Title DPP-9 SOLUTION.pdf ✅

CLASS : XITH SUBJECT : PHYSICS DATE : DPP NO. : 9 1 (c) Gravitational force, F = GM1M2 R2 ⇒ G = FR 2 M1M2 [G] = [MLT −2][L 2] [M2] = [M−1L 3T −2 ] 3 (a) [C] = ( Q V ) = ( Q 2 W ) = [ A 2T 2 ML 2T −2 ] = [M−1L −2T 4A 2 ] 4 (a) Angular velocity = θ t ,[ω] = [M0L 0T 0] [T] = [T −1 ] 5 (c) Given, length of rod A is LA = 3.25 ± 0.01 Of B is LB = 4.19 ± 0.01 Then, the rod B is longer than rod A by a length ∆l = LB − LA ∆l = (4.19 ± 0.01) − (3.25 ± 0.01) ∆l = (0.94 ± 0.02)cm 6 (c) Electric displacement, D = εE Unit of D = C 2 Nm2 N C Topic :-.UNITS AND MEASUREMENTS Solutions OM COACHING CLASES
∴ [D] = ( C m2 ) = [AT] [L 2] = [L −2TA] 7 (a) If E is the intensity of electric field over a small area element dS and θ is angle between E and outdrawn normal to area element. Therefore, electric flux through this element is dφE = (dS)(E cos θ) = E dS cos θ = E. dS Hence, φE = E . S = V d . S ∴ Unit of φE = volt × metre2 metre = volt − metre 8 (d) Diameter = Main scale reading + Circular scale reading × LC + Zero error = 3 + 35 × 1 2 × 50 + 0.03 = 3.38 mm 9 (c) F = −η. A dv dx ⇒ [η] = [ML −1T −1 ] 10 (d) Torque = [ML 2T −2 ], Moment of inertia [ML 2 ] 11 (a) η = F av = [MLT −2 ] [L][LT −1] = [ML −1T −1 ] 12 (a) Required relative error=power × relative error in x. 13 (c) Since for 50.14 cm, significant number = 4 and for 0.00025, significant number = 2 14 (a) Kinetic energy =1 2 mv 2 = M[LT −1 ] 2 = [ML 2T −2 ] 15 (b) T-ratios are dimensionless. So the unit of r is N 2 . 16 (a) 30 VSD = 29 MSD

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