOYEE
GROUP BY DEPTID;

(b) SELECT NAME, DEPTNAME
FROM EMPLOYEE, DEPARTMENT each correct
WHERE query
EMPLOYEE.DEPTID= (4)
DEPARTMENT.DEPTID
AND SALARY>50000;

(c) SELECT NAME FROM EMPLOYEE
WHERE SALARY IS NULL
ORDER BY NAME;

(d) SELECT DISTINCT DEPTID
FROM EMPLOYEE;
12. (i) Advantages
 Ease of service
 Centralized control
 Easy to diagnose faults
 One device per connection ½ mark for
Disadvantages advantage / (2)
 long cable length disadvantage
 difficult to expand
 central node dependency

OR

www: a set of protocols that allow you to access any
document on the internet through the naming 1 mark for
systems based on URLs each correct
Web hosting: Web hosting is a service that definition
allows organizations and individuals to post a
website or web page onto the server, which can be
viewed by everyone on the Internet.
(ii) Packet switching: 1 mark for
 uses store and forward concept to send each correct
messages difference
 no physical path is actually establishes
 message is divided into smaller parts, known (minimum
as packets and then sent forward two points (2)
 tight upper limit on block size should be
 Each data unit knows only the final given)
receiver’s address
[5]

Circuit switching
 physical connection is established between
sender and receiver
 Each data unit knows the entire path from
sender to receiver
 It does not follow store and forward
concept

13.

(a)

BeHappy Corporation (4)
Block B
Block A

Block C Block D

(b)
Repeater : between C and D as the distance
between them is 100 mts.

Hub/ Switch : in each block as they help to share
data packets within the devices of the network in
each block

(c) WAN.

(d) Satellite

[6]