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Measurements: (i) 0.0025(ii) 0.0250(iv) 5.98 x 1024(iii) 0.002500(Ans.: 2, 3, 4, 3)2.2The length of an object measured by vernier calliperis 4.78 cm. If the L.C. of vernier caliper is 0.01 cm.Calculate percentage error.(Ans. : 0.209 %)The diameter of a wire measures by means ofmicrometer screw gauge of least count 0.001 cm is102 mm. Calculate the percentage error inmeasurement.(Ans. : 0.98 %)please guys right answer beta na please |
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Answer» L.C. = 0.01 cmError = + 0.02 cm ⇒ Positive zeroerror = + 0.02 cmmain scale reading (MSR) = 3.60 cm 8 TH vernier scale COINCIDES with MAIN scale⇒ vernier coincidence (VC) = 8We KNOW thatCorrect diameter = [main scale reading+ (L.C × V.C)] - (error)= [3.60 + (0.01 × 8)] - [+ 0.002]= (3.60 + 0.08)- (+0.02)= 3.68 - 0.02= 3.66 cm∴correctradius= 2diameter = 23.66 =1.83cm |
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