JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 18)
Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their
volumes is :
4 : 1
8 : 1
2 : 1
0.8 : 1
Explanation
$${P_{in}} = {P_0} + {{4T} \over {{R_1}}}$$
$$ \Rightarrow 1.01 = 1 + {{4T} \over {{R_1}}}$$
$$ \Rightarrow {{4T} \over {{R_1}}} = 0.01$$
$$1.02 = 1 + {{4T} \over {{R_2}}}$$
$$ \Rightarrow {{4T} \over {{R_2}}} = 0.02$$
$$ \therefore {{{R_2}} \over {{R_1}}} = {1 \over 2}$$
$$ \Rightarrow {R_1} = 2{R_2}$$
$${{{V_1}} \over {{V_2}}} = {{R_1^3} \over {R_2^3}} = {{8R_2^3} \over {R_2^3}} = {8 \over 1}$$
$$ \Rightarrow 1.01 = 1 + {{4T} \over {{R_1}}}$$
$$ \Rightarrow {{4T} \over {{R_1}}} = 0.01$$
$$1.02 = 1 + {{4T} \over {{R_2}}}$$
$$ \Rightarrow {{4T} \over {{R_2}}} = 0.02$$
$$ \therefore {{{R_2}} \over {{R_1}}} = {1 \over 2}$$
$$ \Rightarrow {R_1} = 2{R_2}$$
$${{{V_1}} \over {{V_2}}} = {{R_1^3} \over {R_2^3}} = {{8R_2^3} \over {R_2^3}} = {8 \over 1}$$
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