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Question

(B) The formula for Young's Modulus is $Y = \frac{MgL^3}{4bd^3\delta}$.The fractional error is given by: $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \f

Vedclass · Accepted Answer

$0.0155$

Vedclass · Answer

(B) The formula for Young's Modulus is $Y = \frac{MgL^3}{4bd^3\delta}$.The fractional error is given by: $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{3\Delta L}{L} + \frac{\Delta b}{b} + \frac{3\Delta d}{d} + \frac{\Delta \delta}{\delta}$.Given values:$M = 2 \, kg = 2000 \, g$, $\Delta M = 1 \, g$$L = 1 \, m = 1000 \, mm$, $\Delta L = 1 \, mm$$b = 4 \, cm = 40 \, mm$, $\Delta b = 0.1 \, mm$$d = 0.4 \, cm = 4 \, mm$, $\Delta d = 0.01 \, mm$$\delta = 5 \, mm$, $\Delta \delta = 0.01 \, mm$Substituting these values:$\frac{\Delta Y}{Y} = \frac{1}{2000} + 3 	imes \frac{1}{1000} + \frac{0.1}{40} + 3 	imes \frac{0.01}{4} + \frac{0.01}{5}$$\frac{\Delta Y}{Y} = 0.0005 + 0.003 + 0.0025 + 0.0075 + 0.002 = 0.0155$.

Physical Quantity	Least count and Observed value
Mass $(M)$	$1 \, g$ and $2 \, kg$
Length of bar $(L)$	$1 \, mm$ and $1 \, m$
Breadth of bar $(b)$	$0.1 \, mm$ and $4 \, cm$
Thickness of bar $(d)$	$0.01 \, mm$ and $0.4 \, cm$
Depression $(\delta)$	$0.01 \, mm$ and $5 \, mm$

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