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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 401. |
A current is induced in coil `C_1` due to the motion of current carrying coil `C_2`. (a) Write any two ways by which a large deflection can be obtained in the galvanometer G. (b) Suggest an altermative device to demonstrate the induced current in place of galcanometer. |
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Answer» (a) (i) Two ways are : When coil `C_1 and C_2` held stationary. Coil `C_1` is connected to a sensitive galcanometer G and coil `C_2` is connected to a battery through a tapping key K. (ii) When the key is released, the galvanometer shows again a momentary deflection, but in the opposite direction. The pointer in the galvanmeter returns to zero almost instantly. (b) The galvanometer deflection increases dramatically, when an iron rod is inserted into the coils along their exis. |
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| 402. |
Write the relation for the speed of electromagnetic wave in terms of the amplitudes of electric and magnetic fields. |
| Answer» Correct Answer - `C = C_(0)//B_(0)` | |
| 403. |
Two elelctric bulbs P and Q have their resistance in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. |
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Answer» Let P=x and Q= 2x Given `P : Q = x : 2x - 1:2` For bulb P power loss `P_(1) = (V_(2))/(R_(1)) = (V_(2))/(P)` `P_(1) = (V^(2))/(x)` Similarly `P_(2) = (V^(2))/(Q) = (V^(2))/(2x)` Ratio of power loss `(P_(1))/(P_(2)) = (cancel(V)^(cancel(2))/(cancel(H)))/((cancel(V)^(2))/(2cancel(H))` `P_(1) : P_(2) = 2:1` |
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| 404. |
Write the IUPAC name of the following compounds: |
| Answer» 3-Bromo-2-methyl propene | |
| 405. |
Name the electromagentic readiations used for (a) water puification, and (b) eye surgery. |
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Answer» (a) U.V Rays are used for water purification. (b) Infrared Rays are used for eye surgrey. |
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| 406. |
Name one autosomal dominant and one autosomal recessive. Mendelian disorder in human. |
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Answer» Autosomal dominant mendelian disorder : Poliycystic kidney disease. Autosomal recessive mendelian disorder : Sickle cell anamia |
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| 407. |
Name the defect in the following crystal: |
| Answer» Correct Answer - Schotty defect | |
| 408. |
What is the difference between a complex and a double salt ? |
| Answer» Both double salt as well sas complexes are formed by the combination of two or more stabel compounds in stoichiometic reatio.However they differ in the fact that double salt disssociate ito simple ions cmpletely whin dissolved in water .However complex ions do not simple ions completely. | |
| 409. |
When a coordination compound `CrCl_3`.`6H_2O` is mixed with `AgNO_3` two moles of AgCl are precipitated per mole of the compound .What is the structural formula of the coordination compound ? |
| Answer» `[CrCl(H_2O)_5]Cl_2.H_2O.` | |
| 410. |
Write the IUPAC name of `CH_3-CH-CH_2-CH=CH_2` |
| Answer» `CH_3-CH-CH_2-CH=CH_2` IUOAC Name : 4 - Chloropent -1-ene. | |
| 411. |
Mention the types of evolution that has brought the similarity as seen in potato tuber and sweet potato |
| Answer» Convergent evolution | |
| 412. |
Derive an expression for the magnetic moment `(vec mu)` of an electron revolving around the nucleus in termsof its angular momentum `(vecl)`. What is the direction of the magnetic moment of the electron with respect to its angular momentum? |
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Answer» We have `mu=iA(ev)/(2pir)xxpir^(2)=(evr)/(2)` `"Since "l=mvr` `"or "vr=(1)/(m)" "therefore(vec mu)=-(evecl)/(2m)" "[therefore" electron has a negative charge"]` The direction of `(vec mu)` is opposite to that of `vecl`, because of the negative charge of the electron. |
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| 413. |
Why is the polar region not a suitable habitat for tiny humming birds ? |
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Answer» Small animals have larger surface area relative to their volume, they tend to loss body heat very fast when it is cold outside, then they have to expend much energy to generate body heat through metabolism. So, Polar regions are not a suitable habitat for tiny humming birds. |
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| 414. |
Assign a reasons for the following . (i) Copper (I) ion is not known to exist in aqueous solutions. (ii) Both `O_(2)` and `F_(2)` stabilize high oxidation states of transition metals but the ability of oxygen to do so exceeds that of fluorine. |
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Answer» (i) Cu (I) are unstable in aqueous soluton and under go disproportionation `2Cu^(+) to Cu^(2+) + Cu` `Cu^(2+)` is stable due to more (-) ve `Delta_("Hyd")H` of `Cu^(2+)`(aq) that compensate the 2nd I.E. of Cu. (ii) This is due to the tendency of oxygen to form multiple bonds. |
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| 415. |
When `MnO_(2)`, is fused with KOH in the presence of `KNO_(3)` as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidises Kl to compound (C)whereas an acidified solution of compound (B) oxidises KI to (D), Identity (A), (B), (C), and (D). |
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Answer» `2MnO_(2)+4KOH+O_(2)tounderset((A))(2K_(2)MnO_(4))+2H_(2)O` `underset("(Green)")underset("Manganate")(MnO_(4)^(2-))+H_(2)O+KItounderset("Purple Compound")(2MnO_(4)^(-))+MnO_(2)+2H_(2)O` `2MnO_(4)^(-)+H_(2)O+KIto2MnO_(2)+2OH^( -)+underset((C) )(KIO_(3))` `2KMnO_(4)+3H_(2)SO_(4)+10KItoK_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5I_(2)` |
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| 416. |
Write IUPAC name of the complex [Pt(en),CI]. Draw structures of geometrical isomers for this complex. |
| Answer» Dichloridobis (ethan-1, 2-diamine) Platinum (II). | |
| 417. |
Explain the method of preparation of sodium dichromate from chromite ore . Give the equation representing oxidation of ferrous salts by dichromate ion. |
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Answer» Potassium dischromate is prepared from chromite are `FeCr_2O4` in the following steps. Step : 1 : Preparation of sodium chromate `4FeCr_O_4+16NaOH +7O_2 rarr 8 NaCrO_4+2 Fe_2O_3+8H_2O` Step :2: Conversion of sodium chromate into sodium dichromate `2Na_2CrO_4+Conc.H_2SO_4rarr Na_2Cr_2O_7+Na_2SO_4+H_2 O` Oxidation of ferrous slats by dichromate ion `Cr_2O_7^(2-)+6Fe^(3+)+7H_2O` |
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| 418. |
Calculate the temperature at which a solution containing 54g of glucose, `(C_(6)H_(12)O_(6))` in 250g of water will freeze. (`K_(f)` for water = 1.86 K `mol^(-1)` kg) |
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Answer» Molecular mass of glucose, `M_(B) = 72 + 12 + 96 = 180` `DeltaT_(f)=(K_(f)xxW_(B)xx1000)/(M_(B)xxW_(A)) = (1.86 xx 54 xx 1000)/(180xx250)=(100440)/(45000)=2.23` Freezing point of solution = 0 - 2.23 = - 2.23 `.^(@)C` |
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| 419. |
(a) Calculate the freezing point of solution when 1.9 g of `MgCl_(2)` (M = 95 g `Mol^(-1)`) was dissolved in 50g of water, assuming `MgCl_(2)` undergoes complete ionization. (`K_(f)` for water = 1.86 K kg `mol^(-1)`). (b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why? (ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution? |
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Answer» (a) Molar Mass of `MgCl_(2)` = `95g//mol`. (MB) `W_(B) = 1.9g` `W_(A) = 50g`, `K_(f) = 1.86K kg//mol` i=3 `DeltaT_(f)=iK_(f)xxm` `overset(@)(T)_(A) - T_(S) = (KfxxW_(B))/(M_(B)xxW_(A)(Kg))xxl` `O-T_(s) = (1.86 xx 1.9 xx 1000)/(95 xx 50) xx 3` `-T_(S) = 2.232` `T_(S) = -2.232 .^(@)C` (b) (i) 2M glucose has a higher boiling point than 1M glucose because 2M glucose has 2 moles of glucose, as no, of moles increases, boiling increases. (ii) When external pressure applied, becomes more than the osmotic pressure of solution the solvent will flow from the solution into the pure solvent through semi-permeable membrane. Which is called reverse osmosis. Which is used in the desalination of sea water. |
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| 420. |
Define the following with suritable example of each ( a) Antiseptics ( b) Non-narocotic analgesics (c ) Cationic detegents |
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Answer» see Q.27 (iii) Paper - 2009 Delhi Board ,Set-II [page 45] ( b) See Q 27 (i) Paper -2010 Delhi Board ,Set -I [ page 73] ( c) See Q. 27 ( i) Paper -2009 Delhi Board ,Set-I [ page 36] |
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| 421. |
(i)Write equation for preparation of 1-iodobutane from 1- chlorobutane (ii) Out of 2 - bromopentane 2- bromo-2 methylbutane and 1- bromopentane ,which compound is most reactive towards elimination and why ? ( iii) Give IUPAC name of `CH_3-CH=CH-underset(Br)underset(|)overset(CH_3)overset(|)C-CH_3` |
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Answer» `CH_3CH_2CH_2CH_2-Clunderset("acetone")overset("Nal")rarrCH_3CH_2CH_2CH_2-I` (b) 2-bromopentane is most reactive towards `Beta`- elemention raction with product of pent -1-ene ,pent 2-ene. ( c) 4-bromo -4-Methylpent -2-ene` |
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| 422. |
How will you synthesis the following alcohol from appropriate alkene: Write the mechanism of the following reation : ` `CH_3CH_2OHunderset(443K)overset(H^+)rarrCH_2= CH_2+H_2O` |
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Answer» `(a) CH_2=overset(CH_3)overset(|)C-CH_2-CH_3+H-Ohoverset(H^+)rarrCH_3-underset(OH)underset(|)overset(CH_3)overset(|)C-CH_2-CH_3` ( b) Refer Q.17 (i) Paper - 2018 ( C) [ page 478] |
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| 423. |
Calculate the freezing point of an aqueous solution containing `10.5g` of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of magnesium bromide `=184 g mol ^(-1)`, for water `=1.86K kg mol^(-)`). |
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Answer» `MgBr_2to Mg^(2+)+2Br^(-)` Complete dissociation. Hence `i=3` `W_8=10.5g` `W_(A)=200g` `K_f=1.86g kg//mol` `M_B=184g//mol` `Delta =ixxK_(f)xx(W_(B))/(M_(B)xxW_(A)(Kg))` `T_(A)^(0)-T_(S)=(3xxoverset(0.93)cancel1.86xx10.5xxoverset(cancel5)cancel1000)/(184xxcancel200xxcancelunderset(2)(10))` `Delta T_(f)=0.15K` `T_(S)=T_(A)^(0)-0.15=273.15-0.15` `T_(S)=273K` |
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| 424. |
Two cells of emfs 1.5 V and 2.0v having internal resistance 0.2`Sigma` and 0.3 `Sigma` respectively are connected in parallel . Calculate the emf and internal resistance of the equivalent cell . |
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Answer» `E_(1)= 1.5 V " ", " "E_(2) =2V` `r_(1) = 0.2 Sigma " ", " "r_(2) =0.3 Omega` `(1)/(r_(eq) )=(1)/(r_(1))+(1)/(r_(2))=(1)/(0.2)+(1)/(0.3)=(0.5)/(0.06)` ` r_( e q) =(0.06)/(0.5) =(6)/(50)Omega` ` E_( eq) =[(E_(1))/(r_(1)) +(E_(2))/(r_(2))].r_(eq ).=[(1.5)/(0.2) +(2)/(0.3)] xx (6)/(50) =(0.45+0.4)/(0.06)xx(6)/(50)=(0.85)/(0.06) xx (6)/(50) =1.7V` |
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| 425. |
Write the IUPAC name of the following coordination compound `[NiCl_(4)]^(2-).` |
| Answer» Tetrachlorido nicklete (ii) ion | |
| 426. |
(a) Account for the following : (i) `Cu^(+)` is unstable in an aqueous solution. (ii) Transition metals form complex compounds. (b) Complete the following equation: `Cr_(2)O_(7)^(2-)+8H^(+)+3NO_(2)^(-)rarr` |
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Answer» (ii) Becuase of smaller size of their metal ions. Their high ionic charges. Vacand d-orbital so that these orbitals can accept lone pairs of e-donated byligonads and form coordinate bond. (b) `Cr_(2)O_(7)^(2-)+8H^(+)+3NO_(2)^(-)rarr3NO_(3)^(-)+2Cr_(3)^(+)+4H_(2)O` |
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| 427. |
Out of `CH_(3)OH` and, which one is more acidic ? |
| Answer» Phenol is more acidic that ethanol because of resonance oxygen has positive charge and tends to release Hydrogen radily. Hence phenol is more acidic. | |
| 428. |
Give the IUPAC name of the following compound : `CH_(3)-underset(CH_(3))underset(|)(C) " " = " " underset(Br)underset(|)(C)-CH_(2)OH` |
| Answer» `underset(2-"bromo-3- ethyl - but - 2-en-1-ol")(overset(4)(C)H_(3)-underset(CH_(3))underset(|)overset(3)(C) " " = " " underset(Br)underset(|)overset(2)(C)-overset(1)(C)H_(2)OH)` | |
| 429. |
A pollen grain in angiosperm at the time of dehiscence from an anther could be 2- celled or 3 celled . Explain . How are the cells placed within the pollen grain when shed at -a 2 - celled stage ? |
| Answer» When the pollen grain is mathure , it contains two cells , the vegetative cell and generative cell . The vegetative cell is bigger , has abundant food reserve and a large irregularly shaped nucleus . The generative cell is small and floats in the cytoplasm of the vegetative cell . it is spindle shaped with dense cytoplasm and a nucleus . in over 60 per cent of angiosperms , pollen grains are shed at this 2 - celled stage . in the remaining spescies , the generative cell divides mitotically to give rise to the male gametes before pollen gains are shed (3- celled stage). | |
| 430. |
(i) An a.c. Source of voltage `V = V_(0) " sin " omegat` is connected to a series combination of L,C and R . Use the phasor diagram to obtain expression for impedance of the circuit and phase angle between voltage and current . Find the condition when current will be phase with the voltage . What is the circuit in this condition called ? (ii) In a series LR circuit `X_(L) = R` and power factor of the circuit is `P_(1)`. When capacitor which capacitance C such that `X_(L) =X_(C)` is put in series , the power factor becomes `P_(2)` .Calculate `P_(1)//P_(2)` |
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Answer» (i) from the phasor diagram `V^(2) =V_(4)^(2) + (V_(L) -V_(C ))^(2)` `rArr V = sqrt(V_(R )^(2)+(V_( C) -V_(C ))^(2))` ` V_( R ) =R I , V_( C ) =X_( C ) I , V_(L ) = X_(L ) I` `rArr V = sqrt(R^(2) I^(2) +(X_(C ) -X_(L))^(2)I^(2))=I sqrt(R^(2) (X_( C) -X_(L)))^(2)` ` (V)/(I)= sqrt(R^(2)(X_(C ) -X_(L))^(2)) rArr Z = sqrt(R^(2)( X_(C ) - X_(L)))^(2)` This is required impedance of the circuit . Now ` " tan " phi = (V_(C ) -V_(L))/(V_(R )) =(IX_(C ) -I X_(L))/(IR)` `" tan " phi = (X_(C ) -X_( L))/(R )` When `X_( C) = X_(L)` impedance of circuit beomes minimum and current becomes maximum this condition is called condition of resonance `X_(C ) - X_(L) " " rArr omegaC = 1//omegaL rArr omega^(2) = (1)/(LC ) rArr omega = sqrt((1)/(LC))` this is resonance fruquency (ii) For LR series circuit . , Power factor `P_(1)=( R)/(Z) = (R )/(sqrt(X_(L)^(2) +R^(2))) =(R )/(sqrt(R^(2) +R^(2))) =(R )/(sqrt(2R^(2))) =(1)/(sqrt(2))` For LCr series `P_(2) = (R )/(sqrt((X_(L) -X_(C))^(2)+R^(2)) )=(R )/(sqrt((R-R)^(2) +R^(2)))=1 rArr P_(1): P_(2) =1 : sqrt(2)` |
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| 431. |
State one factor which determines the intensity of light in the photon picture of light. |
| Answer» The number of photons per unit volume at a given point is called photon density. It determines the intensity of light in photon picture of light. Number of photons falling per second per unit area on a surface is called intensity. | |
| 432. |
State the principle of working of a transformer. (b) Define efficiency of a transformer. (c ) State any two factors that reduce the efficiency of a transformer. (d) Calculate the current drawn by the primary of a `90%` efficient transformer which steps down 220 V to 22 V , if the output resistance is 440 `Omega`. |
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Answer» (a) Principle : It works on the principle of mutual induction. (b) Definition of efficiency : It is ratio of output power to input power. `n=("Output power")/("Input power")=(V_(s)I_(s))/(V_(p)I_(p))` (c ) Two factors which reduce the efficiency of transformer : (i) Magnetic flux leakage (ii) Eddy current (iii) Magnetic hysteresis / Resistance of winding (d) `V_(s)=22V,n=90%=0.90,R_(0)=440Omega, V_(s)=I_(s)R_(0)` `I_(s)=(V_(s))/(R_(0))=(22)/(440)=(1)/(20)A=(V_(s)I_(s))/(V_(p)I_(s))rArr0.90=(22xx(1)/(20))/(220xxI_(p))` `I_(p)=(22xx(1)/(20))/(220xx0.90)=(cancel(22))/(20)xx(1)/(cancel(220)_(10)xx0.9)=(1)/(200xx0.9)=(1)/(180.0)` `I_(p)=(1)/(180)A` |
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| 433. |
A porton and an electron travelling along parallel paths entre a regions of unifrom magnetic field, acting prependicular to their paths. Which of them will move in a circular path with higher frequncy? |
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Answer» `e^(-)` will move in circular path with higher frequecy. `a = (F)/(m) = qB (V)/(m)` for `e^(-)` it will be 1. |
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| 434. |
Identify and write the correct statement: (a) In Grasshopper males two sex chromosomes are X and Y types (b) In Grasshopper males there exist XO type of sex-determinants. |
| Answer» Correct Answer - (b) Correct. | |
| 435. |
Give one example each of a fungus which reproduces by : (a) building (b) conidia |
| Answer» (i) buds in yeast, (ii) conidia of penicillium | |
| 436. |
Explain the different phases of menstrual cycle and correlate the phases with the different levels, of ovarian hormones in a human female. |
| Answer» (a) same as Set, Delhi Board-2014 (Comptt). | |
| 437. |
(a) Mention the cause and the body system affected by ADA deficiency in humans. (b) Name the vector used for transferring ADA-DNA into the recipient cells in humans. Name the recipient cells. |
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Answer» (a) The cause is the defective gene not pmducing ADA. The immune system is affected. (b) A retroviral vector is used, recipient cells are lymphocytes. |
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| 438. |
What is amniocentesis? How is it misused ? |
| Answer» Amniocentesis is foetal sex determination test based on the chromosomal pattern in the amniotic find surrounding the developing embryo. It is misused as a tool for female foeticide. | |
| 439. |
Write the level of biodiversity represented by a mangrove. Give another example falling in the same level. |
| Answer» Mangroves represent ecological diversity. Another falling in the same level is rain forests. | |
| 440. |
The two special types of lymphocytes in humans are : |
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Answer» (i) B-lymphocytes : They produce antibodies in response to pathogen in blood stream. (ii) T-lymphocytes :They provide cell mediated immune response and help B-eells produce antibodies. T-eells are also involved in transplant related immune reactions. |
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| 441. |
Name the two gases contributing maximum to the green house effect. |
| Answer» Carbondioxide and Methane. | |
| 442. |
ive reasons for the following: (a) The human testes are located outside the abdominal cavity. (b) Some organisms like honey-bees are called parthenogenetic animals. |
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Answer» (a) The human testes are located outside the abdominal cavity. A pair of testes i: situated outside the abdominal cavity in a sac of skin called the scrotum. Scrotum keeps the testes at a temperature 2-2.5°C lower than the internal body temperature, which is necessary for the synthesis of sperms. (b) The phenomenon of development of female gamete directly into an individual without fertilisation is called Parthenogenesis e.g. rotifers, honey bees, lizards and birds. |
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| 443. |
Why are the enviromentalists worried about the considerable increase in the level of green house gases ? List the different green house gases other than carbon dioxide. |
| Answer» Increase in the level of green house has led to considerable heating of Earth leading to global warming. Scientists believe that this rise in temperature is the cause of harmful changes in the environment and resulting in old climatic changes (example :El Nino effect), thus leading to melting of Himalayan and polar ice caps. Over many years, this can result in a rise in sea level that can submerge many coastal areas. Methane, Nitrous oxide and chloroflurocarbons are the green house gases other than carbon dioxide. | |
| 444. |
Explain the mode of action of Eco RI. |
| Answer» Eco RI is a restriction end nuclease.It functions by inspecting the length of a DNA sequence.Once it finds its finds its specific recognition sequence, it binds to the DNA and cut the two strands of the DNA helix at specific points in their sugar-phosphate backbone. Eco RI recognizes a palindromic nucleotide sequence in the DNA. | |
| 445. |
Explain the laws that Mendel derived from his monohybrid crosses. |
| Answer» See Q.24 (a), Set-III., Outside Delhi, 2015. | |
| 446. |
Explain brood parasitism with the help of an example. |
| Answer» See Q.8, Set-II., Outside Delhi (Comptt.),2015. | |
| 447. |
Find the radius of curvature of the convex surface of a plan-convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5. |
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Answer» Her `f=0.3Omega m=30cm, R_1= oo (" for plane surface "), R_2 =?,mu =1.5` On applying Lensmaker formula , we have `therefore (1)/(f) =(mu-1)((1)/(R_1)-(1)/(R_2)) rArr (1)/(30) =(1.5-1)((1)/(oo) -(1)/(R_2))` `rArr (1)/(30) =0.5 (-(1)/(R_2))rArr R_2=-0.5xx 30 =-15 cm` `therefore` Radius of curved surface =15 cm. |
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| 448. |
How are primary and secondary immune responses carried out in the human body ? Explain. |
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Answer» The primary and secondary immune responses are carried out with the help of two special types of lymphocytes, the B-lymphocytes and the T-lymphocytes. B-lymphocytes : These produce an army of proteins in response to pathogens in blood. These are called antibodies. The different types of antibodies secreted are IgA, Igm, IgE and IgG. This response by the immune system is also called the humoral immune response. T-lymphocytes : These mediate the cell-mediated immunity (CMI).The cell mediated immune response in responsible for the graft refection. |
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| 449. |
Why is secondary immune response more intense than the primary immune response in humans ? |
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Answer» Secondary Immune response is much faster and much more intense than the primary response. This is because the body develops memory for the pathogen during the primary immune response. The secondary immune response is more effective in protecting us from disease. |
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| 450. |
Provide an instance where the population size of species can be estimated indirectly without actually in natural parks is based on pugmarks. |
| Answer» Tiger population in natural parks is based on pugmarks. | |