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NCERT Exemplar · Q11

Q.A rigid rod of length ll and negligible mass hangs horizontally from the ceiling, supported at its two ends by two vertical wires of equal length. The wire at the left end (wire A) is steel and the wire at the right end (wire B) is aluminium. The cross-sectional areas of wires A and B are 1.0 mm21.0\ \text{mm}^2 and 2.0 mm22.0\ \text{mm}^2 respectively, with YAl=70×109 N m−2Y_{Al}=70\times10^{9}\ \text{N m}^{-2} and Ysteel=200×109 N m−2Y_{steel}=200\times10^{9}\ \text{N m}^{-2}. A mass mm is to be hung from the rod at some point along its length. (More than one option may be correct.)

(a) Mass mm should be suspended close to wire A to have equal stresses in both the wires.
(b) Mass mm should be suspended close to B to have equal stresses in both the wires.
(c) Mass mm should be suspended at the middle of the wires to have equal stresses in both the wires.
(d) Mass mm should be suspended close to wire A to have equal strain in both wires.
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Where the mass hangs sets how the load mgmg splits between the two wires (by the lever/torque rule). Requiring the two wires to have equal stress puts the mass nearer the aluminium wire B; requiring equal strain puts it nearer the steel wire A. So statements (B) and (D) are both correct.

Setup

Let the mass hang a distance yy from end A (steel), so a distance l−yl-y from end B (aluminium). The two wire tensions TAT_A and TBT_B support the load:

TA+TB=mg.T_A+T_B=mg.

Taking torques about end A (only TBT_B at distance ll and the weight at distance yy contribute):

TB l=mg y⇒TBmg=yl.(1)T_B\,l = mg\,y \quad\Rightarrow\quad \frac{T_B}{mg}=\frac{y}{l}. \qquad(1)

Areas: aA=1.0 mm2a_A=1.0\ \text{mm}^2 (steel), aB=2.0 mm2a_B=2.0\ \text{mm}^2 (Al).

Case 1 — equal stress

Stress =T/a=T/a. Equal stress means

TAaA=TBaB ⇒ TA1.0=TB2.0 ⇒ TB=2TA.\frac{T_A}{a_A}=\frac{T_B}{a_B}\ \Rightarrow\ \frac{T_A}{1.0}=\frac{T_B}{2.0}\ \Rightarrow\ T_B=2T_A.

With TA+TB=mgT_A+T_B=mg:   TA=mg3, TB=2mg3\;T_A=\tfrac{mg}{3},\ T_B=\tfrac{2mg}{3}.

From (1): yl=TBmg=23\dfrac{y}{l}=\dfrac{T_B}{mg}=\dfrac{2}{3}, so y=2l3y=\tfrac{2l}{3} from A — i.e. only l/3l/3 from B: the mass is closer to wire B. This confirms statement (B) (and rules out A and C).

Case 2 — equal strain

Strain =stressY=TaY=\dfrac{\text{stress}}{Y}=\dfrac{T}{aY}. Equal strain means

TAaAYsteel=TBaBYAl ⇒ TA1.0×200=TB2.0×70.\frac{T_A}{a_A Y_{steel}}=\frac{T_B}{a_B Y_{Al}}\ \Rightarrow\ \frac{T_A}{1.0\times200}=\frac{T_B}{2.0\times70}. …

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