JEE MAIN - Physics (2017 (Offline) - No. 23)
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 $$\times$$ 10-2 m
rise of water, h = 1.45 $$\times$$ 10-2m
Using g = 9.80 m/s2 and the simplified relation T = $${{rhg} \over 2} \times {10^3}N/m$$, the possible error in surface tension is closest to :
diameter of capillary, D = 1.25 $$\times$$ 10-2 m
rise of water, h = 1.45 $$\times$$ 10-2m
Using g = 9.80 m/s2 and the simplified relation T = $${{rhg} \over 2} \times {10^3}N/m$$, the possible error in surface tension is closest to :
10 %
0.15 %
1.5 %
2.4 %
Explanation
Surface tension,
T = $${{rhg} \over 2} \times {10^3}N/m$$
Relative error,
$${{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}$$
Percentage error,
$${{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100$$
$${{\Delta T} \over T} \times 100 = \left( {{{{{10}^{ - 2}} \times 0.01} \over {1.25 \times {{10}^{ - 2}}}} + {{{{10}^{ - 2}} \times 0.01} \over {1.45 \times {{10}^{ - 2}}}}} \right) \times 100$$
= (0.8 + 0.689) = 1.489 % = 1.5 %
T = $${{rhg} \over 2} \times {10^3}N/m$$
Relative error,
$${{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}$$
Percentage error,
$${{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100$$
$${{\Delta T} \over T} \times 100 = \left( {{{{{10}^{ - 2}} \times 0.01} \over {1.25 \times {{10}^{ - 2}}}} + {{{{10}^{ - 2}} \times 0.01} \over {1.45 \times {{10}^{ - 2}}}}} \right) \times 100$$
= (0.8 + 0.689) = 1.489 % = 1.5 %
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