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NADH, NADPH and FADH2 produced in oxidation-reduction reactions in various metabolic pathways like glycolysis, TCA cycle, HMP shunt, beta-oxidation of fatty acids, etc. are called “reducing power” in cellular biology and biochemistry.

Answer (b) 5th June

The complete flower contains both male and female reproductive parts. It has all four parts, that is stamen, pistil, sepals, and petals. It is also called as a bisexual flower. The examples include hibiscus, lily, tomato, and mango. 

Answer (b) Both the statements are true, but Statement-II is not the correct explanation of Statement-I.

The organisms that are capable of preparing their own food from simple inorganic substances like carbon dioxide and water by using sunlight energy in the presence of chlorophyll are called producers. The green plants synthesize their own food through the process of photosynthesis and thus are called the producers.

Functional groups are group of atoms or bonds that define the function of the hydrocarbon that they get attached to. Examples: Alcohol, Aldehyde, Ketone, Carboxylic acid, Halogens, Double & Triple covalent bonds.

We have given,

 \(E^\circ_{Zn^{+2}/Zn}\) = -0.76 V

\(E^\circ_{Cu^{+2}/Cu}\) = + 0.34 V

Zn⁺² + 2e^- \(\longrightarrow\) Zn,  \(E^\circ_1\) = 0.76 V---(1)

Cu⁺² + 2e^- \(\longrightarrow\) Cu, \(E^\circ_2\) = +0.34 V---(2)

(2) - (1) we get

Zn + Cu⁺² \(\longrightarrow\) Cu + Zn⁺²

E^o = \(E^\circ_2-E_1\) 

E^o  = 0.34 - (-0.76)

E^o = 1.10 V

As we know,

Δ G^o = -nFE^o

where,

n = Number of transfered electrons

F = Faraday constant 96500 C mol^-1

E^o = standard electrode potential

\(\therefore\) Δ G^o = -2 x 96500 x 1.10

 = 212.3 KJ

Advantage of fuel cell-

* Fuel cells have 0 higher efficiency than diesel or gas engines.

* Fuel cells can eliminate pollution caused by burning fossil fuels.

* The maintenance of fuel cells is simple.

Due to stability of phenoxide ion by resonance.

Answer is (d) Hg(NO₃)₂

P-Nitrophenol is more acidic than p-methyphenol because P-Nitrophenol has electron with drawing group. Which is more effective, while P-methylphenol has methoxy group. Which is less electron with drawing group.

P-sub-shell of nitrogen has half filled i.e., 2p³. So it is more stable. But sub-shell of carbon is 2p² which is not stable. So electron affinity of nitrogen is less than carbon.

Answer is (d) bcc and hcp

We know that the vapour pressure of the solvent above a solution containing a non-volatile solute (i.e., a solute that doesn't have a vapour pressure of its own) is directly proportional to the mole fraction of solvent in the solution. So, the vapour pressure of a liquid decrease with addition of non volatile

Electrode- An electrode is an electrical conductor used to make contact with a non-metallic part of a circuit. The cathode is the electrode where reduction take place.

Electrode potential- Electrode potential is defined as the potential of a cell consisting of the electrode in question acting as a cathode and the standard hydrogen electrode acting as an anode. Reduction always takes place at the cathode and oxidation at the anode.

(b) Into the collector

(d) None of the above

(b) Lightly doped

(a) Lightly doped

(a) Base-emitter junction

(b) A valence electron

(d) It approximately equals the emitter current.

(a) Less than the base supply voltage

The correct answer is: (a) 0.5 mA

\(S_n = \frac1{1\times 3} + \frac1{3\times 5} + \frac1{5\times 7} + .... + \frac1{(2n - 1 )\times (2n + 1)}\)

\( = \frac12\left( \frac2{1\times 3} + \frac2{3\times 5} + \frac2{5\times 7} + .... + \frac2{(2n - 1 )\times (2n + 1)}\right)\)

\( = \frac12\left( \frac{3-1}{1\times 3} + \frac{5-3}{3\times 5} + \frac{7-5}{5\times 7} + .... + \frac{(2n + 1)- (2n - 1)}{(2n - 1 )\times (2n + 1)}\right)\)

\(= \frac12 \left(\left(1 - \frac13\right) + \left(\frac13 - \frac15\right) + \left(\frac15 - \frac17\right) + ....+ \left(\frac1{2n - 1} - \frac1{2n + 1}\right)\right)\)

\(= \frac12 \left(1 - \frac1{2n + 1}\right)\)

\(\lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty }\frac12 \left( 1 - \frac1{2n + 1}\right) \)

\(= \frac12 \left(1 - \frac1\infty\right)\)

\(= \frac12 (1 - 0)\)

\(= \frac12\)

Eggshells contain calcium carbonate. When nitric acid is added to it, carbon dioxide gas is evolved. The reaction can be given as
CaCO₃ + 2HNO₃   ------>   Ca(NO₃)₂ + H₂O + CO₂

\(\lim\limits_{x\to 0} \frac{f(2 + h)- f(2 -h)}{h}\)

\(= \lim\limits_{x\to 0} \frac{f(2 +h)- f(2 + f(2) - f(2 -h))}{h}\)

\(= \lim\limits_{x\to 0} \frac{f(2 +h) - f(2)}{h} + \lim\limits_{h\to 0} \frac{f(2 -h) - f(2)}{-h}\)

\(= f'(2) + f'(2)\)

\(= 2f'(2)\).

There are three competing factors for which the ionisation potential of sulphur lower than phosphorus: 
1. The effective nuclear charge 
2. The size of the atom and the force of attraction according to Coulomb's law 
3. A pair of electrons in an orbital 

In case of phosphorous and sulfur, phosphorous has the greater ionization energy which means that it takes more energy to remove an electron from a phosphorous atom, which means that the electrons are lower in energy, even though the atom is larger and sulfur has a greater effective nuclear charge. 

This is what we call as stable half filled configuration of phosphorus.

(i) Molecular mass of H2O=2 × Atomic Mass of H+Atomic Mass of O=2+16=18 a.m.u
(i) Molecular mass of CO2 =Atomic Mass of C+2 × Atomic Mass of O=12+32=44 a.m.u
(i) Molecular mass of CH4 =Atomic Mass of C+4 × Atomic Mass of H=12+4=16 a.m.u

(i) H2O:
The molecular mass of water, H2O
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u
(ii) CO2:
The molecular mass of carbon dioxide, CO2
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u
(iii) CH4:
The molecular mass of methane, CH4
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u

1. Molecular mass of H₂O 

= 2(1.008 u) +16.00 u 

= 18.016u

 2. Molecular mass of CO₂ 

= 12.01 u + 2(16.00 u) 

= 44.01 u 

3. Molecular mass of CH₄

= 12.01 u + 4(1.008 u) 

16.042 u

Option (B) and (D) both are correct.

The complexes which have 17 electrons are act as a oxidizing agent and which have 19 or 20 electrons they act as reducing agent.

Here,

\([Fe(Co)_5] = 8 + 2\times 5 = 18 e^-\)

\([Mn(Co)_5] = 7 + 2 \times 5 = 17 e^-\)

\([Co(NO_2)_6]^{4 -} = 9 + (1\times6)+ 4 = 19e^-\)

\([Co(Co)_4] = 9 + 2\times 4 = 17 e^-\)

Hence, \([Mn(Co)_5] \) and \([Co(Co)_4] \) may act as a oxidizing agent.

(1) Electrical energy is converted into mechanical energy. 

(2) Generator.

3 gram atoms.

1 mol of CO2 = 12 + 32 = 44g

Number of oxygen in 1 molecule of CO2 = 2

Therefore, 

44g of CO2 contains 2 mol of oxygen 

18g of CO2 contains = 2/44 × 18 = 0.81 mol of oxygen 

1 mil = 6.022 × 10²³ atoms

0.81 mol = 4.877 × 10²³atoms

\((\frac34\times\frac{4}{3y})(\frac23\times\frac{-3}{2y})\) = \(\frac1y\times-\frac1y=-\frac1{y^2}\)

apply L-h rule 

d/dx of log (x+1)/x we get 1/(x+1) =1/(0+1)=1

Let x = 5 & y = -1

x - 2y = 5 - 2(-1) = 5 + 2 = 7 (satisfied)

∴ x = 5, y = 1 is a solution of equation x - 2y = 7.

Now, 2x + 3y = 2 x 5 + 3 x (-1) = 10 - 3 = 7

∴ x = 5, y = - 1 is also a solution of equation 2x + 3y = 7

Hence, (5, -1) is a solution of the simultaneous equations x - 2y = 7 & 2x + 3y = 7.

yes,

by multiplying the terms we get:

(2x-1)(x-3)=(x+5)(x-1)

2x²-x-6x+3=X²+5x-x-5

2x²-7x+3=X²+4x-5

now mowing RHS side equation to LHS side:

we will get, 2x²-x²-7x-4x+3+5=0

                   = x²-11x+8=0

hence the equation has degree as 2 so it is a quadratic equation 

     (2x-1)(x-3) = (x+5)(x-1) 

=> By multiplying the terms 

=> 2x(x-3)-1(x-3) = x(x-1)+5(x-1)

=> 2x²-6x-x+3 = x²-x+5x-5

=> 2x²-7x+3 = x²+4x-5

=> 2x²-x²-7x-4x+3+5 = 0

=> x²-11x+8 = 0

So, this equation satisfies that it is a Quadratic Equation.

 ax² + bx + c = 0

2[-x¹⁺¹]¹₀

2[-1²-0²]

2[1+0]

2

Given∶

The speed of boat in still water is 7 km / hr.

The speed of stream is 2 km / hr.

The time taken by the man to row upstream is twice the time taken by him to row downstream

Formula used ∶

Let u and v be the speed of boat in still water and speed of stream respectively.

Speed upstream = (u - v)

Speed downstream = (u + v)

Calculations∶

Let the distance travelled upstream = x km

And the distance travelled downstream = y km

Speed upstream = (u - v) = 7 - 2 = 5 km / hr

Speed downstream = (u + v) = 7 + 2 = 9 km / hr

Time taken by the man to row upstream is = \(\frac{{\rm{x}}}{5}\)

Time taken by the man to row downstream is = \(\frac{{\rm{y}}}{9}\)

The time taken by the man to row upstream is twice the time taken by him to row downstream ∶

\(\Rightarrow \frac{x}{5} = \left( {\frac{y}{9}} \right) \times 2\)

⇒ x ∶ y = 10 ∶ 9

Given that

ax + by = c ... (1)

x² + y² = 1 ... (2)

Squaring (1) on both sides, we get

⇒ a²x² + b²y² + 2axby = c²

⇒ a²(1 - y²) + b²(1 - x²) + 2axby = c²  (∵ x² + y² = 1) 

⇒ a² - a²y² + b² - b²x² + 2axby = c²

⇒ a² + b² - c² = a²y² + b²x² - 2axby

⇒  a² + b² - c² = (bx - ay)² 

∴ bx - ay = \( { \pm \sqrt{a^2 + b^2 - c^2} }\)

\(\frac{2\sqrt 3}{\sqrt 6 +2} - \frac{4\sqrt 3}{\sqrt 6 -2}\)

\(= \frac{2\sqrt 3 (\sqrt 6 -2)-4\sqrt 3(\sqrt 6 + 2)}{(\sqrt6 + 2) (\sqrt 6 -2)}\)

\(= \frac{2\sqrt{18}- 4\sqrt 3 - 4\sqrt{18}- 8\sqrt 3}{6-4}\)    \(\left(\because (a + b)(a - b) = a^2 - b^2\right)\)

\(= \frac{-2\sqrt{18} - 12\sqrt 3}2 \)

\(= -\sqrt{118}-6\sqrt 3\)

\(= -\sqrt{18}-6\sqrt 3\)

\(=-3\sqrt 2 - 6\sqrt 3\)

The factors of 55 are 1, 5, 11, and 55. The negative pair factors of 55 are (-1, -55) and (-5, -11).

5√8 is greater            as 3*4=12 where as                 5*2√2=5*2*1.4=14.1

Firstly, we have to Prime Factorise 9216. 

We get 9216 = 2¹⁰ x 3² 

Therefore, 

Equating both the sides, 

We get a = 10 and b = 2.

\(\begin{array}{c|c}2&9216\\\hline 2&4608\\\hline2&2304\\\hline2&1152\\\hline2&576\\\hline2&288\\\hline2&144\\\hline2&72\\\hline2&36\\\hline2&18\\\hline3&9\\\hline3&3\\\hline&1 \end{array}\)

By prime factorization of 9216, we obtain

9216 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

= 2¹⁰ x 3²

= 2^a x 3^b (Given)

So, a = 10 & b = 2 (By comparing)

Concept used:-

Trick to find the square root of a number.

Method and Calculation:-

 • Find the unit digit of the square root by looking at the unit digit of the given number.

For ✓9216, unit digit must be 4 or 6

 • Find tens digit by looking at number formed by the first 2 digits.

81 < 92 < 100 (81 and 100 are squares of 9 and 10, so tens digit of square root will be 9.

 • Find the square root from the narrowed down options.

✓9216 = 94 or 96. To find this, we can find the square of the number between these numbers ending in 5 i.e 95.

95² = 9025 (for left part multiply tens digit with one more than tens digit, the right part is 25)

Clearly, ✓9216 = 96 (9216 is more than 9025)

(2m, m² - 1 , m² + 1) is a pythagorean triplet

(i) Let m² - 1 = 15

⇒ m² = 16 = 42

⇒ m = 4

∴ 2m = 2 x 4 = 8

m² + 1 = 16 + 1 = 17

∴ (8, 15, 17) is a pythagorean triplet.

(ii) Let m² + 1 = 25

⇒ m² = 24 =  4 x 6

∴ m = 2√6

∴ 2m = 4√6

& m² - 1 = 24 - 1

∴ (4√6, 23, 25) is a pythgorean triplet.

(iii) Let 2m = 10

⇒ m² - 1 = 25 - 1 = 24

m² + 1 = 25 + 1 = 26

∴ (10, 24, 26) is a pythagorean triplet.

(iv) Let m² + 1 = 5

⇒ m² = 5 - 1 = 4 = 2²

∴ m = 2

∴ 2m = 4

m² - 1 = 4 - 1 = 3

Hence, (4, 3, 5) is a pythagorean triplet.

correct option:

(D) 0

correct option:

(C) 0

Given A + B + C = 5625 … (1)

And A = \(\frac{1}{2}\)(B + C) ⇒ 2A = B + C … (2)

Also B = \(\frac{1}{4}\)(A + C) ⇒ 4B = A + C … (3)

From 1 and 2 we have

A + B + C = 5625 ⇒ A + 2A = 5625

⇒ 3A = 5625

⇒ A = Rs.1875

Again from 1 and 3 we get B+ 4B = 5625

⇒ 5B = 5625

⇒ B = Rs.1125

∴ from eqn (1) c = 5625 – A – B = 5625 – 1875.1125 = 2625 Rs.

∴ A’s share is 1875 B’s share is 1125, c’s share 2625