Answer
64.8k+ views
Hint: The question has given us all the information we need. First, find the net weight of water in the water in the first case. We all know that the density of water is $1\,g/cc$ . Use this to find the volume of the given density bottle. Then, find the net weight of mercury in the density bottle and use the volume we found out in the first case to find the density of mercury.
Complete step by step answer:
We will be trying to solve the question exactly as told in the hint section of the solution to the question.
Using the first case, we will find the volume of the density bottle. And then, we will use the newly-found volume of the density bottle to find the density of mercury.
Let us first find the volume of the bottle through the first case:
Weight of the empty density bottle as given in the question: ${W_b} = 25\,g$
Total weight after being completely filled by water: ${W_t} = 50\,g$
Hence, weight of the water can be found out as:
$\implies$ ${W_w} = \,{W_t} - {W_b}$
Substituting in the values, we get:
$
\implies {W_w} = 50 - 25 \\
\implies {W_w} = 25\,g \\
$
We already know that the density of water is $1\,g/cc$
Hence, volume of the density bottle is: $V = \dfrac{{25\,g}}{{1\,g/cc}}$ or $25\,cc$
Now, let us consider the second case:
Weight of the empty density bottle as given in the question: ${W_b} = 25\,g$
Total weight after being completely filled with mercury: ${W_t} = 365\,g$
Weight of mercury can be found out as:
$\implies$ ${W_m} = {W_t} - {W_b}$
Substituting in the values, we get:
$
\implies {W_m} = 365 - 25 \\
\implies {W_m} = 340\,g \\
$
We have already found out the value of volume of the given density bottle using the first case as:
$V = 25\,cc$
Using this, we can find the density of mercury as:
$\implies$ $d = \dfrac{{{W_m}}}{V}$
Substituting in the values that we found out:
$
\implies d = \dfrac{{340\,g}}{{25\,cc}} \\
\implies d = 13.6\,g/cc \\
$
Hence, option (C) is the correct option.
Note: A very important thing to do in such questions is to find out the density of the given density bottle or the container in which liquids or fluids are being stored. So always start solving the question by first finding out the volume of the container or density bottle. Also, always check the units of the weight and volume and tick the answer accordingly.
Complete step by step answer:
We will be trying to solve the question exactly as told in the hint section of the solution to the question.
Using the first case, we will find the volume of the density bottle. And then, we will use the newly-found volume of the density bottle to find the density of mercury.
Let us first find the volume of the bottle through the first case:
Weight of the empty density bottle as given in the question: ${W_b} = 25\,g$
Total weight after being completely filled by water: ${W_t} = 50\,g$
Hence, weight of the water can be found out as:
$\implies$ ${W_w} = \,{W_t} - {W_b}$
Substituting in the values, we get:
$
\implies {W_w} = 50 - 25 \\
\implies {W_w} = 25\,g \\
$
We already know that the density of water is $1\,g/cc$
Hence, volume of the density bottle is: $V = \dfrac{{25\,g}}{{1\,g/cc}}$ or $25\,cc$
Now, let us consider the second case:
Weight of the empty density bottle as given in the question: ${W_b} = 25\,g$
Total weight after being completely filled with mercury: ${W_t} = 365\,g$
Weight of mercury can be found out as:
$\implies$ ${W_m} = {W_t} - {W_b}$
Substituting in the values, we get:
$
\implies {W_m} = 365 - 25 \\
\implies {W_m} = 340\,g \\
$
We have already found out the value of volume of the given density bottle using the first case as:
$V = 25\,cc$
Using this, we can find the density of mercury as:
$\implies$ $d = \dfrac{{{W_m}}}{V}$
Substituting in the values that we found out:
$
\implies d = \dfrac{{340\,g}}{{25\,cc}} \\
\implies d = 13.6\,g/cc \\
$
Hence, option (C) is the correct option.
Note: A very important thing to do in such questions is to find out the density of the given density bottle or the container in which liquids or fluids are being stored. So always start solving the question by first finding out the volume of the container or density bottle. Also, always check the units of the weight and volume and tick the answer accordingly.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)