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Kerala Plus One Malayalam Textbook Questions and Answers, Notes, Chapters Summary HSSLive

Unit 1 Kinav

• Chapter 1 Sandarsanam
• Chapter 2 Ormayude Njarambu
• Chapter 3 Verukal Nashtapedunnavar
• Chapter 4 Matsyam

Unit 2 Kalca

• Chapter 1 Kayalarikathu
• Chapter 2 Cinemayum Samoohavum Kalavupoya Cyclum
• Chapter 3 Kazhinjupoya Kalaghattavum
• Chapter 4 Kaippad Kelkkunnundo?

Unit 3 Ullariv

• Chapter 1 Kavyakalaye Kurichu Chila Nireekshanangal
• Chapter 2 Oonjalil
• Chapter 3 Anargha Nimisham
• Chapter 4 Lathiyum Vediyundayum

Unit 4 Uravu

• Chapter 1 Peeli Kannukal
• Chapter 2 Anukampa
• Chapter 3 Muhyadheen Mala
• Chapter 4 Vasanavikrithi
• Chapter 5 Samkramanam
• Chapter 6 Shasthrakriya

Plus One Malayalam Guide has been designed with a view to improving the skill of answering all types of questions quickly and precisely. Plus One Malayalam material will boost your confidence in the subject and help you face examinations with ease.

Salient Features of Plus One Malayalam Material

• A simple and brief summary of each lesson is given with its Malayalam translation.
• All the activities and questions are given after each lesson have been fully answered.
• The answers are given immediately after the questions.
• Besides the questions in the text, Exam-oriented questions and their answers in the new pattern are given.
• Questions on grammar and discourse are also included in this guide.

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Plus One Malayalam Previous Year Question Papers and Answers

HSSLive Plus One

Students can Download Chapter 2 Units and Measurement Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 2 Units and Measurement

Plus One Physics Units and Measurement One Mark Questions and Answers

Question 1.
How many seconds are there in a light fermi?
(a) 10^-15
(b) 3.0 × 10⁸
(c) 3.33 × 10^-24
(d) 3.3 × 10^-7
Answer:
(c) 3.33 × 10^-24
One light fermi is time taken by light to travel a distance of 1 fermi ie. 10^-15m
1 light fermi = \(\frac{10^{-15}}{3 \times 10^{8}}\) = 3.33 × 10^-24s.

Question 2.
Which of the following pairs have same dimensional formula for both the quantities?

1. Kinetic energy and torque
2. Resistance and Inductance
3. Young’s modulus and pressure

(a) (1)only
(b) (2) only
(c) (1) and (3) only
(d) All of three
Answer:
(c) (1) and (3) only

Question 3.
Give four dimensionless physical quantities.
Answer:
Angle, Poisson’s ratio, strain, specific gravity.

Question 4.
The dimensions of plank constant are the same as those of______.
Answer:
Angular momentum

Question 5.
A physical quantity P = \(\frac{\sqrt{a b c^{2}}}{d^{3}}\) measuring a, b, c and d separately with the percentage error of 2% , 3%, 2% and 1% respectively. Minimum amount of error is contributed by the measurement of
(a) b
(b) a
(c) d
(d) c
Answer:
(b) a
P = \(\frac{\sqrt{a b c^{2}}}{d^{3}}\)

The minimum amount of error is contributed by the measurement of a.

Question 6.
The number of significant figures in 11.118 × 10^-6 is
(a) 3
(b) 6
(c) 5
(d) 4
Answer:
As per rules, number of significant figures in 11.118 × 10^-6 is 5.

Question 7.
What is the number of significant figures in 0.06070?
Answer:
4.

Question 8.
If f = x², What is the relative error in f?
Answer:
\(\frac{2 \Delta x}{x}\).

Question 9.
Which of the following measurement is more accu¬rate?
(i) 7000m
(ii) 7 × 10²m
(iii) 7 × 10³m
Answer:
(i) 7000 m

Question 10.
Which of the following measurements is most, accurate?
(a) 5.0 cm
(b) 0.005 cm
(c) 5.00 cm
Answer:
(c) Is most accurate because it has three significant figures. Greater is number of significant figures, more accurate is the measurement.
(a) has 2 significant figures
(b) has 1 significant figure.

Question 11.
Name three physical quantities having same dimension.
Answer:
Work, Energy, and Torque.

Plus One Physics Units and Measurement Tw0 Mark Questions and Answers

Question 1.
Using dimensional analysis derive the relation F = ma. Where the symbols have the usual meaning.
Answer:
Force on a body depends on mass(m), acceleration (a) an
F α m^aa^bt^c
M¹L¹T^-2 = M^a(LT^-2)^bT^c
M¹L¹T^-2 = M^aL^bT^-2a+c
Equating the powers, we get a = 1 ,b = 1, -2b + c = -2, c = 0
F = m¹a¹t⁰ = ma.

Question 2.
Use your definition to explain how simple harmonic motion can be represented by the equation y = a sin ωt
(a) Show that the above equation is dimensionally correct
Answer:
Y = a sin ωt
sin ωt has no dimensions. Hence we get L = L
Hence this equation is dimensionaly correct.

Question 3.
Fill in the blanks.

1. The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_____m² (Write answer in 3 significant digits)
2. Im = ______ ly

Answer:
1. Curved area = 2πl
= 2 × 3.14 (2 × 10²) × 20 × 10²
= 2.51 × 10^-6m²

2. l ly= 9.46 × 10¹⁵ m
lm = \(\frac{l \mathrm{ly}}{9.46 \times 10^{15}}\) ≈ 10^-6ly.

Question 4.

1. Give a physical quantity with a unit and no dimension.
2. Arrange the following in the descending order.
   1 light year, 1 parsec, 1 astronomical unit

Answer:

1. Angle has no dimension. But it has unit.
2. 1 parsec, 1 light year, 1 astronomical unit.

Question 5.
Magnitude of force F experienced by a certain object moving with speed V is given by F = KV². Where K is a constant. Find the dimensions of K.
Answer:
F = KV²

Question 6.
What is the maximum percentage error in the measurement of kinetic energy if percentage errors in mass and speed are 2% and 3% respectively?
Answer:
E = \(\frac{1}{2}\)v²

% error in KE = % error in mass + 2 × % error in speed
= 2% + 2 × 3% = 8%.

Question 7.
Solve the following with regard to significant figures.

1. 5.8 + 0.125
2. 3.9 × 10⁵ – 2.5 × 10⁴

Answer:
1. 5.8 + 0.125 = 5.925
Rounding to first decimal point, we get 5.9

2. 3.9 × 10⁵ – 2.5 × 10⁴
= 3.5 × 10⁵ – 0.25 × 10⁴
= 3.65 × 10⁵
Rounding to first decimal place, we get 3.6 × 10⁵.

Question 8.
What is maximum fractional error in
i) (a + b)
ii) a – b
iii) ab
iv) \(\frac{a}{b}\)
Given ∆ a and ∆ b are absolute errors in measurements a and b.
Answer:

Question 9.

1. What is the fractional error in aⁿ? (Given absolute error in a is ∆ a)
2. What is absolute error in the measurements according to least count?
   • 3.0 kg
   • 25 s
   • 5.62 cm

Answer:
1. n\(\frac{\Delta a}{a}\)

2. The measurements according to least count:

• 0.1 kg
• 1 s
• 0.01 cm

Plus One Physics Units and Measurement Three Mark Questions and Answers

Question 1.
A stone is thrown upwards from the ground with a velocity ‘u’.

1. What is the maximum height attained by the stone?
2. Check the correctness of the equation obtained in (a) using the method of dimensional analysis.

Answer:
1. H = \(\frac{u^{2}}{2 g}\) ___(1)
u = u, v = o, a = -g, h = ?
We can find maximum height using the equation
u² = u² + 2as
0 = u² + 2 × -g × H
2gh = u²
H = \(\frac{u^{2}}{2 g}\)

2. Dimension of H = L
Dimension of u = (LT^-1)
Dimensions of time (t) = T
Dimension of g = (LT^-2)
substituting these values in eq(1) we get
L = \(\frac{\left(L T^{-1}\right)^{2}}{\left(L T^{-2}\right)}\)
L = L.

Question 2.
Derive an empirical relationship for the force experienced on the car in terms of mass of the car m, velocity v, and radius of the track r using dimensional analysis.
Answer:
Centripetal force may depends on mass (m),radius(r) and velocity(v)
F α m^ar^bv^c
M¹L¹T^-2 = M^aL^b(LT^-1)^c
M¹L¹T^-2 = M^aL^bL^cT^-c
M¹L¹T^-2 = M^aL^b+cT^-c
Equating we get a = 1, b + c = 1, c = 2, b = -1
Substituting these values in eq(1),we get
F = \(\frac{M V^{2}}{r}\).

Question 3.
Dimensional formula of a physical quantity indicate how many times fundamental quantity is involved in the measurement of the quantity.

1. What is the dimensional formula of coefficient of viscosity?
2. Write any two drawbacks of dimensional analysis.

Answer:
1. F = ηA\(\frac{d V}{d x}\)

2. The method of dimensional analysis has the following drawbacks:

• It gives no information about the dimensionless constant involved in the equation.
• The method is not applicable to equations involv¬ing trigonometric and exponential functions.
• This method cannot be employed to derive the • exact form of the relationship if it contains sum
  of two, or more terms.
• If the given physical quantity depends on more than three unknown quantities, the method fails.

Question 4.
Principle of homogeneity is based on the fact that two quantities of same nature can be added.

1. What do you mean by principle of homogeneity?
2. Velocity V depends on the time t as V = at² + bt + c. Find dimension of constants a, b, and c.

Answer:
1. For the correctness of an equation, the dimensions on either side must be the same. This is known as the principle of homogeneity of dimensions.

2. V = at² + bt + c
M⁰L¹T^-1 = aT² + bT + c
According to principle of homogenity, we get
aT² = M⁰L¹T^-1
a = \(\frac{\mathrm{M}^{0} \mathrm{L}^{1} \mathrm{T}^{-1}}{\mathrm{T}^{2}}\)
= M⁰L¹T^-3.

Question 5.
If x = a + bt + ct² where x is in meter and t in second.

1. Find the dimensional formula of ‘b’.
2. If error in the measurement of time is 2%. What will be the error in x?

Answer:
1. According to principle of homogeneity, the dimensions of both sides must be same.
ie. L = a + bT + cT²
ie : L = bT, b = L/T

2.

% error in x = 3 × %. error in ‘t’ = 3 × 2% = 6%.

Question 6.
A physical quantity P is related to four observables a, b, c as P = \(\frac{a^{3} b^{2}}{\sqrt{c d}}\). The % error in the measurement of a, b, c, and d are 1%, 3%, 4%, 2% are respectively.

1. What do you mean by error in a measurement?
2. What is the % error in the measurement of P?

Answer:
1. The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called error.

2.

% error in
P = 3 × 1 + 2 × 3 + 1/2 × 4 + 1/2 × 2
= 3 + 6 + 2+ 1
P = 12%

Question 7.
Rahul measured the height of Ramesh in different trials as 1.67m, 1.65m 1.64m, and 1.63m.

1. Find the mean absolute error?
2. Find the percentage error?

Answer:
1. Arithametic mean,

a_mean = 1.645m = 1.65
absolute error,
∆a₁ = a_mean – a₁
∆a₁ = 1.65 – 1.67 = -0.02
∆a₂ = 1.65 – 1.65 = 0
∆a₃ = 1.65 – 1.64 = 0.01
∆a₄ = 1.65 – 1.63 = 0.02
Mean absolute error

= 0.012

2. percentage error = \(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}\) × 100
= \(\frac{0.012}{1.65}\) × 100
= 0.75%.

Question 8.
In a particular experiment Ramu used the relation F = AB + (P + Q) Y to calculate force.

1. Which principle is used to check the correctness of the equation (1)
2. If the dimensional formula of Y is M⁰L¹T^-1, then find the dimensional formula of P

Answer:
1. Principle of homogenity

2. F = AB + (P+Q)Y
F = AB + PY + QY
MLT^-2 = AB + PY+ QY
According to principle of homogeneity
MLT^-2 = PY
M¹L¹T^-2 = P M⁰L¹T^-1
ie. P = \(\frac{M^{\prime} L^{1} T^{-2}}{M^{0} L^{1} T^{-1}}\) = M¹T^-3

Question 9.

1. Which of the following is precise
   • A vernier calliperse with 40 divisions on sliding scale
   • An optical instrument that can measure length of the order of wavelength of light.
2. Is it possible to increase the accuracy of screw gauge by increasing the number of divisions on the head scale?

Answer:
1. (i) L.C of vernier caliperse = \(\frac{1}{40}\) = 0.025mm
= 0.025 × 10^-3m
= 2.5 × 10^-5m.

(ii) L.C of optical instrument = 6000A°
= 6000 × 10^-10m
(Taking λ of visible light = 6000°A)= 6 × 10^-7m

2. Yes. Because L.C proportional to number of division on the headscale. So with the increase in number of divisions, the least count will increase. This leads to increase the accuracy of above screw guage.

Plus One Physics Units and Measurement Four Mark Questions and Answers

Question 1.
In an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate

1. The mean value of the body
2. Mean absolute error
3. Percentage error

Answer:
1. Mean value, M_mean
= \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\)
= 2.5g

2. Absolute error,
Absolute error ∆m₁ = |2.52 – 2.52| = 0
∆m₂ = |2.52 – 2.53| = 0.01
∆m₃ = |2.52 – 2.51| = 0.01
∆m₄ = |2.52 – 2.49| = 0.03
∆m₅ = |2.52 – 2.54| = 0.02
∴ Mean absolute error
\(\frac{0+0.01+0.01+0.03+0.02}{5}\)
∆m_mean = 0.014g

3. Percentage error = \(\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}}\) × 100
= \(\frac{0.014}{2.52}\) × 100 = 0.556.

Question 2.
While discussing the period of a pendulam, one of the student argued that period depends on the mass of the bob.

1. What is your opinion?
2. How will you prove your argument dimensionally?

Answer:

1. Period is independent of mass of the bob
2. The principle of homogeneity of dimensions also helps to derive a relationship between the different physical quantities involved; This method is known as dimensional analysis.

The period of the simple pendulum may possibly depend upon:

• The mass of the bob, m
• The length of the pendulum, I
• Acceleration due to gravity, g
• The angle of swing, q

Let us write the equation for the time period as t = k m^a l^b g^c q^d
where, k is a constant having no dimensions; a, b, care to be found out.
The dimensions of, t = T¹
Dimensions of. m = M¹
Dimensions of, l = L¹
Dimensions of, g = L¹T^-2
Angle q has no dimensions (since, q = arc/radius = L/L)
Equating the dimensions of both sides of the equation, we get,
T¹ = M^aL^b (L¹T^-2)^c
ie. T¹ = M^aL^b+c+ T^-2c.
The dimensions of the terms on both sides must be the same. Equating the powers of M, L and T.

a = 0; b + c = 0; -2c = 1
∴ c = ^–\(\frac{1}{2}\), b = ^– c = \(\frac{1}{2}\)
Hence, the equation becomes,
t = kl^1/2g^-1/2
ie, t = k\(\sqrt{1 / g}\)
Experimentally, the value of k is found to be 2p.

Plus One Physics Units and Measurement Five Mark Questions and Answers

Question 1.
In an experiment with a common balance the mass of a ring found to be 2.52g, 2.5g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate

1. The mean value of the mass of the ring
2. The absolute error in each measurement
3. Mean absolute error
4. Relative error
5. Percentage error

Answer:
1. The mean value of the mass of the ring.
M_mean = \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\) = 2.52g.

2. The absolute error in each measurement.
∆m₁ = M_mean – m₁ = 2.52 – 2.52 = 0.00
∆m₂ = M_mean – m₂ = 2.52 – 2.53 = -0.01
∆m₅ = M_mean – m₅ = 2.52 – 2.54 = -0.02

3. mean absolute error = |∆m₁| + |∆m₂|………..+|∆m₅|
= 0.014

4. Relative error = δm = \(\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}}=\frac{.014}{2.52}\) = 0.00555

5. Percentage error δm × 100 = 0.555%.

Plus One Physics Units and Measurement NCERT Questions and Answers

Question 1.
Fill in the blanks:

1. The volume of a cube of side 1 cm is equal to______m³.
2. The surface area of a solid cylinder of radius 2.0 cm and height 10.0cm is equal to____(mm)².
3. A vehicle moving with a speed of 18km h^-1 covers_____m in 1s.
4. The relative density of lead is 11.3.Its density is_____g cm^-3or_____kgm^-3.

Answer:
1. V = (1 cm)³
= (10^-2m)³
= 10^-6m³
So, answer is 10^-6.

2. Surface area = 2 πrh + 2 × πr²
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 2 × 10(10 × 10 + 2 × 10)mm²
= 1 .5 × 10⁴mm²
So, answer is 1.5 × 10⁴

3. 18kmh^-1 = \(\frac{18 \times 1000}{3600}\)ms^-1
= 5ms^-1
So, answer is 5.

4. 11.3, 11.3 × 10³ or 1.13 × 10⁴.

Question 2.
Fill in the blanks by suitable conversion of units:

1. 1 kgm²s^-2 = _____g cm²s^-2
2. 1 m =_____1 y
3. 3.0 ms² =______kmh^-2
4. G = 6.67 × 10^-11 Nm² (kg)^-2 =_____(cm)³s^-2g^-1

Answer:

1. 10⁷
2. 10^-16
3. 3.888 × 10⁴
4. 6.67 × 10^-8

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J, where 1 J = 1 kgm²S^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β metere, the unit of time is second, show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in terms of the new units.
Answer:
1 cal = 4.2kg m²s^-2

SI	New system
n1 = 4.2	n2 = ?
M1 = 1 kg	M2 = α kg
L1 = 1m	L2 = β meter
T1 = 1s	T2 = γ second

Dimensional formula of energy is [ML²T^-2]. Comparing with [M^aL^bT^c], we find that
a = 1, b = 2, c = -2

Question 4.
Which of the following is the most precise device for measuring length?

1. A vernier callipers with 20 divisions on the sliding scale.
2. A screw guage of pitch 1 mm and 100 divisions on the circular scale
3. An optical instrument that can measure length to within a wavelength of light?

Answer:
The most precise device is one whose least count is the least.
1. Least count = 1SD – 1 VD = 1 SD – \(\frac{19}{20}\) SD

2. Least count

3. Wavelength = 10^-5 cm = 0.00001 cm
Clearly, the optical instrument is the most precise.

Question 5.
State the number of significant figures in the following:

1. 0.007m²
2. 2.64 × 10²⁴kg
3. 0.2370gcm³
4. 6.320 J
5. 6.032 Nm^-2
6. 0.0006032 m²

Answer:

1. 1
2. 3
3. 4
4. 4
5. 4
6. 4

Question 6.
The length, breadth, and thickness of a rectangular sheet of metal are 4.234m, 1.005m and 2.01cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Area of the sheet = 2(l × b + b × t + t × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)m²
= 2 (4.255 + 0.0202 + 0.0851)m²
= 2 × 4.3603m²
= 8.7206m²
= 8.72m²
Volume = lbt
4.234 × 1.005 × 0.0201m³
= 0.0855m³

Question 7.
A Physical qunatity P is related to four observables a, b, c and d as follows:
P = \(\frac{\mathrm{a}^{3} \mathrm{b}^{2}}{\sqrt{\mathrm{c}} \mathrm{d}}\). The percentage errors of measurement in a, b,c, and d are 1 %, 3%, 4% and 2% respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Answer:
P = \(\frac{\mathrm{a}^{3} \mathrm{b}^{2}}{\sqrt{\mathrm{c}} \mathrm{d}}\)

% error in P = 3% + 6% + 2%+2% = 13%
3.763 should be rounded off to 3.8.

Question 8.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of special relativity due to Albert Einstein). Boy recalls the relation almost correctly but forgets where to put the constant c. He writes:
\(\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}\)
Guess where to put the missing e.
Answer:
From the given equation, \(\frac{m_{0}}{m}=\sqrt{1-v^{2}}\)
Since left hand side is dimensionless therefore right hand side should be also dimensionless.

The correct formula is m = m₀ \((\sqrt{1-\frac{v^{2}}{c^{2}}})^{-1 / 2}\).

Students can Download Chapter 6 Data Types and Operators Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Plus One Data Types and Operators One Mark Questions and Answers

Question 1.
___________ is the main activity carried out in computers.
Answer:
Data processing

Question 2.
The data used in computers are different. To differentiate the nature and size of data ___________ is used.
Answer:
Data types

Question 3.
Classify the following data types,
int, array, function, char, pointer, void, float, double, structure
Answer:

Fundamental data types	Derived data types
int	array
float	function
double	pointer
void	structure
char

Question 4.
Sheela wants to store her age. From the following which is the exact data type.
(a) void
(b) char
(c) int
(d) double
Answer:
(c) int

Question 5
Integer data type uses ________ bytes of memory.
(a) 5
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 6.
char data type ________uses bytes of memory.
(a) 1
(b) 3
(c) 7
(d) 8
Answer:
(a) 1

Question 7.
From the following which data type uses 4 bytes of memory.
(a) float
(b) short
(c) char
(d) double
Answer:
(a) float

Question 8.
Full form of ASCII is ___________
Answer:
American Standard Code for Information Interchange

Question 9.
Ramu wants to store the value of From the following which is correct declaration.
(a) char pi = 3.14157
(b) int pi = 3.14157
(c) float pi = 3.14157
(d) long pi = 3.14157
Answer:
(c) float pi = 3.14157

Question 10.
From the following which is not true, to give a variable name.
(a) Starting letter must be an alphabet
(b) contains digits
(c) Cannot be a keyword
(d) special characters can be used
Answer:
(d) special characters can be used

Question 11.
Pick a valid variable name from the following.
(а) 9a
(b) float
(c) age
(d) date of birth
Answer:
(c) age.

Question 12.
To perform a unary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(c) 1 (Unary means one)

Question 13.
To perform a binary operation how many number of operands needed?
(а) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(a) 2 (binary means two)

Question 14.
To perform a ternary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(b) 3 (eg: ternary means three)

Question 15.
In C++ 13 % 26 =
(a) 26
(b) 13
(c) 0
(d) None of these
Answer:
(b) % is a mod operator i.e. it gives the remainder. Here the remainder is 13.

Question 16.
In C++ 41/2 =
(a) 20.5
(b) 20
(c) 1
(d) None of these
Answer:
(b) 20. (The actual result is 20.5 but both 41 and 2 are integers so .5 must be truncated)

Question 17.
++ is a __________ operator.
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(a) Unary.

Question 18.
Conditional operator is _________ operator.
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(c) Ternary

Question 19.
% is a _______ operator
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(b) Binary

Question 20.
State True/False

1. Multiplication, division, modulus have equal priority
2. Logical and (&&) has less priority than logical or ()

Answer:

1. True
2. False

Question 21.
_______is composed of operators and operands.
(a) expression
(b) Keywords
(c) Identifier
(d) Punctuators
Answer:
(a) expression

Question 22.
Supply value to a variable at the time of declaration is known as __________.
Answer:
Initialisation

Question 23.
From the following which is initialisation.
(a) int k;
(b) int k = 100;
(c) int k[10];
(d) None of these
Answer:
(b) int k = 100;

Question 24.
State True/False
In an expression all the operands having lower size are converted(promoted) to the data type of the highest sized operand.
Answer:
True

Question 25.
Classify the following as arithmetic/Logical expression.
(a) x + y * z
(b) x < y && y > z
(c) x / y
(d) x > 89 || y < 80
Answer:
(a) and (c) are Arithmetic, (b) and (d) are Logical

Question 26.
Suppose x = 5 and y = 2 then what will be cout<<(float)x/y
Answer:
2.5 The integer x is converted to float hence the result.

Question 27.
Consider the following.
a = 10;
a* =10;
Then a =
(a) a = 100
(b) a = 50
(c) a = 10
(d) a = 20
Answer:
(a) a = 100. This short hand means a = a*10

Question 28.
Consider the following.
a = 10;
a+ = 10;
Then a =
(a) a = 30
(b) a = 50
(c) a = 10
(d) a = 20
Answer:
(d) a = 20. This short hand means a = a + 10

Question 29.
Pick the odd one out
(a) structure
(b) Array
(c) Pointer
(d) int
Answer:
(d) int, it is fundamental data type the others are derived data types

Question 30.
From the following select not a character of C++ language
(a) A
(b) 9
(c) \
(d) @
Answer:
(d) @

Question 31.
Consider the following
float x = 25.56;
cout<<(int)x;
Here the data type of the variable is converted. What type of conversion is this?
(a) type promotion
(b) type casting
(c) implicit conversion
(d) None of these
Answer:
(b) type casting (explicit conversion);

Question 32.
Identify the error in the following C++ statement and correct it.
short population = 68000;
Answer:
The maximum number that can store in short type is less than 32767. So to store 68000 we have to use long data type.

Question 33.
Consider the following statements in C++ if(mark>=18)
cout<<“Passed”;
else
cout<<“Failed”;
Suggest an operator in C++ using which the same output can be produced.
Answer:
Conditional operator (?:)

Plus One Data Types and Operators Two Mark Questions and Answers

Question 1.
Analyses the following statements and write True or False. Justify

1. There is an Operator in C++ having no special character in it
2. An operator cannot have more than 2 operands
3. Comma operator has the lowest precedence
4. All logical operators are binary in nature
5. It is not possible to assign the constant 5 to 10 different variables using a single C++ expression
6. In type promotion the operands with lower data type will be converted to the highest data type in expression.

Answer:

1. True (sizeof operator)
2. False(conditional operator can have 3 operands)
3. True
4. False
5. False(Multiple assignment is possible)
   eg: a = b = c ==5
6. True

Question 2.
Consider the following declaration.
const int bp;
bp = 100;
Is it valid? Explain it?
Answer:
This is not valid. This is an error. A constant variable cannot be modified. That is the error and a constant variable must be initialised. So the correct declaration is as follows, const int bp = 100;

Question 3.
Consider the following statements in C++

1. cout<<41/2;
2. cout<<41/2.0;

Are this two statements give same result? Explain?
Answer:
This two statements do not give same results. The first statement 41/2 gives 20 instead of 20.5. The reason is 41 and 2 are integers. If two operands are integers the result must be integer, the real part must be truncated.

To get floating result either one of the operand must be float. So the second statement gives 20.5. The reason 41 is integer but 2.0 is a float.

Question 4.
If mark = 70 then what will be the value of variable result in the following
result = mark > 50? ’P’: ’F’;
Answer:
The syntax of the conditional operator is given below
Condition ? Value if true : Value if false;
Here the conditional operator first checks the condition i.e., 70 > 50 it is true. So ’P’ is assigned to the variable result.
So the result is ’P’;

Question 5.
Is it possible to initialise a variable at the time of execution. What kind of initialisation is this? Give an example
Answer:
Yes it is possible. This is known as Dynamic initialisation. The example is given below
eg: int a = 10, b = 5; int c = a*b; here the variable c is declared and initialised with the value 10*5.

Question 6.
Boolean data type is used to store True/False in C++. Is it true? Is there any data type called Boolean in C++?
Answer:
No there is no data type for storing boolean value true/false. But in C++ non -zero (either negative or positive) is treated as true and zero is treated as false

Question 7.
Consider the following
n=-15;
if (n)
cout<<“Hello”;
else
cout<<“hai”;
What will be the output of the above code?
Answer:
The output is Hello, because n = -15 a non zero number and it is treated as true hence the result.

Question 8.
Is it possible to declare a variable in between the program as and when the need arise? Then what is it?
Answer:
Yes it is possible to declare a variable in between the program as and when the need arise. It is known as dynamic initialisation.
eg. int x = 10, y = 20;
int z = x*y;

Question 9.
char ch;
cout<<“Enter a character”;
cin>>ch;
Consider the above code, a user gives 9 to the variable ‘ch’. Is there any problem? Is it valid?
Answer:
There is no problem and it is valid since 9 is a character. Any symbol from the key board is treated as a character.

Question 10.
“With the same size we can change the sign and range of data”. Comment on this statement.
Answer:
With the help of type modifiers we can change the sign and range of data with same size. The important modifiers are signed, unsigned, long and short.

Question 11.
Write short notes about C++ short hands?
Answer:
x = x + 10 can be represented as x + = 10, It is called shorthands in C++. It is faster. This is used with all the arithmetic operators as follows.

Arithmetic Assignment Expression	Equivalent Arithmetic Expression
x+ = 10	x = x + 10
x- = 10	x = x -10
X* = 10	x = x * 10
x/ = 10	x = x /10
x% = 10	x = x % 10

Question 12.
What is the role of ‘const’ modifier?
Answer:
This ‘const’ keyword is used to declare a constant.
eg: const int bp=100;
By this the variable bp is treated as constant and cannot be possible to change its value during execution.

Question 13.
Specify the most appropriate data type for handling the following data.

1. Roll no. of a student.
2. Name of an employee.
3. Price of an article.
4. Marks of 12 subjects

Answer:

1. short Roll no;
2. char name[20];
3. float price;
4. short marks[12];

Question 14.
Write C++ statement for the following,

1. The result obtained when 5 is divided by 2.
2. The remainder obtained when 5 is divided by 2.

Answer:

1. 5/2
2. 5 % 2

Question 15.
Predict the output of the following code. Justify.
int k = 5;
b = 0;
b = k++ + ++k;
cout<<b;
Answer:
Output is 12. In this statement first it take the value of k in 5 then increment it K++. So first operand for + is 5. Then it becomes 6. Then ++k makes it 7. This is the second operand. Hence the result is 12.

Question 16.
Predict the output.
1. int sum = 10, ctr = 5;
sum = sum + ctr–;
cout<<sum;

2. int sum = 10, ctr = 5;
sum = sum + ++ctr;
cout<<sum;
Answer:

1. 15
2. 16

Question 17.
Predict the output.
int a;
float b;
a = 5;
cout<<sizeof(a + b/2);
Answer:
Output is 4. Result will be the memory size of floating point number

Question 18.
Predict the output.
int a, b, c;
a = 5;
b = 2;
c = a/b;
cout<<c;
Answer:
Output is 2. Both operands are integers. So the result will be an integer

Question 19.
Explain cascading of i/o operations
Answer:
The multiple use of input or output operators in a single statement is called cascading of i/o operators.
eg: To take three numbers by using one statement is as follows
cin>>x>y>>z;
To print three numbers by using one statement is as follows
cout<<x<<y<<z;

Question 20.
Trace out and correct the errors in the following code fragments

1. cout<<“Mark =”45;
2. cin<<“Hellow World!”;
3. cout>>”X + Y;
4. Cout<<‘Good,<<‘Morning’

Answer:

1. cout<<“Mark = 45”;
2. cout<<“Hellow World!”;
3. cout<<X + Y;
4. Cout<<“Good Morning”;

Question 21.
Raju wants to add value 1 to the variable ‘p’ and store the new value in ‘p’ itself. Write four different statements in C++ to do the task.
Answer:

1. P=P+1;
2. p++;(post increment)
3. ++p;(pre increment)
4. p+=1;(short hand in C++)

Question 22.
Read the following code
char str [30];
cin>>str;
cout<<str;
If we give the input “Green Computing”, we get the output “Green”. Why is it so? How can you correct that? (2)
Answer:
The input statement cin>> cannot read the space. It reads the text up to the space, i.e. the delimiter is space. To read the text up to the enter key gets() or getline() is used.

Question 23.

Name	Symbol
(a) Modulus operator	(i) ++
(b) Logical Operator	(ii) ==
(c) Relational Operator	(iii) =
(d) Assignment operator	(iv) ?:
(e) Increment operator	(v) &&
(f) Conditional Operator	(vi) %

Answer:

Name	Symbol
(a) Modulus operator	(vi) %
(b) Logical Operator	(v) &&
(c) Relational Operator	(ii) ==
(d) Assignment operator	(iii) =
(e) Increment operator	(i) ++
(f) Conditional Operator	(iv) ?:

Question 24.
Write a C++ expression to. calculate the value of the following equation.

Answer:
x = (-b + sqrt(b*b – 4*a*c)/(2*a)

Question 25.
A student wants to insert his name and school address in the C++ program that he has written. But this should not affect the compilation or execution of the program. How is it possible? Give an example.
Answer:
He can use comments to write this information. In C++ comments are used to write information such as programmer’s name, address, objective of the codes etc. in between the actual codes. This is not the part of the programme. There are two types of comments

1. Single line (//) and
2. Multi-line (/* and *f)

1. Single line (if):
Comment is used to make a single line as a comment. It starts with //.
eg: /./programme starts here.

2. Multi-line (/* and */):
To make multiple lines as a comment. It starts with /* and ends with */.
Eg: /* this programme is used to find sum of two numbers */

Question 26.
Consider the following C++ statements:
char word [20];
cin>>word;
cout<<word;
gets(word);
puts(word);
If the string entered is “HAPPY NEW YEAR”, predict the output and jsutify your answer.
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets () function. Because gets () function reads character upto the enter key.
Hence
gets(word);
puts(word);
Displays “HAPPY NEW YEAR”

Question 27.
Write the difference between x = 5 and x == 5 in C++.
Answer:
x = 5 means the value 5 of the RHS is assigned to the LHS variable x . Here = is the assignment operator. But x == 5, == this is the relational (comparison) operator. Here it checks whether the value of RHS is equal to the value of LHS and this expression returns a boolean value as a result. It is the equality operation.

Question 28.
1. What is the output of the following program?
# include <iostream.h>
void main ()
{
int a;
a = 5 + 3*5;
cout << a;
}

2. How do 9, ‘9’ and “9” differ in C++ program?
Answer:
Here multiplication operation has more priority than addition.
hence
1. a = 5 + 15 = 20

2. Here 9 is an interger
‘9’ is a character
“9” is a string

Question 29.
Read the following C++ program and predict the output by explaining the operations performed.
#include<iostream.h>
void main ()
{
int a = 5, b = 3;
cout<<a++ /–b;
cout<<a/ (float) b;
}
Answer:
Here a = 5 and b = 3
a++ /– b = 5/2 = 2
That is a++ uses the value 5 and next it changes its value to 6
So a/(float) b = 6/(float)2
= 6/2.0
= 3
So the output is 2 and 3

Question 30.
What is the preprocessor directive statement? Explain with an example.
Answer:
A C++ program starts with the preprocessor directive i.e., # include, #define, #undef, etc, are such a preprocessor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
eg. #define x 100. Here the value of x becomes 100 and cannot be changed in the program. No semicolon is needed.

Question 31.
The following C++ code segment is a part of a program written by Smitha to find the average of 3 numbers.
int a, b, c;
float avg;
cin>>a>>b>>c;
avg = (a + b + c)/3;
cout<<avg; .
What will be the output if she inputs 1, 4 and 5? How can you correct it?
Answer:
= (1 + 4 + 5)/3
= 10/3
= 3.3333
Instead of this 3.3333 the output will be 3. This is because if both operands are integers an integer division will be occurred, that is the fractional part will be truncated. To get the correct output do as follows
case 1: int a,b,c; is replaced by float a,b,c;

OR

case 2: Replace (a + b + c)/3 by (a + b + c)/3.0;

OR

case 3: Type casting.
Replace avg = (a + b + c)/3;
by avg = (float)(a + b + c)/3;

Plus One Data Types and Operators Three Mark Questions and Answers

Question 1.
In a panchayath or municipality all the houses have a house number, house name and members. Similar situation is in the case of memory. Explain
Answer:
The named memory locations are called variable. A variable has three important things

1. variable name: A variable should have a name
2. Memory address: Each and every byte of memory has an address. It is also called location (L) value
3. Content: The value stored in a variable is called content.lt is also called Read(R) value.

Question 2.
Briefly explain constants.
Answer:
A constant or a literal is a data item its value doe not change during execution. The keyword const is used to declare a constant. Its declaration is as follows
const data type variable name = value;
eg.const int bp = 100;
\const float pi = 3.14157;
const char ch = ‘a’;
const char[]=”Alvis”;

1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
eg: For decimal 100, 150, etc.
For octal 0100, 0240, etc.
For hexadecimal 0x100, 0x1A, etc.

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals.
eg: 3.14157, 79.78, etc

3. Character literal:
A valid C++ character enclosed in single quotes, its value does not change during execution.
eg: ‘m’, ‘f, etc.

4. String literal:
One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null character(‘\0’)
eg: “Mary’s”, “India”, etc.

Question 3.
Consider the following statements
int a = 10, x = 20;
float b = 45000.34, y = 56.78;
1. a = b;
2. y = x;
Is there any problem for the above statements? What do you mean by type compatibility?
Answer:
Assignment operator is used to assign the value of RHS to LHS. Following are the two chances
(a) The size of RHS is less than LHS. So there is no problem and RHS data type is promoted to LHS. Here it is compatible.

(b) The size of RHS is higher than LHS. Here comes the problem sometimes LHS cannot possible to assign RHS. There may be a chance of wrong answer. Here it is not compatible.
Here
1. a = b; There is an error since the size of LHS is 2 but the size of RHS is 4.
2. y = x; There is no problem because the size of LHS is 4 and RHS is 2.

Question 4.
A company has decided to give incentives to their salesman as perthe sales. The criteria is given below.
If the total sales exceeds 10,000 the incentive is 10%

1. If the total sales >= 5,000 and total sales <10,000, the incentive is 6 %
2. If the total sales >= 1,000 and total sales <5,000, the incentive is 3 %

Write a C++ program to solve the above problem and print the incentive after accepting the total sales of a salesman. The program code should not make use of ‘if’ statement.
Answer:
#include<iostream>
using namespace std;
int mainO
{
float sales,incentive;
cout<<“enter the sales”;
cin>>sales;
incentive = (sales>10000 ? sales*.10: (sales > =5000 ? sales * .06 : (sales >= 1000 ? sales * -03: 0)));
cout<<“\nThe incentive is ” << incentive;
}

Question 5.
A C++ program code is given below to find the value of X using the expression
\(x=\frac{a^{2}+b^{2}}{2 a}\)
where a and b are variables

#include<iostream>
using namespace std;
int main()
{
int a;b;
float x
cout<<“Enter the values of a and b;
cin>a>b;
x = a*a + b*b/2*a;
cout>>x;
}
Predict the type of errors during compilation, execution and verification of the output. Also write the output of two sets of input values

1. a = 4, b = 8
2. a = 0, b = 2

Answer:
This program contains some errors and the correct program is as follows.
#include<iostream>
using namespace std;
int main()
{
int a,b;
float x; .
cout<<“Enterthe values of a and b”;
cin>>ab;
x=(a*a + b*b)/(2*a);
cout<<x;
}
The output is as follows

1. a = 4 and b = 8 then the output is 10
2. a = 0 and b = 2 then the output is an error divide by zero error(run time error)

Question 6.
A list of data items are given below
45, 8.432, M, 0.124,8 , 0, 8.1 × 1031, 1010, a, 0.00025, 9.2 × 10120, 0471,-846, 342.123E03

1. Categorise the given data under proper headings of fundamental data types in C++.
2. Explain the specific features of each data type. Also mention any other fundamental data type for which sample

data is not given
Answer:
1.

2. The specific features of each data type.
(i) int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It con¬sumes 4 bytes (32 bits) of memory.i.e. 232 . numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve ) So a total of 232 numbers. We can store a number in between -231 to + 231-1.

(ii) char data type:
Any symbol from the keyboard, eg. ‘A’ , ‘?’, ‘9’,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Question 7.
Write valid reasons after reading the following statements in C++ and comment on their correctness by give reasons.

1. char num = 66;
   char num = B’;
2. 35 and 35L are different
3. The number 14, 016 and OxE are one and the same
4. Char data type is often said to be an integer type
5. To store the value 4.15 float data type is preferred over double

Answer:

1. The ASCII number of B is 66. So it is equivalent.
2. 35 is of integer type but 35L is Long
3. The decimal number 14 is represented in octal is 016 and in hexadecimal is OXE.
4. Internally char data type stores ASCII numbers.
5. To store the value 4.15 float data type is better because float requires only 4 bytes while double needs 8 bytes hence we can save the memory.

Question 8.
Suggest most suitable derived data types in C++ for storing the following data items or statements

1. Age of 50 students in a class
2. Address of a memory variable
3. A set of instructions to find out the factorial of a number
4. An alternate name of a previously defined variable.
5. Price of 100 products in a consumer store
6. Name of a student

Answer:

1. Integer array of size 50
2. Pointer variable
3. Function
4. Reference
5. Float array of size 100
6. Character array

Question 9.
Considering the following C++ statements. Fill up the blanks

1. If p = 5 and q = 3 then q%p is _______
2. If E1 is true and E2 is False then E1 && E2 will be _______
3. If k = 8, ++k < = 8 will be ________
4. If x = 2 then (10* ++x) % 7 will be ________
5. If t = 8 and m = (n=3,t-n), the value of m will be ______
6. If i = 12 the value i after execution of the expres¬sion i+ = i- – + – -i will be ______

Answer:

1. 3
2. False
3. False(++k makes k = 9. So 9<=8 is false)
4. 2(++x becomes 3 ,so 10 * 3 = 30%7 = 2)
5. 5( here m = (n = 3,8-3) = (n = 3,5), so m = 5, The maximum value will take)
6. Here i = 12

i + = i- – + – -i
here post decrement has more priority than pre decrement. So “i- -” will be evaluated first. Here first uses the value then change so it uses the value 12 and i becomes 11
i + = 12 + – -i
now i = 11.
Here the value of i will be changed and used so “i- -” becomes 10
i + = 12 + 10 = 22
So, i = 22 +10
i = 32
So the result is 32.

Question 10.
The Maths teacher gives the following problem to Riya and Raju.
x = 5 + 3 * 6.
Riya got x = 48 and Raju got x = 23. Who is right and why it is happened? Write down the operator precedence in detail?
Answer:
Here the answer is x = 23. It is because of precedence of operators. The order of precedence of operators are given below.

Here multiplication has more priority than addition

Question 11.
Explain the data types’ in C++. (3)
Answer:
Fundamental data types:
It is also called built-in data type. They are int, char, float, double and void
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory. i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

2. char data type Any symbol from the keyboard, eg. ‘A’,‘9’, …. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having an ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
Void means nothing. It is used to represent a function returns nothing.

• User defined Data types: C++ allows programmers to define their own data type. They are Structure(struct), enumeration (enum), union, class, etc.
• Derived data types: The data types derived from fundamental data types are called Derived data types. They are Arrays, pointers, functions, etc

Question 12.
Predict the output of the following C++ statements.
int a = -5, b = 3, c = 4;
C+ = a++ + –b;
cout<<a<<b<<c;
Answer:
a = -4, b = 2 and c = 1.

Question 13.
Match the following

Answer:

Question 14.
Write any five unary operators of C++. Why are they called so?
Answer:
A unary operator is an operator that need only one operand to perform the operation. The five unary operators of C++ are given below.
Unary +, Unary -, ++, – – and ! (not)

Question 15.
Write C++ examples for the following:

1. Declaration statement
2. Assignment statement
3. Type casting

Answer:

1. int age;
2. age = 16;
3. avg = (float)a + b + c/3;

Plus One Data Types and Operators Five Mark Questions and Answers

Question 1.

• Name: Jose
• Roil no: 20
• Age: 17
• Weight: 45.650

Consider the above data, we know that there are different types of data are used in the computer. Explain different data types used in C++.
Answer:
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory, i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

2. char data type:
Any symbol from the keyboard, eg. A’,’9′,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e, the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Question 2.
Define an operator and explain operator in detail.
Answer:
An operator is a symbol that performs an operation. The data on which operations are carried out are called operands. Following are the operators
1. lnput(>>) and output(<<):
These operators are used to perform input and output operation.
eg: cin>>n;
cout<<n;

2. Arithmetic operators:
It is a binary operator. It is used to perform addition(+), subtraction(-), division (/), multiplication (*) and modulus (%- gives the remainder) operations.
eg: If x = 10 and y = 3 then

x/y = 3, because both operands are integer. To get the floating point result one of the operand must be float.

3. Relational operator:
It is also a binary operator. It is used to perform comparison or relational operation between two values and it gives either true(1) or false(O). The operators are <,<=,>,>=,== (equality)and !=(not equal to)
eg: If x = 10 and y = 3 then

4. Logical operators:
Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operations and it gives either true(1) or false(O). If x=True and y=False then

Both operands must be true to get a true value in the case of AND(&&) operation If x = True and y = False then

Either one of the operands must be true to get a true value in the case of OR(||) operation If x = True and y = False then

5. Conditional operator:
It is a ternary operator hence it needs three operands. The operator is ?:
Syntax: expression ? value if true : value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.
eg: If x = 10 and y = 3 then x>y ? cout<<x : cout<<y;. Here the output is 10

6. sizeof():
This operator is used to find the size used by each data type.
eg. sizeof(int) gives 2.

7. Increment and decrement operator:
These are unary operators.
(a) Increment operator (++): It is used to incre¬ment the value of a variable by one i.e., x++ is equivalent to x = x + 1;
(b) Decrement operator (–): It is used to decre¬ment the value of a variable by one i.e., x-is equivalent to x = x – 1.

8. Assignment operator (=):
lt is used to assign the value of a right side to the left side variable.
eg: x = 5; Here the value 5 is assigned to the variable x.

Students can Download Chapter 10 Applications of Computers in Accounting Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Plus One Accountancy Applications of Computers in Accounting One Mark Questions and Answers

Question 1.
The physical components of computer is called as …………..
(a) Software
(b) Hardware
(c) Liveware
Answer:
(b) Hardware

Question 2.
Set of programs which governs the operation of a computer system is termed …………..
(a) System software
(b) Software
(c) Application of window
Answer:
(b) Software

Question 3.
A centrally controlled integrated collection of data is called ……………
(a) DBMS
(b) Information
(c) Database
Answer:
(c) Database

Question 4.
Tally is a
(a) Utility software
(b) Application software
(c) Operating system
(d) Connecting software
Answer:
(b) Application software

Question 5.
…………… is a software system that manages the creation of use of database.
(a) Database
(b) DBMS
(c) Management system
Answer:
(b) DBMS

Question 6.
Which one of the following is an output device of a computer?
(a) Mouse
(b) Keyboard
(c) Monitor
(d) Barcode reader
Answer:
(c) Monitor

Question 7.

…………. is the storehouse of a computer.
Answer:
Memory

Question 8.
…………… are set of program designed to carry out operations for a specified application.
Answer:
Application software.

Question 9.
The output obtained from VDU (Visual Display Unit) is termed ………..
Answer:
Hard copy.

Question 10.
The part.of the computer which controls the various operations of a computer is called ………..
Answer:
Control unit.

Question 11.
………….. is temporary memory and anything stored in it will remain there a long as the system is on.
Answer:
RAM (Random Access Memory)

Question 12.
Modern computerised accounting systems are based on the concept of ………..
Answer:
Database.

Question 13.
A sequence of actions taken to transform the data into decision-useful information is called ………..
Answer:
Data processing

Question 14.
The joystick is a …….. device of a computer.
Answer:
Input.

Question 15.
VDU is also called ……….
Answer:
Monitor.

Question 16.
Complete the series using the hint given.
Hint: System analyst → Human beings → liveware
a. Windows → Operating system → ?
Answer:
Software.

Plus One Accountancy Applications of Computers in Accounting Two Mark Questions and Answers

Question 1.
Match the following.

Answer:
1-b
2-e
3-d
4-c
5-a

Question 2.
What is a computer?
Answer:
Sp A computer is an electronic device that accepts data and instruction as input, stores them, process the data according to the instructions and communicate the results as output.

Question 3.
Hardware includes different devices. Name any four devices.
Answer:
Keyboard, Mouse, Monitor, Processor.

Question 4.
Redraw the given block diagram of a computer correctly:

Answer:

Question 5.
List out any four features of a computer.
Answer:

1. Highspeed
2. Large volume of data can be stored.
3. Accuracy is very high.
4. Computers are multipurpose information machine ie. versatility.

Question 6.
List out any four limitations of a computer.
Answer:

1. Computers lacks common sense.
2. Lack of decision-making skills
3. Computers have no intelligence.
4. Computers cannot make judgments based on feelings.

Question 7.
What is Accounting Information System?
Answer:
Accounting Information System (AIS) is a collection of resources (people and equipment), designed to transform financial and other data into information. Such information is organised in a manner that correct decisions can be based on it.

Question 8.
What are the basic requirements of a computerised accounting system?
Answer:
Every computerised accounting system has two basic requirements.

1. Accounting Framework: It consists of a set of principles, coding and grouping structure of accounting.
2. Operating procedure: It is a well defined operating procedure blended suitably with the operating environment of the organisation.

Question 9.
State the various essential features of an accounting report.
Answer:
The accounting report must have the following features it.

1. Relevance
2. Timelines
3. Accuracy
4. Completeness
5. Summarisation

Question 10.
Give examples of the relationship between a Human Resource Information System and MIS.
Answer:
There is a relationship between the Human Resource Information System and Management Information System, the following are the example of it.

1. Hiring employees as per the requirement.
2. Evaluating the performance of the workers.
3. Enrolling employees in benefit.

Question 11.
Give examples of two types of Operating System.
Answer:

• DOS – Disk Operating System
• Windows – Windows Operating System

Plus One Accountancy Applications of Computers in Accounting Three Mark Questions and Answers

Question 1.
Explain the term ‘Liveware’.
Answer:
People interacting with computers are called Live-ware of the computer system. It consists of the following three groups.
1. System analysts:
System analysts are the people who design data. processing systems.

2. Programmers:
Programmers are the people who write programs for processing data.

3. Operators:
Operators are the people who participate in operating the computer.

Question 2.
What is the Transaction processing system? Name three components of a Transaction processing system.
Answer:
Transaction processing systems (TPS) are among the earliest computerised systems catering to the requirements of large business enterprises. The purpose of a TPS is to record, process, validate and store transactions that occur in the various functional areas of business for subsequent retrieval and usage.
TPS system has three components:

• Input-Processing-Output
• ATM facility.
• Telephone Account and Airline Seat Reservation System are examples of TPS.

Question 3.
Discuss the different types of accounting packages. The accounting packages are classified into the following categories.
Answer:
1. Ready to use accounting software:
It is relatively easier to learn and people adaptability is very high. It is suited to small/ conventional organisations. The level of secrecy is relatively low. This software offers little scope of linking to other information systems.

2. Customised Accounting software:
Helps to meet the special requirement of the user. It is suited to large and medium organisations and can be linked to the other information system.

3. Tailored:
The accounting software is generally tailored in large business organisations with multi-users and geographically scattered locations. This software requires specialised training for users. The level of secrecy is relatively high and they offer high flexibility in terms of number of users.

Question 4.
“Computers are the servants or masters of human beings.” Elucidate.
Answer:
A computer system have certain special features or advantages which in comparison to human beings become its capabilities.
The advantages of computers are as follows:

1. High speed
2. Accuracy
3. Storage of huge data
4. Versatility
5. Deligence

Even though computers possess the above-mentioned features, it suffers from the following limitations:

1. Computers lack common sense.
2. Lack of IQ
3. Lack of decision making skill
4. No feeling

Question 5.
Find out the odd one and state reasons.

1. Mouse, Monitor, Programmers, Processor.
2. DACEASY, FORTRAN, ALU, LINUX
3. Monitor, Barcode reader, Printer, Plotter

Answer:

1. Programmers, others are hardware components.
2. ALU, others are software
3. Barcode reader, others are output devices.

Plus One Accountancy Applications of Computers in Accounting Four Mark Questions and Answers

Question 1.
Find the odd one and state reason.

1. Keyboard, Mouse, Light pen, Printer
2. System Analysts, Language Processors, System software, Utility Programmes.
3. RAM, Floppy disk, Compact disk, Hard disk
4. COBOL, C++, DOS, BASIC

Answer:

1. A printer is an output unit, all others are input units.
2. System Analyst is a human ware.
3. RAM is the Internal memory unit, all others are. external memory unit.
4. DOS is an operating system, all others are computer languages.

Question 2.
Classify the following into input unit and output unit devices.
Keyboard, Mouse, VDU (Visual Display Unit), Printer, Magnetic tape, Magnetic disk, Light pen, Optical scanner, Plotter, Speech synthesiser, MICR, OCR, Barcode reader, Smart card reader, Speaker, LCD projector.
Answer:

Input devices	Output devices
Keyboard	VDU
Mouse	Printer
Magnetic tape Magnetic	Plotter
disk Light Pen Optical	Speech Synthesiser
scanner MICR OCR	Speaker
Barcode reader Smart	LCD projector
card reader

Question 3.
What are the generic consideration before sourcing accounting software?
Answer:
The following factors are considered before sourcing accounting software:

1. Flexibility
2. Cost of installation and maintenance
3. Size of organisation
4. Ease of adaptation and training needs
5. Utilities / MIS reports
6. Expected level of secrecy (Software and data)
7. Exporting/importing data facility
8. Vendors reputation and capability

Question 4.
Classify the following components as Hardware, soft-ware and liveware.

1. Programmers
2. Keyboard
3. Windows or Linux
4. COBOL or C++
5. Mouse
6. Assembler or Compiler
7. Operators
8. Virus/Antivirus/ Scanners
9. Monitor
10. Processor
11. System Analysts
12. MS-Excel or MS Office

Answer:
a. Hardware:

• Keyboard
• Mouse
• Monitor
• Processor

b. Software:

• Windows or Linux (Operating System)
• COBOL or C++ (Computer Language)
• Assembler or Compilers (Language processor)
• Virus/Antivirus and Scanner (Utility programs)
• MS Excel or MS Office

c. Liveware:

• System analyst
• Programmers
• Operators

Question 5.
Complete the following diagrams showing the functional relationship of the various components of computers.

Answer:
a. Input devices:

1. Keyboard
2. Mouse
3. Light pen

b. Output devices:

1. Monitor
2. Printers
3. Plotters

c. CPU:

1. Memory unit
2. ALU
3. Control unit

d. Secondary storage devices:

1. Floppy disk
2. Hard disk
3. Optical disk

Plus One Accountancy Applications of Computers in Accounting Five Mark Questions and Answers

Question 1.
Computerised Accounting is different from Manual accounting. Explain.
Answer:
Computerised accounting is different from manual accounting, the following are the main difference between these two:

Computerised Accounting	Manual Accounting
1. In computerised accounting data can be easily processed and statements can be prepared with high speed and accuracy.	1. In manual accounting financial statements cannot be prepared with such speed and accuracy.
2. Mass data can be stored in very small space and brought back very easily.	2. Data are stored in large number of books and retrieval of data is a very tedious job.
3. Coding is essential in computerised accounting.	3. Coding is not essential.
4. Closing entries are not necessary.	4. Closing entries are necessary.
5. The possibility of errors are less in computerised accounting.	5. The possibility of errors are more.

Plus One Accountancy Applications of Computers in Accounting Six Mark Questions and Answers

Question 1.
What are the elements of a computer system?
Answer:
A computer system is a combination of six elements. They are as follows:
1. Hardware:
The physical components of a computer system is termed as Hardware. Eg: Mouse, Keyboard, Monitor, Processor, etc.

2. Software:
Set of programs that govern the operations of a computer system is termed as soft-ware. There are six types of software as follows.

1. Application software
2. Operating system
3. Utility programs
4. Language processors
5. System software
6. Connectivity software

3. People:
People interacting with computers are also called the “live-wave” of the computer system. It consists of the following three groups.

1. System analysis
2. Programmers
3. Operators

4. Procedures:
The procedure means a series of operations in a certain order or manner to achieve desired results. There are three types of procedures which constitute part of computer system

1. Hardware oriented
2. Software oriented
3. Internal procedure

5. Data:
These are facts and may consist of numbers, text, etc. These are gathered and entered into a computer system.

6. Connectivity:
Tie manner in which a particular computer system is connected to others says through telephone lines, microwave transmission, satellite, etc. is the element of connectivity.

Question 2.
Define computerised accounting. List out various advantages and limitations of computerised accounting system.
Answer:
A computerised accounting system is an accounting information system that processes financial transactions and events to produce reports as per user requirements.
a. Advantages:

1. Speed
2. Accuracy
3. Reliability
4. Efficiency
5. Storage and Retrieval
6. Automated document production
7. Quality reports
8. Real-time user interface

b. Limitations:

1. Huge training costs
2. Staff opposition
3. System failure
4. Breaches of security
5. Inability to check unanticipated errors

Students can Download Chapter 1 Physical World Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 1 Physical World

Plus One Physics Physical World One Mark Questions and Answers

Question 1.
The word ‘Science’ originated from a Latin verb. Which is that verb?
Answer:
‘Scientia’ means ‘to know’.

Question 2.
The word physics comes from a Greek word______.
Answer:
‘Fusis’ means ‘nature’.

Question 3.
Name the branch of science that deals with

1. Study of stars
2. Study of earth

Answer:

1. Astronomy
2. Geology

Plus One Physics Physical World Three Mark Questions and Answers

Question 1.
Fill in the blanks

Answer: