Title DPP-10 SOLUTION.pdf ✅

CLASS : XITH SUBJECT : PHYSICS DATE : DPP NO. : 10 1 (c) Rs = R1R2 R1 + R2 , ∆Rs Rs × 100 = ∆R1 R1 × 100 + ∆R2 R2 × 100 + ∆(R1 + R2) R1 + R2 = 100 Now, ∆R1 = 10 100 × 4kΩ=0.4kΩ, ΔR2 = 10 100 × 6kΩ = 0.6kΩ Again, ∆Rs Rs × 100 = 0.4 4 × 100 + 0.6 6 × 100 + 0.4 + 0.6 10 × 100 = 10 + 10 + 10 = 30 2 (d) Note carefully that every alterative has Gh and c 5 . [Gh] = [M−1L 3T −2 ][ML 2T −1 ] = [M0L 5T 3 ] [c] = [LT −1 ] ∴ ( Gh c 5 ) 1/2 = [T] 3 (b) C 2LR = [C 2L 2 ] × [ R L ] = [T 4 ] × [ 1 T ] = [T 3 ] As [ L R ] = T and √LC = T 4 (d) Unit of e. m. f. = volt = joule/coulomb 5 (b) % error in g = ∆g g × 100 = ( ∆l l ) × 100 + 2 ( ∆T T ) × 100 EI = 0.1 64 × 100 + 2 ( 0.1 128) × 100 = 0.3125% EII = 0.1 64 × 100 + 2 ( 0.1 64) × 100 = 0.4687% Topic :-.UNITS AND MEASUREMENTS Solutions OM COACHING CLASSES
EIII = 0.1 20 × 100 + 2 ( 0.1 36) × 100 = 1.055% 6 (b) 1 MeV = 106eV 7 (c) [Energy] = [ML 2T −2 ]. Increasing M and L by a factor of 3 energy is increased 27 times. 8 (a) Dimensionally. [ b t ] = [v] or [b] = [vt] = [L]. 9 (a) M =Pole strength × length = amp − metre × metre = amp − metre2 10 (b) ∴ ( ∆R R × 100) max = ∆V V × 100 + ∆I I × 100 = 5 100 × 100 + 0.2 10 × 100 = (5 + 2)% = 7% 11 (c) 0.2 25 × 100 = 0.8 13 (c) [ 1 2 ∈0 E 2 ] = [Energy density] = ML 2T −2 L 3 = ML −1T −2 14 (c) Dimensions of L and R [R] = [ML2T −3A −2 ] [L] = [ML 2T −2 A −2 ] [ L R ] = [ML 2T −2A −2 ] [ML 2T −3A−2] = [T] 15 (d) [E][J] 2 [M] 5[G] 2 [ML 2T −2 ][ML 2T −1 ] 2 [M5][M−1L 3T −2] 2 = [M0L 0T 0 ] 16 (d) As v = 4 3 πr 3 dv v = 3 ( dr r ) ∴ Percentage error in determination of volume = 3 (Percentage error in measurement of radius) = 3(2%) = 6%

ANSWER -KEY Q. 1 2 3 4 5 6 7 8 9 10 A. C D B D B B C A A B Q. 11 12 13 14 15 16 17 18 19 20 A. C A C C D D C A D B