A thin ring of an alloy of inner diameter 20 cm and cross-sectional area 1 mm 2, is heated until it easily slides on a rigid cylinder of diameter 20.025 cm. When the ring cools down, the value of tension in it is found to be a × 102 N. Find the value of a. (For alloy: .α=10-5 / ° C , Y=400 GPa )

Question

A thin ring of an alloy of inner diameter 20 cm and cross-sectional area 1 mm 2, is heated until it easily slides on a rigid cylinder of diameter 20.025 cm. When the ring cools down, the value of tension in it is found to be a × 102 N. Find the value of a. (For alloy: .α=10-5 / ° C , Y=400 GPa ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(T / A/d r / r)=Y T=A Y (d r/r)=1 × 10-6 × 400 × 109 × (0.025/20)=500 N

Q. A thin ring of an alloy of inner diameter 20cm and cross-sectional area 1mm2, is heated until it easily slides on a rigid cylinder of diameter 20.025cm. When the ring cools down, the value of tension in it is found to be a×102N. Find the value of a. (For alloy: α=10−5/∘C,Y=400GPa)

dr/rT/A​=Y T=AYrdr​=1×10−6×400×109×200.025​=500N

$\frac{T / A}{d r / r} = Y$
$T = A Y \frac{d r}{r} = 1 \times 1 0^{- 6} \times 400 \times 1 0^{9} \times \frac{0.025}{20} = 500 N$