Explore topic-wise InterviewSolutions in Current Affairs.

There are 3 surfaces in a cylinder. (top, bottom, side)

Option : (C)

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Volume of a cylinder = πr²h 

Given, 

sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder and the radius of the sphere is r 

Thus, 

height of the cylinder = diameter = 2r and base radius = r 

Volume of the cylinder = π × r² × 2r 

= 2πr³

Option : (A) 

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Volume of a solid cone = \(\frac{1}{3}\)πr²h 

Given, 

solid sphere of radius r is melted and cast into the shape of a solid cone of height r 

Let the base radius be A. 

⇒ \(\frac{4}{3}\)πr³ = \(\frac{1}{3}\)π × A² × r 

⇒ A = 2r

Option : (D) 

Volume of a cube = side³ 

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Given, 

sphere is inscribed in a cube 

Diameter of sphere = side of the cube 

Side of cube = 2r 

Ratio of the volume of the sphere to the volume of cube

= \(\frac{\frac{4}{3}\pi r^3}{(2r)^3}\) 

= π : 6

Given,

Curved surface area of a right circular cylinder = 4.4m²

Radius of base of cylinder = 0.7 m

=2πrh = 4.4

= h = \(\cfrac{4.4\times7}{2\times22\times0.7}\) = 1 m.

∴ Height of cylinder = 1m.

We have,

Total surface area of hollow cylinder = 4620 cm²

Area of base ring = 115.5 cm²

Height of cylinder = 7 cm

Let outer radius be ‘R’ cm , inner radius be ‘r’ cm

Area of hollow cylinder = 2π(R² – r²) + 2πRh + 2πrh

= 2π(R+r)(R-r) + 2πh(R+r)

= 2π(R+r) (h+R-r)

Area of base = πR² – πr²

= π (R² – r²)

= π (R+r) (R-r)

Surface area / area of base = 4620/115.5

{2π(R+r) (h+R-r)} / { π (R+r) (R-r)} = 4620/115.5

2(h+R-r) / (R-r) = 4620/115.5

Let us consider (R-r) = t

2(h+t)/t = 40

2h + 2t = 40t

2h = 38t

2(7) = 38t

14 = 38t

t = 14/38

= 7/19 cm

∴ Thickness of cylinder is 7/19 cm.

Given: 

Total surface area of hollow cylinder = 4620 cm² 

Height of cylinder (h) = 7 cm

Area of base ring = 115.5 cm² 

Thickness of the cylinder Let ‘r₁’ and ‘r₂’ are the inner and outer radii of the hollow cylinder respectively.

Then, πr₂² – πr₁² = 115.5 …….(1) 

And, 2πr₁h + 2πr₂h+ 2(πr₂² – πr₁²) = 4620 

Or 2πh (r₁ + r₂ ) + 2 x 115.5 = 4620 

(Using equation (1) and h = 7 cm) 

or 2π7 (r₁ + r₂ ) = 4389 

or π (r₁ + r₂ ) = 313.5 ….(2) 

Again, from equation (1), 

πr₂² – πr₁² = 115.5 

or π(r₂ + r₁) (r₂ – r₁) = 115.5 

[using identity: a² – b² = (a – b)(a + b)] 

Using result of equation (2), 

313.5 (r₂ – r₁) = 115.5 

or r₂ – r₁ = \(\frac{7}{19}\) = 0.3684 

Therefore, thickness of the cylinder is \(\frac{7}{19}\) cm or 0.3684 cm.

Correct option is D. 4

Number of surfaces in a hollow cylinder = 4

We have,

Inner radius of tap = 0.75 cm

Length of water flowing in 1s = 7m = 700 cm

Volume of water per second derived from tap = πr^‑2l

= 22/7 × 0.75 × 0.75 × 700

= 1237.5 cm³

∴ Volume of water derived in 1 hour (3600 sec) = (1237.5 × 3600)/1000

= 4455 litres

We have,

Diameter of cylinder = 7 cm

So, Radius of cylinder = 7/2 cm

Height of cylinder = 60 cm

By using the formula,

Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 7/2 × 60

= 1320 cm²

Total surface area of cylinder = 2πr (h+r)

= 2 × 22/7 × 7/2 (60 + 7/2)

= 22 (127/2)

= 1397 cm²

We have,

Curved surface area of cylinder = 264 m²

Volume = 924 m³

Volume / Curved surface area of cylinder

πr²h / 2πrh = 924 / 264

r / 2 = 924 / 264

r = 924×2 / 264

= 7m

Radius = 7 m

Diameter of cylinder = 2 × radius = 2×7 = 14m

Curved surface area = 264 m²

2πrh = 264

2 × 22/7 × 7 × h = 264

h = 264×7 / 2×22×7

= 6m

∴Height of cylinder is 6m

Diameter of cylinder is 14m

We have,

Length of solid cylinder = L

Diameter o cylinder = 2 cm

Radius of cylinder = d/2 = 2/2 = 1cm

Volume of cylinder = πr²L ………… (i)

Length of hollow cylinder = 16 cm

External diameter = 20 cm

External radius = 20/2 = 10cm

Thickness = 2.5 mm = 0.25 cm

Inner radius = 10 – 0.25 = 9.75 cm

Volume = π (R² – r²) l …….. (ii)

From (i) and (ii)

πr²L = π (R² – r²) l

π × 1 × 1 × L = π × (10² – 9.75²) × 16

L = 79cm

∴ The length of the solid cylinder should be 79cm.

We have,

Diameter of base = 7 cm

Radius of base = d/2 = 7/2 cm

Height of cylinder = 60 cm

Volume of cylinder = πr²h

= 22/7 × 7/2 × 7/2 × 60

= 2310 cm³

Capacity of cylinder in litres = 2310 / 1000 = 2.31 litres.

We have,

Diameter of base = 21 cm

Radius of base = d/2 = 21/2 cm

Curved surface area = 1320 cm²

2πrh = 1320

2 × 22/7 × 21/2 × h = 1320

h = 1320×7×2 / 2×22×21

= 18480/924

= 20cm

∴ Volume of cylinder = πr²h

= 22/7 × 21/2 × 21/2 × 20

= 6930 cm³

We have,

Diameter of metallic cylinder = 1 cm

Radius of metallic cylinder = d/2 = 1/2 = 0.5 cm

Length of cylinder = 5 cm

Diameter of wire drawn from it = 1 mm = 0.1 cm

Radius of wire = 0.5mm = 0.05cm

Let length of wire be ‘h’ cm

Length of wire drawn from metal = volume of metal/ volume of wire

= πr²h / πr²

= (½)² × 5 / (0.05)²

= (5/4) / 0.0025

= 1.25/0.0025

= 500 cm

= 5m

∴ Length of the wire is 5m.

It is given that

Diameter of the conical pit = 3.5m

Radius of the conical pit = 3.5/2 = 1.75m

Depth of the conical pit = 12m

We know that

Volume of the conical pit = 1/3 πr²h

By substituting the values

Volume of the conical pit = 1/3 × (22/7) × 1.75² × 12

On further calculation

Volume of the conical pit = 38.5 m³ = 38.5 kilolitres

Therefore, the capacity of the conical pit is 38.5 kilolitres.

Height (h) of conical tent = 8 m

Radius (r) of base of tent = 6 m

Slant height (l) of tent = √r² + h²

= (√6² + 8²) m = (√100) m = 10m

CSA of conical tent = πrl

= (3.14 × 6 × 10) m²

= 188.4 m²

Let the length of tarpaulin sheet required be l.

As 20 cm will be wasted, therefore, the effective length will be (l − 0.2 m).

Breadth of tarpaulin = 3 m

Area of sheet = CSA of tent

[(l − 0.2 m) × 3] m = 188.4 m²

l − 0.2 m = 62.8 m

l = 63 m

Therefore, the length of the required tarpaulin sheet will be 63 m.

We have,

Dimensions of roof = 18 m × 16.5 m

Diameter of cylindrical tank = 8 m

Radius of tank = d/2 = 8/2 = 4m

Given that, it rains 10 cm a day or 0.1m a day

Let the rise in level of tank be ‘h’

Volume of tank = volume of roof

πr²h = lbh

22/7 × 4 × 4 × h = 18 × 16.5 × 0.1

h = (18 × 16.5 × 0.1 × 7) / 22×4×4

= 207.9/352

= 0.5906m = 59.06cm

∴ Rise in water level is 59.06cm

We have, 

Radius of conical pit = \(\frac{3.5}{2}\) = \(\frac{7}{4}\)m

Depth of pit 'h' = 12 m 

Volume = \(\frac{1}{3}\)πr²h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 1.75 x 1.75 x 12

= 38.5 m³ 

Since, 

1m³ = 1 kiloliter 

Capacity of the pit = (38.5 × 1) 

= 38.5 kilolitres

ABC is a conical tent. 

Height, AO = h = 8 m. 

base radius, OC = r = 6 cm 

In ⊥∆AOC, 

AC² = AO² + OC² 

l² = (8)² +(6)² 

l² = 64 + 36 

l² = 100 

∴ l = √100

l = 10 m 

Curved surface Area of cone C.S.A. = πrl 

= 3.14 × 6 × 10 = 188.4 m² 

Let the length of tarpaulin required be 1 m. 

In that 20 cm. (0.2 m) is wastage means remaining tarpaulin is (1 – 0.2 m) 

breadth, b = 3m 

∴ Area of tarpaulin = curved surface area of tent. 

(l – 0.2 m) × 3 = 188.4 m² 

l - 0.2m = \(\frac{188.4}{3}\)

l – 0.2 = 62.8 

∴ l = 62.8 + 0.2 

∴ l = 63 m.

We have, 

Radius of right circular cone (r) = 6

Height of cone ( h) = 8 cm

∴ slant height (l) = \(\sqrt{r^2+h^2}\)

= l = \(\sqrt{36+64}\)

= 10 cm

Total surface area of cone = πr²+ πrl= πr(r+l)

= \(\frac{22}{7}\) × 6 × 16 cm² 

= 301.71 cm²

We have, 

Area of canvas = 551 m² 

Area of canvas lost in wastage = 1 m² 

∴ Area of canvas used in making tent 

= (551-1) m² = 550 m² 

= Surface area of the cone = 550 m² 

We have, 

r = radius of the base of the cone = 7 m 

∴ Surface area = 550 m² 

= πrl = 550

= \(\frac{22}{7}\)×7×l = 550 

= l = 25 m

Let h be the height of the cone. 

Then,

l² = r² + h² 

= h = \(\sqrt{l^2-r^2}\)

= \(\sqrt{25^2-7^2}\) 

= 24 m

∴ Volume of the cone = \(\frac{1}{3}\)πr²h

= \(\frac{1}{3}\) ×\(\frac{22}{7}\)×7×7×24m³

= 1232 m³

We have, 

Circumference of circular base of cone = 2πr = 44 m

 Height (h )= 10 m

= 2×\(\frac{22}{7}\) × r = 44 

= r = 7m

Slant height (l ) =√(10²+7²) = √149 = 12.20 m 

Curved surface area of tent = πrl = 22/7 ×7×12.20 

= 268.4 m² 

Hence, 

length of 2m wide canvas needed = 268.4/2 

= 134.2 m

We have, 

Height of tent = 8m 

Base radius = 6m 

Slant height (l) = \(\sqrt{r^2+h^2}\) 

=\(\sqrt{36+64}\) 

= 10 m

Curved surface area of the cone = πrl 

= 3.14 × 6 × 10m² 

∴Length of 3 m wide tarpaulin required

= \(\frac{3.14\times 6 \times 10}{3}\) 

= 62.8 m

Extra length required for stitching and cutting wastage = 0.2 m 

Total length required = 62.8 + 0.2 = 63m

Correct option is B. 80π

Given,

Diameter of cylindrical tunnel = 2 m

So radius = \(\cfrac22\) = 1 m

Length of tunnel = 40 m

Area of iron sheet required = 2πrh

= 2π × 1 × 40 = 80π m²

Radius of cone (r) = 5 cm 

Vertical height (h) = 12 cm 

Slant height of cone (l) = √(r²+h²) 

= √(5²+12²) 

= √25 + 144 

= √169 

= 13 cm 

Curved surface area of cone = πrl 

= 22/7×5×13 

= 204.28 cm²

We have, 

Curved surface area of cone = 60π cm² 

Slant height l = 8 cm 

So, 

πrl = 60π 

rl = 60 

= r = \(\frac{60}{8}\) 

= 7.5 cm

Base radius of the cone(r) = 5.25 cm

Slant height of the cone(l) = 10 cm 

Curved surface area (C.S.A) = πrl 

= 22/7 x 5.25 x 10

= 165 

Therefore, curved surface area of the cone is 165 cm².

For cylindrical part, we have 

Diameter = 24 m 

So, 

radius r = 24/2 = 12m, 

Height = h = 11m 

For conical part, we have 

Height of the cone = (16-11)m = 5m

∴ Slant height = \(\sqrt{12^2+5^2}\) m

= 13 m

Hence, 

Area of the canvas required 

= Curved surface area of cone + Curved surface area of cylinder

= πrl + 2πrh = πr(l+2h) 

= \(\frac{22}{7}\) ×12 × (13 + 22) m² 

= 1320 m²

We have,

Volume of cone = 48π cm³ 

Height of cone h = 9 cm 

= 1/3πr²h 

= 48π 

= r² = 16r 

= 4 cm 

Hence diameter of its base = 2 × radius 

= 2×4 

= 8 cm

Slant height of cone (l) = 60 cm

Radius of the base of the cone (r) = 21 cm 

Now, Curved surface area of the right circular cone = πrl 

= 22/7 x 21 x 60 

= 3960 cm² 

Therefore the curved surface area of the right circular cone is 3960 cm²

We have, 

Base radius of cone = 7 cm

Height of cone = 24 cm

Slant height of cone l = \(\sqrt{(r^2+h^2) }\)

= \(\sqrt{(24^2+7^2)} \)

= \(\sqrt{625}\) 

= 25 cm 

Hence area of metal sheet required = total surface area of cone

= πr(l+r)

= 22/7×7(25 + 7)

= 704 cm²

Option : (D)

We have, 

Height of cone = 12cm 

Area of circular base = 64π cm² 

Height from the vertex of small cone = 9 cm 

πr² = 64π 

r²= 64 

Radius of base = 8cm 

From similarity triangle = \(\frac{12}{8}\) = \(\frac{9}{R}\)

= R = \(\frac{9\times8}{12}\) 

= 6 cm

R = radius of small cone 

Area of small cone = πr² 

= π(6)² 

= 36π cm²

Option : (C)

Let base radius of cone = x 

Let height of cone = y

Volume of cone = \(\frac{1}{3}\)πx²y

New radius of cone = x + x\(\frac{20}{100}\) = \(\frac{6x}{5}\)

New height of cone = y + y\(\frac{20}{100}\) = \(\frac{6y}{5}\)

New volume = \(\frac{1}{3}\)π\(\frac{36x^2}{25}\) x \(\frac{6y}{5}\)

= \(\frac{1}{3}\)π\(\frac{216x^2y}{125}\)

Increase in volume = \(\frac{216x^2y}{125}\) - x²y

= \(\frac{91x^2y}{125}\)

Percentage increase in volume = \(\frac{\frac{91x^2y}{125}}{x^2y}\) x 100

= 72.8 %

Approx 73%

Answer : (d) = \(-\frac{1}{x}\) 

Given f(x) = \(\frac{1+x}{1-x}\) 

f (f (x) = f \(\big(\)\(\frac{1+x}{1-x}\) \(\big)\)  = \(\frac{1+ \big( \frac{1+x}{1-x}\big)}{1- \big( \frac{1+x}{1-x}\big)} = \frac{\frac{1-x+1+x}{1-x}}{\frac{1-x-1-x}{1-x}}\)

= \(\frac{2}{-2x}\) 

= \(-\frac{1}{x}\)

f (f (x)) = \(f\big(\frac{\alpha x}{x+1}\big) = \frac{\alpha . \big( \frac{\alpha x}{x+1}\big)}{\big(\frac{\alpha x}{x+1}\big) +1}\)  

= \(\frac{\frac{\alpha^2 x}{x+1}}{\frac{\alpha x+ x +1}{x+1}} = \frac{\alpha^2 x}{\alpha x +x+1}\) 

Given, f (f (x)) = x  ⇒ \(\frac{\alpha^2 x}{\alpha x+x+1}\) = x 

⇒ α²x =αx² +x² +x 

⇒ α² - 1 = (α+1) x  

⇒ (α – 1) (α + 1) – (α + 1)x = 0 

⇒ (α + 1) (α – 1 – x) = 0 

⇒ α + 1 = 0 

⇒ α = – 1 [∵  α – 1 – x ≠ 0]

g{f (x)} = g (x – 3) = (x – 3)² + 1 

= x² – 6x + 9 + 1 

= x² – 6x + 10. 

Given, g{f (x)} = 10 

⇒ x² – 6x + 10 = 10 

⇒ x² – 6x = 0 

⇒ x (x – 6) = 0 

⇒ x = 0, 6.

Answer : (d) 2x + 1

f (g(x) = f (2x) = 2x + 1 (∵ f (x) = x + 1)

Answer : (c) 3

f(x) = \(\frac{x+1}{x-1}\) 

f f f (2) = f[f{f(2)}] = \(f\big[f\big(\frac{2+1}{2-1}\big)\big]\) = f [ f (3)]

= f \(\big[\frac{3+1}{3-1}\big]\) 

= f(2) = \(\frac{2+1}{2-1}\)

= 3

Answer: (c) 2 

fog (x) = f (g (x) = f (x² + x) 

= sin (x² + x) + cos (x² + x) + 1 

 ∴ fog (0) = sin 0 + cos 0 + 1 

= 0 + 1 + 1 

= 2.

Answer :  (b) f {g (x)} = g{f (x)}

f (x) = e^x and g(x) = log_e x 

∴ f {g (x)} = f (log_e x) = e^{log_e x} = x 

= g{f (x)} = g (e^x ) = log_e^ex = x 

∴ f {g(x)} = g{f (x)}.

Answer : (d) sin x² 

fog (x) = f (g(x)) = f (x²) = sin x².

(fof ) (1 - \(\sqrt{3}\) ) = f { f (1 - \(\sqrt{3}\) )} = f(0) (∵  1 - \(\sqrt{3}\)  is an irrational number)

  = 1 (∵ 0 is rational)

General formula: C_n H_{2n + 1} Cl (where n = 1, 2, 3, …)

i. The groups responsible for inductive effect in these molecules are -Cl, -Br and -I, respectively. All these are halogen atoms which are more electronegative than carbon. Therefore, all of them exert -I effect, that is, electron withdrawing inductive effect. 

ii. The -I effect of halogens is due to their electronegativity. A decreasing order of electronegativity in these halogens follows Cl > Br > I. Therefore, the strongest -I effect is expected in CH₃ – CH₂ – CH₂ – Cl, while the weakest -I effect is expected for CH₃ – CH₂ – CH₂ – I.

A. Metamerism 

B. Functional group isomerism 

C. Tautomerism 

D. Tautomerism

Carbohydrates (sugars), proteins (pulses), fats (edible plant and animal oil) and vitamins are the major classes of organic compounds often used in our daily diet.

The polar organic reactions are brought about by two types of reagents. 

Depending upon the ability to accept or donate electrons from or to the substrate, reagents are classified as

1. Electrophiles (E⁺)

2. Nucleophiles (Nu:)

i. Molecular formula of both (a) and (b) are same i.e., C₃H₆O.

ii. Compound (a), has a ketone (-CO-) functional group (i.e., acetone) and compound.

(b) has an aldehyde (-CHO) functional group (i.e., propionaldehyde). Both the compounds have different functional groups.

iii. Compound (a) and (b) are isomers of each other.

[Note : Aldehydes and ketones are the functional group isomers of each other.]

In above reaction,

The -OH group of the substrate molecule has undergone a change. 

The H-atom of hydroxyl group (-OH) is replaced by sodium forming the product.