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Correct option is D. 88 cm²

Given,

Ratio of length , width and height of rectangular solid = 3 : 2 : 1

Volume of box = 48 cm³

Let length of box = 3x cm

Breadth of box = 2x cm

Height of box = x cm

Volume of box = v = 3x \(\times\) 2x \(\times\) x = 48

= 6x³ = \(\cfrac{48}6\) = 8

= x = \(\sqrt[3]{8}\) = 2 cm

Hence . length = 3x = 3 ×2 = 6 cm

= breadth = 2x = 2 × 2 = 4 cm

= height = x = 2 cm

Total surface area of box = 2(lb + bh + hl) = 2(24 + 8 + 12)

= 88 cm²

Given details are,

Dimensions of tea packet = 10 cm × 6 cm × 4 cm

Dimension of cardboard box = 50 cm × 30 cm × 0.2 m

So,

Number of tea packets can be put in cardboard box =

Volume of cardboard box / volume of tea packet

= (50 × 30 × 20) / (10 × 6 × 4)

= 125 tea packets

Given,

Height of tank = 6 × radius of tank

Let radius of tank = r decimeter

So, height of tank = 6r dm

Total surface area of tank = \(\cfrac{198\times100}{50}\) dm²

= 2πr(h + r) = 198 × 2

= 2πr(7r) = 396

= r² = \(\cfrac{396\times7}{14\times22}\) = 9

r = 3 dm

Height = 6r = 6 × 3 = 18 dm

So, volume of tank = πr²h = \(\cfrac{22}7\times\)9\(\times\)18 = 509.14 dm³

Given details are,

Dimensions of cuboidal block of iron is = 50cm × 45cm × 34cm

Size of small cuboids cutting from it is = 5cm × 3cm × 2cm

So,

Number of small cuboids that can be cut =

Volume of large iron cuboid/ volume of small cuboid

= (50 × 45 × 34) / (5 × 3 × 2)

= 2550 cuboidal blocks

Given,

Ratio of dimensions of rectangular box = 2:3:4

Let dimensions are = 2x ,3x , 4x

So, total surface area of box = 2(6x²+12x²+8x²) = 52 x²

Let, c₁ = cost of covering at rate Rs. 8 per m²

C₁ = 8 × 52 x²

Let c₂ = cost of covering at rate Rs.9.50 per m²

C₂ = 9.50 × 52 x²

= C₂ – C₁ = 1248(given)

= 52x ² × 9.50 – 52x² × 8 = 1248

= 52x² =\(\cfrac{1248}{1.50}\)

= x² = \(\cfrac{1248}{1.50\times52}\)= 16

x = 4

Hence dimensions of box are = 2x = 2 × 4 = 8 cm

= 3x = 3 × 4 = 12 cm

= 4x = 4 × 4 = 16cm

Let ‘r’ be the radius of the tank. 

As per given statement: 

Height (h) = 6(Radius) 

= 6r dm 

Cost of painting for 50 paisa or Rs \(\frac{1}{2}\) per dm² = Rs 198

⇒ 2πr(r + h) x \(\frac{1}{2}\) = 198 

⇒ 2 x \(\frac{22}{7}\) x  r(r + 6r) × \(\frac{1}{2}\) = 198 

⇒ r = 3 dm 

And, h = (6 x 3) dm = 18 dm 

Now, 

Volume of the tank = πr²h 

= \(\frac{22}{7}\) x 9 x 18 

= 509.14 dm³

Given details are,

Dimensions of closed box = 25cm × 0.5m × 15cm 

= 25cm × 50cm × 15cm

We know that,

Area of cardboard required = 2 (lb + bh + hl) cm²

= 2 (25×50 + 50×15 + 15×25)

= 2 (1250 + 750 + 375)

= 2 (2375)

= 4750 cm²

Given,

Length of closed iron tank = 12m

Breadth = 9m

Height = 4m

Total surface area of tank = 2(12 × 9 + 9 × 4 + 4 × 12) = 2(192) = 384 m²

Width of iron sheet = 2m (given)

So, length of iron sheet =\(\cfrac{384}2\)= 192 m

Hence, cost of iron sheet at rate Rs.5 per meter = 5 × 192 = Rs.960

Given details are,

Length of a cuboid = 12 m

Breadth of a cuboid = 10 m

Height of a cuboid = 4.5 m

By using the formula

Volume of cuboid = l × b × h

= 12 × 10 × 4.5

= 540 m³

Let f(x) = x³ sin² x = x³ (sin x)² 

∴ f(- x) = (- x)³ (sin (- x))² = (- x)³ (- sin x)² 

= – x³ sin² x = -f(x) 

f(-x) = -f(x) 

∴ f(x) is an odd function.

∴ ∫x³ sin² x dx for x ∈ [-π/4,π/4] = 0 (by property)

Answer is (c) x cos x

Given that f is continuous and differentiable in the interval [– 5, 5].

It is also given that f’(x) doesn’t vanish anywhere.

According to Rolle’s theorem for a differentiable function on [a, b] will have atleast one c∈(a, b) such that f’(c) = 0, if the following condition had satisfied:

⇒ f(a) = f(b).

According to the problem it is given for any value of x, say r the values never equals to zero.

⇒ f’(r) ≠ 0

This is possible when Rolle’s theorem is not applicable.

Let us Recap the Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a, b].

b) The function ‘f’ needs differentiable on the open interval (a, b).

c) f(a) = f(b)

Then there exists at least one c in the open interval (a, b) such that f’(c) = 0.

First, two conditions are satisfied according to the problem, so the only condition that cannot be satisfied is (c).

So, we can clearly say that f(– 5) ≠ f(5).

Answer is: C. 1/x(log 10)

Given that y = log₁₀ x

Using the property that log_a b = log_eb/log_ea, we get

y = log_ex/log_e10

Differentiating with respect to x, we get

dy/dx = 1/xlog_e10

(b) 5 m

Area of rectangle CEFH + Area of trapezium FHAG 

= (4 × 6) m² + \(\bigg(\frac12\times(6+4)\times6\bigg)\)m² 

= 24 m² + 30 m² = 54 m² 

∴ Area of right angled ΔABC = 84 m² – 54 m² = 30 m² 

⇒ \(\frac12\)x 12 x BD = 30 ⇒ BD = 5 m.

In medieval India, there were 3 strata of people such as the upper class, the middle class, and the lower class. Majority of the people were farmers. When landlords lived luxuriously, the common mass had to struggle with a pathetic life. They were burdened by excessive tax. Majority of the farmers could not own land, oxen, and plough. According to the historical sources of the Mughal period, there were two types of peasants-Khud-Kashta and Pahi-Kashta.

The Zamindars were the dominant class in the agricultural sector during the Mughal period. They were the owners of vast agricultural fields. They did not cultivate directly. They used others to cultivate. They enjoyed high position in the society and collected tax from peasants on behalf of the rulers. They maintained their own castles and troop. Slavery system was also prevalent in medieval India. In short, it can be stated that the medieval period was plagued with inequalities

The condition of women in medieval India was very pathetic. The practice of Sati and child marriage were existed in India. There were restrictions for remarriage. Since girls were married at a very tender age, they did not get the opportunity for education. The role of women in agricultural and nonagricultural sectors was significant. They involved themselves in all the stages of farming from sowing to harvesting.

They were engaged in weaving, pottery, and embroidery too. Even then, there were several women who had adorned higher political and social positions. Noorjahan, the wife of the Mughal Emperor Jahangir and Sultana Rasiya were excellent administrators. Gulbadan Begum (sister of Humayun), Jahanara (daughter of Shah Jahan), and Jeeja Bai (the mother of Shivaji) were women who. held higher positions.

Correct option is (b) Security

Correct option is (d) Outsourcing

Correct option is (d) customization.

Correct option is (a) Line

Correct option is (d) Economic factors

He is in an orchard. There are trees around him. A bird is darting and there are bees.

The ray of the meteor touches the blossom and it stands still in mid-air.

Answer is The flowers

The poet justifies his claim by saying that the bees swarm dilating round the perfect trees.

Correct option is (a) minority influence.

Correct option is (b) to either enhance or impair performance.

Wordsworth has been called greatest poet of Nature. This does not "mean that his poetry is only about trees, forests, hills and valleys. This great romantic poet has a comprehensive philosophic vision.This vision includes the whole cosmos of which man and nature are but parts. Thus wordsworth is more a poet of man than of nature. In any case, he does not see man apart from Nature. Real knowledge and wisdom are born out of man's communion with nature. Thus wordsworth is always the hero of his poem. He explores meaning in man's life and finds it only in man's communion with nature, This poet even reaches god-head through nature.Nature is as much a scripture of God as is the Bible.

True, in the Daffodils the poet gives the picture of nature in a few of its aspects.At the centre are the golden daffodils dancing and tossing their heads along the edge of the Grasmere lake.There is the gentle wind that gives them motion to dance. The waves in the water of the take are also dancing. The there are the comparison of the unending line of daffodils with the unending line milkyway made up of twinkling stars, Then at the focal point is the poet himself with a deep concern:

I gazed and gazed but littler thought

What wealth the-show to me had brought

That is exploring a meaning in Nature. Wordsworth of course finds it.In the school of nature he learns geat truths and wisdom. He developes a philosophic mind that can see beyond the present and even beyond the terrestrial existence. Thus when he is lonely and depressed and the present is shorn of all meaning, the past comes to his rescue. The experience of the past come into the present and the poet is saved from depression and lineliness. The sights of the past as of the daffodils flash on his inner eyes.

And then my heart with pleasure fills.

And dances with the daffodils.

Correct option is (B) 1401

Suppose gof (x₁) = gof (x₂)

⇒ g(f (x₁)) = g(f(x₂)) 

⇒ f(x₁) = f (x₂), as g is one-one

⇒ x₁ = x₂, as f is one-one 

Hence, gof is one-one.

R is reflexive, as 2 divides (a – a) for all a ∈ Z. 

Further, if (a, b) ∈ R, then 2 divides a – b. 

Therefore, 2 divides b – a. 

Hence, (b, a) ∈R, which shows that R is symmetric. 

Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2.

Now, a – c = (a – b) + (b – c) is even.

So, (a – c) is divisible by 2. 

This shows that R is transitive. 

Thus, R is an equivalence relation in Z.

L.H.S. = (1 – sinθ) (1 + sinθ) (1 + tan²θ)

= (1 – sin²θ)sec²θ

= cos²θ · sec²θ

= cos²θ 1/cos²θ = 1 = R.H.S.

(sin⁴ θ – cos⁴ θ + 1)cosec²θ

= [(sin²θ)² – (cos² θ)² + 1]cosec²θ

= [(sin²θ + cos² θ) (sin² θ – cos²θ) + 1]·cosec²θ

= [1(sin² θ – cos²θ) + sin²θ + cos² θ]cosec²θ

= (sin² θ – cos²θ + sin² θ + cos²θ).cosec²θ

= 2 sin²θ x  1/sin²θ = 2

Let tens place digit be y and the units place be x. 

∴ Our number is (10y + x) 

Our given first condition is that sum of the digits of a two-digit number is 15. 

∴ By given condition, 

y + x = 15 …(1) 

Our given second condition is that the number obtained by interchanging its digits exceeds the given number by 9. 

∴ By given condition, 

10y + x + 9 = 10 x + y 

∴ 10y - y + x - 10x = -9 

9y - 9x = -9 

y - x = -1 …(2) 

Solving 1 and 2 simultaneously, we get, 

∴ y = 7 and x = 8 

∴ Our number = (10 × 7 + 8) = 78 

Hence, our number is 78. 

(Answer given is 48 but correct answer is 78)

Let tens place digit be y and the units place be x. 

∴ Our number is (10y + x) 

Our given first condition is that the difference between a 2-digit number and the number obtained by interchanging its digits is 63. 

∴ By given condition, 

(10y + x) - (10x + y) = 63 

∴ 10y - y + x - 10 x = 63 

9y – 9x = 63 

9(y - x) = 63 

y - x = 7 

Hence, the difference between the digits of the number is 7.

Vitamin A, B, C, D, E and K are the most important vitamins.

The different points related to mirrors are 

1) Vertex (P) : 

The point where the central axis touches the mirror is called vertex.

2) Focus or focal point (F): 

The light rays coming from distinct object appear to meet at point in case of concave mirror and tend to meet at point when drawn backward in case of convex mirror. That point is known as focus or focal point. 

3) Centre of curvature (C) : 

It is centre of the sphere to which the mirror belongs.

All kinds of fruits, vegetables, green leafy vegetables, sprouted, pulses, their skin and the bran of cereals and pulses are all sources of vitamins and minerals.

Minerals like iron, calcium, sodium, potassium are essential for the body.

We need very small quantities of vitamins in . our body as lack of or deficiency of any vitamin results in serious disorders.

The various distances related to mirrors are 

1) Focal length (f) : 

The distance between vertex and focus is called focal length.

2) Radius of curvature (R) : 

The distance between vertex and centre of curvature is called radius of curvature.

If there is a deficiency of iron in the blood, the body does not get enough supply of oxygen and one feels constantly weak and tired. This condition is called ‘anaemia’.

Deficiency of vitamin D results in weak and brittle2 bones.

Deficiency of vitamin A leads to night blindness.

We should take the food item and mash it and put it in a test tube and add some drops of iodine solution. If the colour changes to blackish blue, starch is present. If it remains pale yellow . starch is absent.

Correct option is (c) iodine

All the distances are measured from the pole of a spherical mirror.

The sugars obtained from digestion of starch bum slowly in our body and release energy which is necessary for doing various works.

Correct option is (c) iron