Question:
The change in density of mercury when it is heated from 10 °C to 60 °C. (The coefficient of volume expansion of mercury is \(18.2 \times 10^{-5} \, \text{K}^{-1}\))
The change in density of mercury when it is heated from 10 °C to 60 °C. (The coefficient of volume expansion of mercury is \(18.2 \times 10^{-5} \, \text{K}^{-1}\))
Show Hint
For liquids, \(\frac{\Delta \rho}{\rho} = - \beta \Delta T\), where \(\beta\) is volume expansion coefficient.
Updated On: Jul 18, 2026
- 1.82 %
- 0.91 %
- 9.1 %
- 0.45 %
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Recall relation between density and volume expansion.
\[ \rho = \frac{m}{V}, \quad \text{change in density } \frac{\Delta \rho}{\rho} = - \beta \Delta T \]
where \(\beta\) is coefficient of volume expansion, \(\Delta T\) is temperature change.
Step 2: Identify known quantities.
\(\beta = 18.2 \times 10^{-5} \, \text{K}^{-1}, \quad \Delta T = 60 - 10 = 50 \, \text{K}\)
Step 3: Compute change in density fraction.
\[ \frac{\Delta \rho}{\rho} = - 18.2 \times 10^{-5} \times 50 = -9.1 \times 10^{-4} \]
Step 4: Convert to percentage.
\[ |\Delta \rho/\rho| \times 100 = 0.091 \% \approx 0.91 \% \]
Step 5: Verify reasoning.
Negative sign indicates density decreases with heating. Magnitude in percentage matches options.
Step 6: Final conclusion.
Hence, the change in density is:
\[ \boxed{0.91 \%} \]
\[ \rho = \frac{m}{V}, \quad \text{change in density } \frac{\Delta \rho}{\rho} = - \beta \Delta T \]
where \(\beta\) is coefficient of volume expansion, \(\Delta T\) is temperature change.
Step 2: Identify known quantities.
\(\beta = 18.2 \times 10^{-5} \, \text{K}^{-1}, \quad \Delta T = 60 - 10 = 50 \, \text{K}\)
Step 3: Compute change in density fraction.
\[ \frac{\Delta \rho}{\rho} = - 18.2 \times 10^{-5} \times 50 = -9.1 \times 10^{-4} \]
Step 4: Convert to percentage.
\[ |\Delta \rho/\rho| \times 100 = 0.091 \% \approx 0.91 \% \]
Step 5: Verify reasoning.
Negative sign indicates density decreases with heating. Magnitude in percentage matches options.
Step 6: Final conclusion.
Hence, the change in density is:
\[ \boxed{0.91 \%} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Density and Pressure Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions