NARAYANA IIT/NEET Foundation Academy CPT-24 (IX Foundation) ANSWER KEY ( ) SECTION -A

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Janis Morgan
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1 Iq NRYN IIT/NEET Foundation cademy PT-4 (IX Foundation) NSWER KEY (9..07) SETIN - PHYSIS HEMISTRY ILGY MTHEMTIS MT SETIN- PHYSIS HEMISTRY ILGY MTHEMTIS Joule. F.S 0 3. S 0, So W o 4. Use W FS 5. W FScos HINTS/SLUTINS SETIN PHYSIS NRYN IIT/ NEET EMY

2 6. 7. P P 8. Use m m KE mv So x times KE mv So x times HEMISTRY 8 9. Value of e/m for cathode rays is constant gm and it is independent of the nature of gas taken in the discharge tube. 0. harge/mass = zero, for neutron, because charge on Neutron is 0.. Positive rays are positively charges gaseous ion.. Mass of neutron Kg 3. athode ray experiment was done by J.J. Thomson. 4. hange on an electron is coulombs 5. Nucleus of an atom contains protons and neutrons. 6. Neutron was discovered by hadwick. MTHEMTIS 5. In EX and E X = = X = XE = E (alternate interia angles) EX = E (vertically opposite angles) y S criteria EX E E = E E X Now ar( X) = area of parallogram E and parallelgram are on the same base & between the same parallel lines &. ar( E) = ar = ar ( X) = ar( EX) (median of triangle divides the triangle in to two equal area). 6. area ( ) = ar( ) = cm 7. parallelogram and EF are on the same base and between the same parallel lines & F. ar(ef) = ar() = 48 cm 8. ar( ) = ar F ar(f) = 7 = 54 cm 9. Join. P & PQ are on the same P and between the same parallel lines NRYN IIT/ NEET EMY

3 P & Q Q P ar ( PQ) = ar( P) ar 4 (P is the median in the PQ & is the median in the ) ar PQ ar P ar PQ ar( P) = ar( P) = = ar ar ar = ar ar = = ar 3. In a right angle triangle, mid point of the hypotenuse is equidistant from all the vertices. = = = 6.5 cm = = 3 cm = 5 5 ar 5 30cm 3. The centre of the circle lies interior of the circle. 33. The longest chord of the circle is diameter. 34. ngles in semi circle is right angle. 35. Given = 5 cm, = 3 cm. 3 cm5 cm In, = 3 5 = cm. = = = 4 cm. 36. If radius is different and centre is same, then circle is called concentric circle. 37. The mid point of diameter is called centre. 38. Let L = x cm, then M = (4 x) cm Now, x + 64 = r () and (4 x) + 36 = r () From () and (), x + 64 = (4 x ) = 96 8x x = 68 x = 6 From (), r = (6) + 64 = 00 r = 0 Therefore, radius of the circle is 0 cm. 8 cm L 8 cm NRYN IIT/ NEET EMY 3 r r x cm (4 x) cm 6 cm 6 cm M

39. Join. = 90 = 90 [ngle is a semi-circle] [ngle is a semi-circle] Sol + = 90 + 90 = 80 Therefore, is a line. That is lies on the line segment. MT 40. The series is aabbcc/aabbcc/aabbcc.

4 39. Join. = 90 = 90 [ngle is a semi-circle] [ngle is a semi-circle] Sol + = = 80 Therefore, is a line. That is lies on the line segment. MT 40. The series is aabbcc/aabbcc/aabbcc. Thus, the pattern aabbcc is repeated. 4. The series is cabbbb/cabbbb/cabbbb. Thus, the pattern cabbbb is repeated. 4. The series is cabbac/cabbac/cabbac. Thus, the pattern cabbac is repeated. 43. The series is ccc bbb aaa/ccc bbb aaa/c. Thus, the pattern ccc bbb aaa is repeated. 44. ll the three items are partly related to one another. 45. Tiger is a carnivore, while elephant is not. 46. Herring is a type of fish. Fish belong to the class of animals living in water. 47. Nurse and Patient are entirely different. ut, both are parts of hospital. SETIN PHYSIS 48. The force of gravity on the body, is the force exerted by earth on it and is mg (weight) acting vertically down. However, the displacement s=m vertically up. PE 9.8 NRYN IIT/ NEET EMY 4

5 = joule v m 49. mv mv v m m v E E m v If m m,e E m m m = m m m E.E. of rifle is smaller. HEMISTRY 50. Valence electrons are present in outermost shell or orbit. 5. euterium ( ) has proton and neutron. MTHEMTIS 57. raw L Since =, so L is the mid-point of. L= = cm = 6cm L nd, L is the perpendicular bisector of. entre of the circle lies on L. Let be the centre of the circle. Then, lies on L. Join. From right-angled L, we have = L + L L = L = [(0) 6 ] = 64 L = 64 8 cm L = (L ) = (8 r) cm Now, from right angled L, we have = L + L r = (8 r) + 6 r = r 6r r = 00 r = raw the perpendicular bisectors L and M of PQ and RS respectively. PQ RS L and M are in the same line, L and M are collinear Join P and R L P Q R M S NRYN IIT/ NEET EMY 5

6 In right triangle LP, P = L + PL (0) L PQ a circle to a chord bisects the chord) [y Pythagoras theorem] [ The perpendicular drawn from the centre of 00 L.6 00 = L + (8) 00 = L + 64 L = = 36 = (6) L = 6 cm In right triangle MR, R = M + RM [y Pythagoras theorem] M.RS a circle to a chord bisects the chord] [ The perpendicular drawn from the centre of (0) M. (0) = M + (6) M = (0) (6) = = 64 = (8) M = 8 cm, LM = M L = 8 6 = cm Hence the distance between PQ and RS, if they lie on the same side of the centre, is cm. 59. Join. Then, = radius = 7 cm. L 30 cm 5cm. L L L 64 8cm. distance of from = 8 cm NRYN IIT/ NEET EMY 6

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in Chapter - 10 (Circle) Key Concept * Circle - circle is locus of such points which are at equidistant from a fixed point in a plane. * Concentric circle - Circle having same centre called concentric circle.