Physics mock test jmpp

Question

In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

(1) 0.14% 

(2) 0.9% 

(3) 9% 

(4) 1.4%

Answer 1

The correct option is (4) 1.4%

Young's Modulus (Y) is defined as:

\(Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta L/L}\)

where:

- F is the force applied,

- A is the cross-sectional area,

- ΔL is the change in length (elongation),

- L is the original length

The area A of the wire can be calculated using the formula for the area of a circle:

A = πr²

Given that the radius r = 0.2 cm = 0.002 m:

A = π(0.002)² = π × 4 × 10⁻⁶ m²

- Radius r = 0.2 cm with least count δr = 0.001 cm = 0.00001 m

- Length L = 1 m with least count δL = 1 mm = 0.001 m

- Mass m = 1 kg with least count δm = 1g = 0.001 kg

- Elongation ΔL = 0.5 cm = 0.005 m with least count δΔL = 0.001 cm = 0.00001 m

The fractional error in Young's Modulus can be expressed as:

Fractional error due to mass:

\(\frac{\delta m}{m} = \frac{0.001\, kg}{1\, kg} = 0.001\)

Fractional error due to radius:

\(\frac{\delta r}{r} = \frac{0.00001\, m}{0.002\, m} = 0.005\)

Therefore. \(2 \frac{\delta r}{r} = 2 \times 0.005 = 0.01\)

Fractional error due to elongation:

\(\frac{\delta L}{L} = \frac{0.001\, m}{1\, m} = 0.001\)

Now, we can sum up all the fractional errors:

\(\frac{\delta Y}{Y} = 0.001+0.01+0.002+0.001 = 0.014\)

Percentage Error = \(\frac{\delta Y}{Y} \times 100 = 0.014 \times 100 = 1.4\%\)