
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 \[\left( {5.00 \pm 0.05} \right)\,{\text{N}}\]. The weight of the body measured in water is found to be \[\left( {4.00 \pm 0.05} \right)\,{\text{N}}\]. Then the maximum possible percentage error in relative density is
A. 11%
B. 10%
C. 9%
D. 7%
Answer
545.1k+ views
Hint: 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 \[{W_a} = \left( {5.00 \pm 0.05} \right)\,{\text{N}}\] and weight of the body in water \[{W_w} = \left( {4.00 \pm 0.05} \right)\,{\text{N}}\].
Let us determine the apparent loss in the weight of the body as follows,
\[\Delta W = {W_a} - {W_w}\]
Substituting \[{W_a} = \left( {5.00 \pm 0.05} \right)\,{\text{N}}\] and \[{W_w} = \left( {4.00 \pm 0.05} \right)\,{\text{N}}\] in the above equation, we get,
\[\Delta W = \left( {5.00 \pm 0.05} \right) - \left( {4.00 \pm 0.05} \right)\]
\[ \Rightarrow \Delta W = \left( {1.00 \pm 0.1} \right)\,{\text{N}}\]
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,
\[\rho = \dfrac{{{W_a}}}{{\Delta W}}\]
\[\rho = \dfrac{{5.00 \pm 0.05}}{{1.00 \pm 0.1}}\]
Now, the maximum possible percentage error in the relative density is can be written as,
\[\% \rho = \dfrac{{5.00}}{{1.00}} \pm \left( {\dfrac{{0.05}}{{5.00}} + \dfrac{{0.1}}{{1.00}}} \right) \times 100\]
\[ \Rightarrow \% \rho = 5.00 \pm \left( {0.01 + 0.1} \right) \times 100\]
\[ \therefore \% \rho = 5.00 \pm 11\% \]
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.
Complete Step by Step Answer:
We have given the weight of the body in air \[{W_a} = \left( {5.00 \pm 0.05} \right)\,{\text{N}}\] and weight of the body in water \[{W_w} = \left( {4.00 \pm 0.05} \right)\,{\text{N}}\].
Let us determine the apparent loss in the weight of the body as follows,
\[\Delta W = {W_a} - {W_w}\]
Substituting \[{W_a} = \left( {5.00 \pm 0.05} \right)\,{\text{N}}\] and \[{W_w} = \left( {4.00 \pm 0.05} \right)\,{\text{N}}\] in the above equation, we get,
\[\Delta W = \left( {5.00 \pm 0.05} \right) - \left( {4.00 \pm 0.05} \right)\]
\[ \Rightarrow \Delta W = \left( {1.00 \pm 0.1} \right)\,{\text{N}}\]
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,
\[\rho = \dfrac{{{W_a}}}{{\Delta W}}\]
\[\rho = \dfrac{{5.00 \pm 0.05}}{{1.00 \pm 0.1}}\]
Now, the maximum possible percentage error in the relative density is can be written as,
\[\% \rho = \dfrac{{5.00}}{{1.00}} \pm \left( {\dfrac{{0.05}}{{5.00}} + \dfrac{{0.1}}{{1.00}}} \right) \times 100\]
\[ \Rightarrow \% \rho = 5.00 \pm \left( {0.01 + 0.1} \right) \times 100\]
\[ \therefore \% \rho = 5.00 \pm 11\% \]
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

