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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A popular game is ''splatball'', where people use pressurized CO2 cartridges to shoot paint-filled plastic balls at targets and at each other. A typical splatball gun takes a 1.00 L CO_(2) catridge filled to 300 PSI pressure. How many work is done as this cartridge is discharged (to atmospheric pressure) when the gun is fired ? (14.7 PSI = 1 atm) |
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Answer» `-20.41 J` `= -19.4 xx 101.4 = - 1967 J` |
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| 2. |
A polystyrene, having the formula Br_(3)C_(6)H_(3)(C_(3)H_(8))_(n),was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n. |
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Answer» Solution :Let the weight of polystyrene PREPARED be 100 G `therefore` no. of moels of Br in 100 g of polystyrene `=(10.46)/79.9 = 0.1309` From the formula of polystyrene, we have no. of moles of Br = `3 xx` moles of `Br_(2)C_(6)H_(3)(C_(3)H_(8))_(n)` or `0.1309 = 3 xx (WT)/("mol. wt") = (3 xx 100)/(314.7 + 44 n)` `therefore n = 44.9 = 45` |
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| 3. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. Dipole moment of aniline, trifluoromethylbenze and p-trifluoromethyl aniline lies in the following sequence: |
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| 4. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. Dipole moment of trans-2 butenal is greater than butanal: True (a) False(b) |
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Answer» |
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| 5. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. Dipole moment of para-nitrophenol is greater than phenol: True (a) False(b) |
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| 6. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. Dipole moment of methyl alcohol is less than that of water: True (a) False (b) |
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Answer» |
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| 7. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. Dipole moment of methanol is less than ethanol: True (a) False (b) |
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| 8. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. Dipole moment of C-Cl bond is greater than C-F bond: True (a) False(b) |
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Answer» |
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| 9. |
A polar covalent bond is formed between the two elements of different electronegativies. The polarity of a bond depends on the electronegativity difference, the boiling atoms and also on the shape of the molecule. A-B Let x_(A) and x_(B) are the electronegativites of bonding atoms A and B then, percentage ionic character of the bond can be calculated, as % ionic character = 21|x_(A) -x_(B)| +3.5|x_(A)-x_(B)|^(2) Dipole moment of the bond (mu) depends on the bond length and the polarity of the bond. mu = q xx d Here, q = charge of the dipoles d = bond length It is observed that the dipole moment increases with increase in the inductive effect. Answer the following questions as indicated: Select code (a) if the statement is true and code (b) if the statement is false. The dipole moment of CH_(3)-X bond lies in the following sequence: CH_(3)-I lt CH_(3)-Br lt CH_(3)-CI True (a) False (b) |
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| 10. |
A plot of x/m versus log p for the adsorption of a gas on a solid gives a straight line with slope equal to |
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Answer» Solution :x/m `=KP^(1//n) therefore log x/m= log k+(1)/(n) log p` `therefore` plot of log x/m VS log p will be linear with slope `=(1)/(n)`. |
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| 11. |
A plot of the kinetic energy ((1)/(2)mv^(2)) of ejected electrons as a function of the frequency (v) of incident radiation for four alkali metals (M_(1),M_(2),M_(3),M_(4)) is given below: The alkali metals M_(1),M_(2),M_(3),M_(4) are respectively: |
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Answer» Li,NA,K and Rb `v_(0)=(phi)/(h)` where `phi`=work function of the metal. Thus, GREATER is the value of `v_(0)`, more will be the work function of the metal. |
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| 12. |
A plot of kinetic energy of the emittedelectron against frequency of the incident radiation yields a straight line.t he slope of the straight line is: |
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Answer» `0.66xx10^(-35)` `THEREFORE` Slope `=6.6xx10^(-34)=0.66xx10^(-33)` |
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| 13. |
A plot of Gibbs energy of a reaction mixture against the extent of the reaction is : |
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Answer» minimum at EQUILIBRIUM |
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| 14. |
A planewith intercepts 0.5a, 0.25b and 1.5c has the Miller indices |
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Answer» 1 3 6 |
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| 15. |
A plane is parallel to x & z axes and makes unit intercepts alongy-axis. Its Weiss indices are --------. Its Miller indices are --------. The plane is designated as -------. |
| Answer» SOLUTION :`oo , 1 ,oo , 0 , 1 , 0 , (010)` plane | |
| 16. |
A piston filled with 0.04 mol of ideal gas expands reversibly from 50.0mL to 375 mLat a constant temperature of 37.0^(@)C. As it does so, itabsorbs 208J of heat. The value of q and w for the process will be : ( R= 8.314J // mol K , ln7.5 =2.01 ) |
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Answer» `Q= + 208 J , w= + 208 J` |
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| 17. |
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0 ^(@)C. As it does so, it absorbs 208 J of heat.The values of q and w for the process will be (R = 8.314 J mol^(-1)K^(-1)) (In 7.5 = 2.01) |
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Answer» `Q = + 208 J, w = + 208 J` Since the process is isothermal `DELTAU = 0` From first law of thermodynamics, `DeltaU = q + w` `0 = 208 + w or w = -208 J` |
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| 18. |
A piston cylinder device initially contains 0.2m^(3) neon (assume ideal ) at 200kPa inside at T_(1) ""^(@)C. A valve is now opened and neon is allowed to escape until the volume reduces to half the initial volume. At the same time heat transfer with outside at T_(2) ""^(@)C ensures a constant temperature inside. Select correct statement(s) for given process |
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Answer» `Delta U` must be zero To maintain constant `T, q gt 0` |
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| 19. |
A pieceof plumber's solder weighting 3.0 gm was dissolved in dilute nitric acid, then treated with dilute H_(2)SO_(4) . This precipatated the lead as PbSO_(4) which after washing and dryingweighted 2.98 gm. The solution was then neutralized to precipatatestannic acid, which was decomposed by heating, yielding 1.27 gm SnO_(2) . What is the analysis of the solder? (Pb=207.2 amu, Sn=118.7 amu) |
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Answer» 66.7 % PB, 33.3% Sn |
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| 20. |
A piece of metal is 3 inch long. What is its length in cm? |
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| 21. |
A piece of mass 36.81 kg is cut from a block of 4.0 xx 10^2 kg. Calculate the mass of the remaining block upto proper significant figures. |
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Answer» Solution :The mass of the block = `4.0 XX 10^2` kg (one decimal place) The mass of the PIECE cut = 36.81 kg (TWO decimal places) Since, the least precise number has only one decimal place, the final result should also have one decimal place. THEREFORE, the mass of the remaining block: `4.0 xx 10^(2) kg -36.81 kg` `=363.19 kg = 3.6 xx 10^(2)` kg |
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| 22. |
A piece of Cu weighs 0.635 g. How many atoms of Cu does it contain? |
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Answer» SOLUTION :No. Of moles of `Cu =("wt. Of Cu")/("at. Wt. Of Cu")` `=(0.635)/63.5 = 0.01` No. Of atoms of Cu = no. Of moles of atoms `xx` AV. CONST `=0.01 xx 6.022 xx 10^(23) = 6.022 xx 10^(23)` |
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| 23. |
A piece of burning magnesium ribbon continues to burn in sulphur dioxide . Explian . |
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Answer» Solution :A piece of magnesium RIBBON continues to burn in `SO_(2)` since it REACTS to form MgO and S . `2 Mg + SO_(2) overset("Heat") (to) 2 MgO + S` This reaction is so MUCH exothermic that heat EVOLVED keeps the magnesium ribbon burning . |
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| 24. |
A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV In order to have half the velocity in the second case than in the first case, the potential applied should be |
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Answer» 0.5 KV i.e., `("Potential applied in case II")/("Potential applied in case I") = (1)/(4)` or Potential applied in case `II = (1)/(4) xx 1kV` = 0.25 kV |
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| 25. |
A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV The wavelength associated with the electron will be |
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Answer» double in the SECOND case than in the FIRST case |
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| 26. |
A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV The velocity acquired by the electron will be |
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Answer» double in the SECOND CASE than in the FIRST case `= 1000 xx 1.602 xx 10^(-16) J = 1.602 xx 10^(-16)J` K.E. of the electron in case II = 2000 eV `= 2 xx 1.602 xx 10^(-16) J` `:. ((1)/(2) mv_(2)^(2))//((1)/(2) mv_(1)^(2)) = 2 or (v_(2)//v_(1))^(2) = 2` or `v_(2)//v_(1) = sqrt2 = 1.4` |
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| 27. |
A photon with initial frequency 10^(11)Hz scatters off an electron at rest. Its final frequency is 0.9 xx 10^(11)Hz. Calculate the speed of the scattered electron (h = 6.63 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg). |
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Answer» SOLUTION :INCIDENT energy = Scattered energy + K.E. of the scattered electron `hv_(1) = hv_(2) + (1)/(2) MV^(2)` or `(1)/(2) mv^(2) = h (v_(1) - v_(2))` or `V^(2) = (2h(v_(1) - v_(2)))/(m) = (2 xx 6.63 xx 10^(-34) kg m^(2) s^(-1) (10^(11) - 0.9 xx 10^(11)) s^(-1))/(9.1 xx 10^(-31) kg)` `= (2 xx 6.63 xx 10^(-34) xx 0.1 xx 10^(11))/(9.1 xx 10^(-31)) m^(2) s^(-2) = 14571428 m^(2) s^(-2)` or `v = sqrt(14571428) ms^(-1) = 3.8 xx 10^(3) ms^(-1)` |
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| 28. |
A photon of wavelength 1.4 Å collides with an electron. After collision, the wavelength of the photon is found to be 2.0 Å. Calculate the energy of the scattered electron. |
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Answer» Solution :ENERGY of the photon before COLLISON `= hv = (HC)/(lamda)` Energy of the photon after collision `= hv' = (hc)/(lamda')` Difference of energy is imparted to the electron. Hence, energy of the scattered electron `= (hc)/(lamda) - (hc)/(lamda') = hc ((1)/(lamda)- (1)/(lamda'))` `= (6.626 xx 10^(-34) J s) (3 xx 10^(8) m s^(-1)) xx((1)/(1.4 xx 10^(-10)m) - (1)/(2.0 xx 10^(-10) m))` `= (6.626 xx 10^(-34)Js) (3 xx 10^(8) ms^(-1)) xx ((2.0 - 1.4) xx 10^(-10) m^(-1))/((1.4 xx 10^(-10)) (2.0 xx 10^(-10))) = 4.26 xx 10^(-16) J` |
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| 29. |
A photon of wavelength 4xx 10^(-7) m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV) (ii) the kinetic energy of the emission and (iii) the velocity of the photoelectron (1 eV = 1.602 xx 10^(-19) J) |
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Answer» Solution :(i) Energy of the photon `(E) = hv = (HC)/(lamda) = ((6.626 xx 10^(-34) Js) xx(3 xx 10^(8) ms^(-1)))/(4 xx 10^(-7) m) = 4.97 xx 10^(-19) J` `= (4.97 xx 10^(-19))/(1.602 xx 10^(-19)) eV = 3.10 eV` (ii) KINETIC energy of emission `((1)/(2) mv^(2)) = hv - hv_(0) = 3.10 - 2.13 = 0.97 eV` (iii) `(1)/(2) mv^(2) = 0.97 eV = 0.97 xx 1.602 xx 10^(-19) J` i.e., `(1)/(2) xx (9.11 xx 10^(-31) kg) xx v^(2) = 0.97 xx 1.602 xx 10^(-19) J` or `v^(2) = 0.341 xx 10^(12) = 34.1 xx 10^(10) or v = 5.84 xx 10^(5) ms^(-1)` |
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| 30. |
A photon with a wavelength of 4000A^@ is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of I_2 molecule is 246.5 kJ/mol |
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Answer» `8%` Energy supplied by one mole photon in KJ/mole `=3.1 xx 1.6 xx 10^(-19) xx 10^(23) xx 10^(-3) = 297 `KJ `"mol"^(-1)` `therefore` of energy converted to K.E. `=(297 - 246.5)/(297)= 17%` |
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| 31. |
A photon of frequency v has momentum |
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Answer» `hv//c` where , p=momentum h=Plank's contant `rArr P=h/lambda=h/(cv)` |
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| 32. |
(A): Photo chemical smog is Oxidising in nature (R): Photo chemical smog is formed due to the combustion of coal and petroleum products |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 33. |
A petroleum fraction having boiling point range 70-200^(@)C and containing 6-10 carbon atoms permolecule is called |
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Answer» NATURAL gas |
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| 34. |
A person was using water supplied by Municipality. Due to shortage of water he started using underground water. He felt laxative effect. What could be the cause ? |
| Answer» Solution :The laxative effect is observed only when the sulphates present on water have concentration GREATER than `gt` 500 ppm. Otherwise at MODERATE LEVELS it is charmless. | |
| 35. |
A person was using water supplied by corporation. Due to shortage of water he started using underground water. He felt laxative effect. What could be the cause? |
| Answer» Solution :DRINKING water containing MODERATE level of sulphates is HARMLESS. But excessive concentration (`gt ` 500 ppm) of sulphates in drinking water causes laxative effect. | |
| 36. |
A person living in Shimla observed that cooking food without using pressure cooker takes more time. The reason for this observation is that at high altitude : |
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Answer» PRESSURE increases |
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| 37. |
A person living in Shimla observed that cooking food without using presssure cooker takes more time. The reason for this observation is that at high altitude : |
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Answer» PRESSURE INCREASES |
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| 38. |
A person consuming metro water suddenly states consuming well water due to shortage of water supply from municipality . What effect was felt by him ? What could be the cause ? |
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Answer» SOLUTION :He felt laxative effect. The laxative effect is observed when the sulphates present in water have concentration GREATER than 500ppm. OTHERWISE at moderate LEVELS it is harmless. |
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| 39. |
(A) : Peroxide effecti is not observed in the addition of HI to unsymmetrical alkene. (R ) : Free radical is unable to break stronger H-I bond. |
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Answer» A and R are TRUE, R explains A |
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| 40. |
A peroxidase enzyme contains 2% selenium(Se=80). The minimum molecular weight of the enzyme is |
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Answer» 1000 |
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| 41. |
A peroxidase enzyme contains 2% selenium (Se=80). The minimum molecular weight of the enzyme is |
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Answer» 1000 When selenium is 80 G - mol/ wt. peroxidase is? `=(80xx100)/(2)=4000` |
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| 42. |
(A): Pentane and 2-methyl pentane are homologues to each other(R): Pentane is a straight chain alkane, while 2-methyl Pentane is a branched chainalkane |
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Answer» If both (A) and (R ) are correct and (R ) is correct EXPLANATION for (A) |
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| 43. |
A pencil has a length of 9.2 cm. It is broken into two pieces. If the smaller piece has a length of 4.46 cm, what is the length of the larger piece? |
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| 44. |
A patient is said to suffer from acidosis when the pH of his blood |
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Answer» falls below 7.35 |
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| 45. |
A particular water sample has 131 ppm CaSO_(4). What fraction of the water must be evaporated in a container before solid CaSO_(4) begins to deposit ? (K_(SP) of CaSO_(4)= 9.0xx10^(-6)) |
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Answer» |
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| 46. |
A particular solid is very hard and has a very high melting point. In the solid state, it a non conductor and its melt is a conductor of electricity. Classify the solid. |
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Answer» metallic |
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| 47. |
A particular radio station broadcasts at a frequency of 1120 kHz (kilohertz). Another radio station broadcast at a frequency of 98.7 MHz (Megahertz). What are the wavelength of the radiations from each station ? |
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Answer» `lamda = (c)/(v) = (3 xx 10^(8) ms^(-1))/(1120 xx 10^(3) s^(-1)) = 267.85 m` In 2ND case, `v = 98.7 MHz = 98.7 xx 10^(6) " cycles " s^(-1) (1 MHz = 10^(6) " cycles " s^(-1))` |
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| 48. |
A particular form of tribromobenzene (x) forms three mononitrotribromobenzene. The structure of the compound (x) is Br |
Answer» SOLUTION :
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| 49. |
A particular acid rain water water contains sulphite (SO_(3)^(2-)) ions if a 25.0 cm^(3) sample of this water requires cm^(3) of 0.02 M KMnO_(4) solution for titeation what is the amount of SO_(3)^(2-) ions per litre in rain water? |
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Answer» Solution :Step 1 To write the balanced equationfor the redox reaction `MnO_(4)^(-)+8^(+)+5e^(-)rarrMn^(2+)+4H_(2)O]xx2` `SO_(3)^(2-)+H_(2)OrarrSO_(4)^(2-)+2H^(+)+2 e^(-)]xx5` `2MnO_(4)^(-)+5SP_(3)^(2-)+6H^(+)rarr2mn^(2+)+5SO_(4)^(2-)+3H_(2)O` Step 2 To determine the molarity of `SO_(3)^(2-)` ion solution Let `M_(1)` be the molaity of `SO_(3)^(2-)` ions in ACID rain water applying molarity equaiton `(M_(1)V_(1))/(n_(1))(SO_(3)^(2-))=(M_(2)V_(2))/(n_(2))(MnO_(4)^(-))` Thus the molariyt of `SO_(3)^(2-)` ions in acid rain water =0.07 M Mol wt of `SO_(3)_^(2-)` ions =32 +48 =80 `therefore` Amount of `SO_(3)^(2-)` ions in rain water `=0.07xx80=0.56 gL^(-1)` |
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