General Physics Question 18
Question 18
- The edge of a cube is measured using a vernier caliper. (9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is $1 \mathrm{~mm}$ ). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is $2.736 \mathrm{~g}$. Calculate the density in $\mathrm{g} / \mathrm{cm}^{3}$ upto correct significant figures.
(2005, 2M)
Show Answer
Answer:
Correct Answer: 19. $\left(2.66 \mathrm{~g} / \mathrm{cm}^{3}\right)$
Solution:
$$ \begin{aligned} & 1 \mathrm{MSD}=1 \mathrm{~mm} \ & 9 \mathrm{MSD}=10 \mathrm{VSD} \end{aligned} $$
$\therefore \quad$ Least count,
$$ \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=1 \mathrm{~mm}-\frac{9}{10} \mathrm{~mm}=\frac{1}{10} \mathrm{~mm} $$
Measure reading of edge $=$ MSR + VSR $(\mathrm{LC})$
$$ =10+1 \times \frac{1}{10}=10.1 \mathrm{~mm} $$
Volume of cube, $V=(1.01)^{3} \mathrm{~cm}^{3}=1.03 \mathrm{~cm}^{3}$
[After rounding off upto 3 significant digits, as edge length is measured upto 3 significant digits]
$\therefore$ Density of cube $=\frac{2.736}{1.03}=2.6563 \mathrm{~g} / \mathrm{cm}^{3}$
$$ =2.66 \mathrm{~g} / \mathrm{cm}^{3} $$
(After rounding off to 3 significant digits)
