Physics, asked by jurimahanta, 7 days ago

If P= c^3b^2/cd andpercentage error in c,b,d are1% 2% and 3% respectively , calculate the % error in P

Answers

Answered by nivedithasuneesh5
0

Explanation:

Answer:

The percentage error in P is 13%.

Explanation:

Given that,

P=a^3b^2(\sqrt{c}d)P=a

3

b

2

(

c

d)

a = 1%

b= 3%

c=4%

d=2%

The percentage error in P

\dfrac{\Delta P}{P}\times100=[3\times\dfrac{\Delta a}{a}+2\times\dfrac{\Delta b}{b}+\dfrac{1}{2}\dfrac{\Delta C}{C}+\dfrac{\Delta d}{d}]\times100

P

ΔP

×100=[3×

a

Δa

+2×

b

Δb

+

2

1

C

ΔC

+

d

Δd

]×100

\dfrac{\Delta P}{P}\times100=[3\times\dfrac{1}{100}+2\times\dfrac{3}{100}+\dfrac{1}{2}\times\dfrac{4}{100}+\dfrac{2}{100}]\toimes100

P

ΔP

×100=[3×

100

1

+2×

100

3

+

2

1

×

100

4

+

100

2

]\toimes100

\dfrac{\Delta P}{P}\times100=[\dfrac{3+6+2+2}{100}]\times100

P

ΔP

×100=[

100

3+6+2+2

]×100

\dfrac{\Delta P}{P}\times100=[\dfrac{13}{100}]\times100

P

ΔP

×100=[

100

13

]×100

\dfrac{\Delta P}{P}\times100=13\%

P

ΔP

×100=13%

Hence, The percentage error in P is 13%.

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