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A physical parameter a can be determined...

A physical parameter `a` can be determined by measuring the parameters `b, c, d, and e` using the relation `a=b^(alpha)c^(beta)//d^(gamma)e^(delta)`. If the maximum errors in the measurement of `b, c , d, and e are b_(1) %,c_(1)% ,d_(1)% , and e_(1) %`, then the maximum error in the value of `a` determined by the experminent.

A

`(b_1+c_1+d_1e_1)%`

B

`(b_1+c_1-d_1-e_1)%`

C

`(alphab_1+betac_1-gammad_1-deltae_1)%`

D

`(alphab_1+betac_1+gammad_1+deltae_1)%`

Text Solution

Verified by Experts

The correct Answer is:
D

`A=B^alpha(c^beta)/(d^gamma)e^delta`
So maximum error in a is given by `((trianglea)/(a)xx100)_(max)=alpha.(triangleb)/(b)xx100+beta.(trianglec)/©xx100`
`+gamma.(triangled)/(d)xx100+delta.(trianglee)/(e)xx100`
`=(alphab_1+betac_1+gammad_1+deltae_1)%`
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A physical parameter a can be determined by measuring the parameters b,c,d and e using the relation a = b^alpha c^beta //d^gamma e^delta . If the maximum errors in the measurement of b, c, d and e are b_1 % , c_1 % , d_1 % and e_1 % then the maximum error in the value of a determined by the experiment is (b_1+c_1+d_1+e_1)% (b_1+c_1-d_1-e_1)% (alphab_1+betac_1-gammad_1-deltae_1)% (alphab_1+betac_1+gammad_1+deltae_1)%

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