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Correct option is: (A) 0

Most famous centre of learning during the mauryan period was Taxila

(c) inductance

(a) inductive impedance with large phase angle 

(c) for balance, it uses two opposite impedances in opposite arms

(3) E2 > E1 

After decay, the daughter nuclei will be more stable hence binding energy per nucleon will be more than that of their parent nucleus.

Concept:

Let us consider sequence a₁, a₂, a₃ …. a_n is an A.P.

• Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_{n – 1}

• n^th term of the A.P. is given by a_n = a + (n – 1) d

Calculation:

Given series is 2, 6, 10, ...,146

First term, a = 2, last term, a_n = 146, an

Common difference d = 4, so it is an AP

an = a + (n – 1) d

146 = 2 + (n - 1) (4)

⇒ n - 1 = 144/4

⇒ n = 36 + 1 = 37

So, number of terms in given series = 37

Middle term = (37 + 1)/2 = 19th term 

a₁₉ = 2 + (19 - 1) × 4

= 2 + 72 

= 74

Hence, option (3) is correct.

Concept:

The arithmetic mean is the sum of all the numbers in a data set divided by the quantity of numbers in that set.

Calculation:

Given: Arithmetic mean of a, b, c is \(\rm \frac M 3\) 

So, \(\rm \frac M 3 = \frac {a+b+c}{3}\)

⇒ M = a + b + c           .... (1)

Given: \(\rm \frac{1}{a} + \frac{1}{b} = -\frac{1}{c} \)

⇒ \(\rm \frac{1}{a} + \frac{1}{b} +\frac{1}{c} =0\)

⇒ \(\rm \frac {ab + bc+ac}{abc} = 0\)

⇒ ab + bc  + ac = 0      .... (ii)

As we know, (a + b + c)² = a² + b² + c² + 2(ab + bc  + ac)

⇒ M² = a2 + b2 + c2 + 2 × 0             [From equation (i) and (ii)]

∴ a2 + b2 + c² = M²

Now, the arithmetic mean of a2, b2, c2 = \(\rm \frac {a^2+b^2+c^2} 3 = \frac {M^2}3\)

(a) Both the statements are true and Statement II is the correct explanation of Statement I. 

(A) y(t) = A sin πx/6 cos 50πt/3

(C) y(t) = A sin 5πx/6 cos 250πt/3

(D) y(t) = A sin 5πx/2 cos 250πt

Taking y(t) = A f(x) g(t) & 

Applying the conditions: 

1; here x = 3m is antinode & x = 0 is node 

2; possible frequencies are odd multiple of fundamental frequency.

where, V_fundamental = v/4l = 25/3 Hz

Correct option is (D)  L₁ + L₂ - 2M

Current on both the inductor is in opposite direction.

Hence:

L_eq = L₁ + L₂ - 2M

Length of tangent = √S₁₁ 

= √((2)² + (5)² - 5 x 2 + 4 x 5 + k) = 37 = 39 + k 

= – 2 units.

Answer is (b) 5/√6

Answer is (b) 7

Answer is (b) -15

Answer is (c) - 18

In ΔABD,

The angles are 45°, 45°, 90°

As the triangle is equilateral, sides are in the ratio, 1: 1: √2

AB = 6cm

∴ AD = 6cm, BD = 6√2cm

\(CD=\frac{6\sqrt{2}}{2}=3\sqrt{2}\)

\(BC=3\sqrt{2}\times\sqrt{3}=3\sqrt{6}\)

Base perimeter = 48cm

baseedge = \(\frac{48}{4}\) = 12 cm, slant height = 10cm

height = \(\sqrt{(10)^2-(6)^2}\) = \(\sqrt{64}\) = 8cm

Curved siuface area= 2 × 12 × 10 = 240 cm²

Volume = \(\frac{1}{3}\) (12)² × 8 = 384 cm³

Slant height = √3 a

Height = \(\sqrt{(\sqrt{3a})^2-a^2}\) = \(\sqrt{2}a\)

Base edge : Slant height : Height = 2a: √3a : √2 a = 2: √3 : √2

Base edge of the square pyramid = 30 cm 

Height = 30 cm

Slant height = \(\sqrt{30^2+15^2}\)

= \(\sqrt{900+225}\) = \(\sqrt{1125}\) = \(15\sqrt{5}\)

Lateral surface area = \(4\times\frac{1}{2}\times15\sqrt{2}\times30\)

= 60 x 15√5 = 900√5 sq.cm

Total surface area = 900 + 900√5

= 900(1 + √5) sq.cm

Base edge = 8 cm, height = 3 cm

Slant height = \(\sqrt{(\frac{8}{2})^2+(3)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5

Length of lateral edge = \(\sqrt{(slant\,height)^2+(\frac{base\,edge}{2})^2}\)

= \(\sqrt{(5)^2+4^2}{}\)

= \(\sqrt{25+16}\)

= \(\sqrt{41}\) cm

Height = h,

Slant height = l,

Lateral edge = e

\(l^2=h^2+(a/2)^2-(1)\)

\(l^2=e^2-(a/2)^2-(2)\)

(1) + (2) : 2l² = h² + e²

\(l^2=\frac{h^2+e^2}{2}\)

∴ (h², I², e²) are in arithmetic sequence.

Since slant height is 5 cm, such a pyramid can’t be constructed.

Slant height should be greater than half of the base edge. \(\sqrt{13^2-12^2}\) = 5 cm

Slant height = 12 cm

lateral edge = 13 cm

\(\frac{base\,edge}{2}\) = \(\sqrt{(lateral\,edge)^2-(slant\,height)^2}\)

= \(\sqrt{(13)^2-(12)^2}\)

= \(\sqrt{169-144}\)

= \(\sqrt{25}\)

= 5 cm

base edge = 10 cm

Height = \(\sqrt{(12)^2-(5)^2}\) = \(\sqrt{119}\) cm

v₁ = v₂

\(\frac{1}{3}a^2_1h_1\) = \(\frac{1}{3}a^2_2h_2,\) \(\frac{1}{3}a^2h_1\) = \(\frac{1}{3}(\frac{a}{2})^2h_2\)

\(\frac{1}{3}a^2h_1=\frac{1}{3}\times\frac{a^2}{4}\times h_2\)

\(h_1=\frac{h_2}{4},\) 4h₁ = h₂

The height of the second pyramid is 4 times the height of the first pyramid.

slant height = 12 cm

centre angle = \(\frac{360}{60}=60\)

radius of square pyramid = 12 × \(\frac{60}{360}\) = 2 cm

a. Surface area of the sphere = 4 πr²

Here, 4 πr² = 40cm²

πr² = 10cm²

∴ Area of plane surface = πr² = 10cm²

b. Surface area of one piece (hemisphere) = 3 πr² = 3 x 10 = 30cm²

height =12 cm. slant height = 15 cm

a = \(2\sqrt{15^2-12^2}\) = 2 x 9 = 18 cm²

Total surface area = base- area + curved surface area

= (18)² + 2 × 18 × 15 = 324 + 540 = 864 cm²

Volume = \(\frac{1}{3}\)(18)² x 12 = 1296 cm³

Slant height = \(\sqrt{17^2-8^2}\) = \(\sqrt{275}\) = 15 cm

Length of base edge = 6 cm, height = 4 cm, slant height = \(\sqrt{(3)^2+(4)^2}\)

= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 cm

Total surface area = base area + curved 

surface area = (6)² + 2 × 6 × 5 = 36 + 60 = 96 cm²

Sum of edges = 8 x 8 = 64 cm

Slant height = 4√3 cm

Height = \(\sqrt{(4\sqrt{3})^2-4^2}\)

= \(\sqrt{18-16}\) = \(\sqrt{32}\) = \(4\sqrt{2}\) cm

Ratio = \(\sqrt{3}:\sqrt{2}\)

Base perimeter = 80 m, Base edge = 20 m

Lateral edge = 26 m

a. Slant height = \(\sqrt{26^2-10^2}\) = \(\sqrt{576}\) = 24

b. Lateral surface area = \(4\times\frac{1}{2}\times20\times24\)

= 960 sq.metre

Radius of bigger circle = R

Radius of smaller circle = r

Radius of circular base of the pyramid = r = \(\frac{R}{2}\)

Ratio between radius and slant height 

\(r\div R=\frac{R}{2}:R=\frac{1}{2}:1=1:2\)