Explore topic-wise InterviewSolutions in Current Affairs.

Answer is (4)

IUPAC name

[Pt(NH₃)₂Cl(NH₂CH₃)]Cl -Diamminechlorido (methanamine) platinum(II)chloride.

Answer is (4) Diamminechlorido (methanamine) platinum (II) chloride

Answer is (3) +1, +1 and +1

Potasisum has an oxidation of +1 (only) in combined state.

Answer is (3)

Oxidation number of K in K₂O, K₂O₂ and KO₂ = +1, +1, +1

(D) TATA box has a high melting temperature

ALTER TABLE students ADD CONSTRAINT NN_Grade CHECK (adm_no IS NOT NULL);

(D) Requires expression of cellular genes

(D) Operator 

(B) A specific region of the operon preventing transcription of structural genes 

(C) 2MnO^-₄, +3H₂O₂, → 2MnO₂, +30₂, +2H₂O + 2OH^- 

In option (A) and (C) reducing action of hydrogen peroxide is shown. 

In option (A) it is in acidic medium, in option (B) it is in basic medium. 

or 

For reducing ability H₂O₂ changes to O₂ , i.e. oxidize , so in option 'A' & 'C' O₂ is formed but 'A' is in acidic medium so option - C correct.

Messenger RNA, or mRNA, transfer RNA, or tRNA, and ribosomal RNA, or rRNA, are the three main types of RNA.

The newly formed RNA molecule, a precursor of mRNA, is called heterogeneous RNA (hnRNA). The heterogenous RNA bears portions called introns and portions called exons. The hnRNA is processed in many chemical steps, introns are removed and mRNA is created formed only of exons, the biologically active nucleotide sequences.

Correct option (D) GTP

Correct option is (D) Copper

For ZnS, KCN is used as depressant. 

For Gold and silver ⇒ leaching [Cyanide process]

Answer: C. L.H and F.S. H. both 

In the eukaryote cell nucleus RNA can be found dispersed in the nuclear fluid, along with DNA, and as the main constituent of the nucleolus. In cytosol (in eukaryotes or in bacteria) RNA molecules can be found free, as structural constituent of ribosomes (organelles specialized in protein synthesis) or even associated with them in the process of making proteins.

In replication the enzyme DNA polymerase catalyzes the formation of a new polynucleotide chain using free nucleotides in solution and putting them in the new chain according to the DNA template exposed and to the rule A-T, C-G. In transcription the enzyme RNA polymerase makes a new polynucleotides chain according to the DNA template exposed obeying, however, the rule AU, C-G.

In replication the original template DNA a chain is kept bound by hydrogen bonds to the newly formed DNA chain and a new DNA molecule is then created. In transcription the association between the template DNA chain and the newly formed RNA is undone and RNA constituted of only one polynucleotide a chain is liberated.

Correct option is (A) [Xe] 4f¹⁴ 5d⁹ 6s¹ 

₇₈Pt = [Xe] 4f¹⁴ 5d⁹ 6s¹ (Exceptional electronic configuration)

The best method for drying a cloth is Sun shine and air.

Correct option is (C) (A) > (B) > (C) > (D) 

(D) Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.

As equal & similar charge particle will repel each other, hence will never precipitate.

(D) Deionised water

In deionized water no common ion effect will take place so maximum solubility

(a) HF and H₂O

HF = 1 + 9 = 10

H₂O = 2 + 8 = 10

Let the image formed is virtual and erect.

V – u = 30 cm

m = 2

(a) m = - (v/u)

2 = - (v/u)

V = -2u

-2u – u = 30 cm

u = -10 cm

(b) v = -2u = 20 cm

1/v + 1/u = 1/f

1/20 + 1/-10 = 1/f

1/f = -1/20

f = -20 cm

R = 2f = -40 cm

 Concave mirror

Correct option is D) They assume that if a little is good for them a lot must be better

Correct answer is (A) 0 

Correct answer is (A) \(C = \frac{1}{\sqrt{\mu_0 \in_0}}\)

Correct answer is (A) about 10^-7 m 

With increase in the rate of doping, width of depletion region generated at the junction decreases.

The answer is: Decreases

Correct answer is (C) Parallel to \(\vec{B}\) x \(\vec{E}\)

Input frequency = 25 Hz

for a half wave rectification, the output frequency is equal to its frequency.

\(\therefore\) Output frequency = 25 Hz

The answer is : 25Hz

A p-n junction with reverse bias can be used as a photo-diode to measure light intensity.

The answer is : Photodiode

Correct answer is (B) 1.6 x 10^-19 C

Correct answer is (A) 1 ev = 1.6 x 10^-19 J

Correct answer is (A) ML²T^-1

For a p-n junction diode, the dynamic resistance is given by:

r = \(\frac{\Delta V}{\Delta I}\)

Given that:

ΔV = 0.71 - 0.70 = 0.01 V = 10^-2 V

ΔI = 10^-3 x 10 A = 10^-2 A

\(\therefore\) r = \(\frac{10^{-2}}{10^{-2}}\) = 1 Ω.

Dynamic resistance =change in voltage/change in current=1 ohm.

\(\int_0^{\frac{\pi}{2}} sin^2x\,cos^4x\,dx\)

= \(\frac{\Gamma(\frac{2+1}{2})\Gamma(\frac{4+1}{2})}{2\Gamma{(\frac{2+4+2}{2})}}\)

(∵ \(\int_0^{\frac{\pi}{2}} sin^m\theta\,cos^n\theta\,d\theta\) = \(\frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma{(\frac{m+n+2}{2})}}\))

= \(\frac{\Gamma(\frac{3}{2})\Gamma(\frac{5}{2})}{2\Gamma{(4)}}\)

= \(\frac{\frac{1}{2}\Gamma(\frac{1}{2})\times \frac{3}{2}\times \frac{1}{2}\Gamma(\frac{1}{2})}{2\times 3!}\)

(∵ \(\Gamma(n+1)\) = n\(\Gamma\)(n),n>0 & \(\Gamma\)(n+1) = n!, n∈N)

= \(\frac{3}{8}\) x \(\frac{\sqrt{\pi}\times {\sqrt\pi}}{2\times 6}\) 

= \(\frac{\pi}{32}\).

Hence,

\(\int_0^{\frac{\pi}{2}} sin^2x\,cos^4x\,dx\) = \(\frac{\pi}{32}\).

\(\int_0^\frac{\pi}{2} sin^3\theta\,cos^5\theta\,{d}\theta\)

= \(\frac{\Gamma(\frac{3+1}{2})\Gamma(\frac{5+1}{2})}{2\Gamma(\frac{3+5+2}{2})}\) 

= \(\frac{\Gamma(2)\Gamma(3)}{2\Gamma(5)}\)

= \(\frac{1!2!}{2\times 4!}\)

(∵ \(\Gamma\)(n+1) = n!, n ∈ N)

= \(\frac{1}{24}\)

Hence,

\(\int_0^\frac{\pi}{2} sin^3\theta\,cos^5\theta\,{d}\theta\) = \(\frac{1}{24}\) 

Let I = \(\int_0^{2\pi}sin^7\frac{x}{4}{d}x\)

= \(\int_0^{2\pi}(sin^2\frac{x}{4})^3sin\frac{x}{4}{d}x\)

= \(\int_0^{2\pi}(1-cos^2\frac{x}{4})^3sin\frac{x}{4}{d}x\) 

Let cos\(\frac{x}{4}\) = t

⇒ \(-\frac{1}{4}\) sin\(\frac{x}{4}\) dx = dt

⇒ sin\(\frac{x}{4}\)dx = -4dt

Limits changes to t = cos 0 = 1 to t =  cos\(\frac{2\pi}{4}\) 

= cos\(\frac{\pi}{2}\) = 0

∴ I = \(\int_1^0 \) (1-t²) x -4dt

= 4\(\int_0^1 \)(1-t²)³ dt

= 4\(\int_0^1 \)(1-3t²+3t⁴-t⁶)dt

= 4 [t - 3\(\frac{t^3}{3}\)+ 3\(\frac{t^5}{5}\) - \(\frac{t^7}{7}\)\(]^1_0\) 

= 4(1-1+\(\frac{3}{5}\)- \(\frac{1}{7}\))

= 4 x \(\frac{21-5}{35}\) 

= \(\frac{64}{35}\).

Option : (d) \(\sqrt{R^2+(X_L-X_C)^2}\)

Option : (d) Zero

Option : (a) perpendicular to the magnetic field