An object with uniform density ρ is attached to a spring that is known to stretch linearly with applied force as shown below. When the spring object system is immersed in a liquid of density ρ1 as shown in the above figure, the spring stretches by an amount x1(ρρ1). When the experiment is repeated in a liquid of density (ρ2

Question

An object with uniform density ρ is attached to a spring that is known to stretch linearly with applied force as shown below.

When the spring object system is immersed in a liquid of density ρ1 as shown in the above figure, the spring stretches by an amount x1(ρ>ρ1). When the experiment is repeated in a liquid of density (ρ2<ρ1), the spring stretches by an amount x2. Neglecting any buoyant force on the spring, the density of the object is (A) ρ=(ρ1 x1-ρ2 x2/x1-x2) (B) ρ=(ρ1 x2-ρ2 x1/x2-x1) (C) ρ=(ρ1 x2+ρ2 x1/x1+x2) (D) ρ=(ρ1 x1+ρ2 x2/x1+x2)

Tardigrade Admin · Accepted Answer

For equilibrium of block hung from string,
Spring force + Buoyant force

=

Weight of block

So, we have

k x_{1} + ρ_{1} V g = ρ V g

....(i)

k x_{2} + ρ_{2} V g = ρ V g

....(ii)
Eliminating

k

, we get

ρ = \frac{ρ _{1} x _{2} - ρ _{2} x _{1}}{x _{2} - x _{1}}

$ρ = \frac{ρ _{1} x _{1} - ρ _{2} x _{2}}{x _{1} - x _{2}}$

$ρ = \frac{ρ _{1} x _{2} - ρ _{2} x _{1}}{x _{2} - x _{1}}$

$ρ = \frac{ρ _{1} x _{2} + ρ _{2} x _{1}}{x _{1} + x _{2}}$

$ρ = \frac{ρ _{1} x _{1} + ρ _{2} x _{2}}{x _{1} + x _{2}}$

For equilibrium of block hung from string,
Spring force + Buoyant force $=$ Weight of block

So, we have
$k x_{1} + ρ_{1} V g = ρ V g$ ....(i)
$k x_{2} + ρ_{2} V g = ρ V g$ ....(ii)
Eliminating $k$ , we get
$ρ = \frac{ρ _{1} x _{2} - ρ _{2} x _{1}}{x _{2} - x _{1}}$

ρ=x1​−x2​ρ1​x1​−ρ2​x2​​

ρ=x2​−x1​ρ1​x2​−ρ2​x1​​

ρ=x1​+x2​ρ1​x2​+ρ2​x1​​

ρ=x1​+x2​ρ1​x1​+ρ2​x2​​

For equilibrium of block hung from string, Spring force + Buoyant force = Weight of block So, we have kx1​+ρ1​Vg=ρVg....(i) kx2​+ρ2​Vg=ρVg ....(ii) Eliminating k, we get ρ=x2​−x1​ρ1​x2​−ρ2​x1​​

$ρ = \frac{ρ _{1} x _{1} - ρ _{2} x _{2}}{x _{1} - x _{2}}$

$ρ = \frac{ρ _{1} x _{2} - ρ _{2} x _{1}}{x _{2} - x _{1}}$

$ρ = \frac{ρ _{1} x _{2} + ρ _{2} x _{1}}{x _{1} + x _{2}}$

$ρ = \frac{ρ _{1} x _{1} + ρ _{2} x _{2}}{x _{1} + x _{2}}$

For equilibrium of block hung from string,
Spring force + Buoyant force $=$ Weight of block

So, we have
$k x_{1} + ρ_{1} V g = ρ V g$ ....(i)
$k x_{2} + ρ_{2} V g = ρ V g$ ....(ii)
Eliminating $k$ , we get
$ρ = \frac{ρ _{1} x _{2} - ρ _{2} x _{1}}{x _{2} - x _{1}}$