Solveeit Logo

Question

Physics Question on Surface tension

The length of a needle floating on water is 2.5 cm. How much minimum force, in addition to the weight of the needle will be needed to lift the needle above the surface of water? Surface tension of water is 7.2×104N/cm7.2\times {{10}^{-4}}N/cm .

A

7.2×104N7.2\times {{10}^{-4}}N

B

18.0×104N18.0\times {{10}^{-4}}N

C

36.0×104N36.0\times {{10}^{-4}}N

D

none of the above

Answer

36.0×104N36.0\times {{10}^{-4}}N

Explanation

Solution

: Surface tension =Force2(length)=\frac{Force}{2(length)} \therefore 7.2×104Ncm=Force2(2.5) cm7.2\times {{10}^{-4}}\frac{N}{cm}=\frac{Force}{2(2.5)\text{ }cm} \therefore Force =7.2×104×2.5×2N=7.2\times {{10}^{-4}}\times 2.5\times 2\,N =36.0×104N=36.0\times {{10}^{-4}}N