IE IRODOV'S General Physics - Question 1

1. A motorboat going Downstream overcame a raft at a point A ;  T= 60 min later it turned back and after sometime passed the raft at a distance l = 6.0 km  from the point A. Find the velocity,assuming the duty of the engine to be constant.

Answer : Upstream - Downstream motion has to be calculated as per Time.

Unlike Distance based calculations which are two-dimensional ,Time based calculations are made as three-dimensional which includes the direction vector as well

At first, Dimensional Analysis is done to find out the equation strategy. This has been called dimensional analysis of vector equations . This is done as follows:

Since distance based calculation is involved ,   

distance / time = velocity ,which is a vector 

Hence [VT] = a constant (distance)

Hence the length between ' B' and 'P' in the figure ,which is 'BC' has to be found out . 

3-Dimensional approach to this 2-d problem is drawn below

Time at B = Time at P ( for raft (wooden log) and boat respectively)

Time = Distance / Velocity ; 

Therefore,  (l-BC)/Velocity raft  =  AP / Velocity P = 60 min

That means, (6 - BC )/V raft = 60 minutes = 1hour

V raft = (6 - BC) / 1

   As per modern trigonometry,                                                             

 ( Hypotnuese - square ) =( Base - square) + ( Altitude -square)

As per modern Trigonometry, Hypotnuese is 2 times bigger than altitude and base for unity triangle ( which is 1 mm in base )

Now, Indian Mathematics suggests ,PASCAL'S TRIANGLE for solving Trigonometry problems that are unknown to calculation.

In this Triangle, there are fifteen numerals altogether till the fifth degree as shown above. This shows that Indian Approach is based on intense mathematical calculation using a repetitive algorithm ,which is used in modern day, Computer program 

Here ,1 + 4 = 5 , but 6 is given ,Hence, we are showing it as given above in Pascals image .

Now if you calculate diagonally, these numbers can introduce you to Trigonometry ,which is as shown below

                      1  4  5  9  5       , which is known as Pascal's rule in trigonometry.  ( Inclined indicates triangular formation using number digits.)

So,  1 (square ) + 4 ( square ) = square root ( 17) which indicates that square has to be used diagonally in the Pascal's format shown above.

So , 1   2  √3  in triangle calculation becomes 1  2  3 as shown in Pascal's Triangle above ( diagonally)

So, Above BC becomes 3 ( And that is the Solution )

PHYSICS - OBJECTIVE -6

 13. The shown p-V diagram represents the thermodynamic cycle of an engine,operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is 

In the above thermodynamic cycle, input energy is at constant pressure (isentropic) .

So Q1=Heat input ; Q1 = nCpΔT

Cp =5/2 R for monoatomic gas (R is the Avagadro constant)

So, Q1 =5/2nRΔT

Similarly ,Q2 = Heat rejected ; Q2=nCvΔT0

Cp - Cv = R ; So, Cv=5/2R - R =3/2R

Hence, Q2 =3/2nRΔT0

Now , from the graph ΔT = 2ΔT0

Q1 + Q2 = 5/2nR x2ΔT + 3 /2nRΔT = 13/2RΔT

Now , from ideal gas equation PV =nRT

So, RΔT =PΔV

Hence we can deduce , 

Q1 + Q2 = 13/2PV0

So, the answer is (b)

 EXPLANATION: Here, the engine cycle can be approximated to Otto cycle .Here the input is at constant pressure (hence it is isentropic) The output is at constant volume (which is adiabatic).Here the automobile engine undergoes huge pressure difference ,since it is controlled combustion. 

So, my point is the same amount of engine volume has been displaced at the input and output strokes.This shows that the automobile is having fairly low efficiency ( efficiency is low). This problem results in huge pollution since output gases are heavily unfinished(unburned) while undergoing combustion in the piston combustion chamber.This process is completed by addition of muffler to exhaust which reduces the sound of expulsion of output gases ( gases which has particles suspended in it )

PHYSICS - OBJECTIVE -5

12. A block of mass m is on an inclined plane of angle θ.The coefficient of friction between the block and the plane is μ and tan θ > μ.The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive.As P is varied from P1 = mg (sinθ - μ cosθ) to P2 = mg (sinθ + μ cosθ), the frictional force f versus P graph will look like 

 

The above question is from Strength of Materials in engineering.Here there are four choices.One has to find out the correct choice as shown above.

Here we are plotting the stress-strain diagram known as Shear stress diagram .This is of five types. They are as follows:

1. Shear stress diagram of indeterminate nature

Here the Shear (bending moment) is centre-shear .This means the beam is uncontrolled by nature which is end supported with fixed hinges.So here the solution is as in (b)

2. Shear stress diagram of single fixed hinge 

 

Here the solution is a downward straight line in the shear stress diagram 

 

 

3. Shear stress diagram of double fixed type 

 

This type shear stress diagram is two triangles ,both up and down inluded

4. Shear stress diagram with one end fixed and other end roller

Here the shear stress diagram curve is parabolic .

5. Shear stress diagram with both end roller

 

Here one end is hinge roller and other end is full roller.Here is shear stress diagram is extremely complicated 

In the above question of inclined plane one end is fixed hinge and other end is free.This means ,among the above choices diagram which shows shear stress diagram as below X-Y line is correct .The graph should be straight line and not inclined ( inclined is for case 3 where both ends are fixed  and parbolic is for case 4 where one end is fixed and other end is roller supported for the beam type)

So answer is (d)

MATHEMATICS - OBJECTIVE -3

11. If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 

(a) 4/55                 (b) 4/35

(c) 4/33                 (d) 4/1155

 Answer (d)

Solution method: A distinct number that is randomly chosen from a population of big size with a small fixed sample lot is called as Bayer's approximation of probability ratios. This is represented by

          nCr=  n! / r! ( n - r)!

There are 16 digits divisible by 6 ( common factor of 2 and 3 = 2x3 =6) So r=16, N=100

Number of ways in which 3 numbers are selected from these 16 numbers is  16C3

Number of ways in which 3 numbers are selected randomly from 100 numbers is 100C3 

 nCr = ( 16 x 15 x 14 / 3 x 2 x 1 )

         _______________________   = 4 / 1155

         ( 100 x 99 x 98 / 3 x 2 x 1 )

Explanation: This type of analysis is done on a huge set of numbers ,within which ,a smaller subset is selected for probability analysis. This can be considered as the Bayer theorem corollary,since probability theory can be used to negotiate plans for immediate future .This is used in market speculation in current world. As shown above, a definite answer to probability of numbers indicates futuristic ways on market trading in stocks and business prediction in today's competitive world. So here ,it shows that probability is indeed a science of fairly good amount of accuracy .

Chevalier De Mere's problem ( a classic case)shows that Bayer's theory is not fully accurate. This needs approximations as per the above case. 

CHEMISTRY - OBJECTIVE -4

 10. The methods chiefly used for the extraction of lead and tin from their ores, respectively are 

(a) self reduction and carbon reduction

(b) self reduction and electrolytic reduction

(c) carbon reduction and self reduction 

(d) cyanide process and carbon reduction

answer (a)

Answer explanation : metals are extracted from their ores using different chemical processes. Among them the most important is carbon reduction. Coal has different types of reduction methods like carbon reduction, metal oxide reduction, hydrogen reduction and so forth. This has to be analysed as per the valence layer of electrons in the atom of the metal which is oxide coated in nature. Hence the removal of oxide using self reduction has also been involved .In this method, Lead and tin are immersed in water for more than 30 hours ; then it will automatically get removed of the oxide which bubbles out through water. And this type of oxide removal is used in steel manufacturing since it is cheapest and best. 

CHEMISTRY - OBJECTIVE - 3

9. Benzamide on treatment with POCL3 gives 

(a) aniline                     (b) benzonitrile

(c) chlorobenzene          (d) benzyl amine

answer (b)

Answer explanation :Benzamide (C6H5CONH2) on treatment with POCL3 (POCL3 is Phosphorous oxy chloride ) gives C6H5-C≡N (triple bond).

This is called Benzonitrile. nitrile has triple bond for amino group

Nitrogen has electronic configuration 2,2,3 (atomic number of nitrogen is 7). outer 'p' orbital has 3 electrons (1s2,2s2,2p3)

Nitrogen has multiple energy level electrons since it is an inert gas .Inert gas means slow moving electrons inside atoms.So, 'p' orbital has three electrons, since nucleus attracts towards inside more, the inner layer electrons (in s layer) in the valence covalent bond in the case of Nitrogen and Hydrogen .

Hydrogen has 1 atomic number and 1 electron mates with oxygen to form water H2O. Oxygen has atomic number 8,it has 8 electrons .1s,2s,2p levels has 2,2,4 electrons respectively. So, 4 electrons shall part from oxygen atom. Thus, four valence electrons are required for oxide formation for any inert gas .

Carbon has atomic number 6(1S2,2S2,2P2). CN has one single carbon mating with three Nitrogen atoms    ( triple bond). 

 Nitrogen has 3 electrons in the outer layer, out of which 1 mates with carbon .Hence C≡N molecule has Nitrogen atom in which 1 electron is shared with carbon, and 2 free electrons remain free for each Nitrogen atom.So CH3-CH2-C≡N has 6 electrons free for three Nitrogen atoms.

PHYSICS - PROBLEM SOLVING -4

 8. The piston cylinder arrangement shown contains a diatomic gas at temperature 300K.The cross-sectional area of the cylinder is 1m². Initially the height of the piston above the base of the cylinder is 1 m. The

temperature is now raised to 400K at constant pressure. Find the new height of the piston above the base of the cylinder.

 If the piston is now brought back to its original height without any heat loss, find the new eqilibrium temperature of the gas. You can leave the answer in fraction.

Solution :  

At equilibrium,  constant pressure, 

PV =nRT  

P1V1 = P2V2

 But here Pressure is constant ,P

So, Px A xh1 =nRxT1

and Px Ax h2 =nRxT2 ,that is h1T2=h2T1

So, 1 x 400 = h2 x 300

h2 = 1 x 400/300 = 4/3 m

Now let us solve the second part of the above question.

 For adiabatic process.

$T_f{V}_f^{\gamma -1}=T_i{V}_i^{\gamma -1}$
$T_f=T_i(\frac{V_i}{V_f}{)}^{\gamma -1}=(400)(\frac{h_i}{h_f}{)}^{1.4-1}$
$=400(\frac{4}{3}{)}^{0.4}$=448.8 K

(temperature as final and initial has been given)

EXPLANATION:

The Automobile piston has been shown in the figure.This has all the characteristics of the above problem.Hence inorder to achieve perfect combustion ,one needs to achieve gamma ( γ )value of 1.4 ( the value for diatomic air )

Inorder to achieve this value ,a piston ring is attached to the grooves of the cylinder making it air-tight design ,so that atmospheric pressure is maintained inside cylinder at the point of exhaust valve opening and subsequent closure .This design has been used by engineers all around the world in automotive engines with minimum requirement of servicing of piston. 

Aircraft engines uses a special kind of rubber sealant to cover the piston grooves inside cylinder so that when plane takes a right turn, the exhaust valve in the right hand portion of cylinder exhausts dry air which exerts pressure towards right side .Thus the cylinder of aircraft engine has around twelve valve which has inlet and exhaust pairs of valves. The degree turn around is pressed by the co-pilot and corresponding outlet valve opens for pushing down exhaust gas jet which makes it surge ahead with immense power from downward push from the exhaust gas push-out in the direction concerned as per valve emission in the exhaust port as inclined,top,down,forward,rearward,inclined right,inclined left and so forth.