Explore topic-wise InterviewSolutions in Current Affairs.

The correct answer is A) Vitamin – C.

1. A 

2. B 

3. C 

4. scurvy 

5. D 

6. rickets 

7. nerves and blood cells 

8. Fertility disorders 

9. K 

10. C

The correct answer is C) Vitamin – C.

Oil present in the bajji or pakodi, poori, etc. make the paper transparent.

Iodine paper gets discoloured with vitamin C. The substances which made the iodine paper more discoloured contains more vitamin C.

Fats turn the white paper translucent (partially transparent).

1. No, proteins are not present in all food items. 

2. The presence of proteins in food items can be confirm by doing a test with copper sulphate and sodium hydroxide solutions.

1. c) fats

2. b) Iron

3. a) Vitamin A

Answer is (c) y = kx

The correct answer is C) Both A&B.

I support this statement because.

1. Bottled water is nothing but filtered water using reverse osmosis. This not only removes microbes but also all the minerals required for our body from water. If the tap water is boiled and filtered normally harmful microbes will be killed and removed but required minerals will not be removed. So it is better to drink boiled water than bottled water. 

2. Further the empty water bottle and packets also causes pollution to our environment.

The correct answer is C) Potato.

1. y + 3

2. x.x (or) x² 

3. 20.n

After returning from Bangalore after completing his studies, Ranga had deterred from the traditional concepts of arranged marriage that was prevalent among his fellow villagers. He wanted to marry a mature girl who was mature enough to comprehend his words and not misunderstand them. He even gave example of Kalidasa’s play where Dushyantha fell in love with Shakuntala who was very mature, thus, making him convinced that he could only marry a girl he could admire.

Those are dietary fibres also known as roughage. They help in free bowel movement in the digestive tract and prevents constipation.

(i) 8.65

= \(\sqrt[3]{8.65}\) = \(\sqrt[3]{\frac}{865}{100}\) = \(\frac{\sqrt[3]{865}}{\sqrt[3]{100}}\)

We know that value of \(\sqrt[3]{860} = 9.510\) and \(\sqrt[3]{870} = 9.546\)

So by unitary method, 

∵ For difference (870 – 860 = 10 ) difference in cube root values = 9.546 – 9.510 = 0.036 

∴ For difference (865 – 860 = 5) difference in cube root values 

= \(\frac{0.036}{10}\times5 = 0.018\)

= \(\sqrt[3]{8.65}\) = 9.510 + 0.018 = 9.528

We also have,\(\sqrt[3]{100} = 4.642\) (from table)

∴ \(\sqrt[3]{8.65}\) = \(\frac{\sqrt[3]{865}}{\sqrt[3]{100}}\) = \(\frac{9.528}{4.642} = 2.053.\)

Thus the required cube root is = 2.053.

(ii) 7532

= \(\sqrt[3]{7532}\)

We know that value of \(\sqrt[3]{7532}\) will lie between \(\sqrt[3]{7500}\) and \(\sqrt[3]{7600.}\)

From cube root table we get,

= \(\sqrt[3]{7500}\) = 19.57 and \(\sqrt[3]{7600.}\) = 19.66

So by unitary method,

∵ For difference (7600 – 7500 = 100 ) difference in cube root values = 19.66 – 19.57 = 0.09

∴ For difference (7532 – 7500 = 32) difference in cube root values 

= \(\frac{0.09}{100}\times32 = 0.029\)

= \(\sqrt[3]{7532}\) = \(19.57 + 0.029 = 19.599.\)

Thus the required cube root is = 19.599.

(iii) 833

= \(\sqrt[3]{833}\)

We know that value of \(\sqrt[3]{833}\) will lie between \(\sqrt[3]{830} = \sqrt[3]{840.}\)

From cube root table we get,

= \(\sqrt[3]{830}\) = 9.398 and \( \sqrt[3]{840} = 9.435\)

So by unitary method,

∵ For difference (840 – 830 = 10 ) difference in cube root values = 9.435 – 9.398 = 0.037 

∴ For difference (833 – 830 = 3) difference in cube root values 

= \(\frac{0.037}{10}\times3 = 0.011\)

= \(\sqrt[3]{833}\) = 9.398 + 0.011 = 9.409

(iv) 34.2

= \(\sqrt[3]{340} = \) \(\sqrt[3]{\frac}{342}{10}\) = \(\frac{\sqrt[3]{342}}{\sqrt[3]{10}}\)

We know that value of \(\sqrt[3]{342}\) will lie between \(\sqrt[3]{340} \) and \(\sqrt[3]{350.}\)

From cube root table we get,

= \(\sqrt[3]{340} = \) 6.980 and \(\sqrt[3]{350} = 7.047\)

So by unitary method, 

∵ For difference (350 – 340 = 10 ) difference in cube root values = 7.047 – 6.980 = 0.067 

∴ For difference (342 – 340 = 2) difference in cube root values 

=\(\frac{0.067}{10}\times2 = 0.013\)

= \(\sqrt[3]{342}\) = 6.980 + 0.013 = 6.993

We also have, \(\sqrt[3]{10} = 2.154\) (from table)

∴ \(\sqrt[3]{34.2}\) = \(\frac{\sqrt[3]{342}}{\sqrt[3]{10}}\) = \(\frac{6.993}{2.154} = 3.246.\)

Thus the required cube root is = 3.246.

The mean of 9 observations = 38 

The sum of 9 observations = 38 × 9 = 342 

If the observation 27 is mistaken for 72 

then correct observation = 72 – 27 = 45 

Sum of correct observations = 342 + 45 = 387 

∴ Correct Mean = 387/9  = 43

Given, numbers are:

31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30

Arrange in increasing order:

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45

n = 15 (odd)

Therefore, median \(=\frac{n+1}{2}\) value

\(=\frac{15+1}{2}\)

= 8^th value = 35 kg

If the weight 44kg is replaced by 46 kg and 27 kg is replaced by 25 kg

Then new values in order be

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46

n = 15 (odd)

Therefore, new median \(=\frac{n+1}{2}\)

 \(=\frac{15+1}{2}\)

= 8^th value = 35kg

Given:

One says, “Give me a hundred, friend! I shall then become twice as rich as you.”

The other replies, "If you give me ten, I shall be six times as rich as you".

To find:

The amount of their respective capital.

Solution:

Let the capital of two friends be ‘a’and ‘b’ respectively.

Given, one says, “Give me a hundred, friend!

I shall then become twice as rich as you.

It means if one is giving Rs. 100 other is losing the Rs. 100.

Since a is gaining 100 and b is losing 100.

⇒ a + 100 = 2(b – 100)

⇒ a + 100 = 2b – 200

⇒ a - 2b = - 200 - 100

⇒ a - 2b = - 300 ...... (1)

In another condition 2nd friend replies, “Give me a ten, I shall be six times as rich as you".

It means if one person is gaining 10 other person is losing 10.

Here b is gaining Rs 10 and a is losing Rs 10.

⇒ b + 10 = 6(a – 10)

⇒ b + 10 = 6a – 60

⇒ 6a - b = - 60 -10

⇒ 6a - b = - 70 ...... (2)

Now solve equations (1) and (2) to get the amount a and b.

Multiply eq. (1) with 6 and subtract we. (2) from it.

⇒6(a - 2b ) - (6a - b) = 6 (- 300) -70

⇒6a - 12b - 6a+b = - 1800 - 70

⇒ - 11b= - 1870

⇒ b = 170

put the value of b in the eq.(1) to get value of a,

⇒a - 2(170) = - 300

⇒a - 340 = - 300

⇒ a = - 300 + 340

⇒ a = 40

Hence the amount of capital of two friends is Rs 40 and Rs 170.

Let the cost of each bat be ‘a’ and each ball be ‘b’.

Given, coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

Later,

He buys 3 bats and 5 balls for Rs 1750

⇒ 7a + 6b = 3800 ---- (1)

and 3a + 5b = 1750 ---- (2)

Multiplying eq1 by 3 and eq2 by 7 and subtracting eq2 from eq1.

⇒ 3(7a + 6b) - 7(3a + 5b) = 3(3800) - 7(1750)

⇒ 21a + 18b – 21a – 35b = 11400 – 12250 

⇒ -17b = - 850

⇒ 17b = 850

⇒ b = 50

Putting this in eq 1, we get

7a + 6(50) = 3800

⇒ 7a + 300 = 3800

⇒ 7a = 3500

⇒ a = 500

Hence,

Each bat cost a = 500 Rupees and Each ball costs b = 50 Rupees.

Let the cost of 1 pen be ‘a’ and the cost of 1 pencil be ‘b’.

Given,

5 pens and 6 pencils together cost Rs 9.

5a + 6b = 9 --------- (1)

Given,

3 pens and 2 pencils cost Rs 5.

3a + 2b = 5 -------- (2)

Multiplying eq2 by 3 and subtracting eq1 from it.

3(3a + 2b) - (5a + 6b ) = 5(3) - 9

⇒ 9a + 6b – 5a – 6b = 15 – 9

⇒ 4a = 6

⇒ a = 3/2

Substituting value of 'a' in (1)

⇒ 15/2 + 6b = 9

⇒ 6b = 3/2

⇒ b = 1/4

(i) 686/-3456

686/-3456 = -(∛(2×7×7×7)) / (∛(2⁷×2³))

= – (∛(2×7³)) / (∛ (2⁷×2³))

= – (∛(7³)) / (∛(2⁶×2³))

= – 7/(2×2×2)

= – 7/8

(v) -39304/-42875

-39304/-42875 = -(∛(2×2×2×17×17×17)) / -(∛(5×5×5×7×7×7))

= – (∛(2³×17³)) / -(∛(5³×7³))

= – (2×17)/-( 5×7)

= – 34/-35

= 34/35

• As we know that square of an even number is even number.
• Square of an odd number is odd number.

Hence, 

(i) 731

Since 731 is an odd number, the square of the given number is also odd.

(ii) 3456

Since 3456 is an even number, the square of the given number is also even.

(iii) 5559

Since 5559 is an odd number, the square of the given number is also odd.

(iv) 42008

Since 42008 is an even number, the square of the given number is also even.

Answer is (a) e^{-sin x}  

Answer is (C) log x

Answer is (A) sec x + tan x

Answer is (c) (x - y)dy + y².dx = 0

Given differential equation is 

xy(dy/dx) = (x + 2)(y + 2)

∴ (y/(y + 2)) dy = ((x + 2)/x) dx, y ≠ -2, x ≠ 0

or, (1 - (2/(y + 2))) dy = (1 + (2/x)) dy

Integrating,

∫(1 - (2/(y + 2))) dy = ∫(1 + (2/x)) dx

or, y - 2 log|y + 2| = x + 2 log|x| + k ...(i)

Since, curve passes through (1, -1)

So, by(1)

-1 - 2log 1 = 1 + 2log 1 + k

∴ k = -2 

Hence, required equation of the curve is 

y - 2 log|y + 2| = x + 2log|x| - 2

Given differential equation may be written as (dy/dx) + (1/x)y = x²

This is linear differential equation of form (dy/dx) + Px = Q

Here, P = 1/x, Q = x² 

I.F. = e^∫Pdx = e^{∫(1/x) dx} = e^{log x}

= x 

Answer is (a) x log x - x + k

Answer is (b) e^{x^2/2}

Answer is (a) x = ky

Answer is (b) tan^-1 ((x + y)/(1 - xy))

Answer is (A) (1,0,0)

Answer is (D) z = 0

Answer is (D) None of these

To find : Direction cosines of the y- axis

Formula used : If a line makes angles α^o, β^o and γ^o with the positive directions of x-axis, y-axis and z-axis respectively. then direction cosines are given by cos α, cos β, cos γ 

 y-axis makes 90^o with the x and z axes

α = 90^o, β = 0^o and γ = 90^o

Direction cosines of the line is cos 90^o, cos 0^o, cos 90^o

0, 1, 0

Direction cosines of the line is 0, 1 , 0

Given -

A = (2,3,-4)

B = (1,-2,3)

C = (3,8,-11)

To prove – A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((1-2),(-2-3),(3+4))

=(-1,-5,7)

Similarly, the direction ratios of the line BC can be given by

((3-1),(8+2),(-11-3))

=(2,10,-14)

Tip – If it is shown that direction ratios of AB=λ times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(-1,-5,7)

=(-1/2) Χ (2,10,-14)

=(-1/2) Х d.r. of BC

Hence, A, B and C are collinear

Correct option is D. perpendicular to the z-axis

It is perpendicular to z-axis because cos90° is 0 which implies that it makes 90° with z-axis.

Correct option is B.\(\cfrac {a_1}{a_2}=\cfrac{b_1}{b_2}=\cfrac{c_1}{c_2}\) 

We know that if there is two parallel lines then their direction ratios must have a relation

\(\cfrac {a_1}{a_2}=\cfrac{b_1}{b_2}=\cfrac{c_1}{c_2}\)

When the deforming force is removed, the bodies tend to return to their respective original state. It is observed that while some bodies return to their original states, immediately, others take appreciably long time to do so. For example, a quartz fibre returns immediately to its normal state, when the twisting torque acting on it ceases to act. On the other hand, a glass fibre will take hours to return to its original state. This delay in regaining its original state by a body after the removal of the deforming force is called elastic after effect. In galvanometers and electrometers, the suspensions made from quartz and phosphor-bronze are used as the elastic after effect is negligible in the wires of these materials.

The property which resists the relative motion between two surfaces in contact is called friction.

Hardest material : Diamond 

Softest material : Aluminium 

[Note: Material with highest strength is steel whereas material with lowest strength is plasticine clay.]

V_r = 5 cap i - 5 cap j

Maximum range is obtained at an angle of 45^o. 

Correct option is: (D) Horizontal component of velocity

No; Yes; No

One cannot associate a vector with the length of a wire bent into a loop. One can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area. One cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.

No; No
A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.
Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

The direction of the centripetal force is towards the centre of the circle.