A small hole is made at the bottom of a hollow sphere. The water enters into it when it is taken to a depth of $40 c m$ under water. If the surface tension of water is $0.07 N m^{- 1}$ , diameter of the hole is:

1 / 7 m m

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1 / 14 m m

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2 / 15 m m

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3 / 14 m m

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Solution

The correct option is D $1 / 14 m m$
$P_{i n} - P_{o u t} = \frac{4 T}{r}$
given water enters at a depth of $40 c m$
$\Rightarrow P_{i n} - P_{o u t} = h s g$
$40 \times 10^{- 2} \times 1000 \times 10$
$∴= 4000 = \frac{4 T}{r} = \frac{8 T}{d}$
$\Rightarrow d = \frac{8 T}{4000} = \frac{1}{14} m m$

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A small hole is made at the bottom of a hollow sphere. The water enters into it when it is taken to a depth of 40cm under water. If the surface tension of water is 0.07Nm−1, diameter of the hole is:

The correct option is D 1/14 mmPin−Pout=4Trgiven water enters at a depth of 40cm⇒Pin−Pout=hsg40×10−2×1000×10∴=4000=4Tr=8Td⇒d=8T4000=114mm

A small hole is made at the bottom of a hollow sphere. The water enters into it when it is taken to a depth of $40 c m$ under water. If the surface tension of water is $0.07 N m^{- 1}$ , diameter of the hole is:

The correct option is D $1 / 14 m m$
$P_{i n} - P_{o u t} = \frac{4 T}{r}$
given water enters at a depth of $40 c m$
$\Rightarrow P_{i n} - P_{o u t} = h s g$
$40 \times 10^{- 2} \times 1000 \times 10$
$∴= 4000 = \frac{4 T}{r} = \frac{8 T}{d}$
$\Rightarrow d = \frac{8 T}{4000} = \frac{1}{14} m m$