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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.
| 1. |
Objective function of an LPP isA. atleast two of the corner pointsB. all the corner pointsC. atleast one of the corner pointsD. None of the corner points |
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Answer» Correct Answer - C The objective function of LPP defined over the convex set attains its optimum value atleast one of the corner points. |
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| 2. |
The equation of the plane through (-1,1,2), whose normal makes equal acute angles with coordinate axes isA. `r.(hati+hatj +hak) = 2`B. `r.(hati + hatj + hatk) = 6`C. `r.(3hati -3hatj+3hatk) = 2`D. `r.(hati - hatj +hatk) = 3` |
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Answer» Correct Answer - a The equation of plane passing through `a = - hati + hatj + 2hatk` and perpendicular to `n hati +hatj + hatk` is `r. n = a . nr (hati + hatj + hatk)` `= (-hati + hatj + 2hatk) . (hati + hatj + hatk)` `=(-1+ 1 + 2)` `r. (hati + hatj + hatk) = 2` |
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| 3. |
Which of the chief constituent of pyrex glass ?A. `B_(2)O_(3)`B. `SiO_(2)`C. `Al_(2)O_(3)`D. `Na_(2)O` |
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Answer» Correct Answer - B The chief constituetof pyrex glass is `SiO_(2)`. Pyrex glass is obtained by fusing `60`to `80% SiO_(2)`, `10 -25%` `B_(2)O_(3)` and remaining amount of `Al_(2)O_(3)`. |
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| 4. |
Which of the following processes is not used to preserve the food ?A. IrradiationB. Addition of saltsC. Addition of heatD. Hydration |
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Answer» Correct Answer - D For the preservation of foods and enhancing of their appeal, different processes are such as irradation, addition of salts and heat, antioxidants, emulsifiers, etc. |
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| 5. |
Which among the following compounds does not act as reducing agent ?A. `H_(2)O`B. `H_(2)S`C. `H_(2)Se`D. `H_(2)Te` |
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Answer» Correct Answer - A All the hydrides, except water `(H_(2)O)` of group 16 element acts as a reducing agents . The reducing character increases from `H_(2)S` to `H_(2)`Te. This is due to decrease in stability of hydride . |
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| 6. |
Identify the weakest oxidising agent among the following.A. `Li^(+)`B. `Na^(+)`C. `Cd^(2+)`D. `l_(2)` |
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Answer» Correct Answer - A Lithium ion has small size, it has the highest hydration enthalphy, which compensate its high ionisation energy and therefore, its `E^(@)` value is most negative. Thus, Li is the most powerful reducing agent in aqueous solution. |
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| 7. |
The rate constant for a first order reaction is `7.5 xx 10^(-4) s^(-1)`. If initial concentration of reactant is `0.080 M`, what is the half life of reaction ?A. `990 s`B. `79.2s`C. `12375 s`D. `10.10 xx 10^(-4) s` |
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Answer» Correct Answer - A Given , `K = 7 xx 0^(-4)s^(-1),[A]_(0) = 0.08 M` Half life `(t_(1//2))` of a first order reaction is independent of initial concentration and is given as `t_(1//2) = (0.693)/(k) = (0.693)/(7xx 10^(-4)s) = 990 s` |
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| 8. |
`(+2) ` 2-methylbutan -1-ol(-)2-methylbutan -1-of have different values for whichA. Boiling pointB. Relative densityC. Refractive indexD. Specific rotation |
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Answer» Correct Answer - D (+) 2-methyl butan-1-ol and (-) 2-methyl butan-1-ol are enantiomer. They are non-superimposable mirror images of each other. Hence, they are optically active. The optial activity of a compound can be confirmed by the value of specific rotation. It is defined as the rotation produced by solution of 10 cm length and uniti concentration (1g/mL) for the given `lambda` of light at the given temperature. |
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| 9. |
The magnetic moment of electron due to orbital motion is proportional to (n= principle quantum numbers)A. `1/(n^(2))`B. `1/n`C. `n^(2)`D. n |
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Answer» Correct Answer - D Magnetic moment is given as, magnetic moment `(M_(0)) = (e )/(2m_(e)) xx L` where, `e =` electronic charge , `m_(e)` = mass of electron and `L =` orbital angular momentum, As, we know `L = (nh)/(2pi) "…."(i)` where , n = principal quantum number and `h =` Plancks constant. Therefore, from relation (i) and (ii), we get `M_(0) prop n`. |
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| 10. |
The point on the curve `y=sqrt(x-1)`, where the tangent is perpendicular to the line 2x+y-5=0 isA. `(2,-1)`B. `(10,3)`C. `(2,1)`D. `(5,-2)` |
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Answer» Correct Answer - C Let slope of the curve `y = sqrt(x-1)` is `m_(1)` and slope of the line `2x + y - 5 = 0` is `m_(2)` Now, `m_(1) = (dy)/(dx) = d/(dx) sqrt(x-1)` and `m_(2) = (dy)/(dx) = (d)/(dx) (5-2x)` `rArr m_(1) = (1)/(2sqrt(x-1))` and `m_(2) = -2` We know that, lines are perpendicular it and only if `m_(1)m_(2) = - 1` `:. (1)/(2sqrt(x - 1)) = 1 rArr sqrt(x-1) = 1` On squaring both sides, we get `x - 1= 1` `rArr x = 2` On substituting `x = 2` in `y = sqrt(x - 1)`, we get Hence. coordination of the point on the curve, `y = sqrt(x-1)`, where tangent is perpendicular to the line `2x+y = 5` is `(2,1)`. |
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| 11. |
At random variable `X~B(n,p)` if values of mean and variance of X are 18 and 12 respectively, then total number of possible values of X areA. 54B. 55C. 12D. 18 |
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Answer» Correct Answer - B Given, mean `= 18` and variance `= 12` `:. Np = 18` and `npq = 12` `rArr (npq)/(np) = (12)/(18) rArr q = 2/3` Now, `p = 1 - q = 1 - (2)/(3) = 1/3` On substituting `p = 1/3` in `np = 8`, we get `n(1/3) = 18 rArr n = 54` `:.` Values of X are `0,1,2 ".........." 54` Hence, total number of possible values of `X = 55` |
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| 12. |
The equation of the progressive wave is `y = 3 sin[pi(t/3-x/5)+pi/4]` , where x and y are in metre and time in second. Which of the following is correct.A. Velocity `v = 1.5` m/sB. Amplitude `A = 3` cmC. Frequency `f = 0.2 Hz`D. Wavelength `lambda = 10 `m |
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Answer» Correct Answer - D Compare the given equation with the standard equation of wave motion, `Y = A sin[2pi(t/T - x/lambda) + pi/4]` where , A and `lambda` are amplitude and wavelength respectively. Amplitude , `A = 3m` Wavelegth, `lambda = 10m`. |
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| 13. |
Photodiode is a deviceA. which is always operated in reverse biasB. which is always operated in forward biasC. in which photo current is independent of intensity of incident radiationD. which may be operated iin forward or reverse bias. |
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Answer» Correct Answer - A Photodiode is a reversed biased p-n junction made from a photosensitive material, which is used for detecting optical signals. |
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| 14. |
A wheel of moment of inertia `2 kgm^(2)` is rotating about an axis passing through centre and perpendicular to its plane at a speed `60 rad//s`. Due to friction, it comes to rest in 5 minutes. The angular momentum of the wheel three minutes before it stops rotating isA. `24 kg m^(2)//s`B. `48 kg m^(2)//s`C. `72 kg m^(2)//s`D. `96 kg m^(2)//s` |
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Answer» Correct Answer - C Given. `I = 2 kg m^(2)` `omega_(0) = 60` rad/s , `omega = 0` `t = 5` min ` = 5 xx 60 = 300` s From the relation, `omega = omega_(0) + alphat` `alpha = (omega - omega_(0))/(t)` `rArr alpha = (0 -60)/(300) = (-60)/(300) = (-1)/(5) rad//s^(2)` For `t = 3=2` min `= 2 xx 60 = 120s` `omega = omega_(0) + alphat` `= 60 - 1/(5) xx 120 = 60 - 24` `:. omega = 36` rad/s As, angular momentum, `L =komega` Substituting the values in the above relation, we get `L = 2 xx 36 = 72 kg m^(2)//s`. |
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| 15. |
A flywheel at rest is to reach an angular velocity of 24 rad/s in 8 second with constant angular acceleration. The total angle turned through during this interval isA. 24 radB. 48 radC. 72 radD. 96 rad |
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Answer» Correct Answer - D Given, `omega_(0) = 0, omega = 24` rad/s, `f = 8s` From the relation , `omega = omega_(0) + alphat` `rArr alpha = (omega - omega_(0))/(t)` Subsituting the given values, we get `alpha = 28//8 = 3 "rad"//s^(2)` Also, `theta = omega_(0)t + 1/2 alphat^(2) = 0 +1/2 xx 3 xx (8)^(2)` `= (3 xx 64)/(2) = 96` rad |
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| 16. |
Two spherical black bodies of radii `R_(1)` and `R_(2)` and with surface temperature `T_(1)` and `T_(2)` respectively radiate the same power. `R_(1)//R_(2)` must be equal toA. `(T_(1))/(T_(2))`B. `(T_(2))/(T_(1))`C. `((T_(1))/(T_(2)))^(3)`D. `((T_(2))/(T_(1)))^(2)` |
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Answer» Correct Answer - C The rate at which an object radiates energy is given as, `Q/t sigma AT^(4) rArr` Power, `P = sigmaAT^(4)` `:. A prop 1/(T^(4))` It the body radiates same power, then `rArr (A_(2))/(A_(1)) = (T_(1)^(2))/(T_(2)^(4)) rArr (4pir_(2)^(2))/(4pir_(1)^(2)) = (T_(1)^(4))/(T_(2)^(4))` `:. (r_(2))/(r_(1)) = ((T_(1))/(T_(2)))^(2)` |
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| 17. |
A solid sphere of mass 2 kg is rolling on a frictionless horizontal surface with velocity `6 m//s`. It collides on the free and of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring = 36 N/m)A. `sqrt(14) m`B. `sqrt(2.8) m`C. `sqrt(14) m`D. `sqrt(0.7) m` |
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Answer» Correct Answer - B Given, `m = 2` kg, `v = 6` m/s and `k = 36` N/m `= 1/2 mv^(2) + 1/2 lomega^(2)` `= 1/2 mv^(2) + 1/2 xx 2/5 mr^(2) omega^(2)` `=1/2 mv^(2) + 1/5 mv^(2) = 7/10 mv^(2) ` The potential energy of the spring on maximum compression x `:. 1/2 kx^(2) = 7/10 mv^(2)` `rArr x^(2) = 14/10 (mv^(2))/(k)` Substituting the given values, we get `= 14/10 xx (2 xx (16)^(2))/(36) = 2.8` `:. x = sqrt(2.8) ` m |
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| 18. |
The first law of thermodynamics for isothermal process isA. `q = - W`B. `DeltaU = W`C. `DeltaU = q_(v)`D. `DeltaU = - q_(v)` |
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Answer» Correct Answer - A According to first law of thermodynamics, `DeltaU = q+W` where, `DeltaU =` internal energy ` q =` Heat, w = Work done For isothermal process , `DeltaT = 0 , DeltaU = 0` `:. q = - W` |
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| 19. |
Which among the following reactions is an example of pseudo first order reaction ?A. Inversion of cane sugarB. Decomposition of `H_(2)O_(2)`C. Conversion of cyclopropane to propeneD. Decomposition of `N_(2)O_(5)` |
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Answer» Correct Answer - A The reactions in which the molecularity is two but they confirm to the first order are known as pseudo first order are known as pseudo first order reaction or pseudo unimolecular reactions. e.g, inversation of cane sugar, `underset(("Cane sugar"))underset("Sucrose")(C_(12H_(22)O_(11)) + H_(2)O overset(H^(+))rarr underset("Glucose")(C_(6)H_(12)O_(6)) + underset("Fructose")(C_(6)H_(12)O_(6)` The concentration of water remains constant |
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| 20. |
In preparation of sulphuric acid from sulphur dioxide in lead chamber process. What substance is used as a catalyst ?A. Manganese dioxideB. Vanadium pentoxideC. Nitric oxideD. Raney nickel |
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Answer» Correct Answer - C In the preparation of sulphuric acid `(H_(2)O)`. In `2SO_(2) + underset("from air")(O_(2)) + 2H_(2)O + underset("Catalyst")([NO] rarr 2H_(2)SO_(4) + underset("Catalyst")([NO])` |
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| 21. |
Solubility of which among the following solids in water changes slightly with temperature /A. `KNO_(3)`B. `NaNO_(3)`C. `KBr`D. `NaBr` |
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Answer» Correct Answer - D The solubility of NaBr changes slightly with temperature. |
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| 22. |
Which of the following statement(s) is/are incorrect in case of Hofmann bromamide degradation ?A. Reaction is useful for decreasing length of carbon chain by one carbon atomB. It gives tertiary amineC. It gives primary amineD. Aqueous or alco. KOH is used with bromine |
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Answer» Correct Answer - B Statement B is incorrect in case of Hofmann bromamide degradation . This is used to synthesis primary amine. In this reaction, amide on reation with `Br_(2)` in an aqueous or ethanolic solution of `NaOH` give `1^(2)` amine with one C atom less than that present in the amide. `RCONH_(2) + Br_(2) + 4KOHoverset(Delta)rarrunderset(1^(@) "amine")(RNH_(2)) + 2KBr +K_(2)CO_(3) + 2H_(2)O` |
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| 23. |
Which of the following statement(s) is/are incorrect for pair of element Zr-Hf ?A. Both posses same number of valence electronsB. Both have identical sizesC. Both have almost identical radiiD. Both of these belong to same period of periodic table |
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Answer» Correct Answer - D Zirconium (Zr)with atomic number - 40and Hafnium (Hf) with atomic number - 72 belongs of period 5th and 6th respectively. |
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| 24. |
Which among the following equations represents the reduction reaction taking place in lead accumulator at positive electrode, while it is being used as a source of electrical energy ?A. `Pb rarr Pb^(2+)`B. `Pb^(4+) rarr Pb`C. `Pb^(2+) rarr Pb`D. `Pb^(4+) rarr Pb^(2+)` |
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Answer» Correct Answer - D The reaction involved for lead accumulator during discharging i.e, when cell is in the use are At anode : `Pb(s) + HSO_(4)^(-)(aq) rarr PbSO_(4)(s) + H^(+) + 2e^(-)` At cathode : `PbO_(2)(s)+ 3H^(+) +HSO_(4)^(-)(aq)` `+2e^(-) rarr PbSO_4(s) + 2H_(2)O(l)` Overall reaction : `Pb(s) +PbO_(2)(s) + 2H^(+)` `+2HSO_(4)^(-)(aq) rarr 2PbSO_(4)(s) + 2H_(2)O(l)` |
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| 25. |
What is the actual volume occupied by water molecules present in `20 cm^(3)` of water ?A. `20 cm^(3)`B. `10 cm^(3)`C. `40 cm^(3)`D. `24.89 cm^(3)` |
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Answer» Correct Answer - B Half of the volume occupied in water is empty or unoccpied. Therefore, `10 cm^(3)` of the actual volume is occupied by water molecules present in `20 cm^(3)` of water. |
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| 26. |
What is the hybridisation of carbon atoms in fullerene ?A. `sp^(3)`B. `sp`C. `sp^(2)`D. `dsp^(3)` |
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Answer» Correct Answer - C Fullerenes are the allotropes of carbon also called Buckminster fullerenes. They are cage like spheroidal molecules with general formula, `C_(2n)` (e.g, `C_(60), C_(70)` etc). `C_(60)` or the bucky ball consists of 60 C- atoms arranged in 20 six membered and 12 five appearance of a soccer ball. Each C-atom in `C_(60)` in `sp^(2)`-hybridised forming `sigma`-bonds with three other C-atoms . The fourth electron remains free and responsible for its conductivity . |
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| 27. |
Which element among the following does not form diatomic molecules ?A. ArgonB. OxygenC. NitrogenD. Bromine |
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Answer» Correct Answer - A Noble gas (He, Ne, Ar, Kr, Xe and Rn) does not form diatomic molecules. Due to presence of completely filled valence shell, these gases are highly stable. |
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| 28. |
A molecule of stachyose contains how many carbon atoms ?A. 6B. 12C. 18D. 24 |
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Answer» Correct Answer - D Depending upon the number of monosaccharide molecules condensed to form oligasaccharides such as disaccharides (e.g., sucrose) trisaccharides(e.g., raffinose), tetrassaccharides) (e.g, stachyose) atc. Stachyose contains 24 carbon atoms in its structure. |
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| 29. |
Which of the following is not a mineral of iron?A. HaenatiteB. MagnesiteC. MagneticD. Siderite |
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Answer» Correct Answer - B Haematite `- Fe_(2)O_(3)` Magnesite `-MgCO_(3)` Magnetic `- Fe_(3)O_(4)` Siderite `- FeCO_(3)` `:.` Magnesite is the mineral of magnesium (Mg). |
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| 30. |
Which element among the following does form `ppi-ppi` multiple bonds ?A. ArsenicB. NitrogenC. PhosphorusD. Antimony |
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Answer» Correct Answer - B Nitrogen (N) differs from rest of the members of group 15 elements (P, As, Sb, Bi), because of its small size , high electronegativity . High ionisation energy, absence of vacant d-orbitals and capacity to forms `ppi - ppi` multiple bonds `N -= N`. |
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| 31. |
What is the quantity of hydrogen gas liberated when `46 g` sodium reacts with excess ethanol ? (Given atomic mass of `Na = 23`)A. `2.4 xx 10^(-3) kg`B. `2.0 xx 10^(-3) kg`C. `4.0 xx 10^(-3) kg`D. `2.4 xx 10^(-2) kg` |
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Answer» Correct Answer - B The rataction of ethanol with water is as follows. `2C_(2)H_(5)OH (l) + 2Na(s) rarr 2C_(2)H_(5)overline(O) Na^(+) +H_(2)(g) uarr` 2 moles of Na(46 g) = 1 mole of `H_(2)` `= 2g = 2 xx 10^(-3) kg` |
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| 32. |
The colour and magnetic nature of mangante ion `(MnO_(4)^(2-))` isA. green, paramagneticB. purple, diamagneticC. green, diamagneticD. pruple, paramagnetic |
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Answer» Correct Answer - A Oxidation state of Mn in `MnO_(4)^(2-)` is +6. `Mn (+6) 1s^(2) 2s^(2)2p^(6), 3s^(6) 3d^(1), 4s^(0)` Hence , manganate ion `(MnO_(4)^(2-))` is paramagnetic due to presence of unpaired electron. Also, `MnO_(4)^(2-)` is green in colour. |
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| 33. |
Two uniform wires of a the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and radius of the first wire is twice the radius of the second wire, then the ratio of the lengths of the first wire to second wire isA. `1/3`B. `1/4`C. `1/5`D. `1/6` |
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Answer» Correct Answer - A Fundamental frequency of the first wire is given as, `f = 1/(2L_(1)) sqrt(T/m)= (1)/(2L_(1)) sqrt((T)/(pir_(1)^(2)rho)) = 1/(2L_(1)r_(1)) sqrt((T)/(pirho))` The first overtone, `f_(1) = 2f = (2)/(2L_(1)r_(1)) sqrt((T)/(pirho)) = 1/(L_(1)r_(1)) sqrt((T)/(pirho)) "..."(i)` The second overtone of the second wire is given as `f_(2) = 3/(2L_(2)r_(2)) sqrt((T)/(pirho))` Given , `f_(1) = f_(2)` `rArr 1/(L_(1)r_(1)) sqrt((T)/(pirho)) = 3/(2L_(2)r_(2)) sqrt((T)/(pirho))` `:. 3L_(1)r_(1) = 2L_(2)r_(2)` `rArr (L_(1))/(L_(2)) = 2/(3).(r_(2))/(r_(1)) = 2/(3).(r_(2))/(2r_(2))` [given, `r_(1) = 2r_(2)`] `= 1/3` |
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| 34. |
The fundamental frequency of an air column in a pipe closed at one end is 100 Hz. If the same pipe is open at both the ends, the frequencies produced in Hz areA. `100, 200, 300, 400, "….."`B. `100, 300, 500, 700, "……"`C. `200, 300, 400 , 500, "……"`D. `200, 400, 600, 800, "….."` |
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Answer» Correct Answer - D For a closed pipe fundamental frequency, `v_(1) = v/(4L) = 100 Hz` For an open pipe fundamental frequency, `v_(1) = v/(2L) = 200 Hz` In a pipe open at both ends, all multiples of the fundamental are produced. |
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| 35. |
Two parallel plate air capacitance of same capacity C are connected in series to a battery of emf E. Then one of the capacitors is completely filled with dielectric material of constant K. The change in the effective capacity of the series combination isA. `C/2[(K-1)/(K+1)]`B. `2/C [(K-1)/(K+1)]`C. `C/2[(K+1)/(K-1)]`D. `C/2[(K-1)/(K+1)]^(2)` |
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Answer» Correct Answer - A Initial effective capacity of the series combination is `1/(C_(1)) = 1/(C ) + 1/(C ) = 2/C` `= C_(1) = C/2 ` Effective capacity of the series combination with dielectic material is ` 1/(C_(1)) = 1/C + 1/(KC)` `1/(C_(2)) = 1C [1+1/K]` `:. C_(2) = C/(1+1/K) = (CK)/((K + 1))` The change in the effective capacitance is `DeltaC = C_(2) - C_(1)` `= (CK)/((K+1)) - C/2 = C [K/(K+ 1) - 1/2]` `=C [(2K - K - 1)/(2(K+1))] = C/2[(K-1)/(K+ 1)]` |
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| 36. |
For a invertible matrix A if `A(adjA)=[(10,0),(0,10)]`, then |A|=A. `100`B. `-100`C. `-100`D. `10` |
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Answer» Correct Answer - C Given , `A (adj A) = |{:(10,0),(0,10):}| = 10 |{:(1,0),(0,1):}| = 10l` We know that, `A` (adj A) = `|A| l` On comparing both sides , we get `|A| = 10` |
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| 37. |
Two coils P and Q are kept near each other. When no current flows through coil P and current increases in coil Q at the rate `10 A//s`, the emf in coil P is 15 mV. When coil Q carries no current and current of `1.8 A` flows through coil P, the magnetic flux linked with the coil Q isA. `1.4 m Wb`B. `2.2 mWb`C. `2.7 m Wb`D. `2.9 m Wb` |
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Answer» Correct Answer - C Given, `(dl_(theta))/(dt) = 10A//s` `e_(p) = 15 mA` `= 15 xx 10^(-3)V, I_(p) = 18A` Induced emt in coil P is given as, `|e_(p)| = M. (dl_(Q))/(dt)` Substituting the given values, we get `15 xx 10^(-3) = M xx 10` `rArr M = 15 xx 10^(-4) H` Magnetic fluxed linked with coil Q is given as, `phi_(Q) = MI_(p) = 15 xx 10^(-4) xx 1.8` `= 27.0 xx 10^(-4)` `= 2.7 xx 10^(-3) = 2.7 mWb` |
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| 38. |
For a particle moving in vertical circle, the total energy at different positions along the pathA. is conservedB. increasesC. decreasesD. may increases or decreases |
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Answer» Correct Answer - A When a particle is moving a verticle circle it would move from lowest point to the highest point. Its speed would decreases and becomes minimum at the highest point. Thus, the total mechanical energy remains conserved, kinetic energy changes into potential energy and vice-versa. |
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| 39. |
A particle executes `SHM` on a straight line. At two positions, its velocities are `u` and `v` whle accelerations are `alpha` and `beta` respectively `[beta gt alpha gt ]`.The distance between these two positions isA. `(u^(2) - v^(2))/(alpha + beta)`B. `(u^(2) + v^(2))/(alpha + beta)`C. `(u^(2)0 - v^(2))/(alpha - beta)`D. `(u^(2) - v^(2))/(alpha - beta)` |
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Answer» Correct Answer - A Let the distance be p when velocity is u and acceleration `alpha`. Let the distance q when velocity is v and acceleration `beta`. If `omega` is the angular frequency. then `alpha =omega^(2)p` and `beta = omega^(2) q` `:. alpha + beta = omega^(2) (p + q) "..."(i)` Also, `u^(2) = omega^(2)A^(2) - omega^(2) p^(2)` and `v^(2) = omega^(2)A^(2) - omega^(2)q^(2)` `rArr v^(2) - u^(2) = omega^(2) (p^(2)- q^(2))` `v^(2) - u^(2) = omega^(2) (p - q) (p+q) "...."(ii)` By Eqs. (i) and (ii), we get `v^(2) - u^(2) = (p-q) (alpha + beta)` `:. p -q = (v^(2) - u^(2))/(alpha + beta)`or `q- p = (u^(2) - v^(2))/(alpha + beta)` |
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| 40. |
The numbr of principal solution of ` tan 2 theta =1` isA. oneB. twoC. threeD. four |
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Answer» Correct Answer - B We have , `tan 2theta = 1` Here, the value of `tan 2 theta` is positive. So, `theta` lies in 1st and 3rd quadrants. `:.` Number of principal solutions of `tan 2 theta = 1`, is two, |
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| 41. |
A boy tosses faiir coin 3 times. If he gets Rs 2X for X heads, then his expected gain equals to Rs…..A. 1B. `3/2`C. `3`D. 4 |
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Answer» Correct Answer - C Let X be a random variable. That denotes the number of headds in tossing a coin 3 times . Then, X can take value `0,1,2,3`. Again, let `y = ₹ 2x`. Then, Y can taje the values ₹ 0, ₹ 2, ₹4, and ₹ 6 `:. P(y=0) = P` (getting 0 head) `= 1/8` `P(y=2) = P` (getting 1 head) `= 3/8` `P(y = 4) = P` (getting 2 heads) `= 3/8` `P(y = 6) = P` (getting 3 heads) `= 1/8` Expected gain `= sumy_(i) p_(i)` `=0 (1/8) + 2(3/8) + 4(1/8) + 6(1/8)` `= (6 + 12 + 6 )/(8) = 24/8 = 3` |
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| 42. |
If `alpha and beta` are roots of the equation `x^(2)+5|x|-6=0,` then the value of `|tan^(-1)alpha-tan^(-1)beta|` isA. `pi/2`B. `0`C. `pi`D. `pi/4` |
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Answer» Correct Answer - A Given `alph` and `beta` be the roots of the equation `x^(2) + 5|x| - 6= 0` Now, `|x|^(2) + 5|x| - 6 = 0` `|x|^(2 ) + 6|x| - |x|- 6 = 0` [by factorisation] `|x|(|x| +6) - 1(|x|+6) = 0` `(|x|+6)(|x|-1) = 0` `|x| = - 6` or `|x| = 1` (Since, modulus cannot be giving negative values ) `:. |x| = 1 rArr x = +- 1` So, `alpha = 1` and `beta = -1` `:.` Now, `|tan^(-1) alpha - tan^(-1) beta| = |tan^(-1) 1 - tan^(-1)(-1)|` `= |pi/4 - (-pi/4)| = |(pi)/(4) + (pi)/(4)| = |pi/2|` |
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| 43. |
Let `square PQRS` be a quadrilateral. If M and N are the mid-points of the sides PQ and RS respectively, then PS+QR=A. `3MN`B. `4 MN`C. `2 MN`D. `2 NM` |
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Answer» Correct Answer - C Let p,q ,r ,s m and n be the position vectors of P,Q,R,S,M and N be respectively. Given , M and N are the mid-points of PQ and RS respetively of quadrilateral PQRS . `m = (p + q)/(2)` and `n = (p+s)/(2) "…."(i)` Now, `PS " "QR = (PV "of" S - PV "of" P) + (PV "of" R- PV "of" Q)` `= s+ p +r -q` `= (r+s) - (p+q)` `= 2n - 2m` [From Eq. (i)] `= 2(n-m) = 2MN` |
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| 44. |
If slopes of lines represented by `kx^(2)+5xy+y^(2)=0` differ by 1, then k=A. `2`B. 3C. 6D. 8 |
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Answer» Correct Answer - C Given pair of lines be `kx^(2) +5xy +y^(2) = 0"…."(i)` On comparing Eq. (i) with `ax^(2) + 2hx +by^(2) = 0`, we get `a = k, b = 1` and `2h =5` Let `m_(1)` and `m_(2)` be two slopes of pair of lines. `:. m_(1) + m_(2) = (-2h)/(b) = -5` and `m_(1)m_(2) = (a)/(b) = k` Now, `(m_(1) - m_(2))^(2) = (m_(1) + m_(2))^(2) - 4m_(1)m_(2)` `rArr (1)^(2) = (-5)^(2) - 4 k` [given, `m_(1) - m_(2) = 1` or `m_(2) - m_(1) = 1` ] `rArr 1 = 25 - 4k` `rArr 4k = 24 rArr k = 6` |
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| 45. |
If the origin and the points P(2,3,4), Q(1,2,3) and R(x,y,z) are coplanar, thenA. `x - 2y -z = 0`B. `x+2y + z = 0`C. `x- 2y + z = 0`D. `2x- 2y + z = 0` |
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Answer» Correct Answer - C Let `O(0,0,0)` be the origin. It is given that `O(0,0,0), P(2,3,4), Q(1,2,3)` and `R(x,y,z)` are coplanar. `:. [OR" "OP" "OQ]` `|{:(x-0,y-0,z-0),(2-0,3-0,4-0),(1-0,2-0,3-0):}|=0` `|{:(x,y,z),(2,3,4),(1,2,3):}| = 0` `x(9-8) -y(6-4) +z(4-3) = 0` `rArr x-2y + z = 0` |
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| 46. |
if lines represented by equation `px^(2)-qy^(2)=0` are distinct, thenA. `pq gt 0`B. ` pq lt 0`C. `pq = 0`D. ` p +q = 0` |
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Answer» Correct Answer - A Given pair of line is `px^(2) - qy^(2) = 0 "…"(i)` On comparing Eq. (i) with `ax^(2) + 2hxy +by^(2) = 0`, we get `a = p, b = - q, h = 0` We know that, slopes of `ax^(2) +2hxy + by^(2) = 0` are real and distinct if and only if `h^(2) - ab gt 0` `rArr 0 + pq gt0` `rArr pq gt 0` |
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| 47. |
Let `v_(1)` be the frequency of series limit of Lyman series, `v_(2)` the frequency of the first line of Lyman series and `v_(3)` the frequency of series limit of Balmer series. Then which of the following is correct ?A. `v_(1) - v_(2) = v_(3)`B. `v_(1) v_(3) = v_(2)`C. `v_(1) + v_(2) = v_(3)`D. `v_(1) - v_(3) = 2v_(1)` |
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Answer» Correct Answer - A As we know, `v = nlambda`, `rArr 1/lambda = n/v rArr 1/lambda = R ((1)/(n_(1)^(2))- 1/(n_(2)^(2)))` `rArr v = Rc ((1)/(n_(1)^(2)) - 1/(n_(2)^(2)))` `:.` `v_(2) = Rc((1)/(2^(2)) - 1/(3^(2))) = Rc (1/4 - 1/9)"…"(i)` `v_(1) = Rc ((1)/(2^(2))) = (Rc)/(4)` `v_(3) = Rc ((1)/(3^(2))) = (Rc)/(9)` `rArr v_(1) - v_(3) = Rc (1/4 - 1/9)` From Eqs. (i) and (ii), we get `v_(1) - v_(3) = v_(2)` `rArr v_(1) - v_(2) =v_(3)` |
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| 48. |
In Fraunhofer diffraction pattern, slit width is `0.2 mm` and screen is at 2 m away from the lens. If wavelength of light used is `5000 Å`, then the distance between the first minimum on either side of the central maximum is (`theta` is small and measured in radian)A. `10^(-1) m`B. `10^(-2) m`C. `2 xx 10^(-2) m`D. `2 xx 10^(-1) m` |
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Answer» Correct Answer - B Given , `a = 0.2xx 10^(-3)m , D = 2`m `lambda = 5 xx 10^(-7) m` As, `y = (lambdaD)/(a) = (5 xx 10^(-7) xx 2)/(0.2 xx 10^(-3))` `:. Y = (5 xx 10^(-7))/(10^(-4))` `:. Y = 5 xx 10^(-3) m` Distance between 1st minima on either side. `= 5 xx 10^(-3) + 5 xx 10^(-3) = 10 xx 10^(-3) = 10^(-2) m` |
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| 49. |
A radioactive element has rate of disintegration `10,000` disintegrations per minute at a particular instant. After four minutes it becomes `2500` disintegrations per minute. The decay constant per minute isA. `0.2 log_(e) 2`B. `0.5 log_(e) 2`C. `0.6 log_(e) 2`D. `0.8 log_(e) 2` |
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Answer» Correct Answer - B Given `N_(0) = 10,000` disintegeration/min `N = 2500` disintegeration/min `t = 4` min From the radioactive decay law, we have `(N)/(N_(0)) = e^(-lambdat)` `:. (2500)/(10000) = e^(-lambdatxx 4)` `rArr 1/4 = e^(-4lambda) rArr e^(4lambda) = 4` `rArr 4 lambda = log_(e) 4 rArr 4 lambda = log_(e) 2^(2)` `rArr 4 lambda = 2 log_(e) 2` `lambda = 2/4 log_(e) 2` `:. lambda = 0.5 log_(e) 2` |
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| 50. |
The osmotic pressure of solution containing `34.2 g` of cane sugar (molar mass = 342 g `mol^(-1)` ) in 1 L of solution at `20^(@)C` is (Given `R = 0.082` L atm `K^(-1) mol^(-1)` )A. `2.40 atm`B. 3.6 atmC. 24 atmD. 0.0024 atm |
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Answer» Correct Answer - A Osmotic pressure`, pi = CRT` Where, C= concentration of solution. Given `w = 34.3g,V = 1L , M = 342 g mol^(-1)`. `T = 20^(@)C, = 20 + 273 = 293^(@)K`. `:. Pi = (W)/(MV)RT = (34.2)/(342 g mol^(-1)) xx (0.082L atm K^(-1) "mol"^(-1) xx 293 K)/(1L)` `= 2.40` atm |
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