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Water rises upto a height 'h' in a capillary tube on the surface of the earth. The value of ' $\mathrm{h}$ ' will increase if the experimental setup is kept in $[\mathrm{g}=$ acceleration due
to gravity]
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to gravity]
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The correct answer is:
a lift going down with acceleration
(D)
$h=\frac{2 T \cos \theta}{r \rho g} \quad \therefore h \propto \frac{1}{g}$
For a lift going down with acceleration $a$, the effective value of $g$ is $g^{\prime}=g-a$ $g^{\prime}=g-a$
$\mathrm{g}^{\prime} < \mathrm{g} \quad \therefore \mathrm{h}$ will increase.
$h=\frac{2 T \cos \theta}{r \rho g} \quad \therefore h \propto \frac{1}{g}$
For a lift going down with acceleration $a$, the effective value of $g$ is $g^{\prime}=g-a$ $g^{\prime}=g-a$
$\mathrm{g}^{\prime} < \mathrm{g} \quad \therefore \mathrm{h}$ will increase.
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