निम्नांकित चार तार एक ही पदार्थ से बने हैं। य | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Young's modulus,

$Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$

Where $F$

Vedclass · Accepted Answer

तार की लम्बाई $ =50\, cm$, तार का व्यास $=0.5\, mm$

Vedclass · Answer

Young's modulus,

$Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$

Where $F$ is the force applied, $L$ is the length, $D$ is the diameter and $\Delta L$ is the extension of the wire respectively. 

As each wire is made up of same material therefore their $Young's$ modulus is same for each wire.

Foe all the four wires, $Y,F\,(=tension)$ are the same.

$	herefore \Delta L \propto \frac{L}{{{D^2}}}$

$In\,\,\left( a ight)\,\,\frac{L}{{{D^2}}} = \frac{{200\,cm}}{{{{\left( {0.2\,cm} ight)}^2}}} = 5 	imes {10^3}\,c{m^{ - 1}}$

$In\,\,\left( b ight)\,\,\,\frac{L}{{{D^2}}} = \frac{{300\,cm}}{{{{\left( {0.3\,cm} ight)}^2}}} = 3.3 	imes {10^3}\,c{m^{ - 1}}$

$In\,\,\left( c ight)\,\,\frac{L}{{{D^2}}} = \frac{{50\,cm}}{{{{\left( {0.05\,cm} ight)}^2}}} = 20 	imes {10^3}\,c{m^{ - 1}}$

$In\,\,\left( d ight)\,\,\frac{L}{{{D^2}}} = \frac{{100\,cm}}{{{{\left( {0.1\,cm} ight)}^2}}} = 10 	imes {10^3}\,\,c{m^{ - 1}}$

निम्नांकित चार तार एक ही पदार्थ से बने हैं। यदि सभी पर समान तनाव लगाया जाय तो, किसमे सबसे अधिक प्रसार होगा ?

Similar Questions