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Tamil land consisted of five physiographic division viz., kurinji (hilly backwoods), palai (parched zones), mullai (pastoral tract), marutam (wet land), and neital (littoral land).

The 1st FYP stressed the agricultural sector by investing money in huge dams and large scale irrigation projects.

a. iii 

b. i 

c. ii 

d. iv

(a) It was a blueprint for India’s economic future.

Developmental ideas are always full of discussions, arguments and controversies. It is natural that the concept of development will be different to different people. The concept of an industrialist who wants to start a steel industry in a particular place and the concept of an Adivasi living in that area won’t be similar. Therefore, in the first twenty years of independence, a lot of discussions took place on the issue of development.

At this time both experts and ordinary people looked at the development of the Western countries as the model to emulate. They thought that India too should go forward as the Western countries on the path of development. They thought that it would then be necessary to bring about some changes in the social and cultural outlooks of the country. Capitalism and Liberalization would come. Growth, prosperity and technology were related to these things. Based on these criteria counties were categorized as developed, developing and undeveloped.

Our leaders thought that our developmental policies must be different from those of the colonial mindset. The priorities of the government were eradication of poverty and social and economic security. At the time of independence we had two models of development to choose from the capitalist model and the socialist model. In India, the Communist Party, Socialist Party and Nehru preferred the Russian model of development. But some wanted the American model. Therefore we chose a mixed model, amalgamating the two models.

Areas of agreement were: 

1. Development of India should mean both, economic growth and social and economic justice. 

2. The matter of development cannot be left to businessmen, industrialists and farmers only but the government should play a key role. 

3. The task of ‘poverty alleviation and social and economic redistribution’ was being seen as the primary responsibility of Government.

Areas of disagreement were:

Disagreement on the kind of role to be played by the government.

Disagreement over the importance attached to the needs of justice if it differed from the economic growth.

Disagreement on the issue of giving priority to industries vs Agriculture and private vs public sector.

Economic planning in India refers to do a systematic regulation of economic activities by the government to reduce the wastage of time and resources: 

1. Economic planning helps to achieve national goals in a continuous process of development.

 2. It is a rational process to associate with the future needs and goals to evaluate alternate proposals also.

Correct Answer is: (A) Self pollinated 

• Cleistogamous flowers are flowers which always remain closed. Pollination occurring in closed flowers is cleistogamy. 

• It is a condition in which flower does not open. It is an adaptation seen in plants to ensure self-pollination.

1. Some of the largest developmental projects in India’s history were undertaken during this period to include mega dams like Bhakra- Nangal, and Hirakud for irrigation and power generation. 

2. Heavy industries were started in the public sector like steel plants, oil refineries, manufacturing units and defense production etc. 

3. Infrastructure and communication were also improved.

Some of the largest development projects in India’s history were undertaken during this period to include mega dams like Bhakra Nangal and Hirakud for irrigation and power generation. Heavy industries were started in the public sector like oil refineries, steel plants, defence production and manufacturing units, etc. Infrastructure and communication were also improved.

Correct Answer is: (B) Small

• Anaemophilous or wind pollinated flowers, are usually small and inconspicuous, and do not possess a scent or produce nectar. 

• There are also examples of ambophilous (pollinated by two different classes of pollinators) flowers which are both wind and insect pollinated.

Correct Answer is: (C) Wuchereria

• Elephantiasis is also known as lymphatic filariasis. It's caused by parasitic worms, and can spread from person to person through mosquitoes. 

• Elephantiasis causes swelling of the scrotum, legs, or breasts.

The anti-slavery movement began in the northern states of the United States of America.

Correct Answer is: (D) Tigers 

• Kanha National Park is the host of the major animals (mammals) including the Barasingha or swamp deer ((Rucervus duvaucelii), Indian wild dog and the most famed the India Tiger.

• This park is mostly renowned as the “Tiger Reserve” due to the significant amount of population of tigers in the Kanha land.

No. A force can produce torque only along a direction normal to itself as t = r × f . So, when the door is in the xy-plane, the torque produced by gravity can only be along ±z direction, never about an axis passing through y direction.

No change in speed of system as no external force is working.

There is no external force acting on the child and the trolley system, hence the momentum of the entire system is conserved. Therefore the CM keeps moving with its initial speed of V in the same direction.

Velocity = 2.25 x 10⁸ m/s,

Wavelength = 500 x 10^-10m,

Frequency of light in water = 5 x 10¹⁴Hz

Given: µ = 1.5

To find: Brewster angle (i_p)

Formula: µ = tan i_p

Calculation: From formula,

ip = tan^-1(µ) = tan^-1(1.5)

\(\therefore\) i_p = 56.3° = 56°18'

Brewster angle for air to glass transition is 56.3° or 56°18'.

(a)  P^1-γT^γ= constant

x = 3, y = 2 is a solution of 3x - 2y = 5, because L.H.S. = 3x - 2y = 3 × 3 - 2 × 2 = 9 - 4 = 5 = R.H.S. 

 i.e. x = 3, y = 2 satisfied the equation 3x - 2y = 5. 

∴ it is solution of the given equation. 

F = i/B = 1.2 × 0.5 × 2 = 1.2 N

(a) torque is formed

1. \(\overrightarrow F=q(\overrightarrow V \times \overrightarrow B)\)

2. Electric field is due to a charge, either in motion or at rest. Magnetic field is due to the motion of charge. Direction of electric force is coli near to electric field. Direction of magnetic force is perpendicular to magnetic field.

3. If velocity (displacement) is perpendicular to Lorentz force the work done will be zero and hence there will no change in K.E.

1. F = B1l = 1.5 × 7 × \(\frac{20}{100}\)  or F = 2.1 N acting vertically downwards. 

2. Force will again be 2.1 N.

3. F = \(\frac{1.5\times7\times16}{100}\) = 1.68 N

The food we eat provides energy and organic substances for growth and the replacement of worn out and damaged tissues. It regulates and coordinates the various activities that take place in the body.

The storage of excess of body fat in adipose tissue is obesity.

Obesity is caused due to the storage of excess of body fat in adipose tissue.

(a) Emulsification

Hydropower plant

5 * 7 = LCM of 5 and 7 = 35

(L₁, L₁) ∈ R ⇒ L₁ || L₁ which is true

(L₁, L₁) ∈ R, hence reflexive

(L₁, L₂) ∈ R ⇒ L₁ ||L₂

⇒ L₂ ||L₁

⇒ (L₂, L₁) ∈ R, hence symmetric

L₁, L₂) ∈ R and (L₂, L₃) ∈ R

⇒ l₁ || L₂, L₂ ||L₃

⇒ L₁ || L₃ ⇒ (L₁, L₃) ∈ R

hence R is transitive hence it is equivalence relation 

Let L be the required line 

= {L : L is parallel to y = 2x + 4} 

= {L:L is a line whose relation is y : 2x + k where k is any real}

Correct option is (B) gamma rays

Work function is the minimum energy required for the electron in the highest level of the conduction band to get out of the metal. Not all electrons in the metal belong to this level. They occupy a continuous band of levels. Consequently, for the same incident radiation, electrons knocked off from different levels come out with different energies. 

The centripetal force to keep car in circular motion is provided by frictional force (inward to centre).

For DEF

\(m\frac{v^2}{R}\) = mgμ

v_max = \(\sqrt{gμR}\) = \(\sqrt{100}\) = 10 ms^-1

For ABC

\(\frac{v^2}{R}\) = gμ, v = \(\sqrt{200}\) = 14.14 ms^-1

Time for DEF = \(\frac{π}{2}\times \frac{100}{10}\) = 5 πs

Time for ABC = \(\frac{3π}{2} \frac{200}{14.14}\) = \(\frac{300π}{14.14}s\)

For FA and DC = 2 × \(\frac{100}{50}\) = 4s

Total time = 5π + \(\frac{300π}{14.14}\) + 4 = 86.3s

Correct option is (B) c/1.4
 

The maximum number of significant figures of the volumes that can be reported are 2.

The maximum number of significant figures of the volumes that can be reported are 2.

(i) (C) x² + 1

(ii) ƒ : N → N , given by ƒ(x) = x³

for x,y ∈ N ⇒ ƒ(x) = ƒ(y)

x³ = y³ ⇒ x = y

There fore f is injective.

Now 2 ∈ N, but there does not exists any element x in domain N such that ƒ(x) = x³ = 2 their fore f is not surjective.

(iii) (a) Yes

(b) No, because p*q = q; q*p = p

⇒ p*q ≠ q*p

1.
(d) f : R⁺ → R⁺

2.
(a) For x, y ∈ N,

f(x) = f(y) ⇒ x³ = y³ ⇒ x = y

Therefore, f is injective

For 2 ∈ N, there does not exist x in the domain N such that f(x) = x³ = 2.

∴ f is not surjective.

(b) f : R → R given by f(x) = [x]

It seen that f(1.1) = 1 and f(1.8) = 1;

But 1.1 ≠ 1.8;

∴ f is not injective

There does not exist any element x ∈ R

such that f(x) = 0.7

∴ f is not surjective.

Given; A = N × N and

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b)

= (a + c, b + d) = (a, b) * (c, d))

Hence commutative.

Now; (a, b) * [(c, d) * (e, f)] = (a, b) * [c + e, d+f]

= (a + c + e, b + d + f)

[(a, b) * (c, d)] * (e, f) = [a + c, b + d] * (e, f)

= (a + c + e, b + d + f)

Hence associative.

(a, b) * (e, e) = (a, b) ⇒ (a + e, b + e) = (a, b)

⇒ a + e = a, b + e = b ⇒ e=0, e = 0 ⇒ (0,0) ∉ A

So identity element does not exist.

Given that f(x) = (4x + 3)/(6x - 4), x ≠ 2/3

Then (fof)(x) = f(f(x)) = f((4x + 3)/(6x - 4))

= (4((4x + 3)/(6x - 4)) + 3)/(6((4x + 3)/6x - 4) - 4)

= (16x + 12 + 18x - 12)/(24x + 18 - 24x + 16) = 34x/34 = x

Therefore (fof)(x) = x, for all  x ≠ 2/3

Hence, the given function f is invertible and the inverse of f is itself.

Correct option (a) 2n/h

Explanation:

a₁ + a_2n =  a₂ + a_2n–1 =......................= a_n +a_n+1 = a+b and g₁g_2n 

= g₂.g_2n–1 =.....................= g_n .g_n+1 = ab

Also h = 2ab/ a + b

Given expression  = a + b/ab + a + b/ab + ........a+b/ab(n times)

= n a + b/ab = n.2/h = 2n/h

Length of diagonal of a square = √2a

= √2 × 4

= 4√2 m

Given: X and Y are the points on side LN of ∆LMN, such that LX = XY = YN and LM || XZ.

To prove: ar (∆LZY) = ar (quad. MZYX)

Proof: Since ∆XMZ and ∆ZLX are on the same base XZ and between same parallel LM and XZ.

Then, ar (∆XMZ) = ar (∆ZLX) …(i)

On adding ar (∆XYZ) on both sides of equation (i), we get

ar (∆XMZ) + ar (∆XYZ) = ar (∆ZLX) + ar (∆XYZ)

⇒ ar (quad. MZYX) = ar (∆LZY)

Hence proved.

Answer is (C) Perimeter of ABCD > Perimeter of ABEM

Answer is (C) 45.5 cm²

For  f(x) = g (x)

⇒ 2x² – 1 = 1 – 3x

⇒ 2x² + 3x – 2 = 0

⇒ 2x² + 4x – x – 2 = 0

⇒ 2x (x + 2) – 1 (x + 2) = 0

⇒ (2x – 1) (x + 2) = 0

Thus domain for which the function f(x) = g (x) is {1/2, -2}

Data: AP || BQ || CR. 

To Prove: ar.(∆AQC) = ar.(∆PBR) 

Proof: AP || BQ. CR is a line segment. 

∆ABQ and ∆PBQ are on base BQ and in between 

AP || BQ ar.(∆ABQ) = ar.(∆PBQ) ………….. (i) 

∆BQC and ∆BQR are on base BQ and in between BQ || CR. 

∴ ar.(∆BQC) = ar.(∆BQR) ………….. (ii) 

Adding (i) and (ii),

 ar.(∆ABQ) + ar.(∆BQC) = ar.(∆PBQ) + ar.(∆BQR) 

∴ ar.(∆AQC) = ar.(∆PBR).

Answer is (B) 1 : 6