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If $ A = b^4 $ the fractional erroe in A is ......

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Explanation

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 $4 imes rac{ riangle b}{b} = 4 rac{ riangle b}{b}$.

In the experiment of simple pendulum error in length of pendulum (l) is $5 \%$ and that of g is $3 \%$ then find percentage error in measurement of periodic time for pendulum

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Explanation

$ 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 \% $

Acceleration due to gravity is given by $ g = { GM \over R^2 } $ what is the equation of the fractional error $ \triangle g /g $ in measurement of gravity g ? [G & M constant]

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The length of a rod is $ ( 10.15 \pm 0.06 ) cm $ what is the length of two such rods ?

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Explanation

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

For a sphere having volume is given by $ V = { 4 \over 3 } \ pi r^3 $ What is the equation of the relative error $ { \triangle V \over V } $ in measurement of the volume V ?

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Explanation

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.

Kinetic energy K and linear momentum P are related as $ K = { p^2 \over 2m } $ What is the equation of the relative erroe $ { \triangle k \over k } $ in measurement of the K ? (mass in constant)

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Explanation

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 produced in a current carrying conducting wire is $H = I^2Rt$ it percentage error in I, R and t is $2 \% , 4 \% and 2 \%$ respectively then total percentage error in measurement of heat energy ...............

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Explanation

$ 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 \% $

The resistance of two resistance wires are $R_1 = (100 \pm 5) \Omega $ and $ R_ 2 = (200 \pm 7) \Omega $ are connected in series. find the maximum absolute error in the equivalent resistance of the combination.

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The periodic time of simple pendulum is $ T = 2 \pi \sqrt { l \over g} $ relative error in the measurement of T and l are $ \pm a and \pm b $ respectively find relative error in the measurement of g

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Explanation

$ g = 4 \pi ^2 { l \over T^2 } $ $ { \triangle g \over g} = { \triangle l \over l } + 2 \times { \triangle T \over T } $ $ = b + 2 q $

A physical quantity x is given by $ x = { A^4 B^{1 \over 4 } \over C^3 D^{4 \over 3 }} $ due to which physical quantity produced the maximum percentage error in x

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