Hindustan Institute of Technology & Sciences (HITSEE) 2017 Notification Released

What is HITSEE?

HITSEE (Hindustan Institute of Technology & Sciences). Hindustan College of Engineering, started in the year 1985, was conferred the "University Status" by University Grants Commission (UGC), Government of India, Under Section 3 of UGC Act 1956 from the academic year 2008-09 and under the name HITS (Hindustan Institute of Technology and Science).

B.Tech Programmes:

⇒ Candidates seeking admission to the first semester of the 8 - semester B.Tech degree programme should have passed the Higher Secondary examination (12^th) Curriculum (Academic Stream) prescribed by the Government of Tamil Nadu with Mathematics, Physics & Chemistry or equivalent.

⇒ Improvement marks will not be considered for admission

Eligibility Marks: Academic Stream A minimum Average of 60% in Physics, Mathematics and Chemistry put together

For B.Tech Biotech Only:

⇒ A minimum Average of 60% in Physics, Chemistry and Mathematics put together.

Eligibility Marks: Vocational Stream A minimum Average of 60% Marks in the related subjects and vocational subjects (theory and Practical).

B. Arch Programme (5 Years): 

⇒ A pass in the plus two examination with an average of not less than 60% in (10+2) level or its equivalent examination with Mathematics/ Business Mathematics as one of the subjects.

⇒ A pass in 3 year diploma (10+3) examination recognized by the Central/ State Government with Mathematics as a subject of study with an average of not less than 60%. Any other examination of any university or authority recognized by this university as equivalent of plus two. 

⇒ All applicants for B. Arch. course should write the National Aptitude Test for Architecture, India (www.nata.in) conducted by Council of Architecture and the score should not be less than 80 out of 200.

Age Limit: Upper age limit will be 19 yrs as on 30^th April 2017.

Important Date:

Sl.No	Details	Dates
1	Date for Issue of Application form	18th Jan 2017
2	Last Date of Issue HITSEEE Application form	22nd April 2017
3	Last Date for Submission of Filled in Application form	22nd April 2017
4	Online Entrance Exam Dates(Chennai Centres)	30th April to 05th May 2017 (Sun to Fri)
5	Online Entrance Exam Dates (Other Centers)	30th April & 01th May 2017 (Sun & Mon)
6	Publication of Rank list	09th May 2017
7	Counselling	12th May to 25th May 2017
8	Commencement of Classes	10th July 2017

Exam Pattern:

⇒ The question paper will be only in English.

⇒ The duration of the examination will be 2 hours.

⇒ All questions will be of objective type

Part - I: Physics

Part - II: Chemistry

Part - III: Mathematics

Tricky Puzzles (With Answers)

Puzzle 1: A clock with the hours round the face in Roman block numbers, as illustrated in the sketch fell down and the dial broke into four parts. The numerals in each part in every case summed to a total of 20. Can you show how the four parts of the clock face was broken?

Answer:

Puzzle 2: Shown in the sketch are 6 matches. Can you rearrange them to make nothing?

Answer:

Puzzle 3: Here is a sketch: Can you rearrange the position of the numbers 1 to 10 so that the sum of any two adjacent numbers is equal to the sum of the pair of numbers at the opposite ends of the diameter?

Answer:

Puzzle 4: Can you place 10 coins in such a way that they lie in 5 straight lines and on each line there are 4 coins. There are at least two solutions.

Answer:

Puzzle 5: Nine dots are arranged by 3 rows of 3 in the form of a square as shown in the sketch below:

Can you draw 4 straight lines, the second beginning where the first ends, the third beginning where the second ends and the fourth beginning where the third ends so that each dot is or at least one line?

Answer:

Monotonic function

Strictly increasing function: A function f (x) is said to be a strictly increases function on (a, b) if x₁ < x₂ ⇒ f (x₁) < f (x₂) ∀ x₁, x₂ ϵ (a, b)

Ex: aˣ is strictly increasing function on R for a > 1

F(x) = eˣ is increasing on R

Strictly decreasing function: A function f (x) is said to be a strictly decreasing function on (a, b) if x₁ < x₂ ⇒ f (x₁) > f (x₂) ∀ x₁, x₂ ϵ (a, b)

F (x) = aˣ (0 < a < 1) is strictly decreasing function

Monotonic function: A function f(x) is said to be monotonic on an interval (a, b) if it is either increasing or decreasing (a, b)

Definition 1: A function f(x) is said to be increasing (decreasing) at a point x₀, if there is an interval (x₀ - h, x₀ + h) containing  x₀ such that f(x) is increasing (decreasing) on (x₀ - h, x₀ + h)

Definition 2:  A function f(x) is said to be increasing (decreasing) on [a, b] if it is increasing (decreasing) on (a, b) and it is also increasing (decreasing) on (a, b) and it is also increasing (decreasing) at x = a and x = b

Necessary and sufficient condition Monotonicity of functions:

Necessary condition: Let f(x) be a differentiable function defined on (a, b) then f’(x) > 0 or f’(x) < 0 according as f(x) is increasing or decreasing on (a, b)

If f(x) is an increasing function on (a, b) then tangent at every point on curve y = f(x) makes an acute angle in the positive direction of x – axis

tan θ > 0 ⇒ dy/dx > 0 Or f’ (x) > 0 ∀ x ϵ (a, b)

Similarly for decreasing function on (a, b) then the tangent at every point on the curve y = f(x) makes obtuse angle θ with direction of x - axis

tan θ < 0 ⇒ dy/dx > 0 Or f’ (x) < 0 ∀ x ϵ (a, b)

Sufficient condition: Let f be a differentiable real function defined on open interval (a, b)

a) If f’(x) > 0 ∀ x ϵ (a, b), then f is Increasing on (a, b)

b) If f’(x) < 0 ∀ x ϵ (a, b), then f is decreasing on (a, b)

Properties of Monotonic functions:

i) If f(x) is strictly increasing function on an interval [a, b] then f⁻¹ exists and also a strictly increasing function.

ii) If f(x) is strictly increasing function on an interval [a, b] such that it is continuous then f⁻¹ is continuous on [f (a), f (b)].

iii) If f(x) is continuous on [a, b] such that f’< (c) ≥ 0 (f’ (c) < 0) ∀ x ϵ (a, b) then f(x) is monotonically (strictly) increasing function on [a, b].

iv) If f(x) and g(x) are monotonically (or strictly) increasing (or decreasing) function on [a, b], then gof(x) is a monotonically (or strictly) increasing function on [a, b].

v) If one of the two function f(x) and g(x) is strictly (or monotonically) increasing and other a strictly (monotonically) decreasing, the gof (x) is strictly (monotonically) decreasing on [a, b].

If f(x) is monotonically increasing function ∀ x ϵ R such that f’’ (x) > 0 and [f⁻¹ (x₁) + f⁻¹ (x₂) + f⁻¹ (x₃)]/3 < f⁻¹ (x₁ + x₂ + x₃)/3.

Electrode Potential and Standard Electrode Potentials

1. Electrons will flow from the electrode of higher negative charge density to the electrode with lower negative electric charge density.

2. A property closely related to the density of negative electric charge is called the electrode potential.

3. Potential difference between the metal and metal ion in which electrode is dipped is called electrode potential.

Standard electrode potentials: In standard state, i.e., when pressure is 1 atm and concentration is 1M, the electrode potential is called standard electrode potential denoted as E°. Temperature is generally taken as 298 K.

For example:

⇒ E°_{Cl¯|Pt(Cl₂)} is a standard electrode potential of the half-cell Cl¯_(aq)|Pt(Cl₂)
⇒ E°_{AgCl|Cl¯(aq)} is the standard oxidation potential of the half–cell Ag, AgCl|^(-)Cl¯_(aq)
⇒ E°_cell (or) E_cell is the potential difference between the two half cells
⇒ E°_cell = E°_OX + E°_red
⇒ E_cell = E_OX + E_red
⇒ If E°_OX = xV then E°_red = - xV
⇒ E°_Cu²⁺|Cu = 0.34V then E° _Cu|Cu²⁺ = - 0.34V
⇒ E°_cell = E°_right - E°_left.

Note: Both these E°_right and E°_left are reduction potentials of right hand side (cathode) and left hand side (anode) half–cells respectively.

Tricky Puzzles

Puzzle 1: A clock with the hours round the face in Roman block numbers, as illustrated in the sketch fell down and the dial broke into four parts. The numerals in each part in every case summed to a total of 20. Can you show how the four parts of the clock face was broken?

Puzzle 2: Shown in the sketch are 6 matches. Can you rearrange them to make nothing?

Puzzle 3: Here is a sketch: Can you rearrange the position of the numbers 1 to 10 so that the sum of any two adjacent numbers is equal to the sum of the pair of numbers at the opposite ends of the diameter?

Puzzle 4: Can you place 10 coins in such a way that they lie in 5 straight lines and on each line there are 4 coins. There are at least two solutions.

Puzzle 5: Nine dots are arranged by 3 rows of 3 in the form of a square as shown in the sketch below:

Can you draw 4 straight lines, the second beginning where the first ends, the third beginning where the second ends and the fourth beginning where the third ends so that each dot is or at least one line?

Clock Puzzles 4 [With Answers]

Puzzle 1: How many times in a day, are the hands of a clock in straight line but opposite in direction?

A) 20       B) 22

C) 24       D) 48

Answer: 22

Explanation: The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clock only).

So, in a day, the hands point in the opposite directions 22 times.

Puzzle 2: A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:

A) 145        B) 150

C) 155        D) 160

Answer: 155

Explanation: Angle traced by hour hand in 12 hrs. = 360°.

Angle traced by hour hand in 5 hrs. 10 min.  i.e. 31/6 hrs. = [(360/12) x (31/6)]° 155°

Puzzle 3: A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct?

A) 2 p.m. on Tuesday         B) 2 p.m. on Wednesday

C) 3 p.m. on Thursday       D) 1 p.m. on Friday

Answer: 2 p.m. on Wednesday

Explanation: Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.

∴ The Watch gains [2 + 4 (4/5)] min. or 34/5 in 170 hrs.

Now, 34/5 min. are gained in 170 hrs.

∴ 2 min. are gained in (170 x (5/34) x 2) hrs = 50 hrs

∴ Watch is correct 2 days 2 hrs. After 12 p.m. on Monday i.e., it will be correct at 2 p.m. on Wednesday

Puzzle 4: At 3.40, the hour hand and the minute hand of a clock form an angle of:

A) 120 degrees            B) 125 degrees

C) 130 degrees            D) 135 degrees

Answer: 130 degrees

Explanation: Angle traced by hour hand in 12 hrs. = 360°.

Angle traced by it in 11/3 hrs = [(360/12) x (11/3)]° = 110°

Angle traced by minute hand in 60 min. = 360º.

Angle traced by it in 40 min. = [(360/60) x 40]° = 240°

∴ Required angle (240 - 110)° = 130°.

Puzzle 5: How many minutes is it until six o’clock if fifty minutes ago it was four times as many minutes past three o’clock?

Answer: Twenty-six minutes.