{"id":60452,"date":"2024-04-19T10:25:16","date_gmt":"2024-04-19T04:55:16","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=60452"},"modified":"2024-04-19T10:25:16","modified_gmt":"2024-04-19T04:55:16","slug":"ts-10th-class-physical-science-model-paper-set-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ts-10th-class-physical-science-model-paper-set-2\/","title":{"rendered":"TS 10th Class Physical Science Model Paper Set 2 with Solutions"},"content":{"rendered":"

Timed practice with TS 10th Class Physical Science Model Papers<\/a> Set 2 is crucial for improving speed and efficiency during exams.<\/p>\n

TS 10th Class Physical Science Model Paper Set 2 with Solutions<\/h2>\n

Time: 1 Hour 30 minutes
\nMaximum Marks: 40<\/p>\n

General Instructions:\u00a0<\/span><\/p>\n

    \n
  1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.<\/li>\n
  2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.<\/li>\n
  3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.<\/li>\n
  4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.<\/li>\n<\/ol>\n

    Part – A (30 Marks)<\/span>
    \nSection – I (3 \u00d7 2 = 6 Marks)<\/span><\/p>\n

    Instructions :<\/p>\n

      \n
    1. 3 Very short answer questions are there in section – I<\/li>\n
    2. Answer ALL the questions. Each question carries 2 marks.<\/li>\n
    3. Write answer to each question in 3 to 4 sentences.<\/li>\n<\/ol>\n

      Question 1.
      \nWhy do we prefer a convex mirror as a rearview mirror in vehicles?
      \nAnswer:
      \nWe prefer a convex mirror as a rear-view mirror in vehicles because of the following reasons.<\/p>\n

        \n
      1. A convex mirror always forms an erect, virtual and diminished image of an object placed at anywhere in front of it.<\/li>\n
      2. A convex mirror has a wider field of view than a plane mirror of the same size.<\/li>\n
      3. Thus convex mirrors enable the driver to view much larger traffic behind him than would be possible with a plane mirror.<\/li>\n<\/ol>\n

        Question 2.
        \nA rainbow viewed from an airplane may form a complete circle. Where will the shadow of the airplane appear ? Explain.
        \nAnswer:<\/p>\n

          \n
        1. A rainbow viewed from an airplane form a complete circle because the earth does not come along the way of the airplane and rainbow.<\/li>\n
        2. A rainbow is a three dimensional cone of dispersed light it appear as a complete circle.<\/li>\n
        3. The shadow of the airplane appear with in the circle of the rainbow.<\/li>\n<\/ol>\n

          \"TS<\/p>\n

          Question 3.
          \nExplain the reaction of various metals in activity with cold water.
          \nAnswer:<\/p>\n

            \n
          1. From potassium to magnesium displace, hydrogen from cold water with decreasing reactivity. Potassium react with cold water violently but reaction of magnesium is very slow. The reactivity order is given below.
            \nMg < Ca < Na < K<\/li>\n
          2. From Aluminium to gold hydrogen from cold water is not displaced.<\/li>\n<\/ol>\n

            Section – II (3 \u00d7 3 = 9 Marks)<\/span><\/p>\n

            Instructions :<\/p>\n

              \n
            1. 3 Short answer questions are there in section – II.<\/li>\n
            2. Answer ALL the questions. Each question carries 3 marks.<\/li>\n
            3. Write answer to each question in 5 to 6 sentences.<\/li>\n<\/ol>\n

              Question 4.
              \n55<\/sub>Cs has less ionization energy than 3<\/sub>Li. How can you explain it ?
              \nAnswer:<\/p>\n

                \n
              1. It is because of screening effect.<\/li>\n
              2. Atomic number of Cs is 55, means it has more electrons and more inner shells.<\/li>\n
              3. Between nucleus and valence electron, these shells act as screens and decrease nuclear attraction.<\/li>\n
              4. So 55<\/sub>Cs has less ionization energy than 3<\/sub>Li.<\/li>\n<\/ol>\n

                Question 5.
                \nA chemical compound has the formula AB3<\/sub>. Identify the molecule.
                \na) Without any lone pair\u2019s.
                \nb) With one lone pair.
                \nc) Mention the shape of any one of the above molecules.
                \nAnswer:
                \na) BF3<\/sub>, BCl3<\/sub>, AlCl3<\/sub> etc.
                \nb) NH3<\/sub>, PH3<\/sub>, PCl3<\/sub> etc.
                \nc) NH3<\/sub>, PH3<\/sub>, PCl3<\/sub> – Pyramidal Shape
                \nBF3<\/sub> – Traingle Planar Shape<\/p>\n

                Question 6.
                \nDraw the structure of Ethane and electron dot structure of Chlorine.
                \nAnswer:
                \nEthane:
                \n\"TS
                \nChlorine:
                \n\"TS<\/p>\n

                \"TS<\/p>\n

                Section – III (3 \u00d7 5 = 15 Marks)<\/span><\/p>\n

                Instructions :<\/p>\n

                  \n
                1. 3 Essay answer questions are there in section.<\/li>\n
                2. Answer ALL the questions. Each question carries 5 marks.<\/li>\n
                3. Internal choice is given in this section.<\/li>\n
                4. Write answer to each question in 8 to 10 sentences.<\/li>\n<\/ol>\n

                  Question 7.
                  \nHow to make a chemical equation more informative ?
                  \n(OR)
                  \nDistinguish between acids and bases.
                  \nAnswer:
                  \nChemical equations can be made more informative by expressing the following characteristics of the reactants and products.
                  \ni) Physical state
                  \nii) Heat exchange (exothermic or endothermic change
                  \niii) Gas evolved (if any)
                  \niv) Precipitate formed (if any)<\/p>\n

                  i) Expressing the physical state : The physical state of the substances may be mentioned along with their chemical formula. The different states i.e., gaseous, liquid and solid states are represented by the notations (g), (l) and (s) respectively. If the substance is present as a solution in water, it is represented as (aq).
                  \nE.g.: Fe2<\/sub>O3(s)<\/sub> + 2 Al(s)<\/sub> \u2192 2 Fe(s)<\/sub> + Al2<\/sub>O3(s)<\/sub><\/p>\n

                  ii) Heat exchange : Heat is liberated in exothermic reactions and heat is absorbed in endothermic reactions.
                  \n1) C(s)<\/sub> + O2(g)<\/sub> \u2192 CO2(g)<\/sub> + Q (exothermic reaction)
                  \n2) N2(g)<\/sub> + O2(g)<\/sub> \u2192 2NO(g)<\/sub> – Q (endothermic reaction)
                  \n(Or) N2(g)<\/sub> + O2(g)<\/sub> \"TS 2NO(g)<\/sub><\/p>\n

                  iii) Gas evolved : Ifa gas is evolved ma reaction, it is denoted by an upward arrow t or (g).
                  \nE.g.: Zn(s)<\/sub> + H2<\/sub>SO4(aq)<\/sub> \u2192 ZnSO4(aq)<\/sub> + H(2g)<\/sub>\u2191<\/p>\n

                  iv) Precipitate formed: If a precipitate is formed in the reaction, it is denoted by a downward arrow \u2193.
                  \nE.g.: AgNO3(aq)<\/sub> + NaCl(aq)<\/sub> AgCl(s)<\/sub>\u2193 + NaNO3(aq)<\/sub>
                  \n(OR)<\/p>\n\n\n\n\n\n\n\n\n\n
                  Acids<\/td>\nBases<\/td>\n<\/tr>\n
                  1. An acid is a substance whichgives H30+ ions in Water solution<\/td>\n1. A base is a solution which contains OH group in water solution gives hydroxyl ions OH–<\/sup><\/td>\n<\/tr>\n
                  2. Acids are sour in taste<\/td>\n2. Bases are bitter in taste.<\/td>\n<\/tr>\n
                  3. Aciaturns blue litmus to red.<\/td>\n3. Bases turn red litmus in taste<\/td>\n<\/tr>\n
                  4. The orange colour of methyl orange indicator changes to red in acid medium<\/td>\n4. The orange colour of methyl orange indicator changes to yellow in bases medium.<\/td>\n<\/tr>\n
                  5. Acid are formed when non-metal oxides are dissolved in water<\/td>\n5. Bases are formed when metal oxides are dissolved in water<\/td>\n<\/tr>\n
                  6. pH value of acid is less than 7<\/td>\n6. pH value of base is geater than 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                  Question 8.
                  \nWrite characteristics of image formed due to convex lens at various distances.
                  \n(OR)
                  \nDistinguish between emission and absorption spectrum.
                  \nAnswer:<\/p>\n\n\n\n\n\n\n\n\n\n
                  Position of the object<\/td>\nPosition of the image<\/td>\nRelative size of the image<\/td>\nNature of the image<\/td>\n<\/tr>\n
                  1) At infinity<\/td>\nFocal point<\/td>\nHighly diminished<\/td>\nReal and inverted<\/td>\n<\/tr>\n
                  2) Beyond centre of curvature (C2<\/sub>)<\/td>\nBetween F1<\/sub> and C1<\/sub><\/td>\nPoint size diminished<\/td>\nReal and inverted<\/td>\n<\/tr>\n
                  3) At the centre pf curvature (C2<\/sub>)<\/td>\nAt C1<\/sub><\/td>\nSame size<\/td>\nReal, inverted<\/td>\n<\/tr>\n
                  4) Between centre of cur-vature focal point (C2<\/sub> and F2<\/sub>)<\/td>\nBeyond centre of curvature<\/td>\nMagnified<\/td>\nReal, inverted<\/td>\n<\/tr>\n
                  5) At focal point (F2<\/sub>)<\/td>\nAt infinity<\/td>\nHighly magnified<\/td>\nReal and inverted<\/td>\n<\/tr>\n
                  6) Between focal point and pole F2<\/sub> and P<\/td>\nOn the same side of the lens of the object<\/td>\nMagnified<\/td>\nVirtual and erect<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                  (OR)<\/p>\n\n\n\n\n\n\n
                  Emission spectrum<\/td>\nAbsorption spectrum<\/td>\n<\/tr>\n
                  1) The spectrum produced by emitted radia\u00adtion is called emission spectrum.<\/td>\n1) The spectrum produced by absorption of ra\u00addiation is called absorption spectrum.<\/td>\n<\/tr>\n
                  2) The emission spectrum contains bright lines on the dark background<\/td>\n2) The absorption spectrum contains dark lines on the bright background.<\/td>\n<\/tr>\n
                  3) The emission spectrum corresponds the radiation emitted when an excited elec\u00adtron returns back to the ground state.<\/td>\n3) The absorption spfectrum coresponds the radiation absorbed in exciting an electron from lower to the higher energy levels.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                  Question 9.
                  \n\"TS
                  \nObserve the picture. The potential values at A, B, C are 70V, 0V, 10V
                  \na) What is the potential at D ?
                  \nb) Find the ratio of the flow of current in AD, DB, DC.
                  \n(OR)
                  \nList out the apparatus and experimental procedure for the experiment to observe a current carrying wire experiences a magnetic force when it is kept in uniform magnetic field.
                  \ni) pushed into the coil ?
                  \nii) withdrawn from inside the coil?
                  \niii) held stationary inside the coil ?
                  \nAnswer:
                  \n\"TS
                  \na) By following Ohm\u2019s law P.D is (V) = iR
                  \nIn the given circuit, we are applying junction laws.
                  \n\u2018D\u2019 works as junction so, i = i1<\/sub> + i2<\/sub>
                  \nLet P.D at D is V0<\/sub>,
                  \nWe know that, i = \\(\\frac{\\mathrm{V}}{\\mathrm{R}}\\)
                  \n\"TS
                  \n(70 – V0<\/sub>)6 = 5V0<\/sub> – 20
                  \n420 – 6V0<\/sub> = 5V0<\/sub> – 20
                  \n6V0<\/sub> + 5V0<\/sub> = 440 \u21d2 11V0<\/sub> = 440
                  \nV0<\/sub> = \\(\\frac{440}{11}\\) = 40
                  \n\u2234 Potential at D is = V0<\/sub> = 40V<\/p>\n

                  b) Flow of current in AD is
                  \n\"TS
                  \nThe ratio of the flow of current in AD, DB and DC is 3 : 2 : 1.
                  \n(OR)<\/p>\n