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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. |
Two Cu wires of radii `R_(1)` and `R_(2)` are such that `(R_(1) gt R_(2))` . Then which of the following is true ?A. Transverse wave travels faster in thicker wireB. Transverse wave travels faster in thinner wireC. Travels with the same speed in both the wiresD. Does not travel |
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Answer» Correct Answer - B The velocity of a transverse wave `v = sqrt((T)/(rho A))` ` v prop (1)/(sqrtA)` or ` v prop (1)/(R)` Because the velocity of wire depend on the radius . So transverse wave travels faster in thinner wire. |
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| 2. |
The sphere of radii 8 cm and 2 cm are cooling. Their temperatures are `127^(@)C` and `527^(@)C` respectively . Find the ratio of energy radiated by them in the same timeA. `0.06`B. `0.5`C. 1D. 2 |
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Answer» Correct Answer - C Total energy radiated from a body `Q = A epsi sigma T^(4) t` or ` (Q)/(t) prop AT^(4)` `(Q)/(t) prop r^(2) T^(4) " " (because A = 4 pi r^(2))` `therefore (Q_(1))/(Q_(2)) = ((r_(1))/(r_(2)))^(2) ((T_(1))/(T_(2)))^(4)` `= ((8)/(2))^(2) [ (273 + 127)/(273 + 527)]^(4) = 1` |
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| 3. |
The electric intensity of outside a charged sphere of radius R at a distance `r (r gt R)` isA. `(sigma R^(2))/(epsi_(0) r^(2))`B. `(sigma r^(2))/(epsi_(0)R^(2))`C. `(sigma r)/(epsi_(0)R)`D. `(sigmaR)/(epsi_(0)r)` |
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Answer» Correct Answer - A The electric intensity outside a charged sphere , `E = (sigma R^(2))/(epsi_(0) r^(2))` |
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| 4. |
Capacity of a capacitor is `48mu F`. When it is charged from `0.1` C to `0.5` C , change in the energy stored isA. 2500 JB. `2.5 xx 10^(-3) J`C. `2.5 xx 10^(6) J`D. `2.42 xx 10^(-2) J` |
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Answer» Correct Answer - A Change in energy `DeltaU = (1)/(2) [(q_(1)^(2) - q_(2)^(2))/(C)]` `= (1)/(2) [ ((0.5)^(2) - (0.1)^(2))/(48 xx 10^(-6))]` `= (1)/(2)[ (0.25 - 0.01)/(48 xx 10^(-6))]` `= (1)/(2) [(0.24)/(48 xx 10^(-6))]` `= (1)/(2) [ (10^(4))/(2)]` = 2500 J |
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| 5. |
Light travels through a glass plate of thickness t and refractive index `mu.` If c is the speed of light in vacuum, the time taken by light to travel this thickness of glass isA. `t mu c`B. `(tc)/(mu)`C. `(t)/(muc)`D. `(mu t)/(c)` |
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Answer» Correct Answer - D Time taken by light ray to pass through glass slab = `(mu t)/(c)` where t = geometrical path `mut` = optical path |
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| 6. |
`Ce^(4+)` is stable . This is because ofA. half-filled d-orbitalsB. all paired electrons in d-orbitalsC. empty orbitalD. fully filled d-orbital |
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Answer» Correct Answer - C The electronic configuration of Ce is `Ce_(58) = [Xe] 4 f^(1) 5d^(1) 6s^(2)` (predicted) or = `[Xe] 4f^(2) 5d^(0) 6s^(2) `(observed) `Ce^(4+) = [Xe] 4 f^(0) 5d^(0) 6s^(0)` Since , in +4 oxidation state , all (ie, 4f , 5d and 6s) orbitals are empty and Ce gains the stable configuration of nearest inert gas , `Ce^(4+)` is most stable . |
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| 7. |
Find the differential equation of the family of circles whose centres lie on `X`-axis.A. `(d^(2)y)/(dx^(2)) + ((dy)/(dx))^(2) + 1 = 0`B. `y""(d^(2)y)/(dx^(2)) + ((dy)/(dx))^(2) - 1 = 0`C. `y""(d^(2)y)/(dx^(2)) - ((dy)/(dx))^(2) - 1 = 0`D. `y""(d^(2)y)/(dx^(2)) + ((dy)/(dx))^(2) + 1 = 0` |
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Answer» Correct Answer - D The equation of family of circle having centre at x-axis is `x^(2) + y^(2) - 2ax = 0` On differentiating , we get `2x + 2y""(dy)/(dx) - 2a = 0` Again , differentiating , we get `2 + 2 [y ""(d^(2) y)/(dx^2) + ((dy)/(dx))^(2)]`= 0 `iimplies y""(d^(2)y)/(dx^(2)) + ((dy)/(dx))^(2) + 1 = 0` |
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| 8. |
The solution of the differential equation `y (1 + log x) (dx)/(dy) - x log x = 0` isA. y log x = y + cB. x log x = ycC. y (1 + log x) = cD. logx - y = c |
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Answer» Correct Answer - B Given differential equation is `y (1 + log x ) (dx)/(dy) - x log x = 0` `implies int (1 + log x)/(x log x ) dx = int (dy)/(y)` `implies int (1)/(x log x ) dx + int (1)/(x) dx = int (1)/(y) dy` `implies log (log x) + log x = log y + log c ` `implies x log x= y c` |
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| 9. |
If `x^(2) y^(5) = (x + y)^(7) , " then " (d^(2)y)/(dx^(2))` is equal toA. `y //x ^(2)`B. `x//y`C. 1D. 0 |
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Answer» Correct Answer - D Given , `x^(2) y^(2) = (x + y)^(7)` Taking log on both sides , we get 2 log x + 5 log y = 7 log ( x + y) On differentiating , we get `(2)/(x) + (5)/(y) (dy)/(dx) = (7 )/( x + y) = ( 1 + (dy)/(dx))` `implies (dy)/(dx) ((7)/(x+y) - (5)/(y)) = (2)/(x) - (7)/(x + y)` `implies (dy)/(dx) = (y)/(x) " " .... (i)` Again , differentiating , we get `(d^(2) y)/(dx^(2)) = ( x ""(dy)/(dx) - y)/(x^(2))` ` = (x * (y //x) - y)/(x^(2)) `[ from Eq. (i) ] = 0 |
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| 10. |
The value of `lim_(x to 0) (15^(x) - 5^(x) - 3^(x) + 1)/(1 - cos 2x) ` isA. `((log 3) (log 5))/(2)`B. 2 (log 3) (log 5)C. `(log 3 + log 5)/(2)`D. None of these |
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Answer» Correct Answer - A `underset(x to 0) ("lim") (15^(x) - 5^(x) - 3^(x) + 1)/(1 - cos 2x)` `underset(x to 0)("lim") ((3^(x) - 1) (5^(x) - 1))/(1-1 + 2 sin^(2) x)` `= underset(x to 0) ("lim") ((3^(x) - 1)/(x)) ((5^(x) - 1)/(x)) ((x^(2))/(2 sin^(2) x))` `= (1)/(2) (log 3) (log 5) * 1` `[ because underset(x to 0)("lim") ((a^(x) - b^(x))/(x)) = log ((a)/(b))]` |
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| 11. |
If f(x) = x , g (x) = sinx , then `int` f(g(x)) dx is equal toA. sinx + cB. `-cos x + c`C. `(x^(2))/(2) + c`D. `x sin x + c` |
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Answer» Correct Answer - B `int f(g (x) ) dx = int f (sin x ) dx` `= int sinx dx ` `= - cos x + c ` |
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| 12. |
The equation of tangent to the curve given by ` x = 3 cos theta , y = 3 sin theta , at theta = (pi)/(4)` isA. `x + y = sqrt2`B. `3x + y = 3sqrt2`C. `x + y = 3sqrt2`D. ` x + 3y = 3sqrt2` |
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Answer» Correct Answer - C Given x = 3` cos theta , y = sin theta` . On squaring and adding we get `x^(2) + y^(2) = 9` , which represent a circle . `therefore` Equation of tangent at `theta = (pi)/(4)` is `x * (3 cos""(pi)/(4)) + y * (3 sin ""(pi)/(4)) = 9` `implies x + y = 3sqrt2` |
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| 13. |
The value of `int_(0)^(pi//2)` log (sinx) dx isA. `(pi)/(2) ` log 2B. `pi` log 2C. `-(pi)/(2)` log 2D. `2pi` log 2 |
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Answer» Correct Answer - A Let `I = int _(0)^(pi//2) log (cosec x) dx` `= int_(0)^(pi//2) log ((1)/(sinx)) dx` ` = - int_(0)^(pi//2) log sin x `dx `= (pi)/(2) ""log 2` `[because int_(0)^(pi//2) log sin x dx = - (pi)/(2)""log 2`] |
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| 14. |
The value of a and b such that the function `f(x)={(-2sinx"," , -pi le x le -(pi)/(2)),(a sinx+b",", -(pi)/(2) lt x lt (pi)/(2)),(cosx",", (pi)/(2) le x le pi ):}` is continuous in `[-pi,pi]` areA. `-1,0`B. `1,0`C. 1,1D. `-1,1` |
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Answer» Correct Answer - D For continuity in `[-pi , pi]` , we must have At `x = - ""(pi)/(2) , f (-""(pi)/(2)) = underset(x to (-""(pi)/(2))^(-)) ("lim") (-2 sin x )` `= underset( x to (-""(pi)/(2))^(+)) (a sin x + b)` `implies 2 = - a + b " " … (i) ` At `x = pi/2 ,` `f((pi)/(2)) = underset(x to ((pi)/(2))^(-)) ("lim") ( asin x + b) = underset( x to ((pi)/(2))^(+)) ("lim") cos x` `implies 0 = a + b " " .... (ii)` On solving Eqs. (i) and (ii) we get , `a = -1 , b = 1` |
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| 15. |
The heat of formation of water is `260 kJ`. How much `H_(2)O` is decomposed by `130 kJ` heat ?A. `0.25` molB. 1 molC. `0.5` molD. 2 mol |
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Answer» Correct Answer - C The reaction for the formation of water is as `H_(2) + (1)/(2) O_(2) to H_(2) O , Delta H = 260 kJ … (i)` On reversing Eq. (i) , we get `underset(1 "mol")(H_(2)O) to H_(2) + (1)/(2)O_(2) , ` `Delta H = -260 kJ " " … (ii)` `because` By 260 kJ heat water decomposed = 1 mol `therefore` 130 kJ heat will decompose water = `(1 xx 130)/(260)` = 0.5 mol |
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| 16. |
Which of the following relation is true ?A. ` Y = 2eta ( 1 - sigma)`B. `Y = 2 eta (1 + 2 sigma)`C. `Y = 2 eta ( 1 - sigma)`D. `(1 + sigma) 2 eta = Y` |
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Answer» Correct Answer - D We know that `Y = (1 + sigma) 2 eta` |
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| 17. |
Amoung the following , the formula of saturated fatty acid isA. `C_(17)H_(29) COOH`B. `C_(17)H_(35) COOH`C. `C_(17)H_(31) COOH`D. `C_(17)H_(33)COOH` |
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Answer» Correct Answer - B The general formula for saturated fatty acids is `C_(n) H_(2n + 1) COOH` , Among the given acids , only `C_(17)H_(35) COOH` satisfy this formula , hence it is a saturated fatty acid. |
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| 18. |
Reaction for the formation of NaCl isA. Na(g) + `Cl_(2) (g) to NaCl(s)`B. Na(s) ` + (1)/(2) Cl_(2) (g) to NaCl(s)`C. `Na(g) + (1)/(2) Cl_(2) (g) to NaCl(s)`D. `Na(g) + Cl_(2) (g) to NaCl (g) ` |
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Answer» Correct Answer - B The correct reaction for the formation of NaCl is `Na(s) + (1)/(2)Cl(g) to NaCl(s)` (As the compounds are formed from their elements when they are present in their most stable form) . |
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| 19. |
U is the PE of an oscillating particle and F is the force acting on it at a given instant . Which of the following is true ?A. `(U)/(F) + x = 0`B. `(2U)/(F) + x = 0`C. `(F)/(U) + x = 0`D. `(F)/(2U) + x = 0` |
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Answer» Correct Answer - B The potential energy , U = `(1)/(2) k x^(2)` `2 U = k x^(2)` `2U = - Fx" " [because F = -kx]` or `" " (2U)/(F) = -x` or `(2U)/(F) + x = 0` |
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| 20. |
Which of the following is a nucleophile ?A. `BH_(3)`B. `NH_(3)`C. `AlCl_(3)`D. All of these |
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Answer» Correct Answer - B Nucleophiles (ie, nucleus loving are the species that have atleast one lone pair of electrons for donation, ie, they can also be called Lewis bases . Among the given, `BH_(3)` and `AlCl_(3)` , being electron dificient , behave like electrophiles while `overset(..)(N) H_(3)`, due to the presence of a lone pair over N, is a nucleophile. |
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| 21. |
Find the correct equationA. `E_(2) - E_(1) - H_(2) + H_(1) = n_(2)RT - n_(1)RT`B. `E_(2) - E_(1) - H_(2) - H_(1) = n_(2)RT - n_(1)RT`C. `H_(2) - H_(1) - E_(2) + E_(1) = n_(2) RT - n_(1) RT`D. `H_(2) - H_(1) - E_(2) + E_(1) = n_(2) RT - n_(1)RT` |
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Answer» Correct Answer - C We know that , `DeltaH = Delta E + Deltan_(g) RT` `H_(2) - H_(1) = E_(2) - E_(1) + (n_(2) RT- n_(1) RT)` or `H_(2) - H_(1) - E_(2) + E_(1) = n_(2) RT- n_(1) RT` |
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| 22. |
For nuclear fusion reactions , the fusion temperature is of the order ofA. `10^(5) K`B. `10^(3) K`C. `10^(7) K`D. `100` K |
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Answer» Correct Answer - C For the occurrence of nuclear fusion is a very high temperature (ie, 20 million K or `2 xx 10^(7)` K) is as thermonuclear reactions. |
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| 23. |
1 u (amu) is equal toA. `1.492 xx 10^(-10) J`B. `1.492 xx 10^(-7) J`C. `1.492 xx 10^(-13) J`D. `6.023 xx 10^(23) J` |
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Answer» Correct Answer - A `1 u = 1.66 xx 10^(-24)`g `= 1.66 xx 10^(-27)` kg `= 1.66 xx 10^(-27) xx (3 xx 10^(8) )^(2) kg m^(2) s^(-2) (because E = mc^(2))` `= 1.494 xx 10^(-10) J` |
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| 24. |
1 mol alcohol reacts with Na to give what weight of hydrogen ?A. 1 gB. 2 gC. 3 gD. 3.5 g |
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Answer» Correct Answer - A `underset(1 "mol")underset(2"mol")(2C_(2)H_(5)OH) + 2Na to 2 C_(2)H_(5)ONa + underset((1)/(2) "mol")underset(1"mol")(H_(2))` The weight of `(1)/(2)` mol `H_(2) = (2)/(2)`g = 1 g Thus , 1g `H_(2)` is produced when 1 mole of alcohol reacts with Na. |
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| 25. |
LED is a p-n junction diode which isA. forward biasedB. either forward biased or reverse biasedC. reverse biasedD. neither forward biased nor reverse biased |
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Answer» Correct Answer - A When a junction diode is forward biased , energy is released at the junction due to recombination of electrons and holes . In the junction diode made of gallium arsenide or indium phosphide , the energy is released in visible region . Such a junction diode is called a light emitting diode or LED. |
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| 26. |
From which group the fats belong ?A. Carboxylic acidsB. CarbonylC. EsterD. Fatty acids |
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Answer» Correct Answer - C Fats are obtained by the reaction of glyceral (an alcohol) with different long chain carboxylic acids . Hence , chemically these are carboxylic esters or simply esters and are also known as glycerides . |
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| 27. |
Ethyl methyl ketone is obtained by heating calcium salts ofA. formic acid + propionic acidB. acetic acid + propionic acidC. acetic acid onlyD. acetic acid + methanoic acid |
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Answer» Correct Answer - B Ethyl methyl ketone is an unsymmetric ketone , so two different acids (in the form of calcium salt) are used . The structure of ethyl methyl ketone is `C_(2)H_(5) COCH_(3)` The structure shows the presence of 4 carbon atoms , thus the calcium salts must be of acetic acid and propionic acid to form `C_(2)H_(5)COCH_(3)` . `(C_(2)H_(5) COO)_(2) Ca + (CH_(3)COO)_(2) Ca overset("Distill")(to) C_(2)H_(5)COCH_(3) + CaCO_(3)` |
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| 28. |
The orbital frequency of an electron in the hydrogen atom is proportional toA. `n^(3)`B. `n^(-3)`C. `n^(1)`D. `n^(0)` |
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Answer» Correct Answer - B Time period of electron , `T = (4 epsi_(0)^(2) n^(3) h^(3))/(mZ^(2) e^(4))` `therefore T prop n^(3)` `therefore (1)/("frequency" (f)) prop n^(3)` or `f prop n^(-3)` |
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| 29. |
Potassium metabisulphite is a (an)A. preservativeB. antioxidantC. artificial sweetenerD. Both (a) and (b) |
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Answer» Correct Answer - D Potassium metabisulphite , `K_(2)S_(2)O_(5)`, is a white crystalline solid with pungent odour of sulphur . It is used as an antimicrobial preservative , antioxidant and bleaching agent in food . |
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| 30. |
How many isomeric vicinal- dihalides are possible for the compound having molecular formula `C_(3)H_(6) Cl_(2)` ?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A In vicinal-dihalides , the two `-Cl` atoms are present at adjacent positions . In case of `C_(3) H_(6) Cl_(2)` only one such structure is possible . Hence , only one vicinal- dihalide is possible for `C_(3)H_(6)Cl_(2)`. `underset(1,2- "dichloropropane")(CH_(3) - underset(Cl)underset(|)(CH)- underset(Cl)underset(|)(C)H_(2))` |
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| 31. |
Which of the ions is colourless inspite of the presence of unpaired electrons ?A. `La^(3+)`B. `Eu^(3+)`C. `Gd^(3+)`D. `Lu^(3+)` |
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Answer» Correct Answer - C Gadolinium does not absorb light in the visible region of the spectrum , hence it is colourless inspite of the presence of 7 unpaired electrons . Being colourless , gadolium is sometimes used as a support for other rate earth phosphors and luminesce in the knowledge that it will not quench their intense colour. |
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| 32. |
4 L-atom is equal toA. `40.50` calB. `78.75` calC. `95.23` calD. 96 cal |
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Answer» Correct Answer - D `because ` 1 L- atom = 101. 266 J `therefore` 4 L-atm = 101. `266 xx 4` J `= 405.06 J` `= (405.06)/(4.2)"cal" " " [ because` 1 cal = 4.2 J] `= 96.4` cal |
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| 33. |
Chlorine `(Cl^(17))` free radial contains how many electrons around the nucleus ?A. 16B. 17C. 18D. 19 |
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Answer» Correct Answer - B In free radicals , the number of electrons are same as that in isolated neutral atom. `because` In Cl atom, the number of electrons = 17 `therefore` In Cl free radical , the number of electrons = 17 |
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| 34. |
`NH_(4)Cl` is acidic due toA. cationic hydrolysisB. anionic hydrolysisC. its ionic natureD. pH `gt` 7 |
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Answer» Correct Answer - A `NH_(4)Cl + H_(2) O to NH_(4)OH + HCl` or `NH_(4)^(+) + Cl^(-) + H_(2)O to NH_(4) OH + H^(+) + Cl^(-) ` or `underset("cation") (NH_(4)^(+)) + H_(2)O to NH_(4)OH + H^(+)` Due to the presence of `H^(+)` , after cationic hydrolysis , in the solution , the solution of `NH_(4)Cl` is acidic . |
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| 35. |
Given , for `Sn^(+4) // Sn^(2+)` , standard reduction potential is `0.15` V and for `Au^(3+)//Au` , standard reduction potential is 1.5 V. For the reaction , `3Sn^(2+) + 2 Au^(3+) to 3 Sn^(4+) + 2 Au , ` the value of `E_("cell")^(@)` is ,A. `+1.35`B. `+2.55`C. `-1.35`D. `-2.55` |
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Answer» Correct Answer - A `E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@)` = `1.5 - 0.15 = 1.35 V` |
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| 36. |
How many gram equivalents of NaOH are required to neutralize `25 cm^(3)` of decinormal HCl solution ?A. `0.00125`B. `0.0025`C. `0.0050`D. `0.025` |
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Answer» Correct Answer - B For acid-base reactions , g-equivalent of acid (HCL) = g-equivalent of base (NaOH) `therefore` g-equivalent of acid = `(25 xx 10^(-3)) dm^(3) xx (1)/(10)` =` 25 xx 10^(-4)` or = 0.0025 = g-equivalent of NaOH |
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| 37. |
If two dice are thrown together . Then , the probability that the sum of the numbers appearing on them is a prime number, isA. `1//2`B. `3//7`C. `5//12`D. `7//12` |
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Answer» Correct Answer - C Total cases = 36 Prime number are 2 , 3 , 5 , 7 , 11. `therefore` Favourable cases = 15 Required probability = `(15)/(36) = (5)/(12)` |
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| 38. |
If `e_(1)` and `e_(2)` represent the eccentricity of the curves `6x^(2) - 9y^(2) = 144` and `9x^(2) - 16y^(2) = 144` respectively . Then `(1)/(e_(1)^(2)) + (1)/(e_(2)^(2))` is equal to |
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Answer» Correct Answer - B Given curves are `(x^(2))/(9) - (y^2)/(16) = 1` and `(x^(2))/(16) - (y^(2))/(9) = 1` `therefore " " e_(1) = sqrt(1 + (16)/(9)) = (5)/(3)` and `e_(2) = sqrt(1 + (9)/(16)) = (5)/(4)` Now , `(1)/(e_(1)^(2)) + (1)/(e_(2)^(2)) = (9)/(25) + (16)/(25) = 1` |
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| 39. |
Ratio of loss in solvent to gain in `CaCl_(2)` tube isA. `(p^(@))/(p)`B. `(p)/(p^(@))`C. `(p^(@) - p)/(p^(@))`D. `(p-p^(@))/(p)` |
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Answer» Correct Answer - C `because` Loss in the weight of solvent = `p^(@) - p` `because` Gain in the weight of `CaCl_(2) ` tube = `p^(@)` `implies ("Loss in the weight of solvent")/("Gain in the weight of" CaCl_(2) "tube") = (p^(@) - p)/(p^(@))` |
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| 40. |
450 mg of glucose is dissolved in 100 g of solvent. What is the molality of the solution ?A. `0.0025` mB. `0.025` mC. `0.25`mD. 2.5 m |
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Answer» Correct Answer - B Molality = `("moles of solute" xx 1000)/("weight of solvent (in g)")` or `= ("weight of solute" xx 100)/("mol., wt of solute" xx "weight of solvent (in g)")` `= (450 xx 10^(-3) xx 1000)/(180 xx 100)` `[because` Mol. Mass of glucose = 180] `= 0.025` m |
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| 41. |
Oxidation potential of unimoles of calomel isA. `+ 0.25` VB. `0.00`VC. `+0.278 V`D. `-0.28V` |
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Answer» Correct Answer - D Calomel is used as a secondary reference electrode. Since Hg is less reactive than H , its standard reduction potential is positive . Hence , its standard oxidation potential must be negative , ie, `-2.28 V` . [As standard reduction potential = -standard oxidation potential.] |
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| 42. |
Ethanolic KOH givesA. dehalogenation reactionsB. dehydrogenation reactionsC. dehydrohalogenation reactionsD. substitution reactions |
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Answer» Correct Answer - C Ethanolic KOH removes hydrogen and halogen atoms from a haloalkane , thus it is a dehydrohalogenating reagent and gives the same type of reactions . |
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| 43. |
Glucose on reaction with Fehling solution givesA. cupric oxideB. cuprous oxideC. saccharic acidD. both (b) and (c) |
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Answer» Correct Answer - B `C_(5) underset("glucose")(H_(11)O_(5)CH)O + underset("(from Fehling solution)")(CuO) overset(OH^(-))(to) underset("gluconic acid")(C_(5)H_(11) O_(5)COOH) + underset("(red ppt)")underset("cuprous oxide")(Cu_(2)O)` |
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| 44. |
The equation of the plane which passes through (2,-3,1) and is normal to the line joining the points (3,4,-1) and (2,-1,5) is given byA. `x + 5y - 6z + 19 =0`B. `x - 5y + 6z - 19 = 0 `C. `x + 5y + 6z + 19 = 0`D. `x - 5y - 6z - 19 = 0` |
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Answer» Correct Answer - A The equation of plane is `(x - 2) + 5(y + 3) - 6(z-1) = 0` `implies x + 5y - 6z + 19 = 0` |
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| 45. |
Two coins are tossed simultaneously. Then, the value of E(X), where X denotes the number of heads isA. `(1)/(2)`B. `2`C. 1D. None of these |
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Answer» Correct Answer - C Here , n = 2 , p = `(1)/(2)` E(X) = mean = np = `2 xx (1)/(2) = 1` |
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| 46. |
A four digit number is to be formed using the digits 1,2,3, 4, 5,6,7 (no digit is being repeated in any number) . Then , the probability that it is `gt 4000` , isA. `3//2`B. `1//2`C. `4//7`D. `3//7` |
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Answer» Correct Answer - C Total cases = `""^(7)P_(4) = 840` Favourable cases `= 4 xx 6 xx 5 xx 4 = 480` `therefore` Required probability = `(480)/(840) = (4)/(7)` |
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| 47. |
If matrix `A=[(a,b),(c,d)]`, then `|A|^(-1)` is equal toA. ad-bcB. `(1)/(ad-bc)`C. `(1)/(ad-bc) [{:(d,-b), (-c, a):}]`D. None of these |
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Answer» Correct Answer - B Given , A = `[{:(a , b) , (c , d):}]` `|A| `= ad - bc `therefore |A|^(-1) = (1)/(ad - bc)` |
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| 48. |
The position vectors of vertices of a `DeltaABC` are `4hati - 2 hatj , hati + 4hatj - 3hatk` and `-hati + 5hatj + hatk` respectively , then `angleABC` is equal toA. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
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Answer» Correct Answer - D Here , `vec(AB) = - 3hati + 6hatj - 3hatk , vec(BC) = - 2hati + hatj + 4 hatk ` and `vec(AB ) * vec(BC) = 6 + 6 - 12 = 0` `implies angle ABC = (pi)/(2)` |
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| 49. |
The volume of the solid formed by rotating the area enclosed between the curve `y^(2)=4x,x=4` and x = 5 about X-axis is (in cubic units)A. `18 pi`B. `36 pi`C. `9 pi`D. `24 pi` |
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Answer» Correct Answer - A Volume of the solid `= int_(4)^(5) pi y^2 dx` `= pi int_(4)^(5) 4x dx` `= 4pi [ (x^(2))/(2)]^(5)` `= 2x (25-16)` `= 18 pi` cu unit |
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| 50. |
When a resistance of 100`Omega` is connected in series with a galvanometer of resistance R, then its range is V. To double its range, a resistance of 1000`Omega` is connected in series. Find the value of R.A. `700 Omega`B. `800 Omega`C. `900 Omega`D. `100 Omega` |
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Answer» Correct Answer - C When a resistance of `100 Omega` is connected in series current , i= `(V)/(100 + R) " " …. (i)` when a resistance of `1000 Omega` is connected in series , the its range double current , `I = (2V)/(1100 + R) " " …. (ii)` From Eqs.(i) and (ii) `(V)/(100 + R) = (2V) / (1100 +R)` `R = 900 Omega` |
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