A student used a potometer to measure the rate of water uptake by a leafy shoot. An air bubble in the capillary tube moved 36 mm in 4 minutes. The radius of the capillary tube is 0.5 mm.
Volume of water taken up = $\pi r^2 l$
(a) Calculate the rate of water uptake in mm³ per minute. Give your answer to 2 significant figures. [3]
(b) The student then placed a fan next to the shoot. Explain the effect this would have on the rate of water uptake. [2]
(c) Give one reason why the rate of water uptake is not exactly the same as the rate of transpiration. [1]
- (a) volume = π × 0.5² × 36; OR distance moved per minute = 36 ÷ 4 = 9 (mm)(1)
- (a) = 28.3 (mm³) (accept 28.27…); OR π × 0.5² × 9(1)
- (a) = 7.1 (mm³ per minute) (2 s.f. required for this mark; 7.07 scores 2); correct answer with no working scores 3(1)
- (b) (rate increases because) water vapour is blown away from the leaf surface / humidity around the leaf falls(1)
- (b) steeper (water vapour) concentration gradient so faster diffusion (and evaporation) of water vapour out of the stomata, so rate of water uptake increases(1)
- (c) some water is used in photosynthesis / to keep cells turgid (support); accept some water stays in the plant / is not lost by evaporation(1)
- Answer: (a) 7.1 mm³ per minute (b) increases — vapour removed, steeper concentration gradient (c) some water used in photosynthesis / support
- Examiner tip: Square the radius, not the diameter, and remember to divide by the time — 28 mm³ is the total, not the rate.