JEE Main Solved Paper 2017 Question 4
Question: The following observations were taken for determining surface tensiton T of water by capillary method : Diameter of capilary, $ D=1.25\times {10^{-2}}m $ rise of water, $ h=1.45\times {10^{-2}}m $ . Using $ g=9.80,m/s^{2} $ and the simplified relation $ \text{T=}\frac{rhg}{2}\times\text{ 1}{0^{3}},\text{N/m,} $ the possible error in surface tension is closest to: JEE Main Solved Paper-2017
Options:
A) 2.4%
B) 10%
C) 0.15%
D) 1.5%
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \text{T=}\frac{rhg}{2}\times 10^{3} $ $ \frac{\Delta T}{T}=\frac{\Delta r}{r}+\frac{\Delta h}{h}+0 $ $ 100\times \frac{\Delta T}{T}=( \frac{{10^{-2}}\times .01}{1.25\times {10^{-2}}}+\frac{{10^{-2}}\times .01}{1.45\times {10^{-2}}} )100 $ $ =(0.8+0.689) $ $ =(1.489) $ $ 100\times \frac{\Delta T}{T}=1.489% $ $ \simeq 1.5% $
