A student determined Young's modulus of elasticity using the formula = The value of g is taken to be without any significant error, his observation is as following.

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Q.
A student determined Young's modulus of elasticity using the formula $Y = \frac{M g L^{3}}{4 b d^{3} δ} .$ The value of g is taken to be $9.8 m / s^{2},$ without any significant error, his observation is as following. [2021]

Physical quantity Least count of the equipment used for measurement Observed value

Mass (M) 1 g 2 kg

Length of bar (L) 1 mm 1 m

Breadth of bar (b) 0.1 mm 4 cm

Thickness of bar (d) 0.01 mm 0.4 cm

Depression ( $δ$ ) 0.01 mm 5 mm

Then the fractional error in the measurement of Y is

1 0.155

2 0.0083

3 0.083

4 0.0155

Ans.
(4)

Young's modulus of elasticity,

$Y = \frac{M g L^{3}}{4 b d^{3} δ}$

$\frac{∆ Y}{Y} = \frac{∆ M}{M} + \frac{3 ∆ L}{L} + \frac{∆ b}{b} + \frac{3 ∆ d}{d} + \frac{∆ δ}{δ}$

$\frac{∆ Y}{Y} = \frac{10^{- 3}}{2} + \frac{3 \times 10^{- 3}}{1} + \frac{10^{- 2}}{4} + \frac{3 \times 10^{- 2}}{4} + \frac{10^{- 2}}{5} = 0.0155$

Solution

1 0.155

2 0.0083

3 0.083

4 0.0155

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Physical quantity	Least count of the equipment used for measurement	Observed value
Mass (M)	1 g	2 kg
Length of bar (L)	1 mm	1 m
Breadth of bar (b)	0.1 mm	4 cm
Thickness of bar (d)	0.01 mm	0.4 cm
Depression ( $δ$ )	0.01 mm	5 mm

Solution

1 0.155

2 0.0083

3 0.083

4 0.0155

Ans. (4) Young's modulus of elasticity, Y=MgL34bd3δ ∆YY=∆MM+3∆LL+∆bb+3∆dd+∆δδ ∆YY=10-32+3×10-31+10-24+3×10-24+10-25=0.0155

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