Chemistry-81.Two Year CRP-Questions

CHEMISTRY Time: 50 minutes Maximum Marks: 50 Note: 1. This paper contains Two sections A and B each section has two parts objective and subjective. 2. Marks allotted to each question are indicated in the Right Hand Margin. 3. Attempt All questions. 5. Use of Calculator / Logarithm table is NOT PERMITTED. 5. Useful Data: Gas Constant R = 8.314 J K1 mol1 = 0.0821 Lit atm K1 mol1 = 1.987  2 Cal K1 mol1 Avogadro's Number Na = 6.023  1023 1 Faraday = 96500 Coulomb 1 calorie = 4.2 Joule Atomic No: Ca = 20, C = 6, O = 8, K = 19, Cl = 17, N = 7, S = 16, Na = 11. Cu = 29, Co =27, Mn = 25. Atomic Masses: Ag = 108, Mn = 55, C = 12, O = 16, K = 39, N = 14, Na = 23, H = 1, I = 127. Name of the student : Registration Number : Date of Examination : SECTION - A PART - I Each question has four options. Answer the correct one. Each question has 1½ marks. 1. In a closed vessel, a gas is heated from 300K to 600 K , the kinetic energy becomes/remains (A) half (B) double (C) same (D) four times 2. Four rubber tubes are respectively filled with H2,O2, N2 and He. The tube which will be required to be reinflated first is (A) H2 - filled tube (B) O2 - filled tube (C) N2 - filled tube (D) He-filled tube 3. Which gas has the highest partial pressure in atmosphere (A) CO2 (B) H2O (C) O2 (D) N2 4. Equal masses of gaseous H2 and N2 are present in two flasks of same capacity and at constant temperature. What is ratio of number of molecules of H2 and N2? (A) 14 : 1 (B) 1 : 1 (C) 28 : 1 (D) 1.5 :1 5. According to the Maxwell - Boltzman distribution law of gases, the average translational kinetic energy is (A) kT/2 per molecule (B) kT per molecule (C) 3/2 kT per molecule (D) RT per molecule space for rough work 6. Under the same condition of temperature and pressure, two gases ammonia and HCl are introduced from two inlets of 100 cm long tube. A white fumes appears first at (A) the mid-point of tube (B) near the HCl end (C) near the NH3 end (D) in entire tube 7. At 1500 K, the ratio of rms velocity of O2 and O3 is (A) (B) (C) 3/2 (D) 8. Two Vessels of equal volume contain two gases A and B separately at 1 atm and 4 atm respectively. If the vessels are connected through a narrow tube of negligible volume, what is total pressure of gaseous mixture? (A) 1 atm (B) 4 atm (C) 5 atm (D) 2.5 atm 9. Under what conditions of temperature and pressure, the gas have maximum density? (A) 1 atm 0°C (B) 1 atm, 273° C (C) 2 atm, 0°C (D) 2 atm, 273°C 10. An open flask contains air at 27°C and one atm pressure. The flask is heated to 127°C at the same pressure. The fraction of original air remaining in the flask will be: (A) (B) (C) (D) space for rough work PART - II 1. 103 ml of CO2 were collected at 27C and 763 mm pressure. What will be its volume if the pressure is changed to 721 mm at the same temperature? [5] 2. A sealed tube, which can withstand a pressure of 3 atmosphere, is filled with air at 27C and 760 mm pressure. Find the temperature above which it will burst. [5] 3. 5.10 g of O2 were introduced into a evacuated vessel of 5 litres capacity maintained at 27C. Calculate the pressure of the gas in atmospheres in the container. [5] SECTION - B CONCEPT: PHOTO ELECTRIC EFFECT Let us try to analyse the Einstein’s experiment. He irradiated sodium metal with visible/ultraviolet light and observed. 1. Electrons came out as soon as the light was irradiated on the surface. 2. There was no emission of electrons for light having frequency less than a certain frequency, irrespective of the intensity of light used. 3. Photoelectric current increased with increase in intensity of light of same frequency if emission is permitted. On the basis of this Einstein had concluded that light must have streams of energy particles or quanta of energy. K.E.max = Energy of incident photon - Energy required to liberate an electron. 1/2 = h  ho ho = work function = Minimum energy of the photon to liberate an electron. o = Threshold frequency  = Frequency of incident radiation For electrons to be emitted the frequency of incident light should be > Threshold frequency  Threshold frequency 0 work function of the metal 0 = threshold frequency,  = frequency of incident light Also, h = K.E +  K.E = h - h0 = h( - 0) ( = work function) Where h is planck's constant with value 6.626  10–34 J sec Example: Energy required to stop the ejection of electrons from Cu plate is 0.24 eV. Calculate the work function when radiation of  = 253.7 nm strikes the plate? Solution: Energy required to stop the ejection of electron is equal to kinetic energy i.e., (½ mv2) Given K.E = 0.24 eV = 0.24 1.60 10–19J (  1 eV = 1.60  10–19 J)  of incident photon = 253. 7 nm h = ho + K.E Work function (ho) = h – K.E = – ½ mv2 = = 7.835 10–19 – 3.84 10–20 = 7.45 10–19 J = 4.65 eV PART - I Each question has four options. Answer the correct one. [2  5] 1. Kinetic energy of photoelectron depends on (A) Frequency of incident light (B) Intensity of light (C) Both A and B (D) none of the above 2. The work function of a photoelectric material is 3.3 eV. Its threshold frequency will be (A) 8.87  1015 Hz (B) 15  1033 Hz (C) 8  1014 Hz (D) 4  1011 Hz 3. The threshold frequency of a metal is 4  1014 s–1. The minimum energy of photon to cause photo electric effect is (A) 3.06  10–12 J (B) 1.4  10–48 J (C) 3.4  10–19 J (D) 2.64  10–19 J 4. Ultra violet light of 6.2 eV falls on aluminium surface (work function = 4.2 eV). The kinetic (in Joule) of the fastest electron emitted is approximately (A) 3.2  10–21 (B) 3.2  10–19 (C) 3.2  10–17 (D) 3.2  10–15 5. On increasing the intensity of light on metal surface showing photoelectric effect which of the following increase (A) Kinetic energy of photo electron (B) Number of photo electrons (C) Both A and B (D) None space for rough work PART - II Question: Calculate the kinetic energy of an electron emitted from the surface of a metal by light of wavelength 5.510–8 cm. Threshold energy for the metal is 2.6210–9 ergs. [10] 3

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45ยบ) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more