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There are three parties to a bill of exchange:

1. Drawer – He is the creditor who draws a bill of exchange upon the debtor.

2. Drawee – He is the person upon whom the bill of exchange is drawn. He is the purchaser of the goods on credit and the debtor.

3. Payee – He is the person to whom payment of the bill is to be made on the maturity date. The drawer and the payee can be one party when payment is to be made to the drawer.

Selecting the axis as shown in Fig. we can say due to symmetry centroid lies on y axis, i.e. x = 0. 

Now the given T-section may be divided into two rectangles 

A₁ and A₂ each of size 100 × 20 and 20 × 100. The centroid of A₁ and A₂ are g₁(0, 10) and g₂(0, 70) respectively. 

∴ The distance of centroid from top is given by:

\(\bar y\) = \(\frac{100\times20\times10+20\times100\times70}{100\times20+20\times100}\)

= 40 mm

Hence, centroid of T-section is on the symmetric axis at a distance 40 mm from the top.

The wire is divided into three segments AB, BC and CD. 

Taking A as origin the coordinates of the centroids of AB, BC and CD are 

G₁(300, 0); G₂(600, 100) and G₃ (600 – 150 cos 45°, 200 + 150 sin 45°) 

i.e., G₃ (493.93, 306.07) 

L₁ = 600 mm, L₂ = 200 mm, L₃ = 300 mm 

∴ Total length L = 600 + 200 + 300 = 1100 mm 

∴ From the eqn. Lx_c = ΣL_i x_i , we get 

1100 x_c = L₁x₁ + L₂x₂ + L₃x₃ 

= 600 × 300 + 200 × 600 + 300 × 493.93 

∴ x_c = 407.44 mm 

Ly_c = ΣL_i y_i 

1100 y_c = 600 × 0 + 200 × 100 + 300 × 306.07 

∴ y_c = 101.66 mm

Correct option is: (A) - 4

Option : (a) 5

Cij=(-1)^I+j

C₃₁=(-1)*(-3)-2*2=3-4=-1

C₂₃=2*1-(-1)*3=2+3=5

C_{31=(-1)³⁺¹*(-1)=-1}

C_{23=(-1)³⁺¹*5=5}

(-5)*(-1​​​​)=5

Option:(a)

(1) The ideal of what the child should become

(2) The common beliefs of society's members, even though individuals or groups of them may have different beliefs

(1) and (2) are unrelated

First the ball ahead of the other will collide with the wall and reverse its direction and move with same speed.

After that it would collide with the ball approaching it with same speed and exchange the velocities. Hence the rear ball will move backwards with same speed but the ball which was ahead would again move towards wall with same speed.

The ball will again collide with wall and reverse its direction and move with same speed again(behind the other ball) forever.

Hence the total number of collisions is 3.

Erik Erickson' notions of child pedagogy:

• According to Erik Erikson, a prominent developmental theorist of the 1950's, youth must resolve two life "crises" during adolescence. Unlike many other developmental theorists of his era, Erikson's psychosocial theory of human development covers the entire lifespan, including adulthood.
• Erikson used the term "crisis" to describe a series of internal conflicts that are linked to developmental stages. 
• The first crisis typically occurs during early to middle adolescence, and is called the crisis of identity versus identity confusion. This crisis represents the struggle to find a balance between developing a unique, individual identity while still being accepted and "fitting in." Thus, youth must determine who they want to be, and how they want to be perceived by others.
• Erikson believed that when youth successfully navigate this crisis they emerge with a clear understanding of their individual identity and can easily share this "self" with others; therefore, they are healthy and well-adjusted. According to Erikson, an identity crisis is a time of intensive analysis and exploration of different ways of looking at oneself.

Both the statements are related to identity vs role crisis notions of Erik Erickson.

Hence, the correct answer is Option 4.

The heat is transmitted in three types. They are

(1) Conduction

(2) Convection

(3) Radiation.

(1) Conduction : The process of transmission of heat from one place to other without actual movement of the particles of the medium is called conduction.

Ex : When long iron rod is heated at one end, heat transmits to the other end.

(2) Convection : The process of transmission of heat from one place to another by the actual movement of the particles is called convection. 

Ex. : If water in a beaker is heated, the particles of water at the bottom receive the heat first. These particles expand, become lighter and rise up. At the same time colder and denser particles reach the bottom. They get in their turn heated and move up. This process is known as convection.

(3) Radiation : The process of transmission of heat from one place to another without any intervening medium is called radiation.

Ex. : Earth receives heat radiations from the sun.

When an object is placed between F and 2F in front of a convex lens, the image formed is magnified, real, inverted and beyond 2F.

C. 0.16 m s^-2

D. constant.

 The Correct option is A. 825 m

The Correct option is C. 35.3 m

The Correct option is B. 10 m s^-1

The Correct option is C. 3 m s^-2

A. increases

A. acceleration

B. weight of object 

D. acceleration

The Correct option is A. 20 m

Laws of photoelectric effect:

• For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light. 
• Maximum kinetic energy of the photo electrons is independent of intensity of the incident light. 
• Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light. 
• For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
• There is no time lag between incidence of light and ejection of photoelectrons.

Bar x = (∑f_ix_i)/N = 1800/40 = 45

MD (x) = 10.

Correct option is (a) \(\frac12(x^{5/2}+\frac1{x^{5/2}})\)

2 cos θ = \(\sqrt x+\frac1{\sqrt x}\)-----(1)

4 cos² θ = x + 1/x + 2----(2) (By squaring (2))

and 8 cos³θ = x^3/2 + \(\frac1{x^{3/2}}\) + 3\(\sqrt x\) x \(\frac1{\sqrt x}\) (\(\sqrt x+\frac1{\sqrt x}\)) 

 = x^3/2 + \(\frac1{x^{3/2}}\) + 3 (\(\sqrt x+\frac1{\sqrt x}\))---(3)

Now, cos 5θ = cos(4θ  + θ)

 = cos 4θ.cosθ - sin 4θ sin θ

 = (2cos²θ -1) cos θ - 2 sin 2θ cos2θ sin θ

 = (2(2cos²θ  - 1)² - 1) cos θ - 4 sin²θ cos θ(2cos² θ -1)

 = (2(4cos⁴θ  - 4 cos²θ +1)-1) cos θ - 4(1- cos²θ)(2cos³θ - cos θ)

 = (8 cos⁴θ  - 8 cos²θ  + 2 - 1) cos θ - 4(-2cos⁵θ + 2cos³θ  + cos³θ - cos θ)

 = 8 cos⁵θ - 8 cos³θ + cos θ + 8cos⁵θ - 12 cos³θ + 4cos θ

 = 16 cos⁵θ - 20 cos³θ + 5 cos θ

 = \(\frac12(8 cos^3\theta\times4cos^2\theta)-\frac52(8cos^3\theta)\) + \(\frac52\)(2cos θ)

 = \(\frac12((x^{3/2}+\frac1{x^{3/2}}+3(\sqrt x+\frac1{\sqrt x}))\)\((x+\frac 1x+2)\))

 = \(\frac12\Big(x^{3/2}+\frac1{x^{1/2}}+3x^{3/2}+3x^{1/2}+\frac1{x^{5/2}}+\frac3{x^{1/2}}+\frac3{x^{3/2}}\)

\(+2x^{3/2}+\frac2{x^{3/2}}+6x^{1/2}+\frac6{x^{1/2}}-5x^{3/2}-\frac5{x^{3/2}}\)

\(-15x^{1/2}-\frac{15}{x^{1/2}}\Big)\) 

 = \(\frac12(x^{5/2}+\frac1{x^{5/2}}+x^{3/2}(3+2-5)+\frac1{x^{3/2}}(3+2-5)\)

\(+x^{1/2}(3+1+6-15+5)+\frac1{x^{1/2}}(1+3+6-15+5))\)

= \(\frac12(x^{5/2}+\frac1{x^{3/2}}+x^{3/2}(5-5)+x^{1/2}(15-15)+\frac1{x^{1/2}}(15-15))\)

= \(\frac12(x^{5/2}+\frac1{x^{5/2}})\)

sin^-1\(\frac35\) + cos^-1\(\frac{12}{13}\)

 = tan^-1\(\frac34\) + tan^-1\(\frac5{12}\) 

 = tan^-1\(\left(\cfrac{\frac34+\frac5{12}}{1-\frac34\times\frac5{12}}\right)\)

 = tan^-1\((\frac{56}{33})\)

 = cos^-1\((\frac{33}{\sqrt{56^2+33^2}})\) = cos^-1\((\frac{33}{65})\)

Hence Proved

cos²x+1/cosec²x=1

cos²x +sin²x=1

Hence proved.

Explanation:

Given Integral is,

\(I~=~\displaystyle\int_{a+c}^{b+c}\ f(x)dx\)

Put t = x - c i.e. x = t + c

By differentiating we have,

dt = dx

at x = a + c  →  t = a + c - c = a ........(1)

at x = b + c  →  t = b + c -c = b .........(2) 

Now,

\(\displaystyle\int_{a+c}^{b+c}f(x)dx =~\int_a^b f(t+c)dt \)  .........(3)

Also, put t = x then,

t = a →  x = a ,,,,,,from equation (1)

t = b →  x = b ......from equation (2)

From the equation (3)

\(\displaystyle\int_{a+c}^{b+c}f(x)dx =~\int_a^b f(x+c)dt \)

Hence, it is proove.

Concept:

• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \left| \int_{a}^{b}f(x)\ dx \right|\), for curves which are entirely on the same side of the x-axis in the given range.
• Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \int_{a}^{b}f(x)\ dx=[g(x)]_{a}^{b}\) = g(b) - g(a).
•  \(\rm \int{\frac{1}{x}dx}\) = log x + C.

Calculation:

The given equation is of the curve is xy = c, which can also be written as y = f(x) = \(\rm \frac{c}{x}\).

Using definite integrals, the area under the curve from x = 1 to x = 4 and the x-axis, will be given as:

A = \(\rm \left| \int_{1}^{4}{\frac{c}{x}}\ dx \right|\)

Using \(\rm \int{\frac{1}{x}dx}\) = log x + C, we get:

⇒ A = \(\rm c\left[ \log x \right]_{1}^{4}\)

⇒ A = c(log 4 - log 1)

Using log 1 = 0 and log 4 = 2 log 2, we get:

⇒ A = 2c(log 2) sq. units

Explanation:

Given Integral is,

\(\displaystyle∫_0^1 \dfrac{e^t}{1+t}dt=a\)

Considering \(I_1 ~=~\frac{1}{1+t}\) and \(I_2 ~=~e^t \)

Hence by Integration by parts we know that,

∫ [I₁][ I₂ ] dt = I₁∫ I₂ dt - ∫ [I'₁∫ I₂ dt]dt

\(\frac{1}{1+t}\int_0^1 e^t dt-\int_0 ^1 [\frac{d}{dt}(\frac{1}{1+t}).\int_0 ^1 e^t dt]dt~=~a\)

\(\frac{1}{1+t}[e^t ]_0^1-\int_0^1 (\frac{-1}{(1+t)^2}).e^t dt~=~a\)

\([\frac{e^t}{1+t}]_0^1 +\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^t}{1+t}]_0^1\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^1}{1+1}-\frac{e^0}{1+0}]\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e}{2}-1]\)

\(\displaystyle\int_0^1 \dfrac{e^t}{(1+t)^2}dt~=~a+1-\frac{e}{2}\)

Explanation:

Area enclosed by a curve y = f (x) and x axis is given by:

A = ∫ f (x) dx

Since curve is bounded by lines x = a and y = b

i.e Lower limit for x = a and upper limit for  x = b (∵ b > a)

∴ \(A=\displaystyle\int_a^b f(x) \ dx\)

Since f(x) = y

Hence \(A=\displaystyle\int_a^b y \ dx\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ eax dx = \(\rm e^{ax}\over a\)+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

Given f'(x) = \(\rm 4x^3 - 2 {1\over \sqrt x} + e^{-4x}\)

f(x) = ∫ f'(x) dx

⇒ f(x) = \(\rm \int 4x^3 - 2 {1\over \sqrt x} + e^{-4x}\) dx

⇒ f(x) = \(\rm 4\left[{x^4\over4}\right] - 2\left[{x^{1\over2}\over{1\over2}}\right] + \left[{e^{-4x}\over-4}\right] \)

⇒ f(x) = x⁴ - 4\(\rm\sqrt x - {e^{-4x}\over4}\) + C

Now f(0) = \(3\over4\)

⇒  04 - 4\(\rm\sqrt 0 - {e^{-4(0)}\over4}\) + C = \(3\over4\)

⇒ C - \(1\over4\) = \(3\over4\)

⇒ C = 1

⇒ f(x) = x4 - 4\(\boldsymbol{\rm\sqrt x - {e^{-4x}\over4}}\) + 1

Option is (b) obtuse angled triangle.

Explanation

Given ratio of the sides =2:6:7

Let the sides are 2x,6x and 7x unit.

Square of the longest side =49x²

Sum of the square of other two sides are = 4x²+36x²=40x²

If this is 49x² then by Pythagoras theorem the triangle would be right triangle.

But  the sum is less than 49x². So the triangle is obtuse one.

Mnemonic

If a,b,c are three sides of a triangle such that c is the longest side then

When c² = a²+b² -> triangle is rt angled

When c² > a² +b² -> triangle is obtuse angled

When c² < a²+b²-> triangle is acute angled

Hence the correct answer is (b)Obtuse angled triangle.

Lets,

Original length = x m

Original breadth = y m

Original are = l x b

= x m x ym

= xy m²

Each side increased by 20%

Then,

length = 120x/100 = 6x/5 m

breadth = 120y/100 = 6y/5 m

Area = l x b

= 6x/5 x 6y/5 m²

= 36xy/25 m²

Different = 36xy/25 m² - xy m²

= (36xy/25 - xy/1) m²

= (36xy - 25xy)/25 m²

= 11xy/25 m²

Increase% = (Different/ original) x 100%

= ((11xy/25 ) / xy)  x 100%

= (11xy/25) x (1/xy) x 100%

= 44%

13 = 52

1 = 52/13

= 4

5 + 12 + 13 = 30

Total perimeter = 30 x 4

= 120cm

We know diagonals bisects at right angle

∴ In △AOD

AO² + OD² = AD²

(3)² + (4)² = AD²

⇒ AD = 5 cm

The side of rhombus is 5cm.

a, a + d, a + 2d

a + a + d + a + 2d = 207

3a + 3d = 207

3(a + d) = 207

a + d = 207/3 = 69

a x (a + d) = 4623

a x 69 = 4623

a = 4623/69 = 67

a = 67

67 + d = 69

⇒ d = 69 - 67 = 2

a = Rs. 67

a + d = Rs. 69

a + 2d = Rs. 67 + (2 x 2) = Rs. 71

Let a, d and a_n​ be the first term, common difference and nth term of first AP and

A, d and A_n​ be the first term, common difference and nth term of second AP respectively. 

 Given a=2 and A=7 

Also, given that, a₁₀​−A₁₀​=a + 9d − (A + 9d )= a − A = 2 − 7 = −5

and a₂₁​ − A₂₁ ​= a + 20d −(A + 20d) = a − A = 2 − 7 = −5

Hence, the difference between any two corresponding terms of these AP's is equal to difference between first terms of these AP's. 

∴   given statement is true so answer is 1.

Ghulam Ahmed-India and Asif Iqbal-Pakistan

SOLUTION: 
Let the roots of the equation be α and β.  Then  
       α +  β = _1/α²  +  1/β²

or,  α +  β = (α²+β²) / α²β²

or,  α +  β = ( α +  β)² - 2αβ   / α²β²

Putting the values of α +  β and αβ we get 

      -b/a =( (b²/a²) - 2c/a) / (c²/a²)

or,    -b/a = (b²- 2ac)/ c²

or,    -bc² = ab² - 2ca²

or    bc² + ab² = 2ca²

Hence bc² , ca², ab² are in A.P.

Madras Cricket Club

In any metallic conductor the current is due to the flow of the charged particles. Current can flow due to ions or any other charge particle which are freely available for the movement through the conductor.