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Properties of Matter question

2020 · 8 Jan · Shift 1 · Q54
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Properties of Matter question

2020 · 8 Jan · Shift 1 · Q54

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A leak proof cylinder of length 1m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C, relative to the density at T = 0°C is close to :
  1. A
    1.04
  2. B
    1.26
  3. C
    1.01
  4. D
    1.03
View written solutionFree

Correct answer: C

  1. Use floating condition

For a floating body,

weight of cylinder=weight of displaced water.\text{weight of cylinder} = \text{weight of displaced water}.weight of cylinder=weight of displaced water.

Since the cylinder is leak proof and its own expansion is negligible, its weight and total volume remain effectively constant.

So,

ρ0V0=ρ4V4\rho_0 V_0 = \rho_4 V_4ρ0​V0​=ρ4​V4​

where:

  • ρ0\rho_0ρ0​ = density of water at 0∘C0^\circ\text{C}0∘C
  • ρ4\rho_4ρ4​ = density of water at 4∘C4^\circ\text{C}4∘C
  • V0,V4V_0, V_4V0​,V4​ are submerged volumes at the two temperatures.

Because the cylinder has uniform cross-sectional area AAA,

V=A×(submerged length).V = A \times (\text{submerged length}).V=A×(submerged length).
  1. Find submerged lengths

Total length of cylinder = 1 m1\,\text{m}1m.

  • At 0∘C0^\circ\text{C}0∘C, height above water = 20 cm=0.20 m20\,\text{cm} = 0.20\,\text{m}20cm=0.20m

So submerged length is

L0=1−0.20=0.80 mL_0 = 1 - 0.20 = 0.80\,\text{m}L0​=1−0.20=0.80m
  • At 4∘C4^\circ\text{C}4∘C, height above water = 21 cm=0.21 m21\,\text{cm} = 0.21\,\text{m}21cm=0.21m

So submerged length is

L4=1−0.21=0.79 mL_4 = 1 - 0.21 = 0.79\,\text{m}L4​=1−0.21=0.79m

Thus,

V0=A(0.80),V4=A(0.79)V_0 = A(0.80), \qquad V_4 = A(0.79)V0​=A(0.80),V4​=A(0.79)
  1. Apply equilibrium condition
ρ0A(0.80)=ρ4A(0.79)\rho_0 A(0.80) = \rho_4 A(0.79)ρ0​A(0.80)=ρ4​A(0.79)

Cancel AAA:

ρ0(0.80)=ρ4(0.79)\rho_0(0.80) = \rho_4(0.79)ρ0​(0.80)=ρ4​(0.79)

Hence,

ρ4ρ0=0.800.79\frac{\rho_4}{\rho_0} = \frac{0.80}{0.79}ρ0​ρ4​​=0.790.80​ ρ4ρ0≈1.01266\frac{\rho_4}{\rho_0} \approx 1.01266ρ0​ρ4​​≈1.01266

So the required relative density is close to

1.011.011.01
  1. Check options
  • A: 1.041.041.04 ❌
  • B: 1.261.261.26 ❌
  • C: 1.011.011.01 ✅
  • D: 1.031.031.03 ❌

Therefore, the correct option is C.

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