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)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
350 cm3
b
300 cm3
c
250 cm3
d
22 cm3
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
According to Boyle's law, pressure and volume are inversely proportional to each other i.e. P∝1V⇒P1V1=P2V2⇒P0+hρwgV1=P0V2⇒V2=1+hρwgP0V1⇒V2=1+47.6×102×1×100070×13.6×1000V1⇒V2=(1+5)50cm3=300cm3. As P2=P0=70cm of Hg=70×13.6×1000