NEET Practice Questions (MCQs) with Answers & Solutions

Astronomical distances are vast and cannot be measured using direct methods like vernier callipers, spherometer, or screw gauge. Instead, indirect methods such as parallax, standard candles, and redshift measurements are used to determine these distances. These methods involve calculations and observations rather than direct measurement tools.

An electron microscope is used to measure dimensions of particles that are smaller than the wavelength of visible light, which is approximately 4000 Ã… (angstroms). Electron microscopes use a beam of electrons, which have much shorter wavelengths than visible light, allowing them to resolve much smaller details and measure extremely small particles.

In an electron microscope, electrons exhibit wave-like behavior. This wave nature of electrons allows the electron microscope to achieve very high resolution, as the wavelength of electrons is much shorter than that of visible light. This principle is based on the wave-particle duality of electrons described by quantum mechanics.

An optical microscope uses light in the visible range of the electromagnetic spectrum to produce magnified images of small objects. Visible light has wavelengths in the range of approximately 400 nm to 700 nm.

$ \theta = 1.8 ^\circ = 0.01 \pi rad $ $ b = 1.27 \times 10 ^ 7 m $ $ D = { b \over \theta } = 4.06 \times 10^8 m $

To find the distance at which 4 AU subtends an angle of exactly 1" of arc, we can use the formula for angular size: \[ \theta = \frac{d}{D} \\] where \( \theta \) is the angular size in radians, \( d \) is the actual size, and \( D \) is the distance. Given \( d = 4 \times 1.496 \times 10^{11} \) m and \( \theta = 4.85 \times 10^{-6} \) rad: \[ D = \frac{4 \times 1.496 \times 10^{11}}{4.85 \times 10^{-6}} = 1.123 \times 10^{17} \text{ m} \\] So, the correct option is \( 1.123 \times 10^{17} \) m.

The percentage error is calculated using the formula: \[ \text{Percentage Error} = \left( \frac{\text{Absolute Error}}{\text{Measured Value}} \right) \times 100 \\] Given the distance is \( 100 \pm 5 \) cm: \[ \text{Percentage Error} = \left( \frac{5}{100} \right) \times 100 = 5\% \\] So, the correct option is \( 5\% \).

$ density ( \rho ) = { mass (m) \over volume (l^3) } $ Percentage error in density $ = \left[ { \triangle M \over M } + 3 \left( { \triangle l \over l } \right) \right] \times 100 $ $ = 1.75 \% $