---
title: "Quantity of sugar in 48 cm³ solution"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:sugar-rotation-quantity-two-hundred-mm"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/sugar-rotation-quantity-two-hundred-mm
site: BCT Engineering
---

# Quantity of sugar in 48 cm³ solution

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2072 Chaitra, 5 marks; 2068 Chaitra, 5 marks (asked 2 times)
- Video: [A 200 mm long tube containing 48 `cm^(3)` of sugar solution produces an optical rotation of](https://www.youtube.com/watch?v=h2PJhVv0Rnc) — Doubtnut
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/sugar-rotation-quantity-two-hundred-mm
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

A 200 mm long tube containing 48 cm³ of sugar solution produces an optical rotation of 11° when placed in a saccharimeter. If the specific rotation of sugar solution is 66°, calculate the quantity of sugar contained in the tube in the form of a solution.

## Short answer

Concentration C = θ / (l [α]) = 11° / (2 dm × 66° dm⁻¹(g/cm³)⁻¹) = 0.0833 g/cm³. Mass of sugar = C × volume = 0.0833 g/cm³ × 48 cm³ = 4.0 g.

## Answer

Specific rotation [α] = θ / (l × C), where θ is the rotation (degrees), l the tube length in decimetres (dm), and C the concentration in g/cm³.

**Given:** l = 200 mm = 2 dm, θ = 11°, [α] = 66° dm⁻¹ (g/cm³)⁻¹, volume V = 48 cm³.

1. Concentration:

```text
C = θ / (l [α]) = 11 / (2 × 66) = 0.0833 g/cm³
```

2. Quantity (mass) of sugar in the tube:

```text
m = C × V = 0.0833 × 48 = 4.0 g
```

**Answer: the tube contains 4.0 g of sugar.**

## Check yourself

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

1. Which expression gives the specific rotation?
   A. [α] = θ × l × C
   B. [α] = θ / (l × C)
   C. [α] = l × C / θ
   D. [α] = θ C / l
   Answer: B. Specific rotation is rotation per unit length (dm) per unit concentration (g/cm^3). (Key point: Specific rotation formula)

2. What tube length must be used for the 200 mm tube?
   A. 200 dm
   B. 20 dm
   C. 2 dm
   D. 0.2 dm
   Answer: C. The formula needs length in decimetres, so 200 mm = 2 dm. (Key point: Unit of length)

3. What is the concentration C?
   A. 0.0833 g/cm^3
   B. 0.167 g/cm^3
   C. 12 g/cm^3
   D. 0.5 g/cm^3
   Answer: A. C = 11/(2×66) = 0.0833 g/cm^3. (Key point: Concentration)

4. How is the mass of sugar in the tube found, and what is it?
   A. m = C/V = 0.0017 g
   B. m = C × V = 4.0 g
   C. m = C × l = 0.17 g
   D. m = θ × V = 528 g
   Answer: B. Mass = concentration × volume = 0.0833 × 48 = 4.0 g. (Key point: Mass of sugar)

## Close variants set in other exams

- [Quantity of sugar from optical rotation](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asugar-rotation-quantity-thirty-cm) — Engineering Physics, chapter 4, asked 3×

## Related in Engineering Physics

- [Strength of sugar solution](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asugar-strength-twenty-cm-tube) — Engineering Physics, chapter 4