Question

There is a leak proof cylinder of length 1 m, made of a metal that has very low coefficient of
expansion is floating vertically in water at 0
°C such that its height above the water surface is 20 cm.
If the temperature of water is increased to 4° C, the height f the cylinder above the water surface
becomes 21 cm. The density of water at T = 4° C, relative to the density at T =0° C is approximately.​

Answer 1

Question:

There is a leak proof cylinder of length 1 m, made of a metal that has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. If the temperature of water is increased to 4° C, the height f the cylinder above the water surface becomes 21 cm. The density of water at T = 4° C, relative to the density at T =0° C is approximately:

Given:

Length of the cylinder: 1m

Height of the cylinder above the water surface at 0°C = 20cm.

Height of the cylinder above the water surface at 4°C = 21cm.

To find:

The density of water at T = 4° C, relative to the density at T =0° C

Answer:

1.01

Explanation:

As the cylinder is immersed in water, a buoyant force will act over it and only the immersed part of the cylinder will be considered.

We will find the find the heights of the cylinder immersed in water at both the temperatures.

At 0°C,

Immersed height($h$) : 100-20 = 80cm or 0.8 m

At 4°C,

Immersed height($h_0$) : 100-21 = 79 cm or 0.79 m

Now let's calculate the Buoyant force for both the cases.

We have $\boxed{ F_B= \rho A \times h \times g}$

where $\rho$ is density of water, A is cross sectional area of cylinder, h is the immersed height and as usual g is acceleration due to gravity

The density of water differs due to temperature,

At 0°C, let density be $\rho_0$ and density at 4°C be $\rho$

At 0°C,

$F_{B_0}= \rho_0 A \times h \times g \\\\ F_{B_0} = \rho A \times 0.8 \times g - - - - [i]$

At 4°C,

$F_{B_4}= \rho A \times h_0 \times g \\\\ F_{B_4}= \rho A \times 0.79 \times g - - - - [ii]$

Dividing [ii] by [i], we have :

$\frac{ F_{B_4}} { F_{B_0}} = \frac{\rho A \times 0.79 \times g}{\rho_0 A \times 0.8 \times g} \\\\ 1 = \frac{\rho \cancel{A} \times 0.79 \times \cancel{g}}{\rho_0 \cancel{A} \times 0.8 \times \cancel{g}} \\\\ 1 = \frac{\rho}{\rho_0} \times \frac{0.79}{0.8} \\\\ \frac{\rho}{\rho_0} = \frac{0. 8}{0.79} \\\\ \frac{\rho}{\rho_0} \approx 1.01$

Hence, The density of water at T = 4° C, relative to the density at T =0° C is 1.01

Answer 2

Correct option is

Correct option isD

Correct option isD1.01

Correct option isD1.01Let A be the area of cross section of the cylinder.

be the density of water at 0

be the density of water at 0

be the density of water at 0 be the density of water at 4

be the density of water at 0 be the density of water at 4 o

be the density of water at 0 be the density of water at 4 o C.

be the density of water at 0 be the density of water at 4 o C.Total length of the cylinder is 100cm

be the density of water at 0 be the density of water at 4 o C.Total length of the cylinder is 100cmGiven that the length of the cylinder above the surface of water is 20cm

.

C, the weight of the cylinder is balanced by the buoyant force.

C, the weight of the cylinder is balanced by the buoyant force.

C, the weight of the cylinder is balanced by the buoyant force. A×0.8×g=w [Buoyant force =pAlg] .(i)

C, the weight of the cylinder is balanced by the buoyant force. A×0.8×g=w [Buoyant force =pAlg] .(i)At 4^oC$$, thef weight of the cylinder is balanced by the buoyant force.

C, the weight of the cylinder is balanced by the buoyant force. A×0.8×g=w [Buoyant force =pAlg] .(i)At 4^oC$$, thef weight of the cylinder is balanced by the buoyant force.

C, the weight of the cylinder is balanced by the buoyant force. A×0.8×g=w [Buoyant force =pAlg] .(i)At 4^oC$$, thef weight of the cylinder is balanced by the buoyant force. A×0.79×g=w ...(ii)

C, the weight of the cylinder is balanced by the buoyant force. A×0.8×g=w [Buoyant force =pAlg] .(i)At 4^oC$$, thef weight of the cylinder is balanced by the buoyant force. A×0.79×g=w ...(ii)Dividing (i) by (ii)