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17401.

A particle is projected wityh velocity 20 `ms^(-1)` at angle `60^(@)` above horizontal up on an inclined plane having inclination `30^(@)`. Find its shortest distance from the inclined plane when it is moving `53^(@)` above horizontal. (Using `=10ms^(-2)`)A. 11.1mB. 8.2mC. 3.3mD. 14.4m

Answer» `V_("vertical") U_("vertical")-"gt"`
`10 tan 53^(@)=20 sin 60^(@)-10t`
`(4)/(3)=sqrt(3)-1`
`t=sqrt(3)-(4)/(3)`
`s_(y)=u_(y)+a_(y)t^(2)`
`=10xx((3sqrt(3)-4)/(3))-(1)/(2)g cos 30^(@) ((3sqrt(3)-4)/(3))^(2)`
`=10xx((3sqrt(3)-4)/(3))-(1)/(2)xx10sqrt(3)/(2)((3sqrt(3)-4)/(3))^(2)`
`=3.99-2.5xx1.732((3xx1.732-4)/(7))^(2)`
`=3.99-4.33xx0.16`
`=3.99-0.688`
`=3.99-0.69=3.30`
17402.

What is the role of iodine in the refining of titanium ?

Answer» When heated with iodine to about 500 K, titanium metal forms a volatile iodidie `(TiI_(4))` leaving behind the impurities. Upon further heating to 1200 K, `TiI_(4)` decomposes to give pure titatnium.
`Ti+2I_(2) overset(500 K) to underset("Volatile")(TiI_(4)), TiI_(4) overset(1200 K) to underset("Pure")(Ti) +2I_(2)`
17403.

How many set of 4 quantum numbers are possible for last electron of Sc? [Divide your answer by 2]

Answer» `Sc_(21)[Ar]_(18)4s^(2)3d^(1)rArr` Fove d prnotals two types of spin quantum number so toal set `=5xx2=10`
`(10)/(2)=5`
17404.

What is maximum number of electron in an atom whose highest energy electrons have principle quantum number n=5. [Divide your answer by 23]

Answer» `[Ra]7s^(2)5f^(14)6d^(10)7p^(6)8s^(2)5g^(18)`
17405.

Figure shows equi-potential surfaces, at which point on labeled surfaces will an electron have the highest potential energy ?.A. point AB. point BC. point CD. point D

Answer» Correct Answer - B
b. Potential energy of an electron at any point is `U=-eV`, where `V` is the potential at that point. `V` is negative maximum at point `B`, hence `U` is maximum for this point.
17406.

`1.6g of CH_(4)` contains same number of electrons as in :A. `4.48L` of `CH_(4)at NTP`B. `6.02xx10^(22) "molecules of" CH_(4)`C. `1 mol of Na`D. `2.24 L` of `O_(2) at STP`

Answer» Correct Answer - C
`1.6g of CH_(4)` contains same ………….
`1.6g CH_(4)= 0.1mol CH_(4) implies 6.023xx10^(22) "Molecules" implies 10`
`xx6.02xx10^(22)` electrons
(1) `4.48L CH_(4) at N.T.P implies 0.2 mol CH_(4)`
`implies 20xx6.023xx10^(22)` electrons
(2) `6.02xx11^(22)` molecules of `CH_(4)`
`implies 10xx6.023xx10^(22)` electrons
(3) `1 mol Na^(+) implies 10xx6.02xx10^(23)` electrons
(4) `2.24 L of O_(2) at N.T.P. = 0.1 mol O_(2)`
`implies 6.02xx10^(22)` molecules of `O_(2) implies 16xx6.02xx10^(22)` electrons
17407.

`8 g of O^(2-)` ion has amount of charge equal to : `(N_(A) = 6.02xx10^(23) , e = - 1.6xx10^(-19)C)`A. `5N_(A)eC`B. `2N_(A)eC`C. `N_(A)eC`D. `(1)/(2)N_(A)eC`

Answer» Correct Answer - C
`8 g of O^(2-)` ion has amount …………
Moles of `O^(2) = (8)/(16) = (1)/(2)`mole
Number of `O^(2-)` ions `= (1)/(2)xxN_(A)`
Total Change on `O^(2-) = (1)/(2)xxN_(A)xx2e=N_(A)eC`.
17408.

At `27^(@)C` and `3.0` atm pressure, the density of propene gas is :A. `10.1 gL^(-1)`B. `4.03 gL^(-1)`C. `5.12 gL^(-1)`D. `0.506 gL^(-1)`

Answer» Correct Answer - C
At `27^(@)C` and `3.0 atm` ……….
`d = PM//RT = 3xx42//0.082xx300`
`=5.12 gL^(-1)`
17409.

Assertion : For the extraction of iron, haematite ore is used. Reason : Haematite is a carbonate ore fo iron. (1) Both the assertion and reson are correct, but the reason is not hte correct explanation for the assertion. (2) Both the assertion and reason are correct and the reason is the correct explanation for the assertion. (3) Only the reason is correct. (4) Only the assertion is correct. 

Answer»

Correct option (4) Only the assertion is correct.

Explanation:

Extration fo Fe is done from Haematite or this is true but reason is strong as Haematitte is Fe2O3.

17410.

During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, ΔU (in kJ) is : (1) 12 (2) –8 (3) 8 (4) –12

Answer»

Correct option (3) 8

Explanation:

Work done on system = +10 kT

Heat escaped = –2kI 

ΔU = q + w

= 10 – 2 = 8 KJ

17411.

The correct match between Item-I and Item-II is :Item-IItem-)(A) Allosteric effect(P) Molecule binding to the active site of enzyme(B) Competitive inhibitor(Q) Molecule crucial for communication in the body(C) Receptor(R) Molecule binding to a site other than the active site of enzyme(D) Poison(S) Molecule binding to the enzyme covalently(1)  (A) → (R); (B) → (P); (C) → (S); (D) → (Q)(2)  (A) → (P); (B) → (R); (C) → (Q); (D) →(S) (3)  (A) → (R); (B) → (P); (C) → (Q); (D) →(S)(4)  (A) → (P); (B) → (R); (C) →(S); (D) → (Q)

Answer»

Correct option  (3)  (A)  (R); (B)  (P); (C) → (Q); (D) (S)

Informative

17412.

The pair that does NOT require calcination is : (1) ZnO and Fe2O3 .xH2O (2) Fe2O3 and CaCO3.MgCO3 (3) ZnCO3 and CaO (4) ZnO and MgO

Answer»

Correct option:  (4) ZnO and MgO

Explanation:

ZnO and MgO both are in oxide form so there is no change on calcination of ZnO and MgO whereas carbonate decompose into oxide and CO2 on calcination.

17413.

The electrolytes usually used in th electroplating of gold and silver, respectively, are (1) [Au(CN)2] – and [AgCl2] – (2) [Au(CN)2] – and [Ag(CN)2] – (3) [Au(NH3)2] + and [Ag(CN)2] – (4) [Au(OH)4] – and [Ag(OH)2]

Answer»

The correct option is (2).

Explanation:

The electrolytes usually used in the electroplating of gold & silver, respectively are :

[Au(CN)2]θ and [Ag(CN)2]θ

17414.

A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then :(1)  m + n = 68(2)  m = n = 78(3)  m = n = 68(4)  n = m – 8

Answer»

Correct option  (2) m = n = 78

Explanation:

Given : (8 males, 5 females)

Committee to be selected = 11 members

m = no. of ways the committee is formed with at least 6 males.

 (6M, 5F) or (7M, 4F) or (8M, 3F)

= 8C6 × 5C5 + 8C7 × 5C4 + 8C8 × 5C3 = 78

n = no. of ways the committee is formed with atleast 3 female 

(8M, 3F) or (7M, 4F) or (6M, 5F)

= 8C8 × 5C3 + 8C7 × 5C4 + 8C6 × 5C5 

= 10 + 40 + 28 = 78 

m = n = 78

17415.

The major product 'X' formed in the following reaction is : 

Answer»

Correct option: (4)

Explanation:

NaBH4 does not reduces ester functional group.

17416.

In the arrangement given below 20 mole of `N_(2)` and 5 mole of He are present above water. The total pressure above the water is 60 atm. If `K_(H)` of `N_(2)` is `10^(5)` atm. How many moles of `N_(2)` are dissolved in water? give your answer after multiplying by `10^(3)` The vapour pressure of water is 10 atm.

Answer» Correct Answer - 8
17417.

Consider the following compound: The set of compound/s which give at least one same organic product after reductive ozonolysis:A. I and IIB. II andIIIC. I and IVD. II and IV

Answer» Correct Answer - A::B
17418.

Dilithium is crucial to the propulsion system of some starships. Dilithium is fomred by the adhesion of two lithium atoms in the gas phase: `Li(g)+Li(g)toLi_(2)(g)` The enthalpy of formation of dilithium is not easily measurable by direct means. However, th following thermochemicalparameters are known. `DeltaH_(f)^(@)` of `Li_(g)=150kJ//mol` IE of `Li_(g)=5eV//"atom"[1eV//"atom"=100kJ//mol]` `BE` of `Li_((g))^(+)=130kJ//mol` `IE` of `Li_(2(g))=5eV//"molecule"` `[IE=` ionisation energy] `[BE=` bond energy] Q. What is `DeltaH_(f)^(@)` of `Li_(2(g))`?A. `130kJ//mol`B. `170kJ//mol`C. `270kJ//mol`D. `-270kJ//mol`

Answer» Correct Answer - A
17419.

Probability of hitting a target independently of 4 person are 1/2, 1/3, 1/4, 1/8. Then the probability that target is hit, is .....(1) 1/192(2) 5/192(3) 25/32(4) 7/32

Answer»

The correct option (3) 25/32  

Explanation:

P(H) = 1 - P(Not Hitting)

= 1 - 1/2 x 2/3 x 3/4 x 7/8 = 25/32

17420.

The expression ~ (~ p → q) is logically equivalent to : (1) p ∧ q(2) ~ p ∧ q (3) ~ p ∧ ~ q (4) p ∧ ~ q

Answer»

Correct option: (3) ~ p ∧ ~ q 

Explanation:

~(~p → q) 

= ~p ∧ ~q 

17421.

If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th term is : (1) –25  (2) –35  (3) –36  (4) 25 

Answer»

Correct option  (1) –25 

Explanation:

Let the three numbers in A.P. are 

a – d, a, a + d

given that : a – d + a + d = 33

 a = 11

and (a – d)(a)(a + d) = 1155

 a(a2 – d2 ) = 1155

 11(121 – d2 ) = 1155

 d2 = 16

 d = ±4

If d = 4 then first term a – d = 7 

If d = –4 then first term a – d = 15 

T11 = 7 + 10(4) = 47

or T11 = 15 + 10(–4) = –25

17422.

The  focal  chord  to  y^2  =  16 x  is  tangent  to  (x  -  6)^2  +  y^2  =  2,  then  the  possible  values  of  the  slope  of  this chord  are:

Answer» Refers to the below link

http://bit.ly/2Ieimfp
17423.

1° = ...............(a) 600'' (b) 3600'' (c) 180''  (d) 3600'

Answer»

 (b) 3600''

Explanation:

1° = 6'

and     1' = 6''

hence 1° = 60' = 60 × 60'' = 3600''  

17424.

1 Mev = ....... ev(a) 107 (b) 104 (c) 105   (d) 106

Answer»

(d) 10 

Explanation:

MeV = 1 × 106 eV 

17425.

A plank with a uniform sphere placed on it rests on a smooth horizontal plane. The plank is pulled to the right by a constant force `F`. It the sphere does not slip over the plank, then A. Both have the same accelerationB. Acceleration of the centre of sphere is less than that of the plankC. Work done by friction acting on the sphere is equal to its total kinetic energyD. Total kinetic energy of the system is equal to work done by force `F`

Answer» Correct Answer - B::C::D
For pure rolling motion, static friction would act in the forward direction on the sphere. Work done by static friction is change in the kinetic energy of the sphere. Work done by static friction on the complete sphere + Plank system is zero.
17426.

When beats are formed between sound waves of slightly different frequencies, the intensity of the sound heard changes from maximum to minimum in 0.2 s. The difference in frequencies of the two sound waves is (a) 5 Hz (b) 4 Hz (c) 2.5 Hz (d) 2 Hz

Answer»

Correct Answer is: (c) 2.5 Hz

The duration of one beat is from one maximum to the next maximum of intensity. In this case, it is 0.4 s.

∴ the number of beats per second = 2.5 = the difference in frequencies. 

17427.

A sphere of mass m is given some angular velocity about a horizontal axis through its centre, and gently placed on a plank of mass m. The coefficient of friction between the two is μ. The plank rests on a smooth horizontal surface. The initial acceleration of the sphere relative to the plank will be(a) zero(b) μg (c) 7/5 μg (d) 2μg

Answer»

Correct Answer is: (d) 2μg

The only horizontal forces acting on the two bodies are those due to friction of magnitude μmg each, in opposite directions. Hence, they have accelerations μg each, in opposite directions.

17428.

In a Young’s double-slit experiment using identical slits, when only one slit is used, the total energy reaching the screen is E0 and the intensity of light at any point on the screen is I0. When both slits are used, and fringes are formed on the screen, the total energy reaching the screen is E and the maximum intensity on the screen is I. Then, (a) E = 2E0, I = 2I0 (b) E = 4E0, I = 4I0 (c) E = 2E0, I = 4I0 (d) E = 4E0, I = 2I0

Answer»

Correct Answer is: (c) E = 2E0, I = 4I0 

17429.

In a Young’s double-slit experiment using slits of unequal widths, the intensities on the screen due to the slits are in the ratio 4 : 9 when the slits are used separately. When they are used together, the ratio of the intensity at a dark fringe to the intensity at a bright fringe on the screen will be (a) 4 : 9 (b) 1 : 9 (c) 9 : 16 (d) 1 : 25

Answer»

Correct Answer is: (d) 1 : 25

17430.

A mixture of nitrogen and water vapours is admitted to falsk at 760 torr which contains a sufficient solid dryig agent, after long time the pressure reached a stedy value of 722 torr. If the experiment is done at `27^(@)C` and drying agent increaes in weight by 0.9gm, what iws the volume of flask? Neglect any possible vapour pressure of drying agent and volume occupied by drying agent.A. 443.34LB. 246.3LC. 12.315LD. 24.63L

Answer» Vapore pressure of water =760-722=38 torr
`n_(H_(2)O)=(0.9)/(18)=0.05`
`V=(nRT)/(P)=(0.05xx(0.0821)(300))/((38)/(760))=24.63L`
17431.

The ratio of population of 2 energy level is 1.6 x 10-29. Find the wavelength of light emitted by spontaneous emission at 303 K.

Answer»

Given \(\frac{N_2}{N_1}=1.6\times10^{-29}\)

At the thermal equilibrium the ratio

\(\frac{N_2}{N_1} \) = exp(\(\frac{-E_2-E_1}{KT}\))

\(\because\) E2 - E1 = \(\frac{hc}\lambda\) 

1.6 x 10-29 = exp(\(\frac{-hc}{\lambda KT}\))

ln(1.6 x 10-29) = \(\frac{-hc}{\lambda KT}\)

ln(1.6) - 29ln(10) = \(-\frac{hc}{\lambda kT}\) 

0.470 - 29 x 2.30 = \(-\frac{hc}{\lambda KT}\)

0.470 - 66.7 = \(\frac{-hc}{\lambda KT}\)

-66.23 = \(\frac{-hc}{\lambda KT}\) 

\(\lambda = \frac{6.626\times10^{-34}\times3\times10^8}{1.38\times10^{-23}\times303\times66.23}\) 

\(\lambda=\frac{19.87\times10^{-26}}{27693.41\times10^{-23}}\) 

⇒ 0.000717 x 10-3

=  717 x 10-9 m

17432.

Two flask X and Y have volume 1 L and 2 L respectively and each of them contain 1mole of same ideal gas. The temperature of the flask are so adjusted that average speed of molecules in X is twice as those in Y. The pressure in flask X would be:A. same as that in YB. half of that in YC. `(1)/(8)th` of that in YD. 8 times of that in Y

Answer» `PV=(1)/(3)MnU_("rms")^(2)`
Since `(U_(avg))_(x)=2(U_(avg))_(y)`
`P_(x)(I)=4xxP_(y)xx2`
`P_(x)=8P_(y)`
17433.

Select the correct order of mobility in aqueous medium.A. `[Li(H_(2)O)_(x)]^(+)gt[Le(H_(2)O)_(y)]^(+2)`B. `[Li(H_(2)O)_(x)]^(+)]lt[Be(H_(2)O)_(y)]^(+2)`C. `[Li(H_(2)O)_(x)]^(+)=[Be(H_(2)O)_(y)]^(+2)`D. Informations are not sufficient to predict the mobility

Answer» Mobility: `[Li(H_(2)O)_(x)]^(+)gt[Be(H_(2)O)_(y)]^(+2)`
`rarr+ve` charge density increases.
`rarr` Zeft increases.
`rarr` Size of `Be^(+2)` decreaes
`rarr` Surrounding `H_(2)O` molecule increases.
`rarr` Bulky nature increaes.
`rarr` Mobility decreases.
17434.

For gaseous `Ni(CO)_(x)`, what is value of x if under identical condition `CH_(4)` effuses `sqrt(10.5)` times faster than `Ni(CO)_(x)`

Answer» `sqrt(10.5)=(r_(CH_(4)))/(r_(Ni(CO)_(3)))=sqrt((M_(Ni(CO)_(x)))/(16))=M.Wt.`
`Ni(CO)_(x)=168`
`28x=112`
x=4
17435.

A sample of hyrdazine sulphate `(N_(2)H_(6)SO_(4))` was dissovled in 100ml 10ml of this solution was reacted with excess of `FeCl_(3)` solution and warmed to complete the reaction Ferrous ions formed required 20ml of `(M)/(50) KMnO_(4)` solution Given: `4Fe^(+3)+N_(2)H_(4)rarrN_(2)+4Fe^(+2)+4H^(+)` `MnO_(4)^(-)+5Fe^(+2)+8H^(+)rarrMn^(2+)+5Fe^(+3)+4H_(2)O` The amount in gm of hydrazinc sulphate in one litre is:A. 1.30gmB. 6.5gmC. 3.25gmD. 8.66gm

Answer» `N_(kmNo_(4))=(20)/(1000)xx(1)/(50)`
`n_(Fe^(+2))xxn_(KMnO_(4))`
`=5xx(20)/(1000)xx(1)/(50)=0.002`
`n_(N)_(2)H_(4))` is 10 ml `=(1)/(4) n_(Fe^(+2))=(0.002)/(4)`
`n_(N_(2)H_(4))` is 1L`=(0.002)/(4)xx100=(0.2)/(4)`
mass of `N_(2)H_(6)SO_(4)` in `1L=(0.2)/(4)xx130=6.5gm`
17436.

Why equilibrium constant of a reaction does not change in the presence of a catalyst ?

Answer» The catalyst increases the rate of forward and backward reactions to the same extent. Hence, `K_(eq)` remains constant but is attained quickly.
17437.

The element having greatest difference between its first and second ionization energies, is : (1) Ca (2) Sc (3) K (4) Ba

Answer»

Correct option (3) K

Explanation:

K has 4s' configuratioon 

∴ after removal of outermost 

e it aquire inert gas conf. 

∴ Greatest jump in I.E is observes

17438.

What is the effect of adding catalyst on free energy change `(DeltaG)` of a reaction ?

Answer» Since the free energy of reactants and products remains the same, so there is not change in `DeltaG`.
17439.

The correct statements among (a) to (d) are : (a) saline hydrides produce H2 gas when reacted with H2O. (b) reaction of LiAlH4 with BF3 leads to B2H6. (c) PH3 and CH4 are electron - rich and electron - precise hydrides, respectively. (d) HF and CH4 are called as molecular (1) (a), (b), (c) and (d) (2) (a), (c) and (d) only (3) (c) and (d) only (4) (a), (b) and (c) only

Answer»

Correct option (1) (a), (b), (c) and (d)   

Explanation:

Statement a, b, c and d are the correct statements as per the facts

17440.

What are the units of a pseudo unimolecular and pseudo bimolecular reaction ?

Answer» Refer text. `(` see page no. `)`
17441.

The correct statements among (a) to (d) regarding H2 as a fuel are : (a) It produces less pollutants than petrol (b) A cylinder of compressed dihydrogen weighs ~30 times more than a petrol tank producing the same amount of energy. (c) Dihydrogen is stored in tanks of metal alloys like NaNi5 (d) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 kJ, respectively. (1) (b) and (d) only (2) (b), (c) and (d) only (3) (a) and (c) only (4) (a), (b) and (c) only

Answer»

The correct option is (4) (a), (b) and (c) only.

17442.

The correct order of the atomic radii of C, Cs, Al, and S is : (1) S < C < Cs < Al (2) C < S < Al < Cs (3) S < C < Al < Cs (4) C < S < Cs < Al

Answer»

The correct option is (2).

Explanation:

As we move from L --> R, Atomic size decrease and while as we move from top to bottom, atomic size increase. 

17443.

The concentration of dissolved oxygen (DO) in cold water can go upto : (1) 8 ppm (2) 10 ppm (3) 16 ppm (4) 14 ppm

Answer»

The correct option is (2).

Explanation:

In cold water, dissolved oxygen (DO) can reach a concentration upto 10 ppm.

17444.

Which of the following is incorrect for Rf?(1) Rf value depends on the type of chromatography(2) Value of Rf is always between 0 to 1(3) Greater the adsorption, greater will be Rf value(4) Rf is independent of mobile carrier 

Answer»

Correct option: (3)

Explanation:

Low polarity compounds are weekly adsorbed and has greater Rf value.

17445.

The principle of column chromatography is (1) Gravitational force. (2) Capillary action. (3) Differential adsorption of the substances on the solid phase. (4) Differential absorption of the substances on the solid phase. 

Answer»

Correct option (3) Differential adsorption of the substances on the solid phase.

Explanation:

The principle of column chromatography is differential adsorption of substance and hence option on 3 is correct. 

17446.

Which of the following are correct for micelle formation.(a) Enthalpy of system decreases(b) Enthalpy of system increases(c) Entropy of system increases(d) Entropy of system decreases(1) a, d(2) a, c(3) b, c(4) b, d

Answer»

Correct option is (3) b, c

Micelle formation is spontaneous therefore S > 0.

Micelle formation decrease stability of colloidal solution so enthalpy should be positive H > 0

17447.

What are the changes that the disintegration of the Soviet Union brought to the world?

Answer»

The world faced a drastic change after the second world war. America to protect vested interests of imperialism and Soviet Union to protect socialistic ideas. Soviet Union could protect the world from American imperialism. Soviet Union was able to protect the interest of Egypt in the Suez Canal Crisis. Also, Soviet Union stood for justice in the Cuban Crisis and in the conflict on Kashmir between India and Pakistan.

But Soviet Union lost its power in the year 1991. Deviation from the basic principle of socialism and the external intervention as a part of globalization was the reason for the disintegration. The policies of Michael Gorbachev, the President, took Soviet Union to Imperialism.

17448.

For micelle formation, which of the following statements are correct?(A) Micelle formation is an exothermic process.(B) Micelle formation is an endothermic process.(C) The entropy change is positive.(D) The entropy change is negative. (A) A and D only (B) A and C only (C) B and C only (D) B and D only

Answer»

Correct option is (C) B and C only

For micelle formation, DS > 0 (hydrophobic effect) This is possible because, the decrease in entropy due to clustering is offset by increase in entropy due to desolvation of the surfactant, Also DH > 0 6. 

17449.

For the reaction `2X+ Y to` 2Z the rate of disappearance of X si 0.08 `M//s` (a) What is the rate of formation of Z (b) What is the rate of the reaction

Answer» (a) 0.08 `M//s (b) 0.04 M//s`
17450.

Examine the reasons for the disintegration of Soviet Union? How did this lead to a unipolar world?

Answer»
  • Deviation from the basic principles of socialism. 
  • The administrative measures of Mikhail Gorbachev – Glasnost and Perestroika.
  • Corruption and inefficiency of the bureaucracy. 
  • Failure in bringing about changes in economic sector.

In the absence of a socialist bloc, policies and programmes of the capitalist countries remained unquestioned. This had a wide impact on international relations. A unipolar world emerged under the leadership of USA instead of a bipolar world.