JEE Main Previous Year Papers Questions With Solutions Chemistry  Elements of S-Block : Hydrogen,Group 1 and 2

1.K0₂(potassium super oxide) is used in oxygen cylinders in space and submarines. This is because it        (2002)
(1)absorbs C0₂ and increases 0₂ content
(2)eliminates moisture
(3)absorbs C0₂ 4) produces ozone
Ans.(2) 2KO, + 2H₂0 —>2KOH + H₂0₂ + 0₂. Eliminates moisture.

2.A metal M readily forms its sulphate MS0₄which is water soluble. It forms its oxide MO which becomes inert on heating. It forms an insoluble hydroxide M(OH)₂ which is soluble in NaOH solution. Then M is     (2002)
(1)Mg 2) Ba 3) Ca 4) Be
Ans.(4) Beryllium shows anomalous properties with alkaline earth metals, due to its small size.

3.In curing cement plasters water is sprinkled from time to time. This helps in           (2003)
(1)developing interlocking needle-like crystals of hydrated silicates.
(2)hydrating sand and gravel mixed with cement
(3)converting sand into silicic acid
(4)keeping it cool
Ans.(1) Cross links are developed during hydration of calcium aluminosilicates.

4.The substance not likely to contain CaC0₃ is
(1)calcined gypsum 2) sea shells (2003) 3) dolomite 4) a marble statue
Ans.(1) The composition of gypsum is CaS0₄. 2H₂0, while others are carbonates.

5.The solubilities of carbonates decrease down the magnesium group. This is due to a decrease in
(1)hydration energies of cations (2003)
(2)inter – ionic attraction
(3)entropy of solution formation
(4)lattice energies of solids
Ans.(1) The solubility is governed by AH _soIution
AH solution^– AH_lattjce_ AH_Hydration.
Due to increase in size, the magnitude of hydration energy decreases and hence the solubility.

6.Several blocks of magnesium are fixed to the bottom of a ship to          (2003)
(1)make the ship lighter
(2)prevent action of water and salt
(3)prevent puncturing by under-sea rocks
(4)keep away the sharks
Ans.(2) Magnesium provides cathodic protection and prevent rusting or corrosion.

7.Which one of the following processes willproduce hard water ?      (2003)
(1)Saturation of water with MgCO_a
(2)Saturation of water with CaS0₄
(3)Addition of Na₂S0₄ to water
(4)Saturation of water with CaC0₃
Ans.(2) Permanent hardness of water is due to chlorides and sulphates of calcium and magnesium.

8.In hydrogen – oxygen fuel cell, combustion of hydrogen occurs to      (2004)
(1)generate heat
(2)remove adsorbed oxygen from electrode surfaces
(3)produce high purity water
(4)create potential difference between the two electrodes Ans.(4) Heat is converted directlyto electricity.
Ans.(4) Heat is converted directlyto electricity.

9.One mole of magnesium nitride on the reaction with an excess of water gives (2004)
(1)one mole of ammonia
(2)two moles of nitric acid
(3)two moles of ammonia
(4)one mole of nitric acid
Ans.(3) Mg₃N₂ +6H₂0——– » 3Mg(OH)₂ + 2NH₃

10.Hydrogen bomb is based on the principle of
(1)Nuclear fission (2005)
(2) Natural radioactivity
(3) Nuclear fusion
(4)Artificial radioactivity
Ans.(3) Fusion of isotopes of hydrogen.

11.Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point.(2005)
1) LiCl       2) NaCl       3) KC1       4) RbCl
Ans.(2) Although lattice energy of LiCl higher than NaCl, but LiCl is covalent in nature and NaCl ionic. There after, the melting point decreases as we move NaCl because the lattice energy decreases as size of alkali metal atom increases. Higher lattice energy denotes higher melting and boiling points.

12.The ionic mobility of alkali metal ions in aqueous solution is maximum for (2006)
1) K⁺   2) Rb⁺  3) Li⁺   4) Na⁺
Ans.(2) Hydrated Rb⁺ has small size.

13.In context with the industrial preparation of hydrogen from water gas (CO+H₂), Which of  the following is the correct statement ? (2008)
(1)CO and H₂ are fractionally separated using differences in their densities
(2)CO is removed by absorption in aqueous Cu₂Cl₂ solution
(3)H₂ is removed through occulsion with Pd
(4)CO is oxidised to C0₂ with steam in the presence of a catalyst followed by absorption of C0₂ in alkali
Ans.

14. Very pure hydrogen (99.9%) can be made by ’ which of the following processes ?   (2012)
(1)Reaction of methane with steam
(2)Mixing natural hydrocarbons of high molecular weight
(3) Elecrolysis of water
(4)Reaction of salt like hydrides with water
Ans.

15.The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na⁺ will be (2013)
1) -2.55 eV                   2) -5.1 eV
3)-10.2 eV                    4) +2.55 eV
Ans.(2) Inisation potential of alkali metal atom and electron affinity of its monocovalent cation are same in magnitude, but with opposite signs.

16.The metal that cannot be obtained by electrolysis of an aqueous solution of its salt is       (2014)
1) Cr 2) Ag 3) Ca 4) Cu
Ans.(3) Ca canot be obtained by the electrolysis of aqueous solution of its salt.

17.In which of the following reactions H₂0₂ acts as reducing agent ?     (2014)
a.H₂0₂ + 2H⁺ + 2e- -> 2H₂0
b.H₂0₂ – 2e–> 0₂ + 2H⁺
c.H₂0₂ + 2e- -> 20H-
d.H₂0₂ + 20H- 2e- -> 0₂ + 2H₂0
1) (b), (d)    2) (a), (b)      3) (c), (d)    4) (a), (c)
Ans.(1) While acting as reducing agent it gives up electrons and release oxygen gas.

Ans. 4) 2 mol of water softener require one mole of Ca²⁺

19.Form the following statements regarding H₂0₂, choose the incorrect statement (2015)
(1)It can act only as an oxidizing agent
(2)It decompose on exposure to light
(3)It has to be stored in plastic or was lined glass bottles in dark
(4)It has to be kept away form dust
Ans.(1) It is disproportionating agent.

20.Which one of the following alkaline earthmetal sulphates has its hydration enthalpy greater than its lattice enthalpy ?        (2015)
1) CaS0₄ 2) BeS0₄ 3) BaS0₄ 4) SrS0₄
Ans.(2) Among alkaline earth metal sulphates BeS0₄ and MgS0₄ are water soluble due to high hydration enthalpy of Be²⁺ and Mg²⁺.

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JEE Main Previous Year Papers Questions With Solutions Chemistry  Elements of  p-Block: Groups 13,14,15,16,17 and 18

1.When H₂S is passed through Hg₂S we get(2002)
(1)HgS        2) HgS + Hg₂S
3) Hg₂S          4) Hg₂S₂
Ans.

2.Alum helps in purifying water by (2002)
(1)forming Si complex with clay particles
(2)sulphate part which combines with the dirt and removes it
(3)coagulating the mud particles
(4)making mud water soluble
Ans.(3) Alum fumisheds Al³⁺ ions which bring about coagulation of negatively charged clay poarticles bacteria etc.

3.In case of nitrogen, NC1₃ is possible but not NC1_S while in case of phosphorous, PC1₃ as well as PC1₅ are possible. It is due to (2002)
(1)availability of vacant d orbitals in P but not inN
(2)lower electronegativity of P than N
(3)lower tendency of H-bond formation in P than N
(4)occurrence of P in solid while N in gaseous state at room temperature
Ans.(1) ₇N² =ls²2s²3p³, ₁₅P = 1 s²2s²2p⁶3s²3p³ In phosphorous the 3d – orbitals are available

4.Number of sigma bonds in P₄O₁₀ is (2003)
(1)6  2) 1            3) 17            4) 16 Ans.
Ans.

5.What’happens when phosphine gas is mixed with chlorine gas ?    (2003)
(1)PC1₃ an HC1 are formed and the mixture warms up
(2)PC1₅ and HC1 are formed and the mixture cools down
(3)PH₃.C1₂ is formed with warming up
(4)The mixture only cools down
Ans.

6.Graphite is a soft solid lubricant extremely difficult to melt. The reasong for this anomalous behavior is that graphite. (2003)
(1)is an alio tropic form of diamond
(2)has molecules of variable molecular masses like polymers
(3)has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds.
(4)is a non-crystalline substance.
Ans.(1) 4HC/ + 0_2–> 2Cl₂ + 2H₂0  cloud of white fumes.

7.Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that (2003)
(1)oxygen in air reacts with the emitted HC1 gas to form a cloud of chlorine gas
(2)Strong affinity of HC1 gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.
(3)due to strong affinity for water. Concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud.
(4)concentrated hydrochloric acid emits strongly smelling HC1 gas all the time.
Ans.(3) Layers are held together.

8.Glass is a (2003)
(1)super -cooled liquid 2) gel
(3)polymeric mixture
(4)micro-crystalline solid
Ans.(1) Glass is a transparent or translucent super cooled liquid.

9.For making good quality mirrors plates of float glass are used. These are obtained by floating molten glass over a liquid metal which does not solidify before glass. The metal used can be    (2003)
(1)tin           2) sodium
3) magnesium               4) mercury
Ans.(4) It is because mercury exists as liquid at room temperature.

10.Which among the following factors is the mostimportant in making fluorine the strongest oxidizing halogen?(2004)
(1)Electron affinity
(2)Bond dissociation energy
(3)Hydration enthalpy 4) Ionization enthalpy
Ans.(2) Lower dissociation energy of bonds in F₂.

11.Which one the following statement regarding helium is incorrect ?   (2004)
(1) It is used to fill gas balloons instead of H₂because it is lighter and non-inflammable
(2)It is used in gas-cooled nuclear reactors
(3)It is used to produce and sustain powerful superconducting reagents (4)It is used as cryogenic agent for carrying out experiments at low temperatures
Ans.(1) Hydrogen is lightest and is non inflamable.

Ans.(4) Hydrated and coordinated aluminium cation is farmed

13.The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transforma­tion is related to(2004)
(1)an interaction with nitrogen of the air at very low temperatures
(2)an interaction with water vapour contained in the humid air
(3)a change in the partial pressure of oxygen in the air
(4)a change in the crystalline structure of tin
Ans.(4) Transformation of crystal structure.

14.The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is
(2005)
(1)zero 2) two 3) one 4) three
Ans.

15.The correct order of the thermal stability ofhydrogen halides (H – X) is                  (2005)
(1)HI > HBr > HC1 > HF
(2)HF > HC1 > HBr > HI
(3)HC1 < HF > HBr < HI
(4)HI > HC1 < HF < HBr
Ans.(2) HF is most stable.

16.Heating an aqueous solution of aluminiumchloride to dryness will give                (2005)
(1)A1C1₃ 2) A1₂C1₆
3) A1₂0₃    4) Al(OH)CI₂
Ans.

17.In silicon dioxide (2005)
(1)Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms
(2)Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms
(3)Silicon atoms is bonded to two oxygen atoms
(4)There are double bonds between silicon and oxygen atoms
Ans.

18.The structure of diborane (B₂H_fi) contains(2005)
(1)four 2c-2e bonds and two 3c-2e bonds
(2)two 2c-2e bonds and four 3c-2e bonds
(3)two 2c-2e bonds and two 3c-3e bonds
(4)four 2c-2e bonds and four 3c-2e bonds
Ans.

19.Which of the following chemical reactions depicts the oxidizing behaviour of H₂S0₄ ?
(1)2HI + H₂S0₄ -> I₂ + S0₂ + 2H₂0 (2006)
(2)Ca(OH)₂ + H₂S0₄ CaS0₄ + 2H₂0
(3)NaCl + H₂S0₄ NaHS0₄ + HC1
(4)2PC1_S + H₂S0₄ 2POCl₃ + 2HC1 + S0₂C1₂
Ans.

20.A metal, M forms chlorides in its +2 and 44 oxidation states. Which of the following statements about these chlorides is correct ?
(1)MC1₂ is more volatile than MC1₄ (2006)
(2)MC1₂ is more soluble in anhydrous ethanol than MC1₄
(3)MC1₂ is more ionic than MC1₄
(4)MC1₂ is more easily hydrolysed than MC1₄
Ans.

21.Which of the following statements is true ?
(1)H3PO3 is a stronger acid than H₂S0₃ (2006)
(2)In aqueous medium HF is a stronger acid than HC1
(3)HC10₄ is a weaker acid than HC10₃
(4)HNO3 is a stronger acid than HN0₂
Ans.

22.What products are expected from the disproportionation reaction of hypochlorous acid ?     (2006)
(1)HCIO3 and C1₂0 2) HC10₂ and HC10₄
3) HC1 and C1₂0           4) HC1 and HC10₃
Ans.

23.The decreasing values of bond angles from NH₃ (106°) to SbH₃ (101°) down group-15 of the periodic table is due to        (2006)
(!)increasing bp-bp repulsion
(2)increasing p-orbital character in sp³
(3)decreasing lp-bp repulsion
(4)decreasing electronegativity
Ans.(4) Repulsions will be less with lower electronagativity.

Ans.

Ans.

26.Identify the incorrect statement among the fol­lowing (2007)
(1)Ozone reacts with S0₂ to give S0₃
(2)Silicon reacts with aqueous NaOH in the presence of air to give NajS^ and H₂0
(3)Cl₂ reacts with excess of NH₃ to give N₂ and HC1
(4)Br₂ reacts with hot and strong NaOH solu­tion to give NaBr, NaBr0₄ and H₂0
Ans.

Ans.

28.Which one of the following is the correct statement ?          (2008)
(1)Boric acid is a protonic acid
(2)Beryllium exhibits coordination number of six.
(3)Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
(4)B₂H₆.2NH₃ is known as ‘inorganic benzene’.
Ans.

29.In which of the following arrangements, the e sequence is not strictly according to the property written against it ?      (2009)
(1)C0₂ < Si0₂ < Sn0₂ < Pb0₂ : increasing oxidising power
(2)HF < HC1 < HBr < HI : increasing acid strength
(3)NH₃ < PH₃ AsH₃ < SbHj : increasing basic strength
(4)B<C<0<N: increasing first ionization enthalpy
Ans.

Ans.(1) Xe0₃ is more stable than XeF₆. All the remaining reactions are feasible

Ans.

Ans.

33.Which of the following statements regarding sulphur is incorrect ?                           (2011)
(1)At 600°C the gas mainly consists of S₂ molecules
(2)The oxidation state of sulphur is never less than + 4 in its compounds
(3)S, molecule is paramagnetic
(4)The vapour at 200°C consists mostly of S₈ rings
Ans.(2) Sulphur can exhibit – 2, + 2, + 1 oxidation states also.

Ans.(3) Stability of VA group hybrides decreases from NH₃ to BiH₃ due to the decrease in M – H bond energy.
NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
35.Which of the following on thermal decomposition yields a basic as well as an acidic oxide ?          (2012)
1) NaN0₃                      2) KC10₃
3) CaC0₃                       4) NH₄N0₃
Ans.

Ans.(1) ONF and ONO^– are isoelectronic ON Cl and ONS^– are isoelectronic

37.Which one of the following properties is not shown by NO ?                                     (2014)
(1)It’s bond order is 2.5 .
(2)It is diamagnetic in gaseous state
(3)It is a neutral oxide
(4)It combines with oxygen to form nitrogen dioxide
Ans.(2) NO is paramagnetic.

38.Among the following oxoacids, the correct decreasing order of acid strength is (2014)
(1)HC10₂ > HC10₄ > HCIO, > HOC1
(2)HCIO > HC10₂ > HCIO3 > HOCl₄
(3)HC10₄ > HCIO > HC10₂ > HOCl₃
(4)HC10₂ > HCIO3 > HC10₂ > HOC1
Ans.(4) With an increase in unprotonated oxygen atoms, acidic strength increases.

39.In the context of the Hall-Heroult process for the extration of Al, which of the following statements is false? (2015)
(1)CO and C0₂ are produced in this process
(2)A/₂0₃ is mixed with CaF₂ which lowers the melting point of the mixture and brings conductivity
(3)A/³⁺ is reduced at the cathode to form Al
(4)Na₃A/F₆ is serves as the electrolyte
Ans.[4] A/₂0₃ is dissolved in cryolite which serves as electrolyte in molten state.

40.Which among the following is the most reactive ?               (2015)
1) C/₂      2) Br₂    3) I₂ 4) ICI
Ans.[4] Interhalogen compounds are polar and are more reactive.

41.Which one has the highest boiling point ?(2015)
(1) He 2) Ne 3) Kr (4) Xe
Ans.[4] Boiling point increases with an increase in Molecular mass.

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JEE Main Previous Year Papers Questions With Solutions Chemistry  Elements of  d-Block,f-Block and Complexes

Ans.(2) A square planer complex is formed by hybridisation of s, p_x, p_y and d_x2 -d_y2 atomic orbitals.

2.The type of isomerism present in nitro- pentamine chromium (III) chloride is (2002)
(1)optical          2)   linkage
3) ionization    4)   polymerisation
Ans.(2) The nitro group can attach to metal through nitrogen as (-N0₂) or through oxygen as nitrito (-ONO).

3.CH₃-Mg-Br is an organo metallic compound due to  (2002)
(1)Mg-Br bond      2)   C-Mg bond
3) C-Br bond         4)   C-H bond
Ans.(2)Bond between C atom of organic molecule and metal atom.

4.Most common oxidation states of Ce (cerium) are     (2002)
(1)+2, + 3         2)   +2, + 4
3) +3, + 4       4)   +3, +5
Ans.(4)Common oxidation states of Ce(Cerium) are +3 and + 4.

5.Arrange Ce⁺³, La⁺³, Pm⁺³ and Yb⁺³ in increasing order of their ionic radii (2002)
(1)Yb⁺³ < Pm⁺³ < Ce⁺³ < La⁺³
(2)Ce⁺³ < Yb⁺³ < Pm⁺³ < La⁺³
(3)Yb⁺³ < Pm⁺³ < La⁺³ < Ce⁺³
(4)Pm⁺³ < La⁺³ < Ce⁺³ < Yb⁺³
Ans.(1) According to their positions in the periods, these values are in the order :

Ans.(2) A more basic ligand forms stable bond with metal ion, Cl^– is most basic amongst all.

7.Ammonia forms the complex ion [Cu(NH₃)₄]²⁺with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it ? (2003)
(1)In acidic solutions protons coordinate with ammonia molecules forming NH₄ ions and NH₃ molecules are not available
(2)In alkaline solutions insoluble Cu(OH)₂ is precipitated which is soluble in excess of any alkali.
(3)Copper hydroxide is an amphoteric substance
(4)In acidic solutions hydration protects copper ions
Ans.(1) Cu²⁺ forms complex with free NH_y (4)

8.One mole of the complex compound Co(NH₃)₅C1₃, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgN0₃ solution to yield two moles of AgCl_(s). The structure of the complex is          (2003)
(1)[Co(NH₃)₃Cl₃].2NH₃
(2)[Co(NH₃)₄C1₂]C1.NH₃
(3)[Co(NH₃)₄C1]C1₂.NH₃ 4) [Co(NH₃)₅C1]C1₂
Ans.

9.In the coordination compound, K₄[Ni(CN)₄], the oxidation state of nickel is             (2003)
1)0  2) + 1          3) + 2           4) – l
Ans.

10.A reduction in atomic size with increase in atomic number is a characteristic of elements of          (2003)
(1)d-block 2) f-block
3) radioactive series 4) high atomic masses
Ans.(2) f-block elements show a regular decrease in atomic size due to lanthanide/actinide contraction.

11.The coordination number of central metal atom in a complex is determined by (2004)
(1)The number of ligands around a metal ion bonded by sigma bonds
(2)The number of only anionic ligands bonded to the metal ion
(3)The number of ligands around a metal ion bonded by sigma and pi- bonds both
(4)The number of ligands around a metal ion bonded by pi-bonds
Ans.

12.Which one of the following complexes in an outer orbital complex ?      (2004)
(1)[Fe(CN)₆]⁴–              2) [Ni(NH₃)₆]²⁺
3) [Co(NH₃)₆]³⁺             4) [Mn(CN)₆]^4_
Ans.

13.Coordination compound have great importance in biological systems. In this context which of the following statements is incorrect ?     (2004)
‘ 1) Chlorophylls are green pigments in plants and contains calcium
(2)Carboxypeptidase – A is an enzyme and contains zinc
(3)Cyanocobalamin is B₁₂ and contains cobalt
(4)Haemoglobin is the red pigment of blood and  contains iron
Ans.(3)Chlorophyll contains magnesium instead of calcium.

14.Cerium(Z=58)is an important member of the lanthanoids. Which of the following statements about cerium is incorrect ? (2004)
(1)The common oxidation states of cerium are +3 and +4
(2)Cerium (IV) acts as an oxidizing agent
(3)The +4 oxidation state of cerium is not knownin solutions                                        .
(4)The +3 oxidation state of cerium is more stable than the +4 oxidation state
Ans.(3) Oxidation potential of Ce(IV) in aqueous solution is supposed to be -ve i.e. -0.784 V at 25°C.

15.Which one the following has largest number of isomers? (R-= alkyl group, en = ethylenediamine) (2004)
1) [Ru(NH₃)₄C1₂+]      2)         [Co(en)₂Cl₂]⁺
3) [Ir(PR₃)₂ H(CO)]²⁺      4)   [Co(NH₃)₅C1]²⁺
Ans.(2)Geomerical and optical isomerisum.

16.The correct order of magnetic moments (spin only values in B.M.) among is               (2004)
(1)[MnClJ^2- > [CoCl₄]-² > [Fe(CN)₆r
(2)[Fe(CN)₆]’⁴ > [CoClJ²– > [MnClJ^2-
(3)[Fe(CN)J^4- > [MnClJ²– > [CoClJ^2-
(4)[MnClJ^2- > [Fe(CN)J⁴– > [CoClJ^2-
Ans.(1) Number of unpaired e” in Mn²⁺ is 5, Co²⁺ is 3 and Fe²⁺ is zero.

17.The oxidation state of  Cr in  the complex[Cr(NH₃)₄Cl₂]⁺ is  (2005)
(1)+3     2) +2           3) +1           4)0
Ans.

18.Calomel (Hg₂Cl₂) on reaction with ammonium hydroxide gives                                    (2005)
(1)HgNH₂Cl       2)   NH₂-Hg-Hg-Cl
3) Hg₂0                  4)    HgO
Ans.(1) Reaction between calomel and ammonia is Hg₂Cl₂ + 2NH₄OH ->Hg + Hg(NH₂ )C1 + 2H₂

19.The IUPAC name of the coordination compound K₃[Fe(CN)₆] is     (2005)
(1)Potassium hexacyanoferrate (II)
(2)Potassium hexacyanoferrate (III)
(3)Potassium hexacyanoiron (II)
(4)tripotassium hexcyanoiron (II)
Ans.(2) Complex of Fe³⁺ is ferrate(III).

20.Which of the following compounds shows optical isomerism?     (2005)
1) [Cu(NH₃)₄]⁺²               2)    [ZnClJ-²
3) [Cr(C₂0₄)₃]-³                4)    [Co(CN)₆]-³
Ans.(3) The optical isomers are.

Ans.

22.The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct one is       (2005)
(1)d⁴ (in strong ligand filed)
(2)d⁴ (in weak ligand field)
(3)d³ (in weak as well as in strong fields)
(4)d⁵ (in strong ligand field)
Ans.

23.The IUPAC name for the complex [Co(N0₂)(NH₃)₅]C1₂ is (2006)
(1)nitrito-N-pentaamminecobalt (III) chloride
(2)nitrito-N-pentaamminecobalt (II) chloride
(3)pentaammine nitrito-N-cobalt (II) chloride
(4)pentaammine nitrito-N-cobalt (III) chloride
Ans.

24.Nickel (Z = 28) combines with a uninegative monodentate ligand X~ to form a paramagnetic complex [NiXJ^2-. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively       (2006)
(1)one, tetrahedral      2) two, tetrahedral
3) one, square planar   4) two, square planar
Ans.

25.In Fe(CO)₅, the Fe-C bond possesses (2006)
(1)k -character only
(2)both 0 and n characters
(3)ionic character 4) a -character only…
Ans.(2) Because of back bonding both o and kbonds are present.

26.How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca²⁺ ion ? (2006)
(1)Six 2) Three 3) One 4) Two
Ans.(3) EDTA^4- is hexadentate ligand.

27.Lanthanoid contraction is caused due to (2006)
(1)The appreciable shielding on outer electrons by 4f electrons from the nuclear charge
(2)The appreciable shielding on outer electrons by 5d electrons from the nuclear charge
(3)The same effective nuclear charge from Ce to Lu
(4)The imperfect shielding on outer electrons by 4f electrons from the nuclear charge
Ans.(4) Poor slidding by f-electrons.

28.Identify the incorrect statement among the, following                                                (2007)
(1)d-Block elements show irregular and erratic chemical properties among themselves
(2)La and Lu have partially filled d orbitals and no other partially filled orbitals
(3)The chemistry of various lanthanoids is very similar
(4)4f and 5f orbitals are equally shielded
Ans.(4) 4f and 5f belongs to different energy levels, hence the shielding effect is on them is not the same. Shielding of 4f is more than 5f.

Ans.

30.The actinoids exhibits more number of oxida­tion states in general than the lanthanoids. This is because          (2007)
(1)The 5f orbitals are more buried than the 4f orbitals
(2)There is a similarity between 4f and 5f orbit­als in their angular part of the wave function
(3)The actinoids are more reactive than the lanthanoids
(4)The 5f orbitals extend further from the nucleus than the 4f orbitals

Ans.(4) The actinoids exhibit more number of oxi­dation states in general than the lanthanoids. This is because the 5f orbitals extend further from the nucleus than the 4f orbitals.

31.The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)₂(C₂0₄)]N0₂ (where ‘en’ is ethylene diamine) are, respectively,   (2008)
1) 6 and 2 2) 4 and 2 3) 4 and 3 4) 6 and 3

Ans.

32.Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being (2008)
(1)4f orbitals more diffused than the 5f orbitals
(2)Lesser energy difference between 5f and 6d than between 4f and 5d orbitals
(3)More energy difference between 5f and 6d than between 4f and 5d orbitals
(4)More reactive nature of the actinoids than the lanthanoids

Ans.(2) Being lesser energy difference between 5f and 6d than 4f and 5d orbitals.

33.In which of the following octahedral complexes of Co (at. no. 27), will be magnitude of A₀ be the highest ?  (2008)
1) [Co(CN)₆]³“              2) [Co(C₂0₄)₃]³–
3) [Co(H₂0)₆]³⁺             4) [Co(NH₃)₆]³⁺

Ans.

34.Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements in incorrect?(2009)
(1)Because of the large size of the Ln(III) ions the bonding in its compounds is predominantly ionic in character
(2)The ionic sizes of Ln(III) decreases in general with increasing atomic number
(3)Ln(III) compounds are generally colourless
(4)Ln(III) hydroxides are mainly basic in character

Ans.(3) Ln⁺³ compounds are mostly coloured.

35.Which of the following has an optical isomer?(2009)
1) [C0(NH₃)₃C1]\           2) [CO(en) (NH₃)₂]²⁺
3) [C0(H₂0)₄(en)]³⁺        4) [CO(en)₂(NH₃)₂]³⁺

Ans.(4) It is an octahedral complex of the type [M(AA)₂X₂] where AA is bidentate ligand.

36.In context with the transition elements, which of the following statements is incorrect ?(2009)
(1)In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes
(2)In the highest oxidation states, the transition metal show basic character and form cationic complexes.
(3)In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding.
(4)Once the d³ configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.
Ans.(2) In higher oxidation states transition elements show acidic nature.

37.Which of the following pairs represents linkage isomers ?   (2009)
(1)[Cu (NHj )₄ ][PtCl₄ ] and [Pt(NH,)₄][CuCl₄]
(2)[Pd (PPh₃ )₂ (NCS)₂ ] and[Pd(PPh₃)₂(SCN)₂]
(3)[CO(NH₃)₅N0₃]S0₄and [co(NH₃)₅S0₄]N0₃
(4)[PtCl₂(NH₃)₄]Br₂ and [PtBr₂ (NH₃)₄]C1₂
Ans.

38.A solution containing 2.675 g of CoCl₃. 6NH₃ (molar mass = 267.5g mol’¹) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgN0₃ to give 4.78g of AgCl (molar mass = 143.5 g mol^-1). The formula of the complex is (At. Mass of Ag = 108u)                   (2010)
1)   [Co(NH₃)₆]C1₃         2)   [CoC1₂(NH₃)JC1
‘ 3)    [CoC1₃(NH₃)₃]        4)    [CoC1(NH₃)₅]C1₂
Ans.

39.Which one of the following has an optical isomer? (en = ethylenediamine)           (2010)
(1)[Zn(en) (NH₃)₂]²⁺   2)   [Co(en)₃]³⁺
3)[Co(H₂0)₄(en)]³⁺           4)         [Zn(en)₂]²⁺
Ans.

40.The magnetic moment (spin only) of [NiClJ^2]-is     (2011)
1) 2.82 BM           2) 1.41 BM
(3)82 BM             4) 5.46 BM
Ans.

41.Which of the following facts about the complex [Cr(NH₃)₆]Cl₃ is wrong ?    (2011)
(1)The complex is an outer orbital complex
(2)The complex gives white precipitate with silver nitrate solution.
(3)The complex involves d²sp³ hybridisation and is octahedral in shape
(4)The complex is paramagnetic
Ans.

42.In context of the lanthanoids, which of the following statements is not correct ? (2011)
(1)Because of similar properties the separation of lanthanoids is not easy
(2)Availability of 4f electrons results in the formation of compounds in + 4 state for all the members of the series
(3)There is a gradual decrease in the radii of the members with increasing atomic number in the series
(4)All the members exhibit + 3 oxidation state
Ans.(2) All the Lanthanides can not form + 4  oxidation state.

43.Which among the following will be named as dibromidohis (ethylenediamine) chromium (m) bromide ?    (2012)
1) [Cr(en)₃]Br₃₃             2) [Cr(en)₂Br₂]Br
3) [Cr(en)BrJ-                4) [Cr(en)Br₂]Br
Ans.

44.Which of the following complex species is not expected to exhibit optical isomerism ? (2013)
1) [Co(en)₃]³⁺                 2) [Co(en)₂C/₂]⁺
3) [Co(NH₃)C/₃]       4) [Co(en)(NH₃)₂C/₂]⁺
Ans.(3) [Co(NH₃)₃C/₃] do not exhibit optical isomers but othee exhibit.

45.Which of the following arrangements does not represent the correct order of the property stated against it ?  (2013)
(1)V²⁺< CF⁺ < Mn²⁺ < Fe²⁺ :Paramagnetic behaviour
(2)Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺ : Ionic size
(3)Co³⁺ < Fe³⁺ < Cr³⁺ < Sc³⁺ :Stability in aqueous solution
(4)Sc < Ti < Cr < Mn :Number of oxidation states
Ans.

Ans.(3) Order of wavelength of colours Red > Yellow > Green > Blue

Ans.

Ans.(2)Two bidentate and two monodentative – ligands are present.

Ans.(3) Change transfer from oxygen to manganese

Ans. (1) Ferrocyanide is blue.

JEE Main Home Maths Physics Chemistry

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JEE Main Previous Year Papers Questions With Solutions Chemistry  General Organic Chemistry

1.Arrangement of (CH₃)₃– C-, (CH₃)-CH-, CH₃– CH₂-, when attached to benzyl or an unsaturated group, in increasing order of inductive effect is (2002)
1)(CH₃)₃-C – < (CH₃)₂ – CH – < CH₃ – CH₂
2)CH₃-CH₂ – < (CH₃)₂ – < CH – < (CH₃)₃-C –
3)(CH₃)₂ – CH < (CH₃)₃ – C -< CH₃ – CH₂
4)(CH₃)₃ – C -< CH₃ – CH2 < (CH₃)₂ – < CH
Ans.(2) -CH₃ group has + I effect, as number of -CH₃ group increases, the inductive effect increases.

2.Which of the following does not show geometrical isomerism ? (2002)
1)1, 2 – dichloro – 1 – pentane
2)1,3- dichloro – 2 – pentane
3)1,2- dichloro – 1 – pentane
4)1, 4 – dichloro – 2 – pentane
Ans.

3.A similarity between optical and geometrical isomerism is that           (2002)
1)each forms equal number of isomers for a given compound
2)If in a compound one is present then so is the other
3)both are included in stereoisomerism
4)they have no similarity
Ans.(3) Optical and geometrical isomers are stereoisomers.

4.Which of the following compounds has wrong IUPAC name ?           (2002)

Ans.(3) Correct name is 3-methyl – 2-butanol.

5. Racemic mixture is formed by mixing two
1)isomeric compounds (2002)
2)chiral compounds
3)meso compounds
4)optical isomers
Ans.(4) Racemic mixture is formed when enantiomers are mixed in equimolar proportion.

6.The IUPAC name of CH₃COCH(CH₃)₂ is
1)2-methyl-3-butanone (2003)
2)4-methylisopropyl ketone
3)3-methyl-2-butanone
4)Isopropylmethyl ketone
Ans.

7.3The general formula C_nH_2n0₂ could be for open chain (2003)
1) carboxylic acids       2) diols
3) dialdehydes              4) diketones
Ans.(1) Carboxylic acid or ester.

8.Among the following four structures I to IV.it is true that  (2003)

1)only I and II are chiral compounds
2)only III is a chiral compound
3)only II and IV are chiral compounds
4)all four are chiral compounds
Ans.(1) A chiral object or structure has four different groups attached to the carbocation.

9.The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is  (2004)
1) Fe₄[Fe(CN)₆]₃           2) Na₄[Fe(CN)₅NOS]
3) Fe(CN)₃                    4) Na₃[Fe(CN)₆]
Ans.(1) Prussian blue coloured complex compound is ferric ferrocyanide.

10.The IUPAC name of the compound (2004)

1)3,3- dimethyl -1- hydroxycyclohexane
2)1,1 – dimethyl -3- cyclohexanol
3)3,3- dimethyl -1- cyclohexanol
4)1,1 – dimethyl -3- hydroxycyclohexane
Ans. (2)Methyl groups are at position 3.

11.Which one the following does not have sp² hybridized carbon ?  (2004)
1)Acetone                                2) Acetamide
3) Acetonitrile                       4) Acetic acid
Ans.(3) It is having only sp³ and sp hybridized carbon atom.

12.Which of the following will have meso-isomer also ? (2004)\
1)2- chlorobutane
2)2- hydroxyopanoic acid
3)2,3- dichloropentane
4)2-3- dichlorobutane
Ans.

13. Amongst the following compound, the optically active alkane having lowest [ molecular mass is       (2004)

Ans.

14.Which of the following compound is not chiral?  (2004)
1)1-chloropentane’
2)3-chloro-2-methyl pentane
3)1-chloro -2-methyl pentane
4)2-chloropentane
Ans.

15.Due to the presence of an unpaired electron, free radicals are (2005)
1)Chemically reactive   2) Chemically inactive   3) Anions 4) Cations
Ans.(1) Free redicals are chemically active.

16.The decreasing order of nucleophilicity among the nucleophiles         (2005)

Ans.(4) Highest n cloud is present in CN-

17.Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is (2005)
1)n-hexane     2)    2, 3-dimethylbutane
3) 2,2-dimethylbutane    4)   2-methylpentane
Ans.

18.Which types of isomerism is shown by 2,3-dichlorobutane ?   (2005)
1)Diastereo         2)    Optical
3) Geometric     4)    Structural
Ans.

Ans.(3) Double bond is given higher preference

20.The increasing order of stability of the following free radicals is   (2006)

Ans.(1) Triphenyl free radical is most stable.

21.Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is    (2006)
1)Eclipse, Gauche, Anti
2)Gauche, Eclipse, Anti
3)Eclipse, Anti, Gauche
4)Anti, Gauche, Eclipse
Ans.(3) Eclipse is least stable and gauche is most stable.

22.Which of the following molecules is expected to rotate the plane of plane polarized light? (2007)

Ans.(1) The plane of polarized light is rotated by optically active compound. Compound should be chiral.So, option (1) has, chiral C-atom. So, it is optically active.In (2), (3) and (4) plane of symmetry is present.

23.Which one of the following conformation of cyclohexane is chiral ? (2007)
1) Twist boat               2) Rigid
3) Chair                        4) Boat
Ans.(1) Twisted boat is chiral as it does not have plane of symmetry.

Ans.

Ans.

26.The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is(2008)
1)- COOH, – S0₃H, – CONH₂, – CHO
2)- SO3H, – COOH, – CONH2, – CHO
3)- CHO, – COOH, – S0₃H, – CONH,
4)- CONH₂, – CHO, – S0₃H, – COOH
Ans. (2) – S0₃H, – COOH, – CONH₂, – CHO

Ans.(3) 2° carbanion js more stable than 3°. Chlorine causes-I effect

28.The alkene that exhibits geometrical isomerism is      (2009)
1) propene                             2) 2-methyl propene
3) 2-butene                            4) 2-methyl -2-butene
Ans.

29.The number of stereoisomers possible for a compound of the molecular formula CH₃ – CH = CH – CH(OH) – Me is (2009)
1) 3                   2) 2             3) 4           4) 6
Ans.(3) About the double bond, two geometrical v isomers are possible and the compound is m having one chiral carbon.

30.The IUPAC name of  neopentane is (2009)
1) 2-methylbutane  2) 2, 2-dimethylpropane  3) 2-methylpropane  4) 2, 2-dimethylbutane
Ans.

31.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HC1 solution. The excess of the acid required 15mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (2010)
1) 59.0       2) 47.4        3) 23.7        4) 29.5
Ans.

32.Out of the following, the alkene that exhibits optical isomerism is (2010)
1)3 – methyl- 2 – pentane
2)4 – methyl – 1- pentane
3)3 – methyl – 1 – pentane
4)2 – methyl – 2 – pentane
Ans.

33.Identify the compound that exhibits tautomerism (2011)
1) 2 – Pentanone  2) Phenol  3) 2 – Butene            4) Lactic acid
Ans.

34.How many chiral compounds are possible on monocblorination of 2-methylbutane ? (2012)
1) 8      2) 2      3) 4      4) 6
Ans.

35.A solution of (-1) – 1 – 1 – chloro-1- phenylethane in toluene racemises slowly in the presence of a small amount of SbCl₅, due to the formation of    (2013)
1) carbanion                  2) carbene
3) carbocation               4) free radical
Ans.

Ans.(4) Primary carbocation is less stable.

37.For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was M /10 absorbed in 60mL of M/10 sulphuric acid. The unreacted acid required 20 mL of — sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is(2014)
1) 5 %        2) 6%         3) 10%        4) 3%
Ans.

Ans.

Ans.

JEE Main Home Maths Physics Chemistry

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JEE Main Previous Year Papers Questions With Solutions Physics Modern Physics

Multiple Choice with ONE correct answer
1.An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of [1981-2Marks]

Ans.

2.The half life of radioactive radon is 3.8 day. The time at the end of which (l/20)^th of the radon sample will remain undecayed is (given log₁₀e= 0.4343) [1981-2Marks]
a)8 day   b) 16.5 day     c) 33 day    d) 76 day
Ans.

3.Beta rays emitted by a radioactive material are  [1983-1Mark]
a)electromagnetic radiations
b)the electrons orbiting around the nucleus
c)charged particles emitted by the nucleus
d)neutral particles.
Ans.

4.If elements with principal quantum number 4 were not allowed in nature, the number of possible elements would be         [1983-lMark]
a)60        b) 32           c) 4           d) 64
Ans.

5.Consider the spectral line resulting from the transition n = 2 -> n = 1 in the atoms and ions given below. The shortest wavelength is produced by [1983-l Mark]
a)hydrogen atom  b) deuterium atom  c) singly ionized helium  d) doubly ionised lithium
Ans.

Ans.

7.Fast neutrons can easily be slowed down by :
a)the use of lead shielding
b)passing them through water
c)elastic collisions with heavy nuclei
d)applying a strong electric field
Ans.(b) Fast neutrons can easily be slowed down by passing them through water. Heavy water is used as moderator in nuclear reactors.

8.Consider a – particles, b – particles & g rays, each having an energy of 0.5 MeV. Increasing order of penetrating power of the radiations are : [JEE’94]
a)a, b, g b) a, g, b c) b, g, a d) g, b, a
Ans.

9.An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (In eV) required to remove both the electrons form a neutral helium atom is: [JEE’95]
a)2    b) 49.2    c) 51.8    d) 79.0
Ans.

10.A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 year. The time, in year, after which one-fourth of the material remains is [1995]
a)1080    b) 2430    c) 3240   d) 4860
Ans.

Ans.

Ans.

13.As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is
a)51 b) 13.6        c) 40.8        d) 122.4
Ans.

14.The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in Volts is :
a)2          b) 4            c) 6            d) 10
Ans.(b) Stopping potential is the negative potential which stops the emission of (K.E.)_max electrons when applied.

Stopping potential = 4 volt.

15.In Hydrogen spectrum the wavelength of H_a line is 656 nm, where as in the spectrum of a distant galaxy, H_a line wavelength is 706 nm. Estimated speed of the galaxy with respect to earth is,[1999-2Marks]
a)2xl0⁸m/s     b) 2xl0⁷m/s
c) 2xl0⁶m/s    d) 2xl0⁵m/s
Ans.

16.A particle of mass M at rest decays into two particles of masses m1 and m₂, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, l_1/ l₂, is  [JEE ’99]
a)m1/ m₂ b) m₂/m₁ c) 1.0 d) Om₂/Om1
Ans.

17.Which of the following is a correct statement?
a)Beta rays are same as cathode rays
b)Gamma rays are high energy neutrons.
c)Alpha particles are singly ionized helium atoms
d)Protons and neutrons have exactly the same mass
Ans.(a) a) Beta rays are same as cathode rays as they are streams of electrons. The statement is correct.(b)Gamma rays are electromagnetic waves They are not neutrons

(c)Alpha particles are doubly ionised helium atoms, not singly ionised heliumatoms.

(d)Protons have mass slightly less than that of neutrons. Masses are not exactly the same. Option (b), (c) and (d) are incorrect.Option (a) is correct.

Ans.(b) Nuclear density is of the order of 10¹⁷ kg/m³

^19.22Ne nucleus, after absorbing energy, decays into two a-particles and an unknown nucleus. The unknown nucleus is
a)nitrogen b) carbon c) boron d) oxygen
Ans.

Ans.

Ans.

22.The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true? [JEE’ 2000 (Scr)]
a)Its kinetic energy increases and its potential and total energies decrease.
b)Its kinetic energy decreases, potential energy increases and its total energy remains the same.
c)Its kinetic and total energies decrease and its potential energy increases.
d)Its kinetic, potential and total energies decrease.
Ans.

23.Two radioactive materials X₁ and X₂ have decay constants 101 and 1 respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X₁ to that of X₂ will be 1/ e after a time of        [JEE ’ 2000 (Scr)]
a)1/(101)         b) 1/(111)
c) 11/(101)      d) 1/(91)
Ans.

24.Electrons with energy 80 ke V are incident on the tungsten target of an X-ray tube. K shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only                                                    [2000-2Marks]
a)a continuous X-ray spectrum (Bremsstrahlug) with a minimum wavelength of 0.155 A⁰
b)a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
c)the characteristic X-ray spectrum of tungsten
d)a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155 A⁰ and the characteristic X-ray spectrum of tungsten.
Ans.

25.The electron emitted in beta radiation originates from[JEE’2001(Scr)]
a)inner orbits of atoms
b)free electrons existing in nuclei
c)decay of a neutron in a nucleus
d)photon escaping from the nucleus
Ans.

26.The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition [2001-2Marks]
a) 2—>l  b) 3 —>2     c) 4 —>2     d)5—>4
Ans.

Ans.

Ans.

29.The potential difference applied to an X – ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is         [JEE’ 2002 (Scr.)]
a)2 x 10¹⁶ b) 5 x 10¹⁶ c) 1 x 10¹⁷ d) 4 x 10¹⁵
Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

37.In a photoelectric experiment anode potential is plotted against plate current           [2004-2 Marks]
a)A and B will have same intensities while B and C will have different frequencies
b)B and C will have different intensities while A and B will have different frequencies
c)A and B will have different instensities while B and C will have equal frequeancies
d)B and C will have equal intensities while A and B will have same frequencies.
Ans.

38.A 280 day old radioactive substance shows an activity of 6000 dps, 140 day later its activity becomes 3000 dps. What was its initial [2000-2marks]
a) 20000 dps b) 24000 dps
c) 12000 dps d) 6000 dps
Ans.

Ans.

Ans.

41.A photon of 10.2 eV energy collides with a hydrogen atom in ground state inelastically. After few microseconds one more photon of energy 15 eV collides with the same hydrogen atom.Then what can be detected by a suitable detector. [JEE’2005 (Scr)]
a)one photon of 10.2 eV and an electron of energy 1.4 eV
b)2 photons of energy 10.2 eV
c)2 photons of energy 3.4 eV
d)1 photon of 3.4 eV and one electron of 1.4 eV
Ans.

42.A beam of electrons is used in an Young double slit experiment. The slit width is d. When the velocity of electron is increased, then [2005]
a)no interference is observed
b)fringe width increases
c)fringe width decreases
d)fringe width remains same
Ans.

43.If a star can convert all the He nuclei completely into oxygen nuclei, the energy released per oxygen nuclei is [Mass of He nucleus is 4.0026 amu and mass of oxygen nucleus is 15.994 amu]
a) 7.6 MeV        b) 56.12 MeV
c) 10.24 MeV     d) 23.9 MeV
Ans.

44.Given a sample of Radium-226 having half-life of 4 days. Find the probability, a nucleus disintegrates within 2 half lives.            [JEE 2006]
a) 1     b) 1/2     c) 3/4    d) 1/4
Ans.

45.The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is [2007-3Marks]
a) 802 nm            b) 823 nm
c) 1882 nm          d) 1648 nm
Ans.

46.In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is  [2007-3Marks]

Ans.(a) Rest mass energy of U will be greater than the rest mass energy of the nuclei into which it breaks. The constituent nuclei and neutrons will have kinetic energy also, as a result of conservation of linear momentum.

Ans.

48.Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? [2008]
a)Wavelength of characteristic X-rays decreases when the atomic number of the target increases
b)Cut-off wavelength of the continuous X-rays depends on the atomic number of the target.
c)Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube
d)Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the electrons in the X-ray tube.
Ans.

Ans.

Ans.

Ans.

Multiple Choice with ONE or  More Than ONE correct answers
52.The shortest wavelength of X-rays emitted from an X-ray tube depends on [1982-3Marks]
a)the current in the tube
b)the voltagae applied to the tube
c)the nature of the gas in tube
d)the atomic number of the target material
Ans.

53.The threshold wavelength, for photoelectric emission, from a material is 5200 A⁰. Photo­electrons will be emitted when this material is illuminated with monochromatic radiation from a[1982-3Marks]
a)50 watt infrared lamp
b)1 watt infrared lamp
c)50 watt ultraviolet lamp
d)1 watt ultraviolet lamp.
Ans.

Ans.(b),(c) Nuclear fusion reaction occurs when two or more lighter nuclei combine to produce a heavier nucleus.

Ans.

56.The X-ray beam coming from an X-ray tube will be [1985-2Marks]
a)monochromatic
b)having all wavelengths smaller than a certain maximum wavelength
c)having all wavelength larger than a certain minimum wavelength
d)having all wavelengths lying between a minimum and a maximum wavelength
Ans.

57.The mass number of a nucleus is [1986-2Marks]
a)always less than its atomic number
b)always more than its atomic number
c)sometimes equal to its atomic number
d)sometimes more than and sometimes equal to its atomic number.
Ans.

58.Photoelectric effect supports quantum nature of light because     [1987-2Marks]
a)there is a minimum frequency of light below which no photoelectrons are emitted
b)the maximum kinetic energy of photo electrons depends only on the frequency of light and not on its intensity.
c)even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately
d)electric charge of the photoelectrons is quantized.
Ans.(a),(b),(c)

59.During a negative beta decay [1987-2Marks]
a)an atomic electron is ejected
b)an electron which is already present within the nucleus is ejected
c)a neutron in the nucleus decays emitting an electrons,
d)a part of the binding energy of the nucleus is converted into an electron.
Ans.

60.During a nuclear fusion reaction [1987-2 Marks]
a)a heavy nucleus breaks into two fragments by itself.
b)a light nucleus bombarded by thermal neutrons breaks up
c)a heavy nucleus bombarded by thermal neutrons breaks up
d)two light nuclei combine to give a heavier nucleus and possibly other products.
Ans.(d)

61.The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation
a)the intensity increases [1988-2Marks]
b)the minimum wavelength increases
c)the intensity remains unchanged
d)the minimum wavelength decreases
Ans.

62.A freshly prepared radioactive source of half life 2 hr emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is     [1988-2Marks]
a) 6 hr   b) 12 hr    c) 24 hr   d) 28 hr
Ans.

Ans.

64.When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6m away from the photoelectric cell, then   [1992-2Marks]
a)the stopping potential will be 0.2 volt
b)the stopping potential will be 0.6 volt
c)the saturation current will be 6.0 mA
d)the saturation current will be 2.0 mA.
Ans.(b),(d) The stopping potential depends on the frequency of incident radiation and not on the intensity of incident radiation. The source is same, though placed at different distances, consequently the stopping potential remains unchanged. The saturation current depends on the radiation- intensity, incident on cathode of photocell. The radiation-intensity depends on the distance between the source and the cathode. Infact, inverse square law applies to intensity.

Ans.

Ans.

67.Which of the following statement(s) is (are) correct ?  [JEE’94]
a)The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons.
b)The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons.
c)In nuclear fusion, energy is released by fusion two nuclei of medium mass (approximately 100 amu).
d)In nuclear fission, energy is released by fragmentation of a very heavy nucleus.
Ans.(a),(d)

Ans.(c),(d) M, is the mass of ₁₀Ne²⁰
M₂ is the mass of ₂₀Ca⁴⁰ nucleus.
Ne means 10 protons + 10 neutrons.
Ca means 20 protons + 20 neutrons.
Protons and neutrons are constituent particles. Due to mass defect, which is necessary for bind­ing the nucleus, mass of nucleus is always less than the sum of masses of protons and neutrons.

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72.
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73.The half-life period of a radioactive element X is same as the mean-life time of another radioactive element Y. Initially both of them have the same number of atoms. Then     [JEE ’99]
a)X & Y have the same decay rate initially
b)X & Y decay at the same rate always
c)Y will decay at a faster rate than X
d)X will decay at a faster rate than Y

75.Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice (s) given below.

76.Statement-1 : If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-ray do not change.
Statement-2 : When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy.
a)Statement-1 and 2 are true and statement-2 is a correct explanation for statement-1
b)Statement-1 and 2 are true and statement-2 is not a correct explanation for statement-1
c)Statement-1 is true, statement-2 is false
d)Statement-1 is false, statement-2 is true.
Ans.(b)

Comprehension based question 

Read the passage given below and answer the questions that follow

Passage – 1 : In a mixture of H-He⁺ gas (He⁺ is sinagly ionized He atom), H atoms and He⁺ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He⁺ ions (by collisions). Assume that the Bohr model of atom is exactly valid.   [2007-3Marks]

77.The quntum number n of the state finally populated in He⁺ ions is

a) 2    b) 3 c) 4             d) 5

Ans.

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79.The ratio of the kinetic energy of the n = 2 electrons for the H atom to that of He⁺ ion is[2008]
a) 1/4     b) 1/2        c) 1     d) 2
Ans.

Passage – 2 : The key feature of Bohor’s theory of spectrum of hydrogen atom is the quantization of an­gular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.

80.A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the n^th level (n = 0 is not allowed) is

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Passage – 3 : When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as

83.The allowed energy for the particle for a particular value of n is proportional to
a) ar²        b) a^-3/2          c) a^-1     d) a²
Ans.

84.If the mass of the particle is m = 1.0 x 10-³⁰kg and a = 6.6nm, the energy of the particle in its ground state is closest to
a) 0.8 meV        b) 8 meV     c) 80 meV       d) 800 meV
Ans.

85.The speed of the particle, that can take discrete values, is proportional to
a) n-^3/2    b) n^-1    c) n^1/2           d) n
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Ans.
(d) Plasma state is achieved at high temperature

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Matching Type

Ans.a—>p, b —>s, c—>r, d—>q

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Subjective / Numerical integer type    
96.How many electrons, protons and neutrons are there in a nucleus of atomic number 11 and mass number 24? [1982-2Marks]
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97.A uranium nucleus (atomic number 92, mass number 238) emits an alpha particle and the resultant nucleus emits (3 -particle. What are the atomic number and mass number of the final nucleus?         [1982-2Marks]
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101.A double ionised lithium atom is hydrogen-like with atomic number 3.    [1985-6Marks]
i)Find the wavelength of the radiation required to excite the electron in Li from the first to the third Bohr orbit.(Ionisation energy of the hydrogen atom equals 13.6eV).
ii)How many spectral lines are observed in the emission spectrum of the above excited system?
Ans.

102.There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutrons is 700 second, what fraction of neutrons will decay before they travel distance of 10 m? [1986-6 Marks]
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105.A gas of identical hydrogen – like atoms has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level.The atoms of the gas make transition to a higher energy level by absorbing monochromatic light of photon energy 2.7eV. Subsequently, the atoms emit radiation of only six different photon – energies. Some of the emitted photons have energy 2.7eV, some have energy more and some have less than 2.7eV.
i)Find the principal quantum number of the initially excited level B.
ii)Find the ionization energy for the gas atoms.
iii)Find the maximum and the minimum energies of the emitted photons.   [1989-8 Marks]
Ans.

106.Electrons in hydrogen like atoms (Z=3) make transitions from the fifth to the fourh orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photo electrons. The stopping potential for the photo electrons ejected by the shorter wavelength, is 3.95 volt. Calculate the work function of the metal and the stopping potential for the photo electrons ejected by the longer wavelength. (Rydberg constant = 1.094 x 10⁷m-1) [1990-71 marks]
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108.A mono chropiatic point source radiating wave length 6000 A, with powev 2 watt, an aperture A of diameter 0.1 m and a lavge screen SC are placed as shown in fig. A photo emissive detector D of surface area 0.5 cm² is ‘placed at the centre of the screen. The efficiency of the detector for the photo electron generation per incident photon is 0.9.
a)Calculate the photon flux at the centre of the screen and the photo current in the detector.
b)If the concave lens L of focal length 0.6 m is inserted in the aperture as shown, find the new values of photon flux and photo current. Assume a uniform average transmission of 80% from the lens.
c)If the work function of the photo emissive  sur­face is leV, calculate the values of the stopping potential in the two cases (without and with the lens in the aperture). [1991-2+4+2 Marks]

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109.Photo electrons are emitted when 400 nm radiation is incident on a surface of work function 1.9eV. These photo electrons passes through a region containing a-particles. A maximum energy electron combines with an a-particle to form a He⁺ ion, emitting a single photon in this process. He⁺ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4eV range, that are likely to be emitted during and after the combination. [Take, h = 4.14 ’10^-15 eV-s ][JEE ’99]
Ans.

110.A hydrogen – like atom of atomic number Z is in an excited state of quantum number 2 n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40,8 eV is emitted. Find n. Z and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation Ground state energy of hydrogen atom is 13.6 eV.
Ans.

111.When a beam of 10.6 eV photon of intensity 2 W/ m² falls on a platinum surface of area 1 x 10⁴ m² and work function 5.6 ev, 0.53% of the incident photons eject photo electrons. Find the number of photo electrons emitted per sec and their minimum and maximum energies in eV. [JEE’ 2000]
Ans.
“>

112.A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transition between a group of levels. These levels have energies between – 0.85 eV and – 0.544 eV (including both these values) Find the atomic number of the atom.Calculate the smallest wavelength emitted in these transitions.    [JEE’ 2002]
Ans.

113.Two metallic plates A and B each of area 5 x lO”⁴ m², are placed at a separation of 1 cm. Plate B carries a positive charge of 33.7 x 10~¹² A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 10¹⁶ photons fall on it per square meter per second. Assume that one photo electron is emitted for every 10⁶ incident photons. Also assume that all the emitted photo electrons are collected by plate B and the work function of plate A remains constant at the value 2 eV. Determine [JEE’ 2002]
a)the number of photo electrons emitted up to t = 10 sec.
b)the magnitude of the electric field between the plates A and B at t = 10 s and
c)the kinetic energy of the most energetic  photo electron emitted at t = 10 s when it reaches plate B.
(Neglect the time taken by photo electron to reach plate B)
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117.The photons from the Balmer series in Hydrogen spectrum, having wavelengths between 450 nm to 700 nm, are incident on a metal surface of work function 2 eV. Find the maximum kinetic energy of ejected electron. (Given he = 1242 e V nm) [2004-4 Marks]
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“>

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119.Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is (14)^1/3. Find          [JEE 2005]
a)atomic number of the nucleus
b)the frequency of K_a line of the X-ray produced. (R = l.lx 10⁷ nr¹ and c = 3 x 10⁸ m/s)
Ans.

120.In hydrogen-like atom (z = 11), n^th line of Lyman series has wavelength 1 equal to the de-Broglie’s wavelength of electron in the level from which it originated. What is the value of n? [JEE 2006]
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122.Light from a dischargebe containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find –
a)the energy of the photons causing the photoelectric emission.
b)the quantum numbers of the two levels involved in the emission of these photons.
c)the change in the angular momentum of the electron in the hydrogen atom in the above transition, and
d)the recoil speed of the emitting atom assuming it to be at rest before the transition.
(Ionization potential of hydrogen is 13.6 eV)[1992-lO Marks]
Ans.

123.A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest . It is scattered at an angle of 90° with respect of its original direction.
i)Find the allowed values of the energy of the neutron & that of the atom after collision.
ii)If the atom gets de-excited subsequently by emitting radiation , find the frequencies of the emitted radiation. (Given : Mass of he atom = 4x(mass of neutron), ionization energy of H atom =13.6 eV)     [JEE ’93]
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124.A small quantity of solution containing ²⁴Na  radionuclide (half life 15 hours) of activity 1.0 microcurie is injected into the blood of a person. A sample of the blood of volume 1 cm³ taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the personal Curie = 3.7 x 10¹⁰ disintegrations per second) [JEE’94]
Ans.

125.A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV & 17.00 eV respectively. Alternatively , the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV & 5.95 eV respectively. Determine the values of n & Z. (Ionisation energy of hydrogen atom = 13.6 eV)           [JEE’94]
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“>

126.An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of -3.4 eV. Calculate (i) the kinetic energy and (ii) the de Broglie wavelength of the electron. [1996-3Marks]
Ans.

127.At a given instant there are 25% undecyaed radio-active nuclei in a sample. After 10 second, the number of undecayed nuclei reduces to 12.5%. Calculate
(i) mean-life of the nuclei, and
(ii) the time in which the number of undecayed nuclei will further reduced to 6.25% of the reduce number.         [1996-3Marks]
Ans.

128.In an ore containing uranium; the ratio of U²³⁸ to Pb²⁰⁶ nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U²³⁸. Take the half-life of U²³⁸ to be 4.5 x 10⁹ year.        [1997-5Marks]
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132.In a nuclear reaction 235 U undergoes fission liberating 200 Me V of energy . The reactor has a 10% efficiency and produces 1000 MW power, if the reactor is to function for 10 year, find the total¹ mass of uranium required. [2001-5 Marks]
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139.A silver sphere of radius 1 cm and work funtion 4.7 eV is suspended from an insulting thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photo electrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photo elec­trons emitted from the sphere is A x 10^z (Where 1< A <10). The value of “Z” is (2011)
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True / False

140.The kinetic energy of photo electrons emitted by a photosensitive surface depends on the intensity of the incident radiation.          [1981-2 Marks]
Ans.

141.In a photoelectric emission process the maximum energy, of the photo electrons increases with increasing intensity of the incident light. [1986-3 Marks]
Ans.

False : Maximum energy of photoelectrons increases with increasing frequency of incident light. The number of photoelectrons emitted increases with increasing intensity of light.

142.The order of magnitude of the density of nuclear matter is 10⁴ kg m^-2  [1989-2 Marks]
Ans.
False : Nuclear density is of the order of I0ⁱ⁷ kg/m³. Hence the statement is false.

Fill in the blanks

143.To produce characteristic X-rays using a tungsten target in an X-ray generator, the accelerating voltage should be greater than………. volt and the energy of the characteristic radiation is……….. eV.(The binding energy of the innermost electron in tungsten is 40 keV.)  [1983-2 Marks]
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144.The radioactive decay rate of a radioactive element is found to be 10³ disintegration/second at a certain time. If the half life of the element is one second, the decay rate after one second is…………. and after three second is………..     [1983-2 Marks]
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145.The maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the…………. of the incident radiation  . [1984-2Marks]
Ans.

146.In the uranium radioactive series, the initial nucleus is U and the final nucleus is Pb . When the uranium nucleus decays to lead, the number of a -particles emitted are……… and the number of (3 –

particles emitted are………            [1985-2Marks]
Ans.

147.When the number of electrons striking the anode of an X-ray tube is increased the……. of the emitted X-rays increase, while when the speed of the electrons striking the anode are increased the cut-off wavelength of the emitted X-rays………………. [1986-2Marks]
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148.

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149. Atoms having the same…………….but different………. are called isotopes.  [1986-2 marks]
Ans.
Isotopes have same atomic number (Z) and different mass numbers (A).

150.

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151.

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152.

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153.In an X-ray tube, electrons accelerated through a potential differenc e of 15,000 volt strike a copper target. The speed of the emitted X-rays inside the tube is  ……..     rn/s.      [1992-1Mark]
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154.In the Bohr model of the hydrogen atom, the ratio of the kine tic energy to the total energy of the electron in a quantum state n is… [1992-lMark]
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156.A potential difference of 20 KV is applied across an x-ray tube. The minimum wave length of X – rays generated is       [JEE’96]
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158.

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159.The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state is    m/s.      [1997-1Mark]
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JEE Main Home Maths Physics Chemistry

The post JEE Main Previous Year Papers Questions With Solutions Physics Modern Physics appeared first on Learn CBSE.

JEE Main Previous Year Papers Questions With Solutions Chemistry  Metallurgy and  Environmental Chemistry

1.The metal extracted by leaching with a cyanide is        (2002)
1) Mg   2) Ag   3) Cu    4) Na
Ans.(2) Silver ore forms a soluble complex with ¹
NaCl from which silver is precipitated using scrap zinc.
Ag₂S + 2NaCN -> Na[Ag(CN)_?]
2Na[Ag(CN)₂] + Zn ->Na₂[Zn(CN)₄] + 2Ag

2.Aluminium is extracted by the electrolysis of   (2002)
1) bauxite           2) alumina
3)alumina mixed with molten cryolite
4)molten cryolite
Ans.(3) Alumina is mixed with cryolite in the extraction of aluminium metal. Addition of cryolite decreases the melting point and increases the conductivity.

3.Cyanide process is used for the extraction of (2002)
1) barium            2) aluminium
3) boron           4) silver
Ans.(4) Cyanide forms a soluble complex with silver. Finally Ag is precipitated using zinc.

4.When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are         (2002)
cathode                                   anode
1)pure zinc                            pure copper
2)impure sample                 pure copper
3)impure zinc                     impure sample
4)pure copper                    impure sample
Ans.(4) Pure metal always deposits at cathode.

5.When rain is accompanied by a thunderstorm, the collected rain water will have a pH value (2003)
1)slightly higher than that when the thunderstorm is not there
2)uninfluenced by occurrence of thunderstorm
3)which depends on the amount of dust in air
4)slightly lower than that of rain water without thunderstorm
Ans.(4) The rain water after thrunderstorm contains dissolved acid and therefore the pH is less than rain water without thunderstorm.

6.Which one of the following ores is best concentrated by froth-floatation ?       (2004)
1) Magnetite                 2) Malachite
3) Galena                       4) Cassiterite
Ans.(3) Galena is sulphide mineral. Sulphide minerals are usually concentrated by froth floatation method.

7.The smog is essentially caused by the presence of        (2004)
1) 0₂ and 0₃                    2) 0₃ and N₂
3)Oxides of sulphur and nitrogen       4)0₂ and N₂
Ans.(3) Smog is Smoke + Fog. Somg contains mainly S0₂ and NO_x

8.During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’. These are        (2005)
1) Sn and Ag                  2) Pb and Zn
3) Ag and Au                 4) Fe and Ni
Ans.(3) Crude Cu contains Au and Ag as impurities.

9.Heating mixture of Cu_zO and Cu₂S will give  (2005)
1) Cu + S0₂                   2) Cu + S0₃
3) CuO + CuS                 4) Cu₂S0₃
Ans.(1) Self-reduction occurs upon heating a mixture of oxide and sulphide of copper.
2Cu₂0 + Cu₂S——- > 6Cu + S0₂

10.Identify the wrong statements in the following:(2008)
1)Chlorofluorocarbons are responsible for ozone layer depletion
2)Greenhouse effect is responsible for global warming
3)Ozone layer does not permit infrared radiation from the sun to reach the earth
4)Acid rains is mostly because of oxides of nitrogen and sulphur.
Ans.(3) Ozone layer does not allow ultraviolet radiation from sun to reach earth.

11.Which of the following factors is of no significance for roasting sulphide ores to the  oxides and not subjecting the sulphide ores to carbon reduction directly ?         (2008)
1)Metal sulphides are thermodynamically more stable than CS₂
2)C0₂ is thermodynamically more stable than CS₂
3)Metal sulphides are less stable than the corresponding oxides
4)C0₂ is more volatile than CS₂
Ans.(1) Metal slphides are more stable to heat than carbon disulphide.

12. Which method of purification is represented by the following equation      (2012)

Ans.(4) Titanium and zirconium are refined by vapour phase refining called Van Arkels method. Nickel is refined by Mond’s process.

13.The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was        (2013)
1) Methylisocyanate         2) Methylamine
3) Ammonia                   4) Phosgene
Ans.(1) It is called MIC gas.

14.Which series of reactions correctly represents chemical relations related to iron and its compound ?   (2014)

Ans.

15.Assertion (A) : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
Reason (R) : The reaction between nitrogen and oxygen requires high temperature. (2015)
1)Both assertion and reason are correct, and the reason is the correct explanation for the assertion.
2)Both assertion and reason are correct, but the reason is the not correct explanation for the assertion.
3)The assertion is incorrect, but the reason is correct.
4)Both the assertion and reason are incorrect.
Ans.(1) N₂ has triple bond and 0₂ has double bond.

JEE Main Home Maths Physics Chemistry

The post JEE Main Previous Year Papers Questions With Solutions Chemistry Metallurgy and Environmental Chemistry appeared first on Learn CBSE.

NCERT Solutions for Class 10 Science Chapter 12 Electricity

Electric current, potential difference and electric current, Ohms law, Resistance, Resistivity factors on which the resistance of a conductor depends; Series combination of resistors, parallel combination of resistors; and its application on daily life; Heating effect of Electric current, electric Power, Interrelation between P, V, and R.

Formulae Handbook for Class 10 Maths and Science

Page 200

What does an electric circuit mean?
Electric circuit is a continuous and closed path made of conducting wires, through which the electric current flows. It comprises a cell, ammeter, voltmeter, plug key, etc.
Define the unit of current.
SI unit of electric current is ampere (A).
Ampere is the flow of electric charges through an area at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross-section of wire for 1 second, then it would be equal to 1 ampere.

Page 202

Question 1:
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Cell or battery eliminator.

Question 2:
What is meant by saying that the potential difference between two points is 1 V?
Answer:
As we know that V = W / q
Thus, the potential difference between two points is one volt when one joule of work is done to carry a charge of one coulomb between the two points in the electric field.

More Resources for CBSE Class 10

• NCERT Solutions
• NCERT Solutions for Class 10 Science
• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Social
• NCERT Solutions for Class 10 English
• NCERT Solutions for Class 10 Hindi
• NCERT Solutions for Class 10 Sanskrit
• NCERT Solutions for Class 10 Foundation of IT
• RD Sharma Class 10 Solutions

Question 3:
How much energy is given to one coulomb of charge passing through a 6 V battery?
Answer:

Page 209

Question 1:
On what factors does the resistance of a conductor depend?
Answer:
Resistance of a conductor depends upon:
(i) Resistivity of the material.
(ii) Length of the conductor.
(iii) Cross-sectional area of the conductor.

Question 2:
Will current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why?
Answer:
The current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as R ∝ 1/A.

Question 3:
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:

Hence, the current through an electrical component also becomes half of its previous value.

Question 4:
Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The coils of electric toaster and electric iron are made of an alloy rather than a pure metal because of the following reasons;
(i) The resistivity of an alloy is higher than that of a pure metal.
(ii) It has high melting point and does not oxidise.

Question 5:
Use the data in Table 12.2 of NCERT book to answer the following:
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor? ‘
Answer:
(a) Iron because its resistivity is less than mercury.
(b) Silver is the best conductor as it has least resistivity.

Page 213

Question 1:
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, a 8 Ω resistor and a 12 Ω resistor and a plug key, all connected in series.
Answer:

Question 2:
Redraw the circuit of the above question, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the 12 resistor. What would be the reading in the ammeter and the voltmeter?
Answer:

Total resistance of the circuit = R
Since all the three resistors are connected in series, so, the equivalent resistance R is equal to the sum of all resistance.
R = 5 Ω + 8 Ω + 12 Ω = 25 Ω

Page 216

Question 1:
Judge the equivalent resistance when the following are connected in parallel.
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω , 10³ Ω and 10⁶ Ω
Answer:
Equivalent resistance in parallel combination of resistors is always less than the least resistance of any resistor in the circuit.
Hence, in both the given cases, the equivalent resistance is less than 1 Ω.

Question 2:
An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current flows through it?
Answer:

Question 3:
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?
Answer:
Advantages of connecting electrical devices in parallel:
(i) When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that-of battery which is not possible in series connection.
(ii) Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.
(iii) If one appliance fails to work, other will continue to work properly.

Question 4:
How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω?
Answer:
(a) In order to get 4 Ω, resistance 2 Ω should be connected in series with the parallel combination of 3 Ω and 6 Ω.

Question 5:
What is (a) the highest (b) the lowest total resistance that can be secured by combination of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) The highest resistance is secured by combining all four coils of resistance in series.
R_s = 4 Ω+ 8 Ω + 12 Ω + 24 Ω = 48 Ω
(b) The lowest resistance is secured by combining all four coils of resistance in parallel.

Page 218

Question 1:
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also heat produced ‘H’ is

H = I²Rt

Thus, for the same current H oc R, so for more resistance, more heat is produced by heating element and it glows.

Question 2:
Compute the heat generated while transferring 96000 C of charge in one hour through a potential difference of 50 V.
Answer:

Question 3:
An electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Given R = 20 Ω, I = 5 A, t = 30 s
H = I²Rt = (5)² x 20 x 30 = 15000 J = 1.5 x 10⁴ J

Page 220

Question 1:
What determines the rate at which energy is delivered by a current?
Answer:
Electric power determines the rate at which energy is delivered by a current.

Question 2:
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Given I = 5 A, V = 220 V, t = 2 h Power,
p = VI = 220 x 5 = 1100 W
Energy consumed = Vlt = Pt
= 1100 x 2 = 2200 Wh

TEXTBOOK QUESTIONS

Question 1:
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is

Answer:

Question 2:
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V^2/R
Answer:
(b) P = V^2/R = I²R = VI Option (b) does not represent electrical power.

Question 3:
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:

Question 4:
Two conducting wires of same material and of equal lengths and diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer:

Question 5:
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is connected in parallel across any two points in a circuit to measure the potential difference between them with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential of the source.

Question 6:
A copper wire has a diameter 0.5 mm and resistivity of 1.6 X 10^-8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:

Question 7:
The values of the current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:

Question 8:
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:

Question 9:
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer:
Since all the resistors are in series, the same current, 0.67 A flows through the 12 Ω resistor.

Question 10:
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:

Question 11:
Show how you would connect three resistors, each of resistance 6 Ω , so that the combination has a resistance of (i) 9 Ω , (it) 4 Ω .
Answer:
(i) When two 6 Ω resistances are in parallel and the third is in combination to this, the equivalent resistance will be 9 Ω.

(ii) When two 6 Ω resistances are in series and the third is in parallel to them, then it will be 4 Ω.

Question 12:
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Since, N bulbs of power P each connected in parallel will make the total power of NP,

Question 13:
A hot plate of an electric oven connected to 220 V line has two resistance coils A and B, each of 24 Q resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:

Question 14:
Compare the power used in the 2 Ω resistor in each of the following circuits.
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:

Question 15:
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:

Question 16:
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
Energy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh.
Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh.
∴ Energy consumed by TV set is more than the energy consumed by toaster in the given timings.

Question 17:
An electric heater of resistance 8 f2 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:

Question 18:
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminum wires usually employed for electricity transmission?
Answer:
(a) It has high melting point and emits light at a high temperature.
(b) It has more resistivity and less temperature coefficient of resistance.
(c) (i) All appliances do not get same potential in series arrangement.
(ii) All appliances cannot be individually operated.
(d) R ∝ =1 / Area of cross – section.
(e) They are very good conductors of electricity.

Short Answer Type Questions

Question 1:
Three 2 Ω resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.
Answer:

Question 2:
Should the resistance of an ammeter be low or high? Give reason.
Answer:
The resistance of an ammeter should be low so that it will not disturb the magnitude of current flowing through the circuit when connected in series in a circuit.

Question 3:
How does use of a fuse wire protect electrical appliances?
Answer:
The fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains. In this way, fuse wire protects the electrical appliances.

Question 4:
What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?
Answer:
The resistance offered by a metallic wire of unit length and unit cross-sectional area is called electrical resistivity.

Hence, when the length of wire is doubled, the resistance becomes double and current decreases to half.

Question 5:
A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.
Answer:

Question 6:
Why is parallel arrangement used in domestic wiring?
Answer:
Parallel arrangement is used in domestic wiring because
(i) Each appliance gets the same voltage as that of the mains supply.
(ii) If one component is switched off, others can work properly.
(iii) Fault in any branch of the circuit can be easily identified.

Question 7:
B_1, B₂ and B₃ are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B j gets fused?
(ii) What happens to the reading of A₁ ,A₂ , A₃ and A when the bulb B₂ gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
Answer:
(i) Since B₁ ,B₂ and B₃ are in parallel, the potential difference across each of them will remain same. So when the bulb B₁ gets fused, B₂ and B₃ have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.

Long Answer Type Questions

Question 8:
Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same wattage are connected in parallel to the same source.
(а) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
Answer:

(a) The bulbs in the two circuits will not glow equally bright as the current through them is not the same.

(b) As one bulb fuses, the other bulbs in the series circuit will not glow because the circuit becomes an open circuit. While the rest of bulbs in parallel circuit will continue to glow without getting disturbed because in parallel combination, current gets additional paths to flow.

Question 9:
Find out the following in the electric circuit given in figure:


Answer:

(e) No difference, since the ammeters are connected in series and same current will pass through them, so reading of both ammeters will be same.

Multiple Choice Questions (MCQs) [1 Mark each]

Question 1.
To determine the equivalent resistance of two resistors when connected in series, a student arranged the circuit components as shown in the diagram. But he did not succeed to achieve the objective. [CCE 2010]

Which of the following mistakes has been committed by him in setting up the circuit?
(a) Position of ammeter is incorrect
(b) Position of voltmeter is incorrect
(c) Terminals of ammeter are wrongly connected
(d) Terminals of voltmeter are wrongly connected
Answer:
(c) Because positive terminal of ammeter must be connected with positive terminal of cell and negative terminal of an ammeter must be connected to negative terminal of a cell.

Question 2.
For the given circuit, name the components which are connected in parallel. [CCE 2011]

(a) R₁ and R₂
(b) R₁, R₂ and V
(c) R₂ and V
(d) R₁ and V
Answer:
(b) The components R₁, R₂ and V are connected in parallel combination. Because terminals of the resistance and voltmeter are connected together.

Question 3.
A student arranges the following circuit to get equivalent resistance of a series combination of two resistors R₁ and R₂.

Which one of the following statements will be true for this circuit? [CCE 2007]
(a) It gives incorrect reading for current I as well as potential difference V
(b) It gives correct reading for current I but incorrect reading for potential difference V
(c) It gives correct reading for potential difference V but incorrect reading for current I
(d) It gives correct reading for both I and V
Answer:
(b) The voltmeter should be connected across the components of and R₂ to give correct reading for potential difference.

Question 4.
An ammeter has 20 divisions between 0 mark and 2A mark on its scale. The least count of ammeter is
(a) 0.01A
(b) 0.2A
(c) 0.1A
(d) 1A
Answer:
(c) Number of divisions = 20
Maximum reading of ammeter = 2 A
Least count of ammeter = 2/20 = 1/10 = 0.1 A

Question 5.
A student finds that there are 20 divisions between zero mark and 1V mark of a voltmeter. The least count of voltmeter is
(a) 0.1 V
(b) 0.01 V
(c) 0.05 V
(d) 1.0 V
Answer:
(c) Number of divisions = 20
Maximum reading of the voltmeter = 1 V
Least count of voltmeter = 1/20 = 0.05 V

Question 6.
The current flowing through a resistor connected in an electric circuit and the potential difference applied across its ends are shown in figure alongside.

The value of the resistance of the resistor is [CCE2013]
(a) 1 Ω
(b) 5 Ω
(c) 8 Ω
(d) 10 Ω
Answer:
(d) Reading from ammeter (7) = 180 mA= 0.18 A,
reading from voltmeter (V) = 1.8 V
Resistance of the resistor R = V/I = 1.8/0.18 = 180/18 = 10Ω

Question 7.
Which of the following is the correct method to connect the ammeter and voltmeter with resistance in the circuit to verify Ohm’s law? [CCE 2012]
(a) Ammeter and voltmeter in series
(b) Ammeter in series and voltmeter in parallel
(c) Ammeter in parallel and voltmeter in series
(d) Ammeter and voltmeter in parallel
Answer:
(b) In a circuit, ammeter should be connected in series, while voltmeter in parallel.

Question 8.
In an experiment on studying the dependence of the current I flowing through a given resistor on the potential difference V applied across it, a student has to change the value of the current. For doing this, he should change the
(a) number of cells used
(b) resistor itself
(c) ammeter used in the circuit
(d) Voltmeter used in the circuit
Answer:
(a) If we change the number of cells in electric circuit, the potential difference will change and as a result current flowing in the circuit changes.

Question 9.
A milliammeter had graduations marked 0, 100, 200, 300, 400 and 500. The space between 0 mark and 100 mark is divided into 20 divisions. If the pointer of the milliammeter is indicating the seventh graduation after 300 mark, the current flowing in the circuit is
(a) 335 mA
(b) 330 mA
(c) 331mA
(d) 340 mA
Answer:
(a) Number of divisions = 20
Least count of milliammeter = (100-0) / 20 = 5 mA
Milliammeter reading = 300 + 7 x 5 = 335 mA

Question 10.
If a student while studying the dependence of current on the potential difference keeps the circuit closed for a long time to measure the current and potential difference, then
(a) ammeter’s zero error will change
(b) ammeter will give more reading
(c) voltmeter will show constantly higher readings
(d) resistor will get heated up and its value will change
Answer:
(d) If the circuit is closed for a long time, then current flows in it for a long time which results that the resistor is heated.

Question 11.
To determine the eguivalent resistance of two resistors connected in series, a student prepared two electric circuits, correct reading of ammeter in the circuits is [CCE 2015]

(a) In circuit I, 1.0 A and in II, 0.1 A
(b) In both circuits I and II, 1.0 A
(c) In circuit I, 0.1 A and in II, 1.0 A
(d) In both circuits I and II, 0.1 A
Answer:
(b) Equivalent resistance of two resistors 3.5Ω and 1Ω in both the circuits I and II is R = 3.5 + 1 = 4.5 Ω
As, I = V/R = 4.5/4.5 = 1A
Therefore, current in both the circuits I and II is 1.0 A.

Question 12.
When parallel resistors are of three different values, the potential difference across its terminals is [CCE 2015]
(a) greatest across smallest resistance
(b) greatest across largest resistance
(c) equal across each resistance
(d) least across the smallest resistance
Answer:
(c) Potential difference across each resistor is same in parallel combination of resistors.

Hope given NCERT Solutions for Class 10 Science Chapter 12 helpful to you.

If you have any doubts, please comment below. We try to provide online math tutoring for you.

The post NCERT Solutions for Class 10 Science Chapter 12 Electricity appeared first on Learn CBSE.

NCERT Solutions for Class 7 Hindi Chapter 5-कुंती 

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NCERT Solutions for Class 7 Hindi Chapter 37 –  अशवत्थामा

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NCERT Solutions for Class 7 Hindi Chapter 6-भीम 

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NCERT Solutions for Class 7 Hindi Chapter 36 – कर्ण और दुर्योधन भी मारे गए 

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NCERT Solutions for Class 7 Hindi Chapter 35 – भूरिश्रवा ,जयद्रथ और आचार्य द्रोणा का अंत 

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NCERT Solutions for Class 7 Hindi Chapter 7-कर्ण 

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NCERT Solutions for Class 7 Hindi Chapter 34 – यधिष्टर की चिंता और  कामना 

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NCERT Solutions for Class 7 Hindi Chapter 33 – अभिमन्य 

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NCERT Solutions for Class 8th Sanskrit Chapter-9-उपसर्ग: प्रत्यया : च

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NCERT Solutions for Class 8th Sanskrit Chapter 10 – वाक्यरचना – शुद्धरोप्रयोग:

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NCERT Solutions for Class 7 Hindi Chapter 14-शकुनि का प्रवेश

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NCERT Solutions for Class 7 Hindi Chapter 32 – बारहाँवा दिन

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NCERT Solutions for Class 7 Hindi Chapter 9-लाख का घर

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