JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 14)
A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $$\omega $$. If the radius of the bottle is 2.5 cm then $$\omega $$ is close to – (density of water = 103 kg/m3).
2.50 rad s$$-$$1
3.75 rad s$$-$$1
5.00 rad s$$-$$1
7.90 rad s$$-$$1
Explanation
Restoring force due to pressing the bottle with small
amount x,
F = $$ - \left( {\rho Ax} \right)g$$
$$ \Rightarrow $$ ma = $$ - \left( {\rho Ax} \right)g$$
$$ \Rightarrow $$ a = $$ - \left( {{{\rho Ag} \over m}} \right)x$$
$$ \therefore $$ $${{\omega ^2} = {{\rho Ag} \over m}}$$ = $${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}} $$ = 7.90 rad/s
F = $$ - \left( {\rho Ax} \right)g$$
$$ \Rightarrow $$ ma = $$ - \left( {\rho Ax} \right)g$$
$$ \Rightarrow $$ a = $$ - \left( {{{\rho Ag} \over m}} \right)x$$
$$ \therefore $$ $${{\omega ^2} = {{\rho Ag} \over m}}$$ = $${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}} $$ = 7.90 rad/s
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