Practice portal › Mechanical Properties of Solids › Stress, Strain and Hooke's Law

As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45° with the load axis. The length of the wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be x×10⁴ N m⁻². The value of x is ______.

Asked in JEE Main 25th Jan 1st Shift 2023 · Experimental determination of Y

Figure: Experimental determination of Y
Answer: 5

Step-by-step solution

Given: L=62.8 cm=0.628 m and d=4 mm, so r=2×10⁻³ m.

A 45° line through the origin on an extension-versus-load plot means (Δ l)/W=tan 45°=1 m N⁻¹ in SI units.

A=π r²=π(2×10⁻³)²=1.257×10⁻⁵ m².

Y=L/A·W/(Δ l)=(0.628)/(1.257×10⁻⁵×1)=5.0×10⁴ N m⁻².

Hence x=5.

More Stress, Strain and Hooke's Law questionsAll Stress, Strain and Hooke's Law questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer