AP EAPCET 07-Jul-2022 Shift 2 Paper

As, P=

a1∕2⋅b1∕2⋅dα

c1∕2

So % error in P=

∆P

P

×100
=

1

2

(

∆a

a

)×100+

1

2

(

∆b

b

×100)
+α(

∆d

d

×100)+

1

2

(

∆c

c

×100)
⇒% error in P=

1

2

×% error in a+

1

2

×% error in
b+

1

2

×% error in c+α×% error in d ...(i)
Now given,
% Error in P=2%
% Error in a=% error in b
=% Error in c=% error in d=0.5%
Substituting these values in Eq.(i), we get
2=

1

2

×0.5+

1

2

×0.5+

1

2

×0.5+α×0.5
⇒2=

1

2

(

1

2

+

1

2

+

1

2

+α)⇒4=

3

2

+α
or α=4−

3

2

=

5

2

∴α is

5

2