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Question: The relative density of the material of the body is the ratio of its weight in air and to the appare...

The relative density of the material of the body is the ratio of its weight in air and to the apparent loss of its weight in water. By using a spring balance, the weight of the body measured in air is (5.00±0.05)N\left( {5.00 \pm 0.05} \right)\,{\text{N}}. The weight of the body measured in water is found to be (4.00±0.05)N\left( {4.00 \pm 0.05} \right)\,{\text{N}}. Then the maximum possible percentage error in relative density is
A. 11%
B. 10%
C. 9%
D. 7%

Explanation

Solution

Determine the apparent loss in the weight of the body by subtracting weight of the body in water from its weight in air. The relative density of the material of body is the ratio of its weight in air and to the apparent loss of its weight in water. Recall the expression for determining the maximum possible percentage error in the relative density.

Complete Step by Step Answer:
We have given the weight of the body in air Wa=(5.00±0.05)N{W_a} = \left( {5.00 \pm 0.05} \right)\,{\text{N}} and weight of the body in water Ww=(4.00±0.05)N{W_w} = \left( {4.00 \pm 0.05} \right)\,{\text{N}}.
Let us determine the apparent loss in the weight of the body as follows,
ΔW=WaWw\Delta W = {W_a} - {W_w}
Substituting Wa=(5.00±0.05)N{W_a} = \left( {5.00 \pm 0.05} \right)\,{\text{N}} and Ww=(4.00±0.05)N{W_w} = \left( {4.00 \pm 0.05} \right)\,{\text{N}} in the above equation, we get,
ΔW=(5.00±0.05)(4.00±0.05)\Delta W = \left( {5.00 \pm 0.05} \right) - \left( {4.00 \pm 0.05} \right)
ΔW=(1.00±0.1)N\Rightarrow \Delta W = \left( {1.00 \pm 0.1} \right)\,{\text{N}}

As we have given that the relative density of the material of the body is the ratio of its weight in air and to the apparent loss of its weight in water. Therefore, the relative density of the material of the body is,
ρ=WaΔW\rho = \dfrac{{{W_a}}}{{\Delta W}}
ρ=5.00±0.051.00±0.1\rho = \dfrac{{5.00 \pm 0.05}}{{1.00 \pm 0.1}}
Now, the maximum possible percentage error in the relative density is can be written as,
%ρ=5.001.00±(0.055.00+0.11.00)×100\% \rho = \dfrac{{5.00}}{{1.00}} \pm \left( {\dfrac{{0.05}}{{5.00}} + \dfrac{{0.1}}{{1.00}}} \right) \times 100
%ρ=5.00±(0.01+0.1)×100\Rightarrow \% \rho = 5.00 \pm \left( {0.01 + 0.1} \right) \times 100
%ρ=5.00±11%\therefore \% \rho = 5.00 \pm 11\%
Thus, the maximum possible percentage error in the relative density of the material of the body is 11%.

So, the correct answer is option A.

Note: The relative density of liquid is often expressed as the ratio of density of the liquid to the density of water. Since the water is the highest dense liquid, the relative density is always less than 1 for all liquids. Weight is also a force, thus, the unit of weight must be newton.