SATHEE: Properties Of Solids And Liquids Question 282

Properties Of Solids And Liquids Question 282

Question: An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3)[CPMT 1989]

Options:

A)350 cm3

B)300 cm3

C)250 cm3

D)22 cm3

Show Answer

Answer:

Correct Answer: B

Solution:

According to Boyle’s law, pressure and volumeare inversely proportional to each other

i.e. $P \propto \frac{1}{V}$ therefore $P_{1} V_{1} = P_{2} V_{2}$

therefore $(P_{0} + h ρ_{w} g) V_{1} = P_{0} V_{2}$

$\Rightarrow V_{2} = (1 + \frac{h ρ_{w} g}{P_{0}}) V_{1}$

therefore $V_{2} = (1 + \frac{47.6 \times 10^{2} \times 1 \times 1000}{70 \times 13.6 \times 1000}) V_{1}$

$\Rightarrow V_{2} = (1 + 5) 50 c m^{3} = 300 c m^{3} .$ [As $P_{2} = P_{0} = 70 c m$ of Hg $= 70 \times 13.6 \times 1000$ ]

पिछला

अगला

गोपनीयता नीति

द्वारा संचालित Prutor@IITK

sathee@iitk.ac.in

SATHEE का मीडिया कवरेज

खेल स्टोर

ऐप स्टोर

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Properties Of Solids And Liquids Question 282

Question: An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3)[CPMT 1989]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu