Explore topic-wise InterviewSolutions in Current Affairs.

India is facing so many social problems like: 

1. Poverty, unemployment. 

2. Corruption.

Studying sociology will make a change in altitudes and it also gives solution to social problems.

During theological stage all human thoughts were guided by religious ideas and faith in the supernatural powers.

1. Social Statics 

2. Social Dynamics.

1. B. They had not tracked in mud.

2. C. Peggy

3. A. they were waiting for Wanda to come but she didn’t come.

4. B. Making fun of

1. A. Tolerant

2. D. She hitched her left shoulder every now and then.

3. D. Both A and B

4. B. she had an unusual name.

The physical components of a computer are called the hardware.

Viscosity of gases increases while viscosity of liquid decreases.

Velocity of efflux from a hole made at a depth h below the free surface of the liquid (of depth H) is given by v = √2gh

Which is same as the final speed of a free falling object from rest through a height h. This result is known as Torricelli’s theorem.

The given number is 2P. If 2P is divisible by 2, 3 then 2P should be a multiple of 6. 

[ ∵ L.C.M. of 2, 3 is 6]

∴ 2P = 24, 30 …………. 

24 → 2 + 4 → 6/3 (R = 0) 

∴ P = 4

To avoid dehydration of roots.

• Take thin sections of the roots of plantain. Put it in the water in a petridish.
• Transfer the sections to the water taken in a watch glass using a brush.
• Put the thinnest and complete sections in a watch glass containing stain.
• Add a few drops of glycerine on slide to avoid dehydration.
• Place the stained section on the slide after putting it in fresh water.
• Cover it using a cover glass to avoid air bubble.
• Wipe out the excess glycerine using a tissue paper.

A transparent surface is needed to fall sunlight reflected by the mirror on the material to be observed.

The correct answer is 2/3

To avoid drying up it is needed to add a few drops of glycerine on slide. Priya has done it.

17/9+41/9=58/9

17/2+3 1/2= 12

4.55 + 9.73 = 14.28

9 1/4 - 5/4 = 8

Let the edge of the bigger cube be a cm and the edge of the smaller cube be b cm. 

Then, 6a² = 384 and 6b² = 96 

⇒ a² = 64 and b² = 16 

⇒ a = 8 cm and b = 4 cm 

⇒ Volume of bigger cube = a³ = (8)³ cm³ = 512 cm³ and 

Volume of smaller cube = b³ = (4)³ cm³ = 64 cm³.

∴ Total number of smaller cubes = \(\frac{512}{64}\) = 8.

(b) S³ = 216 V²

Let the side of the cube be x units. Then, 

S = 6x² and V = x³ 

∴ S³ = 6³ . (x²)³ = 6³ . (x³)² = 216 V².

(a) 1 cm.

Let the thickness of the pipe be a cm. 

External Radius = 9 cm.

∴ Internal radius = (9 – a) cm. 

Given, Volume of metal = 748 cm³ 

⇒ (π × 92 × 14) – (π × (9 – a)² × 14) = 748

⇒ π × 14 × [81 – (81 + a² – 18a)] = 748 

⇒ –a² + 18a = \(\frac{748}{π\times14}=\frac{748\times7}{22\times14}=17\)

⇒ a² – 18a + 17 = 0 

⇒ (a – 17) (a – 1) = 0 

⇒ a = 1 or 17, but a = 17 is inadmissible.

(c) 20.8 cm

Let the edge of the single cube be ‘a’ cm. 

Then, total volume melted = Volume of cube formed 

⇒ (6)³ + (8)³ + (10)³ = a³ 

⇒ a³ = 216 + 512 + 1000 = 1728 ⇒ a = 12 cm. 

∴ Diagonal of the new cube = \(\sqrt3\,a=(\sqrt3\times12)\) cm = 20.8 cm (approx.)

(c) 450 cm³.

Required difference in capacities = \(\frac{22}{7}\) x (5)² x  21~ (10)² x 21

= (1650 ~ 2100) cm³ = 450 cm³.

(a) Greatest when x is increased

We can check by an example. Let x = 2, y = 3, z = 4. Then 

Volume of box = 2 × 3 × 4 = 24 cu. units. 

If we increase x by 1 unit, keeping the other same i.e, 

x = 3, then new volume = 3 × 3 × 4 = 36 cu. units. 

If we increase y by 1 unit keeping the others same, i.e., 

y = 4, then new volume = 2 × 4 × 4 = 32 cu. units 

If we increase z by 1 unit keeping the other same, i.e., 

z = 5, then new volume = 2 × 3 × 5 = 30 cu. units. 

Hence the increase in volume is greatest, when x increases. 

You can check with other values also.

The correct answer is 15 1/4

(a) 100 hours

Time required (in hours) = \(\frac{\text{Volume of water in the tank}}{\text{Volume of water flowingthrough the pipe per hour}}\)

= \(\frac{150\times100\times3}{0.2\times0.15\times15000}\) = 100 hours

The correct answer is 1/3

Rate of flow = 5 km/hr = 5000 m/hour. 

⇒ Length of cylinder for water flowing in one hour = 5000 m. 

Radius = 7 cm = \(\frac{7}{100}\) m.

∴ Volume of water flowing through the pipe per hour = πr²h 

\(=\frac{22}{7}\times\frac{7}{100}\times\frac{7}{100}\times5000\) m³ = 77 m³. 

Volume of water to be filled in the rectangular tank 

= (50 × 44 × 7/100) m³ = 154 m³ 

∴ Reqd. time = \(\frac{154}{77}\) = 2 hours.

Total surface area of the two halves = 3πr² + 3πr² = 6πr² sq units 

∴ Cost of painting the two halves = 6πr² × Rs 8 = Rs 48 πr².

The correct answer is 7 3/4 

(b) 87 π cm²

Radius and height of cone ABC are 6 cm and 2√7 cm respectively.

∴ Slant height of cone ABC = \(\sqrt{6^2+(2\sqrt7)^2}\)

= \(\sqrt{36+28}=\sqrt{64} = 8\) cm

∴ Curved surface area of cone ABC = π x 6 x 8 = 48π cm² 

Also radius and height of cone DEF are 3 cm and √7 respectively 

∴ Slant height of cone DEF = \(\sqrt{3^2+(\sqrt7)^2}\)

= \(\sqrt{9+7}=\sqrt{16} = 4\) cm.

∴ Curved surface area of cone DEF = π x 3 x 4 = 12π cm² 

∴ Whole surface area of remaining solid 

= Curved surface area of cone ABC + Area of base + Curved surface area of cone DEF 

= 48 π + π(6² – 3²) + 12π 

= 48π + π(36 – 9) + 12π = 87 π cm².

Volume of a right prism = Area of base × height. 

Since the base is an equilateral triangle of side 6 cm, 

Area of base = \(\frac{\sqrt3}{4}\) x (side)² = \(\bigg(\frac{\sqrt3}{4}\times6^2\bigg)\)cm² = \(\frac{\sqrt3}{4}\) x 36 cm² = \(9\sqrt3\) cm²

∴ Volume = (\(9\sqrt3\) x18) cm³ = 162√3 cm³ 

Lateral surface area = Perimeter of the base × Height 

= (6 + 6 + 6) cm × 18 cm = 18 cm × 18 cm = 324 cm² 

Total surface area = Lateral surface area + Area of ends (bases) 

= (324 + 2 × 9√3 ) cm² = (324 + 18√3 ) cm² = (324 + 31.176) cm² = 355.176 cm².

Radius (r) of the base of cone = (10.5/2) cm= 5.25 cm

Slant height (l) of cone = 10 cm

CSA of cone = πrl

= (22/7 x 5.25 x 10) cm²

= (22 x 0.75 x 10) cm² = 165 cm²

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

Diameter of the base of a cone, d = 10.5 cm. 

height h = 10 cm. 

Curved Surface Area, C.S.A. = ? 

d = 10.5 cm, = 10\(\frac{1}{2} = \frac{21}{2}\)cm. 

h = l = 10 cm. 

∴ Curved Surface Area of a cone = πrl

Volume of the prism = Area of the base × height 

⇒ 1056 = \(\frac12(8+14)\times8\times{h}\) ⇒ h = \(\frac{1056\times2}{22\times8}\) = 12 cm.

Regular Polygon: A regular polygon is an enclosed figure. In a regular polygon minimum sides are three.

(i) 3 sides

A regular polygon with 3 sides is known as Equilateral triangle.

(ii) 4 sides

A regular polygon with 4 sides is known as Rhombus.

(iii) 6 sides

A regular polygon with 6 sides is known as Regular hexagon.

(i) Open curve

(ii) Closed curve

(iii) Closed curve

(iv) Open curve

(v) Open curve

(vi) Closed curve

Correct option is (B) √3 cm²

Side of equilateral triangle is a = 2 cm

\(\therefore\) Area of equilateral triangle \(=\frac{\sqrt3}4a^2\)

\(=\frac{\sqrt3}4\times2^2\) = \(\sqrt{3}\) \(cm^2\)

Correct option is  B) √3 cm² 

Consider r and R as the radii of two spheres

We know that

4πr²/ 4πR² = ¼

So we get

(r/ R)² = (1/ 2)²

It can be written as

r/ R = ½

Consider V₁ and V₂ as the volumes of the spheres

So we get

V₁/ V₂ = (4/3 πr³)/ (4/3 πR³)

We can write it as

(r/ R)³ = (1/ 2)³ = 1/8

Therefore, the ratio of their volumes is 1: 8.

Consider x and 2x as the radius of two spheres and S₁ and S₂ as the surface areas

It can be written as

S₁/ S₂ = 4πx²/ 4π (2x)²

On further calculation

S₁/ S₂ = x²/ 4x²

So we get

S₁/ S₂ = ¼

Therefore, the ratio of their surface areas is 1:4.

(i) It is a Simple Closed curve and a concave polygon. This is a simple closed curve and as a concave polygon all the vertices are not pointing outwards.

(ii) It is a Simple closed curve and a convex polygon. This is a simple closed curve and as a convex polygon all the vertices are pointing outwards.

(iii) It is Not a curve and hence it is not a polygon.

(iv) It is Not a curve and hence it is not a polygon.

(v) It is a Simple closed curve but not a polygon.

(vi) It is a Simple closed curve but not a polygon.

(vii) It is a Simple closed curve but not a polygon.

(viii) It is a Simple closed curve but not a polygon.

Correct option is (C) 64 cm³

Let side length of cube is a cm.

\(\therefore\) \(6a^2=96\)    \((\because\) Total surface area of cube is \(6a^2)\)

\(\Rightarrow\) \(a^2=\frac{96}6=16=4^2\)

\(\Rightarrow\) a = 4

\(\therefore\) Volume of cube \(=a^3\)

\(=4^3\) \(=64\,cm^3\)

Correct option is  C) 64 cm³ 

(C) 64 cm³

Explanation:

The surface area of a cube is 96 cm²

Let the length of the cube is l cm.

Thus,

6 × l² = 96

According to formula

l² = 96/6

l² = 16

l = 4

Thus the length of the cube is 4 cm.

Volume of a cube  = l³

= 4³

= 64

Thus volume is 64 cm³ which is option c.

Hence, option C is the correct answer.

(d) None of these

Volume of bigger sphere = \(\frac43\) x π x (10)³ cm³ = \(\frac{4000}{3}\)π cm³

Volume of each smaller sphere = \(\frac{\frac{4000}{3}}{1000}\) cm³ = \(\frac43\)π cm³

∴ If r is the radius of each smaller sphere, then 

\(\frac43πr^3= \frac43π\) ⇒ r = 1 cm. 

Surface area of bigger sphere = 4 x π x (10)² cm² = 400 π cm² 

Surface area of 1000 smaller spheres = 1000 × (4 x π x 1)cm² = 4000 π cm² 

So, the total surface area increases by \(\frac{4000π -400π}{400π}\) times 

= 9 times.

(b) \(26\frac13\)

Given, r₁ + r₂ = 10              ...(i) 

and \(\frac43πr^3_1+\frac43πr^3_2= 880\)

⇒ \(\frac43π(r^3_1+r^3_2) = 880\) ⇒ \(r^3_1+r^3_2=\frac{880\times3\times7}{4\times22}=210\)          ...(ii) 

Taking the cube of both the sides of eqn. (i), we have 

(r₁ + r₂)³ = 1000 ⇒ \(r^3_1+r^3_2+\) 3r₁ r₂ (r₁ + r₂) = 1000 

⇒ 210 + 3r₁ r₂ (10) = 1000 

⇒ 30r₁ r₂ = 1000 – 210 = 790 ⇒ r₁ r₂ = \(\frac{790}{30}=26\frac13.\)

(d) \(\frac12\big(1-\frac1{r+1}\big)\)

Volume of cylinder = πr²h = πr²            (∵h = 1) 

Total surface area of cylinder = 2πr(r + h) = 2πr(r + 1)

∴ Required magnitude = \(\frac{πr^2}{2πr(r+1)}=\frac{r}{2(r+1)}\)

= \(\frac12\big(1-\frac1{r+1}\big)\).

Correct option is (A) 15 cm

\(h=\sqrt{l^2-r^2}=\sqrt{17^2-8^2}=\sqrt{289-64}\)

\(=\sqrt{225}\) = 15 cm

Correct option is A) 15 cm

Correct option is (B) 64 cm³

Let side of cube is a.

\(\therefore6a^2=96\)   \((\because\) Total surface area of cube \(=6a^2)\)

\(\Rightarrow a^2=\frac{96}6=16=4^2\)

\(\therefore\) a = 4 cm

Volume of cube \(=4^3=64\,cm^3\)

Correct option is  B) 64 cm³