Question

no spasm dear,
please help me in solving this question.​

Answer 1

Given:-

• ρ = density of first liquid

• nρ = density of second liquid

• L = length of the cylinder

• d = density of the solid cylinder

To find:-

$\text{The density of the solid cylinder (d).}$

Solution:-

$\text{let the area of cross-section of the cylinder be A, such that, the weight \\}\\\text{of the cylinder is given by :}\\\bold{W_c=Mass*g\\}\\\bold{W_c = density*volume*g}\\\bold{W_c=d*AL*g}\\\\\bullet \text{(g is the acceleration due to gravity.)}$

$\text{Given that, the cylinder floats with its axis vertical and length pL}(p<1)\\\text{ in the denser liquid.}\\\text{The volume of the denser liquid in which the cylinder is immersed is given by :}\\\\\bold{V_1=Cross-sectional\ area\ of\ the\ cylinder*Height\ of\ the\ liquid}\\\bold{V_1=A*pL}\\\\\text{The volume of the less denser liquid in which the cylinder is immersed is given by :}\\\\\bold{V_2 = Cross-sectional\ area\ of\ the\ cylinder*Height\ of\ the\ liquid }\\\bold{V_2=A*(L-pL)=AL(1-p)}$

$\text{The weight of the liquids displaced is given by :}\\\bold{W_1=(Mass\ of\ the\ first\ liquid*g)+(Mass\ of\ the\ second\ liquid*g)}\\\\\bold{\implies W_1=(Density\ of\ first\ liquid*Volume\ of\ first\ liquid*g)}\\\bold{+(Density\ of\ second\ liquid*Volume\ of\ second\ liquid*g)}$

$\bold{\implies W_1=n\rho*V_1*g+\rho*V_2*g}\\\bold{\implies W_1=n\rho*ApL*g+\rho*AL(1-p)*g}\\\bold{\implies W_1=n\rho ApLg+\rho AL(1-p)g}\\\bold{\implies W_1=\rho ALg(np+1-p)}\\\bold{\implies W_1=\rho ALg[1+(n-1)p]}$

$\text{According to the law of floatation,}\\\bold{W_c=W_1}\\\bold{\implies d*AL*g=\rho ALg[1+(n-1)p]}\\\bold{\implies d=\rho [1+(n-1)p]}\\\\\boxed{\implies d=[1+(n-1)p]\rho}$

option (1) is correct.