A hollow aluminium cylinder 20.0 cm deep has an internal capacity of 2.000 L at `20.0^@` C. It is completely filled with tupentine and then slowly warmed to `80.0^@`C a. How much turpentine overflows? B. If the cylindre is then cooled back to `20.0^@`C, how far below the cylinder's rim dows the tupentine's surface recede?
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9.00
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0.72
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Text Solution
Verified by Experts
The correct Answer is:
`0.11`
Volume of overflow `=Delta V _("turpentine") - Delta V _(Al) = gamma _("turpentine") V _(0) Delta theta - gamma _(A l) V _(0) Delta theta` `= (gamma _("Turpentine")- 3alpha _(Al)) V _(0) Delta theta = (9.00xx10^(-4)-2.16xx10^(-4)) V_(0) Delta theta` `= 6.84xx10^(-4) xx 2xx 80L =0.11L`
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VMC MODULES ENGLISH-JEE MAIN REVISON TEST-23-PHYSICS (SECTION 2)
