JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 6)
A bakelite beaker has volume capacity of 500 cc at 30oC. When it is partially filled with Vm
volume
(at 30oC) of mercury, it is found that the unfilled volume of the beaker remains constant as
temperature is varied. If $$\gamma $$(beaker) = 6 × 10–6 oC–1 and $$\gamma $$(mercury) = 1.5 × 10–4 oC–1, where $$\gamma $$ is the
coefficient of volume expansion, then Vm
(in cc) is close to ____.
Answer
20
Explanation
$$\Delta $$V = V$$\gamma $$$$\Delta $$T
$$ \therefore $$ V1$$\gamma $$1 = V2$$\gamma $$2
$$ \Rightarrow $$ 500 $$ \times $$ 6 $$ \times $$ 10-6 = Vm $$ \times $$ 1.5 $$ \times $$ 10-4
$$ \Rightarrow $$ Vm = $${{500 \times 6 \times {{10}^{ - 6}}} \over {1.5 \times {{10}^{ - 4}}}}$$
$$ \Rightarrow $$ Vm = 20 cc
$$ \therefore $$ V1$$\gamma $$1 = V2$$\gamma $$2
$$ \Rightarrow $$ 500 $$ \times $$ 6 $$ \times $$ 10-6 = Vm $$ \times $$ 1.5 $$ \times $$ 10-4
$$ \Rightarrow $$ Vm = $${{500 \times 6 \times {{10}^{ - 6}}} \over {1.5 \times {{10}^{ - 4}}}}$$
$$ \Rightarrow $$ Vm = 20 cc
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