Explore topic-wise InterviewSolutions in Current Affairs.

De Broglie hypothesis : Louis de Broglie (1892-1987), French physicist, proposed (in 1924) that the wave-particle duality may not be unique to light but a universal characteristic of nature, so that a particle of matter in motion also has a wave or periodicity associated with it which becomes evident when the magnitude of Planck’s constant h cannot be ignored.

De Broglie equation : A particle of mass m moving with a speed v should under suitable experimental conditions exhibit the characteristics of a wave of wavelength

λ = \(\cfrac h{mv}\) = \(\cfrac h{p}\)

where p = mv = momentum of the particle. The relation λ = h/p is called the de Broglie equation, and the wavelength λ associated with a particle momentum is called its de Broglie wavelength. The corresponding waves are termed as matter waves or de Broglie waves or Schrodinger waves.

[Note: This hypothesis was revolutionary at that time and accepted by others because Einstein supported it. Erwin Schrodinger (1887-1961), Austrian physicist, formulated a wave equation for matter waves.]

1. Mr. Thorat (Noun Phrase) + nodded (Verb Phrase). 

2. This (NP) +puzzled me (VP). 

3. He (NP) + was a South Indian (VP). 

4. Mr Thorat (NP) + reached the shooting location (VP).

You : Sir, where did you learn film technology? 

Ray : I learned it from Hollywood.

You : Who helped you there?

Ray : My friend Alfred Hitchcock helped me there. 

You : Which film’ attracted you most?

Ray : Adventures of Rin Tin Tin attracted me the most. 

You : What was your wish?

Ray : My wish was to make a film that beat Hollywood.

You : Why did you take such a risk with tigers?

Ray : I took such risk with tigers to make a wonderful film. 

You : How long did it take to shoot the scene? 

Ray : It took two days to shoot the scene.

You : How far away was Notun Gram from here?

Ray : Notun Gram was twenty kilometres away from here.

You : How often do you visit Hollywood?

Ray : I visit Hollywood once in three months.

You : What kind of animals are used in Hollywood films?

Ray : Well trained animals are used in Hollywood films.

You : Thank you very much, Sir! 

Ray : It’s okay!

Statements

• Their training had not gone very bad.
• The shots were too dark.
• Goopy is banished by the king.
• Bharat circus has two tigers with them.
• Mr Thorat was the ring master of the circus.
• Bagha has also been banished.

To take the boat away from the bank of a river, the boatman pushes the bank with an oar because that way he can exert equal and opposite force on the boat making the boat move forward and away from the bank.

1. They are planning something different. 

2. He was reading a book when I visited him.

3. MrThorat had (has) two tigers with him. 

4. A tiger will be impossible to handle. 

5. The shooting will need two more days’ time.

According to this law, the amount of heat radiations' emitted per unit time from a unit area of a body at absolute temperature T is directly proprotional to the fourth power of the temperature.

i.e., E ∝ T⁴

or    E = σT⁴

where σ is a constant of proprotionality and is called Stefan's constant. Its value is = 5.67 X 10^-8 J s^-1 m^-2 K^-4 or Wm^-2 K^-4 in S. I.

Here, Radius of Sun R = 7 × 10⁵ km

= 7 × 10⁸ m

∴ Surface area of sun = 4πR²

= 4π(7 × 10⁸)²

= 6.16 × 10¹⁸m²

σ = 5.67 × 10^-8 Wm^-2 K^-4

Using E = AσT⁴ ,

E = (6.16 × 10¹⁸)(5.67 × 10^-8 )T4

= 3.49 × 10¹¹ × T⁴W

E = 4πr^r × Solar radiant power on earth

Here, Radius of Earth = r = 1.5 × 10²⁵ km = 1.5 × 10⁸ m

E = 4π(1.5 × 10⁸)² × 1400

= 3.96 × 10²⁰ watt 

Then equating

3.49 × 10¹¹ × T⁴ = 3.96 × 10²⁰ 

T⁴ = \(\frac{3.96\times10^{20}}{3.49\times10^{11}}\)

T = 1.83 × 10² K.

(A) Savitri Bai Phule

Answer is (C) Tarabai

(B) Maharashtra

(B) Rama bai

C. the whole week they had given him food to eat.

There are two types of stresses : 

Normal and Shear or tangential stress

(1) Normal stress : Here the force is applied normal to the surface. 

It is again of two types : Longitudinal and Bulk or volume stress. 

(i) Longitudinal stress 

(a) Deforming force is applied parallel to the length and causes increase in length.

(b) Area taken for calculation of stress is area of cross section. 

(c) Longitudinal stress produced due to increase in length of a body under a deforming force is called tensile stress. 

(d) Longitudinal stress produced due to decrease in length of a body under a deforming force is called compressional stress. 

(ii) Bulk or Volume stress 

(a) It occurs in solids, liquids or gases. 

(b) Deforming force is applied normal to surface at all points. 

(c) It is equal to change in pressure because change in pressure is responsible for change in volume. 

(2) Shear or tangential stress : It comes in picture when successive layers of solid move on each other i.e., when there is a relative displacement between various layers of solid. 

(i) Here deforming force is applied tangential to one of the faces. 

(ii) Area for calculation is the area of the face on which force is applied. 

(iii) It produces change in shape, volume remaining the same. 

Golgi complex or Golgi apparatus or Golgi body act as a assembly, manufacturing cum packaging and transport unit of cell. 

1. Structure of Golgi complex: 

a. Golgi complex consists of stacks of membranous sacs called cistemae. 

b. Diameter of cistemae varies from 0.5 to 1pm. 

c. A Golgi complex may have few to several cistemae depending on its function. 

d. The thickness and molecular composition of membranes at one end of the stack of a Golgi sac differ from those at the other end. 

e. The Golgi sacs show specific orientation in the cell. 

f. Each cistema has a forming or ‘cis’ face (cis: on the same side) and maturing or ‘trans’ face (trAnswer:the opposite side). 

g. Transport vesicles that pinch off from transitional ER merge with cis face of Golgi cistema and add its contents into the lumen. 

2. Location of Golgi complex: 

Golgi bodies are usually located near endoplasmic reticulum. 

3. Functions of Golgi complex: 

a. Golgi body carries out two types of functions, modification of secretions of ER and production of its own secretions. 

b. Cistemae contain specific enzymes for specific functions. 

c. Refining (modification) of product takes place in a sequential manner. 

d. For example, certain sugar component is added or removed from glycolipids and glycoproteins that are brought from ER, thus forming a variety of products. 

e. Golgi bodies also manufacture their own products. Golgi bodies in many plant cells produce non-cellulose polysaccharides like pectin. 

f. Manufactured or modified, all products of Golgi complex leave cistemae from trans face as transport vesicles.

Cytoplasm in Eukaryotic cell:

1. The cell contains ground substance called cytoplasmic matrix or cytosol. 

2. This colloidal jelly like material shows streaming movements called cyclosis. 

3. The cytoplasm contains water as major component along with organic and inorganic molecules like sugars, amino acids, vitamins, enzymes, nucleotides, minerals and waste products. 

4. It also contains various membranebound cell organelles like endoplasmic reticulum, Golgi complex, mitochondria, plastids, nucleus, microbodies and cytoskeletal elements like microtubules. 

5. Cytoplasm acts as a source of raw materials as well as seat for various metabolic activities taking place in the cell. 

6. It helps in distribution and exchange of materials between various cell organelles.

While transport vesicles are leaving from the trans face of the Golgi, certain markers get impregnated on their membrane. These markers help them to identify their specific target cell or cell organelle.

Eukaryotic plasma membrane/ Cell membrane/ Biomembrane:

1. It is thin, quasi-fluid structure present both extracellularly and intracellularly. 

2. Extracellularly, it is present around protoplast and intracellularly, it is present around most of the cell organelles in eukaryotic cell. It separates cell organelles from cytosol. 

3. Thickness of bio-membrane is about 75A. 

4. Cell membrane appears trilaminar (made up of three layers) when observed under electron microscope. It shows presence of lipids (mostly phospholipids) arranged in bilayer. 

5. Lipids possess one hydrophilic polar head and two hydrophobic non-polar tails. Therefore, phospholipids are amphipathic. 

6. Lipid molecules are arranged in two layers (bilayer) in such a way that their tails are sandwiched in between heads. Due to this, tails never come in direct contact with aqueous surrounding.

7. Cell membrane also shows presence of proteins and carbohydrates. 

8. Ratio of proteins and lipids varies in different cells. For example, in human beings, RBCs show approximately 52% protein and 40% lipids.

1. The significant function of plasma membrane is transport of molecules across it. Plasma membrane is selectively permeable. 

2. Passive transport: 

a. Many molecules move across the membrane without spending energy. 

b. Some molecules move by simple diffusion along the concentration gradient i.e. from higher to lower concentration.

c. Neutral molecules may move across the membrane by the process of simple diffusion. 

d. Water may also move by osmosis.

3. Active transport: 

a. Few ions or molecules are transported against concentration gradient i.e. from lower to higher concentration. 

b. This requires energy, hence ATP is utilized. As such a transport is an energy dependent process in which ATP is utilized, it is called Active transport e.g. Na /K pump. 

c. Polar molecules cannot pass through nonpolar lipid bilayer. Therefore, they require carrier proteins to facilitate their transport across the membrane.

Nucleus is known as the master cell organelle as it regulates various metabolic activities through synthesis of various proteins and enzymes. 

The nucleus in eukaryotic cell is made up of nuclear envelope, nucleoplasm, nucleolus and chromatin network. 

1. Nuclear envelope: 

a. Nuclear envelope is a double layered delimiting membrane of nucleus. 

b. Two membranes are separated from each other by perinuclear space (10 to 50nm). 

c. Outer membrane is connected with endoplasmic reticulum at places and harbours ribosomes on it. 

d. The inner membrane is lined by nuclear lamina- a network of protein fibres that helps in maintaining shape of the nucleus. 

e. The two membranes along with perinuclear space help in separating nucleoplasm from cytoplasm. However, nuclear membrane is not continuous. 

f. There are small openings called nucleopores on the nuclear membrane. 

g. The nucleopores are guarded by pore complexes which regulate flow of substances from nucleus to cytoplasm and in reverse direction.

Fluid mosaic model: 

1. Fluid mosaic model was proposed by Singer and Nicholson (1972). 

2. This model states that plasma membrane is made up of phospholipid bilayer and proteins. 

3. Proteins are embedded in the lipid membrane like icebergs in the sea of lipids. 

4. Phospholipid bilayer is fluid in nature.

5. Quasi-fluid nature of lipid enables lateral movement of proteins. This ability to move within the membrane is measured as fluidity. 

6. Based on organization of membrane proteins they are of two types, as: 

a. The intrinsic proteins occur at different depths of bilayer i.e. they are tightly bound to the phospholipid bilayer and are embedded in it. They span the entire thickness of the membrane. Therefore, they are known as transmembrane proteins. They form channels for passage of water. 

b. The extrinsic or peripheral proteins are found on two surfaces of the membrane i.e. are loosely held to the phospholipid layer and can be easily removed.

Fluid mosaic model was proposed by Singer and Nicholson (1972).

(i) Given as cot^-1(-√3)

Let y = cot^-1(-√3)

– cot (π/6) = √3

= cot (π – π/6)

= cot (5π/6)

So, the range of principal value of cot^-1 is (0, π) and cot (5 π/6) = – √3

Hence, the principal value of cot^-1 (-√3) is 5π/6

(ii) Given as cot^-1(√3)

Let y = cot^-1(√3)

cot (π/6) = √3

So, the range of principal value of cot^-1 is (0, π) and

Hence, the principal value of cot^-1 (√3) is π/6

(iii) Given as cot^-1(-1/√3)

Let y = cot^-1(-1/√3)

cot y = (-1/√3)

– cot (π/3) = 1/√3

= cot (π – π/3)

= cot (2π/3)

So, the range of principal value of cot^-1(0, π) and cot (2π/3) = – 1/√3

So, the principal value of cot^-1(-1/√3) is 2π/3

(iv) Given as cot^-1(tan 3π/4)

As we know that tan 3π/4 = -1

Substitute these value in cot^-1(tan 3π/4) we get

cot^-1(-1)

Let y = cot^-1(-1)

cot y = (-1)

– cot (π/4) = 1

= cot (π – π/4)

= cot (3π/4)

So, the range of principal value of cot^-1(0, π) and cot (3π/4) = – 1

So, the principal value of cot^-1(tan 3π/4) is 3π/4

Molecular weight of sulphur = 8 × 32 = 256 g. 

256 g of sulphur contains 6.022 × 10²³ molecules. 

Number of molecules present in 16 g of sulphur = \(\frac{16}{256}\,\times\,6.022\,\times\,10^{23}\)

= 3.77 x 10²³ molecules

1. Where is our friend, Amar? 

2. What shall we do now? 

3. Who was there with you just now? 

4. When will you meet me again?

(c) C

√25 = 5 × 5 

= 5² 

= 25

(b) 12 units

We know that area of square = Side × Side

= 12 × 12

= 144 square units

True

We know that, the unit’s digit of the square of a number having digit at unit’s place as 4 or 6 is 6.

Answer is (d) -15

Answer is (B) 5/√6

Correct option is (C) 2

We can draw a line by taking exatly two points.

\(\therefore\) 2 points are required to describe a line.

Correct option is  C) 2

Correct option is (A) 1

A ray has one end point.

Correct option is  A) 1

Correct option is (D) concurrent

If more than two lines passing through a common point then these lines are called concurrent lines.

Correct option is  D) concurrent

From Figure 

(5x + 3)° + 97° = 180° 

{∵ Linear pair} 

5x + 100° = 180° 5x = 180° – 100° 

⇒ 5x = 80°

x = 80°/5 = 16°

∴ x = 16°.

∠AOC + ∠COB + ∠BOP = 270°

To find:

∠AOC, ∠COB, ∠BOD and ∠BOA

Here,

∠AOC + ∠COB + ∠BOD + ∠AOD = 360°(Complete angle)

270° + ∠AOD = 360°

∠AOD = 360° – 270°

= 90°

Now,

∠AOD + ∠BOD = 180°(Linear pair)

90° + ∠BOD = 180°

Therefore,

∠BOD = 90°

∠AOD = ∠BOC = 90°(Vertically opposite angle)

∠BOD = ∠AOC = 90°(Vertically opposite angle)

Correct option is C) infinite

Correct option is (C) intersecting lines

If two lines have a common point then they are called intersecting lines.

C) intersecting lines

∠1 & ∠2 are two interior angles that lie on same side of transversal ‘n’. 

As their sum is less than 180°, the two lines intersect at that side. 

∴ The two lines ‘l’ and ‘m’ are two intersecting lines at the side of the angles ∠1 & ∠2.

Let angles are 4x & 5x. 

∴ Angles are supplementary 

∴ 4x + 5x = 180⁰ 

⇒ 9x = 180⁰ 

⇒ x = 180^o/9 = 20^o 

∴ Angles are 4 × 20⁰ , 5 × 20⁰ ⇒ 80⁰ & 100⁰ 

We are given that AC ⊥ AB

⇒ ∠CAB = 90° (given)

⇒ ∠BAP + ∠PAQ + ∠CAQ = 90°

⇒ x + (x + 5°) + (2x + 5°) = 90°

⇒ 4x + 10° = 90°

⇒ 4x = 80°

⇒ x = 20°

⇒ ∠PAQ = (x + 5°) = 20° + 5° = 25°

Reflex ∠PAQ = 360° – 25° = 335°

Answer is (A) 30°

We are given that EF || CD

EF || CD

∠FEC + ∠ECD = 180° (Sum of the interior angles on the same side of a transversal is 180°)

⇒ 125° = ∠ECD = 180°

⇒ ∠ECD = 180° – 125° = 55°

⇒ ∠BCD = ∠BCE + ∠ECD 

= 25° + 25° = 80°

∠ABC = ∠BCD = 80°

These are alternate interior angles and they are equal.

Hence, AB || CD.

(i) Equal

(ii) Supplementary

(iii) Parallel

(iv) Parallel

(v) Parallel

(vi) Parallel

Given that, ∠LMD = 35°

From the figure we can write

∠LMD and ∠LMC is a linear pair

∠LMD + ∠LMC = 180° [sum of angles in linear pair = 180°]

On rearranging, we get

= ∠LMC = 180° – 35°

= 145°

So, ∠LMC = ∠PLA = 145°

And, ∠LMC = ∠MLB = 145°

∠MLB and ∠ALM is a linear pair

∠MLB + ∠ALM = 180° [sum of angles in linear pair = 180°]

= ∠ALM = 180° – 145°

= ∠ALM = 35°

Therefore, ∠ALM = 35°, ∠PLA = 145°.

∠x + 140° = 180° (linear pair)

⇒ x = 40°

Here it is given that AB || CD

∠x + 80° = ∠BCD (alt. angles)

⇒ ∠x + 80° = 116°

⇒ ∠x = 116° – 80°

⇒ ∠x = 36°

Now at point B

∠y + ∠x + 80° = 180° (straight angle)

⇒ ∠y = 180° – 116°

⇒ ∠y = 64°

Hence, ∠x = 36° and ∠y = 64°

Method I: In the given figure, let us assume that

AOB is a straight line

∴ ∠x + ∠y = 180°

(linear pair of angles) …(i)

and ∠p + ∠q = 180°

(linear pair of angles) …(ii)

From (i) and (ii), we get

∠x + ∠y = ∠p + ∠q

Hence proved

Method II: As we know that sum of the angles around a point is 360°.

∴ ∠x + ∠y + ∠p + ∠q = 360°

or ∠x + ∠y + ∠x + ∠y = 360°

(∵ ∠x + ∠y = ∠p + ∠q)

2 (∠x + ∠y) = 360°

⇒ ∠x + ∠y = 180°

⇒ AOB is a straight line.