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Given

V = 110 V

R = 200 Ω

X_L = 100 Ω

X_C = 40 Ω

f = 50 HZ

C = 25 μF

(a) Impendence of the circuit \(Z = \sqrt{R^2 + (X_L - X_C)^2}\)

\(Z = \sqrt{(200)^2 + (100 - 40)^2}\)

\(Z = \sqrt{40000 - 3600}\)

\(Z = \sqrt{33700}\)

Z = 183.57 Ω

(b) Current V = IZ

I = V/Z

I = \(\frac{110}{183.57}\)

I = 0.59 ampere

(C) The angle

tan ϕ \(=\frac{X_C - X_L}{R}\)

\(=\frac{40 - 100}{200}\)

\(=\frac{-60}{200}\)

tan ϕ \(=\frac{-3}{10}\)

\(\phi = tan^{-1}(\frac{-3}{10})\)

XL=wL

L= X_L/w(frequency)

L=220/(25x25)

L=0.3

We know that X_L = wL

Where,

w = frequency

L  = inductance

Then 

\(L = \frac{X_L}{w}\)

\(L = \frac v{iw}\)

L = \(\frac{220}{25\times25}\) 

L = 0.3 H

(d) Re(z) ≥ |z|

Correct option (B) first or second quadrant

Explanation :

The point (x, y) is equidistant from both axes 

|x| = |y| ⇒ y = ±x.

1.  y = x and (x, y) lies on the line 3x + 5y ⇒ x = 15/8

y = 15/8 ⇒ P lies in the first quadrant.

2.  y = -x and (x, y) lies on 3x + 5y =  15 ⇒ x = 15/2, y = 15/2 ⇒P lies in the second quadrant.

Let the edge of the cube be a. Then, 

6a² = 1734 

a² = 289 => a = 17 cm. 

Volume = a³ = (17)³ cm³ = 4913 cm³.

C. He is uncertain about it and then finds it thrilling.

Volume of the block = (6 x 12 x 15) cm³ = 1080 cm³. 

Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm. 

Volume of this cube = (3 x 3 x 3) cm³ = 27 cm³ . 

Number of cubes = 1080/27 = 40.

B. (2) and (5)

B. contented

C. (2) and (4)

While Roger Skunk was very happy about his own transformation, his mother felt that her baby should smell like a little Skunk and not like roses. So, she took Roger Skunk back to the wizard and hit the wizard over his head with an umbrella. She told him to make Roger Skunk smell bad again.

Jo was worried that now that Roger Skunk smells bad again, no animal would play with him. She wanted the wizard to hit mommy back and refuse to change Roger Skunk.

Length of longest pole = Length of the diagonal of the room 

= √(12² + 8² + 9²  = √(289)

= 17 m.

Volume = (16 x 14 x 7) m³ = 1568 m³. 

Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] cm² 

= (2 x 434) cm² 

= 868 cm² 

D. made some payment to her in advance

The wizard asked Roger Skunk to go to the end of the lane and turn around three times and look down the magic well and there he would find three pennies. Roger followed the instructions, collected the money and handed it over to the wizard as the price for fulfilling his wish.

Volume of the wall = (2400 x 800 x 60) cu. cm. 

Volume of bricks = 90% of the volume of the wall

=((90/100)*2400 *800 * 60)cu.cm. 

Volume of 1 brick = (24 x 12 x 8) cu. cm. 

Number of bricks=(90/100)*(2400*800*60)/(24*12*8) = 45000.

(c) electromagnetic field 

Correct answer is (c) Am

Michael Madhusudan Datta

Correct Answer is GD Birla

The correct answer is Flute.

• Sikkil Mala Chandrasekhar is a noted South Indian Carnatic flutist. Mala Chandrasekhar was born to a musical family.

•  Sikkil Mala Chandrasekhar was born in a reputed musical family, she started learning flute at the age of five from her gurus Padmasree Sangeetha Kalanidhi Smt Sikkil Neela (Mother) and Smt. Sikkil Kunjumani (Aunt). Learning was very informal.
• She has gifted a small flute in Nagercoil.
• They very casually taught her how to create sound, special blowing exercises, and sarali varisais in ragam ‘Harikamboji
• It was during her final year, that she had the opportunity to participate in the “Violin Venu Veena” trio ensemble with Smt. Lalagudi Vijayalakshmi on the violin and her friend Padmavathi on the veena

 \(\frac{x^{e + 1}}{e + 1}\) + e^x - loga.x + C

Answer is Tan (x – 5)

∫xsinxdx = uv - ∫vdu

= x.(-cosx)- ∫-cosxdx = - xcosx + sinx + c

Put 3x² + x + 5 = t. 

(6x + 1)dx = dt 

When x = 0 then t = 5 

When x = 1 then t = 9

\(= \frac{2}{3}(9^\frac{3}{2} - 5^\frac{3}{2}) = \frac{2}{3}(27 - 5 \sqrt5)\)

Concept:

Property of definite integrals:

\(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation:

Given:

\(I=\int^{\frac {\pi} 2}_0 \frac {1}{1 + \sqrt {\cot x}}dx\)

\(I=\int^{\frac {\pi} 2}_0 \frac {1}{1 + \sqrt {\frac{cos\ x}{sin\ x}}}dx\)

 \(I=\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx}}\)   --- (1)

Using the property of definite integrals:

\(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

\({\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} }}{{\sqrt {\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} + \sqrt {\cos \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} }}{\rm{dx}} \)

 \(I= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx\;}}\)   --- (2)

Adding equation (1) and (2), we get

\(2I=\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx}}+ \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx\;}}\)

\(2I = {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx}}\)

\( 2I= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} 1{\rm{dx}}\)

\(2I = {\rm{\;}}\left[ x \right]_0^{\frac{{\rm{\pi }}}{2}} \)

\(2I= \frac{{\rm{\pi }}}{2}\)

 \(\therefore I= \frac{{\rm{\pi }}}{4}\)

\(\int\limits_1^e\int\limits_1^{log y}\int\limits_1^{e^x}logzdz.dx.dy\)

 = \(\int\limits_1^e\int\limits_1^{log y}[zlogz-z]_1^{e^x}dx.dy\)

(\(\because\int log zdz=log z\int1.dz-\int(\frac{d}{dz}log z\int1.dz)dz\) \(=zlogz-\int\frac1z.zdz=zlogz-z\))

 = \(\int\limits_1^e\int\limits_1^{log y}[e^xloge^x-e^x+1]dx.dy\) 

 = \(\int\limits_1^e\int\limits_1^{log y}(xe^x-e^x+1)dx.dy\) 

 = \(\int\limits_1^e[x\int e^x\,dx-\int(\frac{dx}{dx}\int e^xdx)dx-e^x+x]_1^{log y}dy\) 

 = \(\int\limits_1^e(xe^x-e^x-e^x+x)_1^{log y}dy\) 

 = \(\int\limits_1^e(log y.e^{log y}-2e^{log y}+log y-e+2e-1)dy\) 

  = \(\int\limits_1^e(y log y-2y+log y+(e-1))dy\) 

 = [log y\(\int\)y dy -\(\int\)(\(\frac{d}{dy}log y\int ydy\))dy - \(\frac{2y^2}2+y log y-y+(e-1)y]_1^e\) 

 = \([\frac{y^2}2log y-\frac{y^2}4-y^2+y log y+(e-2)y]_1^e\) 

 = \(\frac{e^2}2 log 2-\frac54e^2+elog e+(e-2)e+\frac54-(e-2)\) 

(\(\because log 1 = 0\))

 = \(\frac{e^2}4-\frac54e^2+e+e^2-2e+\frac54-e+2\)

 = \(\frac{e^2}4-2e+\frac{13}4\) = \(\frac14(e^2-8e+13)\)

Let I = \(\int\limits_{-1}^0\)(x² + 1) d (x² + 1)

Let x² + 1 = t

d(x² + 1) = dt

Limit converts from

t = (-1)² + 1 = 1 + 1 = 2

to t = 0² + 1 = 1

∴ I = \(\int\limits_2^1tdt\) = \([\frac{t^2}2]_2^1\) = (1/2 - 1/4) = -3/2

Alternative:

\(\int\limits_{-1}^0(x^2+1)d(x^2+1)\) = \([\frac{(x^2+1)^2}2]_{-1}^0\)

 = \((\frac12-\frac42)\) = -3/2

(i) b. She was created by taking nucleus from undder cell and cytoplasm from unfertilised egg

(ii) c. Dolly sheep

(iii) c. Monkey

(iv) b. Dr. Ian Wilmut

(v) d. Brigitte Boissliar

(i) b. Carcinogen

(ii) a. Oncogenic virus

(iii) d. Cancer is a hereditary disease

(iv) c. Pollen grain and adrenaline

(v) c. Assertion is true but reason is false

1. Cardiac output is the volume of the blood ejected by each ventricle into the arterial system per minute.

2. In response to the sensory impulses and the adrenal medullary hormone, the excitability of the pacemaker is increased and thereby heart rate is increased during exercise. This in turn increases the heart rate.

(b) Repolarisation of atria