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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.
| 1301. |
In which one of the following habitats does the diurnal temperature of soil surface vary most?A. ForestB. DesertC. GrasslandD. Shrub land |
| Answer» Correct Answer - B | |
| 1302. |
The wave described by `y = 0.25 "sin"(10 pi x - 2pit)` , where x and y are in metres and t in seconds , is a wave travelling along the:A. `+ ve` x direction with frequency 1 Hz and wavelength `lambda = 0.2 m`B. `–ve` x direction with amplitude 0.25 m and wavelength `lambda = 0.2 m`C. `– ve` x direction with frequency 1 HzD. `+ ve` x direction with frequency π Hz and wavelength `lambda= 0.2m` |
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Answer» Correct Answer - A Given `y=0.25 sin (10pix-2pit)` Comparing with equation of wave `y=A sin (kx-omegat)` `A=0.25, k=10pi, omega=2pi` `(2pi)/(lambda)=10pi" "2pif=2pi` `f=1Hz` `lambda=(1)/(5)=0.2m` When sign of coefficient of t and x are opposite it means `(dx)/(dt)=V gt 0` i.e., wave is propagating in the direction of growing x. |
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| 1303. |
A wave travelling in the `+ve` x-direction having displacement along y-direction as 1`m`, wavelength `2pi` m and frequency of `1//pi` Hz is represented byA. `y=sin(x-2t)`B. `y=sin(2pi x - 2pi t)`C. `y=sin(10pi x-20 pi t)`D. `y=sin(2pi x+2pi t)` |
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Answer» Correct Answer - A `y=a sin (omegat-kx)` `" "`OR `y=a sin(kx-omegat)` `therefore y=sin[x-2t] " "`...(1) `k=(2pi)/(lambda)=(2pi)/(2pi)` `omega=2pi.v=2pi.(1)/(pi)=2` `a=1m` |
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| 1304. |
Each of the two strings of length `51.6 cm` and `49.1 cm` are tensioned separately by `20 N` force. Mass per unit length of both the strings is same and equal to `1 g//m`. When both the strings vibrate simultaneously, the number of beats isA. 3B. 5C. 7D. 8 |
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Answer» Correct Answer - C `Deltav=(V)/(2l_(1))-(V)/(2l_(2))=(V)/(2)[(1)/(l_(1))-(1)/(l_(2))]` `=(1)/(2)sqrt((T)/(mu)){(1)/(l_(1))-(1)/(l_(2))}` |
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| 1305. |
Which one of the following equations of motion represents simple harmonic motion ?A. Acceleration = kxB. Acceleration `= = k_(0)x + k_(1)x^(2)`C. Acceleration = – k (x + a)D. Acceleration = k(x + a) |
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Answer» Correct Answer - C In SHM, `F_("restoring") prop -x ` |
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| 1306. |
A wave in a string has an amplitude of `2 cm`. The wave travels in the `+ve` direction of x axis with a speed of 1`28 ms^-1` and it is noted that `5` complete waves fit in `4 m` length of the string. The equation describing the wave isA. y = (0.02)m sin (7.58x – 1005 t)B. y = (0.02)m sin (7.85x + 1005 t)C. y = (0.02)m sin (15.7x – 2010 t)D. y = (0.02)m sin (15.7x + 2010 t) |
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Answer» Correct Answer - A `5 lambda=4 ` ` lambda=(4)/(5)` `k=(2 pi)/ lambda=(10pi)/(4)=7.85` wave moves along positive X-direction |
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| 1307. |
The electric field part of an electromagnetic wave in a medium is represented by `E_(x)=0,` `E_(y)=2.5(N)/(C) cos[(2pixx10^(6)(rad)/(m))t-(pixx10^(-2)(rad)/(s))x]` `E_(z)=0`. The wave isA. Moving along –x direction with frequency `10^(6)` Hz and wavelength 200mB. Moving along y direction with frequency `2pi xx10^(6)` Hz and wavelength 200mC. Moving along x direction with frequency `10^(6)` Hz and wavelength 100mD. Moving along x direction with frequency `10^(6)` Hz and wavelength 200m |
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Answer» Correct Answer - D As the coefficient of x is negative , it is moving along +ve x-axis and equating the equation `E_(y)=2.5 cos [(2 pi xx 10^(6))t-(pixx10^(-2))x]` with `y=A cos (omegat-kx)` `omega=2 pi xx 106` `implies f=(omega)/(2pi)=10^(6)Hz ` `k= pi xx 10^(-2)` ` lambda=(2pi)/(k)` `=(2pi)/(pi xx 10^(-2))=200m` |
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| 1308. |
In any `AC` circuit the emf `(e)` and the current `(i)` at any instant are given respectively by `e= E_(0)sin omega t` `i=I_(0) sin (omegat-phi)` The average power in the circuit over one cycle of `AC` isA. `(E_(0)I_(0))/(2)cosphi`B. `E_(0)I_(0)`C. `(E_(0)I_(0))/(2)`D. `(E_(0)I_(0))/(2)sin phi` |
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Answer» Correct Answer - A `e=E_(0)sin omegat` `i=I_(0) sin (omegat-phi)` `P_(av.)=e_("rms").I_("rms").cos phi` `(E_(0))/(sqrt2).(I_(0))/(sqrt2). cosphi = E_(0)I_(0)cosphi` |
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| 1309. |
The rate constant `k_(1)` and `k_(2)` for two different reactions are `10^(16) e^(-2000//T)` and `10^(15) e^(-1000//T)`, respectively. The temperature at which `k_(1) = k_(2)` isA. 2000 KB. `1000//2.303K`C. 1000 KD. `2000//2.303K` |
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Answer» Correct Answer - B `k_(1)=10^(16)e^(-2000//T)` `k_(2)=10^(15)e^(-1000//T)` The temperature at which `k_(1)=k_(2)` will be `10^(16)e^(-2000//T)=10^(15)e^(-1000//T)` `(e^(-2000//T))/(e^(-1000//T))=(10^(15))/(10^(16))` `e^(-1000//T)=10^(-1)` `log_(e)e^(-1000//T)=log_(e)10^(-1)` `2.303xxlog_(10)E^(-1000//T)=2.303xxlog_(10)10^(-1)` `(-1000)/(T)xxlog_(10)e=-1` On solving, we get `T=1000//2,303K` |
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| 1310. |
Number of moles of `MnO_(4)^(-)` required to oxidise one mole of ferrous oxalate completely in acidic medium will beA. 7.0 moleB. 0.2 moleC. 0.6 moleD. 0.4 mole |
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Answer» Correct Answer - D `[5e+MnO_(4)^(-)+8H^(+)rarrMn^(2+)+4H_(2)O…(i)]xx2` `[C_(2)O_(4)^(2-)rarr 2e+2CO_(2)…(ii)]xx5` `MnO_(4)^(-)+16H^(+)+5C_(@)O_(4)^(2-)rarr2Mn^(2+)+10CO_(2)` As 2 moles of `MnO_(4)^(-)` required to oxidize 5 moles of oxalate. So number of moles of `MnO_(4)^(-)` required to oxidize 1 mole of oxalate = 2/5=0.4. |
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| 1311. |
Identify (Z) in the following series. `Ethanol overset(PBr_(3))to (X) overset(Alc.//KOH)to (Y) overset((i)H_(2)SO_(4)//("Room temp")) underset((ii)(H_(2)O,Heat))to(Z)`A. `CH_(3)CH_(2)-OH`B. `CH_(2)=CH_(2)`C. `CH_(3)CH_(2)-o-CH_(2)CH_(3)`D. `CH_(3)CH_(2)-O-SO_(3)H` |
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Answer» Correct Answer - A `{:(" "CH_(3)-CH_(2)-OH overset(PBr_(2))to CH_(3)-CH_(2)-Br),(" "darrAlc.KOH),(CH_(3)-CH_(2)-OH underset("Room temp."(ii) H_(2)O,"heat")overset((i)H_(2)SO_(4))larr CH_(2)=CH_(2)):}` |
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| 1312. |
The correct order of N-compounds, in its decreasing order of oxidation states isA. `HNO_(3), NH_(4)Cl, NO, N_(2)`B. `HNO_(3), NO, NH_(4)Cl, N_(2)`C. `HNO_(3), NO, N_(2), NH_(4),Cl`D. `NH_(4)Cl, N_(2), NO, HNO_(3)` |
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Answer» Correct Answer - C `Hoverset(+2)(N)O_(3), overset(+2)(N)O, overset(0)N_(2), overset(-3)(N)H_(4)Cl` Hence, the correct option is (3). |
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| 1313. |
Which of the following is not true for halogens ?A. All but fluorine show positive oxidation statesB. All are oxidizing agentsC. All form monobasic oxyacidsD. Chlorine has the highest electron-gain enthalpy |
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Answer» Correct Answer - A Due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluroic (I) acid. Oxidation number of F is +1 in HOF. |
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| 1314. |
The total number of octahedral void (s) per atom present in a cubic close packed structure isA. 2B. 4C. 1D. 3 |
| Answer» Correct Answer - C | |
| 1315. |
CsBr crystallizes in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 × `10^(23)` mol–1, the density of CsBr is:-A. `42.5 g/cm^(3)`B. ` 0.425 g/cm^(3) `C. `8.25 g/cm^(3)`D. ` 4.25 g/cm^(3) ` |
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Answer» Correct Answer - D For CsBr no. of formulas/unit cell n=1(like CsCI type) `C.D=(nxxM)/(VxxN_(A))``[{:(M=133+80=213),(V=a^(3)=(436.6xx10^(-12))^(3)):}]` `C.D=(1xx213gm)/(83.22xx10^(-12)cm^(3)xx6.02xx10^(23))=4.25gm//cm^(3)` |
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| 1316. |
In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes areA. n = 21 and x = 7B. n = 7 and x = 21C. n = 21 and x = 21D. n = 21 and x = 14 |
| Answer» Correct Answer - A | |
| 1317. |
A stone falls freely from rest from aheight `h` and it travels a distance `h//2` in the last second. The time of journey isA. `sqrt(2)-1`B. `2+sqrt(2)`C. `sqrt(2)+sqrt(3)`D. `sqrt(3)+2` |
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Answer» Correct Answer - B `h=1/2"gt"^2 " "......(i)` `h/2=1/2g(t-1)^2 " ".........(ii)` `1/4"gt"^2=1/2g(t-1)^2` `t/sqrt2=t-1` `t(1-1/sqrt2)=1` `t=sqrt2/(sqrt2-1)xx(sqrt2+1)/(sqrt2+1)` `t=sqrt2(sqrt2+1)` `t=2+sqrt2` |
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| 1318. |
Two simple harmonic motions of angular frequency `100 rad s^(-1)` and `1000 rad s^(-1)` have the same displacement amplitude. The ratio of their maximum accelerations isA. `1:10^(3)`B. `1:10^(4)`C. `1:10`D. `1:10^(2)` |
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Answer» Correct Answer - D Maximum acceleration `a=omega^(2)A` `(a_(1))/(a_(2))=(omega_(1)^(2)A)/(omega_(2)^(2)A)=((100)^(2))/((1000)^(2))=(1)/(10^(2))` |
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| 1319. |
The ratio of wavelength of the lest line of Balmer series and the last line Lyman series is:A. 2B. 1C. 4D. 0.5 |
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Answer» Correct Answer - C For last Balmer series `(1)/(lambda_(b))=R[(1)/(2^(2))-(1)/(oo^(2))]` `lambda_(b)=(4)/(R)` For last Lyman series `(1)/(lambda_(l))=R[(1)/(1^(2))-(1)/(oo^(2))]` `lambda_(l)=(1)/(R)` `(lambda_(b))/(lambda_(l))=((4)/(R))/((1)/(R))` `(lambda_(b))/(lambda_(l))=4 ` |
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| 1320. |
The phase difference between two waves represented by `y_(1)=10 ^(-6) sin [100 t+(x//50)+0.5]m, y_(2)=10^(-6) cos[100t+(x//50)]m`where x is expressed in metres and t is expressed in seconds, is approximatelyA. 2.07 radiansB. 0.5 radiansC. 1.5 radiansD. 1.07 radians |
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Answer» Correct Answer - D `y_1=10^(-6) sin { 100 t + x/50 +0.5 }` `y_2=10^(-6) cos {100t+x/50}` `=10^(-6) sin { pi/2 + 100t +x/50}` Phase difference between `y_1` & `y_2 =pi/2 -0.5` =1.58-0.5 =1.08 radians |
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| 1321. |
The half-life of radium is about `1600 yr`. Of `100g` of radium existing now, `25 g` will remain unchanged afterA. 6400 yearsB. 2400 yearsC. 3200 yearsD. 4800 years |
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Answer» Correct Answer - C `{:(M_0=100g,"(initial mass )"),("M=25g","(active mass)"):}}rArrN_0/N=100/25` `Delta=2^2rArr n=2` Therefore required time =`2xxT_(1//2)`=3200 yrs |
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| 1322. |
`M_p` denotes the mass of a proton and `M_n` that of a neutron. A given nucleus, of binding energy `B`, contains `Z` protons and `N` neutrons. The mass `M(N,Z)` of the nucleus is given by.A. `M(N,Z)=NM_n + ZM_P+ Bc^2`B. `M(N,Z)=NM_n + ZM_P- B//c^2`C. `M(N,Z)=NM_n + ZM_P+ B//c^2`D. `M(N,Z)=NM_n + ZM_P- Bc^2` |
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Answer» Correct Answer - B Mass of nucleus is slightly less than sum of masses of its constituents. This mass difference is equivalent to binding energy . `therefore Deltam=E/C^2=(ZM_P+NM_n)-M(N.Z)` Hence M (N,Z)=`NM_n+ZM_P-B/C^2` |
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| 1323. |
`M_n` and `M_p` represent mass of neutron and proton respectively. If an element having atomic mass `M` has `N-`neutron and `Z`-proton, then the correct relation will be :A. `M lt {N.M_n+Z.M_p}`B. `M gt {N.M_n+Z.M_p}`C. `M = {N.M_n+Z.M_p}`D. `M = N{M_n+M_p}` |
| Answer» Correct Answer - A | |
| 1324. |
CHOOSE THE INCORRECT PAIR :(a) Furrowed tongue - Down's syndrome(b) Gynaecomastia - Development of breast (c) Rudimentary ovaries - Turner's syndrome(d) No masculine development - Klinefelter's syndrome |
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Answer» Answer is d ...... masculine development means men like appearance which is seen in Klinefelter syndrome .so d is incorrect ....u can also do this question by elimination methodHey , The correct answer is (D) No masculine development - Klinefelter's syndrome regards~ |
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| 1325. |
When a charged particle moving with velocity `vec(V)` is subjected to a magnetic field of induction `vec(B)` the force on it is non-zero. This implies that:A. Angle between `vecV` and `vecB` is necessary `90^(@)`B. Angle between `vecV` and `vecB` can have at value other than `90^(@)`C. Angle between `vecV` and `vecB` can have at value other than zero and `180^(@)`D. Angle between `vecV` and `vecB` is either zero or `180^(@)` |
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Answer» Correct Answer - C F=q`(vecVxxvecB)`=avB sin0 n "therefore when" `0=0^(@) or 0=180^(@)`,F=0 |
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| 1326. |
Two circular coil 1 and 2 are made from the same wire but the radius of the `1^(st)` coil is twice that of the `2^(nd)` coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same-A. 3B. 4C. 6D. 2 |
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Answer» Correct Answer - B Let `R_(1)` and `R_(2)` are the radius of coil 1 & 2.if `B_(1)` and `B_(2)` are magnaetic induction at therir centre,then `B_(1)=(mu_(0)I_(1))/(2r_(1))`, and `B_(2)=(mu_(0)I_(2))/(2r_(1))` Since `B_(1)`=`B_(2)`, and `(R_(1)=2R_(2)` therefore `I_(1)=2I_(2)`Again if `R_(1)=2R_(2)` are resitance of the coil 1 and 2 then `R_(1)=2R_(2)`(as R prop length=2pir) and if `V_(1)and V_(2)`are the potential difference across them respectively,then `(V_(1))/(V_(2))=I_(1R_(1))/(I_(2)R_(2))=((2I_(2))(2R_(2)))/(I_(2)R_(2))=4` |
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| 1327. |
Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` isA. `R_(1)-R_(2)`B. `(R_(1)+R_(2))/2`C. `(R_(1)-R_(2))/2`D. `R_(1)+R_(2)` |
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Answer» Correct Answer - A According to question `E-Ir_(1)=0&I=(E+E)/(r_(1)+r_(2)+R)` `therefore (E)/(r_(1))=(2E)/(r_(1)+r_(2)+R)` `r_(1)+r_(2)=R=2r_(1) implies R=r_(1)-r_(2)` |
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| 1328. |
A black body at `1227^(@)C` emits radiations with maximum intensity at a wavelength of `5000 Å`. If the temperature of the body is increased by `1000^(@)`, the maximum intensity will be observed atA. 4000ÅB. 5000ÅC. 6000ÅD. 3000Å |
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Answer» Correct Answer - D By wiens displacement law `lambda_(m)` T=b we have (5000)(1500)=`(lambda_(m))`(1500+1000) `implieslambda_(m)=((5000)(1500))/(2500)=3000`Å |
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| 1329. |
A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would beA. 225B. 450C. 1000D. 1800 |
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Answer» Correct Answer - D Rate of power loss `r prop R^(1)T^(2)` `(r^(1))/(r^(2))=(R_(1)^(2)T_(1)^(4))/(R_(2)^(2)T_(2)^(4))` `=4 xx (1)/(16)` ` (450)/(r^(2))=(1)/(4)` `r_(2)=1800` watt |
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| 1330. |
The radii of circular orbits of two satellite `A` and `B` of the earth are `4R` and `R`, respectively. If the speed of satellite `A` is `3v`, then the speed of satellite `B` will beA. 3v/4B. 6vC. 12vD. 3v/2 |
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Answer» Correct Answer - B speed of satellite `V=sqrt((GM)/(r ))` `implies (V_(B ))/(V_(A))=sqrt((r_(A))/(r_(B))=sqrt((4R)/(R ))=2` `implies V_(B)=(3v)(2)=6v` |
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| 1331. |
Biosystematics aims atA. The classification of organisms based on broad morphological charactersB. Delimiting various taxa of organism and establishing their relationshipsC. The classification of organisms based on their evolutionary history and establishing their phylogeny on the totality of various parameters from all fields of studiesD. Identification and arrangement of organisms on the basis of cytological characteristics |
| Answer» Correct Answer - C | |
| 1332. |
Which one of the following is a viral disease of poultry?A. PasteurellosisB. SalmonellosisC. CoryzaD. New Castle disease |
| Answer» Correct Answer - D | |
| 1333. |
Ultrasound of how much frequency is beamed into human body for sonography ?A. 45 - 70 MHzB. 30 -45 MHzC. 15 -30 MHzD. 1 - 15 MHz |
| Answer» Correct Answer - D | |
| 1334. |
The partial pressure of oxygen in the alveoli of the lungs isA. equal to that in the bloodB. more than that in the bloodC. less than that in the bloodD. less than that of carbon dioxide |
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Answer» Correct Answer - B The partial pressure of oxygen in alveoli of lungs is less than that in blood. Normal `PO_(2)` in alhcoli is 104 mmttg & `PO_(2)` in oxygenated blood is 95 mm Hg |
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| 1335. |
Osteoporosis, an age related disease of skeletal system, may occue due toA. immune disorder afecting neuromuscular junction leading to fatigueB. high concentration of `Ca^(+ +) and Na^(+)`C. Decreased level of estrogenD. Accumulation of uric acid leading to inflammation of joints |
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Answer» Correct Answer - C Osteoporosis, is an age related disroder/disease and it is more common in elder female. Because of afer menopause secretion of estrogen is stope from ovary. |
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| 1336. |
Which one of the following characters is not typical of the class MammaliaA. Alveolar lungsB. Ten pairs of cranial nervesC. Seven cervical vertebraeD. Thecodont dentition |
| Answer» Correct Answer - B | |
| 1337. |
Which cells of Crypts of Lieberkuhn secrete antibacterial lysozyme ?A. Argentaffin cellsB. Paneth cellsC. Zymogen cellsD. Kupffer cells |
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Answer» Correct Answer - B `implies` Kupffer-cells are phagocytic cells of liver. `implies ` Zymogen cells are enzyme producing cells. `implies` Paneth cell secretes lysozyme which acts as anti-bacterial agent. `implies ` Argentaffin cells are hormone producing cells. |
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| 1338. |
The hapatic partal vein drains blood to liver fromA. HeartB. StomachC. KidneysD. Intestine |
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Answer» Correct Answer - D In hepatic portal system, hepatic portal vein carries maximum amount of nutrients from intestine to liver. |
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| 1339. |
Adult human RBCs are enucleate. Which of the following statement (s) is/are most appropriate explanation for this feature ? (1) They do not need to reproduce (2) They are somatic cells (3) They do not metabolise (4) All their internal space is available for oxygen transport.A. Only (d)B. Only (a)C. (a), (c) and (d)D. (b) and (c) |
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Answer» Correct Answer - A In Human RBCs, nucleus degenerates during maturation which provide more space for oxygen carrying pigment (Haemoglobin). It lacks most of the cell organelles including mitochondria so respires anaerobically. |
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| 1340. |
The population of an insect species shows and explosive increase in numbers during rainy season followed by its disappearance at the end of the season. What does this show ?A. The population of its predators increases enormouslyB. S-shaped or sigmoid growth of this insectC. The food plants mature and die at the end of the rainy seasonD. Its population growth curve is of J-type |
| Answer» Correct Answer - D | |
| 1341. |
Which one of the following statements is correct ?A. Neurons regulate endocrine activity, but not vice verse (B. Endocrine glands regulate neural activity, and nervous system regulates endocrine glands (C. Neither hormones control neural activity nor the neurons control endocrine activityD. Endocrine glands regulate neural activity, but not vice versa |
| Answer» Correct Answer - B | |
| 1342. |
The blue baby syndrome results fromA. Excess of chlorideB. MethemoglobinC. Excess of dissolved oxygenD. Excess of TDS (total dissolved solids) |
| Answer» Correct Answer - B | |
| 1343. |
Praying mantis is a good example ofA. Mullerian mimicryB. Warning colourationC. Social insectsD. Camouflage |
| Answer» Correct Answer - D | |
| 1344. |
Animals have the innate ability to escape from predation Examples for the same are given below. Select the incorrect exampleA. Colour change in chameleonB. Poison fangs in snakesC. Melanism in mothsD. Enlargement of body size by swallowing air in puffer fish |
| Answer» Correct Answer - D | |
| 1345. |
Which one of the following conditions of the zygotic cell would lead to the birth of a normal human female childA. one X and one Y chromosomeB. two X chromosomesC. only one Y chromosomeD. only one X chromosome |
| Answer» Correct Answer - B | |
| 1346. |
Point masses `m_(1) and m_(2)` are placed at the opposite ends of a rigid rod of length `L`, and negligible mass. The rod is to be set rotating about an axis perpendicualr to it. The position of point `P` on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity `omega_(0)` is minimum, is given by : A. `x=(m_(2)L)/(m_(1)+m_(2))`B. `x=(m_(1)L)/(m_(1)+m_(2))`C. `x=(m_(1))/(m_(2))L`D. `x=(m_(2))/(m_(1))L` |
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Answer» Correct Answer - A `I=m_(1)x^(2)+m_(2)(L-x)^(2)` `I=m_(1)x^(2)+m_(2)L^(2)+m_(2)x^(2)-2m_(2)Lx ` `(dI)/(dx)=2m_(1)x+0+2xm_(2)-2m_(2)L=0` `x(2m_(1)+2m_(2))=2m_(2)L` `x=(m_(2)L)/(m_(1)+m_(2))` When I is minimum work done `W=(1)/(2)Iomega_(0)^(2)` will also be minimum. |
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| 1347. |
A common test to find the genotype of a a hybird is byA. Crossing of one F1 progeny with male parentB. Crossing of one F2 progeny with male parentC. Crossing of one F2 progeny with female parentD. Studying the sexual behaviour of F1 progenies |
| Answer» Correct Answer - A | |
| 1348. |
According to MO theory which of thhe following lists makes the nitrogen species in terms of increasing bond order?A. `N_(2)^(-) lt N_(2)^(2-) lt N_(2)`B. `N_(2)^(-) lt N_(2) lt N_(2)^(2-)`C. `N_(2)^(2-) lt N_(2)^(-) lt N_(2)`D. `N_(2) lt N_(2)^(2-) lt N_(2)^(-)` |
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Answer» Correct Answer - C `N_(2)=14e=B.O.=3` `N_(2)^(-)=15e=B.O.=2.5` `N_(2)^(2-)=16e=B.O.=2 ` |
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| 1349. |
Among the following which is the strongest oxidizing agent ? -A. `Cl_(2)`B. `F_(2)`C. `Br_(2)`D. `I_(2)` |
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Answer» Correct Answer - B `F_(2) to` reducing potential very high so strongest oxidizing agent . |
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| 1350. |
In which of the following molecular/ions `BF_(2),NO_(2)^(-),NH_(2)` and ` H_(2)O` the correct atom is `sp^(2)` hybridized ?A. `BF_(3) and NO_(2)^(-)`B. `NO_(2)^(-) and NH_(2)`C. `NH_(2)^(-) and H_(2)O`D. `NO_(2)^(-) and H_(2)O` |
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Answer» Correct Answer - A Both `BF_(3) and NO_(2)^(-)` is `sp^(3)` hybridized . |
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