Explore topic-wise InterviewSolutions in Current Affairs.

Zn(s) → Zn²⁺ (aq) + 2e^-  (✓) 

Zn²⁺ (aq) + 2e^- → Zn(s)   (✘)

(i) Zn

(ii) Cu

Answer is Sodium.

Mark the direction of electron how in the cell illustrated.

Answer is Hydrogen.

Equations (a) and (c) are wrong. 

a. Copper cannot displace hydrogen from acids because it is placed below hydrogen in the reactivity series. 

c. Fe reacts only with super heated steam. Fe does not react with water in liquid state.

Cu(s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag(s) 

Cu(s) → Cu²⁺(aq) + 2e^- (Anode) 

Ag⁺ (aq) + le^- → Ag(s) (Cathode)

2Na + 2H₂O → 2NaOH + H₂

2Mg + O₂ → 2MgO

Chemical Bonding refers to the formation of a chemical bond between two or more atoms, molecules, or ions to give rise to a chemical compound. These chemical bonds are what keep the atoms together in the resulting compound.

Xerophyte do have leaves but their leaves are modified into spines and the function of photosynthesis is carried by stems only because in leaves a great amount of transpiration would occur making the plant unable to produce food and causing it to dry out.

Amenorrhoea is a symptom with many potential causes. Primary amenorrhoea is defined as an absence of secondary sexual characteristics by age 14 with no menarche or normal secondary sexual characteristics but no menarche by 16 years of age.

Inflammation is the body's immune system's response to an irritant. The irritant might be a germ, but it could also be a foreign object, such as a splinter in your finger.

Surgery, branch of medicine that is concerned with the treatment of injuries, diseases, and other disorders by manual and instrumental means. Surgery involves the management of acute injuries and illnesses as differentiated from chronic, slowly progressing diseases, except when patients with the latter type of disease must be operated upon.

Inflammation is the body's immune system's response to an irritant. The irritant might be a germ, but it could also be a foreign object, such as a splinter in your finger.

Concept used:-

Trick to find square root of a number.

Method and Calculation:-

 • Find the unit digit of square root by looking at unit digit of given number.

For ✓24025, unit digit must be 5

 • Find other two digits by looking at number formed by first 3 digit.

225 < 240 < 256 (225 and 256 are squares of 15 and 16, so remaining part of square root will be 15.

Hence, ✓24025 = 155

e^siny cos y \(\frac{dy}{dx}\) + e^siny cos x = cos x

e^siny = t

e^siny. cos y \(\frac{dy}{dx}\) = \(\frac{dt}{dx}\)

\(\therefore\) \(\frac{dt}{dx}\) + cos x t = cos x

\(\therefore\) I.F. = \(e^{\int cosxdx}\) = e^sinx

\(\therefore\) t x I.F. = \(\int\)(I.F.) \(\times\) α dx

⇒ t.e^sinx = \(\int e^{sinx}.cos x dx\)

 = e^sinx + c

⇒ t = 1 + c e^-sinx

⇒ e^siny = 1 + c e^-sinx 

y(0) = 0

⇒ c + 1 = e^o = 1

⇒ c = 0

\(\therefore\) e^siny = 1 is solution of given differential equation.

⇒ sin y = ln 1 = 0

⇒ y = sin^-10 = 0

\(\therefore\) y = 0 is solution of given differential equation.

\(\therefore\) y(π/3) = y(π/4) = y(π/6) = 0

\(\therefore\) 1 + y(π/6) + \(\frac{\sqrt3}2y(\pi/3)\) + \(\frac1{\sqrt2}y(π/4)\) = 1 + 0

= 1

 Symbolic form of the given equation,

(D²+4)y = sin 2x

Corresponding auxiliary equation,

D²+4 = 0

i.e. D = ±2i

Thus, y(C.F.) = (c₁ cos 2x + c₂ sin 2x)

y(P.I.) \(=\frac1{D^2+4}\)sin 2x; Replace

D² = -a² = -4, but here f(-a²) = 0

[In case of failure, \(\frac1{f(D^2)}\)sin(ax+b) \(=x\frac1{f'(-a^2)}\)sin(ax+b)]

Implying y(P.I.)

\(=x\frac{1}{2.D}sin\,2x=\frac{x}{2}(\frac{-cos\,2x}{2})=-\frac{x\,cos\,2x}{4}.\)

Hence the complete solution,

y = (c₁ cos 2x + c₂ sin 2x) \(-\frac{x\,cos\,2x}4.\)

For slope of curve x³ + kxy² = -2, we differentiate

its equation w.r.t. x then

3x² + ky² + 2kxy\(\frac{dy}{dx}=0\)

⇒ \(\frac{dy}{dx}=-\frac{(3x^2+ky^2)}{2kxy}\) = m₁ (Let)--(1)

For slope of curve 3x²y - y³ = 2, we differentiate its equation w.r.t. x then

(3x² - 3y²)\(\frac{dy}{dx}\) + 6xy = 0

⇒ \(\frac{dy}{dx}\) = \(\frac{-6xy}{3x^2-3y^2}\) = m₂(Let)--(2)

Since, both curves are orthogonal to each other

∴ m₁m₂ = -1

⇒ \(\frac{-(3x^2+ky^2)}{2kxy}\times\frac{-6xy}{3x^2-3y^2}=-1\)

⇒ \(\frac{3(3x^2+ky^2)}{k(3x^2-3y^2)}=-1\)

⇒ 3x² + ky² = -kx² + ky²

⇒ -kx² = 3x²

⇒ k = -3

∴ |k| = 3

\(\frac{dy}{dx}=\frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}\) = \(\frac{3e^{2x}(1+e^{2x})e^x}{e^{2x}+1}\)

 = 3e^3x

⇒ dy = 3e^3xdx

⇒ y = \(\frac{3e^{3x}}3+c\)

⇒ y = e^3x + c is general solution of the given differential equation.

Given:

Cost price (C.P.) ∶ Selling price (S.P.)  = 8 ∶ 5.

Formula used:

Loss = C.P. – S.P.

Loss % = (L/C.P.) × 100 

Where

L → Loss

S.P. → Selling price

C.P. → Cost price 

Calculations:

Let the CP be 8x and SP be 5x.

Then, Loss = 8x – 5x = 3x

Loss % = (3x/8x) × 100 = 37.5%

∴ The required Loss% is 37.5%.

y = x + \(\sqrt{a^2 + x^2}\)

\(\frac{dy}{dx}\) = 1 + \(\frac{2x}{2\sqrt{a^2+x^2}}\)

= 1 + \(\frac{x}{\sqrt{a^2+x^2}}\)

\(\frac{d^2y}{dx^2}\) = \(\frac{\sqrt{a^2+x^2}\times 1-x \times 2x/(2\sqrt{a^2+x^2})}{(a^2+x^2)}\)

= \(\frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}}\)

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

⇒ (a² + x²) \(\frac{d^2y}{dx^2}\) = \(\frac{a^2}{\sqrt{a^2+x^2}}\)

Now,

 (a² + x²) \(\frac{d^2y}{dx^2}\)  + x\(\frac{dy}{dx}\) - y = \(\frac{a^2}{\sqrt{a^2+x^2}}\) + x + \(\frac{x^2}{\sqrt{a^2+x^2}}\) - (x + \(\sqrt{a^2+x^2)}\)

= \(\frac{a^2+x^2}{\sqrt{a^2+x^2}}\) - \(\sqrt{a^2+x^2}\)

= \(\sqrt{a^2+x^2}\) - \(\sqrt{a^2+x^2}\) = 0

= R.H.S

Hence proved.

B) were / was / am 

D) were / were / are

C) were / were / in Denmark 

A) were / were / are 

Answer: 2.58 centuries

Answer is (a) Both the statements are true and Statement-II is the correct explanation of Statement-I

A′ = {1,11,9,17}

2 √12 × √3 = 2 √36 = 12 

x² = 4 × 25 =100 

∴ x = 10

Given:

Two metallic spheres of radii = 5 cm and 12 cm

Formula used:

Volume of sphere = 4/3 πr³

Calculation:

Let the radius of the sphere be r cm

Then,

Volume of resulting sphere = sum of the volume of two sphere of radii 5 cm and 12 cm

⇒ 4/3 πr³ = 4/3 π × (5)³ + 4/3 π × (12)³

⇒ r³ = (5)³ + (12)³

⇒ r³ = 125 + 1728

⇒ r³ = 1853 

⇒ r = 12.28 ~ 12 cm

∴ The approximate value of the radius of the resultant sphere is 12 cm

8000 × 3 / 100 = Rs. 240

Class mark = Lower class limit + Upper class limit / 2

= 80 + 90 / 2 = 85 

Given:

x + 3y – 1 = 0; 2x + 2y + 1= 0

Calculation:

x + 3y – 1 = 0 .....(1)

2x + 2y + 1= 0 ....(2)

Multiplying by 2 in equation (1)

⇒ 2x + 6y – 2 = 0 ....(3)

Now, subtracting the equation (2) from (3) 

⇒ (2x + 6y – 2) – (2x + 2y + 1) = 0

⇒ 4y – 3 = 0

⇒ 4y = 3

⇒ y = 3/4

Now, put the value of y in equation (1) we get,

⇒ x + 3 × (3/4) – 1 = 0

⇒ x + 9/4 – 1 = 0

⇒ x + 5/4 = 0

⇒ x = -5/4

∴ The value of x and y is -5/4 and 3/4

The correct option is : (D) 5

m² + 5m + 6 

= m² + 3m + 2m + 6 

= m (m +3)+2(m+3) 

= (m + 2) (m +3)

Calculation:

The least number which is divisible by all the numbers from 1 to 10

LCM of (1 to 10) = 2³ × 3² × 5 × 7

⇒ 2520

∴ The least number is 2520

∠Q and ∠ R is a pair of adjacent angles of parallelogram PQRS. 

∴ ∠Q and ∠ R = 180° 

∴ ∠Q = 180 - 60 = 120 

∴ ∠ R : ∠Q = 60 : 120 = 1 : 2

Given:

a = i + 2j – k and b = -i + j – 2k

Concept used:

Angle between two vectors:

\(\cos θ = \frac{{⃗ a.⃗ b}}{{\left| {⃗ a} \right|\left| {⃗ b} \right|}}\)

For two vectors 

\(⃗ a = \overrightarrow {{a_1}} \hat i + \overrightarrow {{a_2}} \hat j + \overrightarrow {{a_3}} \hat k\)

\(⃗ b = \overrightarrow {{b_1}} \hat i + \overrightarrow {{b_2}} \hat j + \overrightarrow {{b_3}} \hat k\)

\(⃗ a.⃗ b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\)

Calculation:

\(\vec a.\vec b \) = -1 + 2 + 2

⇒ \(\vec a.\vec b \) = 3

⇒ \(\vec a\) = √(1² + 2² + 1²)

⇒ \(\vec a\) = √(1 + 4 + 1)

⇒ \(\vec a\) = √6

⇒ \(\vec b \) = √(1² + 1² + 2²)

⇒ \(\vec b \) = √1 + 1 + 4)

⇒ \(\vec b \) = √6

\(\cos θ = \frac{{⃗ a.⃗ b}}{{\left| {⃗ a} \right|\left| {⃗ b} \right|}}\)

⇒ cosθ = 3/√6 × √6

⇒ cosθ = 3/6

⇒cosθ = 1/2

⇒ cosθ = 60° 

∴ The angle between a and b is 60° 

Given:

Ratio of the ages of father and son 5 years ago = 5 : 3

The sum of their present ages = 90 years

Calculation:

Let the ratio of their ages 5 years ago be 5x and 3x respectively

Present age of father = 5x + 5

Present age of son = 3x + 5

According to the question

⇒ (5x + 5) + (3x + 5) = 90

⇒ 8x + 10 = 90

⇒ 8x = 90 – 10

⇒ 8x = 80

⇒ x = 10

The present age of father = 5x + 5 = (5 × 10 + 5) years

⇒ 55 years

∴ The present age of father is 55 years

Calculation:

If a dice is thrown, sample space (s) = 1, 2, 3, 4, 5, 6

n(s) = 6

Favourable cases (number greater than 2) = 3, 4, 5, 6

Total number of favourable case = 4

Probability = 4/6

⇒ 2/3

∴ Required probability is 2/3

Calculation∶

The series follows following pattern

⇒ 4 × 2 = 8

⇒ 8 + 3 = 11

⇒ 11 × 2 = 22

⇒ 22 + 3 = 25

⇒ 25 × 2 = 50

⇒ 50 + 3 = 53

∴ The value of ? is 53

Given:

p : q = 2 : 3     

q : r = 1 : 2    

Calculation:

q : r = 1 : 2 = 3 : 6     ----eq.1

Also, p : q = 2 : 3     ----eq.2

From eq.1 and eq.2:

p : q : r = 2 : 3 : 6

2p : 2q : r = 2 × 2 : 2 × 3 : 6 = 2 : 3 : 3

∴ 2p : 2q : r = 2 : 3 : 3

Given:

p : q : r = 3 : 4 : 5

Calculation:

Let the value of p, q and r be 3x, 4x and 5x

Substituting the values in the given expression, we get

(3x + 8x + 5x)/(9x + 4x – 5x)

⇒ 16x/8x

⇒ 2 ∶ 1

∴ The required ratio is 2 ∶ 1

Correct answer is: (b) 6

(2,6) = ((3p-2)/2, (4+2q)/2)

3p-2=4, 4+2q=12

P=2, q = 4 hence p + q = 6

Correct answer is: (a) 1/6

P(same no on each die) = 6/36 = 1/6