XI STD_PMT_LJ TEST-06 (MEDICAL) : TEST NO. : 06

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2 CPT-.6/XI-STD _PMT ( LJ) [].. 4 The area under velocity time graph for a particle to a given interval of time represents displacement.. 3 Acceleration of a particle is given as a a Given 0 Integrating eq. (i) we get dx C Again integrating...(i) x Ct C`...(ii) Comparting eq. (ii) with equation of a straight line Y mx c 3. 4 By the definition, the slop of displacement time graph is veloity i.e. v tan 0 v tan tan 30 v 0 tan tan 45 v v 3 PHYSICS 6. 3 for curve 4. Given Distance from 0 to 5s = 40 m Ditance from 5 to 0 s = 0 Distance from 0 to 5 s = 60 m Distance 5 to 0s = 0 m so, net distance = = 0 m Total time taken = 0 min Hence, average speed Distance( m) 0 6m min Time(min) The acceleration of a moving body is found from the slope of velocity time graph a = slope of v t curve = positive It represents uniformly accelerated motion for curve II, = negative It represents uniformly retarted motion. 7. Velocity in general means instantaneous velocity and its magnitude is always equal to speed v u as v as as u 0 The graph between v and s will be of the from parabola which will be symmetric with respect to v - axis. so curve (d) is the right answer. 9. velocity of particle = Area of a- t graph m / s 0. 4 In the positive region the velocity decreases linearly ( during rise) and in the negative region velocity increases linearly ( during fall) and the direction is opposite to each other during during rise and fall, hence fall is shown in the negative region.. Taking the motion from 0 to s. u 0, a 3 ms, t s, v? v u at 0 3 6ms Taking the motion from s to 4s ms v So option a is correct We know that the velocity of body is given by the slope of displacement time graph. So it is clear that initially slope of the graph is positive and after some time it becomes zero ( corresponding to the peak of graph) and then it will become negative.

3 [3] 3 3 From acceleration time graph, acceleration is constant for first part of motion so, for this part velocity of body increases uniformly with time and as a = 0 thent he velocity becomes constnat. Then again increased because of constant acceleration XI-PMT-LJ-/Medical/ otal length Time 4 sec Re lativevelocity Total distance m m / s Relative velocity Hence t 50/ 50 5s. When two particles moves towards each other then. v v 6 when these particles moves in the same direction then v v 4 By solving v 5 and v m / s

4 CPT-.6/XI-STD _PMT ( LJ) [4]

5 [5] XI-PMT-LJ-/Medical/ Given line have positive intercept but negative slope. so its equation can be written as v mx v 0...(i) [ where m = v tan 0 ] x0 By differentiating with respect to time we get dx m mv Now substituting the value of v from eq. (i) we get m mx v0 m x mv0 a m x mv 0 i.e. the graph between a an should have positive slope but negative intercept on a- axis. so graph (a) is correct. CHEMISTRY Co : Ar3d 4s No. of unpaired electrons Cu : Ar 3d 9 Ni : Ar 3d 8 3 Fe : Ar 3d Cr : Ar 3d 3 3 Co 3 Ar 3d 6 Correct order of unpaired electrons 3 3 Fe Cr Ni Cu 59. Within a group E decreases from top to bottom with negative sign the chorine has highest electron gain enthalpy ( in magnitude) 65. Nitrogen has half filled stable configuration so that its ionization enthalpy is greater than that of oxyge. So, correct order of increasing first ionization enthalpy is B C O N The configuration corresponds to that of Cl, which has the highest nagative electron gain enthalpy.

6 CPT-.6/XI-STD _PMT ( LJ) [6] Group 7 > Group Group 7 Cl F Br I Group 6 S Se Te Po O The correct order is O S F Cl 70. For isoelectronic ions, the ionic radius decreases with incease in nuclear charge. Correct order is 3 O 8 F 9 N a M g A l Cesium and fluorine form ionic bonds ionic q d has zero dipole moment e. s. u..750 cm e. s. u. cm 6.D % ionic character Observed dipolemoment 00 Dipole moment for completeionic character % Compounds having same shape involving same hybridisation of the central atom are isostructural. Both XeF and IF are linear species involving Next Exam Date : 30/0/7 PHASE TEST- 3 sp d hybridisation of the central atom SCl 4 has sea saw geometry while all other have tetrahedral geometry. 83. Syllabus : SYLLABUS COVERED FROM SL NO. 6-0 CBSE PATTERN PAPERPAPER FROM THE SAME SYLLABUS WILL BE PROVIDED AS HOME ASSIGNEMENT NO3 : trigonal planar, CO 3 : trigonal planar, ClO3 : pyramidal and SO 3 : trigonal planar, NO3 and CO 3 are isoslectronic also. 84. Because it has linear and symmetrical structure SiF 4 is tetrahedral while SF 4 has see- saw geometry XeOF 4 involves 3 sp d hybridisation and has one lone pair of electrons. O Acetophenone, C6 H 5C C H 3 has maximum dipole moment ( about 3D) SF molecule has permanent dipole moment because it is not symmetrical. PAPER SETTER PHYSICS CHEMISTRY BOTANY ZOOLOGY AURANGABAD MR. DURGESH SINGH MOBILE: MR. DEVDATTYA ZHA MOBILE: DR. ALOK KUMAR MOBILE: DR. D. MUKATI MOBILE:

Chapter 8 Test Study Guide AP Chemistry 6 points DUE AT TEST (Wed., 12/13/17) Date:

Chapter 8 Test Study Guide AP Chemistry 6 points DUE AT TEST (Wed., 12/13/17) Date: Chapter 8 Test Study Guide Name: AP Chemistry 6 points DUE AT TEST (Wed., 12/13/17) Date: Topics to be covered on the December 13, 2017 test: bond bond energy ionic bond covalent bond polar covalent bond