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Answer: three codons

Meiosis was discovered and described for the first time in sea urchin eggs in 1876 by the German biologist Oscar Hertwig. It was described again in 1883, at the level of chromosomes, by the Belgian zoologist Edouard Van Beneden, in Ascaris roundworm eggs.

1- Plasma____4- Lymph

Correct option is (a) B ∝ r

\(B=\frac{\mu_{0}}{2\pi},\frac{ir}{R^2}\)

B ∝ r

Mass of glucose = 2.82 g 

No. of moles of glucose = \(\frac{2.82}{180}\) = 0.016

Mass of water = 30g = \(\frac{20}{18}\) = 1.67

X_H20 = \(\frac{1.67}{1.67 +0.016}\) = \(\frac{1.67}{1.686}\) = 0.99

∴x_H20  + xx_glucose = 1 

0.99 + x_glucose = 1 

x_glucose = 1 – 0.99 = 0.01

Answer is (a) Ammonia is the weakest reducing agent and the strongest base among Group 15 hydrides.

The reducing character of hydrides increases down the group due to decrease in bond dissociation enthalpy.

Given:

Marked price of table(MP) = Rs. 6000

Selling price of table(SP) = Rs. 5520

Formula used:

Discount = Marked Price – Selling Price

Discount% = (MP – SP)/MP × 100

Calculation:

Discount = 6000 – 5520 = 480

Discount% = 480/6000 × 100

⇒ 8%

GIVEN:

Product of the two numbers = 4107 and H.C.F of these two numbers = 37

FORMULA USED:

Product of the two number = their (H.C.F × L.C.M)

CALCULATION:

Let L.C.M be x

Product of the two numbers = 4107 and H.C.F of these two numbers = 37

⇒ Product of the two number = their (H.C.F × L.C.M)

⇒ 4107 = 37 × x

⇒ x = 4107/37

∴ L.C.M of these two numbers is 111

Given A.P is

117,104,91,78,.....

first term of this A.P is  a₁​ = 117

second term of this A.P is  a₂ ​= 104

 common difference of an A.P is given by

(i.e., d = a_n+1​−a_n​)

putting n = 1 in above equation

d = a₂​−a₁ ​= 104−117 = −13

hence common difference for this A.P is d = −13

the nth term of an A.P is given by.

a_n​=a₁​+(n−1)d

⟹a_n​=117+(n−1)(−13)

⟹a_n​=117−13n+13

⟹a_n​=130−13n....eq(1)

for finding 10th term (a₈​)of this A.P put n=8 in above eq(1)

⟹a₈​=130−13×8

⟹a₈​=130−104

⟹a₈​=26

GIVEN:

Product of the two co-prime = 117

CONCEPT USED:

Two number are said to be a co-prime when their H.C.F is 1

CALCUATION:

Product of the two co-prime = 117 = 13 × 9

⇒  their H.C.F is 1

⇒ L.C.M. of (13 and 9) = 117

∴ L.C.M of the two numbers are 117

Given:

Numbers = 26, 91 and 117

Concept used:

HCF can be found by prime factorization method

Calculation:

26 = 2 × 13

91 = 7 × 13

117 = 3 × 3 × 13

The common factor in all of these is 13

∴ The HCF of the 26, 91 and 117 is 13

Given:

Distance travelled by the car = 10 m

Time taken = 1 sec

Formula used:

(i) Speed = Distance/time

(ii) 1 m/s = 18/5 km/hr

Calculations:

Speed = Distance/Time

⇒ 10/1

⇒ 10 m/s

⇒ 10 m/s × (18/5)

⇒ 36 km/hr

∴ The speed of car is 36 km/hr.

In this type of problem, we find the LCM of five numbers first and then we look for what makes the LCM a perfect square.

1st term = 10 = 5 × 2

2nd term = 12 = 2 × 2 × 3

3rd term = 15 = 3 × 5

4th term = 18 = 2 × 3 × 3

5th term = 20 = 2 × 2 × 5

Hence, LCM ( 12,10,15,18,20)

⇒  180

Since the LCM= 180 is not a perfect square , we make it a perfect square by multiplying 5 ( which is the least value to be multiplied)

Then 180 × 5 = 900 = 30²

Therefore, 900 solider can be drawn up in the form of a square.

Given:

The ratio between the age of Father and Mother is 8 : 7.

After 8 years, the total age of father and mother is 91 years

Calculation:

Let present age of father and mother be a and b years respectively.

⇒ a : b = 8 : 7

⇒ a = 8(b/7)

Then,

⇒ (a + 8) + (b + 8) = 91

⇒ a + b = 75

After solving,

a = 40 years and b = 35 years

∴ The present age of father is 40 years.

Given:

Let present age of Ram and Raj be a years and b years respectively.

Calculation:

⇒ a + b = 42

⇒ a = b/2 + 3

⇒ 2a - 6 = b

Solving,

The present age of Ram = 16 years

The present age of Raj = 26 years

∴ Required ratio = 16 : 26 = 8 : 13

Given:

A person sells a cooker for Rs. 2,568

He bought cooker for Rs. 2,400

Formula used:

P = SP - CP

Profit percent = (P/CP) × 100

where, 

P is Profit

SP is Selling price 

CP is Cost price

Calculations:

According to the question, we have

Profit gained by the shopkeeper is

P = SP - CP

⇒ P = Rs. 2,568 - Rs. 2,400

⇒ P = Rs. 168

Now, Profit percentage is

Profit percent = (P/CP) × 100

⇒ Profit percent = (168/2400) × 100

⇒ Profit percent = 168/24

⇒ Profit percent = 7%

∴ The profit percentage gained by the shopkeeper is 7%.

(x² - x) + (y² - y)

 = x(x - 1) + y(y - 1)

=  even

\(\because\) x(x - 1) is always even for an integer.

\(\therefore\) x(x - 1) is even & y(y - 1) is even

⇒ x(x - 1) + y(y - 1) is even

⇒ (x² - x) + (y² - y) is even

GIVEN:

Total CP of dogs = Rs. 2600

loss on the first dog = 28%

profit on the second dog = 36%

SP of both dogs is the same

EXPLANATION:

Let the cost price of the dogs be x and y.

∵ SP is equal ⇒ \(72\%\space of \space x= 136\% \space of \space y \)

\(\frac xy=\frac{17}{9} \\x : y = 17: 9 \\total\space CP= 17+9=26\space ratio \\CP\space of \space first\space dog=\frac{17}{26}\times2600=Rs1700\)

Given:

Bhavin sold two dogs at same price.

In 1 he got a profit of 10% and in the other he got a loss of 10%

Formula Used:

Loss% = (CP – SP)/CP × 100

Calculation:

Here, selling price are same, profit and loss percent are same.

In such transactions, there is always loss,

Loss percent = 10 × 10/100 = 1%

5⁴ = 5 x 5 x 5 x 5
5⁴ = 25 x 5 x 5
5⁴ = 125 x 5
5⁴ = 625

The given equation of circle is

2x² + 2y² + px + py = 0

⇒ x² + y² + p/2 x + p/2 y = 0 (Dividing both sides by 2)

⇒ (x² + p/2 x + p²/16) + (y² + p/2 y + p²/16) - p²/16 - p²/16 = 0

(Adding and subtracting p²/16 in L.H.S)

⇒ (x + p/4)² + (y + p/4)² = 2p²/16 = p²/8

By comparing (x - h)² + (y - k)² = r², we get

h = \(\frac{-p}4\) and k = \(\frac{-p}4\)

and r² = p²/8

⇒ r = r/2√2

Hence, radius of the circle is r = p/2√2

Let equation of circle be 

x² + y² + 2gx + 2fy + c = 0 …. (1) 

Here circle passing through (0,2)(3,0) and (3,2) 

6g + C = -9 …. (2) 

4f + C = 4 …. (3) 

6g + 4f + c = -13 …. (4) 

6g – 4 = -13 6g = -9 g = \(\frac{-3}{2}\)

4f – 9 = 13 

4f = -4 f = -1 

- 4 + c = -4 c = 0 (1) 

⇒ x² + y² – 6x – 2y = 0

(2) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) 

(3) λ_{radio waves} > λ_{micro waves} > λ_visible > λ_x-rays 

Information based

λ_{radio waves} > λ_{micro waves} > λ_visible > λ_x-rays 

When a forward bias is applied to a p-n junction, the size of the depletion layer decreases. The movement of the majority carriers takes place across the junction, resulting current, known as forward current, which increases rapidly with increase in forward voltage.

It is due to that such waves, in which the direction of oscillation of particles in perpendicular to the direction of its propagation are polarised,so longitudinal wave are not polarised.

Band width: The modulated signal lies in the frequency range from ω_c -ω_m to + ω_m i.e., 2ω_m. It is called the band width of the modulating signal.It is defined as the twice of the frequency of modulating signal. The frequency range of a modulated signal is called band width.

Base band: It is the original frequency range of a transmission signal before it is converted to modulated to a different frequency range.

Fading: The disappearance of the signals for a short time due to the variation in the height and density of ionisation of the ionosphere is called fading.

Correct answer is Rome

Correct option is A) Turning suddenly, with tears in his eyes

Correct option is C) couldn’t help laughing

Correct option is B) Until

Correct option is D) in

Correct option is C) on

Correct answer is transportation

Correct option is (b) Tb⁴⁺, Yb²⁺

Electronic configuration of Tb⁴⁺ = [Xe] 4f⁷ and for Yb²⁺ = [Xe] 4f^I4

A = {1,2},

B = {1,2,3,4}

C = {5,6}

D = {5,6,7,8}

A x C = {(1,5),(1,6),(2,5),(2,6)}

B x D = {(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}

Clearly,

A x C \(\subseteq\) B x D

Alternate : 

A \(\subseteq\) B & C \(\subseteq\) D

∴ A x C \(\subseteq\) B x D.

π = C.R.T.

7.47 = C x 0.083 x 300

C = 0.3 M

= 0.3 x 180 gL^-1

 = 54 gL^-1

A complete hydatidiform mole (CHM) is a type of molar pregnancy and falls at the benign end of the spectrum of gestational trophoblastic disease.

In 1991 Soviet Union disintegrated

Correct Answer is : (c) 60 – 80%

(c) Hydroponics

Abnormality of Apo-E

Correct option is A) to 

Correct option is B) with