SATHEE: Properties of Matter 1 Question 5

Properties of Matter 1 Question 5

5. A boy’s catapult is made of rubber cord which is $42 c m$ long, with $6 m m$ diameter of cross-section and of negligible mass. The boy keeps a stone weighing $0.02 k g$ on it and stretches the cord by $20 c m$ by applying a constant force. When released the stone flies off with a velocity of $20 m s^{- 1}$ . Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to

(Main 2019, 8 April I)

(a) $10^{6} N m^{- 2}$

(b) $10^{4} N m^{- 2}$

(d) $10^{3} N m^{- 2}$

Show Answer

Answer:

Correct Answer: 5. (a)

Solution:

When rubber cord is stretched, then it stores potential energy and when released, this potential energy is given to the stone as kinetic energy.

So, potential energy of stretched cord

$=$ kinetic energy of stone

$\Rightarrow \frac{1}{2} Y {(\frac{Δ L}{L})}^{2} A \cdot L = \frac{1}{2} m v^{2}$

Here, $Δ L = 20 c m = 0.2 m, L = 42 c m = 0.42 m$ , $v = 20 m s^{- 1}, m = 0.02 k g, d = 6 m m = 6 \times 10^{- 3} m$

$∴ A = π r^{2} = π {(\frac{d}{2})}^{2} = π {(\frac{6 \times 10^{- 3}}{2})}^{2}$

$= π {(3 \times 10^{- 3})}^{2} = 9 π \times 10^{- 6} m^{2}$

On substituting values, we get

$Y = \frac{m v^{2} L}{A (Δ L)^{2}} = \frac{0.02 \times (20)^{2} \times 0.42}{9 π \times 10^{- 6} \times (0.2)^{2}} \approx 3.0 \times 10^{6} N m^{- 2}$

So, the closest value of Young’s modulus is $10^{6} N m^{- 2}$ .

पिछला

अगला

Properties of Matter 1 Question 5

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 5

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

Menu