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 \[\ell =\frac{H{{\rho }_{B}}}{{{\rho }_{L}}}\] \[=\frac{650\times 54}{900}=39cm\]You need to login to perform this action.
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