Question

║⊕QUESTION⊕║
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CLASS 11
UNITS AND MEASUREMENTS
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The density P of a piece of metal of a mass m and volume V is given by the formula P = m/v; If m = 375.32 ± 0.01 g, and V = 136.41 ± 0.01 cm³. Find % error in P

Answer 1

Explanation:

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The density P of a piece of metal of a mass m and volume V is given by the formula P = m/v; If m = 375.32 ± 0.01 g, and V = 136.41 ± 0.01 cm³. Find % error in P

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We know,

$P = \frac{m}{v}$

So, percentage error in density will be=>

$\implies \frac{ \Delta P}{P} = ( \frac{ \Delta m}{m} + \frac{ \Delta v}{v} ) \times 100 \\ \implies \frac{ \Delta P}{P} =( \frac{0.01}{375.32} + \frac{0.01}{136.41} ) \times 100 \\ \implies \frac{ \Delta P}{P} = ( \frac{1}{37532} + \frac{1}{13641} ) \times 100 \\ \implies \frac{ \Delta P}{P} = ( \frac{13641 + 37532}{ {(37532)}(13641)} ) \times {100} \\ \implies \frac{ \Delta P}{P} = \frac{ \cancel{51173}}{ \cancel{511974012}} \times 100 \\ \implies \huge \boxed{ \frac{ \Delta P}{P} =0.01 }$

• For error,

1. If there is formula like this a = b. c we use ∆a/a = ∆b/b + ∆c/c
2. If there is formula like this a = b/c then we use ∆a/a= ∆b/b +∆c/c This is because we want maximum error...
3. For percentage error just refer (1) and (2) and then multiply it by 100

Answer 2

SOLUTION.

As you got your solution but this solution miss a point which is very important and hence after this point you do not need to learn all the formula of calculating % error.

Let's start.....

Density = Mass/Volume

P = M / V

Take log on both side.

Log(P) = Log(M) - Log(V)

Differentiate it and ignore all the negative sign.

∆P/P = ∆M/M + ∆V/V

Multiply by 100

% error =[(0.01/375.32) + (0.01/136.41)]×100

% error = [0.00002 + 0.00007] × 100

% error = 0.00009 × 100

% error = 0.009

% error = 0.01 % (Approx)

As I want you to use this method this will reduce your effort.

DIFFERENTIALS USED....

Differential of Log(P) = ∆P/P

Differential of Log(M) = ∆M/M

Differential of Log(V) = ∆V/V

Please ignore all the negative sign . While solving error question

#answerwithquality

#BAL