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Mining causes pollution because if produces large amount of slag which is discarded for every tonne of metal that is extracted.

False.

Let us assume number of boys be 5y and the number of girls be 4y.

From the question,

5x – 4x = 9

x = 9

∴ Number of boys = 5x = 5 × 9 = 45 boys

Number of girls = 4x = 4 × 9 = 36 girls.

Let the common ratio between the number of boys and numbers of girls be x.

Number of boys = 7x

Number of girls = 5x

According to the given question,

Number of boys = Number of girls + 8

∴ 7x = 5x + 8

On transposing 5x to L.H.S, we obtain

7x − 5x = 8

2x = 8

On dividing both sides by 2, we obtain

2x/2x = 8/2

x = 4

Number of boys = 7x = 7 × 4 = 28

Number of girls = 5x = 5 × 4 = 20

Hence, total class strength = 28 + 20 = 48 students

Let the no. of girls be 5x
No. of boys=7x
According to the question
5x+8=7x
7x-5x=8
2x=8
x=8/2
x=4
No. of girls=5*4=20
no. of boys=7*4 or 5*4+8=28
Hence the total strength of the class is 20+28=48

The measure of biodiversity of an area is the number of species found there. Since, in a forest we can find a range of different life,forms of plants and animals the forests are the biodiversity hot spots.

Biodiversity hot-spot means a place where large number of species are found. 

The range of different life forms i.e., bacteria, fungi, ferns, flowering plants, variety of animals likes, insects, birds, reptiles, aves, mammals etc are all found in the forest.

Just behind the cornea is a dark coloured muscular diaphragm called iris which has a small circular opening in the middle called pupil. The pupil appears black because no light is reflected from it.

The iris regulates the amount of light entering the eye. It regulates the light by adjusting the size of the pupil.

Light enters the eye through cornea. The space behind the cornea is filled with a liquid called aqueous humour.

1. Atrophy means degeneration. Atrophy of thyroid gland will result in deficient secretion of thyroid hormones leading to hypothyroidism. Deficiency of thyroid hormones [T₃ and T₄ ] and thyrocalcitonin will cause following effects on the body.

2. Decrease in BMR i.e. basal metabolic rate, decrease in the blood pressure, heart beat, body temperature, etc. 

3. Occurrence of myxoedema in which there is abnormal deposition of fats under the skin giving puffy appearance in adults. 

4. Irregularities in menstrual cycle in case of female patients. 

5. Hair become brittle and fall. 

6. Calcium metabolism also disturbs due to lack of thyrocalcitonin.

1. The symptoms observed in Krishna were due to sympathetic nervous system. Emergency conditions trigger sympathetic nervous system to stimulate adrenal medulla.

2. The cells of adrenal medulla secrete catecholamines like adrenaline and nor¬adrenaline. 

3. These hormones have direct effect on the pacemaker of the heart which causes increase in the heart rate and other associated symptoms. 

4. This is a typical fright reaction caused by intervention of sympathetic nervous system

The output produced here is unidirectional.

The slip rings are replaced with split rings to achieve this.

To get a direct current (DC) generator a split-ring type commutator must be used. In this arrangement, one brush is at all times in contact with the arm moving up in the field while the other is in contact with the arm moving down. Thus a unidirectional current is produced in such a generator. The output produced here is unidirectional.

The slip rings are replaced with split rings to achieve this.

To get a direct current (DC) generator a split-ring type commutator must be used. In this arrangement, one brush is at all times in contact with the arm moving up in the field while the other is in contact with the arm moving down. Thus a unidirectional current is produced in such a generator.

No. of H₂ Molecules in 22.4 liters or 22400 cm³ at NTP = 6.02 x 10²³.

∴ No. of H₂ Molecules in 1 cm³ at NTP = (6.02 x 10²³)/22400 = 2.6875 x10¹⁹.

No. of degrees of freedom associated with each H₂ (a diatomic) molecule = 5

∴ Total no. of degree of freedom associated with 1cm³ gas

= 2.6875 x 10¹⁹ x 5 = 1.3475 x 10²⁰.

Correct Answer is: (c) Modem

Every object will remains at rest or in a state of uniform motion, unless compelled to change its state by the action of a net force.

(i) P=1/f, f=100/5=20 cm

An object at 18 cm, 22 cm, and 30 cm, the image can be magnified.

(ii) At 22 cm and 30 cm, the image can be obtained on a screen.

(b) Na < Al < Mg < Si < P

The energy required to remove the most loosely held electron from an isolated gaseous atom is called ioniiation energy.

M_(g) + energy M⁺_(g) + electron 

The unit of ionization energy is KJ mole^-1 .

1. N (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ . It has exactly half filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high. O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ . It has incomplete electronic configuration and it requires less ionization energy. 

I.E₁ N > I.E₁ O

2. C (Z = 6) 1s² 2s² 2p_x¹ 2p_y¹ . The electron removal from p orbital is very difficult. So carbon has highest first ionization potential. B (Z = 5) 1s² 2s² 2p¹ . In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential. 

I.E₁ C > I.E₁ B

But it is reverse in the case of second ionization energy. Because in case of B⁺ the electronic configuration is 1s² 2s² , which is completely filled and it has high ionization energy. But in C⁺ the electronic configuration is 1s² 2s² 2p¹ , one electron removal is easy so it has low ionization energy. 

I.E₂ B > I.E₂ C

3. Be (Z = 4) 1s² 2s² 

Mg (Z = 12) 1s² 2s² 2p⁶ 3s² 

Noble gases has the electronic configuration of ns² np⁶ . All these are completely filled and are more stable. For all these elements Be, Mg and noble gases, addition of electron is unfavorable and so they have zero electron affinity.

Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ . It has half filled electronic configuration. So addition of electron is unfavorable and it has very low electron affinity value of 0.02 eV. Phosphorus (Z = 15) 1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹ . It also has half filled electronic configuration. Due to the symmetry and more stability, it has very low electron affinity value of 0.80 eV.

4. F_(g) + e^- → F_(g)^- exothermic 

F (Z = 9) 1s² 2s² 2p⁵ . It is ready to gain one electron to attain the nearest inert gas configuration. By gaining one electron, energy is released, so it is an exothermic reaction.

O_(g) + 2e^- → O^2-_(g) endothermic 

O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ . It is the small atom with high electron density. The first electron affinity is negative because energy is released in the process of adding one electron to the neutral oxygen atom. Second electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative, therefore it must overcome the repulsion.

Mg (Z = 12) 1s² 2s² 2p⁶ 3s² . 

Al (Z = 13) 1s² 2s² 2p⁶ 3s² 3p¹ . 

Although the nuclear charge of aluminium is greater than that of magnesium, I.E of Mg is greater than that of Al. It is because Mg atom has more stable configuration than Al atom. IE₁ of Mg > IE₁ of Al.

1. The electronic configuration of Be and B are 1s²2s² and 1s² 2p² 2p¹. 

2. The element Boron has less ionization energy due to less penetration power of 2p compared to 2s.

Be (Z= 4) 1s² 2s² . it has completely filled valence electrons, which requires high IE₁ . 

B(Z =5) 1s² 2s² 2p¹ . It has incompletely filled valence electrons, which requires comparatively less IE₁. 

Hence I.E₁ Be > I.E₁ B.

Functions of Thyroxine hormone: 

• It increases oxidative metabolism so the rate of energy production increases our speed of life increases. 
• It increases protein synthesis, gluconeogenesis, the temperature of body and functions of nerves.
• It increases the glucose absorption, by intestine oxygen consumption and basic metabolic rate (BIR). 
• It increases the heartbeat. 
• It plays a significant role in the metamorphosis of tadpole larva into a frog.
• It plays a role in osmotic regulation in cold-blooded vertebrates. 
• It also increases the activity of neurosecretion adrenalin and non-adrenalin.

• An endocrine gland secretes its products, for example hormones, directly into the blood. 
• An exocrine gland secretes its products for example enzymes, into ducts that lead to the target tissue. 
• For example the salivary gland secretes saliva into the collecting duct which leads to the mouth.

(a) Yellow.

(b) Blue.

(c) Green.

(a) Backups, access authorised are the techniques to secure data from damage and getting lost.

Hacking refers to attempts to gain information from otherwise undisclosed areas.

(b) Integrity means information is accurate and complete.

There are three major issues as follows: 

1.  Confidentiality 

2.  Integrity 

3.  Availability

Security and Integrity of information ensures that information and other resources are modified only by those users, who have the authority of making changes in the data.

An electronic page in a presentation is called slide

PowerPoint is a presentation graphics software.

From Layout page, it is possible for you to choose slide designs.

Option : (b) 2g H₂

According to the atomic model of Thomson, electrons are embedded in a gel of positive charge.

Option : (a) 4.5

All atoms, except normal hydrogen contain neutrons in their nuclei.

Option : (a) 6.022 × 10²⁰

\(^{12}_{6}C\) and \(^{14}_{6}C\) are the isotopes of carbon.

Except hydrogen, the nuclei of all atoms contain neutrons.

(A) \(6.022 × 10^{20}\)

As we know,

\(1\,cm^3=1\,ml\)

\(\therefore 22.4\,cm^3=22.4\,ml\)

The volume of 1 mole of a gas at STP = 22400 ml/mol

\(\therefore\) Number of moles of \(O_3\) gas = \(\frac{22.4\,ml}{22400\,ml/mol}\)

= \(10^{-3}\) moles.

\(\because\) 1 mole of \(O_3\) molecule contain = \(6.022\times10^{23}\) \(O_3\) molecules.

\(\therefore 10^{-3}\) moles of \(O_3\) molecule contain = \(6.022\times10^{23}\times10^{-3}\) \(O_3\) molecules.

= \(6.022\times10^{20}\) \(O_3\) molecules.

Option : (A) 6.022 × 10²⁰

\(^{12}_{6}C\) , \(^{13}_{6}C\) and \(^{14}_{6}C\) are isotopes of carbon.

(a). 5 mg carbon = 5 × 10 g carbon 

Atomic mass of carbon = 12 u 

∴ Molar mass of carbon 12 g mol^-1

Number of moles = \(\frac{Mass\,of\,a\,subtance}{Molar\,mass\,of\,a\,subtance}\)

= \(\frac{5\times 10^{-3}g}{12\,g\,mol^{-1}}\)

= 4.167 x 10^-4 mol

(b).12 mg carbon = 12 × 10^-3 g carbon

Number of moles = \(\frac{Mass\,of\,a\,subtance}{Molar\,mass\,of\,a\,subtance}\)

= \(\frac{12\times 10^{-3}g}{12\,g\,mol^{-1}}\)

= 1 x 10^-3 mol

Number of atoms = Number of moles × Avogadro’s constant 

Number of atoms of carbon = 1 × 10^-3 mol × 6.022 × 10²³ atoms/mol 

= 6.022 × 10²⁰ atoms

∴ Number of moles of carbon in his homework writing = 4.167 × 10^-4 mol

Number of atoms of carbon in 12 mg homework writing = 6.022 × 10²⁰ atoms

(D) \(1.505 × 10^{23}\)

Molar mass of oxygen gas = 32g/mol.

\(\therefore \) Number of moles oxygen molecule in \(8g=\frac{8g}{32g/mol}=0.25\) mole

\(\therefore \) Number of moles oxygen molecules in 0.25 mole = \(0.25\times6.022\times10^{23}\)

= \(1.505\times10^{23}\) oxygen molecule.

Option : (D) 1.505 × 10²³

(B) 0.36 g

As we know,

Molar mass of \(C_6H_{12}O_{6}=180\,g\)

\(\therefore \) weight of 0.002 mole of \(C_6H_{12}O_{6}=0.002\times180\)

= 0.36 g

Option : (B) 0.36 g

An atom has 11 protons and 12 neutrons and hence its atomic mass number is 23.

Given : 

Mass of urea = 250 g, 

Cost for 250 g glucose = Rs 40, 

Molecular formula of glucose = C₆H₁₂O₆

To find : Cost per mole of glucose.

Calculation : Molecular formula of glucose is (C₆H₁₂O₆). 

Molecular mass of glucose 

= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O) 

= (6 × 12 u) + (12 × 1 u) + (6 × 16 u) 

= 180 u 

∴ Molar mass of glucose = 180 g mol^-1

 Number of moles = \(\frac{Mass\,of\,a\,subtance}{Molar\,mass\,of\,a\,subtance}\)

 = \(\frac{250\,g}{180\,g\,mol^{-1}}\)

= \(\frac{250}{180}\) mol

Now,

\(\frac{250}{180}\) mol of glucose cost = Rs 40

1 mol glucose cost = x

∴ x = \(\frac{40\times 180}{250}\)

= Rs 28.8/mol of glucose

∴ The cost of glucose per mole is Rs 28.8.

[Calculation using log table :

\(\frac{40\times 180}{250}\)

= Antilog₁₀[log₁₀(40) + log₁₀(180) + log₁₀(250)]

= Antilog₁₀[1.6021 + 2.2553 – 2.3979] 

= Antilog₁₀[1.4595] 

= 28.80]

1. There are 12 neutrons in the sodium (\(^{23}_{11}Na)\).

(N = A – Z) 23 – 11 = 12

2. Atomic mass number of \(^{14}_{6}C\) is 14.

3. There are 17 protons in chlorine \((^{17}_{37}Cl)\)

According to the question,

f and g functions defined by f (x) = 3x² – 1 and g (x) = 3 + x

For what real numbers x, f (x) = g (x)

To satisfy the condition f(x) = g(x),

Should also satisfy,

3x² – 1 = 3 + x

⇒ 3x² – x – 3 – 1 = 0

⇒ 3x² – x – 4 = 0

Splitting the middle term,

We get,

⇒ 3x² + 3x – 4x–4 = 0

⇒ 3x(x + 1) – 4(x + 1) = 0

⇒ (3x – 4)(x + 1) = 0

⇒ 3x – 4 = 0 or x + 1 = 0

⇒ 3x = 4 or x = –1

⇒ x = 4/3, –1

Hence, for x = 4/3, –1, f (x) = g (x),

i.e., given functions are equal.

(D) \(3.6 × 10^{24}\)

Number of moles of \(C_6H_{12}O_6=\frac{180}{180}=1\,\text{mole}\)

\(\therefore\) one molecule of \(C_6H_{12}O_6\) contain = 6 carbon atoms.

\(\therefore\) 1 mole of \(C_6H_{12}O_6\) molecule contain = \(6\times6.022\times10^{23}\) carbon atom

= \(3.6\times10^{24}\) carbon atom.

Option : (D) 3.6 × 10²⁴