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Correct option is 1440

The series starts with 2. We multiply first term with 6 to get second term 12. Then we multiply second term with 5 to get third term 60. Then we multiply third term with 4 to get fourth term 340.Then we multiply fourth term with 3 to get fifth term 720.Then we multiply fifth term with 2 to get fifth term 1440.Then we need to multiply with 1 and we get 1440 again.

0.25 molal aqueous solution of urea means that 0.25 mole of urea are present in 1 kg of water  
Moles of urea = 0.25mole 
Mass of solvent (water) = 1 kg = 1000 g 
Molar mass of urea (NH₂CONH₂) = 60 g mol
1. 0.25moleofurea=0.25mol×60gmol1=15g 
Total mass of solution=1000+15g=1015g=1.015kg

 Thus; 1.015 kg of solution contain urea = 15g×2.5kg of solution will require urea

\(\frac{15g}{1.015kg}\times2.5kg=37g\)

Concept:

As we know that after decimal there is no significance of zeros after a non-zero digit 

Calculation:

So, we have to calculate for

⇒ √0.0001

⇒ √(1/10000)

⇒ 1/100

= 0.01

∴ The value of √0.0001 is 0.01.

(a) To estimate the maximum height of a mountain. 

(b) In minimizing of the bending of loaded beam 

(c) In selecting metallic rope for crane

The Correct option is A. 15.6 W

Correct option  (b) 45

Correct option   (c) √(− 1)

\(3\sqrt x - \frac1{4\sqrt x} = \sqrt 7\)

Squaring on both sides, we get

\(9x + \frac1{16x} - 2(3\sqrt x) \left(\frac1{4\sqrt x}\right) = 7\)    \((\because (a-b)^2 = a^2 + b^2 - 2ab)\)

⇒ \(9x + \frac1{16x} = 7 + \frac32 = \frac{17}2\)

Again squaring on both sides

\(81x^2 + \frac1{256x^2}+ 2(9x)\left(\frac1{16x}\right)= \frac{17}2\)

⇒ \(81x^2 + \frac1{256x^2} = \frac{17}2- \frac98= \frac{68-9}{8}= \frac{59}{8}\)

Hence, the value of \(81x^2 + \frac1{256x^2} = \frac{59}{8}\).

Correct option (b) 90

Correct option is (C) \(|\vec a \times \vec b |^2\)

\([\vec a + \vec b \,\,\,\,\vec b\,\,\,\, \vec a \times \vec b]\)

\(= (\vec a + \vec b).(\vec b \times (\vec a \times \vec b))\)

\(= (\vec a + \vec b).((\vec b.\vec b)\vec a - (\vec b.\vec a)\vec b)\)

\(= (\vec b.\vec b).((\vec a + \vec b).\vec a)- (\vec b .\vec a)((\vec a + \vec b).\vec b)\)

\(= (\vec b.\vec b).(\vec a .\vec a + \vec b. \vec a)- (\vec b .\vec a)(\vec a .\vec b + \vec b.\vec b)\)

\(= (\vec b.\vec b)(\vec a .\vec a)+ (\vec b.\vec b)(\vec b . \vec a) - (\vec b . \vec a)(\vec a.\vec b)- (\vec b .\vec b) (\vec b .\vec a)\)

\(= |\vec b|^2 \,\,|\vec a|^2 - (\vec a. \vec b)^2 \)     \((\because \vec a. \vec b = \vec b . \vec a)\)

\(= |\vec b|^2 \,\,|\vec a|^2 - (|\vec a| \,|\vec b|\,cos\theta)^2 \)

\( = |\vec a|^2 \,|\vec b|^2 - |\vec a|^2 \,|\vec b|^2 cos ^2 \theta\)

\(= |\vec a|^2 \,|\vec b|^2 (1 - cos^2\theta)\)

\(= |\vec a|^2\, |\vec b|^2 \, sin^2\theta\)

\(= (|\vec a| \,|\vec b|\, sin\theta )^2\)

= \(|\vec a \times \vec b |^2\) 

Correct option  (C) (3, 4/3)

Explanation :

A point inside a triangle divides the triangle into three triangles of equal areas if and only if the point is the centroid of the triangle. Hence, R must be the centroid of ΔOPQ. Therefore

R(0 + 3 + 6/3 , 0 + 4 + 0/3) = (3,4/3)

X² - 7x + 10

\(sin h(mx) = m \,sin x\)

⇒ \(\frac{sin(i \, mx)}{i} = m\, sin x \)     \((\because sin(ix) = i \,sin\,hx)\)

⇒ \(- i \,sin(i\, mx) = m \,sin x\)

If \(m = i\)

Then equation satisfies 

Hence, \(m = i\).

Concept:

Line in 3D: 

Equation of line passing through a point P(x1, y1, z1) with direction cosine l, m, n:

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{\rm{l}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{\rm{m}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{\rm{n}}}\)

Equation of line passing through two points P(x1, y1, z1) and Q(x2, y2, z2):

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{{{\rm{y}}_2} - {{\rm{y}}_1}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{{{\rm{z}}_2} - {{\rm{z}}_1}}}\)

Note:

Let direction ratios of a line be (l1, m1, n1) and for another line be (l2, m2, n2) then:

If lines are perpendicular, l1l2 + m1m2 + n1n2 = 0

If lines are parallel, \(\frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} = \frac{{{{\rm{m}}_1}}}{{{{\rm{m}}_2}}} = \frac{{{{\rm{n}}_1}}}{{{{\rm{n}}_2}}}\)

Calculation:

Given:

A = (1, 0, 3)

B = (4, 7, 1)

C = (3, 5, 3)

Direction ratio of line BC is (-1, -2, 2)

Equation of BC,

\(\frac{{{\rm{x}} - 4}}{{3 - 4}} = \frac{{{\rm{y}} - { 7} }}{{5-7}} = \frac{{{\rm{z}} - 1}}{{3-1}}\)

\(\Rightarrow \frac{{{\rm{x}} - 4}}{{-1}} = \frac{{{\rm{y}} - { 7} }}{{-2}} = \frac{{{\rm{z}} - 1}}{{2}} =\rm k\)

⇒ x = -k + 4, y = -2k + 7 and z = 2k + 1

General point on BC be D = (-k + 4, -2k + 7, 2k + 1) and let D be foot of perpendicular.

Direction ratio of line AD will be (-k + 4 - 1), (-2k + 7 - 0), (2k + 1 - 3)

⇒ (-k + 3, -2k + 7, 2k - 2) 

Now, since the line BC and AD are perpendicular, l1l2 + m1m2 + n1n2 = 0

So, (-k + 3) × -1 + (-2k + 7) × -2 + (2k - 2) × 2 = 0

⇒ k - 3 + 4k - 14 + 4k - 4 = 0

⇒ 9k = 21

⇒ k = \(\frac 73\)

So, D = \(\left( \frac 5 3 , \frac 7 3 , \frac {17}{3}\right)\)

Calculation:

The equation of the plane is 5y + 8 = 0, i.e.,

0x + 5y + 0z = -8

The direction ratio are (0, 5, 0)

Let the co-ordinates of the foot of the perpendicular be (x, y, z)

So equation of the perpendicular line will be 

\(\rm {x-0\over 0}= {y-0\over 5}= {z-0\over 0}\) = r (say)

x = 0, y = 5r and z = 0

As the point satisfy the equation of the plane, 

∴ 5y + 8 = 0

5(5r) + 8 = 0

r = \(-8\over25\)

So y = 5r = \(-8\over5\)

∴ The coordinates are (0, \(-8\over5\), 0)

Concept:

If three points are collinear then the area of the triangle formed from these 3 points will be 0.

[as the collinear points lie on the same straight line]

Analysis:

Area of triangle passing through the points

(x, y), (x₂, y₂) and (x₃, y₃) is given by:

\(A = \frac{1}{2}\left[ {x,\;\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\)

(x₁, y₁ = α, β); (x₂, y₂ = 5, 0); (x₃, y₃ = 0, 5)

∴ \(A = \frac{1}{2}\left[ {\alpha \left( { - 5} \right) + 5\left( {5 - \beta } \right) + 0} \right]\) 

-5α + 25 – 5β = 0

α + β = 5

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Calculation:

Given lines are  \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{-{(\rm y - {\rm{4})}}}{-2} = \frac{2{{(\rm{z}} - 2)}}{{\rm 2k}} \Leftrightarrow \frac{{\left( {{\rm{x}} +4 } \right)}}{{\rm{2}}} = \frac{{{\rm{y}} - 4}}{{ 2}} = \frac{{{\rm{z}} - 2}}{{ \rm k}}\)

So, the direction ratio of the first line is (2, 2, k)

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3

Concept:

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d and a point whose coordinate is, (x1, y1, z1).

Then the distance between the point and the plane is given by

 \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Quadratic equation with roots α and β:

Quadratic equation with roots α and β is given by

(x - α)(x - β) = 0  

Calculation:

We know that the distance between the point and the plane is given by

\(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

So, the distance of the point (2, 3, -5) from the given plane

 \(α\ =\ \left| {\frac{{2\; \times \;1\; + \ {2}\ \times \;3\; + \;(-\ 2)\ \times \ (-\ 5)\; - \;9}}{{\sqrt {{1^2}\; + \;\;{{\left( {-2} \right)}^2}\; + \;{{\left( 2\right)}^2}} }}} \right|\ =\ 3\)

Similarly, the distance of the point (3, 1, 1) from the given plane 

\(β \ =\ \left| {\frac{{3\; \times \;1\; + \ {2}\ \times \;1\; + \;(-\ 2)\ \times \ 1\; - \;9}}{{\sqrt {{1^2}\; + \;\;{{\left( {-2} \right)}^2}\; + \;{{\left( 2\right)}^2}} }}} \right|\ =\ 2\)

Therefore, quadratic equation with roots 3 & 2 will be

(x - 2)(x - 3) = 0

⇒ x² - 3x - 2x + 6 = 0

⇒ x2 - 5x + 6 = 0

Hence, option 3 is correct.

(a) IV, II, III, V, I

Fair conductors are the materials, which have less conductivity than that of semiconductor. Fair conductor gives more opposition to the flow of free electrons than that of semiconductors.

Examples for fair conductors are 

a. Charcoal 

b. Coke 

c. Carbon 

d. Plumbago

(a) Andhra Pradesh, Jharkhand, Maharashtra and Orissa

The area with annual rainfall less than 50 cmIn a year is Leh in Kashmir.

The bridge of sand and rock in the Palk Strait between India and Sri Lanka is  Adam's Bridge.

The bridge of sand and rock in the Palk Strait between India and Sri Lanka is Adam's Bridge.

(d) 1, 2, 3 and 4

(d) shifting cultivation

(A) Methionine

In a slanting hilly Indian terrain experiencing more than 200 cms of annual rainfall Tea crops can be cultivated best.

Indian desert is called Thar.

Nicobar is Indian islands lies between India and Sri Lanka.