Question

Ratio of the weights of a $1\text{kg}$ block of iron and $1\text{kg}$ block of wood as measured by a spring balance is:
Given: Density of iron $=7800{\text{kg/m}}^3$, density of wood $=800{\text{kg/m}}^3$ and density of air $=1.293{\text{kg/m}}^3$

• A. $1.5$
• B. $2.0015$
• C. $1.0015$
• D. $3.0015$

Answer 1

The correct option is C $1.0015$

FBD of the block:


Applying equilibrium condition i.e $\sum F=0$ along vertical direction. Considering air as a fluid, it will exert buoyant force on block.
$\Rightarrow T+F_b-mg=0$
$\Rightarrow T=mg-F_b$
$\Rightarrow T=mg-{\rho }_{a}gV...\left(1\right)$
$[{\rho }_a\to \text{density of air}]$
From Eq.$\left(i\right)$ for iron block:
$T_1=mg-{\rho }_{a}g{V}_i$
where $V_i\to \text{volume of iron block}$
$\Rightarrow T_1=(1\times 10)-1.293\times 10\times \frac{1}{7800}$
$[\because V_i=\frac{m}{{\rho }_i}]$
$\Rightarrow T_1=9.9983\text{N}$
Similarly, for wooden block ,
$T_2=mg-{\rho }_{a}g{V}_w$
$[V_w=\frac{m}{{\rho }_w}]$
$\Rightarrow T_2=(1\times 10)-1.293\times 10\times \frac{1}{800}$
$\Rightarrow T_2=9.9838\text{N}$
On dividing $T_1\&T_2$,
$\frac{T_1}{T_2}=\frac{9.9983}{9.9838}$
$\therefore T_1:T_2=1.0015$
Ratio of weights of iron and wooden block of mass $1\text{kg}$ is $1.0015$.