44727.Which of the following hormones can play a significant role in osteoporosis?
Estrogen and Parathyroid hormone
Progesterone and Aldosterone
Aldosterone and Prolactin
Parathyroid hormone and Prolactin
Explanation:
Estrogen promotes the activity of osteoblast and inhibits osteoclast. In an ageing female osteoporosis occurs due to deficiency of estrogen. Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.
Estrogen promotes the activity of osteoblast and inhibits osteoclast. In an ageing female osteoporosis occurs due to deficiency of estrogen. Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.
44728.Which of the following structures or regions is incorrectly paired with its functions?
Hypothalamus : production of releasing hormones and regulation of temperature, hunger and thirst.
Limbic system : consists of fibre tracts that interconnect different regions of brain; controls movement.
Medulla oblongata : controls respiration and cardiovascular reflexes.
Corpus callosum : band of fibers connecting left and right cerebral hemispheres.
Explanation:
Limbic system is emotional brain. It controls all emotions in our body but not movements.
Limbic system is emotional brain. It controls all emotions in our body but not movements.
44729.The amnion of mammalian embryo is derived from
mesoderm and trophoblast
endoderm and mesoderm
ectoderm and mesoderm
ectoderm and endoderm
Explanation:
The extraembryonic or foetal membranes are amnion, chorion, allantois and Yolk sac. Amnion is formed from mesoderm on outer side and ectoderm on inner side. Chorion is formed from trophoectoderm and mesoderm whereas allantois and Yolk sac membrane have mesoderm on outerside and endoderm in inner side.
The extraembryonic or foetal membranes are amnion, chorion, allantois and Yolk sac. Amnion is formed from mesoderm on outer side and ectoderm on inner side. Chorion is formed from trophoectoderm and mesoderm whereas allantois and Yolk sac membrane have mesoderm on outerside and endoderm in inner side.
44730.Hormones secreted by the placenta to maintain pregnancy are
hCG, hPL, progestogens, estrogens
hCG, hPL, estrogens, relaxin, oxytocin
hCG, hPL, progestogens, prolactin
hCG, progestogens, estrogens, glucocorticoids
Explanation:
Placenta releases human chorionic gonadotropic hormone (hCG) which stimulates the Corpus luteum during pregnancy to release estrogen and progesterone and also rescues corpus luteum from regression. Human placental lactogen (hPL) is involved in growth of body of mother and breast. Progesterone maintains pregnancy, keeps the uterus silent by increasing uterine threshold to contractile stimuli.
Placenta releases human chorionic gonadotropic hormone (hCG) which stimulates the Corpus luteum during pregnancy to release estrogen and progesterone and also rescues corpus luteum from regression. Human placental lactogen (hPL) is involved in growth of body of mother and breast. Progesterone maintains pregnancy, keeps the uterus silent by increasing uterine threshold to contractile stimuli.
44731.The difference between spermiogenesis and spermiation is
In spermiogenesis spermatozoa from sertoli cells are released into the cavity of seminiferous tubules, while in spermiation spermatozoa are formed.
In spermiogenesis spermatozoa are formed, while in spermiation spermatids are formed.
In spermiogenesis spermatids are formed, while in spermiation spermatozoa are formed.
In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules.
Explanation:
Spermiogenesis is transformation of spermatids into spermatozoa whereas spermiation is the release of the sperms from sertoli cells into the lumen of seminiferous tubule.
Spermiogenesis is transformation of spermatids into spermatozoa whereas spermiation is the release of the sperms from sertoli cells into the lumen of seminiferous tubule.
44732.The contraceptive ‘SAHELI’
is an IUD.
increases the concentration of estrogen and prevents ovulation in females.
blocks estrogen receptors in the uterus, preventing eggs from getting implanted.
is a post-coital contraceptive.
Explanation:
Saheli is the first non-steroidal, once a week pill. It contains centchroman and its functioning is based upon selective Estrogen Receptor modulation.
Saheli is the first non-steroidal, once a week pill. It contains centchroman and its functioning is based upon selective Estrogen Receptor modulation.
44733.Ciliates differ from all other protozoans in
using pseudopodia for capturing prey
having a contractile vacuole for removing excess water
using flagella for locomotion
having two types of nuclei
Explanation:
Ciliates differs from other protozoans in having two types of nuclei. eg. Paramoecium have two types of nuclei i.e. macronucleus & micronucleus.
Ciliates differs from other protozoans in having two types of nuclei. eg. Paramoecium have two types of nuclei i.e. macronucleus & micronucleus.
44734.Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system
Aves
Reptilia
Amphibia
Osteichthyes
Explanation:
The digestive tract of Aves has additional chambers in their digestive system as crop and Gizzard. Crop is concerned with storage of food grains. Gizzard is a masticatory organ in birds used to crush food grain.
The digestive tract of Aves has additional chambers in their digestive system as crop and Gizzard. Crop is concerned with storage of food grains. Gizzard is a masticatory organ in birds used to crush food grain.
44735.Which of the following features is used to identify a male cockroach from a female cockroach?
Forewings with darker tegmina
Presence of caudal styles
Presence of a boat shaped sternum on the 9th abdominal segment
Presence of anal cerci
Explanation:
Males bear a pair of short, thread like anal styles which are absent in females. Anal/caudal styles arise from 9th abdominal segment in male cockroach.
Males bear a pair of short, thread like anal styles which are absent in females. Anal/caudal styles arise from 9th abdominal segment in male cockroach.
44736.Which one of these animals is not a homeotherm?
Camelus
Chelone
Macropus
Psittacula
Explanation:
Homeotherm are animals that maintain constant body temperature, irrespective of surrounding temperature. Birds and mammals are homeotherm. Chelone (Turtle) belongs to class reptilia which is Poikilotherm or cold blood.
Homeotherm are animals that maintain constant body temperature, irrespective of surrounding temperature. Birds and mammals are homeotherm. Chelone (Turtle) belongs to class reptilia which is Poikilotherm or cold blood.
44737.Which of the following animals does not undergo metamorphosis?
Moth
Tunicate
Earthworm
Starfish
Explanation:
Metamorphosis refers to transformation of larva into adult. Animal that perform metamorphosis are said to have indirect development. In earthworm development is direct which means no larval stage and hence no metamorphosis.
Metamorphosis refers to transformation of larva into adult. Animal that perform metamorphosis are said to have indirect development. In earthworm development is direct which means no larval stage and hence no metamorphosis.
44738.Which of the following organisms are known as chief producers in the oceans?
Cyanobacteria
Diatoms
Dinoflagellates
Euglenoids
Explanation:
Diatoms are chief producers of the ocean.
Diatoms are chief producers of the ocean.
44739.Which one of the following population interactions is widely used in medical science for the production of antibiotics?
Parasitism
Mutualism
Commensalism
Amensalism
Explanation:
Amensalism/Antibiosis (0, –)
• Antibiotics are chemicals secreted by one microbial group (eg : Penicillium) which harm other microbes (eg : Staphylococcus)
• It has no effect on Penicillium or the organism which produces it.
Amensalism/Antibiosis (0, –)
• Antibiotics are chemicals secreted by one microbial group (eg : Penicillium) which harm other microbes (eg : Staphylococcus)
• It has no effect on Penicillium or the organism which produces it.
44740.All of the following are included in ‘ex-situ conservation’ except
Botanical gardens
Sacred groves
Wildlife safari parks
Seed banks
Explanation:
• Sacred groves – in-situ conservation.
• Represent pristine forest patch as protected by Tribal groups.
• Sacred groves – in-situ conservation.
• Represent pristine forest patch as protected by Tribal groups.
44741.Match the items given in Column I with those in Column II and select the correct option given below :
a b c d
| Column-I | Column-II |
|---|---|
| a. Eutrophication | i. UV-B radiation |
| b. Sanitary landfill | ii. Deforestation |
| c. Snow blindness enrichment | iii. Nutrient |
| d. Jhum cultivation | iv. Waste disposal |
a b c d
iii iv i ii
i iii iv ii
ii i iii iv
i ii iv iii
Explanation:
| a. Eutrophication | iii. Nutrient enrichment |
| b. Sanitary landfill | iv. Waste disposal |
| c. Snow blindness | i. UV-B radiation |
| d. Jhum cultivation | ii. Deforestation |
44742.In a growing population of a country,
reproductive and pre-reproductive individuals are equal in number.
reproductive individuals are less than the post-reproductive individuals.
pre-reproductive individuals are more than the reproductive individuals.
pre-reproductive individuals are less than the reproductive individuals.
Explanation:
Whenever the pre-reproductive individuals or the younger population size is larger than the reproductive group, the population will be an increasing population.
Whenever the pre-reproductive individuals or the younger population size is larger than the reproductive group, the population will be an increasing population.
44743.Which part of poppy plant is used to obtain the drug “Smack”?
Roots
Latex
Flowers
Leaves
Explanation:
‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant.
‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant.
44744.All of the following are part of an operon except
an enhancer
structural genes
an operator
a promoter
Explanation:
• Enhancer sequences are present in eukaryotes.
• Operon concept is for prokaryotes.
• Enhancer sequences are present in eukaryotes.
• Operon concept is for prokaryotes.
44745.A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by
Only grandchildren
Only sons
Only daughters
Both sons and daughters
Explanation:
• Woman is a carrier
• Both son & daughter inherit X–chromosome
• Although only son be the diseased
• Woman is a carrier
• Both son & daughter inherit X–chromosome
• Although only son be the diseased
44746.According to Hugo de Vries, the mechanism of evolution is
Phenotypic variations
Saltation
Multiple step mutations
Minor mutations
Explanation:
As per mutation theory given by Hugo de Vries, the evolution is a discontinuous phenomenon or saltatory phenomenon/ saltation.
As per mutation theory given by Hugo de Vries, the evolution is a discontinuous phenomenon or saltatory phenomenon/ saltation.
44747.AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?
ACCUAUGCGAU
UGGTUTCGCAT
AGGUAUCGCAU
UCCAUAGCGUA
Explanation:
Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil in mRNA.
Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil in mRNA.
44748.Match the items given in Column I with those in Column II and select the correct option given below :
a b c
| Column I | Column II |
|---|---|
| a. Proliferative Phase | i. Breakdown of endometrial lining |
| b. Secretory Phase | ii. Follicular Phase |
| c. Menstruation | iii. Luteal Phase |
a b c
ii iii i
i iii ii
iii ii i
iii i ii
Explanation:
During proliferative phase, the follicles start developing, hence, called follicular phase.
Secretory phase is also called as luteal phase mainly controlled by progesterone secreted by corpus luteum.
Estrogen further thickens the endometrium maintained by progesterone.
Menstruation occurs due to decline in progesterone level and involves breakdown of overgrown endometrial lining.
During proliferative phase, the follicles start developing, hence, called follicular phase.
Secretory phase is also called as luteal phase mainly controlled by progesterone secreted by corpus luteum.
Estrogen further thickens the endometrium maintained by progesterone.
Menstruation occurs due to decline in progesterone level and involves breakdown of overgrown endometrial lining.
44749.Match the items given in Column I with those in Column II and select the correct option given below :
a b c d
| Column I | Column II |
|---|---|
| a. Glycosuria | i. Accumulation of uric acid in joints |
| b. Gout | ii. Mass of crystallised salts within the kidney |
| c. Renal calculi | iii. Inflammation in glomeruli |
| d. Glomerular | iv. Presence of nephritis glucose in urine |
a b c d
ii iii i iv
i ii iii iv
iii ii iv i
iv i ii iii
Explanation:
Glycosuria denotes presence of glucose in the urine. This is observed when blood glucoselevel rises above 180 mg/100 ml of blood, this is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals in the joint.
Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney.
Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and haematuria.
Glycosuria denotes presence of glucose in the urine. This is observed when blood glucoselevel rises above 180 mg/100 ml of blood, this is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals in the joint.
Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney.
Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and haematuria.
44750.Match the items given in Column I with those in Column II and select the correct option given below:
a b c d
| Column I | Column II |
|---|---|
| (Function) | (Part of Excretory system) |
| a. Ultrafiltration | i. Henles loop |
| b. Concentration of urine | ii. Ureter |
| c. Transport of urine | iii. Urinary bladder |
| d. Storage of urine | iv. Malpighian corpuscle |
| v. Proximal convoluted tubule |
a b c d
v iv i ii
iv i ii iii
iv v ii iii
v iv i iii
Explanation:
Ultrafiltration refers to filtration of very fine particles having molecular weight less than 68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water absorption from glomerular filtrate as a result of hyperosmolarity in the medulla created by counter-current mechanism in Henles loop.
Urine is carried from kidney to bladder through ureter.
Urinary bladder is concerned with storage of urine.
Ultrafiltration refers to filtration of very fine particles having molecular weight less than 68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water absorption from glomerular filtrate as a result of hyperosmolarity in the medulla created by counter-current mechanism in Henles loop.
Urine is carried from kidney to bladder through ureter.
Urinary bladder is concerned with storage of urine.
44751.Which of the following gastric cells indirectly
help in erythropoiesis?
help in erythropoiesis?
Goblet cells
Mucous cells
Chief cells
Parietal cells
Explanation:
Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis.
Intrinsic factor is essential for the absorption of vitamin $B_{12}$ and its deficiency causes pernicious anaemia.
Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis.
Intrinsic factor is essential for the absorption of vitamin $B_{12}$ and its deficiency causes pernicious anaemia.
44752.Match the items given in Column I with those in Column II and select the correct option given below :
a b c
| Column I | Column II |
|---|---|
| a. Fibrinogen | (i) Osmotic balance |
| b. Globulin | (ii) Blood clotting |
| c. Albumin | (iii) Defence mechanisms |
a b c
(i) (iii) (ii)
(i) (ii) (iii)
(iii) (ii) (i)
(ii) (iii) (i)
Explanation:
Fibrinogen forms fibrin strands during coagulation. These strands forms a network and the meshes of which are occupied by blood cells, this structure finally forms a clot.
Antibodies are derived from $\gamma$-Globulin fraction of plasma proteins which means globulins are involved in defence mechanisms.
Albumin is a plasma protein mainly responsible for BCOP.
Fibrinogen forms fibrin strands during coagulation. These strands forms a network and the meshes of which are occupied by blood cells, this structure finally forms a clot.
Antibodies are derived from $\gamma$-Globulin fraction of plasma proteins which means globulins are involved in defence mechanisms.
Albumin is a plasma protein mainly responsible for BCOP.
44753.Which of the following is an occupational respiratory disorder?
Botulism
Silicosis
Anthracis
Emphysema
Explanation:
Silicosis is due to excess inhalation of silica dust in the workers involved grinding or stone breaking industries.
Long exposure can give rise to inflammationleading to fibrosis and thus causing serious lung damage.
Anthrax is a serious infectious disease caused by Bacillus anthracis. It commonly affects domestic and wild animals. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.
Botulism is a form of food poisoning caused by Clostridium botulinum.
Silicosis is due to excess inhalation of silica dust in the workers involved grinding or stone breaking industries.
Long exposure can give rise to inflammationleading to fibrosis and thus causing serious lung damage.
Anthrax is a serious infectious disease caused by Bacillus anthracis. It commonly affects domestic and wild animals. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.
Botulism is a form of food poisoning caused by Clostridium botulinum.
44754.Calcium is important in skeletal muscle contraction because it
Detaches the myosin head from the actin filament.
Activates the myosin ATPase by binding to it.
Binds to troponin to remove the masking of active sites on actin for myosin.
Prevents the formation of bonds between the myosin cross bridges and the actin filament.
Explanation:
• Signal for contraction increase Ca++ level many folds in the sarcoplasm.
• Ca++ now binds with sub-unit of troponin (troponin "C") which is masking the active site on actin filament and displaces the sub-unit of troponin.
• Once the active site is exposed, head of the myosin attaches and initiate contraction by sliding the actin over myosin.
• Signal for contraction increase Ca++ level many folds in the sarcoplasm.
• Ca++ now binds with sub-unit of troponin (troponin "C") which is masking the active site on actin filament and displaces the sub-unit of troponin.
• Once the active site is exposed, head of the myosin attaches and initiate contraction by sliding the actin over myosin.
44755.Nissl bodies are mainly composed of
Nucleic acids and SER
DNA and RNA
Proteins and lipids
Free ribosomes and RER
Explanation:
Nissl granules are present in the cyton and even extend into the dendrite but absent in axon and rest of the neuron. Nissl granules are in fact composed of free ribosomes and RER. They are responsible for protein synthesis.
Nissl granules are present in the cyton and even extend into the dendrite but absent in axon and rest of the neuron. Nissl granules are in fact composed of free ribosomes and RER. They are responsible for protein synthesis.
44756.Which of these statements is incorrect?
Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms
Glycolysis occurs in cytosol
Enzymes of TCA cycle are present in mitochondrial matrix
Oxidative phosphorylation takes place in outer mitochondrial membrane
Explanation:
Oxidative phosphorylation takes place in inner mitochondrial membrane.
Oxidative phosphorylation takes place in inner mitochondrial membrane.
44757.Select the incorrect match :
Submetacentric – L-shaped chromosomes chromosomes
Allosomes – Sex chromosomes
Lampbrush – Diplotene bivalents chromosomes
Polytene – Oocytes of chromosomes amphibians
Explanation:
Polytene chromosomes are found in salivary glands of insects of order Diptera.
Polytene chromosomes are found in salivary glands of insects of order Diptera.
44758.Which of the following terms describe human dentition?
Pleurodont, Monophyodont, Homodont
Thecodont, Diphyodont, Heterodont
Thecodont, Diphyodont, Homodont
Pleurodont, Diphyodont, Heterodont
Explanation:
In humans, dentition is
• Thecodont : Teeth are present in the sockets of the jaw bone called alveoli.
• Diphyodont : Teeth erupts twice, temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth.
• Heterodont dentition : Dentition consists of different types of teeth namely incisors, canine, premolars and molars.
In humans, dentition is
• Thecodont : Teeth are present in the sockets of the jaw bone called alveoli.
• Diphyodont : Teeth erupts twice, temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth.
• Heterodont dentition : Dentition consists of different types of teeth namely incisors, canine, premolars and molars.
44759.Which of the following events does not occur in rough endoplasmic reticulum?
Cleavage of signal peptide
Protein glycosylation
Protein folding
Phospholipid synthesis
Explanation:
Phospholipid synthesis does not take place in RER. Smooth endoplasmic reticulum are involved in lipid synthesis.
Phospholipid synthesis does not take place in RER. Smooth endoplasmic reticulum are involved in lipid synthesis.
44760.Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as
Plastidome
Polyhedral bodies
Polysome
Nucleosome
Explanation:
The phenomenon of association of many ribosomes with single m-RNA leads to formation of polyribosomes or polysomes or ergasomes.
The phenomenon of association of many ribosomes with single m-RNA leads to formation of polyribosomes or polysomes or ergasomes.
44761.In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?
Ringworm disease
Ascariasis
Elephantiasis
Amoebiasis
Explanation:
Elephantiasis is caused by roundworm, Wuchereria bancrofti and it is transmitted by Culex mosquito.
Elephantiasis is caused by roundworm, Wuchereria bancrofti and it is transmitted by Culex mosquito.
44762.Which of the following is not an autoimmune disease?
Alzheimers disease
Rheumatoid arthritis
Psoriasis
Vitiligo
Explanation:
Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune whereas Alzheimers disease is due to deficiency of neurotransmitter acetylcholine.
Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune whereas Alzheimers disease is due to deficiency of neurotransmitter acetylcholine.
44763.Among the following sets of examples for divergent evolution, select the incorrect option :
Brain of bat, man and cheetah
Heart of bat, man and cheetah
Forelimbs of man, bat and cheetah
Eye of octopus, bat and man
Explanation:
Divergent evolution occurs in the same structure, example - forelimbs, heart, brain of vertebrates which have developed along different directions due to adaptation to different needs whereas eye of octopus, bat and man are examples of analogous organs showing convergent evolution.
Divergent evolution occurs in the same structure, example - forelimbs, heart, brain of vertebrates which have developed along different directions due to adaptation to different needs whereas eye of octopus, bat and man are examples of analogous organs showing convergent evolution.
44764.Conversion of milk to curd improves its nutritional value by increasing the amount of
Vitamin $B_{12}$
Vitamin A
Vitamin D
Vitamin E
Explanation:
• Curd is more nourishing than milk.
• It has enriched presence of vitamins specially Vit-$B_{12}$.
• Curd is more nourishing than milk.
• It has enriched presence of vitamins specially Vit-$B_{12}$.
44765.The similarity of bone structure in the forelimbs of many vertebrates is an example of
Convergent evolution
Analogy
Homology
Adaptive radiation
Explanation:
In different vertebrates, bones of forelimbs are similar but their forelimbs are adapted in different way as per their adaptation, hence example of homology.
In different vertebrates, bones of forelimbs are similar but their forelimbs are adapted in different way as per their adaptation, hence example of homology.
44766.Which of the following characteristics represent ‘Inheritance of blood groups’ in humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
b, d and e
a, b and c
b, c and e
a, c and e
Explanation:
• $I^{A}I^{O}, I^{B}I^{O}$ - Dominant–recessive relationship
• $I^{A}I^{B}$ - Codominance
• $I^{A},I^{B}$ & $I^{O}$ - 3-different allelic forms of a gene (multiple allelism)
• $I^{A}I^{O}, I^{B}I^{O}$ - Dominant–recessive relationship
• $I^{A}I^{B}$ - Codominance
• $I^{A},I^{B}$ & $I^{O}$ - 3-different allelic forms of a gene (multiple allelism)
44767.The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is
$\dfrac{1}{3}$
$\dfrac{2}{3}$
$\dfrac{2}{5}$
$\dfrac{2}{7}$
Explanation:
Given process is isobaric
dQ n $C_{p}$ dT
dQ n $\left(\dfrac{5}{2}R\right)$ dT
dW = P dV = n RdT
Required ratio = $\dfrac{dW}{d Q}= \dfrac{nRdT}{n\left(\dfrac{5}{2}R\right)dT}= \dfrac{2}{5}$
Given process is isobaric
dQ n $C_{p}$ dT
dQ n $\left(\dfrac{5}{2}R\right)$ dT
dW = P dV = n RdT
Required ratio = $\dfrac{dW}{d Q}= \dfrac{nRdT}{n\left(\dfrac{5}{2}R\right)dT}= \dfrac{2}{5}$
44768.The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
12.5 cm
8 cm
13.2 cm
16 cm
Explanation:
For closed organ pipe, third harmonic = $\dfrac{3v}{4l}$
For open organ pipe, fundamental frequency = $\dfrac{v}{2\acute{l}}$
Given, $\dfrac{3v}{4l} = \dfrac{v}{2\acute{l}}$
$\Rightarrow\acute{l}=\frac{4l}{3*2}= \dfrac{2l}{3}$
= $\dfrac{2*20}{3}$ = 13.33cm
For closed organ pipe, third harmonic = $\dfrac{3v}{4l}$
For open organ pipe, fundamental frequency = $\dfrac{v}{2\acute{l}}$
Given, $\dfrac{3v}{4l} = \dfrac{v}{2\acute{l}}$
$\Rightarrow\acute{l}=\frac{4l}{3*2}= \dfrac{2l}{3}$
= $\dfrac{2*20}{3}$ = 13.33cm
44769.At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earths atmosphere?
(Given : Mass of oxygen molecule (m) = 2.76 × 10^{-26}kg
Boltzmanns constant $k_{B} = 1.38 × 10^{-23} JK^{-1})$
(Given : Mass of oxygen molecule (m) = 2.76 × 10^{-26}kg
Boltzmanns constant $k_{B} = 1.38 × 10^{-23} JK^{-1})$
5.016 × $10^{4} K$
8.360 × $10^{4} K$
2.508 × $10^{4} K$
1.254 × $10^{4} K$
Explanation:
$V_{escape}$ = 11200 m/s
Say at temperature T it attains $V_{escape}$
So, $\sqrt{\dfrac{3K_{B}T}{m_{O_{2}}}}=11200m/s$
On solving,
T= 8.360 * $10^{4}K$
$V_{escape}$ = 11200 m/s
Say at temperature T it attains $V_{escape}$
So, $\sqrt{\dfrac{3K_{B}T}{m_{O_{2}}}}=11200m/s$
On solving,
T= 8.360 * $10^{4}K$
44770.The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
6.25%
20%
26.8%
12.5%
Explanation:
Efficiency of ideal heat engine, $\eta = \left(1-\dfrac{T_{2}}{T_{1}}\right)$
$T_{2}$: Sink temperature
$T_{1}$: Source temperature
%$\eta = \left(1-\dfrac{T_{2}}{T_{1}}\right)$ * 100
= $\left(1-\dfrac{273}{373}\right)*100$
$\left(\dfrac{100}{373}\right)*100$=26.8%
Efficiency of ideal heat engine, $\eta = \left(1-\dfrac{T_{2}}{T_{1}}\right)$
$T_{2}$: Sink temperature
$T_{1}$: Source temperature
%$\eta = \left(1-\dfrac{T_{2}}{T_{1}}\right)$ * 100
= $\left(1-\dfrac{273}{373}\right)*100$
$\left(\dfrac{100}{373}\right)*100$=26.8%
44771.A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be
Yellow – Green – Violet – Gold
Yellow – Violet – Orange – Silver
Violet – Yellow – Orange – Silver
Green – Orange – Violet – Gold
Explanation:
(47 ± 4.7) kΩ = 47 ×$10^{3}$± 10%
$\therefore$ Yellow – Violet – Orange – Silver
(47 ± 4.7) kΩ = 47 ×$10^{3}$± 10%
$\therefore$ Yellow – Violet – Orange – Silver
44772.A set of n equal resistors, of value R each, are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of n is
20
11
10
9
Explanation:
$I = \dfrac{E}{nR+R} \cdot\cdot\cdot\cdot\left(i\right)$
$10I=\dfrac{E}{\dfrac{R}{n}+R} \cdot\cdot\cdot\cdot\left(ii\right)$
Dividing (ii) by (i),
$10=\dfrac{(n+1)R}{\left(\dfrac{1}{n}+1\right)R}$
After solving the equation, n = 10
$I = \dfrac{E}{nR+R} \cdot\cdot\cdot\cdot\left(i\right)$
$10I=\dfrac{E}{\dfrac{R}{n}+R} \cdot\cdot\cdot\cdot\left(ii\right)$
Dividing (ii) by (i),
$10=\dfrac{(n+1)R}{\left(\dfrac{1}{n}+1\right)R}$
After solving the equation, n = 10
44773.A battery consists of a variable number n of identical cells (having internal resistance r each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?
Explanation:
I = $I=\dfrac{n\epsilon}{nr}= \dfrac{\epsilon}{r}$
So, I is independent of n and I is constant.
$\therefore$
I = $I=\dfrac{n\epsilon}{nr}= \dfrac{\epsilon}{r}$
So, I is independent of n and I is constant.
$\therefore$
44774.Unpolarised light is incident from air on a plane surface of a material of refractive index $\mu$. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
$i=sin^{-1}\left(\dfrac{1}{\mu}\right)$
Reflected light is polarised with its electric vector perpendicular to the plane of incidence
Reflected light is polarised with its electric vector parallel to the plane of incidence
$i=tan^{-1}\left(\dfrac{1}{\mu}\right)$
Explanation:
When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.
Also, tan i = $\mu$ (Brewster angle)
When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.
Also, tan i = $\mu$ (Brewster angle)
44775.In Youngs double slit experiment the separation d between the slits is 2 mm, the wavelength $\lambda$ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same $\lambda$ and D) the separation between the slits needs to be changed to
2.1 mm
1.9 mm
1.8 mm
1.7 mm
Explanation:
Angular width = $\dfrac{\lambda}{d}$
0.20° = $\dfrac{\lambda}{2mm} \cdot\cdot\cdot\left(i\right)$
0.21° = $\dfrac{\lambda}{d} \cdot\cdot\cdot\left(ii\right)$
Dividing we get, $\dfrac{0.20}{0.21}= \dfrac{d}{2mm}$
$\therefore$ d = 1.9 mm
Angular width = $\dfrac{\lambda}{d}$
0.20° = $\dfrac{\lambda}{2mm} \cdot\cdot\cdot\left(i\right)$
0.21° = $\dfrac{\lambda}{d} \cdot\cdot\cdot\left(ii\right)$
Dividing we get, $\dfrac{0.20}{0.21}= \dfrac{d}{2mm}$
$\therefore$ d = 1.9 mm
44776.An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
Large focal length and large diameter
Large focal length and small diameter
Small focal length and large diameter
Small focal length and small diameter
Explanation:
For telescope, angular magnification = =$\dfrac{f_{0}}{f_{E}}$
So, focal length of objective lens should be large.
Angular resolution = $\dfrac{D}{1.22\lambda}$ should be large.
So, objective should have large focal length ($f_{o}$) and large diameter D.
For telescope, angular magnification = =$\dfrac{f_{0}}{f_{E}}$
So, focal length of objective lens should be large.
Angular resolution = $\dfrac{D}{1.22\lambda}$ should be large.
So, objective should have large focal length ($f_{o}$) and large diameter D.