Question 17: Properties of Matter

Question

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $P 	imes 10^{11} \, 	ext{Nm}^{-2}$, where the value of $P$ is: (Take $g = 3\pi \, 	ext{m/s}^2$)

Accepted Answer

LetÃ¢ÂÂs break down the solution step by step:
First, note the given values:
Length of the wire, $$L = 3 \, 	ext{m}$$
Radius of the wire, $$r = 3 \, 	ext{mm} = 3 	imes 10^{-3} \, 	ext{m}$$
Extension of the wire, $$\Delta L = 0.1 \, 	ext{mm} = 1 	imes 10^{-4} \, 	ext{m}$$
Mass attached, $$m = 50 \, 	ext{kg}$$
Acceleration due to gravity, $$g = 3\pi \, 	ext{m/s}^2$$
The force (weight) acting on the wire is:
$$F = mg = 50 	imes 3\pi = 150\pi \, 	ext{N}$$
Next, calculate the cross-sectional area, $$A,$$ of the wire:
$$A = \pi r^2 = \pi (3 	imes 10^{-3})^2 = \pi 	imes 9 	imes 10^{-6} = 9\pi 	imes 10^{-6} \, 	ext{m}^2$$
YoungÃ¢ÂÂs modulus, $$E,$$ is given by:
$$E = \frac{FL}{A \Delta L}$$
Substitute the values into the formula:
$$    E = \frac{(150\pi) 	imes 3}{(9\pi 	imes 10^{-6}) 	imes (1 	imes 10^{-4})}    $$
Simplify step by step:
Multiply in the numerator:
$$150\pi 	imes 3 = 450\pi$$
Multiply in the denominator:
$$9\pi 	imes 10^{-6} 	imes 1 	imes 10^{-4} = 9\pi 	imes 10^{-10}$$
Cancel the common factor $$\pi$$ in the numerator and denominator:
$$    E = \frac{450}{9 	imes 10^{-10}}    $$
Divide 450 by 9:
$$    E = \frac{450}{9} 	imes \frac{1}{10^{-10}} = 50 	imes 10^{10} = 5 	imes 10^{11} \, 	ext{Nm}^{-2}    $$
The problem states that Young's modulus can be expressed as $$P 	imes 10^{11} \, 	ext{Nm}^{-2}.$$ Here we have:
$$    P = 5    $$
Thus, the correct option is:
Option D (5).

Answer

2.5

Answer

25

Answer

10

Answer

5

Properties of Matter

Choose the correct answer: