Question

Q3 Homework. Unanswered You are given a sphere of diameter Ds, and a cylinder of diameter Dc and length L. The dimensions of these objects are such that Ds = 4Dc = L. The volume specific surface area of the sphere, [V SA]s, and volume specific surface area of the cylinder, [V SA]c, are related as: O A [V SA]c = 3[V SAJS O B VSA]c = [V SAJs o c [VSA]c = 4[V SAJS O D [V SA]c=2[V SAJS Unanswered Submit Q4 Homework. Unanswered The smallest division on the main scale of a slide caliper is 1 mm. If 25 divisions of the vernier scale on this vernier caliper are equal to 24 divisions of the main scale, the instrumental limit of error for this caliper is: O A 0.04 mm o B 0.01 mm o C 25 mm OD 24 mm Unanswered Submit

Answer 1

diameter of sphere Ds diameter of cylinder Dc Length of rylinder 4 De =L Ds Volm specific susface area sphere LV SAJ, is 4JTR susface area 3 EvSal 송고요리 radius of where R Sphere EVSA1 d cylinder VSAJ= volm specific susface area LVSA Surfece area VoJm 27R2+ 2TR2 L TR2L where R-radius st cylinder De 2 2 L De 2(R+L) [VSAE RL De L 2 4 Dc L Ds Prom De +41 LVSA 4 Dc Dc 2 2 Ds from 4 vSA Ds Ds

VSA1 fram L4DE EVSA 2 Σナ4D。 1 2 De 4 De De 2 fram O Ds DC 4 18 Ds C 2 Ds ram Ds But LUSAJS SATS vSAe 18 +8 [V SAJe Q.4) Instrumental limit of error less than is shich is 4ven by or equal to least count value of smallest divi sion on main scale No. of divisions on vexnier sca le 1 mm 25 O. 04 mm |/