Question

A cargo ship carrying iron floats in a casual locks. An accident in unloading the ship causes the iron core block to be dumped into the water. Density of iron block is ${\rho }_i$, mass of ship= M, density of water is ${\rho }_w$, mass of iron block=m, surface area of canal lock=A, surface area of ship=a. What is change in level of water?

• A. $\frac{m}{A-a}[\frac{1}{{\rho }_w}-\frac{1}{{\rho }_i}],levelfalls$
• B. $\frac{m}{A-a}[\frac{1}{{\rho }_w}-\frac{1}{{\rho }_i}],levelrises$
• C. $\frac{m}{A}[\frac{1}{{\rho }_w}-\frac{1}{{\rho }_i}],levelfalls$
• D. $\frac{m}{A}[\frac{1}{{\rho }_w}-\frac{1}{{\rho }_i}],levelrises$

Answer 1

Answer: (D)

$\frac{m}{A}[\frac{1}{{\rho }_w}-\frac{1}{{\rho }_i}],levelrises$

Before,

We have to find $h_4-h_1$

B= weight of two block

$Pwa{h}_{2}g=(m+M)g$

$Pwa{h}_2=(m+M)$

After

B'= weight of M mass block

$Pwa{h}_{3}g=M_g$

$Pwa{h}_3=M$

Initial volume of water = final value of water

$A{h}_1-a{h}_z=A{h}_4-a{h}_3=\frac{m}{f_1}$

We know $h_2=\frac{m+M}{Pwa}{h}_3=\frac{M}{Pwa}$

$A(h_1-h_4)=\frac{a}{A}(h_2-h_3)-\frac{m}{P_i}$

$h_1-h_4=\frac{a}{A}(\frac{M+m}{Pwa}-\frac{M}{Pwa})-\frac{m}{P_{i}A}$

$=\frac{1}{A}\left[\frac{m}{Pw}\right]-\frac{m}{P_{i}A}$

$=\frac{1}{A}[\frac{m}{Pw}-\frac{m}{P_i}]$

$=\frac{m}{A}[\frac{1}{Pw}-\frac{1}{P_i}]$

$P_i>P_w$

So level raises.