Physics MCQs for NEET — Practice Questions with Answers

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 \% $

The fractional error in a quantity raised to a power can be found by multiplying the relative error of the base quantity by the exponent. Here, $A = b^4$, so the fractional error in $A$ is $4 imes rac{ riangle b}{b} = 4 rac{ riangle b}{b}$.

$ T = 2 \pi \sqrt { l \over g } $ $ { \triangle T \over T } \times 100 = \left[ { 1 \over 2 } \times { \triangle l \over l } \times { 1 \over 2 } \times { \triangle g \over g } \right] \times 100 $ $ = 4 \% $

Length of two rods = 2l $ = 2 ( 10.15 \pm 0.06 ) m $ $ = ( 20.30 \pm 0.12 ) cm $

The volume of a sphere is given by $V = rac{4}{3} \pi r^3$. The fractional (or relative) error in $V$ due to the measurement of $r$ is $3 rac{ riangle r}{r}$ because the exponent of $r$ is 3, thus multiplying the relative error in $r$ by 3.

To find the relative error in the measurement of kinetic energy (K), we start with the given equation: $ K = { p^2 \\ 2m } $. The relative error in kinetic energy can be found by differentiating this equation. The relative error formula is $ \frac{dK}{K} = 2 \frac{dp}{p} $. Thus, the correct answer is $ 2 \frac{\Delta p}{p} $.

$ Heat energy H = I^2 RT $ $ { \triangle H \over H } \times 100 = \left[ 2 { \triangle I \over I } + { \triangle R \over R } + { \triangle T \over T } \right] \times 100 $ $ = 10 \% $