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A vernier callipers is used to measure the width and the length of a rectangular plate. The measured values are 1.38 cm and 4.02 cm respectively.Find the uncertainly in the value of its area. |
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Answer» Given that, width w=1.38 cm length , l=4.02 cm Since the least count of a vernier callipers is 0.01 cm, each of the above mentioned measurements has an uncertainly of `pm`0.01 cm. hence, the values can be written as w=1.38 cm `pm`0.01 cm `l=4.02 cm pm 0.01 cm` The area obtained using the measured value is A=w x l = 1.38 x 4.02 `A=5.55 cm^2` ....(i) If we take the lower limits of both the measured values i.e. `l_"min"`=4.02-0.01 =4.01 cm and `w_"min"`=1.38-0.01 =1.37 cm The minimum possible area (lower limits of the area ) comes out to be `A_"min"=4.01xx1.37 cm^2` `A_"min"=5.49 cm^2`...(ii) The upper limit can also be calculated similarly. `therefore A_"max"=4.03xx1.39 cm^2` `A_"max"=5.60 cm ^2` ...(iii) using equation (i),(ii) and (iii) , we find that `A_"min"` is lower than A (the nominal measurement ) by the value `0.06 cm^2` And `A_"max"` is higher than A by the value `0.05 cm^2` As a practical rule, we choose the higher of these two deviations (from the measured value ) as the uncertainly, in our result. Therefore, in this particular example, we can write the area of plate as `5.55 cm^2 pm 0.06 cm^2` |
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