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They possess the same functional group.

The general formula is C_nH_2n+2

Homologous series 

A group or a series of organic compounds in which each member contains the same characteristic functional group and differs from each other by a fixed unit form a homologous series and therefore its members are known as homologous. The members of the homologous series can be represented by a general formula and the successive members differ from each other in the molecular formula by a – CH₂ unit. There are a number of homologous series in organic chemistry such as alkanes, alkenes, alkynes, haloalkanes, alkanols, amines etc.

During noon the rate of transpiration becomes higher than the rate of water absorption by plant. It causes loss of turgity or wilting.

Wilting occurs when the loss of water by evaporation exceeds the rate of uptake by roots.

When wilting is permanent, water present in soil is largely in unavailable form. The soil contains 10-15% water depending upon its texture.

Honeybees.

Detailed answer:

In honey bees as also in wasps, ants and termites, there is haplodiploid sex determination. The sex of the individual is determined by the change in number in genome. In honey bees the females are diploid (2n) and formed through fusion of sperm with egg, while males are haploid and develop, directly from the egg through haploid parthenogenesis.

Collenchyma .

Excessive use of fertilizers increases the solute Concentration of the soil solution causing exosmosis resulting in the wilting of plants.

Monocot leaf is called isobilateral because both the surfaces of the leaf are equally green. 

(a) Haemophilia 

(i) It is a sex-linked recessive disorder. 

(ii) Patient continues to bleed even with a minor cut because of a defect in blood coagulation. 

(iii) The gene for haemophilia is located on X chromosome. 

(iv) More males suffer from haemophilia than females because in males single gene for the defect is able to express as males have only one X chromosome. 

(v) The defective alleles produce non-functional proteins which later form a nonfunctional cascade of proteins involved in blood clotting. 

(vi) Females suffer from this disease only in homozygous condition, i.e., Xc Xc . 

(vii) Queen Victoria was a carrier of this disease and produced haemophilic offsprings. 

(b) Sickle-cell anaemia 

(i) It is an autosome-linked recessive trait. 

(ii) The disease is controlled by a single pair of allele HbA and HbS . 

(iii) Only the homozygous individuals for HbS , i.e., HbS HbS show the diseased phenotype. 

(iv) The heterozygous individuals are carriers (HbAHbS ). 

(v) Due to point mutation, glutamic acid (Glu) is replaced by valine (Val) at the sixth position of β-globin chain of haemoglobin molecule. 

(vi) HbS behaves as normal haemoglobin except under oxygen stress where erythrocytes lose their circular shape and become sickle-shaped. As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affect blood supply to different organs.

a. Both are caused due to mutation or alteration in a single gene, and follow Mendelian inheritance, therefore, they are called Mendelian disorders. Symptoms of colour blindness: unable to discriminate between red and green colours. Symptoms of thalassemia: formation of abnormal haemoglobin resulting in Anaemia. 

b. ▪ Thalassemia 

a. It is an autosome-linked recessive disease. 

b. It occurs due to either mutation or deletion resulting in reduced rate of synthesis of one of globin chains of haemoglobin. 

c. Anaemia is the characteristic of this disease. 

d. Thalassemia is classified into two types: 

a. α-thalassemia—Production of α-globin chain is affected. It is controlled by the closely linked genes HbA1 and HbA2 on chromosome 16. It occurs due to mutation or deletion of one or more of the four genes. 

b. β-thalassemia—Production of β-globin chain is affected. It occurs due to mutation of one or both HbB genes on chromosome 11. 

▪ Colour blindness 

a. It is a sex-linked recessive disorder. 

b. It results in defect in either red or/and green cone of eye, resulting in failure to discriminate between red and green colour. 

c. The gene for Colour blindness is present on X chromosome. 

d. It is observed more in males (X^cY) because of presence of only one X chromosome as compared to two chromosomes of females.

Symptoms of haemophilia: Patient continues to bleed through a minor cut as the patient does not possess natural phenomenon of blood clotting. 

Symptoms of sickle-cell anaemia: Erythrocytes lose their circular shape and become sickleshaped. As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affects blood supply to different organs.

(c) Rs 1,20,000

S.P. of the whole field = \(\frac{360000\times110}{100}\) = Rs 396000

S.P of the \(\frac{1}{3}\)rd part of the field = \(\frac{360000}{3}\) x \(\frac{80}{100}\)

= Rs 96000

S.P. of the \(\frac{2}{5}\)th part of the field

= 360000 x \(\frac{2}{5}\) x \(\frac{125}{100}\)= Rs 180000

S.P. of the remaining field

= 396000 – (96000 + 180000)

= Rs 120000.

(d) 20%

Let C.P. = Rs c

\(\therefore\) C.P. of \(\frac{1}{4}th\) of the goods = \(\frac{\frac{c}{4}\times90}{100}\) = Rs \(\frac{9c}{40}\)

C.P of \(\frac{3}{4}th\) of the goods = Rs \(\frac{3c}{4}\)

Let profit on this remaining part = P%. Then,

S.P. of \(\frac{3}{4}th\) of the goods = \(\frac{\frac{3c}{4}\times(100+P)}{100}\)

= \(\frac{3c}{400}\)(100+P)

Profit on the whole transaction = 12.5%

\(\therefore\) S.P. of the whole = Rs \(\frac{c\times112.5}{100}\)

\(\therefore\) \(\frac{9c}{40}\) + \(\frac{3c}{4}\) + \(\frac{3c\times P}{400}\) = \(\frac{112.5c}{100}\)

\(\Rightarrow\) \(\frac{90+300+3p}{400}\) = \(\frac{112.5}{100}\)

\(\Rightarrow\) \(\frac{390+3p}{4}\) = 112.5  \(\Rightarrow\) 390 + 3P = 450

\(\Rightarrow\) 3P = 60  \(\Rightarrow\) P = 20.

(d) Rs 2400

Let the C.P. of the wholesale dealer = Rs x. Then, 

S.P. of the wholesale dealer = \(\frac{X\times 112.5}{100}\)

= Rs \(\frac{112.5X}{100}\)

C.P. of the retail dealer = \(\frac{112.5X}{100}\) , Gain = 20%

\(\therefore\) S.P. of the retail dealer = \(\frac{112.5X}{100}\) x \(\frac{120}{100}\)

Given, \(\frac{112.5X}{100}\) x \(\frac{12}{10}\) = 3240

\(\Rightarrow\) x = Rs \(\frac{3240\times1000}{112.5\times12}\) = Rs 2400.

(d) Rs 80

Let C.P. = Rs x, S.P. = Rs 144, Profit = x%

\(\therefore\) x + x% of x = 144

\(\Rightarrow\) x + \(\frac{X}{100}\) x x = 144

\(\Rightarrow\) x² + 100x - 14400 = 0

\(\Rightarrow\) x² + 180x – 80x – 14400 = 0

\(\Rightarrow\) x (x +180) – 80 (x + 180) = 0

\(\Rightarrow\) (x +180) (x – 80) = 0

\(\Rightarrow\) x = – 180 or 80

Neglecting negative values, x = 80.

Correct option is C) A and B

(c) Rs 12

Let the S.P. of one dozen pens be Re x. Then, 

S.P of 16 dozen pens = Rs 16x 

Loss = S.P. of 4 dozen pens = Rs 4x

\(\therefore\) C.P. = S.P. + Loss = Rs 16x + Rs 4x = Rs 20x

\(\therefore\) Loss per cent =\(\big(\frac{4X}{20X}\times100\big)\) = 20%

C.P. of 16 dozen pens = Rs 240, Loss = 20%

\(\therefore\) S.P. of 16 dozen pens = Rs\(\frac{240\times80}{100}\) = Rs 192

\(\therefore\) S.P. of 1 dozen pens = Rs \(\frac{192}{16}\) =Rs 12

(2) 12 1/2%

We know that the CP = ₹120

SP = ₹105

Since SP is less than CP it’s a loss

Loss = CP – SP

= 120 – 105

= 15

∴ Loss% = (loss × 100) /CP

= (15 × 100) / 120

= 12.5%

Correct option is A) oxygen

We know that the VAT = 1%

SP = ₹15756

So let’s consider the original price of shirt as x

∴ VAT amount = 1% of x

= 1x/100

∴ x + x/100 = 15756

(100x + x)/100 = 15756

101x /100 = 15756

x = (15756 × 100) / 101

x = 15600

∴ The original price of the 10g Gold excluding VAT is ₹15600

Correct option is C) space

(a) \(16\frac{2}{3}\)%

1st radio. 

S.P. = Rs 5000, Gain = 25%

\(\therefore\) C.P. = Rs \(\frac{5000\times100}{125}\) = Rs 4000

2nd radio

S.P. = Rs 5000, Loss = x% (say)

\(\therefore\) C.P. =\(\frac{5000\times100}{100-X}\)

Since the person neither gained nor lost, Total C.P. = Total S.P.

\(\Rightarrow\) 4000+ \(\frac{500000}{(100-X)}\) = 10000

\(\Rightarrow\) \(\frac{500000}{(100-X)}\) = 6000

\(\Rightarrow\) 500000 = 600000 – 6000x

\(\Rightarrow\) 6000 x = 100000

\(\Rightarrow\) x = \(\frac{100}{6}\)% = \(\frac{50}{3}\)% = 16\(\frac{2}{3}\)%

B) 80°C, 10°C

Correct option is A) increases

A) force of attraction

(3) 33 1/3%

We know that the CP = 75

SP = 100

Since SP is more than CP it’s a gain

Gain = SP – CP

= 100 – 75

= 25

∴ Gain% = (Gain × 100) /CP

= (25 × 100) / 75

= 33.33%

B) carbon dioxide

CP of 1 Apple = 75/10 

= Rs.7.5 

CP of 1 Dozen Apple = 7.5 × 12 

= Rs.90 

SP of 1 Dozen Apple = Rs.75 

Loss = CP – SP 

= 90 - 75 

= 15

\(Loss\%=\frac{Loss\times100}{CP}\)

\(=\frac{15\times100}{90}\)

= 16.66%

Let us consider the number of eggs as x

The CP of 10 eggs = ₹16x

CP of 1 egg = ₹16x/3

SP of an egg = ₹36x/5

Since, SP is more than CP so it’s a gain.

Gain = SP – CP = (36x/5) – (16x/3) = 168

By taking the LCM 5 and 3 we get, (108x-80x)/15 = 168

∴ 28x =168 × 15

x = (2520) /28

x = 90

∴ The number of eggs is 90.

1. Particles of matter have space between them. 
2. The rate of diffusion of gases is higher than that of liquids and solids.

Procedure: 

Open a water tap and allow the water to reach the ground. Now try to break the stream of water with our finger. But we cannot break the stream permanently. We cannot also break a piece of iron nail with our hands. But we can break a piece of chalk with our hands. Observations: From the above observations we can say that particles of the matter have forces acting between them that keeps the particles together.

It is also clear that this force is not equally strong and different in different forms of matter.

It is required. But in some cases like natural evaporation of water, it is not necessary.

A) Both statements -1 & II are true.

The temperature at which solid melts to become a liquid is called the ‘melting point’.

D) aquatic animals

Correct Answer is: (b) ice

We know that the CP of 10 apples = ₹75

CP of 1 apple = ₹75/10 = ₹7.5

CP of 1 dozen apples = ₹7.5 × 12 = ₹90

SP of 1 dozen bananas = ₹75

Since, SP is less than CP so it’s a loss.

Loss = CP – SP = 90 – 75 = 15

Loss % = (loss × 100) / CP

= (15 × 100) / 90

= 16.66%

∴ The Loss percent is 16.66%

D) Both B & C

Correct option is B) increases

The states of matter can be changed by changing the temperature or pressure.

The particles are arranged orderly in the case of solids, while particles move randomly in gases

Water present around cell wall has more diffusion pressure than inner cell sap hence water moves in the cell through freely permeable cell wall by diffusion.

Change of state takes place by change in temperature.

The hypertonic

Correct option is B) dry ice