question_answer

A cylindrical block of wood (density \[=650\,kg\,{{m}^{-3}}\]), of base area \[30\text{ }c{{m}^{2}}\]and height 54 cm, floats in a liquid of density \[900\text{ }kg{{m}^{-}}^{3}.\]The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) : [JEE Main Online Paper (Held On 11 April 2015)]  

A)  65 cm

B)  52 cm

C)  39 cm

D)  26 cm

Correct Answer: C

Solution :

Let block is floating with dissolve depth h. Then about equilibrium \[{{M}_{Block}}g={{F}_{up}}\] \[\left( AH{{\rho }_{B}} \right)g=\left( Ah \right){{\rho }_{L}}g\_(1)\] When block depressed by distance x then \[{{F}_{Net}}=F{{'}_{up}}-{{M}_{Block}}g\] \[=A\left( H+x \right){{\rho }_{L}}g-AH{{\rho }_{B}}g\] from equation (1) \[{{F}_{Net}}=Ax{{\rho }_{L}}g\] \[{{F}_{Net}}=-Ax{{\rho }_{L}}g\] \[\cancel{A}H{{\rho }_{Block}}.\frac{{{d}^{2}}x}{d{{t}^{2}}}=-\cancel{A}x{{\rho }_{L}}g\] \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-\frac{{{\rho }_{L}}g}{H{{\rho }_{Block}}}x\] \[{{\omega }^{2}}=\frac{{{\rho }_{L}}g}{H{{\rho }_{B}}}\] For simple pendulum \[{{\omega }^{2}}=\frac{g}{\ell }\] Equating \[\ell =\frac{H{{\rho }_{B}}}{{{\rho }_{L}}}\] \[=\frac{650\times 54}{900}=39cm\]