SATHEE: Properties of Matter 1 Question 1

Properties of Matter 1 Question 1

1. In an experiment, brass and steel wires of length $1 m$ each with areas of cross-section $1 m m^{2}$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produce a net elongation of $0.2 m m$ is

[Take, the Young’s modulus for steel and brass are respectively $120 \times 10^{9} N / m^{2}$ and $60 \times 10^{9} N / m^{2}$ ]

(2019 Main, 10 April II)

(a) $1.2 \times 10^{6} N / m^{2}$

(b) $0.2 \times 10^{6} N / m^{2}$

(d) $4.0 \times 10^{6} N / m^{2}$

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Answer:

Correct Answer: 1. (*)

Solution:

In given experiment, a composite wire is stretched by a force $F$ .

Net elongation in the wire $=$ elongation in brass wire + elongation in steel wire

Now, Young’s modulus of a wire of cross-section $(A)$ when some force $(F)$ is applied, $Y = \frac{F l}{A Δ l}$

We have,

$Δ l = elongation = \frac{F l}{A Y}$

So, from relation (i), we have

As wires are connected in series and they are of same area of cross-section, length and subjected to same force, so

$Δ l_{net} = \frac{F}{A} (\frac{l}{Y_{brass}} + \frac{l}{Y_{steel}})$

Here,

$\begin{aligned} Δ l_{net} = 0.2 m m = 0.2 \times 10^{- 3} m \\ and \begin{matrix} l = 1 m \\ Y_{brass} = 60 \times 10^{9} N m^{- 2}, Y_{steel} = 120 \times 10^{9} N m^{- 2} \end{matrix} \end{aligned}$

On putting the values, we have

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Properties of Matter 1 Question 1

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Properties of Matter 1 Question 1

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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