To determine the least count of the travelling microscope, we need to understand the concept of least count and how it applies in this context.
The least count (LC) of an instrument is the smallest measurement that can be taken accurately with the instrument. For a travelling microscope, the least count can be calculated using the formula:
\(LC = \frac{\text{Value of 1 main scale division}}{\text{Number of vernier scale divisions}}\)
\(\text{Value of 1 main scale division} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}\)
The total length covered by 25 divisions on the vernier scale is equivalent to 24 divisions on the main scale. Therefore, one division on the vernier scale is:
\(1 \text{ vernier division} = \frac{24 \text{ main scale divisions}}{25}\\)
\(LC = \frac{0.05 \text{ cm}}{25} = 0.002 \text{ cm}\)
Therefore, the least count of the travelling microscope is 0.002 cm.
The correct option is 0.002.
To determine the least count of the travelling microscope, we need to understand the relationship between the main scale and the vernier scale.
\[ \text{One division on main scale} = \frac{15 \, \text{cm}}{300} = 0.05 \, \text{cm} \]
\[ 24 \times 0.05 \, \text{cm} = 1.2 \, \text{cm} \]
\[ \text{One division on vernier scale} = \frac{1.2 \, \text{cm}}{25} = 0.048 \, \text{cm} \]
\[ \text{LC} = \text{Main scale division} - \text{Vernier scale division} = 0.05 \, \text{cm} - 0.048 \, \text{cm} = 0.002 \, \text{cm} \]
Therefore, the least count of the travelling microscope is 0.002 cm, making the correct option:
0.002
This approach allows us to calculate the precision of measurements that can be made using the travelling microscope.
Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.

Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: