A block of weight $$100N$$ is suspended by copper and steel wires of same cross sectional area $$0.5cm2$$ and, length $$3m$$ and $$1m$$, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $$30$$º and $$60$$º, respectively. If elongation in copper wire is ΔlC and elongation in steel wire is ΔlS, then the ratio ΔlCΔlS $$is___$$[Young’s modulus for copper and steel are $$1$$×$$1011N/m2$$ and $$2$$×$$1011N/m2$$, respectively.]

Question

A block of weight $$100N$$ is suspended by copper and steel wires of same cross sectional area $$0.5cm2$$ and, length  $$3m$$ and $$1m$$, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $$30$$º and $$60$$º, respectively. If elongation in copper wire is ΔlC and elongation in steel wire is  ΔlS, then the ratio ΔlCΔlS $$is___$$[Young’s modulus for copper and steel are $$1$$×$$1011N/m2$$ and  $$2$$×$$1011N/m2$$, respectively.]

Infinity Learn · Accepted Answer

Horizontal, TScos⁡$$60=TCu$$ $$cos$$⁡$$30$$⇒$$TS2=TCu32$$⇒$$TS=3TCU$$ Vertical, TSsin⁡$$60+TCusin$$⁡$$30=100$$⇒$$3TCu$$×$$32+TCu2=100$$⇒$$2TCu=100$$⇒$$TCu=50N$$∴$$TS=3$$×$$50N$$ Elongation in Cu wire ⇒Δ$$ICu=ICuTCuAYCu$$ Elongation in Steel wire ⇒Δ$$IS=ISTSAYS$$ Ratio $$=$$ΔICΔ$$IS=IcTCuAYCuIsTSAYsteel=YSICuTCuYCuISTS=2$$×$$3$$×$$501$$×$$3$$×$$50=2$$

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!