Explore topic-wise InterviewSolutions in Current Affairs.

Correct answer is (A) L_m = 5/6 L_g

Correct answer is (B) \(\frac{q}{\in_0}\)

Correct answer is (B) q x ε₀

Correct answer is (B) -10 cm

Correct answer is (C) I sin ϕ 

Given,

Mechanical advantage = 2

Load = 15 Kgf

Mechanical advantage = load/Effort

Effort = Load/Mechanical advantage

= 15/2 = 7.5 kgf

जेनर डायोड का व्यापक रूप से प्रयोग एक सर्किट में वोल्टेज को नियंत्रित रखने के लिए किया जाता है। रिवर्स-बायस्ड करने के लिए जब इसे एक परिवर्ती वोल्टेज के स्रोत के साथ समानांतर रूप से जोड़ा जाता है, तो वोल्टेज, डायोड के रिवर्स ब्रेकडाउन वोल्टेज तक पहुंचने से जेनर डायोड संचालित होता है। उस बिंदु से वह वोल्टेज को उसी मूल्य पर रखता है।

Let æ_i  represents initial angular speed and æ_f represents final angular speed. let % represent pie.

let f_i &f_f are frequencies initially and finally.

f_i = 2 rev/sec         f_f  = 8 rev/sec

æ_i = 2%f_i = 2% rad/sec        æ_f = 2%f_f = 16%  rad/sec          del æ=d æ =æ_f - æ_i = 14% rad/sec

let a_c is angular acceleration

a_c = d æ / dt = 14% /  20 = 0.7 rad/sec²

^Proof:   æ_f = æ_i + a_c * t = 2% + 0.7*20 = 16% rad/sec

Concave surface of the spoon forms the virtual image of the real image from by the spoon (which here acts as the concave mirror). Since spoon is a metallic surface, reflection forms the inverted image.

9b) Full-wave signal

The correct answer is: (a)  40.7 V

The correct answer is: (d) 35.4 V

(b) it has large cohesive energy

(a) frequency

Kinetic energy of emitted electron depends on the frequency of incident radiation.

(c) 180 degrees

 (c) 1:4:9

E_n = \(\frac{-13.6 \times z^2}{n^2}\)

n = 1; E =- 13.6 eV/ atom 

n = 2; E = – 3.4 eV/ atom 

n = 3; E = -1.51 eV/atom     

The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

(c) 0° – 180°

(d) the fine structure of even hydrogen spectrum

 Bohr theory could not explain the five structure of hydrogen spectrum.

The amount of heat(H)produced is given by the joule’s law of heating as, H = VIt, where 

Current, I = 5A 

Time, t = 30s 

∴ Voltage, V = I * R = 5 * 20 = 100 V 

∴ H = 100 * 5 * 30 = 1.5 * 104 J 

Therefore, the amount of heat developed in the electric iron is 1.5 * 104 J.

According to optical property, it is classified into two groups:

1. refracting telescope

2. reflecting telescope

\(2x - y = 5\)

⇒ \(4x - 2y = 10\)    ....(1)

\(4x + 2y = 10\)        .....(2)

By adding both equations, we get 

\(8x = 20\)

⇒ \(x = \frac{20}8 = \frac52\)

\(\therefore y = 2x - 5 = 5 - 5 = 0\)

Hence, x = \(\frac52\), y = 0 is solution of given system of equations.

\(f: A \to B\) is a surjection and defined by \(f(x) = cos x\)

Then \(f(A) = B\)

But \(A = \left\{0, \frac\pi6, \frac\pi4,\frac\pi3,\frac\pi2\right\}\)

\(\therefore f(A) = \left\{cos 0, cos\frac\pi6, cos\frac\pi4, cos\frac\pi3, cos\frac\pi2\right\}\)

\(= \left\{1, \frac{\sqrt3}{2}, \frac1{\sqrt2}, \frac12, 1\right\}\)

\(\therefore B = \left\{1,\frac{\sqrt3}2, \frac1{\sqrt2},\frac12, 1\right\}\)

 (2 + √x + 4/(√x - 2))^-40

 = \((\frac{x-4+4}{\sqrt x-2})^{-40}\)

 = x^-40(√x - 2)⁴⁰

 = x ^-40(\(\sum\limits_{r=0}^{40}\) ⁴⁰C_r x ^{\(\frac{40-r}2\)}(-2)^r), 0\(\leq\)r \(\leq\) 40

 = \(\sum\limits_{r=0}^{40}\) ⁴⁰C_r x^{\(-(\frac{40-r}2)\)}(-2)²

For independent term-(\(\frac{40+r}2=0\))

⇒ r = -40 which is not positive.

Hence, independent term in expansion of (2 + √x + \(\frac4{\sqrt4 - 2}\))^-40 is 0.

\(\left(2+\sqrt x - \frac 4 {\sqrt x -2}\right)^{-40} = \left(\frac{x- 4 - 4}{\sqrt x - 2}\right)^{-40}\)

\(= (x - 8) ^{-40} ( \sqrt x - 2)^{40}\)

\(= x^{-40} \left(1 - \frac8 x\right)^{-40} (\sqrt x - 2^{40})\)

\(= x^{-40} \left( \displaystyle\sum^{40}_{r=0} \,^{40}C_r \;x^{\frac{40 - r}2} \;(-2)^r\right) \left(1 - \frac8x\right)^{-40}\)

\(= \left(\displaystyle\sum^{40}_{r = 0} (-1)^r 2^r \,^{40}C_r\; x^{\left(\frac{40 + r}2\right)}\right)\left(1 - \frac8x\right)^{-40}\)

\(= \left(\displaystyle\sum^{40}_{r = 0} (-1)^r 2^r \,^{40}C_r\; x^{\left(\frac{40 + r}2\right)}\right)\left(\displaystyle\sum^\infty_{k = 0}\frac{n(n + 1)...(n + k - 1)}{k!} x^k\right), n = 40;k\in Z\)

\(= \displaystyle\sum^{40}_{r = 0} \displaystyle\sum^\infty_{k =0} ( -1)^r 2^r \,^{40}C_r \frac{n (n + 1)....(n + k - 1)}{k!} x^{-\frac{40+r}2+ k} , n = 40;k\in Z\)

For independent term 

\(-\frac{40 + r}{2} + k = 0\)

⇒ \(k = \frac{40 + r}{2}\)

and \(0 \le r\le 40\)

∴ Independent term = \(\displaystyle\sum^{40}_{r = 0\\r = even} 2^r \, ^{40} C_r \frac{40. 41.42. ......\left(\frac{r + 118}2\right)}{\left(\frac{40 + r}{2}\right)!}\)

\(\vec{B} - \vec{A} = 3\hat{i} - 3\hat{k}\) (direction)

= \(\vec{A}.\vec{B} = (2.5 + \sqrt{7}\sqrt{7}) = 17\) (magnitude)

= \(\frac{17}{\sqrt{2}} (\hat{i} - \hat{k})\)  Ans Option B

As the vector \(\vec A+\vec B \) and \(\vec A-\vec B\) are perpendicular to each other, therefore

\((\vec A+\vec B).(\vec A-\vec B)=0\)

\(\vec A.\vec A-\vec A.\vec B+\vec B.\vec A-\vec B.\vec B=0\)

or  A² - B² = 0

A = B  (∵ \(\vec A.\vec B=\vec B.\vec A\))

The principle of superposition states that, the total response (in the form of motion or force etc) will be a result of the algebric addition of all stimuli at a given point in space-time.

 A microscope has two converging lenses. This is because it is easier to get higher magnification with two lenses rather than just one.

Polarization, property of certain electromagnetic radiations in which the direction and magnitude of the vibrating electric field are related in a specified way. A polarizing filter has its molecules all aligned in the same direction.

Brewster's law, relationship for light waves stating that the maximum polarization (vibration in one plane only) of a ray of light may be achieved by letting the ray fall on a surface of a transparent medium in such a way that the refracted ray makes an angle of 90° with the reflected ray.

Polarization, property of certain electromagnetic radiations

Mixed anhydride of HClO₂​ and HClO₃​

ClO₂​ reacts with KOH forming KClO_3​ and KClO₂​

2KOH+2ClO_2​→KClO₂​+KClO₃​+H₂​O

This reaction indicates that ClO₂​ is a mixed anhydride of HClO₂​ and HClO₃​.

1) childish 

2) pronunication 

3) ill thinking 

4) successful 

5) crowdy

6) championship 

7) likelihood 

8) weakness 

9) stoppage 

10) combination

The self linking property of an atom known as Catenation.

Correct Answer is Lichchavis

Father of Indian Ophiology

1. RBS are produced in red bone marrow and destroyed in spleen 

2. Haemoglobin

Correct option: 1 Merkel cell

Correct option: 1. horny layer

The B atom in BCl₃ has only six electrons in the valence shell and hence is an electrondecient molecule. It easily accepts a pair of electrons donated by water and hence BCl₃ undergoes hydrolysis to form boric acid (H₃BO₃) and HCl.

BCl₃ + 3H₂O → H₃BO₃ + 3HCl

In contrast, C atom in CCl₄ has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule. As a result, it neither accepts nor donates a pair of electrons. In simple words, it does not accept a pair of electrons from H₂O molecule and hence CCl₄ does not undergo hydrolysis in water.

1. No. Generally, non metallic oxides are acids.

2. The binary compounds formed by the combination of oxygen with metals or nonmetals are called oxides. 

3. Acidic oxides, basic oxide, amphoteric oxide, neutral oxide. 

4. Acidic oxide → SO₂ , NO₂ 

Basic oxide → MgO 

Amphoteric oxide → Al₂O₃ , ZnO 

Neutral oxide → CO, N₂O

1. Graphite, diamond and fullerene. 

2. I agree with this statement. Because, CO has strong and reversible binding with haemoglobin resulting in the formation of carboxy haemoglobin. This reduces the amount of haemoglobin available in blood for oxygen transport. This causes laboured respiration, muscle weakness and even death.

(a) 5' – GAATTC – 3' 

3' – CTTAAG – 5' 

(b) EcoRI is the restriction endonuclease that recognizes this palindrome. 

(c) When the restriction enzyme cuts the DNA strands a little away from the center of the palindromic sequence, between the same two bases on both the strands, sticky ends are produced.

• The stickiness of the ends facilitates the action of the enzyme DNA-ligase.