---
title: "Strength of sugar solution"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:sugar-strength-twenty-cm-tube"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/sugar-strength-twenty-cm-tube
site: BCT Engineering
---

# Strength of sugar solution

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2068 Shrawan, 5 marks
- Video: [Specific rotation of Sugar Solution using Half Shade Polarimeter | OMLABS #cbse #science](https://www.youtube.com/watch?v=ddFtwGL3UlQ) — OM LABS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/sugar-strength-twenty-cm-tube
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

A 20cm long tube having sugar solution rotates the plane of polarization by 20°. If specific rotation of sugar is 66°, calculate the strength of the solution.

## Short answer

Concentration c = θ/(l[α]) with l in dm. l = 2 dm, θ = 20°, [α] = 66°, so c = 20/(2×66) ≈ 0.15 g/cm³ (≈ 151.5 kg/m³).

## Answer

Specific rotation [α] is the rotation produced by a 1 dm tube containing the solution at 1 g/cm³ concentration:

[α] = θ / (l × c)

so c = θ / (l × [α]), with *l* in decimetres and *c* in g/cm³.

**Given:** l = 20 cm = 2 dm, θ = 20°, [α] = 66° dm⁻¹ (g/cm³)⁻¹.

```text
c = θ / (l [α])
  = 20 / (2 × 66)
  = 0.1515 g/cm³
```
In SI units this is 0.1515 × 1000 = 151.5 kg/m³.

**Answer: strength (concentration) of the solution ≈ 0.15 g/cm³ (≈ 151.5 kg/m³, about 15 g per 100 cm³).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which expression gives the concentration c from the observed rotation θ, tube length l and specific rotation [α]?
   A. c = θ l [α]
   B. c = θ / (l [α])
   C. c = l [α] / θ
   D. c = θ [α] / l
   Answer: B. Since [α] = θ/(l c), rearranging gives c = θ/(l [α]). (Key point: Formula for concentration)

2. In the specific rotation formula, in what unit must the tube length be put for [α] = 66° dm^-1 (g/cm^3)^-1?
   A. Centimetres, so l = 20
   B. Decimetres, so l = 2
   C. Metres, so l = 0.2
   D. Millimetres, so l = 200
   Answer: B. Specific rotation is defined for a 1 dm tube, so 20 cm = 2 dm is used. (Key point: Unit of length)

3. What strength results for l = 2 dm, θ = 20°, [α] = 66?
   A. 0.0165 g/cm^3
   B. 0.66 g/cm^3
   C. 0.1515 g/cm^3
   D. 1.515 g/cm^3
   Answer: C. c = 20/(2 × 66) = 0.1515 g/cm^3, about 151.5 kg/m^3. (Key point: Final result)

## Related in Engineering Physics

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