Question

In an experiment, the following observation's
were recorded : L = 2.820 m, M = 3.00 kg, 1 =
0.087 cm, Diameter D = 0.041 cm Taking g =
9.81 m/s using the formula, Y= *mg, the
maximum permissible error in Y is
(b) 4.56%
(a) 7.96%
(d) 8.42%
(c) 6.50%
​

Answer 1

Answer:

Here we used the formula:

Y=4MgL/πD^2l

L= 2.820 m M= 3.00 kg, l = 0.087 cm , Diameter D = 0.041 cm

g = 9.81 m/s

Now

so maximum permissible error in Y =

ΔY/Y ×100 = (ΔM/M+Δg/g+ΔL/L+2ΔD/D+Δl/l)×100

                  =(1/300+1/981+1/2820+2×1/41+1/87)×100

                  =0.065×100

                    =6.5

Therefore option (c) i.e 6.50% is the correct answer.

Answer 2

Answer:

Y=4MgL/πD^2l

L= 2.820 m M= 3.00 kg, l = 0.087 cm , Diameter D = 0.041 cm

g = 9.81 m/s

Now

so maximum permissible error in Y =

ΔY/Y ×100 = (ΔM/M+Δg/g+ΔL/L+2ΔD/D+Δl/l)×100

                 =(1/300+1/981+1/2820+2×1/41+1/87)×100

                 =0.065×100

                   =6.5

Therefore option (c) i.e 6.50% is the correct answer.