Explore topic-wise InterviewSolutions in Current Affairs.

Correct answer is Physics

Correct answer is Geography

Correct answer is Chemistry

The correct option is (B).

α   β−keto acid having it's α carbon at bridged head can't under go tautomerisation and it has a fixed hybridisation of sp³.

⇒ can't undergo decarboxylation.

(i) (d) 

(ii) (b) 

(iii) (d)

(iv) (c) 

(v) (b)

• Exine is made up of sporopollenin and intine is made up of pectin and cellulose.
• The exine provides protection to the pollen grains; the sporopollenin of exine is one of the most resistant organic material which can withstand very high temperature and strong acids and alkalies.

• Hymen is the membrane that partially covers the vaginal opening.
• The hymen gets torn during the first coitus.
• But it can also be torn by active participation in sports like horseriding, cycling, or by a sudden fall or jolt and insertion of vaginal tampon; so, its presence or absence cannot indicate virginity.

  \(H=mL \)

∵  H = KA \((\frac{△t}{△x})t\)

KA \((\frac{△t}{△x})+=ML \)

0.03 × 6 × (0.45)² × \((\frac{20}{0.03})×3600=M×80×4.2×1000\)

\(M=\frac{87480}{336000} \)

\(M=0.26kg \)

Let two gas of sample is A and B then pressure of A is less than that of B

(c) 4 m 

Pressure in water column of height ‘h’.

P = \(\rho\)gh 

3.924 × 10⁴ = 10³ × 9.81h

\(\therefore\) h = \(\frac{39.24}{9.81}\) = 4m

(c) Atmospheric cooling 

In reciprocating engine to achieve high pressure ratio the condition of minimum work is required which is possible in isothermal compression. To do isothermal compression intercooler is installed in between stages of compressor at outside of cylinder since intake is at atmospheric in general (except aircraft and all), so, temperature is at inlet is atmospheric. Since, process is isothermal so, intercooling will be isothermal atmospheric intercooling (the above is for ideal case not for real).

By providing appropiate dimension to cylinder we can increase volumetric efficiency not achieve the isothermal cooling.

(d) Area of the quadrilateral whose side and one diagonal are given can be calculated by dividing the quadrilateral into two triangles and using the Heron's formula.

(b) Sides of white coloured triangular plastic sheet are 4 cm, 6 cm and 6 cm but sides of grey coloured triangular sheet are also 4 cm, 6 cm and 6 cm. It is clear they must have equal areas. 

Hence, difference in areas of shaded and unshaded colour sheets is (zero) 0 cm².

(c) Area of the field 1/2 (18 x 40) = 360 m² 

Father divided the field in the ratio 2 : 1 : 1 by keeping bigger part for himself. 

He devided it into two equal parts. One part he kept for himself and other part he further divided between the sons. 

Total area distributed to the sons = 360 ÷ 2 = 180 m²

Slit separation, d = 1 mm = 1 x 10^-3 m

wavelength \(\lambda\) = 6.5 x 10^-7 m

screen distance D = 1m

Distance b/w 3^rd dark fringe and 5^th bright fringe

Dark fringe y_n = \(\frac{(2n-1)\lambda D}{2d}\)

Bright fringe y_n = \(\frac{n\lambda D}d\)

y₅ - y₃ = \(\frac{5\times6.5\times10^{-7}\times1}{10^{-3}}\) - \(\frac{(2\times3-1)\times6.5\times10^{-7}}{2\times10^{-3}}\)

 = 32.5 x 10^-4 - 16.25 x 10^-4

= 16.25 x 10^-4 m

The distance b/w the third dark fringe and fifth bright fringe is 1.625 mm.

Given:

The number of terms in an A.P. -3, 3, 9, 15,………..,201

Formula used:

d = t₂ – t₁

a + (n – 1)d

where, a = first term, n = last term and d = common difference

Calculation:

a = -3

d = 3 – (-3)

⇒ d = (3 + 3) = 6

n^th term = 201

According to the question

⇒ a + (n – 1)d

⇒ -3 + (n – 1) × 6 = 201

⇒ -3 + 6n – 6 = 201

⇒ 6n = 210

⇒ n = 35

∴ Total number of term is 35

Given:

x = 8, y = 27

Calculation:

According to the question

\(\left({X^{\frac{4}{3}}+y^{\frac{2}{3}}}\right)^{\frac{1}{2}}\)

⇒ (8^4/3 + 27^2/3)^1/2

⇒ (2^{3 × 4/3} + 3^{3 × 2/3})^1/2

⇒ (2⁴ + 3²)^1/2

⇒ (16 + 9)^1/2

⇒ (25)^1/2

⇒ 5

∴ Required value is 5

Given line is \(\frac{x}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) . 

The direction ratios of the line are 2, 3 and 4. 

Now,  \(\frac{x}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) = \(\lambda\) (Let)

⇒ x = 2\(\lambda\), = 3\(\lambda\) + 2 and = 4\(\lambda\) + 3.

Hence, the points on the line \(\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) is of the form Q (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3).

The direction ratios of the line joining points (3, −1, 11) and (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3) are

2\(\lambda\) − 3, 3\(\lambda\) + 2 − (−1) and 4\(\lambda\) + 3 − 11 = 2\(\lambda\) − 3, 3\(\lambda\) + 3 and 4\(\lambda\) − 8.

Let the foot of the perpendicular is Q (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3).

Let line PQ is perpendicular to the given line \(\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}.\)

Therefore, 2(2\(\lambda\) - 3)+3(3\(\lambda\) + 3) + 4(4\(\lambda\) - 8) = 0.

(Because, the sum of product of direction ratios of two perpendicular lines is equal to zero.) 

⇒ 4\(\lambda\) − 6 + 9\(\lambda\) + 9 + 16\(\lambda\) − 32 = 0

⇒ 29\(\lambda\) − 29 = 0 ⇒ \(\lambda\) = \(\frac{29}{29}= 1\).

Now, putting the value of = 1 in the equation of points on the given line. 

Therefore, the foot of the perpendicular is Q (2\(\lambda\), 3\(\lambda\) + 2, 4\(\lambda\) + 3) = Q(2, 5, 7).

The equation of line PQ is \(\frac{x-3}{2-3} = \frac{y-(-1)}{5-(-1)}\) = \(\frac{z-11}{7-11} ⇒ \frac{x-3}{-1}\) = \(\frac{y+1}{6}\) = \(\frac{z-11}{-4}\)

The equation of the perpendicular from the point (3, −1, 11) to the line \(\frac{x}{2} = \frac{y-2}{3}\) = \(\frac{z-3}{4}\) is 

\(\frac{x-3}{-1} = \frac{y+1}{6} = \frac{z-11}{-4}\).

Now, length of perpendicular is = \(\sqrt{(2-3)^2 + (5-(-1))^2+ (7-11)^2}\)

= \(\sqrt{(-1)^2 + 6^2 + (-4)^2}\) = \(\sqrt{1+36+16}\) = \(\sqrt{53}\) units.

We know that the direction cosines of a line making \(\alpha\), \(\beta\) and \(\gamma\) angles with x, y and z-axes, respectively are l = cos \(\alpha\) ,m = cos \(\beta\) and n = cos \(\gamma\). 

Now, given that a line makes equal angles with the co-ordinates axes.

i.e.,  \(\alpha\) = \(\beta\) = \(\gamma\).

Therefore, the direction cosines of given line are l = cos \(\alpha\) , m = cos \(\alpha\) and n = cos\(\alpha\). 

We also know that the sum of squares of direction cosines is equal to 1. i.e., l² + m² + n² = 1. 

Therefore, cos² \(\alpha\) + cos² \(\alpha\) + cos² \(\alpha\) = 1 

⇒ 3 cos² \(\alpha\) = 1 

⇒ cos² \(\alpha\) = \(\frac{1}{3}\) . 

⇒ cos = ± \(\frac{1}{\sqrt{3}}\).

Therefore, the direction cosines of given line are l = ±\(\frac{1}{\sqrt{3}}\),m = ±\(\frac{1}{\sqrt{3}}\) and n = ±\(\frac{1}{\sqrt{3}}\).

We have to find smallest number which is divisible by 306 and 657

306 = 2 × 3² × 17

657 = 3 × 3 × 73 

LCM of 306 and 657 = 2 × 3² × 17 × 73

= 2 × 9 × 17 × 73

= 18 × 17 × 73

= 306 × 73

= 22338

Hence, smallest number is 22338 which is divisible by 306 and 657.

Correct answer is: (b) 3√2 cm

CD/BD = BD/AD

BD² = CDXAD = 6X3

BD=3√2 cm

Given:

Ram calculates profit percent on C.P and Amit calculates on S.P

Profit earned by Ram + profit earned by Amit = 360

Profit percentage earned by Ram and Amit each = 25%

Selling price of Ram = Selling price of Amit

Formula used:

% Profit = (profit/cost price) × 100

Cost price = 100/(100 + a%) × selling price

Where, a% → percent profit

Concept used:

Profit/loss is always calculated on C.P

Calculation:

Let S.P of Ram and Amit be X

% profit earned by Ram = 25%

C.P of Ram = 100/(100 + 25) × X

⇒ (100/125) × X

⇒ 4/5 × X

⇒ 4X/5

Ram calculates his profit percent on C.P

Profit earned by Ram = 25% of C.P

⇒ 25% of 4X/5 = X/5      ----(1)

Similarly, S.P of Amit = X

Amit calculates his profit percent on S.P

Profit earned by Amit = 25% of X

⇒ (25/100) × X = X/4   ----(2)

Profit earned by Ram + profit earned by Amit = 360

⇒ X/5 + X/4 = 360

⇒ (4X + 5X)/20 = 360

⇒ 9X = 360 × 20

⇒ X = 7200/9

⇒ X = 800

Given:

Difference between their profits = 200

Profit gain by Mahesh and Dinesh both = 20%

Formula used:

Gain or loss percent = (Loss or gain/Cost price) × 100

Concept used:

Profit = Selling Price – Cost Price

Loss = Cost Price – Selling Price

Calculation:

Let the selling price for both of them be Rs. x.

Cost price of Dinesh = x × {(100 - 20)/100}

⇒ x × {80/100}

⇒ 4x/5

Cost price of Mahesh = x × {100/(100 +20)}

⇒ x × {100/120}

⇒ 5x/6

Dinesh’s profit = x – (4x/5)

⇒ x/5

Mahesh’s profit = x – (5x/6)

⇒ x/6

Difference of their profits = (x/5) – (x/6) = 200

⇒ x/30 = 200

⇒ x = 6000

∴ The selling price for both of them is Rs. 6000.

Given:

Discount 1 percent = 22%

Discount 2 percent = 17%

Discount 3 percent = 11%

Calculation:

Let the initial price be 100

Discount 1 = 22% of 100 

Discount 1 = 22

Price after discount 1 = 100 – 22 = 78

Discount 2 = 17% of 78

Discount 2 = 13.26

Price after discount 2 = 78 – 13.26 = 64.74

Discount 3 = 11% of 64.74

Discount 3 = 7.1214

Price after discount 3 = 64.74 – 7.1214 

Final price after third discount = 57.6186

Total discount = Initial price – Final price = 100 – 57.6186 = 42.3814

Total discount ≈ 42

Discount% ≈ (Total discount/Initial price) × 100

⇒ Discount% ≈ (42/100) × 100

⇒ 42%

∴ Discount percentage is approximately 42%

Given:

The selling price of each article = Rs. a

Gain on one article = b%

Loss on another article = b%

Concept used:

Where there is a profit of x% and loss of y% in a transaction then the resultant profit or loss per cent is given by 

[x - y - xy/100]

Calculations:

Here x = y = b

(b - b - b × b/100) = -b²/100

⇒ -(b/10)²

-ve sign means loss occurred 

∴ The loss is (b/10)²

Given y = x² + x, x = 10 and ∆x = 0.1 

∆y = f(x + ∆x) – f(x) 

= (x + ∆x)² + (x + ∆x) – (x² + x) 

= x² + 2x ∆x + (∆x)² + x + ∆x – x² – x 

= 2x ∆X + (∆x)² + ∆x 

= 2(10) (0.1) + (0.1)² + 0.1 

= 2 + 0.01 + 0.1 

= 2.11 

dy = f¹ (x) ∆x 

= (2x + 1) ∆x 

= {2.(10) + 1} (0.1) 

= (21) (0.1) 

= 2.10.