JEE MAIN - Physics (2011 - No. 20)
Work done in increasing the size of a soap bubble from a radius of $$3$$ $$cm$$ to $$5$$ $$cm$$ is nearly (Surface tension of soap solution $$ = 0.03N{m^{ - 1}},$$
$$0.2\pi mJ$$
$$2\pi mJ$$
$$0.4\pi mJ$$
$$4\pi mJ$$
Explanation
$$W = T \times \,\,$$ change in surface area
$$W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$$
$$ = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J$$
$$ = 0.4\pi \times {10^{ - 3}}\,J$$
$$ = 0.4\pi mJ$$
$$W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$$
$$ = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J$$
$$ = 0.4\pi \times {10^{ - 3}}\,J$$
$$ = 0.4\pi mJ$$
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