Explore topic-wise InterviewSolutions in Current Affairs.

D) Both A and R are correct and R is correct reason for A

Correct option is A) basic oxide

B) Non-metals

B) Acidic oxide

Correct option is C) Bromine

Correct option is  A) Sonority

A) basic oxide

B) Malleability

B) Dropping the metal on a concrete floor

A) Litmus paper

Curd and other sour substances contains acids which can react with the metals of brass and copper vessels to form toxic (poisonous) metal compounds which can cause food poisoning and damage our health.

Curd and sour substances contain acids which react with copper vessels and form poisonous substances. So curd and sour substances should not be kept in copper vessels.

Curd and sour substances contain acids. Acids reacts with metals to give salt and hydrogen gas. So, if such substances are kept in copper container, the acid will react and the container will be corroded.

(d) Angular momentum

L = mvr = \(\frac{nh}{2π}\)

(b) voltage applied to the tube

λ_min = \(\frac{12375}{V (volt)}\)Å; λ_min ∝\(\frac{1}{V}.\)

Correct answer is (b) 10^-15 m

(c) λ ∝ \(\frac{1}{Z^2}\) 

According to moseley’s law, √V = a(Z – b) or V = \(\frac{c}{λ}\) = a² (Z – b)²

∴ (c) λ ∝ \(\frac{1}{Z^2}\) 

(b) Neon street sign

1. Drawbacks of Rutherford model: 

Rutherford atom model helps in the calculation of the diameter of the nucleus and also the

2. Size of the atom but has the following limitations:

(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom. According to classical electrodynamics, any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy.

Hence, it can no longer sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen. Hence, Rutherford model could not account for the stability of atoms.

(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.

Excitation potential is defined as excitation energy per unit charge.

The neutrino has the following properties: 

1. It has zero charge 

2. It has an antiparticle called anti-neutrino. 

3. Recent experiments showed that the neutrino has very tiny mass. 

4. It interacts very weakly with the matter. Therefore, it is very difficult to detect. In fact, in every second, trillions of neutrinos coming from the sun are passing through our body without any interaction.

The total number of neutrons and protons in the nucleus is called the mass number and it is denoted by A. Hence, A = Z + N.

The number of neutrons in the nucleus is called neutron number (N)

The number of protons in the nucleus is called the atomic number and it is denoted by Z.

Let the number of fissions per second be n. 

Then, Energy released per second = n x 200 MeV

= n x 200 x 1.6 x 10^-13 J

Energy required per second = Power x Time 

= 1kW x 1 s = 1000 J 

Energy released = Energy required n x 200 x 1.6 x 10^-13 = 1000

n = 3.125 x 10^-13

Electric field lines start from positive charge and terminate at negative charge. If there is a single positive charge, the field lines start from the charge and terminate at infinity. So, the electric field lines do not form closed loops.

(a) material of the wire

Explanation: 

The stress is force per unit area. The breaking stress of a wire is a property of its material.

The correct answer is (d) 2000 N.

Explanation: 

The wire breaks when the maximum permissible stress in the cross-section of the wire is attained. Here 500 N gives maximum permissible stress to a 1 cm diameter wire. Since the area of the cross-section of a 2 cm diameter wire will be four times that of the 1 cm diameter wire, its load carrying capacity will also be four times, i.e. 4*500 N =2000 N.  

(d) none of them.

EXPLANATION:  

Though the particle repeats its motion but not at a regular interval (Since the speed is continuously increasing), so (a) is not true.  (b) and (c) are not true because the motion is not a to and fro motion. Since none of the (a), (b) and (c) is true hence (d) is true. 

(a) periodic

EXPLANATION:  

Since a uniform circular motion repeats itself at a regular time interval.        

(b) not true because it is not a to and fro motion.            

(c) not true because SHM is a special type of oscillation.         

(d) not true because the particle is not oscillating about a fixed axis.

New charge on A is \(\cfrac{q_A}2\) and new charge on B is \(\cfrac{q_a+2q_B}4\).

Charge on each  = \(\cfrac{{q_A}-3{q_A}}2=-q_A\)

No. of electrons per second = 2 x10¹⁶ electrons / sec.
Charge passing per second = 2 x10¹⁶ x 1.6 x 10^–19 coulomb/sec
= 3.2x 10^–3 Coulomb/sec
Current = 3.2x10^–3 A.

Correct Answer is: (d) 10²⁸

Correct Answer is: (a) 1 cm/s

(c) the average velocity of a free electron over a large period of time is zero
(d) the average of the velocities of all the free electrons at an instant is zero.

Correct Answer is:(a, d)

The speed of the electrons increases as they are accelerated. The current I remains constant, as the number of electrons emitted per second by the gun is constant. Now, I = Avne. As I, A and e are constant, while v increases, n must decrease.

The correct answer is (a) zero

The correct answer is (c) W_A = W_B = W_C.  

(d) Stark effect

Splitting of spectral lines in magnetic field is called Zeeman effect and splitting of spectral lines in electric field, is called Stark effect.

(i) Two lines intersects each other

∴ ∠x = ∠55°, (Vertically opposite angles)

∴ x + z = 180° (Linear pair angle)

⇒ 55° + ∠z = 180°

⇒ ∠z = 180°- 55° = 125°

Clearly ∠z = ∠y, (Vertically opposite angles)

⇒ ∠y = 125°

∠x = 55°, ∠y = 125°

and ∠z = 125°.

(ii) 40° + ∠x + 25° = 180° (Straight angle)

⇒ 65 + ∠x = 180°

⇒ ∠x = 180° – 65° = 115°

and ∠y + 40° = 180°, (Linear pair angle)

⇒ ∠y= 180° – 40° = 140°

and ∠y + ∠z = = 180°, (Linear pair angle)

∠z = 180°- ∠y

= 180°- 140° = 40°

∴ ∠x = 115°, ∠y = 140° and ∠z = 40°

(i) In figure, alternate angles are equal

∴ l ॥ m

(ii) In figure, l ॥ m and t is a transversal

∴ x + 70° = 180°,

Sum of angles made by a transversal on its one side is 180°,

∴ x = 180°- 70° = 110°

(iii) In figure, x = 180° – 130° = 50°

∴ Corresponding angles are equal

∴ l ॥ m

(i) In figure l ॥ m and a is transversal line ∠x = 110°
(Corresponding angle)

(ii) In figure l ॥ m and a is transversal line

∴ ∠x = 100° (Corresponding angle)

(i) In figure, l ॥ m

∠x = 60° (Corresponding angles)

(ii) In figure, l ॥ m

∴ ∠x = 50° (Alternate angle)

(iii) In figure, l ॥ m (Sum of internal angles = 180°)

∴ x+ 110° = 180°

∴ x = 180°- 110° = 70°

(iv) In figure, l ॥ m ∠x = 115° (External alternate angle)

(i) Obtuse angle

(ii) Right angle

(iii) Acute angle

Let ABC is an equilateral triangle.

∵ All angles of equilateral triangle are equal.

∴ ∠A =∠B =∠C = x (Let)

⇒ x + x + x = 180°

⇒ 3x = 180°

⇒ x = 60°

∴ ∠A =∠B =∠C = 60°