Explore topic-wise InterviewSolutions in Current Affairs.

We know that, F = ma

⇒ α = \(\frac{F}{m}\)

= \(\frac{5N}{50kg}\)= 0.1 ms^-1.

Since there is no change in velocity the a = 0.

⇒ F = ma = (25 × 10^-3 ) × (0) = 0 N.

Given m = 3 kg

u = 2 m/s

v = 3.5 m/s

t = 25 s.

F = ma

but we know that a = \(\frac {v-u}{t}\)

∴ F = m \(\frac {(v-u)}{t}\) = 3 \((\frac {3.5-2}{25})\)

= 0.18 N

since direction acceleration ‘a’ is positive force is acting in the direction of motion.

Peter’s mother called him to go and give cakes to his friend who was blind.

The work assigned to Peter’s father was tending the sluice. Opening and closing the gates when ships entered the sea from Holland.

As the drop of rain is falling with constant speed, in accordance with first law of motion, the net force on the drop of rain is zero.

The little children of Holland are aware the dikes must be watched every moment and that even a small hole can be very dangerous. It would bring the sea into Holland

Radian is the S.l. unit of plane angle. It is defined as the angle made by an are at the centre of the circle whose length is equal to the radius of the circle.

Correct option is (d) stimuli for a receiver.

A) has positive charge

Correct option is (d) verbal and nonverbal listener responses.

Correct option is (c) distorts or interferes with a message.

Correct option is (b) the person who decodes a message.

एक घन के 6 फलक होंगे। इनमें से दो फलक y-z समतल के, दो z-x समतल के तथा दो x-y समतल के समान्तर होंगे। विद्युत-क्षेत्र E = 3 x 10³ i N/C x-अक्ष के अनुदिश है; अत: यह z-x तथा x-y समतलों के समान्तर फलकों के समान्तर होगा। इन चारों फलकों से गुजरने वाला फ्लक्स शून्य होगा। 

विद्युत-क्षेत्र एकसमान है; अतः y-z समतल के समान्तर फलकों में से जितना फ्लक्स एक फलक से अन्दर प्रविष्ट होगा उतनी ही फ्लक्स दूसरे फलक से बाहर आएगा। अतः घन से गुजरने वाला नेट फ्लक्स शून्य होगा।

Area = π × radius² 

Circumference = 2πr or πd 

(where radius = r and diameter = d) 

Area of ring = \(π(r^2_2 - r^2_1)\), where r₂ and r₁ are the outer and inner radius respectively.

(c) 16√3 cm²

Because,

Area of the equilateral triangle = ((√3)/4) × (sides)² sq. Units

= (√3 /4) × (8)²

= (√3/4) × 64)

= (√3 × 16)

= 16√3 cm²

(b) 48 m²

Because,

Given,

base = 12 m

height = 8 m

∴ Area of the triangle = (1/2) × base × height

= (1/2) × 12 × 8

= 1 × 6 × 8

= 48 m²

(b) 4 cm

Because,

From the question is given that,

Area of an equilateral triangle is (4√3) cm²

Area of the equilateral triangle = ((√3)/4) × (sides)² sq. Units

= (4√3) = ((√3)/4) × (sides)²

= (sides)² = (4√3)/ (4/(√3)

= (sides)² = (4√3 × 4)/ √3

= (sides)² = 4 × 4

= Sides = √16

= Sides = 4 cm

Hence, the length of each side of the triangle is 4 cm.

(a) 84 cm²

Because,

Let a, b, c be the sides of the triangle

Then, s = (1/2) (a + b + c)

= (1/2) × (13 + 14 + 15)

= (42/2)

= 21

Area of triangle = √(s(s-a) × (s-b) × (s-c))

= √(21 × (21 – 13) × (21 – 14) × (21 – 15))

= √(21 × (8) × (7) × (6))

= √(3 × 7 × 2 × 2 × 2 × 7 × 2 × 3)

= √(3² × 7² × 2² × 2²)

By cancelling square and square root we get,

= (3 × 7 × 2 × 2)

= 84 cm²

(i) 18 cm

Area of the equilateral triangle = ((√3)/4) × (sides)² sq. Units

= (1.73/4) × (18)² [because √3 = 1.73]

= (1.73 × 324)/4

= 140.13 cm²

(ii) 20 cm

Area of the equilateral triangle = ((√3)/4) × (sides)² sq. Units

= (1.73/4) × (20)² [because √3 = 1.73]

= (1.73 × 400)/4

= 173 cm²

Correct option is  C) 4 : 5

Correct option is  A) 49 cm

Correct option is  B) 9 cm

Correct option is  A) 64 sq.m

(a) 60 cm

Perimeter of the given figure = AB + BC + CD + DE + EF + FG + GH + IH + IJ + AJ

= (AJ + IH + GF + BC = DE) + 3 + 5 + 2 + 20 + 4 + 6

= 20 + 40

= 60 cm

Correct option is  D) l.b

Correct option is A) Area of outer square – Area of inner square

(a) 4 cm

From the question it is given that, the largest circle that can be cut out of the rectangle measuring 10 cm in length and 8 cm in breadth.

Diameter of circle = 8 cm

Then, radius = diameter/2

= 8/2

= 4 cm

Perimeter of a regular polygon = length of one side × (number of sides)

(b) 18πr

Let us assume R be the radius of circle

We know that, area of circle = πR²

81πr² = πR²

R = √81 r

R = 9r

Circumference = 2 πr

= 2 π (9r)

= 18πr

Given: Side of the square = 20 cm

Area of the shade region

= (Area of the square) – (Area of unshaded region)

= (Square ABCD) – (ΔEAF + ΔFBC + ΔEDC)

= side × side – (1/2 AF × AE + 1/2 FB × BC + 1/2 DE × DC)

= 20 × 20 – (1/2 × 10 × 1/2 + 1/2 × 10 × 20 + 1/2 × 8 × 20)

= 400 – [60 + 100 + 80] = 400 – 240 = 160cm²

C) square units

(b) 10 π

Let us assume the side of the square be x.

Form the question it is given that, area of a square is 100 cm²

We know that, area of square = 100 cm²

X² = 100 cm²

X = √100 cm²

X = 10 cm

So, the largest circle in the square, diameter of the circle must be equal to the side of the square.

Diameter = 10 cm = side of square

2r = 10 cm

r = 10/2

r = 5 cm

then, circumference of circle = 2 πr

= 2 × π × 5

= 10 π

(a) 8πr²

We know that, area of semicircle = πr²/2

Then, the area of a semicircle of radius 4r is = (π(4r)²)/2

= 16πr²/2

= 8πr²

Correct option is  B) Area

(a) 9 times

Area of circle = πr²

From the question, the radius of a circle is tripled = 3r

Then, area = π(3r)²

= 9πr²

Let the length of the square be a.

Perimeter of square = 4 x side 

= 4 x a = 4a 

Side of new square = 3 x a = 3a 

Perimeter of new square = 4 x side 

= 4 x 3a = 3 x 4a = 3x perimeter of original square. 

∴ The perimeter of new square will be three times the perimeter of original square.

Let two sides of an isosceles triangle are of same measure “a”.

Base = b = 80 cm

Area of triangle = 360 cm². (given)

We know, Area of isosceles triangle = 1/4 × b√(4a² – b²)

360 = 1/4 x 80 x √(4a² – (80)²)

360 = 20 x √(4a² – 6400)

360 = 20 x 2√(a² – 1600)

9 = √(a² – 1600)

Squaring both sides

81 = a² – 1600

a = 41

=> a = 41 cm

Perimeter of triangle = Sum of all the sides = 41 + 41 + 80 = 162

Perimeter of triangle is 162 cm.

Correct option is   D) 2(l + b)

Correct option is  D) 6 times

Correct option is  C) 5 cm

Correct option is  B) 15 cm

Let each side of an equilateral triangle be “a” cm

Area of an equilateral triangle = 36√3 cm² (Given)

Area of an equilateral triangle = √3/4 (side)²

36√3 = √3/4 x a²

a² = 144

or a = 12 cm

Perimeter = 3 (side) = 3 x 12 = 36 cm.

(i) Consider a as the side of equilateral triangle i.e. a = 8cm

Area of an equilateral triangle = (√3 × a²)/4 sq units

So we get

Area of the equilateral triangle = (√3 × 8²)/4

On further calculation

Area of the equilateral triangle = (√3 × 64)/4

So we get

Area of the equilateral triangle = √3 × 16

By substituting the value of √3

Area of the equilateral triangle = 1.732 × 16 = 27.712

By correcting to two places of decimal

Area of the equilateral triangle = 27.71 cm²

(ii) We know that

Height of an equilateral triangle = √3/2 a

By substituting the value

Height of an equilateral triangle = √3/2 × 8

So we get

Height of an equilateral triangle = √3 × 4

By substituting the value of √3

Height of an equilateral triangle = 1.732 × 4 = 6.928

By correcting to two places of decimal

Height of an equilateral triangle = 6.93 cm

Let h be the height, then

Each of the equal sides of an isosceles triangle = (h + 2) cm and

Base = 12 cm

Area of a triangle = 1/2 × Base × Height

= 1/2 x 12 x h ….(1)

Area of isosceles triangle = 1/4 x b x √(4a² – b²)

= 1/2 x 12 x √(4(h+2)² – 144) …(2)

From (1) and (2), we get

1/2 x 12 x h = 1/2 x 12 x √(4(h+2)² – 144)

6h = 3√(4h2 + 16h + 16-144)

2h = √(4h² + 16h-128)

4h² = 4h² + 16h – 128

16h = 128

or h = 8

Height is 8 cm.

Now,

Area of a triangle = 1/2 × Base × Height

= 1/2 × 12 cm × 8 cm

= 1/2 × 96 cm²

= 48 cm². Answer!!

Let the parallel sides be 5x and 3x respectively.

Then, \(\frac{1}{2}\times\) 24 x (5x + 3x) = 1440 ⇒ 8x = \(\frac{1440\times2}{24}\) ⇒ x = 15 m.

∴ The longer parallel side = 5 × 15 m = 75 m.

Height of an equilateral triangle = 6 cm

Let x be a side of triangle.

Height = √3/2 (side) = √3/2 x

6 = √3/2 x

x = 12/√3

x = 12√3/3 = 6.92 cm

Now,

Area of a triangle = √3/4 (side)² 

= √3/4 x 6.92 x 6.92 

= 20.76 cm²

From the question is given that,

Area of an equilateral triangle is (16 × √3) cm²

Area of the equilateral triangle = ((√3)/4) × (sides)² sq. Units

= (16 × √3) = ((√3)/4) × (sides)²

= (sides)² = (16 × √3)/ (4/(√3)

= (sides)² = 16 × 4

= (sides)² = 64

= Sides = √64

= Sides = 8 cm

Hence, the length of each side of the triangle is 8 cm.

Consider a as the side of an equilateral triangle

We know that

Height of an equilateral triangle = √3/2 a

It is given that height = 9cm

So we get

√3/2 a = 9

On further calculation

a = (9 × 2)/ √3

Multiplying both numerator and denominator by √3

We get

a = (9 × 2 × √3) / (√3 × √3)

So we get

Base = a = 6√3cm

We know that the area of an equilateral triangle = ½ × b × h

By substituting the values

Area of the equilateral triangle = ½ × 6√3 × 9

On further calculation

Area of the equilateral triangle = 27√3 cm²

By substituting the value of √3

Area of the equilateral triangle = 27 × 1.732 = 46.764

By correcting to 2 places of decimal

Area of the equilateral triangle = 46.76 cm²

Therefore, the area of an equilateral triangle = 46.76 cm².

Correct option is C) 3.5 cm

Let radius of the circle is r cm.

Circumference of circle is P = 22cm(given)

⇒ 2\(\pi r\) = 22

⇒ 2 x 22/7 x r = 22

⇒ r = 22 x 7/22 x 1/2 = 7/2 = 3.5 cm

Hence, radius of the circle is  r = 3.5 cm