A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density $(1 / 3)^{rd}$ that of mercury. Initially h was 228 cm.

Through a small hole in the bulb gas leaked assuming pressure decreases as $\frac{d P}{d t} = - k p .$ If value of h is 114 cm after 14 minutes, what is the value of k (in $hour^{- 1}$ )?
[ Use : In (4/3) =0.28 and density of Hg =13.6 g/mL.].

Question

A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density

(1 / 3)^{rd}

that of mercury. Initially h was 228 cm.

Through a small hole in the bulb gas leaked assuming pressure decreases as

\frac{d P}{d t} = - k p .

If value of h is 114 cm after 14 minutes, what is the value of k (in

hour^{- 1}

)?
[ Use : In (4/3) =0.28 and density of Hg =13.6 g/mL.].

Filo · Accepted Answer

Answer: bSolution: (b) P=P0e−kt​,P0​=3228​+76=152cmHg At  t=14min,P=3114​+76=114cmHg  ⇒   152e−kt=114 ⇒k=t1​In114152​=141​×0.28  =0.02×60hr−1=1.2hr−1

Question Text	A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density $(1 / 3)^{rd}$ that of mercury. Initially h was 228 cm. Through a small hole in the bulb gas leaked assuming pressure decreases as $\frac{d P}{d t} = - k p .$ If value of h is 114 cm after 14 minutes, what is the value of k (in $hour^{- 1}$ )? [ Use : In (4/3) =0.28 and density of Hg =13.6 g/mL.].
Answer Type	Text solution:1
Upvotes	150

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