Calculate the maximum possible error in Y in the following relation- Y - 4MgL-D2lWhere mass M - 1000 g- original length L - 200-00 cm- diameter D - 0-075 cm- extension l - 0-325 cm-

Y = 4MgLπD^2l Δ YY = Δ LL + 2D/D + Δ ll 0.1200 + 2 ×0.0010.075 + 0.001.325 Δ YY = 0.03024 Δ YY% #160; = 0.03024 × 100 = 3.024% ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Area of a Rectangle

Calculate the...

Question

Calculate the maximum possible error in $Y$ in the following relation:
$Y = \frac{4 M g L}{π D^{2} l}$
Where mass $M = 1000 g :$ original length $L = 200.00 c m;$ diameter $D = 0.075 c m;$ extension $l = 0.325 c m .$

Open in App

Solution

$Y = \frac{4 M g L}{π D^{2} l}$
$\frac{Δ Y}{Y} = \frac{Δ L}{L} + \frac{2 D}{D} + \frac{Δ l}{l}$
$\frac{0.1}{200} + 2 \times \frac{0.001}{0.075} + \frac{0.001}{.325}$
$\frac{Δ Y}{Y} = 0.03024$
$\frac{Δ Y}{Y} % = 0.03024 \times 100$
$= 3.024 %$

Suggest Corrections

Similar questions

Q. In an experiment, the following observations were recorded:

L = 2.820 m, M = 3.00 k g, l = 0.087 c m

, diameter

D = 0.041 c m

. Taking

g = 9.81 m s^{- 2}

and using the formula, Y =

\frac{4 M g}{π D^{2} l}

, find the maximum permissible error in Y.

Q. In Searle's experiment to find Young's modulus the diameter of wire is measured as

d = 0.05 c m

, length of wire is

l = 125 c m

and when a weight,

m = 20.0 k g

is put, extension in wire was found to be

0.100 c m

. Find the maximum permissible error in Young's modulus

(Y)

. Use:

Y = \frac{m g l}{(π / 4) d^{2} x}

Q. In Searle's experiment to find Young's modulus, the diameter of wire is measured as

D = 0.05 c m

, the length of wire is

L = 125 c m,

and when a weight,

m = 20 k g

is the put, extension in wire was found to be

0.100 c m

. Find the maximum permissible error in Young's modulus (Y).

Q. In Searle's exp to find Young's modulus, the diameter of wire is measured as

D = 0.05 c m,

length of wire is

L = 125 c m

, and when a weight,

m = 20.0 k g

is put, extension in the wire was found to be

0.100 c m .

Find a maximum permissible error in young's modulus (Y).

In the determination of Young’s modulus $(Y = \frac{4 M L g}{π l d^{2}})$ by using Searle’s method, a wire of length L=2m and diameter d=0.5mm is used. For a load M=2.5kg, an extension l=0.25mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement is:

Join BYJU'S Learning Program

Calculate the maximum possible error in Y in the following relation:Y=4MgLπD2lWhere mass M=1000g: original length L=200.00cm; diameter D=0.075cm; extension l=0.325cm.

Y=4MgLπD2lΔYY=ΔLL+2DD+Δll0.1200+2×0.0010.075+0.001.325ΔYY=0.03024ΔYY%=0.03024×100=3.024%

Calculate the maximum possible error in $Y$ in the following relation:
$Y = \frac{4 M g L}{π D^{2} l}$
Where mass $M = 1000 g :$ original length $L = 200.00 c m;$ diameter $D = 0.075 c m;$ extension $l = 0.325 c m .$

$Y = \frac{4 M g L}{π D^{2} l}$
$\frac{Δ Y}{Y} = \frac{Δ L}{L} + \frac{2 D}{D} + \frac{Δ l}{l}$
$\frac{0.1}{200} + 2 \times \frac{0.001}{0.075} + \frac{0.001}{.325}$
$\frac{Δ Y}{Y} = 0.03024$
$\frac{Δ Y}{Y} % = 0.03024 \times 100$
$= 3.024 %$