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योजक कड़ियाँ विशेष रूप से जीवित जीवधारियों से  संबंधित हैं जो जीवों के दो अलग अलग समूहों की विशेषता को लिए हुए होते है। ऐसे जंतु जिनमे दो समूहों के लाक्षणिक गुण उपस्थित हो, योजक कड़ी कहलाते हैं। 

We have given, 

Constant external pressure 

= 1.5 bar or 1.5 x 10⁵ Pa 

or 1.5 x 10⁵ N m^-2

Initial volume of gas = 20dm³ or 0.020m³

Workdone = -700 J

W = \(\int\limits_{V_1}^{V_2} \)- P_extd V

W = -P_ext \(\int\limits_{V_1}^{V_2} \)d V

W = -P_ext x(V₂ -V₁)

-700 J = - 1.5 X 10⁵ ( V₂ - 0.0.02 m³)

= 700/1.5 x 10⁵ = V₂-0.02

= 0.00467 = V₂ - 0.002

= V₂ = 0.00467 + 0.020

V₂ = 0.02467 m³

or V₂ = 24.67 dm³

Polynomial is passes through 4 collinear points.

So least degree polynomial be of degree 3.

Let P(x) = Ax³ + Bx² + Cx +D be required polynomial.

∵ P(x) passes through (0, -1).

∴ -1 = D        ........(1)

∵ P(x) passes through (1, -1).

∴ 1 = A + B + C + D

⇒ A + B + C = 2          ........(2)     (From (1))

∵ P(x) passes through (2, 1)

∴ 1 = 8A + 4B + 2C + D

⇒ 8A + 4B + 2C = 2     ........(3)     (From (1))

By Applying equation (3) - 2 x equation (2), we obtain

(8A + 4B + 2C) - (2A + 2B + 2C) = 2 - 4

⇒ 6A + 2B = -2

⇒ 3A + B = -1        .......(4)

∵ P(x) passes through (3, -2)

∴ -2 = 27A + 9B + 3C + D

⇒ 27A + 9B + 3C = -1      ........(5)     (From (1))

By applying equation (5) - \(\frac32\) x equation (3), we obtain 

(27A + 9B + 3C) - (12A + 6B + 3C) = -1 - 3

⇒ 15A + 3B = -4        ........(6)

By applying equation (6) - 3 x equation (4), we obtain 

(15A + 3B) - (9A + 3B) = -4 - (-3)

⇒ 6A = -1

⇒ A = \(\frac{-1}6\)

From equation (4), we get

\(\frac{-3}{6} + B = -1\)

⇒ \(B = -1 + \frac12 = \frac{-1}{2}\)

Put A = \(\frac{-1}6\) & B = \(\frac{-1}2\) in equation (2), we get

\(-\frac{1}{6} -\frac{1}{2} + C = 2\)

⇒ \(C = 2 + \frac16+ \frac12 = \frac{12 + 1 + 3}{6} = \frac{16}{6} = \frac83\)

∴ \(P(x) = \frac{-x^3}{6}-\frac{x^2}{2} + \frac{8}{3}x - 1\)

is the least degree polynomial passing through the points (0, -1) , (1, 1) , (2, 1) & (3, -2).

P(x) = (a + 1)x² + (2x + 3)x + (3x + 4)

= (a + 3)x² + 6x + 4

Sum of zeros  = -1    (Given)

⇒ \(\frac{-6}{a + 3} = -1\)

⇒ a + 3 = 6

⇒ a = 6 - 3 = 3

∴ Product of zeroes = \(\frac{4}{a + 3} = \frac{4 }{3 +3} = \frac46 = \frac23\).

Azole antifungals consist of two primary classes: imidazoles and triazoles. Both classes are fungistatic agents and share similar mechanisms of action. The azoles interfere with the synthesis and permeability of fungal cell membranes by inhibiting cytochrome P450-dependent 14-alpha-sterol demethylase.

A. the amount of protein nitrogen that is retained by the body from a given amount of protein nitrogen that has been consumed

General anesthesia works by interrupting nerve signals in your brain and body. It prevents your brain from processing pain and from remembering what happened during your surgery.

Given,\(\text{y=e}^{\text{tan(2x)}}\)

Then ,\(\frac{dy}{dx}=\frac{d}{dx}(e^{tan2x})\)

                 \(=e^{tan2x}.\frac{d}{dx}(\small{tan2x})\\ =e^{tan2x}.\small{sec^22x}.\frac{d}{dx}(\small{{2x}})\\=\mathbf{2sec}^{\mathbf{2}}\mathbf{2x}.{\mathbf{e}^{\mathbf{tan2x}}}\)

Given : 

• Density(d)→1.2g/cc
• Percentage by mass(w/w%)→45%
• Weight of solute →20g 

solution :

\(\text{percentage by volume}(w/V\%)=(w/w\%)\times{d}\)

                      (w/V%)=45×1.2

                  (w/V%)=54

We know that ,

\((w/V\%)=\frac{\text{weight of solute}}{\text{volume of solution}}×100\)

        \(54\%=\frac{20}{x}\times{100} \\x=37.03\text{ml}\)

Volume of solution will be 37.03ml

By relation,

\(\frac{1}{m}=\frac{d}{M}-\frac{1000}{M_o}\)

Where,

• m→molality
• M→molarity
• M_o→molar mass of solute 
• d→density

Given that M=0.1M ,d=1.11g/cc ,M_o= 98

  \(\frac{1}{m}=\frac{1.11}{0.1}-\frac{1000}{98}\)

   \(\frac{1}{m}=11.1-10.2 \\ \frac{1}{m}=0.9 \\ m=1.11m\)

Correct option is A) watch carefully

P. I. = \(\frac1{D^2-D+1}(3x^2-1)\)

= (1 - (D² - D) + (D² - D)² - (D² - D)³+...)(3x² - 1)

(\(\because\) \(\frac1{1+x}\) = 1 - x + x² - x³ + ....)

 = (3x² - 1) - (D² - D)(3x² - 1) + (D⁴ - 2D³ + D²)(3x² - 1)

 = 3x² - 1 - D.D(3x²) - D3x² + D.D(3x²)

 = 3x² - 1 - D(6x) - 6x + D.6x (\(\because\) D(3x²) = \(\frac{d}{dx}3x^2=6x\))

 = 3x² - 1 - 6 - 6x + 6

 = 3x² - 6x - 1

Correct option is C) as a group

To change order of integration, we need to write an integral with order dydx. This means that x is the variable of the outer integral. Its limits must be constant and correspond to the total range of x over the region D

I = \(\int\limits_0^2\frac{\sqrt x}{\sqrt x+\sqrt{2-x}}dx\)----(1)

\(\therefore\) I = \(\int\limits_0^2\frac{\sqrt {2-x}}{\sqrt {2-x}+\sqrt{2-(2-x)}}dx\) 

⇒ I = \(\int\limits_0^2\frac{\sqrt {2-x}}{\sqrt x+\sqrt{2-x}}dx\) 

 By adding equations (1) and (2) we get,

2I = \(\int\limits_0^2\frac{\sqrt x+\sqrt{2-x}}{\sqrt x+\sqrt{2-x}}dx\) 

 = \(\int\limits_0^2dx\) = \((x)_0^2\) = 2 - 0 = 2

\(\therefore\) I = 2/2 = 1

⇒ \(\int\limits_0^2\frac{\sqrt x}{\sqrt x+\sqrt{2-x}}dx\) = 1

B. 3’-OH group is absent

Electric field is force per unit positive charge.

Given : \(\frac{d}{dx}\bigg[\Big(\frac{dy}{dx}\Big)^4\bigg]\)=0

By chain rule ,

 \(\implies\)\(4\Big(\frac{dy}{dx}\Big)^3\Big(\frac{d^2y}{dx^2}\Big)\)=0

Order :order of highest derivative

order=2

Degree: degree of derivative of highest order 

Degree=1

By using chain rule to evaluate the derivative on the left hand side, we get:

\(\frac{d}{dx} {(\frac{dy}{dx})^4} = 0\)

⇒ 4 \((\frac{dy}{dx})^3 \frac{d^2y}{dx^2} = 0\)

The perimeter of rectangle = 16x³-6x²+12x+4

We know that , perimeter of rectangle =2(​​​​​I​​​ + b)

Given that one of it's side is 8x²+3x

Let, l ​​​=8x²+3x

\(\implies\)2(l +b) = 16x³-6x²+12x+4

\(\implies\) b = (8x³-3x²+6x+2)-(8x²+3x)

\(\implies\) b = 8x³-11x²+3x+2

The other side = 8x³-11x²+3x+2 

Perimeter of rectangle=16x³ − 6x² + 12x + 4

2(l + b) = 16x³ − 6x² + 12x + 4

l + b = 8x³ − 3x² + 6x + 2

b = (8x³ − 3x² + 6x + 2) − (8x² + 3x)

= 8x³ − 3x² + 6x + 2 − 8x² − 3x

= 8x³ − 11x² + 3x + 2

Answer is A. hydrogen

Let us convert 22/15,55/6 into like fractions.​ 

​​​​​​Steps for conversion:

1) Find the LCM of the denominators .LCM of 15 and 6 is 30.

2) Calculate their equivalent fractions with the same denominator, that is, the LCM.

\(\frac{22}{15}=\frac{22\times2}{15\times2}=\frac{44}{30}\)

\(\frac{55}{6}=\frac{55\times5}{6\times5}=\frac{275}{30}\)

\(\frac{22}{15},\frac{55}{6}\) are unlike fractions, which can be represented as \(\frac{\textbf{44}}{\textbf{30}} \,\text{and} \,\frac{\textbf{275}}{\textbf{30}}\) which are like fractions.

\((n+1)^2-n^2=( n^2+2n+1)-n^2\\ \qquad \quad \qquad \quad =2n+1\)

Answer : (n+1)²-n²= 2n+1 

\(Given : x=a(θ+sinθ) \) and \(y=a(1-cosθ)\)

Differentiating w.r.t. t we get       

\({dx \over dt} =a({dθ \over dt}) + cosθ({dθ \over dt })\)••••• (1) And  \({dy \over dt } = asin({dθ \over dt} )\)   •••••(2)

\({dy \over dx} = {{dy \over dt} \over {dx \over dt} } = {a{dθ \over dt}(sinθ) \over a{dθ \over dt} (1+cosθ) }\) ••••from (1) and (2)

       \( = { sinθ \over 1+cosθ }\) 

          \(= {2sin({θ \over 2}) cos({θ \over 2 }) \over 2 cos²({θ \over 2}) }\) 

      \(= {sin({θ \over 2}) \over cos( {θ \over 2}) }\) 

       \(= tan({θ \over 2})\)

The environment is affected by biotic and abiotic factors such as temperature, pressure, humidity, and organisms like human activity.

Potassium hydroxide solution (KOH aqueous) is a colourless inorganic liquid that acts as a strong base (alkali). KOH solution is also known as caustic potash or potash lye and has many different applications.

B. 5’ CATCGCTAGT 3

B. Sticky ends

The correct option is (a) 1.

\(\int\limits_0^{\frac{41\pi}2}sin\,x\,dx=-[cos\,x]_0^{\frac{41\pi}2}\) 

 = \(-(cos\frac{41\pi}2-cos0)\) = -(0- 1) = 1

Answer is A. Primers

Answer is B. pBR 322

Let I = \(\int\limits_1^5(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}})dx\) 

Let x - 1 = t²

then dx = 2tdt and limit converts to t = 0 to t = 2

\(\therefore\) I = \(\int\limits_0^22t(\sqrt{t^2+1+2t}+\sqrt{t^2+1-2t})dt\) 

 \(=\int\limits_0^22t(\sqrt{(t+1)^2}+\sqrt{(t-1)^2})dt\) 

\(=\int\limits_0^22t({(t+1)}+{(t-1)})dt\) 

\(=\int\limits_0^22t\times2tdt = \int\limits_0^24t^2dt\) 

 = \(\frac43(t^2)_0^2=\frac43(263 - 0) = \frac{32}3\)

B. chromosome 22

Answer is A. SNPs

B. Max Perutz

A. elevation of a tripeptide glutathione in cells

C. Browning,

A .Glutathione

Soups are considered healthy food, two reasons are:

1.Easy to digest 

2.Sufficient amount of fluids