Physics ms of this year, Exams of Chemistry

Download Physics ms of this year and more Exams Chemistry in PDF only on Docsity! Class: XII SESSION : 2022-2023 MARKING SCHEME CBSE SAMPLE QUESTION PAPER (THEORY) SUBJECT: PHYSICS Q.no Marks SECTION A 1 (ii) q1q2<0 1 2 (iv) zero 1 3 (ii) material A is germanium and material B is copper 1 4 (iv) 6A in the clockwise direction 1 5 (iii) 4:3 1 6 (i) decreases 1 7 (ii) increase 1 8 (iv) Both electric and magnetic field vectors are parallel to each other. 1 9 (ii) the circular and elliptical loops 1 10 (iv) 0.85 1 11 (iii) 3000 Å 1 12 (iv) 4.77 X 10-10m 1 13 (ii) The nuclear force is much weaker than the Coulomb force . 1 14 (i) 30 V 1 15 (i) 1 16 c) A is true but R is false 1 17 c) A is true but R is false 1 18 a) Both A and R are true and R is the correct explanation of A 1 SECTION B 19 λ1 -Microwave λ2 - ultraviolet λ3- infrared Ascending order - λ2<λ3<λ1 ½ ½ ½ ½ 20 A - diamagnetic B- paramagnetic The magnetic susceptibility of A is small negative and that of B is small positive. ½ ½ ½ ½ 21 From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus R Fe/RAl = (AFe/AAl)⅓ = (125/27)⅓ R Fe = 5/3 RAl =5/3 x 3.6 = 6 fermi OR Given short wavelength limit of Lyman series ½ ½ ½ ½ 1  = 𝑅 1 1 − 1 ∞ 1 913.4 Å = 𝑅 1 1 − 1 ∞ L = = 913.4 Å For the short wavelength limit of Balmer series n1=2, n2 = ∞ 1  = 𝑅 1 2 − 1 ∞  = = 4 x 913.4 Å = 3653.6 Å ½ ½ ½ ½ 22 1 𝑓 = (µ − 1) 1 𝑅 − 1 𝑅 1 𝑓 = µ µ − 1 1 𝑅 − 1 𝑅 µ µ = 1.25 1.33 µ µ = 0.98 The value of (µ – 1) is negative and ‘f’ will be negative. So it will behave like diverging lens. ½ ½ ½ ½ 23 To keep the reading of ammeter constant value of R should be increased as with the increase in temperature of a semiconductor, its resistance decreases and current tends to increase. OR B - reverse biased In the case of reverse biased diode the potential barrier becomes higher as the battery further raises the potential of the n side. C -forward biased Due to forward bias connection the potential of P side is raised and hence the height of the potential barrier decreases. 1 1 ½ ½ ½ ½ 24 Angular width 2φ = 2λ/d Given λ = 6000 Å In Case of new λ (assumed λ’ here), angular width decreases by 30% New angular width = 0.70 (2 φ) 2 λ’/d = 0.70 X (2 λ/d) ∴λ’= 4200 Å ½ ½ ½ ½ b) c) 1 1 30 For a transition from n=3 to n =1 state, the energy of the emitted photon, h𝑣 = E2 –E1 = 13.6 [ − ] eV = 12.1 eV. From Einstein’s photoelectric equation, h𝑣 = 𝐾 + W0 ∴ 𝑊 = ℎ𝑣 − 𝐾 = 12.1 − 9 = 3.1 𝑒𝑉 Threshold wavelength, λth = = . . . = 4x10-7 m 1 ½ ½ 1 SECTION D 31(a) (b) (a) Here, A =6 x 10-3 m2, d =3mm=3x10-3m (i) Capacitance, C=∈0A/d = (8.85x10-12 x 6x10-3/3x10-3) = 17.7x10-12 F (ii) Charge, Q = CV =17.7 x10-12x100=17.7x10-10C (iii) New charge Q’= KQ=6x17.7 x10-10 =1.062 x 10-8 C OR Diagram 1 + 1 1 1 1 1/2 (b) (i) (ii) ( ) + ( ) + ( )( ) = 0 + = 0 or = q = 4Q or = Electric field due to a uniformly charged thin spherical shell: When point P lies outside the spherical shell: Suppose that we have calculate field at the point P at a distance r (r>R) from its centre. Draw Gaussian surface through point P so as to enclose the charged spherical shell. Gaussian surface is a spherical surface of radius r and centre O. Let ?⃗? be the electric field at point P, then the electric flux through area element of area 𝑑?⃗? is given by 𝑑φ = ?⃗?. 𝑑?⃗? Since 𝑑?⃗? is also along normal to the surface 𝑑φ= E dS ∴ Total electric flux through the Gaussian surface is given by φ = ∮ Eds = E ∮ ds Now, ∮ ds = 4 𝜋 r2 ...(i) = Ex4 𝜋 r2 Since the charge enclosed by the Gaussian surface is q, according to the Gauss’s theorem, φ = ∈ ........(ii) From equation (i) and (ii) we obtain E x 4 𝜋r2 = ∈ E= ∈ . (for r>R) 1 ½ ½ ½ ½ ½ 1 32(a) (b) (a) (b) Drift velocity: It is the average velocity acquired by the free electrons superimposed over the random motion in the direction opposite to electric field and along the length of the metallic conductor. Derivation I = ne A Vd Here, I = I1 + I2 ...(i) Let V = Potential difference between A and B. For cell 1 Then, V= 1 – I1 r1  1 1 1 V I r    Similarly, for cell 2 2 2 2 V I r    Putting these values in equation (i) 1 2 1 2 V V I r r       or 1 2 1 2 1 2 1 1 I V r r r r                or 1 2 2 1 1 2 1 2 1 2 r r r r V I r r r r               ...(ii) Comparting the above equation with the equivalent circuit of emf ‘eq’ and internal resistance ‘req’ then, eq eqV Ir   ...(iii) Then (i) 1 2 2 1 eq 1 2 r r r r       (ii) 1 2 eq 1 2 r r r r r   (iii) The potential difference between A and B eq eqV Ir   OR Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero Derivation ½ 1½ 3 1 1 3