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701.

Green alga spirogyra have type of sexual reproduction.

Answer»

Spirogyra, (genus Spirogyra), any member of a genus of some 400 species of free-floating green algae (division Chlorophyta) found in freshwater environments around the world. Named for their beautiful spiral chloroplasts, spirogyras are filamentous algae that consist of thin unbranched chains of cylindrical cells. They can form masses that float near the surface of streams and ponds, buoyed by oxygen bubbles released during photosynthesis. They are commonly used in laboratory demonstrations.

Spirogyra species can reproduce both sexually and asexually. Asexual, or vegetative, reproduction occurs by simple fragmentation of the filaments. Sexual reproduction occurs by a process known as conjugation, in which cells of two filaments lying side by side are joined by outgrowths called conjugation tubes. This allows the contents of one cell to completely pass into and fuse with the contents of the other. The resulting fused cell (zygote) becomes surrounded by a thick wall and overwinters, while the vegetative filaments die.

702.

Which among the following statements is/are incorrect regarding significance of biopatents?(a) Permit private, monopoly rights over cells, genes, animals and plants(b) People would not research in such areas, which are dominated by patents(c) Lead to research programmes dominated by potentiability and profitability(d) Philosophy and social commentary that deal with the biological sciences and their potential impact on society

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Answer is : (d)

Philosophy and social commentary that deal with the biological sciences and their potential impact on society

The statement (d) is incorrect regarding significance of biopatents. It can be corrected as

Bioethics is the branch of ethics, philosophy and social commentary that deals with the biological sciences and their potential impact on society.

703.

The effect(s) of ozone depletion include(s)(a) inhibition of photosynthesis in phytoplanktons(b) damage in nucleic acids of living organisms(c) increase in UV-radiations reaching the earth surface(d) All of the above

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Answer is : (d)

The depletion of ozone layer leads to an increase in ground level ultraviolet radiation, because ozone is an effective absorber of ultraviolet radiation. The effects of ozone depletion include inhibition of photosynthesis in phytoplanktons which affects the food chains. It damages nucleic acids in living organisms (mutation). The ozone depletion causes an increase in UV radiations reaching the earth’ surface which can cause skin cancer.

704.

In which of the following structures insulin does not increase the glucose uptake?(a) Cardiac muscles and skeletal muscles(b) Renal tubules and intestinal mucosa(c) Adipose tissues and cardiac muscles(d) Smooth muscles and skeletal muscles

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Answer is : (b) Renal tubules and intestinal mucosa

Insulin increases glucose uptake in liver, leucocytes and smooth, cardiac and skeletal muscles. It does not do so in brain, renal tubules, intestinal mucosa and RBCs.

705.

Kepler space telescope generated a plot of intensity of radiation versus wavelength of three stars in Andromeda galaxy. What is the relation between their temperature?(a) TA > TB > TC(b) TC > TB > TA(c) TA > TC > TB(d) TB > TC > TA

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Answer : (c) TA > TC > TB

Wien’s law, 

λm ∝ \(\frac{1}{T}\) 

And from the figure,

m)< (λm)< (λm)B.

Therefore, 

TA > TC > TB.

706.

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10 s it will decrease to α times its original magnitude, then α is equal to :(a) 0.7 (b) 0.81 (c) 0.729 (d) 0.6

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Answer : (c) 0.729 

The amplitude of damped oscillator is,

A = A0e-kt

Where k = \(\frac{b}{2m}\) 

At t = 5s, 

0.9 A0 = A0e-5k

⇒ e-5k = 0.9

At t = 15s, 

αA0 = A0e-15k

⇒ e-15k = α

⇒ e(-5k)3 = α

⇒ α = (0.9)3

= 0.729

707.

Identify the incorrectly matched pair.(a) Chlorophyceae–Chlorophyll-a and b, carotenoids(b) Phaeophyceae–Chlorophyll-a and c, fucoxanthin(c) Rhodophyceae–Chlorophyll-a, phycocyanin, phycoerythrin, xanthophyll(d) None of the above

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Answer is : (d)

All the pairs are correctly matched. The photosynthetic pigments in Chlorophyceae are chlorophyll-a and b carotenoids.

The photosynthetic pigments in Phaeophyceae are chlorophyll-a and c, fucoxanthin.

The photosynthetic pigments in Rhodophyceae are chlorophyll-a, phycocyanin, phycoerythrin and xanthophyll.

708.

The MN blood group in humans is an example of(a) partial dominance(b) codominance(c) incomplete dominance(d) epistasis

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Answer is : (b) codominance

The MN blood group in humans is an example of codominance. In codominance, there is joint expression of both alleles in a heterozygote. In this situation, two alleles of a single gene are responsible for producing two distinct and detectable gene products.

709.

Which of the following category of genes found in normal cells have an antiproliferative function?(a) Oncogenes(b) Proto-oncogenes(c) Retro genes(d) Tumour-suppressor genes

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Answer is : (d) Tumour-suppressor genes

Tumour-suppressor genes (e.g. p53 gene) are very important to maintain the genomic integrity. They help in stopping division of cells having damaged DNA.

710.

In the glomerulus of nephron, the afferent arteriole is(a) longer than efferent arteriole(b) of same diameter as efferent arteriole(c) wider than efferent arteriole(d) narrower than efferent arteriole

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Answer is : (c) wider than efferent arteriole

In glomerulus, the afferent arteriole is shorter and wider than efferent arteriole which is longer and narrower.

711.

K-T boundary refers to :(a) mass extinction of 60 million years ago when dinosaurs disappeared (b) deposits of iridium which are rare on earth (c) Both (a) and (b)(d) None of these

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Answer is : (c) Both (a) and (b) 

K-T boundary refers to the mass extinction of 60 million years ago when dinosaurs disappeared. 

It is connected with deposits of iridium which are rare on the earth.

712.

Match the Column I with Column II and choose the correct option from the codes given below.Column I Column IIA. Phycophages1. Fungal virusesB. Mycophages2. Carry enzyme reverse transcriptaseC. Retroviruses3. Benign and malignantD. Tumour viruses4. Viruses which are parasitic on algae CodesABCD(a)1234(b)4321(c)4123(d)1324

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Answer is : (c)

Phycophages are the viruses which are parasitic on algae. 

Mycophages are fungal viruses. 

Retroviruses carry enzyme reverse transcriptase. 

They have RNA genome. 

Tumour viruses cause benign and malignant tumours in animals and humans.

713.

Which of the following cranial nerves is incorrectly matched with its nature? (a) Olfactory – Sensory (b) Trochlear – Motor (c) Glossopharyngeal – Mixed (d) Hypoglossal – Sensory

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Answer is : (d) Hypoglossal – Sensory

Among the given options, option (d) is incorrectly matched. 

Hypoglossal is a motor nerve which controls the movements of the tongue. It originates from the ventral side of the medulla oblongata.

714.

Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen ?(1) Pseudomonas(2) Mycoplasma(3) Nostoc(4) Bacillus

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The correct answer is (2)

Mycoplasmas are smallest, prokaryotes lacking cell wall and are pleomorphic in nature. These are pathogenic to both plants and animals.

715.

Which of the following represents order of Horse' ?(1) Perissodactyla(2) Caballus(3) Ferus(4) Equidae

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The correct answer is (1)

Horse belongs to 

Orders - Perissodactyla
Family - Equidae
Genus - Equus
Species - ferus
Subspecies - caballus

716.

A solid cylinder of radius 5 cm and mass 300 g rolls down an inclined plane (1 in 20). The velocity of cylinder after 5s will be(a) 1 63. ms−1(b) 1 56. ms−1(c) 2 ms−1(d) 3 26. ms−1

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Answer is : (a) 1 63. ms−1

Acceleration with which the cylinder rolls down,

\(a=\cfrac{g\,sin\theta}{1+\frac{l}{MR^2}}=\cfrac{g\,sin\theta}{1+\frac{\frac{1}{2}MR^2}{MR^2}}\) \(=\frac{2}{3}g\,sin\theta\) \(=\frac{2}{3}\times 9.8\times \frac{1}{20}\) \(=0.326\,ms^{-2}\)

Using first equation of motion,

v = u + at \(= 0 + 0.326\times 5\) \(=1.63\,ms^{-1}\)

717.

The work done by a gas is maximum, when it expands(a) isothermally(b) adiabatically(c) isentropically(d) isobarically

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Answer is : (d) isobarically

Work done is maximum, when process involved is isobaric.

718.

When a charged particle moving with velocity v is subjected to a magnetic field of induction B, the force on it is non-zero. This implies that(a) angle between them is either zero or 180º(b) angle between them can have any value other than zero or 180º(c) angle between them is necessary 90º(d) angle between them can have any value other than 90º

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Answer is : (b) angle between them can have any value other than zero or 180º

As, F = q(v x B\(\Rightarrow\) F = qvB \(sin\,\theta\)

At \(\theta\) = 0°, F = 0

At \(\theta\) = 90°, F = qvB

 At \(\theta\) = 180°, F = 0

So, for non-zero force angle between v and B can have any value other than zero or 180°.

719.

Radius of an air bubble at certain depth of Indian ocean is r and it becomes 18r, when air bubble rises to the top surface of the ocean. If t cm of water be the atmospheric pressure, then the depth of the ocean is(a) 3835 t cm(b) 3400 t cm(c) 4852 t cm(d) 5831t cm

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Answer is : (d) 5831t cm

Let the depth of Indian ocean is x cm

\(\therefore p_1V_1=p_2V_2(tdg+xdg)\)\(\left(\frac{4}{3}\pi r^3\right)\) \(=tdg\left[\frac{4}{3}\pi (18r)^3\right]\)

(t + x) = t x 183

183t - t = x 

So, x = t 5831 cm = 5831 t cm

720.

A battery of emf 8V and internal resistance r is connected to a load resistance of 5Ω. Then, power at load resistance will be maximum if the value of internal resistance r is(a) 5 Ω(b) 10 Ω(c) 15 Ω(d) 20 Ω

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Answer is : (a) 5 Ω

Output power will be maximum, if internal resistance of battery is equal to external resistance or load resistance, hence r = 5\(\Omega\).

721.

The magnetic flux through a circuit of resistance R changes by an amount ∆φ in a time ∆t. Then, the total quantity of electric charge Q that passes at any point in the circuit during the time ∆t is represented by(a) \(\frac{\triangle \varphi}{R}\)(b) \(\frac{1}{R}\frac{\triangle \varphi}{\triangle t}\)(c) \(R\frac{\triangle \varphi}{\triangle t}\)(d) \(\frac{\triangle \varphi}{\triangle t}\)

Answer»

Correct option is : (a) \(\frac{\triangle \varphi}{R}\)

Induced emf is given by

Eind = \(\frac{\triangle \varphi}{\triangle t}\)

Current, i = \(\frac{Q}{\triangle t}\)

\(=\frac{\triangle \varphi}{\triangle t}\times \frac{1}{R}\)

\(\Rightarrow Q=\frac{\triangle \varphi}{R}\)

722.

A well with vertical side and water at the bottom resonates at 3 Hz and at no other lower frequency. The air in the well has density 1.10 kg m-3 and bulk modulus of 1.32 x 105 Nm-2 Nm, then the depth of well is(a) 30 m(b) 29 m(c) 25 m(d) 32 m

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Answer is : (b) 29 m

For air, \(\rho\) = 1.10 kgm-3

K = 1.32 x 105 Nm-2

Speed of sound in air,

v = \(\sqrt{\frac{K}{\rho}}\)

\(\sqrt{\frac{1.32\times 10^5}{1.10}}\)

= 3.46 x 102

= 346 ms-1

It is similar to a closed organ pipe, hence fundamental frequency of well is

v = \(\frac{v}{4L}=\frac{346}{4\times L}\)

\(\Rightarrow L=\frac{346}{4\times 3}\) = 29 m

723.

In the case of free expansion,(i) ∆W = 0(ii) ∆Q = 0(iii) ∆U = 0(iv) ∆T = 0Correct statements are(a) (iii) and (iv)(b) (i), (ii) and (iii)(c) (i) and (iv)(d) (i), (ii), (iii) and (iv)

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Answer is : (d) (i), (ii), (iii) and (iv)

All thermodynamic changes are zero.

724.

A brass sphere of mass 5 kg is heated in a furnance to a temperature 500°C and then placed on a large ice block. The mass of ice that will melt in this process will be (specific heat of brass = 500 J kg-1 °C-1 and heat of fusion of ice = 336 kJ kg-1 )(a) 5.25 kg(b) 3.86 kg(c) 2.56 kg(d) 3.72 kg

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Answer is : (d) 3.72 kg

Fall in temperature of brass sphere,

\(\triangle T\) = 500 - 0 = 500°C

Heat loss by sphere,

Q = \(ms\triangle T\) = 5 x 500 x 500

= 1250 kJ

Heat for melting m2 kg of ice,

Q2 = m2L = m2 x 336

From principle of calorimetry,

= 1250 kJ = m2 x 336

m2\(\frac{1250}{336}\) = 3.72 kg

725.

A 6 µF capacitor is charged to 360 V. If its plates are joined through a resistance, then heat produced in the resistor is(a) 0.78 J(b) 0.68 J(c) 1.2 J(d) 0.39 J

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Answer is : (d) 0.39 J

Heat produced in the resistor

= Energy of capacitor = \(\frac{1}{2}CV^2\)

\(\frac{1}{2}\) x 6 x 10-6 x 360 x 360

= 0 39 J

726.

A planet of mass m revolves around the sun of mass M in an elliptical orbit. The maximum and minimum distances of the planet from the sun are r1 and r2, respectively. The time period of the planet in terms of r1 and r2 is(a) T ∝ (r1 + r2)2(b) T ∝ (r1 + r2)3(c) T ∝ (r1 + r2)1/2(d) T ∝ (r1 + r2)3/2

Answer»

Answer is : (d) \(T\propto (r_1+r_2)^{\frac{3}{2}}\)

The semi-major axis of the elliptical orbit around the sun is given by

\(r=\frac{r_1+r_2}{2}\)

According to Kepler’s third law,

\(T^2\propto r^3\)

\(\Rightarrow T^2\propto (\frac{r_1+r_2}{2})^3\)

\(or\: T\propto (\frac{r_1+r_2}{2})^\frac{3}{2}\)

\(or\:T\propto (r_1+r_2)^{\frac{3}{2}}\)

727.

The gravitational field due to a mass distribution is I =  k/r3 in the x - direction (k is a constant). The gravitational potential is taken to be zero at infinity, then its value at a distance x is(a) k/x(b) k/2x(c) k/x2(d) k/2x2

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Answer is : (d) \(\frac{k}{2x^2}\)

As, I = \(\frac{-dV}{dr}\)

⇒ dV = - ldr

So, \(\int\limits^v_0 dV\) = \(\int\limits^x_0\) - ldr = \(\int\limits^x_0\) - kr-3 dr

⇒ V = - k\(\left(\cfrac{r^{-3+1}}{-3+1}\right)^x_0\)

\(\frac{k}{2x^2}\)