Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is  D) 2b – 5a

Correct option is  C) -x/4

Correct option is  A) 2y – 8

Correct option is  B) p – 3

Correct option is  B) 6

Correct option is  C) -2

Correct Answer is  :

Correct option is  C) 3y + 1

Correct option is  C) 5x

Correct option is  D) 8

Correct option is  A) 3x

Correct option is  A) Constant

Correct option is  B) Variable

Given that Radius of flower bed = 2 m

Area of flower bed = πr² = π(2)² = 4π

Radius of the quarter circular plot = 2 m

Area of the quarter circular plot = (πr²)/4

Area of 4 quarter circular plots = 4 × (πr²)/4

= πr² 

= π(2)² 

= 4π

We know that area of the rectangular region = Length x Breadth

Area of the rectangular region = 8 x 6 = 48 m²

Area of the remaining field = Area of the rectangular region – (Area of 4 quarter circular plots + Area of the flower bed)

Area of the remaining field = 48 – (4π + 4π)

= 48 – 25.14

= 22.86 m²

It is given that, diameter of the big circle = 18 cm

Radius of the big circle = 9 cm

Area of the big circle, A = πr² = π(8)² = 81π cm²

Let d₁ = (2/3) × 18 = 12 cm

r₁ = 6 cm

Area of the circle, A₁ = πr² = π(6)² = 36π cm²

d₂ = (1/3) × 18 = 6 cm

r₂ = 3 cm

Area of the circle, A₂ = πr² = π(3)² = 9π cm²

Area of the shaded portion = A – (A₁ + A₂)

Area of the shaded portion = 81π – (36π – 9π) = 36π cm²

Let A₁ = the area of the circular field whose radius is 5 m [given]

A₂ = the area of the circular field whose radius is 13 m [given]

Now we have to find the area of circular field such that area is the difference of the areas of first and second field

A₃ = A₂ – A₁

⇒ πr² = π(13)² – π(5)²

⇒ πr² = π(13)² – π(5)²

⇒ πr² = π[(13)² – (5)²]

⇒ r² = 169 – 25

⇒ r² = 144

⇒ r = 12 m

Hence, the radius of the circular field is 12 m.

Let A₁ = the area of the circle whose radius is 20 m [given]

A₂ = the area of the circle whose radius is 48 m [given]

Now we have to find the radius of third circle such that whose area is equal to the sum of areas of two fields.

Hence,

A₃ = A₁ + A₂

⇒ πr² = π(20)² + π(48)²

⇒ πr² = π[(20)² + (48)²]

⇒ r² = 400 + 2304

⇒ r = 52 m

Therefore radius = 52 m

Let the area of the circle whose radius is 14 cm be A₁.

We know that area of the circle = πr²

Therefore,

A₁ = π (14)²

Let A₂ and r₂ be the area and radius of the second circle respectively whose area is double the area of circle A₁.

A₂ = 2 A₁

⇒ π (r₂)² = 2 × π (14)²

⇒ (r₂)² = 2 × (14)²

⇒ r₂ = 14√2 cm

Hence the radius of the circle A₂ is 14√2 cm.

It is given that, Perimeter of the square = 5024 m

⇒ 4 × side = 5024

⇒ Side = 5024/4

⇒ Side = 1256 m

The same wire is converted into the form of a circle. Therefore,

Circumference of the circle = Perimeter of the square

⇒ 2πr = 5024

⇒ 2 × π × r = 5024

⇒ r = 2512/π

We know that area of the square: Area of the circle = (side)² : πr²

Area of square/ area of circle = (side)²/ πr²

Area of square/ area of circle = (1256 × 1256)/ [π × (2512/ π) × (2512/ π)]

= (1256 × 1256 × 22)/ (2512 × 2512 × 7)

= 11/14

Area of the square: Area of the circle = 11: 14

Diameter of park = 7 m

so, radius of park = r₁ = 3.5 m

Width of park = 0.7 m

Bigger radius of park = r₂ = 0.7 + 3.4 = 4.2 m

Now, Area of path = area of bigger circle -area of smaller circle

= Pi r₂² – Pi r₁²

= Pi(r₂² – r₁²)

= 22/7 (4.2²– 3.5 ²)

= 22/7(17.64 – 12.25)

=16.94 m²

Also,

Cost of expenditure = rate × area

= 110 × 16.94

= 1864.40

Cost of Expenditure of cementing the path is ₹ 1864.40.

Given perimeter of circle = 4πr cm

Which can be written as = 2π(2r)

We know that the perimeter of a circle = 2πr

Therefore we have radius = 2r

We also know that area of circle = πr²

= π(2r)²

= 4πr² cm²

Let radius of circular region = r

Radius of circular path of uniform width h surrounding the circular region of radius,

r = r + h

Therefore area of path = π(r + h)² – πr²

= πr² + πh² + 2πrh – πr²

= πh (2r + h)

Hence the proof.

Given circumference of park = 352 m

But we know that circumference of circle = 2πr = 352 m

2 × (22/7) × r = 352

r = (352 × 7)/ 44

r = 56 m

Radius of the path including the 7 m wide road = (r + 7) = 56 + 7 = 63 m

Therefore area of the road = π × (63)² – π × (56)²

A = 22/7 × 63 × 63 – (22/7) × 56 × 56

A = 22 (9 × 63 – 8 × 56)

A = 22 (567 – 448)

A = 2618 m²

(c) old profit sharing ratio of partners

Any partner take retirement under the following ways:

• On the basis a agreement between partners.
• With the connect of partners.

Any partner can retire from his firm due to old age, poor health, mutual dispute etc. When any partner voluntarily or due to any other reason separates him self from the firm, then this is called retirement of a partner.

According to Section 37 of Indian Partnership Act, 1932, if remaining partners continue to carry on the business without payment of final settlement amount to the retiring partner then there are two options available to the retiring partner, he is entitled

• get interest 6% per, annul on the outstanding amount till receipt of the final payment or,
• to get share in earned profit for that period in the ratio of capital, whichever is beneficial to him can be accepted by him.

Correct answer is (a) 2F

∆l ∝ F

Answer is Sublimation

(a) sublimation

Surface energy per unit area is equal to surface tension. 

E = increase in surface area × ST

= 4π (22 – 1²) × 2.5 

= 4π × 3 × 2.5 

= 1.02 × 10³ erg

The students may discuss with their teachers and find answer.

Correct option is B) a copper rod

1. Initial velocity, u > 0. Also, velocity is constant with time. Hence, acceleration is zero.

2. As finite initial velocity is increasing with time, acceleration, a > 0 and is constant.

3. Initial velocity, u = 0. Velocity is increasing with time so, acceleration a is positive. But it is decreasing in magnitude with time.

4. Initial velocity, u = 0. Velocity is linearly increasing with time. Hence, starting from rest acceleration is constant.

5. Initial velocity, u = 0. Acceleration and velocity is increasing with time.

6. Initial velocity u > 0. Velocity decreases and ultimately comes to rest. Hence, acceleration a < 0.

Correct option is B) plate

1. Consider two trains A and B moving on two parallel tracks in the same direction.

2. Case 1: Train B overtakes train A.

For a passenger in train A, train B appears to be moving slower than train A. This happens because the passenger in train A perceives the velocity of train B with respect to him/her i.e., the difference in the velocities of the two trains which is much smaller than the velocity of train A.

3. Case 2: Train A overtakes train B.

For a passenger in train A, train B appears to be moving faster than train A. This happens because the passenger in train A perceives the velocity of the train B w.r.t. to him/her i.e., the difference in the velocities of the two trains which is larger than the velocity of train A.

4. If \(\overset\rightarrow{v}_A\) and \(\overset\rightarrow{v}_B\) be the velocities of two bodies then relative velocity of A with respect to B is given by \(\overset\rightarrow{v}_{AB}=\overset\rightarrow{v}_A-\overset\rightarrow{v}_B.\)

5. Similarly the velocity of B with respect to A is given by, \(\overset\rightarrow{v}_{AB}=\overset\rightarrow{v}_B-\overset\rightarrow{v}_A.\)

Thus, relative velocity of an object w.r.t. another object is the difference in their velocities

6. If two objects start form the same point at t = 0, with different velocities, distance between them increases with time in direct proportion to the relative velocity between them.

Given: v_A = 10 m/s, v_B = 6 m/s,

To find: (i) Velocity of A relative to B (v_A – v_B)

(ii) Velocity of B relative to A (v_B – v_A)

Formulae: (i) v_AB = v_A – v_B

(ii) v_BA = v_B – v_A

Calculation: From formula (i),

v_AB = 10 – 6 = 4 m/s

From formula (ii),

v_BA = 6 – 10 = -4 m/s -ve sign indicates that driver of car A sees the car B lagging behind at the rate of 4 m/s.

∴ v_AB = 4 m/s, v_BA = -4 m/s

(i) Velocity of A relative to B is 4 m/s.

(ii) Velocity of B relative to A is -4 m/s.

The process of transfer of negative or positive charges is called electrical discharge. 

The electrical discharge can occur between two or more clouds, or between clouds and the earth.

Given that at first the object 20 cm from the mirror. So the image is also at a distance of 20 cm from mirror (Since object distance = image distance). Now the mirror is moved 15 cm away from the object. So the image also moves 15 cm away from actual image distance before. So now the image distance is 35 cm. Therefore the image is 70 cm from the object.

The valve present between the left atrium and the left ventricle is Bicuspid valve.

When the blood flows in the artery it exerts a pressure on the elastic wall of the artery which is called blood pressure. 

It is greater during systole (contraction) than during diastole (relaxation) of the heart. In a normal adult the blood pressure is 120/80 where systolic is 120 and diastolic pressure is 80 mm of Hg (mercury).

Protein, acetone, creatinine

Physical properties of urine:

(a) Colour: Straw-yellow colour (due to presence of urochrome).

(b) Volume: 1 -1- 5 litre per day but varies.

(c) pH: 5 to 8, i.e., it is slightly acidic (pH = 6).

(d) Odour: On standing, the smell of urine becomes strong, ammonia-like due to bacterial activity otherwise faint smell.

(e) Specific gravity: 1003 to 1035.

Constituents of Urine: The normal human urine consists of about 95% of water and 5% of solid wastes dissolved in it. The percentage of the solid wastes may slightly vary according to the food taken and according to the time after taking food but usually these are approximately as follows:

Besides the normal constituents the urine may pass out certain hormones and also certain medicines like the antibiotics and the excess vitamins.

It is the mechanism by which the osmotic pressure is maintained and regulated between the body cells and the intercellular fluid by taking in the fluid or by giving out excess of fluid in the cell.

The Bowman’s capsule is also called the Nephric capsule and represents the free end of the nephron. The Bowman’s capsule is a double-walled cup-like structure which lies in the cortex of the kidney. The glomerular filtrate which leaves the blood capillaries of the glomerulus during ultrafiltration enter the capsule and then passes to the first part of the nephric tubule.

The ureter carries urine from the kidneys to the urinary bladder.

The urethra carries urine from the urinary bladder to the outside of the body.

The efferent arteriole is narrower than the afferent arteriole. These are further divided into many narrow capillaries thereby increasing the pressure of the blood flowing in it which is also known as hydrostatic pressure.