We know that Young’s modulus γ⇒ΔL=Δ4 LFL=πD24 L4FL=πD2 L4FL According to questions. ΔLCΔLS=FCFS⋅LCLS⋅DS2 LCDC2 LS Here FS=(5 m+2 m)g=7mgFC=5mg,=?,DCDS=P,LCLS=S Putting in eq. (ii) ΔLCΔLS=5mg7mg×q+(P1)2( S1)=5p2 S7q
In the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are p, q and s respectively, then the…
NEET UG 2013 — Physics Mechanics
In the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are p,q and s respectively, then the corresponding ratio of increase in their length would be:
Held on 30 Apr 2013 · Verified 9 Jul 2026.
Options
- A
(7sp2)7q
- B
(5sp2)7q
- C
(5sp)2q
- D
(5sp)7q
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