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Exam
NEET
Subject
Physics
Chapter
Thermal Properties of Matter

NEET · Physics · Thermal Properties of Matter

A temperature-compensated pendulum is designed such that its...

A temperature-compensated pendulum is designed such that its effective length is the difference between the length of a central steel rod and the effective length of parallel brass rods. To ensure the time period remains constant at all temperatures, this effective length must not change with temperature. If the required constant effective length is 30 cm30 \text{ cm}, what should be the actual length of the central steel rod?

(Given: Coefficient of linear expansion of steel αs=1.2×10−5 K−1\alpha_s = 1.2 \times 10^{-5} \text{ K}^{-1} and of brass αb=1.8×10−5 K−1\alpha_b = 1.8 \times 10^{-5} \text{ K}^{-1})

Options

  1. A90 cm90 \text{ cm}
  2. B60 cm60 \text{ cm}
  3. C45 cm45 \text{ cm}
  4. D18 cm18 \text{ cm}

Correct answer

A. 90 cm90 \text{ cm}

Step-by-step solution

For the effective length (Ls−LbL_s - L_b) to remain constant at all temperatures, the change in length of the steel rod must equal the change in length of the brass rods.

ΔLs=ΔLb\Delta L_s = \Delta L_b

LsαsΔT=LbαbΔTL_s \alpha_s \Delta T = L_b \alpha_b \Delta T

Lsαs=LbαbL_s \alpha_s = L_b \alpha_b

Lb=Ls(αsαb)=Ls(1.2×10−51.8×10−5)=23LsL_b = L_s \left(\dfrac{\alpha_s}{\alpha_b}\right) = L_s \left(\dfrac{1.2 \times 10^{-5}}{1.8 \times 10^{-5}}\right) = \dfrac{2}{3} L_s

Given the effective length is 30 cm30 \text{ cm}:

Ls−Lb=30 cmL_s - L_b = 30 \text{ cm}

Ls−23Ls=30 cmL_s - \dfrac{2}{3} L_s = 30 \text{ cm}

Ls3=30 cm\dfrac{L_s}{3} = 30 \text{ cm}

Ls=90 cmL_s = 90 \text{ cm}

Answer: 90 cm90 \text{ cm}
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