2 Marks Questions

Question 12 Marks

The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5mm, calculate the minimum inaccuracy in the measurement of distance.

Answer

Here, parts on Vernier scale = n = 50 parts No. of division of M.S. coinciding with n parts of V.S. = (n - 1)$\therefore$ L.C. of instrument $=\frac{\text{L.C. of Mainscale}}{\text{No. of parts on V.S.}}=\frac{0.5\text{mm}}{50}$
Or minimum inaccuracy = 0.01mm.

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Question 22 Marks

Moon is seen to be of $\Big(\frac{1}{2}\Big)^\circ$ diameter from the earth. What must be the relative size compared to the earth?

Answer

According to the problem, moon is seen as $\Big(\frac{1}{2}\Big)^\circ$ diameter from earth and earth is seen as $2^\circ$ diameter from moon.
As $\theta$ is proportional to diameter,
Hence, $\frac{\text{Diameter of earth}}{\text{Diameter of moon}}=\frac{2}{\Big(\frac{1}{2}\Big)}=4$

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Question 32 Marks

Calculate the solid angle subtended by the periphery of an area of $1cm^2$ at a point situated symmetrically at a distance of 5cm from the area.

Answer

Solid angle, $\Omega=\frac{\text{Area}}{(\text{Radial distance})^2}$$=\frac{1\text{cm}^2}{(5\text{cm})^2}=\frac{1}{25}=4\times10^{-2}\ \text{steradian}$
($\because$ Area = $1cm^2$, distance = $5cm$)

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Question 42 Marks

Name the device used for measuring the mass of atoms and molecules.

Answer

Deflection of a charge particle or ionized atom or molecule depends on the magnitude of either magnetic or electric field. Mass and Charge of a particle by using this principle can be measured by using spectrograph or spectrometer which measures the mass of atoms and molecules.

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Question 52 Marks

The distance of a galaxy is of the order of $10^{25}m$. Calculate the order of magnitude of time taken by light to reach us from the galaxy.

Answer

According to the problem, distance of the galaxy $= 10^{25}m$. Speed of light $= 3 \times 10^8m/s$ Hence, time taken by light to reach us from galaxy is,$\text{t}=\frac{\text{Distance}}{\text{Speed}}=\frac{10^{25}\text{m}}{3\times10^8\text{m/s}}=\frac{1}{3}\times10^{17}$
$=\frac{10}{3}\times10^{16}=3.33\times10^{16}\text{s}$

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