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NCERT Exemplar · Q22

Q.A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young's modulus for steel is 2×10112 \times 10^{11} Nm−2^{-2}.)

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The cable stretches by 4.6×10−44.6 \times 10^{-4} m (0.46 mm). This is found using Young’s modulus: stress = F/AF/A, strain = ΔL/L\Delta L/L, and Y=stress/strainY = \text{stress}/\text{strain}.

The key idea here is that a material under tension stretches in a way described by Young’s modulus — a measure of stiffness. For a cable, the stretch is tiny because steel is very stiff. The problem gives you all the numbers; you just need to connect them through the definition of Young’s modulus.

Young’s modulus YY relates the tensile stress (force per area) to the tensile strain (fractional change in length). The formula is:

Y=stressstrain=F/AΔL/LY = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}

where FF is the tension, AA is the cross-sectional area, LL is the original length, and ΔL\Delta L is the stretch. Rearranging gives ΔL=FLAY\Delta L = \frac{F L}{A Y}.

So the plan is: compute the cable’s cross-sectional area, then plug everything in.

  1. Find the cross-sectional area. The cable has a radius r=5r = 5 mm = 5×10−35 \times 10^{-3} m. Area of a circle: A=πr2=π(5×10−3)2=π×25×10−6=7.854×10−5A = \pi r^2 = \pi (5 \times 10^{-3})^2 = \pi \times 25 \times 10^{-6} = 7.854 \times 10^{-5} m2^2 (roughly).

  2. List the given values.

    • F=800F = 800 N
    • L=9.1L = 9.1 m
    • Y=2×1011Y = 2 \times 10^{11} N/m2^2
    • A≈7.854×10−5A \approx 7.854 \times 10^{-5} m2^2
  3. Plug into the stretch formula.

ΔL=FLAY=800×9.1(7.854×10−5)×(2×1011)\Delta L = \frac{F L}{A Y} = \frac{800 \times 9.1}{(7.854 \times 10^{-5}) \times (2 \times 10^{11})}

First, numerator: 800×9.1=7280800 \times 9.1 = 7280 N·m.

Denominator: 7.854×10−5×2×1011=15.708×106=1.5708×1077.854 \times 10^{-5} \times 2 \times 10^{11} = 15.708 \times 10^{6} = 1.5708 \times 10^{7} N.

So ΔL=72801.5708×107=4.635×10−4\Delta L = \frac{7280}{1.5708 \times 10^{7}} = 4.635 \times 10^{-4} m. …

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