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Work energy theorem states that the change in kinetic energy of an object is equal to the net work done on it by the net force.

Let us suppose that a body is initially at rest and a force \(\vec{F}\) is applied on the body to displace it through \(d\vec{S}\)along the direction of the force. Then, small amount of work done is given by

dW = \(\vec{F}\).\(d\vec{s}\) = FdS

Also, according to Newton's second law of motion, we have

F = ma

where a is acceleration produced (in the direction of force) on applying the force. Therefore, 

dW = MadS = M\(\frac{dv}{dt}\)dS

OR dW = M\(\frac{dS}{dt}\)dv = Mvdv

Now, work done by the force in order to increase its velocity from u (initial velocity) to v (final velocity) is given by

W = \(\int^v_uMvdv\) = M\(\int^v_uvdv\) = M\(\Big|\frac{v^2}{2}\Big|^v_u\)

W = \(\frac{1}{2}Mv^2-\frac{1}{2}Mu^2\)

Hence, work done on a body by a force is equal to the change in its kinetic energy.

In ΔPQR, PQ = 24 cm

QR = 26 cm

∠QPR = 90°

By using pythagoras theorem,

⇒ QR² = QP²+PR²

⇒ 26² = 24²+PR²

⇒ 676 = 576+PR²

⇒ PR² = 100

⇒ PR = 10 cm  ...(1)

In ΔPKR, KR = 8 cm

∠PKR = 90°

By using pythagoras theorem,

⇒ PR² = PK²+KR²

⇒ 10² = PK²+8²

⇒ 100 = PK²+64

⇒ PK² = 36

⇒ PK = 6 cm

Correct option  (B) has no solutions

Explanation :

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

then  a_1/a_{2  =}  b₁/b₂  ≠ c₁/c₂ shows equation has no solution.

(b) 2008

The 2008 marks the 250th anniversary of Cox & Kings, the longest established travel company in the world.

Correct option  (1) 4x – 3y – 4 = 0 

Explanation:

x – 2y + kz = 1 …. (i)

2x + y + z = 2 … (ii)

3x – y – kz = 3 … (iii)

for locus of (x, y)

equation (i) + (iii)

4x – 3y = 4 

4x – 3y – 4 = 0

2\(\frac{dx}{dt}\) + \(\frac{dy}{dt}\) = 5e^t .....(1)

\(\frac{dy}{dt}\) - 3\(\frac{dx}{dt}\) = 5 .....(2)

By subtracting equation (2) from (1), we get

5\(\frac{dx}{dt}\) = (e^t - 1)

\(\frac{dx}{dt}\) = e^t-1

dx = (e^t - 1)dt

x = e^t- t + c (By integrating both sides)

at t = 0, x = 0

∴ 0 = -e⁰ - 0 + c

c = -e⁰ = -1

∴ x = e^t - t -1

By putting x = e^t - t -1 in equation (2), we get

\(\frac{dy}{dt}\) = 5 + 3\(\frac{d}{dt}\) (e^t- t -1)

= 5 + 3e^t - 3

= 2 + 3e^t

dy = (2+3e^t)dt

y = 2t + 3e^t + c (By integrating both sides)given that at t = 0, y = 0

∴ 0 = 2 x 0 + 3e⁰ + c

c = -3e⁰ = -3

∴y = 2t + 3e^t - 3

Hence, solution of given system of differential equation is x = e^t - t - 1 and y = 2t + 3e^t - 3

Concept:

Homogenous equation are of the form \(\frac{{dy}}{{dx}} = \frac{{f\left( {x,y} \right)}}{{\emptyset \left( {x,y} \right)}}\)

Where f(x, y) and ∅(x, y) Homogenous functions of the same degree in x and y.

To solve a homogenous equation

1. Put y = vx, then \(\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\)
2. Separate the variable v and x and integrate.
3. Put \(v = \frac{y}{x}\)

Calculation:

\(3{x^2}dy + \left( {{y^2} - 2xy} \right)dx = 0\)

\(\frac{{dy}}{{dx}} = \frac{{2xy - {y^2}}}{{3{x^2}}}\)

Put y = vx, then \(\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\)

\(v + x\frac{{dv}}{{dx}} = \frac{{2v{x^2} - {v^2}{x^2}}}{{3{x^2}}}\)

\(x\frac{{dv}}{{dx}} = \frac{{ - v - {v^2}}}{3}\)

\(\frac{{dv}}{{{v^2} + v}} = \frac{{ - dx}}{{3x}}\)

\(\left( {\frac{1}{v} - \frac{1}{{v + 1}}} \right)dv = \frac{{ - dx}}{{3x}}\)

By integrating both sides we get

\(\smallint \left( {\frac{1}{v} - \frac{1}{{v + 1}}} \right)dv = \smallint \frac{{ - dx}}{{3x}}\)

\(\ln v - \ln \left( {v + 1} \right) = \frac{{ - 1}}{3}\ln x + \ln c\)

\(\ln \frac{v}{{v + 1}} = \ln \left( {c{x^{\frac{{ - 1}}{3}}}} \right)\)

\(\frac{v}{{v + 1}} = c{x^{\frac{{ - 1}}{3}}}\)

Put v = y/x in the above equation we get

\(\frac{{\frac{y}{x}}}{{\frac{y}{x} + 1}} = c{x^{\frac{{ - 1}}{3}}}\)

Monday: Given 6 letters 

∴ number of arrangement = ⁶P₆ = 6! 

(i) Taken 4 letters at a time 

= 6P₄ = 6!/2! = 6 × 5 × 4 × 3 = 360 ways 

(ii) all letters at a limt = 6P₆ = 6! = 720 ways. 

(iii) start with 0 ……… = 5P₅ = 120 

Start with A ……… = 5P₅ = 120 

240 ways.

CONCEPT:

Point P is uniquely determined by the ordered pair of real numbers (r, θ), called the polar coordinates of the point P.

P represent the nonzero complex number z = x + iy.

Here \(r = \sqrt {{x^2} + {y^2}} = \left| z \right|\) is called the modulus of the given complex number.

The argument of Z is measured from positive x-axis only.

Let z = r (cosθ + i sinθ) is polar form of any complex number then following ways are used while writing θ for different quadrants –

For the first quadrant, \({\rm{\theta }} = {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

For the second quadrant \({\rm{\theta }} = {\rm{\pi }} - {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

For the third quadrant \({\rm{\theta }} = - {\rm{\pi }} + {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

For the fourth quadrant \({\rm{\theta }} = - {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

CALCULATION:

Given complex number \(z = \frac{{ - 4}}{{1 + i\sqrt 3 }}\)

On rationalization we get –

\(z = \frac{{ - 4}}{{1 + i\sqrt 3 }} \times \frac{{1 - i\sqrt 3 }}{{1 - i\sqrt 3 }} = \frac{{ - 4\left( {1 - i\sqrt 3 } \right)}}{{{1^2} - {{\left( {i\sqrt 3 } \right)}^2}}} = \frac{{ - 4\left( {1 - i\sqrt 3 } \right)}}{4} = - 1 + i\sqrt 3 \)

rcosθ = -1, rsinθ = √3 

By squaring and adding, we get –

\({{\rm{r}}^2}\left( {{\rm{co}}{{\rm{s}}^2}{\rm{\theta }} + {\rm{si}}{{\rm{n}}^2}{\rm{\theta }}} \right) = 1 + 3\)

∴ r = 2

\(\Rightarrow cos\theta = \frac{{ - 1}}{r} = \frac{{ - 1}}{2}\) and \(sin\theta = \frac{{\sqrt 3 }}{r} = \frac{{\sqrt 3 }}{2}\)

Since it is in the second quadrant, \(\theta = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}\)

Concept:

Consider a complex number z = a + ib,

Polar form is given by z = r(cosθ + isin θ), where

 r = \(\rm \sqrt {a^2 +b^2}\) and θ = tan^-1 \(\left(\rm\frac{Im(z)}{Re(z)} \right)\)

Calculation:

Given complex number, -1 - i

|z| = r = \(\rm \sqrt {{(-1)}^2 +{(-1)}^2}\)

= √2

Now,   -1 - i can be represented as point P(-1, -1),which lies in 3rd quadrant.

Now, θ  = -π + tan-1\(\left(\rm\frac{Im(z)}{Re(z)} \right)\)

= -π + π/4 = -3π/4

∴ Polar form: z = √2[cos(-3π/4) + i sin(-3π/4)]

= √2[cos(3π/4) - i sin(3π/4)]    [∵ cos (-θ) = cos θ and sin (-θ) =- sin θ]  

Concept:

Consider a complex number z = a + ib,

Polar form is given by z = r(cosθ + isin θ), where r = \(\rm \sqrt {a^2 +b^2}\) and θ = tan-1 \(\left(\rm\frac{Im(z)}{Re(z)} \right)\)

Calculation: 

We have, z = -1 - i√3

Here,  Re(z) = -1 and Im(z) = -√3.

∴ tanθ = \(\frac{-\sqrt3}{-1}\)

⇒ θ = tan^-1 (\(\frac{\sqrt3}{1}\)) = π/3

Now,  z = -1 - i√3, represented by the points, P(-1,-√3), which lies in the 3rd quadrant.

∴ arg(z) = θ = (-π + θ)

= (-π + π /3)

= -2π/3

Hence, option (3) is correct.

Concept:

The imaginary unit i is defined as i = √-1.

i^{4n + 1} = i, i4n + 2 = -1, i4n + 3 = -i, i4n = 1, etc.

Calcultion:

The given expression can be written as:

i⁰ + i2 + i4 + i6 + ... + i40

= (i⁰ + i4 + i⁸ + ... + i40) + (i2 + i6 + i10 + ... + i38)

= [1 + 1 + ... (11 times)] + [-1 + -1 + ... (10 times)]

= 11 - 10

= 1.

Concept:

Power of i:

• i = \(\sqrt{-1}\)
• i2 = -1
• i3 = -i × i2 = -i
• i4 = (i2)2 = (-1)2 = 1
• i4n = 1

Calculation:

To Find: Value of i3 + i4 + i⁵

As we know, i2 = -1

So, i3 = -i × i2 = -i

i4 = (i2)2 = (-1)2 = 1

i⁵ = i⁴ × i = 1 × i = i

Now,

i3 + i4 + i ⁵ = -i + 1 + i = 1

Securities Pay-in: The process of delivering securities to the clearing corporation to effect settlement of a sale transaction.

Securities Pay-out: The process of receiving securities from the clearing corporation to complete the securities settlement of a purchase transaction.

1. Extreme loss margin is additional upfront margin on Va R. 

2. The Extreme Loss Margin is collected/ adjusted against the total liquid assets of the member on a real time basis

1) advertising 

2) sales promotion 

3) public relations 

4) direct mail 

5) personal selling

1. They take positions in financial markets to earn riskless profits. 

2. The arbitrageurs take short and long positions in the same or different contracts at the same time to create a position.

Derivatives can either be traded over the counter (OTC) or on an organized exchange..

1. Usually forwards, some types of options, swaps exotic products are OTC derivatives. 

2. Futures, exchange-traded options are exchange traded derivatives.

Given:

The ratio of the present age of Mahesh and Ajay is 3 : 2 respectively.

Ratio of their age will after 8 years 11: 8.

Calculation:

Let The present age of Mahesh be 3x years.

And The present age of Ajay be 2x years.

According to the question;

The ratio of their age will 8 years from now will be 11: 8,

⇒ (3x + 8)/(2x + 8) = 11/8

⇒ 8 × (3x + 8) = (2x + 8) × 11 

⇒ 24x + 64 = 22x + 88

⇒ 24x – 22x = 88 –64

⇒ 2x = 24

⇒ x = 12

So, The present age of Ajay = 2x years = 2 × 12 = 24 years

∴ Mahesh's Son age = 24/2 = 12 years.

Given:

A + V + S = 65

A : S = 5x : 3x

A - S = 10

Calculation:

From above equations.

2x = 10 years.

1x = 5 years.

Ajay = 25 years.

Sakshi = 15 years.

Vijay = 25 years.

Sakshi : Vijay = 3 : 5

Option : (c)

\(A^2=3I,\)

\(\Rightarrow\begin{bmatrix}\alpha^2+\beta \gamma&0\\0&\beta \gamma+\alpha^2\end{bmatrix}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\)

\(\Rightarrow3-\alpha^2-\beta \gamma=0\)

Concept used:

⇒ 2nd term = (1st term × 1 + 2)

⇒ 3rd term = (2nd term × 2 + 3)

⇒ 4th term = (3rd term × 3 + 4)

And so on..........

∴ 5 is the wrong answer.

Given that,

2, 9, 30, 93, 282, 846

 2

⇒ 2 × 3 + 3 = 9

⇒ 9 × 3 + 3 = 30

⇒ 30 × 3 + 3 = 93

⇒ 93 × 3 + 3 = 282

⇒ 282 × 3 + 3 = 849

∴ 846 is the wrong number in the given pattern 

Given:

The given number series is:

1, 5, 20, 57, 121, 221, 365

Concept Used:

A number series, with the relation between successive terms.

Calculation:

The given number series follows the relation:

1 + 2² = 5,

5 + 4² = 21,

21 + 6² = 57,

57 + 8² = 121, 

121 + 10² = 221,

221 + 12² = 365

∴ The wrong number in the given series is 20

Gan’s Permutit Method:

In this method, sodium aluminum ortho silicate known as permutit or zeolite is used to remove the permanent hardness of water.

Reaction: Na2 Al2 Si2 O8.KH2O + Ca++→ 2Na+ + Ca Al2 Si2 O8.xH2O

Calgon’s Process:

In this method, sodium-hexa-meta-phosphate (NaPO3)6 known as Calgon is used. The hardness in water is removed by the adsorption of Ca++ and Mg+

Ion Exchange Resin Method:

In this method, the permanent hardness of water is removed by using resins. Ca++/Mg++ ions are exchanged with Cl–, SO4-2 ions are exchanged with anion exchange resin (RNH2OH). Demineralized water is formed in this process.

⇒ 2RCOOH + Ca++ → (RCOO)2Ca + 2H+

⇒ RNH2OH + Cl– → RNH2Cl + OH–

⇒ H+ + OH– → H2O+ ions

Washing soda reacts with soluble chloride and sulphates of calcium and magnesium present in hard water of from insoluble carbonates which can be removed by filtration

CaCl₂​+Na₂​CO_3​→CaCO₃​↓2NaCl

MgCl₂​+Na_2​CO₃​→MgCO₃​↓+2NaCl

CaSO₄​+Na₂​CO₃​→CaCO₃​↓+Na₂​SO₄​

MgSO_4​+Na₂​CO₃​→MgCO₃​↓+Na₂​SO₄​

Given E = -3.4 ev

Then

E = \(\frac{-13.6}{n^2}\) 

-3.4 = \(\frac{-13.6}{n^2}\) 

n² = 4

n = 2

(a) r = \(\frac{e^2}{8\pi\varepsilon_0E_k}\)

r = \(\frac{e^2}{4\pi\varepsilon_02E_k}\) 

r = \(\frac{9\times10^9\times(1.6\times10^{-19})^2}{2\times3.4}\)

r = \(\frac{23.04\times10^{-29}}{6.8}\) 

r = 3.38 x 10^-29m

(b) Angular momentum L = \(\frac{nh}{2\pi}\)

 = \(\frac{2\times h}{2\pi}\) 

L = h/π

2.1 x 10^-34 JS

(c) For hydrogen atom

Kinetic energy = -(total energy)

Kinetic energy = 3.4 ev

Potential energy = -2 x K.E.

= - 2 x 3.4

P.E. = -6.8 ev

Answer is  NO

a. Concentration of the ore methods are: 

• Levigation/hydraulic washing 
• Froth floatation 
• Magnetic separation 
• Leaching 

b. Extraction of metal from the concentrated ore has 2 stages. 

• Conversion of ore into its oxide. The different methods are Calcination, Roasting etc. 
• Reduction of oxidised ore using suit-‘ able reducing agents.

c. Refining of metals. Different methods are: 

• Liquation 
• Distillation 
• Electrolytic refining

High availability 

Extraction of metal should be easy The percentage of metal content in the mineral should be comparatively high. Cost of production should below.

a. Heat conductivity, lightweight, can be molded in any shape, low price, etc. 

b. Malleability 

c. Ductility

2Al(OH)₃ → Al₂O₃ + 3H₂O

Bauxite (Al₂O₂H₂O)

Copper, Zinc

Correct option (1)  T₁ >T₂

Explanation:

Resistance temperature α Slope of V-I graph = resistance greater slope, more will be resistance greater will be temperature. Hence T₁ > T_2.